THE

STATE SCHOOL ARITHMETIC

ARRANGED TO SUIT THE REQUIREMENTS OF

©lie programme of Instruction

ISSUED BY

THE EDUCATION DEPARTMENT

BY

JOHN J. BURSTON

AUTHOR OF “ THE STATE SCHOOL GRAMMAR ”

\

REVISED EDITION, ENLARGED

GEORGE ROBERTSON & CO., LIMITED,

MELBOURNE, SYDNEY, AND ADELAIDE

1884


PREFACE.

THE STATE SCHOOL ARITHMETIC aims m supplying a text-book suited at once to the requirements of the State School Programme of Instruction and to the several Examination Programmes prescribed by the Victorian Education Department for the classification of both teachers and pupil-teachers.

The exercises, which are both graduated and exhaustive, have been compiled chiefly from Teachers’ Examination Papers ; and as these papers have always been sought for and prized, as affording the best means of preparation available to pupil-teachers and candidates for scholastic honors, the adoption of this Arithmetic as a class-book offers the twofold advantage of exercising the scholar and preparing the young teacher for his own examinations.

Chapters I. to IV. are written respectively to suit the Third, Fourth, Fifth, and Sixth Classes.

Chapter V. is more especially adapted to more advanced scholars and teachers.

The success which has attended the sale of The State School Grammar leads me to look with confidence to the teachers of Victoria and the neighbouring colonies for a fair trial for this, the First Australian Arithmetic ; and I am vain enough to flatter myself that it needs but this trial to obtain for itself a place amongst the most popular of the many excellent treatiseson the subject already published.

As many new features will be found throughout the work, special attention is directed to the matter contained in the following pages:—Pages 32 to 42, 68, 69, 70, 124 to 129,

136 to 142, 148, 150,161, 170 to 175, 180, 181, 182, and 199 to 224.

From Head Teachers of State Schools I have received numerous letters complimenting me, in flattering terms, upon the suitability of my Arithmetic to the requirements of advanced classes, pupil-teachers, and candidates for examinations.

The first thousand copies sold in the short space of one month, thus showing an urgent demand for a new Arithmetic in our State Schools.

In revising my Arithmetic for the present edition (the fourth thousand), I have critically reviewed, and reconstructed, the definitions on pages 3 to 7 ; have carefully , revised the answers, and made some verbal alterations in the various rales, &c., through the book.

The now almost universal use of this Arithmetic as a pupil-teacher’s class-book, and the uniformly favourable opinion of my felk>w-teachers as to the eminent suitability of the S. S. A. to advanced scholars, lead me to think that the rapidly-increasing sale is indicative of a growing popularity ; and I hopefully look forward to its general adoption in public and private schools throughout the Australian colonies.

My best thanks are due to my brother, Mr. B. J. Burston, for his valuable assistance in this and in my new work,

“ Milton Parsed,” now at press.

JOHN J. BUBSTON,

State School No. 1267, Sandhurst.

Sandhurst, 31st January, 18SO.

Introduction ...

*

« •

PAGE

1

Definitions * • *

-

.

2

Arithmetical Symbols - -

-

- •

8

Notation and Numeration - »

.

*

9

Suggestions upon Teaching Tables

a

* -

13

Addition (Simple) - • «

*

*

16

Subtraction - -

-

« -

18

Multiplication - - •

-

»

24

Short Division - *

- -

28

Long Division - - -

-

-

30

Hints upon Cancelling by Inspection

-

«

32

Principles in connexion with the Simple

Rules

-

34

Proof by Casting out the Nines -

-

- -

41

Proof by Casting out the Elevens -

-

.

42

Exercises upon Notation, Numeration, and the Simple Rules -

43

Tables ....

-

.

56

The Compound Rules and Reduction

-

68

Compound Addition - -

70

Compound Subtraction - -

- -

71

Compound Multiplication - .

.

«

74

Compound Division - -

«

• -

81

Exercises upon the Compound Rules -

-

-

84

Reduction - - -

.

99

Exercises on Reduction • •

.

104

Practice ...

«

107

Table of Aliquot Parts - -

-

114

Exercises on Practice    -    -    •    •

PAGH

114

119

124

130

136

143

147

151

156

159

163

167

169

174

176

181

183

188

191

195

199

200 202 205 207 209 214 218 222

225


243


Ratio and Proportion .    -    .    •

Proportion (Direct and Inverse)    *    •    •

Exercise on Simple Proportion ... Compound Proportion    -    .    .

Exercise on Compound Proportion    -    *

Interest

Compound Interest ....

Exercises on Interest, from Teachers’ Examination Papers The Greatest Common Measure    -    -    -

The Least Common Multiple    ...

Notation and Numeration of Fractions    -    -

Vulgar Fractions -    -    -    -    -

Directions to be observed in Working Fractions -Decimal Fractions .....

Explanation of Recurring Decimals    -    -

Exercises on Vulgar Fractions    ...

Exercises on Decimal Fractions    -    -    -

Complex Fractions, from Teachers’ Examination Papers -Discount and Present Worth    -    -    -

Proportional Parts    ....

Exercise on Proportional Parts    -    -    -

The Reciprocal of Numbers ....

Profit and Loss    ....

Exercise on Profit and Loss ....

Arithmetical Equations    -    -    -

Average ......

Examples Worked out -    -    -    -

Miscellaneous Questions, from Teachers’ Examination Papers

ANSWERS .....

APPENDIX -

THE STATE SCHOOL ARITHMETIC.

Arithmetic, both as an abstract science and as applied to magnitudes of various kinds, treats of the value and properties of numbers ; the proper method of expressing them ; the various methods of making calculations; and the course of reasoning to be followed in order to arrive at the desired result.

Numbers are capable of increase or decrease, and all operations tend to affect numbers in one or other of these ways.

There are two methods of increase, Addition and Multiplication ; and two methods of decrease, Subtraction and Division.

These are in general use and are termed “ The Four Simple Rules.”

Most calculations are made by combinations of “ The Four Simple Rules.” It is frequently very difficult, however, to decide which of these processes is to be employed and, in some cases, a choice may be made, and the answer obtained in various ways. In fact, most questions requiring the employment of any two or more of the simple rules mav

be worked variously. Hence, no fixed rule can be given as the sine qud non to the solution of such questions.

The office of an Arithmetic is to describe the plan of working these “ simple rules,” to point out some of the simplest methods of applying them, and to give other rules for guidance in employing the most concise and convenient methods of solving ordinary practical questions.

DEFINITIONS.

1.    Arithmetic, the science of numbers.

2.    Axiom, a well-known or self-evident truth.

3.    Unit or Unity, a single one ; the number 1.

4.    Digits, the figures 1, 2, 3, 4, 5, 6, 7, 8, 9.

5.    Cypher, Zero, or Nought, the figure 0.

6.    Integer, a whole or entire number: as 7, 1760, 4840.

• 7. Fraction, a part or parts of unity ; as, £, f, |, §

(read one-half, two-thirds, three-fourths, three-fifths).

8.    Denominator, or Namer, the lower part of a fraction which gives a name to the parts ; thus, in §, 5 gives the name fifths.

9.    Numerator, or Numberer, the upper part of a fraction which tells how many parts ; thus, in 3 shows how many fifths.

10.    Vulgar Fraction, one whose numerator is placed above its denominator ; as, y’r, V, y5.

11.    Decimal Fraction, one whose Denominator is a power of 10 and is expressed by a dot called a decimalpoint; thus, -5 -75 -625 -6.

12.    Repeating, Circulating, or Recurring Decimal, one having figures which constantly recur : as, -33333, &c. (expressed as -3); -0*125, or -0128. See page 179 et seq.

13.    Finite Decimal, one having no recurring figures.

14.    A Mixed Number consists of an integral number and a fraction : as, 5|, 30\.

15.    Reciprocal, that which any number must be multiplied by in order to produce unity. The quotient obtained by dividing unity by the number of which the reciprocal is required.

Note.—The reciprocal of a fraction is that fraction with its terms inverted ; thus, the reciprocal of Tsr is y or 2^. The reciprocal of a whole number is the fraction formed by placing the figure 1 above it; thus, | is the reciprocal of 8, Ty of 12, &c.

16.    Concrete Numbers refer to particular things : as, ten men,five days.

17.    Abstract Numbers do not refer to particular things.

18.    A Positive Number expresses an absolute amount.

19.    Negative Numbers are purely imaginary numbers, and are expressed by prefixing the sign, —, to denote a deficiency. This deficiency is supposed to represent as much less than nothing as the corresponding positive number represents more than nothing.

20.    Absolute or Intrinsic Value, the value of a di<rit

o

in itself; i.e., its value when standing alone.

21.    Relative or Local Value, the value of a digit from the place it occupies with respect to the units’ place.

In the number 1728, the local value of the figure 7 is seven hundred ; the intrinsic value is simply seven.

22.    Notation, expressing numbers in figures : as, 365.

23.    Numeration, the art of expressing numbers in words: as, Three ham We l and sixty-fi ve.

24.    Addends, numbers or quantities to be added together.

25.    Addition, the method of adding numbers together.

26.    Total, or Sum, the amount obtained by adding; the answer of addition ; also, an arithmetical problem.

27.    Subtraction, the method of subtracting one number from another, or of taking a part from the whole.

28.    Minuend, the number from which another number is to be taken. The upper line in subtraction.

29.    Subtrahend, the number which is to be subtracted.

30.    Difference or Excess, the remainder after part has been taken away ; the answer of subtraction ; the result of taking one number from another; or the amount by which one number exceeds another.

31.    Multiplication, a short method of adding the same number repeated any number of times.

32.    Multiplicand, the number to be multiplied.

33.    Multiplier, the number by which we multiply.

34.    Product, the result or answer of multiplication.

35.    Continued Product, the final result of successive multiplications. Earlier products are Partial Products.

36.    Division, the method of finding how often one number is contained in another; how often one number may be taken from another ; or of repeatedly subtracting the same number from another number.

37.    Dividend, the number which is to be divided.

38.    Divisor, the number by which we divide.

39.    Quotient, the result or answer of division. The quotient shows how often the dividend contains the divisor.

40.    Remainder, that which still remains in division after the integral part of the quotient has been obtained.

41a. Measure, an exact divisor, or one which leaves no remainder ; called also a factor or an aliquot part.

41 b. Aliquot Part, a measure, or a part of a concrete number represented by a simple fraction having 1 for its numerator : thus 10s., 6s. 8d., and 2s. 6d. are aliquot parts of a £, being    respectively.

42.    Common Measure or Common Factor, one which measures all the given numbers.

43.    Greatest Common Measure, the greatest number which measures the given numbers : thus, while 2, 3, 4, and 6 are all measures common to 24 and 36, 12 is the greatest common measure, or the Highest Common Factor.

44.    Factors, measures of a number; or numbers which taken any exact number of times will make the number : thus, 3 and 8, 4 and 6, and 2 and 12 are all factors of 24.

45.    Composite Number, a number having factors : as, 24.

46.    Prime Number, a number having no factors other than itself and unity : as, 5, 7, 11, 13, 17, 19, 23, 29, &c.

47.    Power, a product of a number multiplied by itself any number of times; or a product obtained by using the same number as a factor any number of times.

Note.—The number itself is called the first power or base. If the number enter as a factor twice, the product is called the square ; if thrice, the cube ; if four times, the fourth power, &c., &c. Thus 5 is the first power of 5 ; 5 times 5 or 25 is the square ; 5 x 5 x 5 or 125 is the cube, and 625 the fourth power.

48.    A Multiple of a number is the product of that number and any other number : thus, 9, 12, 15, and 18 are all multiples of 3.

49.    A Common Multiple contains each- of the given numbers an exact number of times. See page 163 et seq.

50.    The Least Common Multiple is the least multiple common to the given numbers; thus, 24 is the L. C. M. of 3, 6, 4, and 8. See page 163.

51.    Loot, the number which is multiplied by itself to produce any power ; thus, 4 is the square root of 16, or the cube root of 64 ; because it must be used as a factor twice to give 16, and thrice to give 64. See page 247 et seq.

52.    Ratio is the comparative magnitude of two abstract numbers, or of two quantities of the same kind; thus \ is the ratio of 3 to 6, or of 6d. to Is.

53.    Proportion, the equality of ratios. In proportion, the same result arises from two distinct comparisons, the ratios produced being identical. See page 119 et seq.

54.    Reduction, the process by which we reduce concrete numbers of one name to their equivalent in another name.

55.    Compound Quantities, quantities embracing things of different names ; thus, 2 cwt. 3 qrs. 14 lbs.

56.    Practice, a method of working compound multiplication by using aliquot parts and applying division.

57.    Interest, money paid for the loan of other money.

58.    Discount, a reduction allowed upon paying money; an allowance made on account of prepayment.

59.    Principal, the money lent, or invested in business.

60.    Amount, the amount to be repaid ; the sum comprising the principal together with its interest.

61.    Present Value or Present Worth, the‘principal less the discount; or the actual worth at the present time.

62.    Mean or Average, the medium of numbers; the result of taking one with another or with others. Of two numbers it is called the Mean.

Note.—The mean of two numbers is J their sum. The average of three numbers is £ of their sum ; of four, \ of their sum, &c.

63.    Proportional Parts, parts which bear a given relation to each other, and together make up the whole.

64.    Per Centage, parts of 100 ; thus 5 p. c. is T(nyths.

65.    Stocks, Funds, Debentures, &c., shares in the public debt or in joint-stoclc companies.

66.    Surds, roots which cannot be exactly obtained, but are indicated by the signs f, 2\/> 3\/> &c-; as, ^3, 3\/9.

Note.1 is the only prime number whose roots can be found ; all roots of 1 are unity itself. Roots of all other prime numbers and of all negative numbers are surds. See pages 252, 253.

67.    The Complement of a number is its difference from the power of 10 next above it; thus, the complement of 639 is (1000 — 639) = 361. The complement of 78726 is (100000 — 78726) = 21274.

68.    Involution, the method of finding a power of a number.

69.    Evolution, the method of finding a root of a number.

70.    Unit of Measurement, the unit used as the standard of comparison in estimating the value of a concrete number ; thus, an inch is the “ unit of measurement ” in Long Measure.

71.    Logarithm, the exponent or index of the power to which another number must be raised in order to produce a given number. Thus, 103 - 1000, where the index, 3, is the logarithm of 1000.

72.    Progression, numbers proceeding by equal differences, or by equal ratios, whether increasing or decreasing.

Note.—When numbers proceed hv equal differences, as 3, 6, 9, 12, they are said to be in Arithmetical Progression ; if they proceed by equal ratios, as 3, 9, 27, 81, they are said to be in Geometrical Progression.

ARITHMETICAL SYMBOLS.

+ (plus), means adcl, the sign of addition.

— (minus), means less, the sign of subtraction.

x (into), * means multiplied by.

-i- (by), means divided by.

^ (difference), means take the smaller from the greater, (therefore), means consequently.

V    (wherefore), means because.

; (is to), means compared with.

;: (so is), means is the same as.

As 2 : 8;: 3 : 12, reads, as 2 is to 8 so is 3 to 12 ; and means that 2 is the same part of 8 which 3 is of 12.

= (equality), denotes that the amounts on each side of it are equal in value.

V    (root), a corruption of r, the initial letter of radix, a root.

2 ^} 3^    &c., the square root, cube root, fourth

root, &c.

102, 103, 104, (index), the small figures 2, 3, 4, denote the power,_the square, cube, and fourth power respectively.

()> \ }->[]> (brackets), enclose numbers to be acted upon as a whole; or denote that what is enclosed is to be treated collectively.

Note.—A straight line drawn over numbers has the same effect as enclosing them in brackets; thus, 6 — 8    5 = 6 — (8    5).

* The words “by ” and “ into” are more frequently used to mean « multiplied by ” and “ divided into ” than as the equivalent of the signs X and That is, such an expression as 6 X 8 is read as by 8 ; and 12 -i- 3 as 3 into 12.

NOTATION AND NUMERATION.

All numbers are expressed by the nine significant figures, 1, 2, 3, 4, 5, 6, 7, 8, 9 (called digits), and the 11011-significant figure U (called a cypher, zero, or nought).

When used alone, or on the right of other figures, a digit is in the units place, and expresses the number which its name indicates ; thus, in the numbers, 9, 59, and 1009, the figure 9 denotes nine units, or single ones.

The units’ place is that on the right (except in decimals, when it is immediately 011 the left of the decimal point).

In the units’ place a figure has its ordinary or name value —called its nominal or intrinsic value.

In every other place the value of a figure depends upon its position, and this value varies in proportion to its distance from the units’ place.

The number 1, which, in the units’ place, simply means seven units, denotes seven tens (or seventy) when removed from the units place by a nought, and seven hundreds when removed two places by two noughts ; thus, 70, 700.

In like manner, the figure 5, which in 5000 indicates five thousands, signifies one-tenth of five thousands (or five hundred) when removed one place nearer to the units’ place, and one-tenth of five hundreds (orfifty) when removed two places; thus, 500, 50.

The value a figure thus receives from its position with regard to the units' place is called its local value.

The local value of a digit increases tenfold each remove to the left, and decreases tenfold each remove to the rifrht It is thus seen that the figure 0, which has no value in itself, affects the value of the significant figures by removing them from the units’ place.

A figure in the first place on the left of the units’ place will represent ten times as much as when in the units’ place; but only one-tenth as much when in the first place

on the riglbt of the units’ place.

Note.—Figures on the right of the units’ place are called

decimals.

Seeing that the value of digits increases or decreases at a tenfold rate with each remove of one place to the left or right, it is necessary to give names to the various places to the left or right of the units.

Let the cypher below occupy the units’ place, then the significant figures will occupy the places named :—

9876543210-123456789

CD ,3

3    a

a    o

03 .2 S

All places between any particular place and the units’ place (together with the units’ place itself) must be filled up with significant figures or cyphers before any number of that particular place can be expressed : thus, for a digit to express millions, six other figures must follow it; while to express hundreds of millions, eight other figures must follow. In like manner any particular place in decimals is denoted by filling the places between it and unity with other digits or cyphers; but here the position of the units’ place is simply indicated by a dot, as -5, "25, -0125, -0002.

Children will do well to remember how many figures are required for each particular place : thus, for tens, 2 ; for hundreds, 3; for thousands, 4 ; for tens of thousands, 5 ; for hundreds of thousands, 6; for millions, 7, Ac.; for tenths, 1 (in addition to the decimal point); for hundredths, 2; for thousandths, 3; for tens of thousandths, 4; for hundreds of thousandths, 5, Ac., Ac.

Remark.—Decimal Notation would be less confusing if a plan of placing the decimal point over the units’ place, instead of after the units, were adopted. This would necessitate the filling of the units’ place with a cypher, when no whole number preceded the decimal, and thus make the number of figures necessary to express a millionth coincide with the number required for a million ; instead of, as at present, requiring one less.

Notation.—To Notate a Number is to Express it in Figures.

Numeration.—To Numerate a Number is to Express it in Words.

In order to enable us to notate and numerate more readily, ’ the figures which express large numbers are divided into groups, called periods.

There are two systems of notation, differing from each other in the number of figures composing a period :—

I.    The French or Italian Method, having three places in each period : units, tens, and hundreds.

II.    The English Method, having six places in each period : units, tens, hundreds, thousands, tens of thousands, and hundreds of thousands.

The names of the periods in the English System are (1) Units’ Period, (2) Millions, (3) Billions, (4) Trillions, (5) Quadrillions, (6) Quintillions, (7) Sextillions, (8) Septillions, (9) Octillions, (10) Nonillions, (11) De-

CEMTILLIONS, &C., &C.

In the French System, a thousands’ period occurs between the units’ and millions’ period; while the billions follow directly upon the hundreds of millions,—there being no thousands of millions.

It will be observed that numbers of less than ten figures will read similarly in both systems; the first change occurring on the tenth figure.

Note 1.—The French or Italian System of Notation and Numeration has recently been adopted in England, in lieu of the English method above described. See Arithmetic by W. and R. Chambers.

Example: Numerate 785,010,011,012,900.

Old English Method.—Seven hundred and eighty-five billions, ten thousand and eleven millions, twelve thousand nine hundred. -

French or Continental Method. —Seven hundred and eighty-five trillions, ten billions, eleven millions, twelve thousand nine hundred.

Note 2.—It is customary to mark off the periods in English Notation by semicolons, and to subdivide them by commas.

In the French System the periods are marked off' by commas.

left; while after

means on the right.

OLD ENGLISH METHOD.

Trillions.

Billions.

Millions.

Units.

000000

000000

000000

000000

CONTINENTAL METHOD.

Billions.

Millions.

Thousands.

Units.

Hund eds. Tens. Units.

Hund. Tens. Units.

Hund. Tens. Units.

Hund. Ten9. Units.

0 0 0

0 0 0

0 0 0

0 0 0


Note 3.—In speaking of a figure as before others, we mean on the

SUGGESTIONS UPON TEACHING TABLES. 13

SUGGESTIONS UPON TEACHING TABLES.

The following method of writing the Addition and Subtraction Table is here recommended, as being :—

1.    Analogous to Addition and Subtraction ; and hence

an introduction to them.

2.    A more effective plan than that usually adopted.

3.    A check upon the habit of learning by sequence.

4.    An excellent plan for ensuring ready replies to the pro

miscuous method of questioning known as “dodging."

How often do we hear a child who blunders at the first irregular question, affirm that he can say his tables “straight on !”

How many others hesitate to reply until they have mentally repeated from the beginning of the table !

The reason of this is to be found in the consecutive order employed in teaching the tables. The remedy would be to arrange them in an alternate order, as below :—

Example.—-Table of 8, Addition.

1

3

5

7

9

11

2

4

6

8

10

12

, 8

8

8

8

8

8

8

8

8

8

8

8

9

11

13

15

17

19

10

12

14

16

18

20

Table

OF 8,

Subtraction.

9

11

13

15

17

19

10

12

14

16

18

20

8

8

8

8

8

8

8

8

8

8

8

8

___

_

1

f>

O

5

7

9

11

O

4

6

8

10

12

Note. —It is a good plan, in teaching the Subtraction Tables, to proceed thus S from 9 leaves 1, because 8 and 1 are 9 ; 8 from 11 leaves 3, because 8 and 3 are 11. Again, the Addition and Subtraction Tables may be taught together, thus : 8 and 1 are 9, 8 from 9 leaves 1 ; 8 and 3 are 11, 8 from 11 leaves 3.

The reciprocal connexion between the Addition and Subtraction Tables, and between the Multiplication and Division Tables, should be well impressed on the youthful mind.

If the ordinary method of arrangement be considered preferable, the Addition and Subtraction Tables might be combined by writing them side by side, and the alternate order still preserved. Thus : -

Table of 9, Addition and Subtraction.

9 and 1 are 10 ; then 9 from 10 leaves 1

9 „

3 ,

, 12

9 „

. 12 „

3

9 „

5 ,

, 14

9 „

. 14 „

5

9 „

7 ,

, 16

9 „

. 16 „

7

&c.    &c.

MULTIPLICATION TABLE.

This may be taught in connexion with the Pence Table, with this advantage :—The two tables combined can be learnt in very little more time than each takes up singly; while the pernicious habit of learning the Pence Table by sequence is prevented. Who of us, if asked how many shillings there are in 87 pence, does not say to himself mentally, 84d. = 7s., then 87d. = 7s. 3d. i If the number of pence which are represented by the composite numbers of the Multiplication Table be well known, the whole Pence Table up to 150d. will be easily mastered. Another good feature in this plan is, that the scholar observes what any number of threepences, fourpences, sixpences, Ac., amount to.

Example.—Multiplication Table of 8 and 10.

d.

d.

S.

d.

8 times 1

=

8

=

0

8

3

=

24

=

2

0

5

=

40

=

3

4

7

=

56

r=

4

8

9

=

72

=

6

0

11

=

88

==

7

4

2

=

16

=

1

4

4

=

32

=

2

8

6

=

48

=

4

0

8

=

64

=

5

4

10

=

80

=

6

8

12

=

96

8

0


d.

d.

s.

d.

10 times 1 =

10 =

0

10

3 =

30 =

2

6

5 =

50 =

4

2

7 =

70 =

5

10

9 =

90 =

7

6

11 =

110 =

9

2

2 =

20 =

1

8

4 =

40 =

3

4

6 =

60 =

5

0

8 =

80 =

6

8

10 =

100 =

8

4

12 =

120 =

10

0


In repeating the above, it will be simply necessary to say :_

8 ones are 8, 8d. are 8d. ; 8 threes are 24, 24d. are 2s. ; 8 fives are 40, 40d. are 3s. 4d., &c., &c. The repetition of the multiples 8, 24, 40, &c., impresses them upon the memory.

It is not thought necessary to give the Addition, Subtraction, and Multiplication Tables in full.

Another good combination of tables is that of the Multiplication and Division Tables, written thus :—

9 times 1

are 9

9

J 9

3

,,

27

9

99

5

,,

45

9

99

7

9 9

63

9

9 9

11

,,

99

9

9 9

2

9 9

18

9

99

4

9 9

36

9

9 9

6

9 9

54

9

9 9

8

9 9

72

9

99

10

99

90

9

99

12

99

108


9 into 9 goes 1

9

9 9

27

„ 3

9

9 9

45

„ 5

9

9 9

63

„ 7

9

99

99

„ 11

9

9 9

18

„ 2

9

99

36

„ 4

9

99

54

„ 6

9

99

72

„ 8

9

99

90

„ 10

9

99

108

„ 12


ADDITION AND SUBTRACTION.

To place numbers for Addition or Subtraction :—

Rule.—Place the numbers one under the other, so that the units shall be under units, tens under tens, hundreds under hundreds, thousands under thousands, dec., dec., in vertical columns, and then draw a straight line beneath them.

To add numbers

Rule.—Add up the figures in the units’ column; cut off the last figure and put it under tlbe units. Add up the number formed by the remaining figures with the digits in the next column ; cut off the last figure, as before, and carry the number expressed, by the others to the next column. Add each column in the same way ; putting down the ivhole result of the last column.

Examples :—    I.    II.

376495    1234567

729843    20796

26S594    379

726487    7005

2101419    1262747

In Example /., the sum of the first column is 19,—put down 9 and carry the one to the next column ; the 1 carried, with the figures of the second column, gives 31,—put down the 1 and carry the 3 to the hundreds’ column ; 3 added to the figures in this column gives 24,— put down the 4 and carry the 2 ; 2 added to the fourth column gives 31,—put down the 1 and carry the 3 ; 3 added to the fifth column gives 20,—put down the 0 and carry 2 ; 2 added to the sixth column gives 21,—put down the whole number, because there is no other column to add up.

Note.—It is usual to say, “ Put down the units and carry the tens.” Note.—In putting down the addends, it matters not in what order they are put: that is, which is put as the top line, which as the second line, third line, &c.

In Example II., the three last columns give sums each less than 10 ; and in this case there is no figure to carry.

Method of Proof :—

Rule.Add the answer to the addends to obtain twice the

answer.

Note.—The following plan of reasoning Mull be found useful to prevent children from counting instead of adding, when learning addition :—

Suppose in adding a vertical column consisting of the figures 8, 7, 5, 9, 7 and 9, a boy hesitated at the last step of the addition to tell what 36 and 9 came to.

Place the number 9 under 6, 16, 26, 36, 46, 56, respectively, and then cause him to repeat—“Because 9 and 6 are 15, 19 and 6 are 25 ; 29 and 6 are 35 ; 39 and 6 are 45,” &c. He will then observe that any two numbers ending in 9 and 6 respectively M ill give a sum ending in 5.

The proof of the simple rules by casting out the nines is treated of at page 41. It will be sufficient, here, to point out how the teacher may construct addition sums, the answers of which he may see at a glance.

This will be best done by examples.

Examples:— I.

154327614 \

845672385 i 628541237 371458762 )

514273826 )

485726173 }

795286089 key-line


II.

2376594285 key-line 1235432683 j * 3421213104 [ 5343354212 ) 3251264107 ) 6213600291 [ 535135601 )

22376594283


, 3795286086

In I., the figures of the first and second lines, added, will give a line of 9’s, thus, 999999999 ; the figures of the third and fourth lines also give a line of 9’s; as also do the figures of the fifth and sixth lines. N o vv, three such lines of nines will be but 3 less than 3000000000, and this added to the bottom line will make 37952S6089 all but 3' Hence, the answer may be at once written by placing 3 before the bottom line (called the key-line) and subtracting 3 from the units’ figure of that line. Hence, A ns. =3795286086.

In II., the top is taken as the key-line ; while the second, third, and fourth lines give ten nines ; and the fifth, sixth, and seventh give another line of ten nines. Now, two lines of ten nines make but 2 short of 20000000000. Hence, if we take 2 from the key-line, and place 2 before it, we obtain the answer at once, =22376594283, Ans.

Note—Any line may be taken as the key-line; while the sum may be constructed with any number of lines.

SUBTRACTION.

Subtraction is the method of finding the difference between two numbers by taking one from the other.

It is also used to take part of any number from the whole.

We take one number from another when we want to find (1) by how much the one exceeds the other; (2) by how much the less is in defect of the other.

If the former be the object of Subtraction, the difference is called the Excess ; if the latter, it is called the Defect.

When Subtraction is performed with the object of taking away part of a number from the whole, the difference is called the Remainder. The term Remainder, however, is generally applied to the number which remains over after division has been performed.

Difference is, therefore, a general term for the Excess, Defect, or Remainder.

The number from which another number has to be taken is called the Minuend ; the number to be taken away is called the Subtrahend.

Note—Minuend, from Minuo, I diminish or lessen; and Subtrahend, from Subtraho, I subtract. As observed by Cornwall and Fitch, in their “ Science of Arithmetic,” words ending in and or end from Latin participles in andus and endue, signify an action which has to be done to a thing ;—'“ Thus the Minuend is that which has to be diminished ; Subtrahend, that which has to be subtracted ; Multiplicand, that which has to be multiplied; Dividend, that which has to he divided.”

Illustration.—If 12 balls be placed on the top wire of a ball-frame, or Arithmetican, and then 7 be taken away, there will be 5 left. The whole 12 is the Minuend ; the 7 taken away the Subtrahend ; and the 5 left the Remainder.

Again, if 12 balls be placed on the top wire and 9 on the next, we see that there are 3 more on the top wire than on the second wire ; or that the number 12 exceeds the number 9 by 3 ; so that 3 is the excess. We also observe that 3 more require to be placed on the bottom wire to make up 12, so that 3 is the defect of 9 as compared with 12.

To find the difference between two numbers by considering how much one exceeds or is less than the other, is very easy in the case of small numbers; but with large numbers the case is different.

I. We observe that any number of tens, or hundreds, or thousands, taken from another number of tens, hundreds, or thousands, will leave the same number of tens, hundreds, or thousands, as the cor responding number of units as Subtrahend from the corresponding number of units as Minuend will leave units.

For example, because 3 from 8 leaves 5 ; 3 tens from 8 tens leave 5 tens ; 3 hundreds from 8 hundreds leave 5 hundreds ; 3 thousands from 8 thousands leave 5 thousands, &c., &c.

Or, 8

80

800

8000

80000

3

30

300

3000

30000 „ 7 7.,

&c. y ad Libitum.

5

50

500

5000

50000


On account of this correspondence between the remainders in the case of units, tens, hundreds, See., we say, when dealing with the hundreds or thousands, &c., 3 from 8 leaves 5, instead of 3 hundreds from 8 hundreds leave 5 hundreds ; 3 thousands from 8 thousands leave 5 thousands, &c., &c., taking care, however, to place the diffeience under the hundreds, or thousands, or whatever we may be dealing with.

11. Again, v e observe that the aijfevence between two numbers is not altered by adding the same number to each.

Thus, if one boy have 9 marbles and another boy 5 marbles, the difference, 4, will not be altered by my giving 10 marbles to each j for they will then have 19 and 15 respectively, or the one will still have 4 more than the other.

From I. and II. we obtain the ordinary method of working.

Rule. — Place the number to be subtracted under the other, units under units, tens under tens, dec. Take each figure from the one above. If the upper figure be the less, add 10 to it, and then subtract. Add 1 to the next figure of the bottom line before subtracting it from the figure above it.

The rule more fully expressed is :—

Rule.After placing the numbers as directed (page 16), commence with the units, and take the figures in the Subtrahend from those directly above them in the Minuend. If this cannot be done on account of the lower figure being greater than the upper, add 10 to the upper figure, subtract the lower, and put down the remainder. To prevent the difference being altered hy the addition of 10 to the Minuend, add 1 to the next figure in the Subtrahend before subtracting.

Should nothing be left when taking the lower figure from the upper, put down a nought ; unless it is seen that no significant figure will come before it, in which case put nothing at all down.

Note.—The local value of the 1 added to the figure of the subtrahend is equal in value to the 10 added to the minuend, since its position is one place farther to the left. Hence, in adding 10 to the upper figure, and carrying 1 to the next figure in the minuend, we are simply adding an equal amount to two unequal numbers (viz., to the subtrahend and minuend) ; and it has been shown that an equal amount added to two unequal numbers does not alter tlieir difference.

In working Subtraction by the above method, the terms “ borrow” and “ carry ” are generally used to remind us where to add 1 ten to the bottom line after having added 10 to the top line. These are excellent mnemonical aids ; but since they have been condemned on the grounds that they are inappropriate and apt to mislead as to the real nature of the process employed, they are not used here.

That the terms “ borrow ” and “ carry ” are inappropriate will be seen by applying them to the subtraction of 18 from 24. Here we say S from 4 cannot be taken, “borrow 10.'’ Whence are we to borrow? If the 10 were borrowed from the subtrahend 18 there would be no ten left; yet we say “return the, 10 we borrowed, 1 ■and 1 are two tens” Now, if the ten were borrowed and returned, there would be but one ten. Hence, we should rather say, add 10 to the minuend (of course meaning 10 units, 10 tens, 10 hundreds, &c., according to the local value of the figure with which we are dealing), and then, after subtracting, say, add 1 ten to the subtrahend, so that the difference shall be unaltered.

Examples.—From 207368, take 193295 ; and take 990998 from

1001001.

I.

207368 minuend. 193295 subtrahend.


II.

1001001 minuend. 990998 subtrahend.


14073 difference.    10003 difference.

I.    Take the number 5 from the number 8, leaves 3 ; 9 from 6 we cannot (take), add 10 to the 6, 9 from 16 leaves 7 ; add 1 to the 2, 3 from 3 leaves nothing, put down 0 (nought) ; 3 from 7 leaves 4 ; 9 from 0 we cannot, add 10, 9 from 10 leaves 1 ; add 1, 2 from 2 leaves 0, and since a nought is of no value on the left of a number, leave it out.

II.    8 from 1 we cannot, add 10, 8 from 11 leaves 3 ; add 1, 1 and 9 are 10, and 10 from 0 we cannot, add 10, then 10 from 10 leaves 0 ; add 1, 1 and 9 are 10, 10 from 0 we cannot, add 10, 10 from 10 leaves 0 ; add 1, 1 from 1 leaves 0 ; 9 from 0 we cannot, add 10, 9 frpm 10 leaves 1 ; add 1, 1 and 9 are 10, and 10 from 10 leaves 0, do not put it down, as it is seen that the next figure will be a nought also.

Remember that when we say “ add 10,” it is to the minuend ; while “ add 1 ” refers to the subtrahend.

Note.—The subtrahend can in no case be greater than the minuend ; for it is impossible to take a greater number from a less.

Second Method :—

Rule II.—Proceed as before to subtract the under digit from the upper. If it cannot be taken, remove 1 from the next significant figure of the Minuend, and change this into its corresponding value in the position which the figure we are dealing with occupies, and put a stroke through the figure from which 1 has been removed to show that it now denotes 1 less.

Example I.—Take 4128237, from 9243672.

9243672 minuend.

4128237 subtrahend.

5115435 difference.

Process. —7 from 2 we cannot, remove 1 from the 7, the 1 removed = 10 units, 10 and 2 are 12, and 7 from 12 leaves 5. Next, 3 from 6 leaves 3 (not 3 from 7, as 1 was removed from the 7). Then 2 from 6 leaves 4 ; 8 thousand from 3 thousand we cannot, remove 1 ten thousand from the 4 tens of thousands, 10 and 3 are 13 thousands and 8 thousands from 13 thousands leaves five thousands. Next, 2 from 3 (one having been removed from the 4), leaves 1. 1 from 2 leaves 1 ; and 4 from 9 leaves 5.

When the next figures of the Minuend are noughts, go to the next significant figure, as before, substitute 9 for every nought and add 10 to the figure being dealt with.

Example II.—Find the difference between 30002 and 12345.

999

30002 minuend.

12345 subtrahend.

17657 difference.

Pi •ocess.—5 from 2 we cannot, go to the next significant figure in the minuend, viz., 3 ; cross it out and change the 1 taken from it into 9 thousands, 9 hundreds, 9 tens and 10 units. Then 10 and 2 are 12, 5 from 12 leaves 7. Then 4 from 9 leaves 5 ; 3 from 9 leaves 6 ; 2 from 9 leaves 7 ; and 1 from 2 leaves 1.

It is necessary,' in working by this method, to remember that any digit from which 1 has been removed now represents 1 less than its original intrinsic value.

Third Method :—

In working Subtraction by the first, or ordinary method, a plan is sometimes adopted by which subtraction from any number greater than 10 is avoided.

Rule III.—When the lower figure cannot he taken from the upper, take it f rom 10, add the result to the upper figure, and put down the sum. Carry 1 to the next figure in the Subtrahend, and proceed as usual.

Example.—By how much does 30165940 exceed 18139400 ?

30165940 minuend.

18139400 subtrahend.

12026540 excess.

Process.—In dealing with the thousands, say 9 from 5 we cannot, take 9 from 10 leaves 1 ; then, 1 and 5 are 6, put it down. Add 1 to the 3 in the subtrahend, and continue the subtraction.

Note 1.—This method is very suitable for young children, as the difficulty which they experience in subtracting from large numbers is entirely avoided.

Note 2 —In practice, the second method presents more difficulties than either the first or third ■. but the principles upon which it is worked are more easy of comprehension. The third method, which is hut a modification of the first, is the easiest both to work by and to teach.

Method of Proof :—

The Excess + the Subtrahend — the Minuend.

For, the excess of the greater number is also the defect of the less, and if the defect be added to the less number the result will equal the greater. Hence—

Rule I.—Add together the two bottom lines to obtain the top'line.

Again, The Minuend — the Difference = the Subtrahend.

For if the excess be taken from the larger number (or Minuend) that number will no longer exceed the smaller, but will be equal to it. Hence—

Rule II.—Take the answer from the Minuend to obtain the Subtrahend.

MULTIPLICATION.

To place Numbers for Multiplication :—

Rule.—Place the Multiplier under the Multiplicand with the last significant figure of the former beneath the units of the latter. If the Multiplier end in cyphers, place them to the right of the figure under the units.

Example:—Multiply 307695by 2307 and 870900 respectively.

307695 Multiplicand.    307695 Multiplicand.

2307 Multiplier.    870900 Multiplier.

Note.—In speaking of a figure as first or last in a number, first means farthest on the left, and last, farthest on the right.

RULE FOR MULTIPLICATION.

Rule.—Multiply the whole of the figures in the Multiplicand, beginning at the units, by the last significant figure of the Multiplier. Multiply the Multiplicand by the next significant figure, and then by the next, and so on ; taking care to place the first obtained figure of each partial product directly under the figure you are Multiplying by.

When the product of any two figures exceeds 10, put down the units and carry the tens, as in addition.

Wherever a cypher occurs in the multiplier, put a nought directly under it, and proceed to the next significant figure.

The partial products are placed beneath each other and finally added up, as shown in the following example :—

Example I.—Find the product of 150672 and 504200.

Note.—When the product of two abstract numbers is required, either numbers may be used as the multiplier.

150672 Multiplicand.    504200

504200 Multiplier.    150672

I. - IL --

30134400 \    1008400

602688    > Partial Products.    3529400

7533600    )    3025200

--- 25210000

75968822400 Product.    504200

75968822400

Example II.—Find the continued product of 8070, 30500, and 370.

Here we first find the product of any two of the given numbers, and multiply this product by the third number.

8070

30500

4035000

242100 246135000

370

17229450000

738405000 91069950000 Continued Product.

Multiplication by Factors :—

When the multiplier is a composite number, the product is the same as the continued product of the multiplicand and the factors of the multiplier. Hence, to multiply by factors :—

Rule.—Multiply the Multiplicand by the First

Factor; multiply the Product thus obtained by the next Factor, and so on, using a complete set of Factors.

Example III.—Multiply 76809504 by 1728.

The Factors of 1728 are 12, 12, and 12

76809504

12

921714048

12

11060568576

12

132726822912 Answer.

Multiplication by Prime Numbers is often effected:—

I.    By multiplying by the factors of the next lower composite number and adding to the result the product of the multiplicand and the difference between the composite number selected and the proper prime multiplier.

II.    By multiplying by the factors of the next higher composite number, and subtracting from the result the product of the multiplicand and the difference between the composite Dumber selected and the proper prime multiplier..

Examples.—Multiply 3928 by 89.

1st Plan.    2nd Plan.

3928 X 89    3928 x 89

8    9

31424 = 8 times 3928    35352 = 9 times top line

11    10

345664 =    88 times 3928    353520 =    90 times top line

add 3928 = 1 time 3928    subtract 3928 = 1 time top line

349592 = 89 times 3928.


349592 = 89 times 3928.


In such a sum as the preceding the second plan is the better, since the multiplication by 10 can he effected at once by adding a cypher to 9 times the top line : that is, we might have multiplied by 90 in one line, and then have subtracted the top line, thus—

3928 X 89 90

353520    90 times 3928

3928 = 1 time 3928

319592 = 89 times 3928.

To multiply by 10, 100, 1000, 10000, &c., at sight:—

Rule. Simply add as many cyphers to the multiplicand as there are in the multiplier :

Thus, 87 X 1000 = 87000.

To multiply by 25 :—

Rule.—Add two noughts and divide by 4 :

Thus, 64 x 25 = 6400 4- 4 = 1600.

To multiply by 99 :—

Rule.—Add two noughts and subtract the multiplicand :

Thus, 789 X 99 = 78900 - 789 = 78111.

To multiply by 125 :—

Rule.—Add three noughts and divide by 8 :

Thus, 848 X 125 = 848000 4- 8 = 106000.

To divide by 10, 100, 1000, &c., at sight:—

Rule.—Cut off as many figures as there are cyphers in the divisor: Thus, 370 -4- 10 = 37; 3700 -P 100 = 87; 365000 -p 1000 = 365.

To divide by 50, 500, or 5000, &c. :—

Rule.—Multiply by 2, and divide by 100, 1000, 10000, <tc.

Thus, 72900 -P 50 = 145800 -4- 100 — 1458.

To divide by 25, 75, 125, 175, 225, or 275 : —

Rule.Multiply by 4, and divide by 100, 300, 500, 700, 900, or 1100. Thus, 5625 -4- 225 = 5625 x 4- 900 = 25.

To divide by 15, 35, 45, or 55, respectively :—

Rule.Multiply by 2, cut off the last figure of the product, and divide the result by 3, 7, 9, or 11.

Thus, 360 -4- 15 = 72/0 -4- 3 = 24.

SHORT DIVISION.

Division by any number less than 13 is called Short Division.

Short Division is worked as under :—

Divisor. 9) 763848 Dividend.

84872 Quotient.

Here, 9 is not contained in 7 ; so we see how often it is contained in 76. It is contained 8 times and 4 over. Put down the 8 and place the 4 before the next figure 3; then 9 into 43 = 4, and 7 over. Place the 7 before the next figure 8 ; then 9 into 78 = 8, and 6 over. Place the 6 before the 4 ; then 9 into 64 = 7, and 1 over. Place the 1 before the 8 ; then 9 into 18 = 2.

To divide by any composite number greater than 12 :—

Rule.—Divide the dividend by one factor, and then divide result by the next factor, and so on, until each factor has been used.

Example I.—Divide 132726822912 by 1728.

Here the factors of the divisor are 12, 12, and 12.

12) 132726822912 Dividend.

12) * 11060568576 Partial Quotient.

12)    * '921714048 Partial Quotient.

76809504 Quotient, or Answer.

Compare the working of the above sum with the example of multiplication by factors on page 26, Example 111.

Example II.—How many times can 378 be taken from 9094680 %

Note.—Division is a short method of finding how often one number can be taken from another.

The factors of 378 are 9, 7, and 6.

9 ) 9094680 7 ) 1010520 6 ) 144360

24060

When remainders occur at the several steps of the operation in dividing by factors, the true remainder is obtained thus:—

Rule.—Multiply each remainder by the factors preceding that which produced the remainder, and add together the several results.

Example.—Divide 31255 into (1) 240, and (2), 5760 equal parts.

Here, 12 and 20 are convenient factors of 240; and 2), 3, 8, and 12 are the most convenient factors of 5760.

I.

12)31255


II.

20)31255

20)2604 and 7 over. 130 and 4 over.


3)1562 and 15 remainder.

8)520 and 2 remainder.

12)65 and 0 remainder.

5 and 5 remainder.

I. True Remainder (4 X 12) + 7 = 55.

II. True Remainder (5 X 8 X 3 X 20) + (2 x 20) + 15 = 2455.

To the scholar who can work Reduction, the reason for the above will be made clear by considering the first example as one of reduction of pence to pounds; and the second example as one of the reduction of grs. to lbs. Troy. The remainders 4 and 7 will then represent 4 shillings and 7 pence, or 55 pence ; while the remainders 5, 0, 2, 15, in example II., will represent 5 ozs., 0 dwts., 2 scr. 15 grs., or 2455 grains.

To find the True Remainder when two factors only are used :_

Rule. —Multiply the last remainder hy the first divisor, and add in the first remainder.

LONG DIVISION.

To Place Numbers for Division :—

Place the numbers thus :

Divisor.) Dividend. (Quotient.

To divide numbers :—

Rule.—Mark off from the left of the Dividend as many figures as will express a number not less than the Divisor. See how often the Divisor is contained in the number cut of, and put the result in the Quotient. Multiply the Divisor by the figure placed in the Quotient, and subtract the product thus obtained from the part of the Quotient cut off. Bring down the next figure of the Dividend, and divide the number thus formed in the same way, putting the result in the Quotient,— multiplying and subtracting, as before. Continue in this way until all the figures of the Dividend have been brought down.

Important Note.—Whenever the divisor is not contained in the number formed by bringing down the next figure of the dividend, put a nought in the quotient, bring down the next figure, and continue the division. Should the divisor still not be contained, put another nought in the quotient, bring down the next figure, &c. Remember, there will be one more figure in the quotient or answer than there are figures following the part at first marked off.

Example I.—Divide 349592 by 89.

Divisor Dividend Quotient.

89 ) 349/592 ( 3928 267

825

801

249

178 712

712

In this example there is no remainder.

Example II.—How often is 8070 contained in 246135600 ?

Divisor. Dividend. Quotient.

8070) 24613/5600 (30500 24210

40356

40350

• • ■ -600 Remainder.

When the product of the divisor and first figure is subtracted and the figure 5 brought down from the dividend, the number formed does not contain the divisor ; hence a nought is placed in the quotient and the next figure, 6, is brought down. The divisor is now contained 5 times, leaving 6 as a remainder, which becomes 60 when the next figure is brought down. 8,070 is not contained in 60, hence we put a nought in the quotient and bring down the next figure of the dividend. The divisor is still not contained, so we place another nought in the quotient ; and, as there is no other figure to bring down, the sum is finished. The answer is 30500, and the remainder 600.

The remainder is frequently placed after the quotient, with the divisor beneath it; thus—Quotient 30500 -g^^.

Method of Proof :—

To prove Division :—Rule.Multiply together the Divisor and Quotient, and add the Remainder to obtain the Dividend.

8070 x 30500 = 246135000 (see example of multiplication), and 246135000+ 600 = 246135600 = the Dividend.

The Dividend is thus the product of the Divisor and Quotient, + the Remainder. Hence, the Divisor may always be found by subtracting the Remainder from the Dividend and dividing by the Quotient.    .

Thus (246135600-600) + 30500 = 8070, the Divisor.

HINTS UPON CANCELLING BY INSPECTION.

1.    All even numbers are exactly divisible by 2.

2.    3 will measure every number the sum of whose digits is a multiple of 3.

3.    4 will measure any number whose two last figures denote a number divisible by 4.

4.    5 will measure every number ending in 5 or 0.

5.    6 will measure every even number the sum of whose digits is a multiple of 3.

6.    8 will measure every number whose three last figures denote a number divisible by 8.

7.    9 will measure every number whose digits give a sum divisible by 9.

8.    10 will measure every number ending in 0.

9.    11 will measure any number whose alternate figures added give the same sum as the other figures ; as, 785268, 374 ; or, whose alternate figures give a sum which is 11, 22, or some multiple of 11, in excess or defect of the sum of the other figures, as 8170943.

10.    12 will measure every number whose last two figures denote a number divisible by 4, and whose digits added give a multiple of 3 as a sum.

11.    15 will measure every number answering to the requirements of (2) and (4) above, and 18 will measure every even number whose digits give a sum divisible by 9.

12.    20 will measure every number answering to the requirements of (3) and (4).

13.    25 will measure every number ending in 25, 50, 75, or 00.

14.    Numbers divisible by 14 are also divisible by 2 and 7; numbers divisible by 16 are divisible by 4 and 4, or by 2 and 8, &c.; numbers divisible by 18 are divisible by 9 and 2, or by 3 and 6, &c. ; and, generally, numbers divisible by any composite number are divisible by the factors of that number

15.    Odd numbers multiplied by odd give an odd product.

16.    Even numbers multiplied by odd or even give an even product. The product of two or more odd numbers is odd ; all other products are even.

17.    Even numbers divided by even give an even remainder, if any;

HINTS UPON CANCELLING BY INSPECTION. 33

odd by even always give an odd remainder. In all other cases the remainders may be either odd or even.

18.    Numbers divisible by 7 can be told by trial only.

19.    Every prime number greater than 7 ends in 1, 3, 7, or 9 ; for all even numbers, and numbers ending in 5, are composite.

20.    Every prime number greater than 3 is either 1 more or 1 less than a multiple of 6 : thus, 13 is 1 more than twice 6, 17 is 1 less than three times 6.

21.    Every measure of two numbers is also a measure of their sum, or difference.

22.    Since 13, 17, and 19 are prime numbers, and their products, together with those of the figure 7, can be told by trial only, it is necessary to learn the multiplication tables of these numbers in order to facilitate cancelling.

23.    Any number consisting of two figures repeated in the same order an even number of times is divisible by 101 ; as, 5454.

For 5454 := 4 times 101 + 50 times 101, or 54 times 101.

24.    Any number consisting of three figures repeated an even number of times is divisible by 1001 ; as, 159159, 240240240240, 29029.

For 159159 = 9 times 1001 + 50 times 1001 + 100 times 1001 ; or, 159 times 1001.

Hence, (by 14), such a number will also be divisible by the factors of 1001 ; that is, by 7, 11, and 13.

25.    Any number consisting of four figures repeated twice, four times, &c., is divisible by 10001 ; or, by its factors, 73 and 137.

Note. (23), (24), and (25) are useful in cancelling a circulating decimal; while the following will enable us to measure a pure repeater of one figure

26 Any number made up of the same digits three times repeated —as 222, 555, 444666,—is divisible by 111; and, consequently, by 3 and 37, the factors of 111.

Cancellation.—Cancelling is a method of shortening the labour of multiplication and division when both operations have successively to be performed.

The process depends upon the principle that to multiply by a certain part of one number, and to divide by the same part of a second number, produces the same result as multiplying and dividing by the numbers themselves.

PRINCIPLES IN CONNECTION WITH THE FOUR SIMPLE RULES.

Axiom I.—Of any two numbers

(a)    The sum + the difference = twice the larger.

(b)    The sum — the difference — twice the smaller.

Reason.—The difference of two numbers is the excess of the larger over the smaller, or the deficiency of the smaller as compared with the larger.

Hence, The smaller -I- the difference gives the larger.

And, The larger - the difference leaves the smaller.

Now, since the sum is the larger and the smaller, by adding the defect of the smaller to the sum we obtain twice the larger; and, by taking away the excess of the greater from the sum, we get twice the smaller.

For, adding the difference to the sum is equivalent to adding the defect to the smaller and then adding the result to the larger, thus making twice the larger.

Again, taking the difference from the sum is equivalent to taking the excess from the larger and then adding the result to the smaller, thus making twice the smaller.

Having the sum and difference of two numbers, we thus obtain the following rule for finding the numbers :—

1.    To find the larger. Rule.—Add together the sum and difference

and divide by 2.

2.    To find the smaller. Rule.—From the sum take the difference

and divide by 2.

Or, as expressed by Algebraists,

\ sum + £ difference = the larger number.

| sum - | difference = the smaller number.

Problem 1.—The sum of two numbers is 20659 and their difference is 6077 ; find the numbers.

S;lution.—The largerof (20659 + 6077) = ^ of 26736 = 1336S.

The smaller = y of (20659 - 6077) = j of 14582 = 7291.

Problem 2.—The sum of the ages of a father and son is 37, and the father was 23 when the son was born ; find their ages.

Solution.—Here the difference between the ages is 23 years.

The father’s age = \ of (37 + 23) = 30 years.

The son’s age = £ of (37 - 23) = 7 years.

Axiom II.—The sum of numbers is the same in whatever order they are added:

Thus, 100 + 15 + 75 + 2 gives the same sum as 75 + 100 + 2 + 15.

Axiom III.—When addition and subtraction have severally to be 'performed, the residt is the same in whatever order the operations are performed:

Thus, 100 — 6 + 12 — 80 + 34 gives the same result as 100 + 34 + 12 — 80 — 6 = 146 — 80 — 6 = 66 — 6 = 60.

Axiom IV.— When several numbers have to be successively taken away, the result will be the same, if the sum of the numbers to be taken be subtracted :

Thus, 146 - 80 - 6 = 146 - (80 + 6) = 146 - 86 = 60.

Note.—When addition and subtraction occur promiscuously, it is frequently convenient to subtract the sum of the numbers preceded by the sign — from the sum of the other numbers.

The algebraic rule is—add together separately the positive and negative numbers, find the difference of these sums, and prefix the sign of the greater sum:

Thus, 8 + 9- 19 -8 + 10 - 5 = 27 - 32    - 5 ans.

Note.—Negative numbers belong more properly to algebra; but, since questions with negative answers have been proposed at recent examinations, an explanation of a negative number may not be inappropriate here.

Negative numbers, then, are always preceded by a minus sign ( —), and represent a deficiency. It is easy to understand that 12-7 signifies the number twelve together with a deficiency of seven, which gives the positive number five as the result ; but that 7-12 leaves a negative nifinber — 5, as the deficiency, is not so easily realized.

A negative number standing alone (thus, - 2) is purely an imaginary number, expressing (if the analogy may be pardoned), as much less than nothing, as its corresponding positive number expresses more than nothing. 20-35 = - 15, for since taking 20 from 20 would leave nothing, we can imagine that taking 15 more away would leave 15 less than nothing. In the same way, the property of a man who has £1000, and owes £1250, may be represented by - £250, since he is £250 worse off than if he simply had nothing, and were clear of debt. Negative numbers are subject to the same laws of increase and decrease as ordinary or positive numbers ; and are frequently useful to assist in solving difficult problems.

Negative numbers X negative numbers give a positive answer : thus, - 6 X - 8 = + 48.

Negative numbers X positive numbers give a negative answer; thus, - 6 X 8 zr - 48.

Negative numbers 4- negative numbers give a positive answer : thus, - 6 4- - 2 — 3.

Negative numbers 4- positive numbers give a negative answer : thus, - 6 4 2 ~ - 3.

Positive numbers 4- negative numbers give a negative answer : thus, 6 4- - 2 — - 3.

The product and quotient of two numbers being given, to find the numbers.

I.    To find the number which was divided by the other :—

Rule.Multiply the product by the quotient, and extract

the square root of the result.

II.    To find the number used as divisor :—

Rule.Divide the product by the quotient, and extract the square root of the result.

Axiom V.To multiply or divide by the factors of a number is to multiply or divide by the number itself :

Thus, 240 divided by 3 and 4, or by 2 and 6, gives the same as 240 4- 12.

Axiom VI.—The product of two numbers divided by either gives the other :

Thus, if S7 is the product of 29 and another number, that other number must be 87 divided by 29, or 3.

Axiom "VII.—Every dividend is the product of the divisor and quotient :

Hence, the dividend divided by the quotient will give the divisor.

Note.—If there be a remainder, subtract it from the dividend before dividing.

Axiom VIII.—The product of the sum and difference of two numbers equals the difference of their squares:

Thus, (8 + 3) x (8 - 3) = 82 - 32 = 55.

Axiom IX.—The square of the sum of two numbers equals the sum of their squares plus twice their product :

Thus, (8 + 3)2 = 82 + 32 + (8 x 3) x 2, or 64 + 9 + 4S = 121.

Axiom X.—Concrete numbers cannot be multiplied together :

Thus, we cannot say £5 times £8 ; but simply five times £8, or eight times £5. Such a sum as £19 19s. 9|d. x £19 19s. 9|d. is therefore without meaning. The multiplier must always be an abstract number.

Axiom XI.—Concrete numbers can be added or subtracted <mly when they refer to the same kind of thing :

Thus, if we consider horses and cows under the common head, animals, we can say that 5 horses and 8 cows make 13 animals ; but we cannot say that 5 horses and 8 cows make 13 horses or 13 cows.

Axiom XII.—The value of a number is not altered by removing one part and adding it to another part:

4

Hence, in Subtraction, when we cannot take the under number from the upper, we remove one of the figures in the place next on the left and add its equivalent to the number we wish to subtract from :

Thus, in taking 57 from 92, we say 7 from 2 we cannot, remove one ten from the 9 tens (leaving 8 tens), 10 and 2 are 12 and 7 from 12 leaves 5. We then take the 5 tens from the 8 tens, and thus obtain 3 tens. Then 3 tens and 5 = 35.

Axiom XIII.—The difference between tivo numbers is not altered by adding an equal amount to each:

Thus, the difference between 57 + 10 and 92 + 10 is the same as the difference between 57 and 92. Hence, Subtraction is often worked thus, 7 from 2 we cannot, add ten to the upper number (7 from 12 leaves 5) ; then, to prevent the difference from being altered, add 1 ten to the bottom line, and say 1 ten and 5 tens are 6 tens, 6 from 9 leaves 3.

Axiom XIV.—The product of two numbers plus the square of half their difference = the square of their average :

Thus, 12 x 8 + 22 = 102 = 100.

Hence, the product and difference of two numbers being given, to find the numbers :—

Rule.To the product, add the square of half the difference and extract the square root for their half sum.    Then, the

half-sum + the half-difference gives the larger number, and the half-sum - the half-difference gives the smaller number.

This rule, algebraically expressed, is :—

Rule.To four times the product, add the square of the difference and extract the square root for the sum. Then, having the sum and difference, find the numbers by Axiom /.

To find two numbers, whose sum and product are given:—

Rule.—From the square of half the sum (which is then-average) take the product, and extract the square root of the remainder to obtain the half-difference. To this half-difference add the half-sum for the larger, and take the half-difference from the half-sum for the smaller.

To prove Axiom XIV., take any numbers in succession, as, for instance, 4, 5, 6, 7, 8, 9, 10, 11, 12 ; square one of them (say 8), then the product of any two equidistant from it will be less than this square by the square of half the difference of these numbers. Thus, 7 x 9, 6 times 10, 5 times 11, and 4 times 12 will be 1, 4, 9, and 16 respectively less than 82 or 64.

Axiom XV.—In every multiple of 9, the digits when added make an exact number of nines. For illustration see Multiplication Table of 9.

Axiom XVI.—Every number, minus the sum of its digits, leaves a midtiple of nine.

Thus, 100 - 1 = 99, and 524 - 11 = 513 = 57 times 9 ; because, taking away the figure in the units’ place leaves 0 or no times 9, and taking away as many ones as there are tens will leave so many nines, and taking away as many units as there are hundreds will leave so many 99’s, and taking away as many units as there are thousands will leave so many 999’s, &c., &c.

Upon Axioms XV. and XVI. the rules for proving all the simple rules by casting out the nines depend. See page 41.

Axiom XVII.—Of any two numbers, the products of each part of one into all the parts of the other give a sum equal to the product of the numbers themselves.

If 10    7 + 3, and 12 = 6 + 4 + 2, then, 7x6+7x4+7x

2 + 3 X 6 + 3X4 + 3X 2 = 42 + 28 + 14 + 18 + 12 + 6 = 120 ; or 10 times 12.

Again, 16 X 15    (10 + 6) X (10 + 5) = 100 + 50 + 60 + 30

= 100 + 30 + 60 + 50 = 240.

We thus obtain the following rule to multiply one binominal by another.

Rule.—Add together the products of the first terms, the last terms, the means, and the extremes :

Thus (10 + 7) x (12 - 3) zz 120 - 21 + 84 - 30 zz 204 - 51 = 153 ; or, 17 times 9.

Note.—A binominal is a bracket enclosing two numbers or terms. The means are the inner of four numbers, the extremes the outer.

Corollary of Axiom XIY.—Of any two numbers with equal sums, the two which are nearest equal have the greatest ‘product:

Thus, 8 times 8 zi 64; while 7 times 9, 6 times 10, 5 times 11, &c., give products each less than 64, and less than the preceding product.

Corollary of Axiom XIV.—The square of any number = the product obtained by doubling the unit and multiplying the number thus formed by the tens and adding the square of the unit:

Thus, 63 x 63 zz 66 x 60 + 9 zz 3969.

Corollary of Axiom VIII.—The square of any number exceeds the square of the number next below it by the sum of the two numbers.

Thus, 13" zz 12| + (13 and 12), or 144 -f- 25 zz 169, which is the square of 13.    99'- zz 1002 - 199 zz 10000 - 199 zz 9801.

Corollary of Axiom VIII.—Of any two numbers whose difference is one, their sum = the difference of their squares.

For, since the sum X the difference zz the difference of their squares (Axiom 8), their sum X unity (or the unaltered sum) is the difference of their squares. Of any two fractions whose sum is unity, their difference zz the difference of their squares ; for the sum (or 1) X the difference zz difference of their squares, that is one time (the sum being 1) the difference zz the difference of their squares.

Note.—Multiplying or dividing by unity does not alter a number.

Axiom XVIII.—The sum of the dividend and divisor contains the divisor once more than the dividend itself does.

Axiom XIX.—The dffere'nce of the dividend and divisor contains the divisor once less than the dividend itself does.

Note.—Hence the rules, on page 55, for finding two numbers whose sum and quotient, or difference and quotient, are given.

PROVING BY CASTING OUT THE NINES. 41

RULES FOR PROVING BY CASTING OUT THE NINES.

Addition.—Rule.Cast the nines out of all the digits in the Adderids (in any order whateverJ, and the remainder will be the same as that produced by casting the nines out of the Sum or answer.

Subtraction.—Rule.Cast the nines out of the Minuend and Subtrahend; subtract the latter remainder from the former. This should give the same as casting the nines out of the Difference or answer.

Note.—If the remainder from the minuend cannot be taken from the remainder of the subtrahend, add 9 to the latter and then subtract.

Multiplication.—Rule.Cast the nines out of the Multiplicand and Multiplier ; multiply the remainders together, a,nd cast out the nines. This should give the same as cays ting the nines out of the Product.

Division.—Rule.Cast the nines out of the Divisor and Quotient; multiply the remainders together, and add in the excess of nines in the Remainder of the sum. This should give the same as casting the nines out of the Dividend.

Examples:—

1. Addition. 79890 Leaving out the 9’s and 0’s we have 7 and 36542    8 are 15 = 9 and 6 over ; 6 and 3 are 9 ;

90970    6 and 5 are 11 =9 and 2 over ; 2 and 4

-    and 2 and 7 are 15 = 9 and 6 over.

207402

Casting the nines out of the sum, we get, 2 + 7 + 4 + 2 = 15 = 9 and 6 over.

2.    Subtraction. 79890 Casting out nines leaves 6 over.

36542    ,,    ,,    ,,    2 over.

43348    ,,    ,,    ,,    4 over    = 6-2.

3.    Multiplication. 36542 Casting out nines leaves 2.

90970    „    ,,    ,,    7.

2357940

328878    7 x 2 = 14 = 9 and 5 over.

32887S0

3324225740 Casting out nines leaves 5 over.

4.    Division.    36542) 3324225748 (90970

2 over. 4 over. 7 over.

Remainder 8.

Then 7x2 = 14 + 8 = 22 = 2 nines and 4 over, and casting the nines out of the dividend leaves 4 over.

Note.—When an error of 9 or 0, or of misplacement of any digit or partial product, &c., has been made, casting out the nines will not detect the error.

PROOF BY CASTING OUT THE ELEVENS.

To cast out the elevens from a number : —

Rule.First cast out the elevens from the sum of the digits in the odd places, and then from the sum of the digits in the even places. Subtract the smaller remainder from the greater.

To prove by casting out the elevens :—

Rule.—Cast out the elevens, and proceed as in proof after casting out the nines.

Note.—The final results should either coincide or else give 11 as a sum.

The above rules depend upon the following principle :

Every odd power of 10 (as 10, 1000, 100000, &c.), is 1 less than a vnidtiple of 11, and every even power of 10 (as 100, 10000, 1000000, ¿cc.), is 1 more than a multiple of 11.

Thus, (10 + 1), (1000 + 1), (100 - 1), (10000 - 1), are all divisible by 11.

Exercise I.—NOTATION, NUMERATION, AND THE SIMPLE RULES.

1.    In the English method of notation, state what are the values of the figures which occur in the places named below :—the 1st, 3rd, 7th, 8th, and 13th.

2.    Write in the English method one billion sixty-seven millions five hundred and nine.

3.    Write down ten millions one hundred, and one million ten thousand; and find the product of their sum and difference.

4.    Find the product of the sum and difference of seventy millions six hundred and eighty thousand and forty, and nine millions nine hundred and nine thousand and fifty-four. Express the result in words.

5.    Find the sum, difference, and product of the following numbers. Sixty millions thirty-six thousand, eight hundred and five millions four hundred thousand and nine. Numerate the product according to the English method.

6.    If three thousand nine hundred and eighty-one millions seventy-three thousand seven hundred and twenty be diminished by seven thousand nine hundred and forty-eight, and the remainder again diminished by the same number, and so on ; find how many times the operation must be performed before there is no longer any remainder.

7.    Find the product of ten millions one thousand and ten, and one million eleven thousand, and express the result in words.

8.    In a division sum the divisor is seventy thousand nine hundred and eight, the quotient five hundred and eight thousand nine hundred and seventy, and the remainder six thousand and twenty. Find the dividend, and express it in words.

9. The dividend of a sum being four billions three hundred and thirty-nine thousand three hundred and forty-six millions two hundred and forty-two thousand two hundred and forty; and the quotient five hundred and seventy thousand eight hundred and ninety-six. Find the divisor.

10 (a). Distinguish between the intrinsic and local value of a figure.

(b) . Set down at full length the local value of each figure in 9072'06.

(c) . What is meant by the power of a number ? Give an instance, and explain it.

11.    Find the quotient of six hundred and thirty millions seventy-eight thousand and four by eighty-six thousand five hundred and forty-nine; and prove the correctness of

the operation.

12.    Find the quotient of five hundred millions six hundred and eighty thousand and forty-nine, by seventy-eight thousand five hundred and ninety-six. Prove the correctness of the operation.

13.    The product of two numbers increased by four thousand seven hundred and eighty-four (which is just one half of one of the numbers) is six thousand seven hundred and four millions two hundred and six thousand seven hundred and four. Find the numbers.

14.    Find the product of ten billions one thousand and ten and one million eleven thousand, and express the result in words.

15.    Find tlie sum, difference, and product of the following numbers :—Sixty millions thirty-six thousand eight hundred and seven, and five millions four hundred thousand and nine. Numerate the product according to the English method.

16.    Divide two hundred and thirty-eight billions four hundred and eighteen thousand six hundred and nine millions two hundred and twenty thousand nine hundred and forty-six by four millions three hundred and seven. Numerate the quotient according to the English method.

17.    Multiply twelve millions seven thousand and fifteen by one hundred and seven, and state in words the difference between the product and one thousand millions.

18.    Divide ten thousand one hundred and five millions twenty thousand and seven by fifty thousand and thirteen, and numerate the quotient and remainder.

19.    Division bears the same reference to Subtraction as Multiplication does to Addition. Explain this statement.

20.    What number multiplied by six hundred and seventy-nine will give the same result as five thousand six hundred and seven multiplied by four hundred and eighty-five 1

21.    Divide 47139 by 35, making use of factors. Explain how the full remainder is arrived at, and give reasons for the process.

22.    Divide 16798 by 48, making use of factors.

23.    Define the terms notation, addend, quotient. Find how often seven thousand six hundred and nine can be subtracted from thirty-eight millions six hundred and forty thousand and five, and state what the remainder will be.

24.    The sum of two numbers is 7410967470, and the greater exceeds the less by 7290785910. Find their product and numerate the result, stating what system of notation you adopt.

25.    Explain the process in working a sum in Long Division, and show that Division is but a shorter method of performing Subtraction.

26.    Find the product of the sum and difference of 746969164 and 38068794 and numerate the result after the English method.

27.    Define the terms Numeration, Minuend, Product. Find how often eight thousand nine hundred and seven must be added before it exceeds seventy millions nine thousand six hundred and four.

28.    Find the number which, when multiplied by 8405, will give 293839455590. Express the answer in words.

29.    Find and write down in words the number which, when divided by six hundred and eighty-five, will give for a quotient seventy-three thousand four hundred and thirty-six, with a remainder equal to three hundred and forty-seven.

30.    Find and write down, in words, the number which, when multiplied by seven hundred and sixty-five, will give such a product as, when increased by seven hundred and forty-two, will amount to sixty millions fifty-four thousand and seven.

31.    The product of two numbers is fifty-eight billions nine hundred and eighty-two thousand eight hundred and ninety-nine millions three hundred and ninety-five thousand and sixty, and the less of them is six millions eighty thousand and ninety-seven : find the greater.

32.    The greater of two numbers is five hundred and forty millions three hundred and sixty-four thousand eight hundred, and the less is one-sixth of the greater. Numerate their product according to the English method.

33.    What number divided by seven hundred and nine thousand five hundred and thirty is greater by five hundred and three thousand nine hundred and ninety-eight than nine hundred and sixty-three thousand and seven Numerate it.

34.    How many times must eight thousand seven hundred and nine be added to itself to give the sum of eight hundred and thirty-six billions seven hundred and sixty thousand seven hundred and twenty millions ?

Note.—8709 is contained in 836760720000000 exactly 96080000000 times. See Note to Sum 38.

35.    How many times can the difference between one million, and three hundred, be subtracted from seven thousand and forty millions nine thousand and eighteen What number must be added to this latter number that the above difference may be subtracted just once more1?

36.    The divisor is six hundred and nine millions four hundred and twenty-six thousand and seven, and the quotient is three hundred and six thousand and ninety-seven, and the remainder seven millions eight hundred and seventy-nine thousand six hundred and four. Write the dividend in words according to the English method.

37.    The quotient of a sum is seven millions nine thousand eight hundred and thirty and the divisor exceeds the ninth part thereof by 1240. What is the dividend 1

38.    How many times must eight thousand seven hundred and nine be added to itself to give the sum of eight hundred and thirty-six millions seven hundred and sixty thousand seven hundred and twenty 1

Note.—8709 is contained in 836,760,720 exactly 96080 times ; but, since the question is how many times must 8709 be added to itself, the correct answer is 96079, or one less than the number of times which it is contained. If 8 be added to itself once, the sum will be 16, or twice 8 ; if twice, the sum will be 24, or thrice 8; if three times, the sum will be 32, or four times 8, &c., &c.

In the answer given to sum (3) of I. of Ex. IX., Barnard Smith’s Arithmetic (page 31), the principle that a number added to itself 398 times will give 399 times that number has been overlooked. If the words “ to itself were omitted, the answer given would be correct.

39.    Write down in words the number which will give four thousand and forty as a remainder when eight thousand and fifteen has been subtracted from it seventy-five thousand and twenty-two times.

40.    Write down in words the number which will give three thousand one hundred and eighty-nine as remainder after forty thousand nine hundred and eighty have been subtracted from it five thousand and seven times.

41.    The sum of two numbers is six millions four hundred and eight thousand four hundred and thirty-two, and the greater is fifteen times the less. Find their product.

42.    The sum of two numbers is 2986900, and the greater is to the less as 7 : 3. Find their product.

43.    The sum of two numbers is 61589979, and their difference 61588441. Find how often the less is contained in the greater.

44.    Find a number which, on being multiplied by 769078 will give a product 5230337971620.

45.    Multiply 6784096 by 8726481, using only four multipliers.

Note.—This is done by first multiplying by the 8 in the millions’ place ; next, this product by 9 for the 72 ; this again by 9 for the 648 ; and, lastly, the top line by 1. Of course, the partial products must be carried to the right instead of to the left, and the proper number of cyphers appended before they are added.

46.    By what number must eight thousand and one be multiplied that the product may exceed fifty-six millions seventy-eight thousand and eight by one thousand and one 1

47.    What number multiplied by four thousand seven hundred and nine will give the same result as two millions three hundred and seven thousand four hundred and ten multiplied by ninety-four thousand and seventy ?

48.    How many times must seven hundred and eight thousand nine hundred and sixty-five be used as an addend before it exceeds 800,000,000 1

49.    What number diminished by three hundred and ninety millions twenty-three thousand seven hundred will be equal to the product of fifty millions nine hundred thousand six hundred and eighty, and seven millions eight hundred and nine thousand and four ? Express the answer in words.

50.    Write down one million and ten, and ten millions one thousand ; and prove that the product of their sum and difference is ecpial to the difference between their squares

51.    The divisor being 6809507, the quotient 786090, and the remainder 40967, find the dividend and enumerate it, according to the English method.

52.    What number, increased by four hundred and eight millions two thousand and sixteen will be equal to the product of (a) sixty-nine millions eighty thousand and seventy, and (b) eight millions seven hundred thousand nine hundred and five 1 Express your answer in words.

Exekcise II.—EXERCISE UPON THE SIMPLE

RULES.

1.    Add together 1300, 2750, 97, 6394 and 27942.

2.    Find the sum of 30709, 5625, 73297, 1200 and 605.

3.    Find the difference between 1700960 and 799990.

4.    By how much does 305060 exceed 290006 1

5.    How much less is 3706209 than 9190909 1

6.    What number must be added to 37609 to make 100000 1

7.    What is the defect of 9100905 compared with 100000001

8.    What number taken from a million will leave 2990091

9.    What is the amount of 500, 2001, 976, and 131

10.    Which is the greater—10000 diminished by 9654, or 3901

11.    When 50976 is taken from 59076, what is left 1

12.    From ten millions five hundred, take seven millions and ninety.

13.    Take 806398 from one million ten thousand and ten.

14.    What remains when fifty thousand nine hundred is taken from 568701

15.    To 50607, add 5689, and take 55689 from the sum.

16.    From 82000 take seventy thousand nine hundred and nine.

17.    How many years from the beginning of 1847 to the end of 19001

18.    Take eleven hundred and eleven from one thousand two hundred.

19.    What is the difference between thirteen hundred and one thousand two hundred 1

20.    By how much does seventeen thousand seventeen hundred exceed 18650 1

21.    What number added to 70286 will make 90495 1

22.    What number taken from 30706 will leave 260901

23.    From what number must 30706 be taken to leave 260901

24.    If 3026, 509 and 7000, successively taken from a certain number left 6095, what was the original number 1

25.    Add together the sum and difference of 20609 and 80964.

26.    From the sum of 8090 and 9090, take the difference.

27.    Add together ¿£1248, ¿£295, ¿£864, ¿£2900, and ¿£87.

28.    How many more days in a year of 365 days than in seven months of 31 days each ?

29.    The months of the year contain 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 days respectively. Find the number of days in twelve months.

30.    Find the value of 8600 + 32609 — 726 + 3727 — 9000.

31.    Find the difference between the sum of the positive and negative numbers contained in the following expression:—

36 - 40 + 509 - 67 + 837 + 497 - 364 - 246.

32.    To what number must 12976 be added to give 59688 ?

33.    By how much is the difference between 80096 and 70984 less than their suin'?

34.    What is the remainder of 706072 ^ 907062?

35.    Find the value of 3968 - 495 + 826 - 963 + 120096.

36.    Multiply 6806009 by 489700,and numerate theanswer.

37.    Find the product of seven thousand six hundred and eighty-nine and ninety-seven.

38.    Multiply six thousand seven hundred and eighty-nine by sixty-nine.

39.    Add together eight thousand seven hundred and ninety-six, seven thousand six hundred and eighty-nine, nine hundred and fifty-seven, and five thousand eight hundred and seventy-four.

40.    From five hundred and sixty thousand and forty-nine lake fifty-nine thousand nine hundred and sixty-four. Numerate the difference.

41.    Express in figures five millions eighty thousand and seven.

42.    From three hundred and seventy thousand and thirty-seven take fifty-nine thousand nine hundred and forty-three.

43.    Write down in words—7009030, and express in figures four millions eight hundred thousand and five.

44.    Express in figures nine millions seventy thousand and six.

45.    Divide eight hundred thousand one hundred and sixty-seven by seventy-nine.

46.    What is the least number which, taken from 764982, leaves a remainder divisible by 96 ; and what is the least number which, added to 70609, gives a sum divisible by 87?

47.    Find the product of 12345670 and 245497 by means of three lines of multiplication, and numerate the answer.

48.    Divide 357357 by 21021, using the factors 1001, 3 and 7.

49.    The difference between two numbers is 936 and the larger is 1267. Find the other.

50.    The difference between two numbers is 936 and the smaller is 1267. Find the other.

51.    The sum of two numbers is 2648 and one of them exceeds the other by 170. What are the numbers 1

52.    Find a number such that, if it be added to 768953, the sum will be 986742.

53.    The 96th part of a number is 2798. What is the number ?

54.    The product of two numbers is 15580656, and one of them is 13104. What is the other number1?

55.    The sum of two numbers is 987653, and their difference is 4781. Find the numbers.

56.    The quotient of a division sum is 2908, the divisor is 546, and the remainder 297. Find the dividend.

57.    The quotient arising from the division of 868000 by a certain number is 879, and the remainder 427. Find.the divisor.

58.    The product of two numbers is 1587768, and their difference is 2362. Find the numbers.

59.    The product of two numbers is 10137600, and their sum is 7520. What are the numbers 1

60.    Three numbers are such that the third is the same number of times the second that the second is of the first. The second number is 48, and the difference between this and the third is 96. Find the three numbers.

61.    Three numbers are such that the third exceeds the second by as much as the second exceeds the first—namely, by 9. If the sum of the numbers be 78, what are they1?

62.    The quotient of a division sum equals five times the divisor, and the divisor is five times the remainder. If the three together amount to 217, find the dividend.

63.    Multiply 987654321 by 576729, using three multipliers only. Numerate the answer.

64.    What number exceeds the sum of its half and 399

by 777 ?    _

65.    What number is less than the sum of its half and 4840 by 2200 1

66.    The continued product of 17, 7, and another number is 2261. Find the other number.

67.    Divide the sum of six hundred millions five hundred thousand and twelve and five hundred and ninety-nine millions nine thousand and nine by their difference.

68.    Divide 3795462 by 198, using factors.

69.    Find the sum of all prime numbers less than 100.

70.    Find the sum of all composite numbers from 40 to 100 inclusive.

71.    The less of two numbers is 567 and their difference 453. Fmd their product, and numerate it.

72.    Find the difference between 780 multiplied by itself and 67900. Numerate it.

73.    The quotent of two numbers is 71 and the larger number is 7029. Fmd the other number.

7 4. Find a number which is seventeen times as great as seventy-five times nine score.

75.    What number multiplied by 57 and divided by 79 will equal 56943 1

76.    What number will equal 108737 when 507 has been added to it and the sum multiplied by 97 1

77.    What number will make 1000 when it has been divided by 9, the quotient thus obtained multiplied by and 237 added to the product 1

78.    What number is it from which, if 32 be taken, 900 be added to the remainder, and the sum thus obtained be multiplied by itself will amount to a million 1

79.    By what number must we multiply one thousand in order that, after dividing the product by 1080, we shall obtain 6001

80.    What number must be multiplied by 20 in order that, by continued division by 4, 7, 8 and 10, the quotient will be 11

81.    What number must be divided by 15, and the quotient thus obtained multiplied by 7, in order to give 1611

82.    If 74 be added to a certain number and 90 taken from the sum, the result will be 100. What is the number!

Note —In working the last nine sums, it will be well to remember that, to undo what has been done, the opposite operation to that which is mentioned must be performed ; thus, if the number 240 multiplied by 17, and the product divided by 12 equals 340, then 340 x 12 —17 will equal 240.

Addition and Subtraction, Multiplication and Division, Involution and Evolution are operations the converse of each other.

83.    Find and numerate the continued product of 90070 X 5090 x 3060800. Before performing the operation required, say how many figures the answer will contain.

84.    Without performing the operation of multiplication, say how many cyphers the product of 375 multiplied by 76000 will contain.

85.    Increase 270960 ninety thousand and fifty-two times. Numerate your answer.

86.    Square 1728 ; that is, nudtiply 1728 by itself.

87.    Cube 365; that is, multiply 365 times 365 by 365.

88.    Find the least number that will contain 13, 17, 19, 23, and 29, each an exact number of times. This is done by finding the continued product of the numbers.

89.    What number taken from the square of 165 will leave 175 times 155 1

90.    What number is that to which, if 5000 be added and 4926 be taken from the sum, will equal 864741 See note to sum 82.

91.    Divide the fourth part of 31416 by 66 both by short and by long division. In the latter case prove the correctness of the answer by adding the three subtrahends as they occur one above the other. This should give the dividend. By what number of one digit is the answer divisible 1

92.    Divide 277274 by 79 and prove the correctness of the answer by adding the remainder to the various subtrahends as they appear above one another.

93.    Divide the sum of 1760, 5280, 4840, and 1728, by the difference between 365 and 313. Prove by casting out the nines.

94.    The product of two numbers is 78 times the larger;

what is the larger number, if the eighth part of the product equals 8780851    _    ^

95.    Find the value of the expression: 6372+-18 + 37 x

34    2 — 639 -r- 71 + 3065 - 3729 + 70007 x 13 x 11 -

350. Numerate the answer.

96.    What does (1224 — 2560 -h 4 + 336 -f 84 x 7 — 512 + 126 r 6 x 9 — 100) x 189 amount to?

97.    The product of two numbers is 40320000 and their difference is 1240. Find the numbers.

98.    The product of two numbers is 1728 and the quotient of the larger divided by the smaller is 6|. Find the numbers.

99.    A father was 35 years of age when his eldest son was born, and the sum of their present ages is 51. Find their ages.

100.    Find the sum, difference, product, and quotient of 450723000 and 635. Numerate the answers according to both the English and the Continental method.

101.    Multiply 12345679 by 9, 18, 27, 36, 45, 54, 63, 72, and 81 respectively, and find the sum of the products.

102.    Multiply 12345679 by 171 and by 153, and find and numerate the differerice of the products.

103.    The minuend is 9876543210 and the difference is 8758832033. Find the subtrahend, and numerate it.

104.    The product of two numbers is 605377, and the quotient obtained by dividing the larger by the smaller is

97. Find the numbers.

105.    The difference of two numbers contains the smaller once less than the larger number contains it. Hence : To find two numbers whose difference and quotient are given.

Rule.Divide the Difference by one less than the Quotient to obtain the smaller number, and Divide the Product of the Quotient and Difference by one less than the Quotient to find the larger number. From this find the numbers whose difference is 76, and quotient 5.

106.    The sum of two numbers contains the smaller once more than the larger number contains it. Hence : To find two numbers whose sum and quotient are given.

Rule.Divide the Sum by one more than the Quotient to obtain the smaller number ; and the Product of the Quotient and Sum by one more than the Quotient to find the larger number. By these rules find the numbers whose sum is 776, and quotient 7.

CHAPTER II.

TABLES.

Money Table.

1    Farthing = One Quarter of a Penny, written |d.

2    Farthings = One Half-Penny, written |d.

3    Farthings = Three Quarters of a Penny, written |d.

4    Farthings = 1 Penny, written Id.

12 Pence = 1 Shilling, written Is.

20 Shillings = 1 Pound or Sovereign, written XI.

ADDITIONAL.

A Florin == 2s. Od. A Half-Crown = 2s. 6d. A Crown = 5 s. Od.


A Half-Sovereign = 10s. Od. A Half-Guinea = 10s. 6d. A Guinea    =    21s.    Od.


Pounds, shillings, and pence, together with the other coins of the realm, are called Sterling Money.

The coins not mentioned above bear names significant of the number of pence they contain: they are, the threepence, fourpence, and sixpence.

An Alloy is a mixture of gold or silver with an inferior metal.

Standard Gold is an alloy of 22 parts of pure gold and 2 parts of copper.

A Carat means a twenty-fourth part; so that if 22 parts out of 24 be pure gold, the alloy is said to be 22 carats fine.

Standard gold is 22 carats fine.

Jewellers’ gold is 18 carats fine.

A Sovereign weighs 123-hyl grains, while 5,607 sovereigns contain 110 lbs. Troy of pure gold and 10 lbs of copper.

£, s., d., and q., signifying pounds, shillings, pence, and farthings, are the initial letters of Libra, Solidus, Denarius, and Quadrans, Latin words, names of Roman coins.

Obsolete coins—A Noble = 6s. 8d. ; an Angel = 10s. ; a    Mark =    13s.    4d. ; a Guinea =    21s. ;    a    Carolus    =    23s.    ;

a    Jacobus    =    25s. ; a Moidore =    27s. j    a    Groat    =    4d.;    a

Tester = 6d. ; a Half-Guinea = 10s. 6d.

Decimal coinage—

10 Mils = 1    Cent.

10 Cents = 1    Florin

10 Florins = 1    £

Note.—A Cent is the one hundredth part of a £, or 2|d. A Mil is rather less than a farthing.

Note.—Bronze coins are not legal payment for more than a shilling; nor is silver a legal tender for more than forty shillings.

Lineal Measure.

12    Lines    =

12    Inches    =

3    Feet    =

51    Yards    =

40    Poles    =

8    Furlongs    =

3    Miles    =


1 Inch (in.)

1 Foot (ft.)

1 Yard (yd.)

1 Rod, Pole, or Perch 1 Furlong 1 Mile 1 League


The unit of measurement is a straight line an inch long; thus, ___    .

ADDITIONAL.

7 Yards = 1    Irish Perch    22    Yards    =    1    Chain

3 Inches = 1    Palm    4    Inches    =    1    Hand

9 Inches = 1    Span    18    Inches    =    1    Cubit

Links and Chains are used in measuring land, and the Hand in measuring the height of a horse.

Note.—Lineal Measure, or Long Measure, is used to find the length, breadth, depth, or height of anything.

Cloth Measure.

2^ Inches 4 Nails

4    Quarters

5    Quarters


11 Irish Miles


: 1 Degree


30 Inches = 1 Pace


6 Feet = 1 Fathom


(Military)


1760 Yards = 1 Mile


7‘9 2 Inches = 1 Link 100 Links — 1 Chain


6080 Feet = 1 Knot

(Nautical)


792 Inches = 1 Chain 80 Chains = 1 Mile


1 Nail (na.)

1 Quarter (qr.)

1 Yard (yd.)

1 English Ell (En. ell.)

ADDITIONAL.

3 Quarters = 1 Flemish Ell; 6 Quarters = 1 French Ell.

N0TE-—This Measure was used in measuring cloth ; but the nail is now very seldom employed, being supplanted by its equivalent, one-sixteenth of a yard.

Square Measure, or Surface Measure.

144 Square Inches = 1 Square Foot (sq. ft.)

9 Square Feet = 1 Square Yard (sq. yd.)

30-| Square Yards = 1 Square Pole or Perch (sq.po. or sq. per.) 40 Perches    =    1 Pood (ro. or rd.)

4 Poods    =    1 Acre (ac.)

640 Acres    =    1 Square Mile (sq. ml.)

ADDITIONAL.

49 Sq. Yds. =1 Sq. Perch Irish 484 Sq. Yds.    = 1 Sq. Chain

62-7264 Sq. In.    = 1 Sq. Link

10,000 Sq. Links = 1 Sq. Chain 100,000 Sq. Links = 1 Acre 10 Sq. Chains = 1 Acre 272^ Sq. Feet = 1 Square Foci (The Sq. Fod is used in Brickwork.)

4,840 Sq. Yards = 1 Acre 121 Irish Acres = 196 English Acres

The Unit of Measurement is a Square Inch ; that is, a square measuring am inch each way ; thus,—

A Square Inch.

The Feet, Yards, and Perches of Square Measure are obtained from Lineal Measure by squaring the numbers representing (a) the No. of inches in a foot, (b) the No. of feet in a yard, and (c) the No. of yards in a perch.

Note—To square a number is to multiply it by itself.

12 In. = 1    Ft. ; then, 12 times 12 (or 144) Sq.    In.    =    1    Sq.    Ft.

3 Ft. =1    Yd. ; then, 3 times 3 (or 9) Sq. Ft.    =    1    Sq.    Yd.

Yds. = 1    Per.; then, 5|times5.) (or30£) Sq. Yds.    =    1    Sq.    Per.

Note—Square Measure is used for measuring    the    area of    any

surface, whether land; the walls, floor, or ceiling of a room; the surface of a plank, or of a roll of paper; or anything where the extent of surface is required. The area of any surface is obtained by multiplying together its two dimensions ; whether length and breadth, length and height, or length and depth. No account is taken of the third dimension, which is generally the thickness of the thing measured. Links and chains are usually employed in measuring land. They afford a much readier means of calculation than yards, perches, roods, &c.; for acres may be at once expressed in links or chains, and links or chains as acres, by merely removing the decimal point. Furthermore, the work of Addition, Subtraction, Multiplication, and Division, requires the application of the simple rules only.

Examples:—

35-760 acres = 357’60 chains = 3576000 square links.

3265000 square links = 326'5000 square chains = 32'65 acres.

528 acres = 5280 square chains = 52800000 square links.

From these examples it will be seen that acres are expressed as square chains by removing the decimal point one place to the right ; and as square links by removing it five places. Again, to express square links as acres we have merely to remove the decimal point five places to the left; or, in case there is no decimal point, to cut off the last five figures of the number expressing the links.

This matter will be treated of more fully under Decimal Fractions.

Cubic Measure.

1728 Cubic Inches = 1 Cubic Foot (c. ft.) 27 Cubic Feet = 1 Cubic Yard (c. yd.)

ADDITIONAL.

= 1 Ton of Shipping — 1 Cord of Firewood


40 Cubic Feet of unhewn timber, or 50 Cubic Feet of hewn timber

42 Cubic Feet 128 Cubic Feet

The unit of measurement is a cube measuring an inch each way 3 that is a solid—an inch long, an inch broad, and an inch deep.

A Cubic Inch is represented thus :—

Cubic Inch.

Cubic Measure is obtained from Long Measure by finding the cube of the numbers expressing (a) the number of inches in a foot, (b} the number of feet in a yard.

Thus :—Since 12 inches = 1 lineal foot,

12 x 12 x 12, or 1728 cubic inches = 1 cubic foot. And, since 3 feet = 1 lineal yard,

3    3 x 3, or 27 cubic feet = 1 cubic yard.

Cubic Measure is used for finding the cubical contents of any mass, whether solid, fluid, or aerial.

Note.—A Cube is a figure having six equal square surfaces.

Each surface of a cubic inch is a square inch.

Each surface of a cubic foot is a square foot.

Each surface of a cubic yard is a square yard.

Hence, in finding the area of the sides of a cube, or of any rectangular parallelopiped, Square Measure must be employed.

A Rectangular Parallelopiped is a figure the shape of a room or box. It differs from a cube in having different dimensions for its length, breadth, and thickness.

TABLES OF WEIGHT.

Avoirdupois Weight.

16 Drams 16 ozs.

28 lbs.


4 qrs. 20 cwts.


1 Ounce (oz.)

1 Pound (lb.)

1 Quarter (qr.) 1 Hundredweight (cwt.) 1 Ton

ADDITIONAL.

14 lbs. = 1 Stone; 7000 grains Troy = 1 lb. Avoirdupois.

144 lbs. Av. = 175 lbs. Troy ;

Hence, 1 lb. Av. = 1 lb. 2 ozs. 11 dwts. 16 grs. Troy. Note.—Avoirdupois weight is used for all common goods.

Troy Weight.

24 Grains 20 dwts. 12 ozs.


1 Pennyweight (dwt.) 1 Ounce (oz.)

1 Pound (lb.)

ADDITIONAL.

3^ Grains = 1 Carat (used in weighing diamonds).

The term grain originated in the use of a grain of wheat as a unit of weight.

Troy weight is used for gold, silver, jewels, and some precious stones. It is also used in philosophical experiments, and for ascertaining the specific gravity of anything.

Apothecaries’ Weight.

20 Grains    =    1    Scruple (1 sc. or 13.)

3 Scruples    =    1    Dram (1 dr. or 1 3.)

8 Drams    =    1    Ounce (1 oz. or 1 5.)

12 Ozs.


1 Pound (1 lb. or ft>.)

Liquid Measure.

Apothecaries’ 60 Minims = 8 Fluid Drams = 20 Fluid Ounces = 8 Pints    =


1 Fluid Dram (fl. dr.) 1 Fluid Ounce (fl. oz.) 1 Pint (0.)

1 Gallon (C.)

A minim is about two drops; a dram, a teaspoonful; and an ounce, two tablespoonfuls.—H. Evers’s Arith.

Apothecaries’ weight is used for mixing medicine. Medicines are bought and sold, however, by Avoirdupois weight.

MEASURES OF CAPACITY.

Liquid Measure.

4 Gills

=

1 Pint (pt.)

2 Pints

=

1 Quart (qt.)

4 Quarts

=

1 Gallon (gall.)

ADDITIONAL.

Of Wine-

-42 Gall. ==

1 Tierce

63 Gall. = 1 Hogshead

2 Hhds. =

1 Pipe

2 Pipes — 1 Tun

Of Beer-

-36 Gall. =

1 Barrel

54 Gall. = 1 Hogshead

2 Hhds. =

1 Butt

2 Butts = 1 Tun

Note.—A gallon of water weighs about 10 lbs. (Avoir.)

A gallon of water measures 277 ‘274 cubic inches A cubic foot of water weighs 1000 ozs. nearly.

Coals are sold by the bag of 140 lbs., sixteen bags making 1 ton.

Dry Measure or Corn Measure.

4 Quarts 2 Gallons

4    Pecks

8 Bushels

5    Quarters 2 Loads


1 Gallon (gal.) 1 Peck (pk.)

1 Bushel (bus.) 1 Quarter (qr.) 1 Load (Id.)

1 Last (last).

ADDITIONAL.

2 Quarts = 1 Pottle    2 Bushels = 1 Strike

4 Bushels = 1 Coomb    2 Coombs = 1 Quarter

20 lbs. = 1 Bush, of Bran or Pollard 40 lbs. = 1 Bush, of Oats 50 lbs. = 1 Bush, of Barley 60 lbs. = 1 Bush, of Wheat.

Measure

60 Seconds    =

60 Minutes    =

24 Hours    =

7 Days    =

4 Weeks    =

12 Calendar Months * =

Note.—Seconds and Minutes a 1 second, T for 1 minute.


of Time.

1 Minute (1 min.)

1 Hour (hr.)

1 Day (dy.)

1 Week (wk.)

1 Lunar Month 1 Year

e frequently marked thus, 1" for


The Common Year contains 365 days.

Leap Year contains 366 days, the extra or intercalary day being added to February, which month then contains 29 days.

* The number of days in the Calendar Months are

January,

31

April,

30

July,

31

October,

31

February,

28

May,

31

August,

31

November,

30-

March,

31

June,

30

September, 30

December,

31

The Solar Year, which contains 365-242218 days, is the time which the earth takes to make one complete revolution around the sun.

The Average or Julian Year contains 365^ days.

The difference between the Solar and Julian Year is thus, 365-242218 — 365-25, or -007782 days.

Now, -007782 days x 400 = 3-1128; whence we see, that taking every fourth year as leap year, causes an error of rather more than three days in every 400 years. To avoid this, three leap years are omitted out of every 400 years, viz.,— those which complete a century the name of which is expressed by a number not divisible by 4; thus, while the year 1600 is leap year together with every fourth year in general, the years 4700, 1800, 1900, 2100, &c., are not leap years. See Barnard Smith’s Arithmetic, page 94.

The subjoined well-known rhyme assists the memory in learning the number of days in the Calendar Months.

Thirty days hath September,

April, June, and November ;

While all the rest have thirty-one,

Save February twenty-eight alone,

Except in leap-year, at which time February’s days are twenty-nine.

Measure of Angles, and Division of the Circle.

60 Seconds (")    1 Minute (')

60 Minutes (') = 1 Degree (°)

90 Degrees (°) = 1 Quadrant 360 Degrees = 1 Circle.

ADDITIONAL.

An angle of 90° is called a Right Angle.

A right angle, better known to children as a square corner (i.e. the corner of a square), is the angle formed by two lines, one perpendicular to the other; thus, J, |, or |_.

Acute angle, an angle of less than 90°; as, ¿T .

Obtuse angle, an angle of more than 90°; as -y .

It will be seen, by drawing one straight line perpendicular to another, that two right angles, or an angle of 180° is, in reality, not an angle at all, for the lines which form it must be part of one and the same straight line, 90^j^o^

DEFINITIONS.

I.    The circumference of a circle is the line which bounds it.

II.    The diameter is the distance across from side to side through the centre.

III.    The radius is the half-diameter, or distance from the centre to the circumference.

The circumference of a circle is made the unit of measurement for the measure of angles and the division of the circle.

Every circle may be supposed to be divided into 360 equal parts, by 360 lines drawn from the centre to the circumference (that is, by 360 radii). The angles formed at the centre by these radii would each contain 1°, and the parts of the circumference cut off by two adjacent radii would each contain 1°.

Paper Measure.

1 Quire 1 Ream 1 Bale


24 Sheets 20 Quires 10 Reams

ADDITIONAL.

The principal varieties of sheets are :—

Foolscap,

161

inches by 13^ inches

Post,

19

15i „

Large Post,

21

U

16* „

Medium,

22

yy

17* „

Demy,

23

yy

18 „

Poyal,

25

yy

20

Imperial,

30f

yy

22

Double Elephant,

40

yy

26f „

Antiquarian,

53

yy

31 „

Miscellaneous.

12 Articles

=

/

1 Dozen

20 Articles

==

1 Score

144 Articles

=

1 Gross

8 lbs. of Meat

==

1 Stone

100 Per cent.

=

The whole

50 Per cent.

=

One-half

33^ Per cent.

=

One-third

25 Per cent.

=

One-fourth

20 Per cent.

=

One-fifth

16| Per cent.

=

One-sixth

12|- Per cent.

=

One-eighth

10 Per cent.

=

One-tenth

8^ Per cent.

=

One-twelfth

5 Per cent.

=

One-twentieth

THE COMPOUND RULES AND REDUCTION.

The Compound Rules.

The Simple Rules deal exclusively with abstract numbers, and concrete numbers of one denomination only.

In the Compound Rules, many of the principles explained in connection with the simple rules are extended to concrete numbers of two or more denominations.

In Compound Addition and Compound Subtraction, concrete numbers only are dealt with; and, further, the things to be operated upon must be of the same kind.

In Compound Multiplication, a compound concrete number is multiplied by an abstract number, and a concrete product is the result.

In Compound Division a compound concrete number is divided by either (1) an abstract number, or (2) a concrete number of the same kind as itself.

Remark.—The division of a concrete number by an abstract number, produces, as quotient, a concrete number of the same name as the dividend. The division of one concrete number by another of the same name and kind produces an abstract number as quotient.

It has been shown that one concrete number cannot be multiplied by another; for, the multiplier must always be an abstract number. Further, in Division, which is the converse operation of Multiplication, either the divisor or the quotient must be an abstract number ; for the dividend is the product of divisor and quotient, and no product could be obtained from them if both were concrete.

Compound Addition. The addends and sum are concrete numbers of the same kind.

Compound Subtraction. The subtrahend, minuend, and difference are concrete numbers of the same hind.

Compound Multiplication. The multiplicand and product are concrete numbers of the same hind. The midtiplier is an abstract number.

Compound Division. The dividend is a concrete number. The quotient is abstract when the divisor is concrete; and, vice versd, concrete when the divisor is abstract.

There is an apparent exception to the rule that the multiplier can never be concrete; for, in order to produce square inches, square feet, square yards, or square perches, we appear to multiply lineal inches by lineal inches, lineal feet by lineal feet, &c. A little reflection, however, will convince us that the anomaly is merely apparent, and that, in reality, we treat the numbers representing the inches, feet, &c., as abstract. For what possible meaning can we attach to the expression “ eight inches times eight inchesany more than to the phrase “five shillings times five shillings ?” Again, the meaning is not “8 times 8 lineal inchesfor the product would then be 64 lineal inches.

The fact is that the base of Square Measure (or the “unit of measurement ”) is a square whose side is an inch in length ; and that a rectangle contains as many of these units of measurement as are expressed by the product of the abstract numbers representing in inches the length and breadth of the figure.

In like manner, if the sides be given in feet, yards, or perches, we multiply together the abstract numbers representing these. Suppose, for example, we require to find the number of sq. ft. in a rectangle 4 ft. long and 2 ft. broad. We say,—“ 4 times 2 are 8; ” and then call these 8 square feet.

In Cubic Measure, likewise, the expression “feet multiplied by feet multiplied by feet” is merely a convenient method of expressing the multiplication of the abstract numbers representing these feet.

Note.—The apparent irregularity of the multiplication of two or more concrete numbers exists only where the product obtained is a power (as the square, cube &c.) of the quantities to be multiplied. Thus, we say, “ 3j ft. multiplied by 2 ft. equals 6 sq. ft.; ” but not “£3 multiplied by £2 equals 6 square pounds ; ’’ for, we have no such name as square pounds. For a similar reason we do not say

“ 4 fur. x 3 fur. = 12 sq. furlongs,” because we have no measurement termed a sq. furlong.

The convenience arising from the use of such expressions as multiply feet by feet, ifc.," is sufficient reason for retaining them in treating of square and cubic measure.

In support of what has been said, we may quote the following from “Cornwall & Fitch’s Science of Arithmetic,” page 44 : “The multiplier is an abstract number, for it expresses the number of times that the multiplicand is repeated. It can never be concrete. ”

We may also append Bishop Colenso’s note to an example of the multiplication of the length and breadth of a room in order to find the area of the floor. See “Colenso’s Arithmetic,” page 26.

“ It might at first sight appear that we are here multiplying inches by inches, contrary to the statement in (15); but, in reality, it is only the numbers 212 and 108 that we multiply, not the quantities 212 in. and 108 in.: so also the resulting product is only the number 22896, to which we append sq. in., because we know from the above, that this is the number of square inches in the given area. A similar remark applies to all such c ases, and to all such expressions as multiplying the length by the breadth, &c. The Student’s attention should be strongly drawn to this.”

Some arithmeticians confine the term Compound Rules to addition, subtraction, &c., of Money. In the following chapter the term is extended to concrete quantities of all kinds.

COMPOUND ADDITION.

To place numbers for addition or subtraction :—-

Rule.—Place them one under the other so that units of the same name shall he in vertical columns.

RULE FOR ADDITION.

Rule.—Add up the numbers in each column, commencing with that of the lowest denomination ; divide the sum by the number which expresses one of the next higher denomination ; put down the remainder, if any, under the column just added, and carry the result to the next column. Add in the number carried when finding the sum of the next cohcmn.

Example I-

Example II.

£

s.

d.

Cwts. qrs.

lbs.

206

15

9f

3

614

2

13

1040

2

Hi

1

27

3

14

30

16

84

2

500

1

20

954

18

6i

1

45

2

25

2476

12

n

2

720

3

8

4709    6    71    9    1909    1    24

Ans. I.—£4709 6s. 7¿d.    Ans. II.—1909 cwt. 1 qr. 24 lbs.

Process :—

Example I. — Here we first add the farthings, obtaining 9 farthings, which equal 2^d. ; we put down the |d. and carry the 2 pence to the pence’ column, which then gives 43d. as the sum. Now 43d. equal 3s. 7d.; we put down the 7d. and carry the 3s. to the shillings’ column, which then gives a sum of 66s. Now 66s. equal £3 6s.; so we put the 6s. under the shillings’ column, and carry the £3 to the pounds’ column, which then gives £4709 as its sum.    '

Note.—The farthings, pence, and shillings tables being known, it becomes unnecessary to use division in order to reduce the sum of any column to the next higher name in the addition of money.

“ With young beginners it is as well to put small figures opposite the farthings, as above, until the scholar has become more familiar with the fractional forms

ia, ia., |fi.

In Example II, after having added the lbs. and divided by 28 to reduce them to qrs., and having added the qrs. and divided by 4 to reduce them to cwts., the sum of the cwts. might have been divided by 20 to reduce them to tons. The answer would then be expressed as—95 tons, 9 cwt., 1 qr., 24 lbs.

COMPOUND SUBTRACTION.

FIRST METHOD.

Rule.Commencing at the lowest denomination, take the units of each name in the subtrahend from those of the same nam,e in the minuend. If this cannot be done, take the number in the subtrahend from the number of units which would make one of the next higher denomination, and add the result to the number above in the minuend. Put the sum as that part of the answer, and proceed to treat the numbers of the next denomination in the same way, taking care to add 1 to the subtrahend (as in simple subtraction), if the number in the subtrahend was greater than that in the minuend in the denomination last dealt with.

Example I.—From 326 cwt. 3 qrs. 20 lbs. 14ozs. 12 drs., take 169 cwt. 3 qrs. 26 lbs. 6 ozs. 8 drs.

cwt.

qrs.

lbs.

ozs.

drs.

326

3

20

14

12

Minuend

169

3

26

6

8

Subtrahend

156

3

22

8

4

Answer

Process.—8 drs. from 12 drs. leaves 4 drs. ; 6 ozs. from 14 ozs. leaves 8 oz. ; 26 lbs. from 20 lbs. we cannot ; 28 lbs. make 1 qr., take 26 from 28 leaves 2, which, added to the 20, gives 22 lbs. ; carry 1, 1 and 3 are 4 ; 4 from 3 we cannot, 4 qrs. make 1 cwt., take 4 from 4 leaves 0, which, added to the 3, gives 3 qrs. ; carry 1, then 170 cwt from 326 cwt. leaves 156 cwt. The 156 cwt. might be brought to tons, when the answer is 7 tons 16 cwt. 3 qrs. 22 lbs. 8 ozs. 4 drs.

Example II.—Subtract 17lbs. 11 ozs. 23grs., from 20lbs. lOoz. 3dwt. 20 grs.

lbs.

ozs.

dwt.

grs.

20

10

3

20

Minuend

17

11

0

23

Subtrahend

2

11

2

21

Answer

Process.—23 from 20 we cannot, 23 from 24 (the number of grs. in dwt.) leaves 1, which, added to 20, gives 21; carry 1, 1 from 3 leaves 2 ; 11 from 10 we cannot, 11 from 12 (the number of ozs. in a lb.) leaves 1, 1 and 10 are 11 ; carry 1, 18 from 20 leaves 2.

Answer.—2 lbs. 11 oz. 2 dwt. 21 grs.

SECOND METHOD.

Rule.—When any number in the subtrahend cannot be taken from the number above it, change one of the units of the next higher name in the minuend into its equivalent in

this denomination, add this to the upper number and then subtract1 In dealing with the next column, remember that the upper number expresses one less on account of the removal

of one of its units.

Example I.

Example

II.

£. s. d.

Ac.

ro. per.

sq. yds.

2468 11 6f Minuend

325

2 13

261 Minuend

1359 12 8| Subtrahend

112

1 6

28|- Subtrahend

1108 18 10£ Difference

213

1 6

28 Difference.

Process :—

Example I.—Two farthings from 3 farthings leaves 1 farthing. 8d. from 6d we cannot ; change one of the eleven shillings into pence ; Is. = 12d., then 12d. added to 6d. = 18d. ; and 8d. from 18d. leaves lOd. Next, 12s. from 10s. we cannot; remove a pound from £2468 and change into shillings ; £1 = 20s., then 20s. and 10s. = 30s., and 12s. from 30s. leaves 18s. (Here note that we have treated the 11s. as 10s. because Is. was removed and changed into pence. For a similar reason we now treat the £2468 as £2467.) Next, £1359 from £2467 leaves £1108. Answer.—£1108 18s. 10^d.

Note.—When we changed one of the eleven shillings into pence and obtained 12d., we might have first subtracted the 8d. from this and then have added the result, 4d., to the 6d., as in the first method. Thus, 4d. + 6d. = lOd.

Example II.—Here we cannot take 28j yards from 26| yards, so we change one of the 13 perches into yards ; 1 perch == 30J yards ; 30| yards •+- 26^ yards 56£ yards, and 28£ yards from 56 £ yards leaves 28 yards. Next, 6 perches from 12 perches (1 perch having been removed from the 13 perches) leaves 6 perches. Then, 1 rood from 2 roods leaves 1 rood; and 112 acres from 325 acres leaves 213 acres. Answer.—213 acres 1 rood 6 perches 28 square yards.

COMPOUND MULTIPLICATION.

Rule.—Multiply the units of the lowest denomination. If the product will make one or more of the next denomination, reduce it to the higher name, as in Addition, put down the remainder and carry the result. Multiply the units of the next denomination and add in the number carried. Reduce the result to units of the next denomination, putting down the remainder as before. Treat the numbers of the other denominations in a similar way.

Example I.    Example II,

£

s.

d.

Miles.

Fur.

Per.

Yds.

Ft.

20

18

9f

36

6

30

3

2

9

12

£188

9

3f Ans.

442

1

8

0

0 Ans.

Process :—

Example I.—Nine times three farthings — 27 farthings = 6£d., put down the fd. and carry the pence. Nine times 9 pence = 81 pence add the 6d. carried makes 87d. = 7s. and 3d.; put down the 3d. and carry the 7s. Nine times 18s. = 162s.; add the 7s. = 169s = £8 9s.; put down the 9s. and carry the £8. Nine times £20 — £180 ; add the £8 carried = £188. Ans. £188 9s. 3|d.

Example II. —Twelve times 2 ft. = 24 ft. ; divide by 3 to bring to yds. = 8 yds. Twelve times 3 yds. = 36 yds. ; add in the 8 yds., gives 44 yds., divide by 5^ to bring the yards to perches, (we cannot conveniently divide by the mixed number oh, unless we happen to see that 8 times 5§ makes 44, so we double both the 44 yds. and the divisor 5s, and thus obtain 88 yds. -f- 11), = 8 per. Twelve times 30 per. = 360 per., add in the 8 per., gives 368 per. ; divide by 40 to bring the per. to fur., = 9 fur. and 8 per. ; put down the 8 per. and carry the 9 fur. Twelve times 6 fur. are 72 fur. and 9 fur. =81 fur.; divide by 8 to bring to miles = 10 miles 1 fur. Twelve times 36 miles = 432 miles, add 10 miles = 442 miles.

TO MULTIPLY BY A NUMBER GREATER THAN 12.

When the multiplier is a composite number greater than 12, it is usual to break it up into factors, and to find the continued product of the multiplicand and these factors. Thus, to multiply 20 ac. 2 rds. 18 per. 25 sq. yds. by 504, we multiply continuously by the factors of 504, viz. by 9, 8, and 7.

Ac.

20

rds.

2

per.

18

sq. yds.

25 x 9

504 = (9 x 8 x 7).

185

2

9

13i -8

^ 9 times top line.

1484

1

35

151 = 7

: 72 times top line.

10391

1

8

16 =

: 504 times top line.

If the multiplier be a prime number, the factors of the nearest or most convenient composite number may be used. The top line must then be multiplied by the difference between this number and the multiplier, and the result thus obtained added to or taken from the continued product, according as the composite number selected is less or greater than the proper multiplier.

Thus, to multiply £20 12s. 6|d. by 143, we may multiply by the factors of 132, and add 11 times the top line to the continued product • or we may multiply by the factors of 144, and subtract once the top line.

£ s. d.    £ s. d.

20 12 6Jx (12x11 + 11).    20 12 6J x (12 x 12-1).

12    12

247 10 3 =12 times top line. 247 10 3 =12 times top line.

11    12

2722 12 9 =132 times top line. 2970 3 0 = 144 times top line, add 226 17 8| = 11 times ,, sub. 20 12 61 = once top line.

£2949 10 5| Ans.    £2949 10 5f = 143 times top line.

Perhaps the easiest method of working this sum is to

multiply by 11 and 12, and then to add once the first product, which is 11 times the top line.

£

s.

d.

20

12

6£ X

143, or (11 X 12 + 11).

11

226

17

8f =

11 times top line.

12

2722

12

9 =

132 times top line.

226

17

8| -

once 2nd line, or 11 times top line.

£2949 10 5f = Ans. = 143 times top line.

When the multiplier is a large number, whose factors are not easily found, it is customary to adopt the following method :—

Rule.—Multiply by as many tens, less one, as there are figures in the Multiplier, and then by the figure of the highest local value. To the continued product thus obtained, add the various products of (1) the top line by the figure in the units' place, (2) the 2nd line by the figure in the tens’ place, (3) the 3rd line by the figure in the hundreds' place, and so on, using all the figures below that of the highest name.

Example I.—Multiply £5 12s. 3|d. by 6543.

£ s. d.    Times top line.

5 12 3£

10

56

2

8*

=

10

561

7

10

1 =

10 times 2nd line =

100

5613"

10

10

10 = 6

0 =

10 times 3rd line =

1000

33681

5

6 times 4th line —

6000

16

16

9f =

3 times 1st line =

3

224

10

10 =

4 times 2nd line =

40

2806

15

5 =

5 times 3rd line =

500

36729

8

II

°

Answer =

6543

Example II.—Multiply 12 bus. 3 pks. 1 gall. 3 qts. 1 pt. by 654321.

bus.

4pks.

2gall.

4qts.

2pt.

Times top line.

12

3

1

3

1

10

129

3

0

3

0

= 2nd line ...

=

10

10

1298

1

1

2

0

= 3rd line ...

100

10

12984

1

1

0

0

= 4th line ...

1000

10

129843

3

0

0

0

= 5th line ...

!-

10000

10

1298437

2

0

0

0

= 6th line ...

100000

6

7790625

0

0

0

0

= 6 times 6th line

=

600000

12

3

1

3

1

= 1 time top line

=

1

259

2

1

2

0

= 2 times 2nd line

—-

20

3895

1

0

2

0

= 3 times 3rd line

=

300

51937

2

0

0

0

= 4 times 4th line

=

4000

649218

3

0

0

0

= 5 times 5th line

=

50000

8495949

0

1

3

1

= Answer

654321

Note.—For beginners, it is a good plan to place small figures above the multiplicand, indicating the relation between each denomination and the next higher. See the small figures 4, 2, 4, 2, above the last example.

The products to be added are sometimes placed in order of their magnitude ; that is, the additive products obtained by multiplying the top line by the units, the second line by the tens, &c., are placed one under the other in the reverse order.

Example III.—Multiply 8 tons 12 cwt. 3 qrs. 9 lbs. by 2345.

tons 20cwt. 4qrs. “lbs.

8

12 3

9    X 2345.

10

86

8 1

6 = 10 times 1st line = 10

10 times

top line

864

3 0

4 =10 times 2nd line = 10

100

99

, 8641

10 1

12 =10 times 3rd line = 2

1000

99

17283

0 2

24 = twice 4th line =

2000

2592

9 0

12 =3 times 3rd line =

300

345

13 0

24 =4 times 2nd line =

40

43

4 0

17 =5 times top line =

5

99

20264

7 0

21 Ans. =

2345

99

The following plan of arrangement is more compact :—

tons. 20cwt. 4qrs.

“lbs. tons. 20cwt. 4qrs.

“lbs.

8

12 3

9    x 5 = 43 4 0

10

17=5 times top line

86

8 1

6 x 4 = 345 13 0 10

24=40

99

864

3 0

4 x 3 = 2592 9 0 10

12=300

99

8641

10 1

12 x 2 = 17283 0 2 Ans. 20264 7 0

24=2000 21 2345

99

99

Ans. 20264 tons 7 cwt. Oqrs. 21 lbs.

Another method of multiplying a compound quantity by a large number.

Hule.—Multiply the units of the lowest name by the whole multiplier, bring the result to the next 'name, and carry as usual. Multiply, at once, the units of the next oíame, add the number carried, and reduce to the next higher denoimnation ; and so on,—thus obtaining the result by one line of multiplication.

Example :—

Bus. pks. gal. qts. pt.

12    3    1    3    1    2) 654321

654321    -

____    327160 qts. 1 pt.

8495949    0131    1962963

4) 2290123

572530 gals. 3 qts 654321

2) 1226851

613425 pks. 1 gal. 1962963

4) 2576388

644097 bus. 0 pk. 7851852 8495949 busbels.

Pkocess :—

Multiplying 1 pt. at once by 654321 we obtain 654321 pts., which equal 327160 qts. 1 pt. Multiplying the 3 qts. at once by 654321 (or, what amounts to the same thing, multiplying 654321 by the abstract number 3, and calling the product qts.), we have 1962963 qts. which, added to the 327160 qts. gives 2290123 qts. =572530 gals. 3 qts. Multiplying 1 gal. by 654321 and adding to the 572530 gals., we obtain 1226851 gals. = 613425 pks. 1 gal. Again multiplying 654321 by 3, and calling the result pks. gives us 1962963 pks. to add to the 613425 pks. = 2576388 pks. = 644097 bus. Multiplying 654321 by 12 and calling the result bushels, we obtain 7851852 bus. which added to the bushels carried from the pks. gives 8495949 bus.

The best and safest method of working such an example as the last is to reduce the compound quantity to the lowest name it contains, multiply, and return the result to the higher names.

Thus, in working the last example, we should bring the bushels, etc., to pints, multiply, and bring back the result to bushels, pecks, etc.

Note.—When, as in this case, the multiplier is much greater than the multiplicand, we may, for convenience in multiplying, consider the latter as abstract, and use it as the multiplier, as shown in the following example. Here it may be remarked that, in all cases of the multiplication of a concrete number of one name by a large number, the multiplier may be treated as a concrete number of that name, while the concrete number itself may be considered as abstract and used as multiplier. Upon this principle depend many of the rules for Mental Arithmetic and the process of working Simple Practice.

Example:—

Bus. pks. gal. qts. pt.

12    3    1    3    1    654321

4

51

2

103

4

415    .

2

831 pints


831

654321 1962963 5234568

2)543740751

4)271870375 qts. 1 pt.

2)67967593 gals. 3 qts.

4)33983796 pks. 1 gal.

8495949 bushels.

Ans.—8495949 bushels 1 gal. 3 qts. 1 pt.

This method anticipates Reduction, but we regard this more as an advantage than otherwise, since this process, with its application, will have become familiar to the pupil without his being conscious that he has in reality learnt a new rule.

Note.—Compound Multiplication is proved by Compound Division.

When the multiplier is a mixed number, as 4§, 5f, or we first multiply by the whole number (4, 5, or 6), and then multiply by the fraction and add the result.

To multiply by a fraction :—

Rule.Multiply by the numerator' and divide the result by the denominator.

COMPOUND DIVISION.

To divide a compound quantity by an abstract number.

Rule.—Divide the part of the highest name, and put the result in the quotient, with the same name over it. Reduce the remainder, if there be any, to the next lower name, add in that part of the dividend bearing the same name, and then divide the number thus formed, putting the result in the quotient with the same name above it. Reduce the remainder to the next denomination, add in and divide as before. Continue in this way until the lowest denomination has been dealt with, and then should there still be a remainder, place it after the last part of the quotient with the divisor beneath it.

Example I.—Divide £679 2s. 6d. by 72.

£    s.    d. £    s.    d.

72 ) 679    2    6    ( 9    8    71    Answer.

648    12    '

31    113 3 9

20 6

622    £679 2 6 Proof.

576

46

12

558

504

54

4

216

216

By Factors :—

£

s.

d.

8) 679

2

6

9) 84

17

9f

£9

8

H

9

84

17

9f

8

£679

2

6

Note.—Compound Division is proved by Compound Multiplication, as shown above.

Example II.—Divide £376 12s. 4|d. by 12.

£ s. d.

12)376 12 4£

31    7    8£ and or £ q. over, Quotient.

12

376 12 41 Proof.

To divide one compound quantity by another of the same kind * :—

Pule.—Reduce both quantities to the same name f and then divide.

Note.—The answer will be an Abstract number.

• This is done when we wish to find how often one quantity is contained in the other.

t We generally reduce to the lowest name which is mentioned in either.

Example I.—How often is 28 ac., 3 rds., 32 per., 2| sq. yds., contained in 8916 ac., 3 rds., 4 per. ?

28 ac. 3 ro. 32 per. 2| yds. 4

115

40

4632

30£

138962|

1158

140120|

4

560483 quarters of yards.


8916 ac. 3 ro. 4 per. 4


35667

40

1426684

301

42800520

356671

43157191

4

17262S764 quarters of yards.


560483) 172628764 (308. Answer. 1681449 Subtrahend.

4483864

44S3S64 Subtrahend.

172628764. Proof of working.

The proof of division here employed is that of adding the several subtrahends used in working to reproduce the dividend. As this is by far the simplest method of proof, we here give the rule.

Rule.—Add up the remainder and the various subtrahends used in the working as they occur above each other. This should give the dividend.

In working division, whether simple or compound, it will be well to bear in mind—

1. —That subtrahends must not be greater than minuends.

2. —That remainders must be less than the divisor.

3. —That there must be one figure more in the quotient than there are figures to be brought dovm after the first subtraction.

4-When the remainder is greater than the divisor a larger number must be. placed in the quotient.

PROOF OF THE COMPOUND RULES BY CASTING OUT THE NINES.

To cast the nines out of a compound number :—

Cast the nines out of the part bearing the highest name ; reduce the remainder to the next lower name, add in the units of that name ; cast out the nines and reduce the remainder to the next name ; and so on, until the lowest denomination has been dealt with.

To prove the compound rules by casting out the nines :—

Rule.—Cast out the nines, and proceed as in the simple rules.

Exercise III.—COMPOUND ADDITION.

MONEY.

5.

£ s.

d.

£ s.

d.

£ s.

d.

£ s.

d.

12 7

6

2.

16 6

11

3. 95 14

81

4. 39 17

61

4 8

9

19 8

7

72 16

81

83 11 111

37 10

6

14 9

3

87 14

3

94 19

3

59 9

8

20 16

10

25 19

4*

47 14

71

40 12

4

14 3

4

63 18

71

39 17

81

£ s.

d.

£ s.

d.

£ s.

d.

£ s.

d.

27 18

11

6.

14 19

3

7. 206 19

9f

8. 728 14

4f

72 19 10*

15 12

8*

768 14

81

295 17

5*

30 8

4

38 12

5

142 11

2

402 8

8

89 19

4*

76 14 Ili

846 10 10*

767 19

ID

34 11

7*

84 12

3i

289 9

91

628 18

81

97 13

6*

29 19

104

377 11

111

327 17

71

£ s.

d.

£ s.

d.

£ s.

d.

£ s.

d.

594 19

7f

10.

877 17

2*

11.6576 2

91

12. 3698 12

41

378 12

4

355 9

n

7623 14 10*

7964 19

91

987 6

51

876 5

4

6543 12

8795 14

44

876 5

4*

735 4

2*

7062 17

8

7373 13

111

397 6

2f

847 19

7*

42S7 6

91

3789 18


£ s.

d.

£ s. d.

£

s.

d.

£ s.

d.

13.

246 15

71

14. 624 14 111

15.

363

18

10

16.

829 14

11

723 13

81

372 15 9f

789

19

7

372 18

41

363 18 111

767 19 91

254

13

31

246 16

3*

789 19

4

538 12 2f

372

18

lié

375 18

824 11 101

533 13 7

897

12 101

418 19

2

724 18

11

279 14 101

787

17

7f

978 14

^2'

£ s.

d.

£ s. d.

£

s.

d.

£ s.

d.

17.

377 17

7

18. 912 19 51

19. 7294

6

31

20.

49 18

7*

723 13

31

42 12 7

8924

10

7

24 6

81

264 4

111

370 10 10J

2467

19

91

72 12

4|

764 13

21

462 14 51

7635 18

81

98 14

5t

857 18

5f

299 18 71

2763 15

n

76 15

£ s.

d.

£ s. d.

£

s.

d.

£ s.

<L

21.

2046 13

31

22. 2679 11 41

23.

372

16

91

24.

291 12

4

7902 15

41

8240 12 5f

895

13

21

708 19

2

2768 12

H

3769 19 91

726 18

51

263 14

7

1027 18

7

2728 4 8

284

13

7

764 13

4

4008 9

91

7265 11 31

375 17

8f

700 8

6

2759 4

41

2756 15 10

978

11

Hi

908 11

9

£ s.

d.

£ s. d.

£

s.

d.

£ s.

d.

25. 3759 19

71

26. 3995 14 111

27. 9786 15

41

28.

718 13

8252 12

3727 11 101

7527 17

7

379 14

4f

7289 18

91

7845 17 41

8219

13

101

873 19

n

8547 15

in

7223 14 51

7264

16

81

279 18

62

7285 10

3784 11 101

4297

12

n

557 12

8

/

Exercise IV.—COMPOUND ADDITION.

WEIGHTS AND MEASURES.

lbs.

ozs.

drs.

cwt.

qrs.

lbs.

tons

cwt.

qrs.

qrs.

lbs.

ozs.

1. 16

12

14

2. 7

3

20

3. 29

18

2

4. 3

20

7

56

15

15

15

2

14

17

10

3

2

26

15

75

14

12

17

0

24

75

12

1

1

14

12

24

9

10

30

1

27

27

19

3

3

23

9

cwt.

qrs.

lbs.

cwt.

qrs.

lbs.

lbs.

ozs.

drs.

tons cwt. lbs.

5. 18

3

20

6. 23

3

14

7. 35

14

14

8. 37

19

17

19

2

25

35

1

18

18

12

10

49

17

20

9

1

16

26

2

12

17

7

0

52

10

12

17

3

14

33

3

13

25

15

11

45

0

8

lbs.

ozs.

dwt.

lbs.

ozs.

dwt.

9. 13

5

17

10. 24

7

19

20

10

14

15

8

13

32

8

10

17

6

12

16

11

19

20

10

10

drs.

scr.

grs.

drs.

scr.

grs.

13. 3

1

8

14. 6

2

19

5

2

5

6

0

17

7

2

17

5

1

10

4

0

14

7

2

8

lbs. ozs.

dwt.

lbs.

ozs.

dwt.

17. 15

10

18

18. 8

7

12

17

11

14

5

11

10

16

9

19

12

5

8

13

3

13

10

8

16

cwt.

qrs.

lbs.

cwt.

qrs.

lbs.

21. 17

3

20

22. 3

3

13

15

2

15

17

0

15

20

1

18

18

0

12

16

0

22

19

3

27

yds.

ft.

in.

yds.

ft.

in.

25. 27

2

5

26. 8

2

6

16

1

6

3

1

10

35

0

11

11

0

11

12

2

5

17

2

8

ozs. dwt. grs. ozs. dwt. grs. 11. 3 16    3 12. 9 17 17

7

18

23

6

13

15

9

15

20

5

12

14

11

12

12

7

19

23

ozs.

drs.

scr.

OZS. 1

drs.

scr.

15. 4

7

2

16. 11

5

2

5

5

1

10

7

0

2

0

2

5

4

2

8

7

0

9

3

1

lbs.

ozs.

dwt.

ozs. dwt.

grs.

19. 35

7

16

20. 34

19

23

17

5

13

17

17

17

25

11

19

13

15

9

50

8

15

12

10

12

cwt.

qrs.

lbs.

tons cwt.

qrs.

23. 15

0

18

24. 36

15

3

11

2

15

42

18

2

14

3

14

59

19

0

12

2

12

72

13

3

yds.

ft.

in.

yds.

ft.

in.

27. 16

1

9

28. 26

0

8

12

0

2

14

2

6

14

2

5

13

1

9

13

1

7

19

2

6

tons cwt.    qrs.    lbs.

29. 223    18    3    15

162    14    2    12

576    13    0    13

729    15    3    14

832    18    2    10

tons cwt. qrs lbs.

30. 355    17    0    12

726    12    3    14

928    14    0    12

600    13    1    14

366    18    2    12

tons cwt qrs. lbs.

31. 995    0    0    0

72    18    3    18

520    17    2    13

202    19    1    20

419    10    3    26

lbs.

ozs.

dwt

grs.

lbs.

ozs.

dwt.

grs.

32. 729

10

19

20

33. 824

8

16

22

540

8

12

23

176

11

10

10

288

6

15

12

325

5

17

14

654

3

12

15

866

10

17

18

838

9

4

22

297

3

3

20


lbs. ozs. dwt. grs.

34. 640 11    0    10

716 10 19 19 126    3    10    13

428 11 IS 18 954    8    14    21

lbs.

ozs.

dwt.

grs.

lbs.

OZS. I

Iwt.

grs.

lbs.

ozs.

dwt.

grs.

15

10

15

20

36. 39

5

14

16

37. 84

5

9

18

17

7

17

7

72

11

5

19

44

3

7

19

27

8

19

22

64

10

19

10

90

10

10

4

35

10

14

23

35

11

6

8

36

4

18

17

64

7

18

16

22

4

6

7

23

7

17

22

m.

fur.

per.

yds

fur

. per

■ yds.

ft.

per.

yds

. ft.

in.

38.

17

5

35

4

39.

3

17

2

2

40.

35

4

2

10

32

7

28

3!

7

15

4

0

13

5

1

6

75

6

30

5

3

29

5

1

27

2

0

11

28

4

16

4!

6

39

3

2

36

0

2

8

12

3

0

2i

7

20

1

1

19

4

2

9

ac.

ro.

per.

s. yds.

a.

ro.

per. s

• yds

per.

s.yd.

s.ft.

s.in.

4L

316

2

30

30

42.

212

2

23

13

43.

17

15

7

132

214

3

15

25

415

3

14

10

19

12

8

112

712

1

17

m

726

2

28

18

26

7

2

45

512

2

25

18

214

0

19

17

10

4

0

27

428

3

37

22!

313

3

12

16

30

11

5

100

gais.

qts.

pts. gills

qrs.

bush.pks.

gal.

bush.

. pks.

gal.

qts.

44.

13

3

0

3

45.

52

3

3

1

46.

24

2

1

3

26

3

1

2

16

7

0

0

41

3

0

3

32

0

1

1

38

2

3

0

32

3

1

0

28

3

0

2

46

6

0

1

23

2

1

3

c. yds.

c. ft.

c. in.

c. yds

c. ft.

c. in.

c. yds.

c. ft.

c. in.

47. 20

20

120

48. 35

12

854 49.

30

25

125

39

26

140

89

22

216

54

18

729

72

14

108

36

26

343

87

20

100

36

18

132

48

10

512

95

15

625

wks. dys. hrs.

min.

wks. dys.

hrs.

min.

dys.

hrs.

min. sec.

50. 4

3 20

35

51. 12

5

17

37

52. 12

5

34

42

14

6 15

45

10

2

14

34

24

20

20

20

27

4 13

28

15

6

18

50

14

19

37

49

65

2 18

27

20

3

18

48

23

23

40

56

Exercise Y.—COMPOUND SUBTRACTION.

MONEY.

£    s.    d.

4.    12    4    9j

9    14    4J


£    s.    d.    £

1. 2365    17    9    2.    1270

1234    15    6    1224


s.    d.    £    s.    d.

13    5è    3.    9765    11    3*

12    4|    8237    10    10|


£    s.    d.    £    s.    d.

5. 1027    13    3f    6.    7840    11    7è

794    11    4    2355    17    4£


£ s. d.

7. 370 12 2

199 19 9f


£ s. d.

8. 798 16 4! 770 16 4!


£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

9. 3020

17

9f 10. 3576

12

9*

11. 7962 14

2

12.

273

13

44

1254

12

5>-

1777

17

71

7925 14

8*

199

19

94

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

13.

723

15

9£ 14.

92

18

8f

15. 3762

8

4*

16.

455

13

74

399

15

94

82

8

8*

2468

16

5f

299

19

7*

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

17.

109

12

Hi 18.

374

13

10|

19.

789 10

7

20.

908

12

4

78

19 104

372 10

2f

587

17

64

907

18

2*

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

21.

375

17

4i 22.

507

13

85

23.

902

18

10

24.

845

12

6*

275

12

73 t 4

425

17

7t

827

19

H

333

13

74

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

25.

725

18

10* 26.

376

11

7*

27.

798

10

74

28.

125

15

54

375 19

»4

222

12

8*

378 18

8f

72

16

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

29.

837

12

7* 30.

940

11

11*

31.

628 17

3*

32.

750 18

84

235

10

10|

920

13

9f

600

14

2f

066

19

9f

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

33.

990 17

7 34.

789

18

n

35.

927

14

8*

36.

545

13

9*

877

19

8|

279 18

8*

922

12

7f

212

10

104

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

37.

787

15

7 38.

374 12

5i

39.

729

19

54

40.

335

14

7*

528

18

4*

255

15

7f

347

11

6f

273

15

9*

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

41.

375

18

9* 42. 2760 17

6

43.

370

13

4

44.

758 12

74

299

19

81

1375

13

8

279

11

9

254

14

6*

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

45.

378

19

9* 46.

717

17

7*

47.

879

18

8f

48.

257

13

54

228

18

9f

225

17

8

389

19

94

145 17

8

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

49. 1175

0

0 50.

372

0

51.

57

0

0*

52.

1000

0

0

999

19

9

237

17 Ilf

28

11

Ilf

987

13

8*

Exercise VI.—COMPOUND SUBTRACTION.

WEIGHTS AND MEASURES.

lbs.

oz.

drs.

cwt.

qrs.

lbs.

qrs.

lbs.

oz.

lbs.

oz.

drs.

1. 5

12

10

2.

17

2

20 3.

2

14

11

4. 13

14

11

3

13

7

15

3

17

2

12

8

9

15

15

cwt.

qrs.

lbs.

tons

cwt.

qrs.

qrs.

lbs.

oz.

tons

cwt.

lbs.

5. 12

1

15

6.

234

18

3 7.

2

20

12

8. 35

13

17

10

3

27

177

19

1

1

22

14

17

17

15

lbs.

oz.

dwt

oz.

dwt.

grs.

lbs.

oz.

dwt.

oz.

dwt.

grs.

9. 27

8

13

10.

11

13

20 11.

30

8

12 :

12. 5

16

16

14

9

14

2

9

23

15

9

18

1

17

17

dr.

scr.

grs.

lbs.

oz.

drs.

oz.

drs.

scr.

dr.

scr.

grs.

13. 7

2

14

14.

14

5

7 15.

11

2

2 :

16. 6

1

15

3

1

19

12

11

2

9

7

0

1

2

19

yds.

. ft.

in.

mis.

fur.

po.

fur.

per.

yds.

per.

yds.

ft.

17. 4

2

10

18.

44

6

30 19.

7

30

4 20. 27

4

2

3

1

7

27

4

35

3

38

5

19

41

2

ac.

ro.

per.

ac.

ro.

per.

ro. '

per. s

.yds.

s. per.s

• yds.

s.ft.

21. 35

3

30

22.

17

2

12 23.

3

20

27 24. 25

15

5

17

2

38

12

3

25

1

30

30

15

15

, 7

s.yds. ;

s. ft.

s. in.

s.

per. s.yds

s. ft. s. yds

s. ft.

s. in.

ac.

ro.s.

yds.

25. 30

6

132

26.

30

16

7 27.

23

2

90 !

28. 37

1

26

15

8

140

27

16

8

17

7

137

17

0

29

c. yds. c

:. ft.

c. in.

c. yds. c

. ft. c

. in. c. yds. c. ft. c

:. in. c

. yds. c

. ft. c

:. in.

29. 37

20

1624

30.

509

22

122 31.

76

3

100 32. 178

15

899

17

17

1717

199

19

999

57

8

89

135

19

900

gals.

qts

. pts

qts.

pts.

, gills.

bus

. pks.

• gal-

pks.

gal.

qts.

33. 3

2

l

34.

3

1

2 35.

29

2

0 36. 17

0

1

1

3

1

2

1

3

17

3

1

10

1

2

qrs.

bus.

, pks

bus.

pks.

gal. bus.

gals,

. qts.

qrs.

bus.

pks.

37. 3

7

0

38.

13

2

1 39.

27

12

2 -

40. 12

5

2

1

7

3

3

3

0

17

20

3

7

7

3

dys.

hrs.

min.

hrs.

min.

sec.

dys.

hrs.

sec.

wks.

dys.

hrs.

41.225

13

57

42.

15

40

40 43

. 17

1

55

44. 12

5

17

137

17

59

10

50

50

12

23

59

11

3

19

deg.

min.

sec.

deg.

min

sec.

Id.

bus.

gal

. qrs

bus

gal.

45. 51°

42'

43"

46.

27u

15'

15"

47. 11

2

7

48. 12

2

2

41°

27'

49"

19°

29'

38"

7

3

5

7

7

7

mis.

ch. links.

ch.

yds.

ft.

ac.

ch.

Iks.

ac.

ch.

Iks.

49. 37

70

70

50.

47

20

1

51. 95

6

1277

52. 457

3

850

19

75

95

17

21

2

55

7 :

9990

220

9

969

Exercise YII.—COMPOUND MULTIPLICATION.

MONEY.

Multiply—

1.    £12    6s. 8d. by    2, 4, 6, 8, 10, and 12, respectively.

2.    £83    17s.    11 pi.    by 3, 5, 7, 9, and 11, respectively.

3.    £365 9s. 9pl. by 8, 10, 12, and 11, respectively.

4.    £12 6s. oid. by 14, 16, 18, and 20, using factors.

5.    £17 16s.    4fd.    by    24, 27, 28, and 30, using factors.

6.    £23 15s.    5|d    by    32, 35, 36, and 40, using factors.

7.    £175    18s.    8gd.    by 42,    44,    45, and 49, using factors.

8.    £260    12s.    2pi.    by 50,    54,    55, and 56, using factors.

9.    £827    13s.    3fd.    by 60,    63,    64, and 66, using factors.

10.    £150    15s.    10$d. by 70, 72, 77, and 80, using factors.

11.    £300    17s.    7pi.    by 81,    84,    88, and 90, using factors.

12.    £19    19s.    llfd.    by 96, 98, 99, and 100, using factors.

13.    £10    10s.    10pl.    by 17, 19, 29, and 31, using factors for 16, 18,

28, and 30, and adding in the top line.

14.    £14 14s. 4fd. by 29, 37, 57, and 97, using factors for 28, 36, 56, and 96, and adding in the top line.

15.    £15 15s. 5pl. by 26, 38, 58, and 98, using factors for 24, 36, 56, and 96, and adding in twice the top line.

16.    £123 13s. Ill d. by 23, 39, 59, and 79, using factors for 24, 40, 60, and 80, and subtracting the top line.

17.    £421 3s.    7fd.    by    729, 792, and 297, using factors.

18.    £60 19s.    9Id.    by    336, 504, and 720, using factors.

19.    £11 11s. 1 Ipd. by 34658, and 27594, using the factors 10, 10, 10, See., as in Example III., page 78.

20.    £117 7s. 7fd. by 24680, and 13579, using no factors.

21.    £842 17s. 6d. by 3257, and by 47959.

22.    £2846 6s. 8id. by 9786, and by 63702.

23.    £2958 14s. lDd. by 37265, and by 86579.

24.    £41620 16s. 6|d. by 72987, and by 65973.

25.    416 guineas by 76895, and by 72698.

26.    £317 17s 7fd. by 104, 105, and 108, using factors.

27.    £711 19s. llpl. by 110, 112, and 120, using factors.

28.    £987 18s. 9id. by 121, 124, and 125, using factors.

29.    £1907 10s. 10|d. by 126, 128, and 130, using factors.

30.    £2706 4s. llfd. by 132, 140, and 144, using factors.

EXERCISE VIII.-MULTIPLICATION BY FACTORS. 91

Exercise VIII.—MULTIPLICATION BY FACTORS. Multiply, using three factors in each case—

1.    £125 15s. 5Id. by 210, 216, and 220, respectively.

2.    £795 18s. 2£d. by 252, 264, and 280, respectively.

3.    £1265    13s. 3fd. by 2S0,    308,    and    315, respectively.

4.    £6247    12s. 24d, by 343,    378,    and    396, respectively.

5.    £7859    18s. 8fd. by 432,    441,    and    462, respectively.

6.    £2357 19s. llfd. by 512, 504, and 576, respectively.

7.    £8593    17s. 7fd. by 616,    648,    and    672, respectively.

8.    £7526    14s. 4^d. by 729,    768,    and    792, respectively.

9.    £3325 17s. lOfd. by 864, 847, and 891, respectively.

10.    £9876 16s 8^d. by 924, 929. and 972, respectively.

11.    £3784 11s. llfd. by 693, 792, and 840, respectively.

12.    £9999 19s. 9fd. by 625, 968, and 990, respectively.

13.    £7654 3s. 2fd. by 567, 448, and 810, respectively.

14.    £8765 4s. 3|d. by 770, 880, and 588, respectively.

15.    £2340 5s. 6fd. by 756, 1056, and 1320, respectively.

16.    £5760 18s. 9^d. by 1089, 1331, and 1728, respectively.

Exercise IX.—MULTIPLICATION BY FACTORS.

Multiply, using the factors of the extended Multiplication Table—

1.    £117 17s. 8^d. by 26, 39, 52, and 65, respectively.

2.    £475 16s. 2fd. by 78, 91, 104, and 117, respectively.

3.    £327 14s. 4fd by 28, 42, 56, and 70, respectively.

4.    £272 19s. 5^d. by 84, 98, 112, and 126, respectively.

5.    £725 18s 8fd. by 30, 45, 60, and 75, respectively.

6.    £235 14s. 5£d. by 90, 105, 120, and 135, respectively.

7.    £625 19s. 8fd. by 32, 48, 64, and 80, respectively.

8.    £717 17s 7fd. by 96, 112, 128, and 144, respectively.

9.    £928 16s. 10gd. by 34, 51, 68, and 85, respectively.

10.    £437 12s. 9fd. by 102, 119, 136, and 153, respectively.

11.    £626 16s. 2^d by 36, 54, 72, and 90, respectively.

12 £415 2s 9Id. by 108, 126, 144, and 162, respectively.

13.    £625 15s 5fd. by 38, 57, 76, and 95, respectively.

14.    £232 18s. 7^d. by 114, 133, 152, and 171, respectively.

15.    888 18s. 6fd, by 130, 170, 190, and 198, respectively.

Note.—It is recommended that the above sums be worked by multiplication, by two factors only.

Exercise X.—COMPOUND MULTIPLICATION.

WEIGHTS AND MEASURES.

1. 15 cwt. 3 qrs. 12 lbs.    X 12

3. 12 lbs. 14 ozs. 15 drs.    X 16

5. 3 qrs. 8 lbs. 9 ozs. 14 drs. X 36

7. lOlbs 11 ozs 8dwt. 6grs X 64

9. 51bs. lOozs. 18dwt. 7grs X 100

11. 3lbs. 8 ozs. 6 drs. 2 scr. X 144

13. 17 yds. 2 ft. 10 in. X 216 15. 27 mis. 5 fur. 35 per. x 512

17. 20 ac. 3 r. 35 sq per. X 1000

19. 3sq p.25sq.yds.8sq ft.X 1728

21. 5 c. yds. 20c ft 720c. in x 213

23. 16 E. ells 3 qrs. 3 na. X 177

25. 17 FL ells 2 qrs 2 na. X 797

27. 23a. Ssq.chs. 880sq. Iks X 750

29. 35 lea 2 mis 75 chs. X 720

31. 57 gals. 3qts. 1 pt. X 235

33. 72 qrs. 4 bus. 3pks. X 522

35. 13 yds. 1 qr. 1 na.    X 616

37. 35 c ft. of rough timber X 39

39. 115 75 29 19grs X 119

41. 76hhd 23gal 3qt (wine) X 126

43. 16bars. 13gals. lqt (beer) X 91 45. £15 9 flor. 8 c. 7 mils X 987

47. 3 days 20 hrs. 15 min. X 104

49. 3 deg. 14 min. 15 sec. X 25 51. 11 C. 7 0. 16 fl. oz. x 67

53. 15ac 3r. 14p. 16s. yds X 76

55. 123a. 3r. 18p. 5£ sq yds X 11

57. 17 mis 7 fur. 7 ch. 7 yds X 33

59. ltunlbuttllihd.27gal. X 54 61. 3pks lgal. 3 qts lpt. X 614 63. 8 lun months 13 days X 132 65. 18 chain 5 yds 2 ft. X 648 67. 3 miles 3 ch. 3 per. X 115 69. 15 ac. 5 s ch. 15 s per. X 212 71. 19 Eng. ells 4 qrs. 3 na. X 89 73 18 cwt. 3 qrs. 14 lbs. X 65

75. 56 lbs. 7 ozs 19dwts. X 39 77. 15 fathoms 2 feet    X 88

79. 18 c.yds. 1700 c in.    X 66

81. 35 c ft. of rough timber X 55

83. 6 ba. 10 rms. 15 quires X 72

2

2 tons 17 cwt. 3 qrs.

x 11

4.

17 cwt. 2 qrs. 18 lbs.

X 25

6.

7tns. 18cwt. 2qrs. 201bs

. >

< 49

8.

4lbs. 9ozs. 19dwt. 8grs. x 81

10.

10 ozs. 19 dwt. 23 grs.

X

121

12.

8 lbs. 3 ozs. 7 drs. 1 scr.

X

125

14.

27 po. 4 yds. 2ft. 10 in.

X

343

16.

7 fur. 15 per. 4£yds.

X

729

18.

10 ac. 12 sq. per >

: 1331

20.

30sq.yds.7sq.ft.13sq.in.

, X

275

22.

7 c. yds 7 c. ft. 7 c. in.

X

517

24.

18 Fr. ells 5 qrs 2 in.

X

187

26.

12 E. ells 3 qrs. 3 na.

X

792

28.

18 a. 9 s ch. 40 sq. Iks.

X

880

30.

95 s mis 180 acres

X 1100

32.

12 qrs. 7 bus. 3 gals.

X

78

34.

36 bus. 23 quarts

X

127

36.

12 gals. lqt. lpt.

X

252

38.

37 c ft. of hewn timber

X

52

40.

21b. 83 33 19

X

133

42.

84 hhds. 35 gals, (beer)

X

37

44.

14 pipes 80 gals (wine)

X

136

46.

£37 7fl. 8 c. 9 mils

X

789

48.

23 hrs. 52 min. 48 sec.

X

112

50.

5° 15' 10"

X

36

52.

10 fl. oz. 6 fl. dr. 40min.

X

68

54.

12ac.2ro 18per. 10yds

X

98

56.

16sq per. 16 s.yds 6ft.

X

144

58.

15ml.5fur. 5ch.5|yds.

X

48

60.

Stuns lbutt lhhd.7gal.

X

16

62.

13bus. 3pks. lgal 2qts.

X

981

64.

9 com. years 313 days

X

345

66.

7 miles 75 chains 18 yds.

X

62

68.

13 chains 20 yds 2 feet

X

363

70.

8 s. ch. 8 s per. 8 s yds

X

272

72.

3 qrs. 3 na 1$ m-

X

98

74.

3 qrs. 26 lbs. 14 ozs.

X

144

76.

lOozs. 19dwts. 23grs.

X

222

78»

6 fur. 35 per 5 yds.

X

735

80.

26 c ft. 1660 c.in

X

625

82.

44 c. ft. of hewn timber

X

515

84.

7 rms. 12qirs. 9 sheets

X

100

Exercise XL—COMPOUND SHORT DIVISION.

MONEY.


Divide—

1.    £58 7s. 6d. by 2, 3, 4, 5, 6, 8, 10, and 12, respectively.

2.    £10 16s. 6fd. by 3, 5, 7, 9, 10, 11, and 12, respectively.

3.

£

75

s.

6

d.

104

by

2

5.

146

12

104

by

3

7.

312

10

8

by

4

9.

378

11

H

by

5

11.

940

7

3

by

7

13.

1391

7

6

by

8

15.

1242

13

4

by

10

17.

2020

1

101

by

11

19.

1546

7

0

by

12

21.

3002

10

51

by

9

23.

196

14

U

by

11

25.

600

0

1

by

11

27-

591

18

4

by

8

29.

8539

3

9

by

12

31.

999

19

94

by

10

£

s.

d.

4.

17

5

21

by

7

6.

313

19

44

by

9

8.

191

10

9

by

11

10.

1411

17

n

by

6

12.

95

15

6

by

12

14.

41

16

81

by

11

16.

127

11

3f

by

11

18.

75

18

6

by

8

20.

88

11

6

by

12

22.

6541

11

104

by

10

24.

239

19

9

by

12

26.

455

6

3f

by

9

28.

2096

8

8*

by

7

30.

989

16

6f

by

11

32.

1067

4

6

by

12


Exercise XII.—DIVISION BY FACTORS. Divide.—

1. £47 5s. Od. by 14, 16, 18, and 20, respectively, using factors.

2. £48 2s. 6d. by 21, 22, 24, and 25, respectively, using faotors.

3.    £433 2s. 6d. by 25, 27, 28, and 30, respectively, using factors.

4.    £423 10s. Od. by 30, 32, 33, and 35, respectively, using factors.

5.    £375 7s. 6d. by 36, 40, 42, and 44, respectively, using factors.

6.    £367 10s. Od. by 45, 48, 49, and 50, respectively, using factors.

7.    £173 5s. Od. by 54, 55, 56, and 60, respectively, using factors.

8.    £1617 0s. Od. by 63, 64, 66, and 70, respectively, using factors.

9.    £1443 15s. Od. by 72, 75, 77, and 80, respectively, using factors.

10.    £519 15s. Od. by 81, 84, 88, and 90, respectively, using factors.

11.    £4042 10s. Od. by 96, 98, 99, and 100, respectively, using factors.

12.    £1126 2s. 6d. by 104, 105, 108, and 110, respectively, using factors.

13.    £13128 10s. Od. by 112, 120, 121, and 132, respectively, using factors

14.    £13650 0s. Od. by 125, 126, 128, and 130, respectively, using factors.

15.    165 guineas by 132, 135, 1'40, and 144, respectively, using factors.

16.    3850 guineas by 147, 150, 154, and 160, respectively, using ractors.

Exercise XIII.—COMPOUND LONG DIVISION.

MONEY.

1.

£

316

s.

8

d.

10

23

2.

£

445

s.

17

d.

6

29

3.

511

0

1

-i-

29

4.

657

15

n

4-

31

5.

871

2

u

-f

37

6.

1250

12

6f

-f

41

7.

1665

7

111

-r

43

8

4963

2

04

-f

47

9.

1423

16

■f

53

10.

13104

10

34

59

11.

2304

4

oa

-f

61

12.

4018

10

81

-f

67

13.

402

6

3f

-f

17

14.

13330

0

5f

-i.

19

15.

1387

0

34

•f

71

16.

7067

12

4

-f

73

17.

3343

13

6

•f

79

18.

24343

19

2i

81

19.

4228

1

51

-f .

83

20.

S4312

10

44

86

21.

4236

2

51

-f

89

22.

19057

1

14

4.

93

23.

3571

7

6 è

-f

97

24.

20199

1

Of

101

25.

752

12

8

•f

67

26.

11695

14

9f

4-

167

27-

5238

17

U

-f

89

28.

1966

7

2f

4-

73

29.

5702

3

64

-f

95

30.

2333

1

74

4-

59

31.

31200

4

104

-f

78

32.

30967

17

ill

4-

98

33.

9163

16

6f

-f

135

34.

112496

9

81

4-

375

35.

30954

10

Hi

555

36.

99889

11

104

4-

999

37.

122052

5

7

-f

646

38.

635896

3

2

4-

818

39.

59124

12

3

4-

222

40.

377499

7

04

•f

614

Exercise XIV.—COMPOUND DIVISION.

MONEY.

1. How often is £12 3s. 7d. contained in £158 6s. 8d. ?

2.

9 9

99

£5 17s. 6^d. „

£111 13s. 34d. ?

3.

9 9

99

£1 10s. 54d. „

£775 3s 34d. ?

4.

9 9

9 9

£3 18s. Od. ,,

£3116 2s. Od. ?

5.

9 9

9 9

£10 Ils. 104d. „

£826 6s 3d ?

6.

9 9

9 9

£13 13s. Od. „

£3003 Os. Od. ?

7.

99

99

£8 14s. 3|d. „

£1542 5s. lljd. ?

8.

9 9

9 9

£2 8s. 8d. ,,

£42 11s 8d ?

9.

9 9

9 9

£23 0s. 11 fd. „

£2281 12s. 9fd. ?

10.

9 9

9 9

£1 15s. 9fd. „

£782 10s. Ofd. ?

11.

9 9

13s. 74d. „

£747 4s. 14d. ?

12.

9 9

9 9

£1 Is. lid.

£1013 Os. Od.?

13.

9 9

99

£17 3s. lljd. „

£17,196 17s. 6d. ?

14.

9 9

99

19 guineas ,,

£18,134 11s. Od.?

15.

99

99

164 guineas ,,

£13,669 8s. 6d. ?

16

99

99

£29 7fl. 8c. 5rn. ,,

£18,287 9fl. 9c. ?

Exercise XV.—COMPOUND DIVISION. Bind, by division, how often we may subtract—

£

s.

d.

£

s.

d.

1.

17

10

4

from

1839

5

0

2.

30

17

H

9 9

370

10

6

3.

10

19

6

*9

406

1

6

4.

325

14

9f

99

18892

19

14

5.

755

15

0

9 9

73307

15

0

6.

233

12

84

9 9

18457

3

114

7.

12

12

6

99

3067

17

6

8.

28

16

n

99

6025

10

34

9.

802

11

5f

9 9

82665

2

4.4

10.

77

3

11

f 9

77813

8

0

11.

24

19

104

99

4948

15

3

12.

117

13

4

9 9

57774

6

8

13.

709

16

8

9 9

12066

9

2

14.

315

2

6

9 9

22689

0

0

15.

810

10

5f

9 9

226946

14

2

16.

375

18

44

9 9

108264

12

0

17.

757

17

71

99

842007

4

6i

18.

313

13

114

9 9

542070

0

0

19.

888

8

8

99

5117376

0

0

20.

9 4 guineas

9 9

48279

0

0

Exercise XVI.—COMPOUND DIVISION. '

WEIGHTS AND MEASURES.

Divide—

1. 23tons 15cwt. 3 qrs. Sib bv8 3. 53 cwt. 2 qr. 12 lb 8 oz. by 12 5. 197. ac. 1 ro. 36 per. by 11 7. 284 a. 3r 32pr. 20£s. y.bjrbS 9. 5099 mil.3 fur. 30 per by 638 11. 240 miles 2 fur. 32 per. by 528 Í3. 7311a. 2s. ch. 6480 s lk by 272 15. 664 qrs. 6 bus. 3 pks by 37 17. 69271b.4oz. 14dw. 9gr by579 19. 3489 weeks 2 days 1 hr. by 253 21. 4144 loads 2 qrs. by 208 23. 5115 yds. 0 qrs 0 nls by 99 25. 1395hhd. 60gal. (wine)by 715 27. 484 tierces (wine) by 42 29. 687 sq. miles    by 458

31. 651 ac. 3 ro. 3 per. by 1089


2. 2 cwt.Sqrs. 181bs. 13ozs.by7

4. 69tonsllcwt.2qr. 141bs.by9

6. 53a 3ro. 34per.3£s yd by 12

8. 1000a lro 26pr.9^s.ys.by78

10. 480 ac. 2 ro. 32 per. by 528

12. 2032 mil. 71 ch 351ks.by357

14. 5 sq miles 512 ac 2ro by 432 16 5036 qrs 1 bus. 5 gal by 177

18. 7121b 9oz 3dwt. I8gr.by957

20. 353days 15hrs. 24min.by360

22. 298 bar. 7 gal 2 qts. by 68

24. 2555 En. ells 2 na. by 162

26. 85 hhd. 10 gal. (beer) by 115

28. 1210 barrels (beer) by 132

30. 713 lbs. 3 oz. troy by 720 32. 829 C. 7 0. 15 fl. ozs. by 360


Exercise XVII.—COMPOUND DIVISION.

WEIGHTS AND MEASURES.

How often is—

1.    18 lbs. 10 ozs. 12 drs. contained in 2 cwt. 0 qr. 0 lbs. 1 oz. ?

2.    2 cwt. 3 qrs. 16 lbs. contained in 8 tons 4 cwt. 3 qrs. 16 lbs. ?

3.    3 cwt. 2 qrs. 16 lbs. contained in 139 tons 14 cwt. 0 qrs. 8 lbs. ?

4.    12 cwt. 3 qrs. 14 lbs. contained in 418 tons 8 cwt. 3 qrs. ?

5.    2 ro. 25 per. 26 yds. contained in 49 ac. 2 ro. 19 per. 14 yds. ?

6.    26 sq. yds. 6 sq ft. 132 sq. in. contained in 4 ac. 1 ro. 31 per. 13j sq yds. 8 sq. ft. ?

7.    2 gals.    3 qts.    1 pt.    contained    in    4 hkds 35 gals 2 qts. of wine ?

8.    5 gals    2 qts.    1 pt.    contained    in    100 hhds. of beer ?

9.    2 oz. 2    dwt. 2 grs.    contained    in    19 lbs. 11 ozs. 17 dwts. 12 grs. ?

10.    33 po.    3 yds    2 ft.    contained    in    89 miles 7 fur. 22 percbes ?

11.    3 oz. 4 dwt. 10 grs. contained in 11 lbs. 3 ozs. 5 dwt, 12 grs. ?

12.    3 yds. 1 qr. 2 nails contained in 243 yds. ?

MISCELLANEOUS EXERCISE (XVIII.) UPON THE COMPOUND RULES.—FROM EXAMINATION PAPERS.

1.    Add together £64 18s. 4|d , £78 14s. 2|d., £53 17s. 6|d., £36 13s. 4|d., £72 18s. 4|d, and £40 16s. 7£d-

2.    Find the sum of £5060 16s 8d , £376 13s. 4d, £999 17s. 6d., £976 18s. 4d., £1005 17s. 3fd , and £640 18s 9^d.

3.    On Hospital Sunday, the collections at the following places were :—Melbourne, £6132 19s ll|d ; Sandhurst, £2578 12s. 6£d. ; Ballarat, £2015 16s. 2d ; Geelong, £1215 7s. 8id. ; Ararat, £870 12s. 9d.; Castlemaine, £990 5s. Sd. ; Beechworth, £715 8s. lid. ; Warrnambool, £517 12s. 6d. ; Hamilton, £297 17s 5Id.; and Avoca, £250 17s. 6d. Find the total amount collected in the ten towns named.

4.    What do the following quantities amount to :—11 cwt. 3 qrs. 16 lbs. 11 ozs., 27 cwt. 2 qrs 13 lbs. 14 ozs., 12 cwt Oqrs 13 lbs., 18cwt. 27 lbs. 15 ozs.?

5.    Find the sum of 28lbs. 11 ozs. 19dwt. 23 grs., 6 lbs. 8 ozs. 12dwt., 20 lbs. 13 dwt., lOozs, 14 dwt. 17grs., 171bs 11 ozs. 12 grs, and 14 lbs. 3 ozs. 10 dwt 10 grs.

6.    Find, by Compound Multiplication, the weight of 250 packages, each weighing 10 lbs. 6 oz. 5 drs. 2 scruples.

7.    Find the esitent of an estate which contains 1090 areas, each measuring 7 ac. 3 ro. 19 per.

EXERCISE FROM EXAMINATION PAPERS. 97

8. If the Government lines of railway open in Victoria on the 1st of May, 1S77, be as under, find the total length of railway :—

Melbourne to Echuca ...

miles

156

fur.

2

per.

36

yds.

4

Sandhurst to Inglewood ...

30

1

20

5

Melbourne to Stawell ...

176

7

14

3

Geelong to Birregurra ...

38

4

35

2f

Ararat to Dunkeld... ...

47

4

39

4

Castlemaine to Ballarat ...

76

6

20

2*

Maryborough to Avoca ...

15

5

15

5

Maryborough to Dunolly ...

13

3

29

2

Melbourne to Wodonga ...

187

5

30

3

Beechworth to Wangaratta ...

26

1

25

4*

Melbourne to Williamstown ...

9

2

0

0

North Melbourne to Essendoli...

3

1

34

3

9 From the information given in 8, find how much farther it is from Melbourne to Stawell than from Melbourne to Echuca.

10.    From the information given in 8, find by how much the distance from Maryborough to Avoca is less than the difference between the following distances (1) Melbourne to VVodonga, and (2) Castlemaine to Ballarat

11.    Find the weight of 147 packages, each weighing 2 lbs. 3 oz. 2 scr.

12.    By how many times does the number of grains in 159 lbs. loz. 0 drs. 1 scr. 4 grs. exceed the number of ozs. in 1 ton 19 cwt. 1 qr.

9 lbs 8 ozs ?

13.    By how many times does the number of sq. feet in 17 ac. 3ro. 37 per. 20| sq yds. 6 sq. ft. exceed the number of pints in 604 hhds.

28 gals of beer ?

14.    Multiply £79 16s. 7sd. by 409^.

15    If 9 packages of cloth, each containing 17 pieces, each piece containing 36 yds 2 qrs. 3 nls., be divided equally amongst 97 persons, how much will each receive ?

16    Find, by Compound Multiplication, the distance traversed by a person in performing 1070 times the circuit of a circular path measuring 3 fur 17 po. 2 yds.

17    A field containing 24 acres 2 ro. 30 per. is divided into 79 equal areas Find the number of square links in each.

18    What should be my daily expenditure if I intend to save £229 13s 9d per annum, when my income is £200 per quarter ?

19    Multiply £509 0s. 8|d. by 98^-.

20.    Multiply £388 8s. 3fd. by 87$.

21.    The contents of 106 bags of sugar, each weighing 2 cwt. 3 qrs. 17 lbs., are to be divided amongst ■ 175 persons: what weight will each receive ?

22.    A cubic foot of water weighs 1,000 ozs. : find the number of cubic inches in a tank which holds 6 tons of water.

23.    A block of land containing 308 ac. 2 ro. 15 per. was cut up into allotments of 2 ac. 1 ro. 35 per. each : required, the number of allotments.

24.    Amongst how many persons may 350 guineas be divided, 114 of them receive each £2 12s. 9d., and each of the remaining persons £2 Os. 6d. ?

25.    A merchant, at the end of seven years, possessed £13,000. During his first year he gained £364 19s. 7d., during the three next years he gained £586 4s. 9d. yearly, and during the three last years he gained £873 14s. 6d. yearly. What sum did he begin with ?

26.    What will it cost to dig a garden 132 yards long and 165 feet wide, at 2s 6d. per square rod ?

27.    A merchant mixes together 467 gallons of brandy at 18s. 3d. per gallon, 189 gallons at 22s. 9d. per gallon, and 63 gallons at 25s. 4d. per gallon : at what must he sell the mixture per gallon to clear £214 17s. 6d. on the whole ?

28.    A grocer mixes 3 cwt. of sugar at 4|d. per lb. with 1^ cwt. at 3|d. per lb., and 2f cwt. at 5^d. per lb.: at what rate must he sell the mixture per lb to gain £6 Os. 9d. on the whole ?

. 29. Show that there are 5 times as many gills in 3 pipes 1 hhd. 29 gals, of wine as there are quarts in 3 tuns 1 hhd. 50 gallons of beer.

30.    A spirit dealer purchases 100 gallons of brandy at 11s. 4d. per gallon. He sold 48 gallons at the rate of Is. 4d. per pint. At what price per pint must he sell the rest to gain £5 6s. 8d. on the whole transaction ?

31.    A gentleman purchases a farm for £798 6s., the amount of his savings during seven years. His expenses had averaged 25 guineas per calendar month. What was his daily income, allowing for one leap year, during the seven years ?

32.    At the end of three years, of which one was a leap year, the profits of a certain steamer were found to be £9295 14s. Its expenses had averaged 205 guineas per calendar month. What were the average daily receipts ?

33.    A cubic foot of    water    weighs 1000 ozs.    Find the number    of

cubic inches in a tank which holds 3 tons 6    cwt. 3 qrs. 24 lbs.    of

water.

34.    A cubic foot of    water    weighs 1000 ozs.    Find the number    of

cubic inches in a tank    which    holds 1 ton 2 cwt.    1 qr. 8 lbs. of water.

35.    A cubic foot of water contains 6£ gallons nearly : on this supposition, find the number of pints that will be required to fill a vessel that contains 328 cubic inches.

36.    A coach wheel, whose circumference is 16 feet, revolves twice in 3 seconds. Find the number of miles traversed by it in an hour and a-half.

37.    A coach wheel, whose circumference is 12^ feet, revolves 27 times in 32 seconds. In what time would it traverse a mile ?

38.    Of two vessels, one can accomplish 137 miles in 12 hours, and the other 96 miles in 9 hours. Supposing them to start from a place at the same time and in opposite directions, how far apart will they be at the end of 25 minutes ?

39.    How much good cloth in 56 pieces, each 27 yards 1 nail, supposing 23 yds. 3 qrs. 3 nls. to be damaged ?

REDUCTION.

Reduction is the process by which we bring things of one name to their equivalent in another name.

By Reduction we find how many things of one name are equal in value to a given number of things of a different name, but of the same kind.

Thus, we bring pounds to shillings, shillings to pence, pence to farthings, ounces to lbs., lbs. to cwts., miles to yards, &c., &c., all by Reduction.

All Reduction is worked by Multiplication and Division.

To find how many things of a less value are equal in worth to a given number of a greater value, we must employ Multiplication; for the less the value of a thing the more of them are required to make up a given amount,-—so that in order to obtain the equivalent of a number in a lower name we must increase it by Multiplication.

To find how many things of a greater value are equal in worth to a given number of things of a less value, we must employ Division : for the greater the value of a thing, the less of them are required to make up a given amount,—so that in order to obtain the equivalent of a number in a higher name we must diminish it by Division.

In short:—

To reduce to a lower name, we Multiply,

To reduce to a higher name, we Divide,

Reduction to a lower name is called DESCENDING REDUCTION.

Reduction to a higher name is called ASCENDING REDUCTION.

Rule for Descending Reduction :—

Multiply by the number which expresses how many units of the lower name make one of the higher 'name.

Rule for Ascending Reduction :■—

Divide by the number which expresses how many units oj the lower name make one of the higher name.

In order to change a concrete number to its equivalent in a lower name, we have thus to increase it as many times as the value of a unit of the lower name is less than that of one of the higher name. Suppose, for example, we wish to know how many pence are equivalent to five shillings. We observe that a penny is twelve times less in value than a shilling, so that we shall require, in exchange, twelve times as many pence as we have shillings so as to compensate for the less value of a penny. Hence, five shillings = twelve times five, or sixty, pence.

Again, in order to change a concrete number to its equivalent in a higher name we have to diminish it as many times as the value of a unit of the lower name is less than that of one of the higher name. Suppose, for example, we wish to know how many shillings are equivalent to seventy-two pence. We observe that a shilling is of twelve times greater value than a penny, so that we shall require, in exchange, one shilling only for every twelve pence ; and since 72d. — six times twelve pence, six shillings will be the equivalent of 72d.

Examples.—Reduce (1) £288 to shillings ; and (2) 45540 pence to

shillings.

(1)

£288

20

(2.) 12)45540 pence.

5760 shillings.

3795 shillings.

To reduce a compound quantity to a single denomination:—

Rule.Bring to the higher or lower name, step by step, adding in the part bearing the same name as each denomination is reached.

Note.—Since beginners generally experience most difficulty in distinguishing when they should multiply and when divide* we repeat the rule in short.

Rule :—

Higher to lower, multiply ) by as many of the lower as

Lower to higher, divide    f make one of theldgher name.

Examples of Descending Reduction:—

Example I.—Reduce £325 18s. 9fd. to farthings; and lOozs. 6drs


2 scr. 18 grs. to grains.

£20 s.12 d. 325 18 9f 20

6518 shillings.

12

78225 pence.

4

ozs 8 drs.3 scr.20 grs. 10 6 2 18 8

86 drams.

3

260 scruples.

20

312903 farthings Answer.    5218 grains. Answer.

Example II—Bring 182 tons 15 cwt. 3 qrs. 22 lbs. to pounds ; and 288 lbs. 4 ozs. IS dwt 20 grs. to grains.


tons20 cwt. 182    15

20


qrs.

3


lbs.    lbs 12 ozs.20 dwt.24 grs.

22    288    4    18    20

12


3655 cwt. 4


3460 ozs 20


14623 qrs. 28


69218 dwt. 24


117006

29246


276892

138436


409466 lbs. Answer.


1661252 grains. Answer.


Example III.—Reduce 75 miles 7 fur. 36 per. 3 yds. 2 ft. to feet ; and 48 ac. 3 ro. 32 per. 26 sq. yds. 4 ft. to square feet.


miles8 fur.40 per. ^ yds.3 ft. 75    7    36    3    2

8


ac.

48

4


per.“î 32


sq yds.9 sq. ft. 26    4


607 furlongs. 40

24316 perches. 5 b

195 roods. 40

7832 perches. 304

121583

12158


234986

1958


133741 yards. 3


236944 sq. yards. 9


401225 feet. Answer.


2132500 sq. feet. Answer


Example IV.—Reduce 36gals. 3qts. 1 pt. 3gills to gills; and 45 qrs. 7 bus. 3 pks. 1 gal. to gallons.

gals.4 qts.2 pt.4 gills. 36    3    1    3

4

147 quarts.

2

295 pints.

4

1183 gills. Answer.


qrs.8 bus.4 pks.2 gal. 45    7    3    1

8

367 bushels.

4

1471 pecks.

2

2943 gallons. Answer.


Examples of Ascending Reduction :—

Example I.—Reduce 1320903 farthings to pounds ; and 3717 grams Apoth. to ounces.

4)1320903 farthings. 2,0)371,7 grains.

8)61 drs. 2 scr. 17 grs.


12)330225 | pence.

2,0)2751,8s. 9|d.


3)185 scr. 17 grs.


Ana., £1375 18s. 9£d. Ans., 7 ozs. 5 drs. 2 scr. 17 grs.

Example II.—Reduce 1637866 lbs. to tons ; and 6645008 grains to lbs. Troy.

28=


l 4)1637866 lbs ( 7)409466+2 j 4)58495+1 j


24 =


( 4)6645008 grains.


- 6 lbs.


( 6)1661252 +0 2,0)27687,5 + 2


= 8 grs.


2,0)1462,3 cwt 3 qrs 6 lbs. 12)13843 ozs. 15 dwt. 8 grs

Ans , 731 tons 3 cwt. 3 qrs. 6 lbs. Ans., 1153 lbs. 7 ozs. 15 dwt. 8 grs.

Example III.—Reduce 13123 gills to gallons ; and 32373 gallons to quarters.

4)13123 gills.    2)32373 gallons.

2)3280 pints 3 gills. 4)1640 quarts 3 gills. Ans ,410 gals. 3 gills.


4)16186 pecks 1 gal.

8)4046 bus. 2 pks. 1 gal.

Ans, 505 qrs. 6 bus. 2 pks. 1 gal.

5j) 100298 yards 2 feet.

Example IV.—Reduce 3610755 inches to miles ; and 19192500

sq. feet to acres.

12)3610755 inches.    9)19192500 sq. feet.

3) 300896 feet 3 inches.    30j 2132500 yards.

-- 4


11)8530000 quarter yards.


2

--- 121 =

11)200596 half yds.    ( 11) 775454 + 6 ) = 1Q5 qr y(Jg

4,0)1823,6 per.    4,0)7049,5 + 9 )    26? yds.

8) 455 fur. 36 per.    4)1762 ro. 15 sq. per.

Ans.,56 miles 7 fur. 36 per.2 ft.3in. Ans., 440 ac. 2ro.l5 per.26| yds.

12

18 14 15

6,0)110245,5 seconds.

24

6,0)1837,4 min. 15 seo.

66

24

( 6)306 hrs. 14 min. 15 sec.

24= \ -—-

306

hrs.

( 4) 51 + 0 )

60

--> =18 hrs.

18374

min.

12+3 )

60

Ans., 12 days 18 hrs 14 min. 15 sec.


Example V.—Reduce 12 days 18 hrs. 14 min. 15 sec. to seconds, and bring the result back to days, days hrs. min. sec.

1102455 seconds. Answer.

It is sometimes necessary to reduce to a lower name common to both quantities before dividing.

Example VI.— Reduce £188538 to gs. ; and 493800 per. to chains. £188538 to guineas.    493800 perches.

20    5J

( 3)3770760 shillings.

21---

( 7)1256920

Ans., 179560 guineas. 4)493800 perches


2469000

246900


( 2)2715900 yards.

22 \--

( 11)1357950


Ans., 123450 chains


Ans., 123450 chains.


Note.—Since 1 chain = 4 per., the latter answer may be more easily obtained by simply dividing the perches by 4, as shown to the left of the answer.

Exekcise XIX.— REDUCTION OF MONEY.

I.—Reduce the following sums to Farthings :—

II.—Reduce the following to £ s. d.


1. 3876 pence

3. £187 15s. 2-2~d.

5. £2064 19s 4£d.

7. £375 18s. 4£d.

9- £9001 11s. lHd.

11. £8604 2s. llfd.

13 125 guineas

15. 70509 half-crowns

17. £9564 13s. 7£d.

19. £2222 12s. 2£d.

I.    3062759 farthings

3. 7625406 farthings

5. 732905 farthings

7. 1260981 farthings

9. 859658 shillings

II.    295627 shillings

13. 32760 sixpences

15. 56278 fourpences

17. 30625 crowns

19. 95862 half-pence

III. —Reduce—

1. £315 to farthings

3. 3279 guineas to half-pence

5. £8652 to half-crowns 7. 972640 farthings to crowns

9. 123456789farthingstopounds

II.    326784 pence to pounds

13. 3024 sixpences to pounds 15. 1441 threepences to shillings 17. 896 crowns to threepences 19. £5062 to guineas

21. 5760 threepences to crowns

23. 1921 sovereigns to guineas

25. 1760 half-sovs. to guineas

27. £21 16s. 81 d. to half-pence

29. 217911 marks to pounds

31. 7560 crowns to florins 33. 277274 farthings to guineas 35. 314159 farthings to crowns 37. 991 guineas to £ s. d.

39. 1000 guineas to pounds.

2. 9458 shillings

4. £374 17s. llfd

6. £965 18s. 9£d.

8. £639 18s. 8f-d.

10. £999 9s. 94d.

12. £6402 19s llfd. 14. 7321 half-guineas

16. 8265 crowns

18. £3264 15s. 5£d.

20. £3333 13s. 3|d.

2. 826092 pence

4. 814862 pence

6. 275943 pence

8. 222222 pence

10. 426598 shillings

12. 596824 shillings

14. 76249 threepences

16. 92857 half-crowns

18. 68932 half-sovereigns

20. 706000 half-pence 2. 63209787 shillings to guineas

4. 9076240 sixpences to pounds

6. 20560 guineas to sixpences

8. 2806500pence to half-crowns

10. 987654321 farth’gs to guineas

12. 826735 farthings to pounds

14. 95082 half-pence to pounds 16. 78624 fourpences to crowns 18. 95640 crowns to guineas 20. 762590 guineas to £ s. d.

22. 7000 half-crowns to florins

24. 4840 groats to sixpences

26. 7920 guineas to sovereigns

28. 137580 half-pence to pounds

30. 37^ guineas to £ s. d.

32. 175 florins to half-crowns

34. 7782 pence to marks

36. 242218 half-crowns to florins

38. £17 19s. 1 l^d. to half-pence

40. 1000 pounds to guineas

Exercise XX.—REDUCTION OF WEIGHTS AND MEASURES.


Reduce—

1. 366 cwts. to lbs.

3. 3 tons 15 cwt. to lbs.

5. 20600 lbs. to cwts.

7. 73206 oz. to cwts.

9. 13 lbs. 15 drs. to drs.

11. 360804 drams to tons.

13. 37264 grs. to lbs. Troy

15. 76259 grs. to lbs. Troy

17. 12 lb. 9 oz. 23 gr. to gr. Troy

19. 2 lbs. 11 oz.7 drs.togrs.Apo.

21. 276 miles to yards

23. 86 miles 3 fur. to yards

25. 37689 feet to miles

27. 726082 inches to yards

29. 78492 feet to furlongs

31. 30 ac. 2 ro. 15 per. to per.

33. 3769032 sq. yds to acres

35. 5 sq. mis. 400 ac. to sq. yds.

37. 30 sq.yd. 5 ft. 132 in. to sq. in.

39. 26047800 sq. inches to acres

41. 17260980 sq. yds. to acres 43. 1760 inches to Eng. ells 45. 48400 inches to yards 47. 32680 nls. to Eng. ells 49. 369 French ells to Eng. ells 51. 29 c. yards to c. inches 53. 367287 c. inches to c. yards 55. 6408 c. ft. to Ids. rgh. timber 57. 72604 c. ft. to tons shipping 59. 1824 c. yards to c. inches 61. 474624 seconds to days 63. 17 days 5 hrs. 50 min to sec. 65. 36072 minutes to days 67. 39546 days to years 69. 37 yrs. 11 cal. mo. to cal. mos. 71. 36800 pints to gallons 73. 78062 pecks to bushels 75. 26 barrels beer to pints 77. 7282 pints to gallons 79. 136 bushels to quarts 81. 86542 lbs. wheat to bushels 83. 182 bushels bran to lbs.

85. 268 fathoms to feet 2. 59 tons to lbs.

4. 12 cwt. 3 qrs. 20 lbs. to oz.

6. 836720 oz. to tons

8. 37268 drs. to lbs. Avoir.

10. IStons 13cwt. 12 lbs. to lbs.

12. 764098 oz. to cwt.

14. 76094 grs. to dwts.

16. 131b.lloz.19dwt.togr.Troy

18. 161b. 9oz. 8 dwt. 14gr.togr.

20. 6540 drs. Apoth. to lbs.

22. 37698 feet to miles 24. 40 mis. 2 fur. 35 per. to ft. 26. 1786095 yds. to miles 28. 886294 inches to miles 30. 76350 per. to miles 32. 378 ac. 3 r. 14 per. to sq. yd. 34. 76240 sq. per. to acres 36. 35 ac. 2 ro. 15 per. to sq. ft. 38. 20per. 18 sq.yd. 8 sq.ft, to ft. 40. 37678905 sq. ft. to acres 42. 695804 acres to sq. miles 44. 36480 in. to French ells 46. 57 Flemish ells to nails’

48. 72 English ells to nails

50. 786 FI. ells to Eng. ells 52. 94 c. yds. 26. ft. to c. feet

54. 78642 c. in. to c. yards

56. 37248 c. ft. to Ids. hn. tmb.

58. 3682 Ids. unhn. tm. to c. ft.

60. 56720 c. ft. to Ids. of unh.tm. 62. 13 weeks 6 days to minutes 64. 16 days 15 hours to minutes 66. 762840 seconds to hours 68. 16 years 3 lunar mo. to days 70. 76S54 cal. months to years 72. 84 gall. 2 qts. 1 pt. to pints 74. 31 bush 3pks. 2 gall, to qts.

76. 65 hhds. wine to quarts 78. 3240 quarts to pecks 80. 32064 galls, to quarters 82. 64500 lbs to bush, (oats)

84. 34 bushels pollard to lbs.

86. 62 hhds. beer to pints

87. 100 pipes of wine to pints 89. 63870 nails to qrs.

91. 7525 lbs. Troy to lbs. Avoir. 93. 623 cwt. 3 qrs. to stones 95. 484000 sq. yards to acres 97. 72640 yards to chains 99. 86488 sq. chains to acres 101. 76205 miles to feet 103. 42 ac. 3rds. 14 per. to sq.yds. 105. 56 11)6. Apoth. to grains 107. 68725 sq. yards to sq. perches 109. £642 13s. 4d. to marks

111. 37204 days to lunar months 113. 76 loads to pints 115. 70 furlongs to feet 117. 80 sq. miles 72 ac. to sq. yds. 119. 836 leagues to yards 121. 88648 minims to gallons 123. 5280 knots to miles 125. 395rms.5qrs. ISsht.tosheet 127. 9° 57' 56" to seconds 129. April 1st to Dec.31st tohrs. 131. 208708^ florins to halfpence 133. 81220 drams to stones 135. 7041 moidores to pence 137. 212076 inches to nails 139. 488840 sq. yards to acres 141. 31423 paces to miles 143. 78900 links to perches 145. 34816 perches to chains 147. 5712 sq. perches to sq. chains 149. 41407 stones to ozs.

151. 1214875 dwt. to lbs.

153. 81200 lbs. Troy to lbs. Av. 155. 40320 lbs. Av. to lbs. Troy 157. 775 acres to square links 159. 72650000 sq. links to acres 161. 19215 yards to paces 163. 14610 days to Julian years 165. 67560 bus. (wneat) to lbs. 167. 32 bushels wheat to lbs. 169. 175620 lbs. (bran) to bus. 171. 192 bushels pollard to lbs. 173. 45 bushels (oats) to lbs.

175. 817250 lbs. (oats) to bushels 177. 333760 sq. per. to sq. miles 179. 15 yds. 2 ft. to English ells 181. 87 s. mi. 480 ac. to sq. yds. 183. 1678 dwts. 15 grains to lbs. 185. 3 E. ells 1 yd. 2 na. to inches

88. 26084 yards to chains 90. 164872 grs. Troy to lbs. A\. 92. 84726 lbs. Av to lbs. Troy 94. 54260 stones to cwt.

96. 1256000 links to chains 98. 70628 chains to miles 100. 76290 acres to sq. links 102. 762000 yards to miles 104. 786000 sq. feet to roods 106. 56 lbs. troy to grains 108. £8654000 to nobles 110. 3040 guineas to sovs.

112. 760870 dwts. to lbs.

114. 762 pecks to qrs.

116. 764025 days to years 118. 7654 galls, to minims 120. 4840 Eng. ells to Fr. ells 122. 1443^ stones to cwt. qrs. lbs. 124. 968 sq. ml. Irish to s.m.Eng. 126. 78868 sheets t<* reams 128. 42035 seconds to degrees 130. 8888880 seconds to days 132. 10000 gallons to loads 134. 27048 c. ft. to tons (shipping) 136. 7240773 sq. inches to acres 138. 1919 acres to square yards 140. 81854 weeks to years 142. 3 C. 7 O 18 fl. ozs. to minims 144. 61440 qts. of peas to qrs. 146. 5 bus. 3 pks. 1 gal. to qts. 148. 6 oz. 5 drs. 2 scr. to scruples 150. 379 reams to sheets 152. 3996 crowns to florins 154. 99 groats to threepences 156. 82 yds. 2 qrs. 3 na. to nails 158. 567 qrs. of corn to pecks 160. 33 butts (beer) to pints 162. 22500 pints (beer) to hhds. 164. 1234 qrs. to pecks 166. 17 ds. 17 hrs. 17 min. to sec. 168. 48° 56' 37" to seconds 170. 307008 seconds to degrees 172. 37 yds. 1 ft. 6 in. to Eng. ells 174. 55 E. ells 4 qrs. 3 nls. to yds. 176. 89 French ells to Flem. ells 178. 7 hhd. 1 tierce (wine) to gals. 180. 2 loads 2 bus. 2 pks. to qts. 182. 506880 inches to miles 184. 13 ac. 17 per. 2f yds. to yds. 186. 774400 sq. per. Ir. to acres

CHAPTER III.

PRACTICE AND PROPORTION.

PRACTICE.

PRACTICE is a short method of working Compound Multiplication by using aliquot parts.

This rule is chiefly employed to find the cost of any given quantity when the rate is known.

An Aliquot Part is a part which makes up the given quantity when taken an exact number of times; thus, a crown, a half-crown, a shilling, a penny, are all aliquot parts of a pound, since <£1 contains an exact number of each.

To find the aliquot parts of a quantity, we divide it by a whole number, as, 2, 3, 4, 5, 6, &c. We thus observe that an aliquot part of a quantity must be exactly a half, a third, a fourth, a fifth, dec., of that quantity ; thus, 10s., 6s. 8d., 5s., and 4s., are aliquot parts of <£1.    t

Aliquot Parts are therefore concrete measures of a quantity. The term 44 Factor,” on the other hand, is usually applied to the measures of an abstract number.

Practice is SO named from being the method most generally used in actual practice.

Thus—A tradesman in computing the cost of 36 articles at 17s. 6d. each, would first find the cost at 10s., then at 5s., and lastly at 2s. 6d., and then add these costs together. He would say—“36 at 10s. each cost 36 half-pounds, or one-half of £36, i.e., £18; at 5s. each they unit cost one-half as much as at 10s., i.e., £9; and at 2s. 6d.

they will cost one-half as much as at 5s., i.e., £4 10s.; so that, at 17s. 6d. each, 36 will cost £18 + £9 + £4 10s. ; or, £31 10s.

Again, in reckoning the value of 1 cwt. 2 qrs. 14 lbs. of sugar, at 56s. per cwt., he would proceed thus :—1 cwL cost 56s., then 2 qrs. will cost one-half of 56s., i.e., 28-s. ; and 14 lbs. (being the fourth part of 2 qrs.) will cost 7s. ; so that the whole quantity, 1 cwt. 2 qrs. 1) lbs., will cost 56s. + %8s. + 7s. ; or, £4 11s.

The process he would thus employ is that of Practice, and would be exhibited thus :—

10s. = £ of a £


36 articles at 17s. 6d. each.

5s. = 5 of 10s. 2s. 6d. = è of 5s.


£18 cost at 10s. each.

£9 cost at 5s. each.

£4 10s. cost at 2s. 6d. each.


Answer. £31 10s. cost at 17s. 6d. each.

2 qrs. ~ '2 cwt.    56 shillings per cwt.

-    cwt. qrs. lbs.

56s.    cost    of    1    0    0

14 lbs. ~ I of 2 qrs.    28s.    cost    of    0    2    0

7s.    cost    of    0    0    14

Answer. £4 11s. cost of 1    2 14 lbs.

When the given quantity is of one denomination only, or the value of a given number of articles is required (as in the first of the above examples), the process is termed Simple Practice.

When the value of a compound quantity is required (as in the second example), it is termed Compound Practice.

SIMPLE PRACTICE. —In Simple Practice the number of articles is treated as money, while the number of pounds or shillings in the price is treated as abstract, and used as multiplier. See page 80.

Simple Practice.Pule.—Consider the number of articles as pounds (or shillings *) and multiply by the number oj

jjote.—As shillings when the price is very small; as. Is. 5äd.

pounds (or shillings) in the price. Take aliquot parts for the remaining part of the price.

Examples

1.    Find the value of 325 articles at £2 16s. 8d. each.

2.    What will 132 things cost at £6 18s. 4d. each ?

Example I.

Example II.

Of £1

325

@ £2 16s. 8d.

Of £1

132

@ £6 18s. 4d.

10s. = i

2

10s. =

6

Of 10s.

650

Of 10s

>.

792

cost at £6 0s. Od.

each.

5s. = $

162

10s.

5s. -

1

66

cost at

10s. Od.

each.

Of 5s.

81

5s.

Of £1

33

cost at

5s. Od.

each.

II

•"d

00

27

Is. 8d.

3s. 4d. =

1

6

22

cost at

3s. 4d.

each.

£920

16s. 8d. Answer.

£

913

cost at £618s. 4d

each.

In the selection of aliquot parts there is good scope for the display of a discreet choice ; for any set of parts may be taken, so long as they together with the pounds used as multiplier make up the whole of the given price.

Thus, in Example II. we might have taken the two aliquot parts, 2s. 6d. as the half of 5s., and then lOd. as the third of 2s. 6d.; instead of taking 3s. 4d. as the sixth of £1.

After a measure of a pound has been selected as the first aliquot part, each subsequent part should, as far as possible, be a measure oj the part last used ; thus, in Example I., one-half of 10s. (i.e. 5s.) is taken as the second aliquot part, and then Is. 8d., or one-third of this 5s., as the next part.

When, however, it is seen that the remaining part of the price is an aliquot part of a pound, or of an aliquot part previously used, while it is not a measure of the last aliquot part, labour may be saved and the sum completed in one line ; as in Example II., where the 3s. 4d. is an aliquot part of £1, but not of 5s.

Actual practice will be required to gain sufficient skill to distinguish the most suitable aliquot parts.

In dividing by the denominator of the fraction representing the aliquot part chosen, it is important that we remember that it is the line which shows the cost at the price of which we are taking a part which must he divided. That is, if we take an aliquot part of a pound, the top line must be divided (as in obtaining the last line of

Example II.) ; if we take an aliquot part of the last used aliquot part, the last line must be divided ; and so on.

When it is observed that the amount by which the given price is short of £1, is itself an aliquot part of a pound, we may find the value at the rate of this deficiency and subtract. Thus, in the last examples the deficiency of a pound of 16s. 8d., is 3s. 4d. ; and of 18s. 4d., Is. 8d. Now, 3s. 4d. is one-sixth of a £, and Is. 8d. is one-twelfth ; so that we may work the last examples by finding the value at £3 and £7 respectively, and then subtracting the cost at 3s. 4d. and Is. 8d. respectively; thus,

Example I.

Of a £. 3s. 4d. = i


From

Subtract


325 at £2 16s. 8d. 3

975 0 0 cost at £3 0 0 54 3 4 cost at 0 3 4

Answer, £920 16 8 cost at £2 16 8

Example II.

Of a £.

Is. 8d. = -jL-


132 at £6 18s. 4d. 7'

From 924 cost at £7 0 0 Subtract 11 cost at 0 1 8

Answer, £913 cost at £6 18 4

COMPOUND PRACTICE.—In Compound Practice, the money is multiplied by the part of the given quantity bearing the name of which the price of one is known, aliquot parts being taken for the other parts of the quantity.

Compound Practice.Pule.—Multiply the yiven price by the number representing that part of the given quantity which is of the same name as that for which the rate of one is given.    (If higher names are mentioned, reduce them to

this name before multiplying.) Take aliquot parts for all lower names.

Examples:—

1.    2 tons 10 cwt. 2 qrs. 21 lbs. of tea at £9 10s. per cwt.

2.    1 lb. 8 ozs. 17 dwt. 12 grs. of gold at £3 18s. per oz.

Example I.


Of a cwt. 2 qrs. = \


£ s. d.    Of an oz.

9 10 0 price of 1 cwt. 10 dwt. = ^ 50


Example II.


£ s. d.

3 18 0 price of 1 oz. 20


78 0 0 price of 20 oz. 1 19 0 ,,10 dwt. 0 19 6 ,,    5 dwt.

0 9 9 ,, 2£ dwt.


475 0 0 cost of 50 cwt.

14 lbs. = ^    4 15 0 ,, 2 qrs. 5 dwt. = \

7 lbs. = |    13 9,, 14 lbs. 2^ dwt. = |

0 11 104 „    7 lbs.

£481 10 7^ Answer.    £81    8 3 Answer.

When, as in the above examples, the aliquot parts chosen are measures of the aliquot part immediately above them, it is not necessary to show this in full (thus, 14 lbs. is \ of 2 qrs., 7 lbs. is ^ of 14 lbs.), it being understood that each is the part named of the last quantity, except when shown to be otherwise.

It is advisable to keep the parts taken opposite the line which is to be divided, as this shows the working more satisfactorily.

Examples:—

3. Find the dividend upon £835 16s. at 19s. l|d. in the £.

4 Find the value of 1 acre 3 roods 37 poles 15£ yards at £16 16s. per acre.

£

s.

d.

acre

835

16

0

2 ro. = \

1 ro. = a

417

18

0

20 po. = i

208

19

0

10 po. = i

104

9

6

5 po. = \

52

4

9

5 2^ po. = \

13

1

2*

( or, 2po. 15i yds.

2

12

2f

*q.

£799 4 8 |q.


Example IV.


Example III.


£

10s. = 4

5s. i 2s. 6d. = Is. 3d. = 3fd. = *

fd. = *


£ s. d.

16 16 0 cost of 1 acre 8    8    0    cost of 2 ro.

4 4 0 cost of 1 ro.

2 2 0 cost of 20 po. 110 cost of 10 po. 10 6 cost of 5 po.

5 3 cost of 2j po.

£33 6 9 Answer.


In Example III. we get 17 farthings, to be divided by 5 ; we say 5 into 17 goes 3 times and 2 over, and call the 3, farthings, placing the 2 over, above the 5, and calling it two-fifths of a farthing,

or \ q-

We now proceed to give another example yielding a fraction in the answer, with instructions how to deal with the fractions which occur. These instructions will be found more fully explained in the chapter on Fractions.

Example V.—Find the value of 347 acres 2 roods 35 poles at £6 14s. 2d. per acre.

Example V.—First Plan.

Ac.

£

s.

d.

2r.

=

i

6

14

2

... Value of one acre.

347

2327

15

10

... ,, 347ac.

20po.

=

i

3

7

1

... „ 2r.

lOpo.

=

*

16

4

... „ 20po.

5po.

=

h

8

IO

... „ lOpo.

4

2* s

.. ,, 5po.

£2332

12

3ts

... Value of 347ac. 2r. 35po.

In dividing the 16s. 9^d. by 2, we find that we have 5 farthings, or fd., to divide by 2. Now, the rule is—Divide the numerator (or top number of the fraction), and when this cannot be done without leaving a remainder, multiply the denominator instead: we therefore multiply the denominator 4, and thus obtain fd. Again, in dividing by 2 in the next line, we do the same thing, and thus obtain -^d. Now, taking the last denominator 16, we divide it by each of the preceding denominators, 8 and 4, and multiply the result by the numerators 5 and 1, we thus obtain the small figures placed to the right, which, added, give nineteen-sixteenths (or ||), which we find, by dividing the numerator by the denominator, to be Id. and ^ths of a penny.

Example V.—Second Plan.

347 acres 2 roods 35 poles at £6 14s. 2d. per acre.

Rule.—Consider the acres as pounds, and the roods and

poles as fractions of a pound ; then multiply by the pounds,

and take parts for the shillings and p>ence.

347ac., considered as pounds = £347 0 0 2r., or £ac.    ,,    =    10 0

35po., or -n^ac.    ,,    =    4 4^

99


347ac. 2r. 35po.


= £347 14 4\


£

£

s.

d.

£

s.

d.

10s. =

h

347

14

\\ = Value at 6

1

0

0

2086

6

3 = „

6

0

0

2s. 6d. =

i

173

17

2i 2 4 = »»

10

0

Is. 3d. =

i

43

9

3-j9fT - 8 4 — 9 9

2

6

5d. =

4

21

14

7M-78= „

1

3

7

4

10^=87= »

5


£2332 12 3tV = Value at £6 14 2

Another and better plan of -working the last example is to find the value of 347 acres at £6 14s. 2d. by Simple Practice, and the value of the 2 ro. 35 po. at the same rate, by Compound Practice, and then add the answers together for the value of the whole. Thus,

£

s.

d.

6

14

2

3

7

1

16

9i 4

8

H 10

4

4

16

PC 3

2327

15

10

Third Plan—

10s. =

1

2

347 @ £6 14s. 2d.

2 ro. =

1

2

6

20 po. =

i

2082

10 po. =

4

2s. =

X

5

173 10 0

5 po. =

4

2s. =

1

5

34 14 0

2d. =

t2

34 14 0

2 17 10

2327 15 10 cost of 347

acres.


Cost of 347 ac. 2 ro. 35 po. £2332 12 3^ Ans.

Example VI.—Find the value gold at £12 18s. 4d. per oz.

3 lbs. 4 oz. = 40 ozs.


10

dwt.

II

toW

of 1 oz.

5

dwt.

.= i

of 10

dwt.

2

dwt.

- 1

5

of 10

dwt.

1

dwt.

1 2

of 2

dwt.

12

grs.

_ 1

2

of 1

dwt.

6

grs.

- i

of 12

grs.


of 3 lbs. 4 oz. 18 dwts. 18 grs. of


£ s. d.

12 18 4 value of 1 oz.

40

- oz dwt. grs.

516    13    4    value    of    40    0    0

6    9    2    value    of    0    10    0

3    4    7    value    of    0    5    0

1    5 10    value    of    0    2    0

12 11 value of 0 1 0 6 51 value of 0 0 12 3 2| value of 0 0 6


Answer £528 15 6£ value of 40 18 18

9

TABLE OE ALIQUOT PAETS.

Parts of a £.

Parts of a ton.

10s. Od.

_ 1

- 2

10 cwt.

1

- 2

6s. 8d.

_ 1

- 3

5 cwt.

= i

5s. Od.

= i

4 cwt.

4«. Od.

i

2c. 3q 12lbs

_ 1

- 7

3s. 4d.

— J. - 6

2 c. 2 q. 0 lbs

= i

2s. 6d.

_ JL

Parts of a cwt.

2s. Od.

— _i_ - IO

2 qrs.

= 4

Is. 8d.

_ 1

- 12

1 qr.

l

- 4

Is. 4d.

_ 1

- 1 5

16 lbs.

- A

7

Is. 3d.

_L_ - 1 6

14 lbs.

== 8"

Is. Od.

_ 1

- 2 O

Parts of a qr.

8d.

_ 1

- 3 O

14 lbs.

- 2

6d.

_ I

- 40

7 lbs.

JL - 4

4d.

_ I

- 6ÏÏ

4 lbs.

_ 1

7

3d.

= 80

3 lbs. 8 ozs.

2d.

— 1 — T2Ö

2 lbs.

— A

Id.

- 1

2 40

1 lb. 12 ozs.

= T6


Parts of lib Av.

Parts of 1 foot

8 oz.

è

6 in. == £

4 oz.

= i

4 in. =j

2 oz.

_ 1

— «

3 in. =1

Parts of 1 lb. Tr.

2 in. =}

6 oz.

1

- 2

1 in- = A

4 oz.

_ 1

- 3

Parts of 1 s. ft.

3 oz.

= i

72 s in. = a

2 oz.

_ 1

- 6

48 s in. = %

1 oz.

_ 1

- TÔT

36 s. in. = £

Parts of 1 oz. Tr.

24 s. in. = a

10 dwt.

_ 1

- 2

18 s. in. = a

5 dwt.

_ 1

- 4

12 s. in. =

4 dwt.

Parts of 1 c. ft.

2.j dwt.

= i

864 c. in. = £

2 dwt.

_ 1

10

576 c in. = J

U dwt

1 1 6

432 c. in. = £

1 dwt.

= JL 20

288 c. in. = J


Exercise XXI.—PRACTICE.

Find, by Practice, the cost of the following :—

1. 144 articles at £2 2s. 2d., and at £4 6s. Sd.

2. 640 articles at £7 7s. 7d., and at £6 13s. 4d.

3.    2440 articles at £9 9s. 9d., and at £8 11s. 8d.

4.    7624 articles at £11 11s. lid., and at £5 17s. 6d.

5.    326 articles at £9 2s. 8d., and at £12 13s.

6.    2880 articles at £3 7s. 6d., and at £15 16s. 3d.

7.    7250 articles at £12 17s. 9d., and at £19 8s. 4d.

8.    720 articles at £9 13s. 8id., and at £8 18s. 4d.

9.    65700 articles at £11 11s. lid., and at £7 14s. 6gcL

10.    540 articles at £16 14s. 2|d., and at £50 18s. 9fd.

11.    960 articles at £17 18s. 5d., and at £75 13s. 7R1.

12.    1728 articles at £65 19s. llfd., and at £63 15s. ll|d.

What will be the cost of—

13.    736 things at £1 16s. 5d., and at £9 14s. 2]d.

14.    68 things at £10 4s. ll£d., and at £12 10s. llfd

15.    2365 things at 17s. 6d., and at 19s. 5^d.

16. 1774 things at 14s. 2|d., and at 17s. 10fd.

17. 3333 things at 19s. 9}d., and at 18s. 8fd.

18.    88881, things at 14s. 2£d., and at 6s. 9}d.

What will be tlie cost of—

¿ 19. 444f things at £3 16s. 4d., and at £13 14s. 6d.

20.    Ill things at £4 11s. 9|d., and at £9 11s. 4Jd.

21.    333| things at £9 4s. 8fd., and at £16 17s. 4d.

; 22. 224| things at £11 11s. 4^d., and at £17 6s. 5^d.

23.    1002§ things at £10 10s. ll^d., and at £19 17s. 4|d.

24. 466f things at £5 19s. 7fd., and at £13 14s. 6|d.

25. 999£ things at £4 18s. 2£d., and at £6 6s. 6§d.

Find, by Practice, the dividend on—

26.    £2634 12s. 6d. at 2s. 6d. in the £, and at 7s. 6d.

27.    £1224 at 17s. 7|d. in the £, and at 11s. 3fd.

28.    £956 18s. 4d. at 14s. llfd. in the £, and at 17s. 6|d.

29.    £21146 17s. 6d. at 13s. 2|d. in the £, and at 12s. 9fd.

30.    £9276 15s. 8d. at 14s. 7|d. in the £, and at 19s. 2d.

31.    Find, by Practice, the deficiency of a bankrupt who owes £2575 18s. 6d., and pays 11s. 8d. in the pound.

32.    If 1284 selectors each take up 320 acres 3 roods 26 perches of land, how much land was selected ?

33.    If 268 casks contain each an average of 8 gallons 3 quarts

1    pint : find, by Practice, how many gallons the whole contained.

34.    If 65408 mats of sugar each weigh 1 cwt. 2 qrs. 21 lbs. : find by Practice, the total weight of the sugar they contain.

35.    In a shire containing 37775 inhabitants, one-fifth of whom held property, the average rates collected amounted to £3 17s. 6d. per ratepayer : find the total rates collected.

36.    If 24 railway trucks carry each an average of 4 tons 1 cwt. 3 qrs. 18 lbs. per truck : find, by Practice, the total weight carried."

37.    Find, by Practice, the cost of erecting 156 miles of telegraph

at £30 18s. 7d. per mile.    ° P

38.    Find, by Practice, the product of 50 English ells 3 qrs. 2 nls

2    in. multiplied by 17280.

39.    Find, by Practice, what 1042 times 3 miles 3 furlongs 3 perches

equals.    '

40.    Find, by Practice, the amount of gold in 2856 ozs. of amalgam,

if each ounce of amalgam contains 17 dwt. 20 grains of gold.    ’

, 41. If 365 days 5 hours 48 minutes 48 seconds make a solar year : find, by Practice, the number of days, hours, &c., in 1878 years.

42.    Find, by Practice, the number of inches in 10000 links a'link being equal to 7 inches 11 ^ lines.

43.    Find, by Practice, the product of 16 gallons 3 quarts 1 pint

multipled by 4840.    1

44.    Find the weight, in tons, of 798 fodders of lead, a fodder being

19 cwt, 2 qrs.    s

45.    What is the value of 2007 marks, each worth 13s. 4d. ?

46.    A pound avoirdupois contains 14 ozs. 11 dwt. 16 grs. trov •

how many lbs., ozs., &c., troy in 1720 lbs. avoirdupois? '    * '

47.    What do 1576 bags of potatoes, each 1 cwt. 1 qr. 25 lbs., weigh’

Exercise XXII.—PRACTICE FROM TEACHERS’ EXAMINATION PAPERS.

1.    Find, by Practice, the price of 7 ozs. 2 scruples 14 grs., at £2 7s. 3d. per lb.

2.    Find, by Practice, the value of 37 qrs. 3 bus. 2| pks. of wheat, at £2 11s. 9d. per quarter.

3.    Find, by Practice, the value of 13 yds. 3 qrs. 3 nls. of cloth, at £17 18s. 6d. per piece of 36 yds.

4.    Find, by Practice, the value of a gold snuff-box weighing 7 ozs. 11 dwts. 18 grs., the gold being worth £4 3s. lOd. per oz., and 5s. 6d. per oz. being allowed for workmanship.

5.    426 acres 2 roods 16 poles, at £7 16s. 3d. per acre.

6.    14 lbs. 6 ozs. 13 dwts., at £3 17s. 5d. per oz.

7.    Find, by Practice, the cost of 26 acres 3 roods 18 perches, at £1 17s. 4d. per acre.

8.    Find, by Practice, the value of 2 tons 3 qrs. 11 lbs., at £2 5s. lOd. per cwt.

x 9. Find, by Practice, the value of 375 ozs. 17 dwts. 20 grs. of gold, at £3 17s. lid. per oz.

10.    48 boxes of raisins weighed, each of them, in the gross, 2 cwt. 1 qr, 15 lbs. ; 22 lbs. is allowed for the weight of each box. Find, by Practice, the value of the nett weight of the whole, at £3 16s. 7d. per cwt.

11.    Find, by Practice, the value of 579 lbs. 10 ozs. 13 dwts., at £7 15s. lid. per lb.

12.    Find, by Practice, the value of 64 lbs. 4 ozs. 3 dwts. 12 grs.

V of gold, at £3 14s. lOd. per oz.

13.    Find, by Practice, the value of 347 acres 2 roods 35 poles, at £6 14s. 2d. per acre.

14.    Find, by Practice, the value of 65 acres 3 roods 16 perches, at £3 16s. 10^d. per acre.

15.    Find, by Practice, the value of 367 qrs. 25 lbs. 13 ozs., at £5 16s. 5d. per quarter.

16.    Into how many allotments of 5 acres 3 roods 19 perches each may 281 acres 2 roods 32 perches be cut up ? Find, by Practice, the cost of the land, at £3 17s. 9d. per acre.

17.    Find, by Practice, the value of a man’s work during 7 lunar months 2 weeks 5 days, at £13 17s. 8d. per month.

18.    Find, by Practice, the cost of 39 ozs. 16 dwts. 18 grs. of silver, at 9s. lOd. per oz.

19.    Find, by Practice, the value of 3079 acres 3 roods 19 perches, at £7 15s. 9d. per acre.

20.    Find, by Practice, the value of 13 acres 2 roods 27 poles of

land* at £17 18s. 9d. per acre.    ia

21.    Find, by Practice, the cost of repairing 6 miles 3 furlongs 18 perches of road, at £57 15s. lOd. per mile.

22. Find, by Practice, the value of 103 yds. 3 qrs. 1 nail, at £5 15s. 6d. per yard.

J 23. Find, by Practice, the value of 37 miles 92 acres 3 roods, at £■47 13s. 7|d. per mile.

v 24. Find, by Practice, thé value of a farm containing 207 acres

2    roods 11 perches 3 square yards, at £20 3s. 4d. per acre.

25.    Find, by Practice, the cost of 308 qrs. 5 lbs. 11 oz. at £4 13s. fid. per quarter.

26.    Find, by Practice, the value of 4 square nerches 3 yards 2 feet 120 niches at £2 6s. l^d. per square yard.

27.    Find, by Practice, the cost of gilding 37 square yards 1 square foot 23 square inches at £3 7s. 6d. per square yard.

28.    Find, by Practice, the cost of gilding 26 square yards 2 square feet 19 square inches at £6 Is. 6d. per square yard.

29.    Find, by Practice, the cost of 93 cubic yards 11 cubic feet 720 inches at £3 0s. 9d. per cubic yard.

30.    Find, by Practice, the cost of 37 cubic yards 10 cubic feet 624 inches at £6 Is. 6d. per cubic yard.

31.    Find, by Practice, the value of 465 miles 7 furlongs 17 poles at £4 15s. 7d. per mile.

32.    Find, by Practice, the value of 804 quarters 3 bushels 3 pecks of wheat at £2 17s. 9d. per quarter.

33.    Find, by two methods of Practice, the value of 4 tons 12 cwt.

15 lbs. at £3 15s. 7èd. per cwt.

34.    Find, by Practice, the cost of 5 lbs. 10 ozs. 16 grs. of silver at £3 6s. 9d. per lb.

35.    Find, by Practice, the cost of repairing 3 miles 3 furlongs 120 yards of a road, at £47 18s. 6d. per mile.

36.    Find, by Practice, the value of 38 quarters 3 bushels 3 pecks of wheat at £2 13s. lOd. per quarter.

37.    Find, by Practice, the price of 3 lbs. 14dwt. 18 grs. of gold at £3 18s. 2d. per oz.

38.    Find, by Practice, the value of a block of stone containing 17 ■cubic yards 5 feet 100 inches at £2 13s. 6d. per cubic yard.

39.    Find, by Practice, the cost of repairing 28 miles 5 furlongs 12 poles of road at £95 18s. 6d. per mile.

40.    Find, by Practice, the value of 567 cwt. 3 qrs. 19 lbs. at £3 18s. 9d. per cwt.

41.    Find, by Practice, the cost of 43 qrs. 27 lbs. 10 ozs. at „ £3 0s. 8d. per qr.

42.    Bought 35 quarters 1 bushel of wheat. For waste, &c., the seller allowed me 2 bushels 1 peck, and for the remainder I paid at the rate of £2 5s. 4^d per qr. : find, by Practice, the sum I paid him.

43.    Find, by Practice, the value of 9 silver teaspoons, each ■weighing 2 lbs. 11 ozs. 13 dwt., at £3 6s. lid. per lb.

44.    Find, by Practice, the value of 14 silver teapots, each weighing

3    lbs. 5 ozs. 7 dwt., at £3 7s. 5d. per lb.

45.    Find, by Practice, the value of 34 quarters 8 lbs. 11 oz., at £1 10s. 4d. per quarter.

46.    Bought 44 quarters 1 bushel of maize. For waste, &c., the seller allowed me a bushel and a half, and for the remainder I paid him at the rate of £2 2s. 8gd. per quarter : find, by Practice, the sum I paid him.

47.    Find, by Practice, the value of 97 acres 2 roods 27 poles, at £5 17s. 9d. per acre.

48.    Find, by Practice, the value of 57 ozs. 13 dwt. 17 grains, at £3 17s. (id. per oz.

49.    Find, by Practice, the sum which should be paid for 4 casks of brandy, at £1 Os. 1 Id per gallon. Each cask should contain 29 gallons, but from each there has leaked an average of 1 gallon 2 quarts 1 pint 1 gill.

50.    Find, by Practice, the sum which should be paid for 5 casks of brandy, at £1 Is. 9d. per gallon. Each cask should contain 28 gallons, but it is found that from each there has leaked an average of 1 quart 1 pint.

51.    Find, by Practice, the value of 26 quarters 11 lbs. 9 ozs., at £1 19s. 8d. per. quarter.

52.    Find, by Practice, the value of 24 square yards 1 square foot 52 square inches, at £2 7s. 3d. per square yard.

53.    Find, by Practice, the value of 37 square yards 2 square feet 10 square inches, at £1 13s. 9d. per square yard.

54.    Find, by Practice, the value of 23 bushels 3 pecks 1 gallon at £2 9s. 8d. per bushel.

55.    Given 36 bales of silk, each bale weighing 3 cwt. 0 qrs. 17 lbs.; 18 lbs. per bale allowed for the weight of the package : find, hy Practice, the value of the nett weight of the whole at £3 17s. 8d. per cwt.

56.    Find, by Practice, the value of 749 oz. 17 dwt. 19 grs. at £3 16s. lOd. per oz.

57.    Find, by Practice, the rent of 29 acres 3 roods 18 perches at £3 16s. 4d. per acre.

58.    Find, by Practice, the value of 109 oz. 11 dwt. 13^ grs. of gold at £3 15s. 10^d. per oz.

59.    A merchant ordered 36 pieces of silk, containing each 28 yds.

0 qrs. 1§ nls. to be made into dresses of three different sizes ; those of the largest size contained each 15f yds., those of the next size 12 yds. 0 qr. 1 nl., while the smallest contained only 9 yds. 2 qrs. nls. There was to be an equal number of each size. Find the number made, and the value of the silk at 16s. 8d. per ell, French.

60.    Divide 336 yds. 3 qrs. 3f nls. of broadcloth between A, B, and C, so that for every 5 yards A has, B shall have 3, and 0 If yds. Find the value of the cloth at 17s. 9d. per ell, English.

61.    What is the total area of 124 gardens, each containing 3 roods 35 perches 22 square yards?

62.    Find, by practice, the value of 18200 dollars, each worth 4s 3d. J

63.    Multiply 7 bushels 2 pecks 1 gallon 3 quarts by 57600, and work the same by Practice.

RATIO AND PROPORTION.

Ratio is the relation which one number bears to another with respect to magnitude : thus, the ratio of 4 to 8 is one-half; that of 4 to 12 is one-third, and that of 6 to 3 is two.

The ratio of one number to another is expressed thus :—

As 3 : 6 (read as 3 is to 6).

Of two numbers that are so compared, the first is called the Antecedent and the second the Consequent of the ratio.

In order to compare the magnitude of any two things, they must be of the same kind ; for no number of times, or parts of a thing of one kind, can be equal to, or make up, another thing of a totally different kind; thus, for example, no amount of money, or time, or land, can be any part of a certain number of horses or cows.

Hence, ratio can only subsist between abstract numbers and concrete numbers of the same kind.

Proportion.When two things bear the same relationt to each other as two other things, the four are said to be in

proportion.

For four numbers to be in proportion, the first must be as large, compared with the second, as the third is compared with the fourth. That is, the ratio of the first and second numbers must be the same as that of the third and fourth numbers.

Proportion is therefore defined the equality of ratios

Proportion thus shows that one thing is the same part or number of times of a second that a third is of a fourth.

Definition.—Ratio is the comparative magnitude of two things of the same kind.

In questions on ¿proportion, the ratio of two given numbers is identical with that of two other numbers of which one is given or known, and the other required to be found.

Note.—The ratio of one number to another may always be expressed by making the first number the numerator of a fraction, and the other number the denominator of the same. Thus, the ratio of 3 to 5 is that of 15 to 20 is -£4, or

In order to express proportion as subsisting between four numbers, they are placed thus :

As 6 : 8 :: 9 : 12 (read, As 6 is to 8, so is 9 to 12).

This is called stating’. The numbers so stated are called the terms of the statement.

The inner terms, 8 and 9, are called the means.

The outer terms, 6 and 12, are called the extremes.

In the statement, As 5 cwt. : 15 cwt. :: 6s. : 18s., 5 cwt. is the first term, 15 cwt. the second term, 6 shillings the third term, and 18 shillings the fourth term.

Since ratio cannot subsist between things of different kinds, the, first and second terms of a proportion statement must he of the same kind, while the third and fourth terms must also be of like kinds.

Example.—As 5 horses : 20 horses :: 3 days : 12 days.

Here the comparison of 5 horses with 20 horses, and that of 3 days with 12 days, produce the same result—viz., that one is one-fourth of the other.

But if we say, “ As 5 horses is to 10 days, so is 15 horses to 30 days,” the statement is void of sense ; for, although 5 is one-half of 10, and 15 one-half of 30, yet 5 horses is no part of 10 days, nor is 15 horses any part of 30 days. The mind refuses to draw a comparison between a number of horses and a number of days.

Principle.—In every proportion statement,

The product of the means = the product of the extremes.

In other words, the product obtained by multiplying together the second and third terms is the same as the product of the first and fourth terms. Again, by dividing

the product of two numbers by one of them, the other number is obtained ; so that, when any three terras of a proportion are known, the fourth or unknown term may be found.

In three terms of a proportion we necessarily have either (1) the two means and one extreme, or (2) the two extremes and one mean. Hence, by multiplying together the two fellow terms, we can obtain the product of the missing term and its fellow. Then, by Ax. VI., page 36, the product of two numbers divided by either number gives the other.

From these principles we obtain the Rule of Three» which enables us from the first three terms of a proportion to find the fourth.

Rule.—Multiply the 2nd and 3rd terms together, and divide the product hy the 1st term.

The principles of proportion are of such universal importance throughout the more advanced rules of arithmetic, that we may be pardoned a little repetition in order to make our explanation full and clear.

In every proportion statement, the product of the 2nd and 3rd terms is the same as that of the 1st and Ifh terms. Example.—As 4 I 12    5 I 15.    ,

Here, 5 times 12 = 60 = 4 times 15.

This is but another form of stating the following principle of Multiplication :—

If one number be increased as many times as another is diminished, the product of the numbers thus obtained is the same as the product of the original numbers themselves.

Now, if we take the four terms of any proportion, we have in the fourth term a number which is as many times greater or less than the third term as the first term is less or greater than the second term.

Thus, in the above example, 15 is three times greater than 5, while 4 is three times less than 12. Hence, if 4 be missing, it may be found by dividing the product of 12 and 5 by 15; if 15 be missing, it may be found by dividing the product of 12 and 5 by 4.

Again, if either of the means he missing, it may he found by dividing the product of 4 and 15 by the given mean.

I.    To find either extreme. Rule.—Divide the product of the means by the given extreme.

II.    To find either mean. Rule.—Divide the product oj

the extremes by the given mean.

Examples

In the following proportion statements, let x mark the missing term :—

As    3    I    4 ; *    9    i    x.    Then, the fourth term — ( 4 X    9 3) = 12

As 35    :    20::    x    :    4.    Then, the third term = (35 X    4 20) = 7.

As    7    x :3    :    12.    Then, the second term = ( 7 X    12-r 3) =28.

As    x    15 *. 1    6    I    18.    Then, the first term = (15 X    6 -f-18) = 5.

Find the fourth proportional of the following :—

(l.)As    3!    5'.:    9’. x.    Here the fourth term = ( 5X    9-r-    3) =15.

(2.)As    8 :    12::    O'.x.    Here the fourth term = ( 12 X    6-j-    8)= 9.

(3.) As 15 :    10: *.    9 : x.    Here the fourth term = (10 X    9-=-15)=6.

(4.)As    7 :    11:: 14 : SB.    Here the fourth term = (11 X    14 —7)=22.

In all questions on proportion, we have three of the terms given, and require to find the fourth. We have, therefore, so to arrange the given terms that, by dividing the product of the means by the given extreme, we shall obtain the fourth term or answer.

This proper arrangement of the first, second, and third terms is called Stating the sum, and as this stating is by far the most difficult part of the work, we shall endeavour to make the rules and explanations as full, clear, and simple as it is possible to make them.

Problems on proportion consist of two distinct parts :—

1.    The Data. Tlie information given to work upon.

2.    The Question.* That which it is asked to find.

Example.“If 6 books cost 15s., what will 8 books cost?” The fact that 6 books cost 15s. is the information given to enable us to solve the question, and is, therefore, the Data or Suppo* sition; “ what Will 8 books cost ?” is the Question or Demand, which shows what is sought from the information given.

• These two parts of a problem are sometimes termed the Data (things given)

and the Qnsesita (things sought).

Note.—The third term is always to be found in the Data.

The Data usually comes before the Question, and begins with “If,” “Suppose,” &c., or else makes a direct assertion ; while the Demand or Question always begins with “How,” “What,” “How many,” “ How much,” or some other interrogative word.

In every question on proportion, the answer is regarded as the 4til term; and that quantity in the data which is of the same nature as the answer as the 3rd term.

Ordinary Rule for Stating.

Rule.—Place in the Third term that which corresponds to the answer. Consider, then, whether, from the nature of the question, the answer should he greater or less than the Third term; if greater, place the greater, and if less, place the less, of the other numbers in the Second term ; and the remaining number in the First term.

Remark.—Remember that the Second term regulates the answer. If it be greater than the First, the answer will be greater than the Third. If it be less, the answer will be less than the Third.

Cancelling.—The First term may be cancelled with the Second or Third ; but never, in any case, cancel the Second with the Third.

Example.—If 300 men could do a piece of work in 12 weeks, how many men could do it in 15 weeks 1

The question is, how many men, &c., hence the answer will be so many men, so we place the “ 300 men ” in the third term.

Next consider whether more or less men will be required to do it in 15 weeks than in 12 weeks. Answer, less ; then place the less number of weeks in the Second term, and the greater number (15 weeks) in the First term. This gives us the statement,

As 15 weeks : 12 weeks :: 300 men :    men. Ans.

Then the numbers 12 and 300 multiplied give 3600, which, divided by 15, gives 240 men. Answer.

Note.—The statement being made, the numbers must be regarded as Abstract; otherwise the Second and Third terms cannot be multiplied together when both are concrete numbers. Vide Axiom X., page 37.

Had the question taken the form “300 men could finish a contract in 12 weeks, how many men lefit, if it took those who remained 15 weeks to finish it ?” We should find the number of men that would do it in 15 weeks, as above, and subtract the answer, 240, from the original number of men, 300, since the others must have left. And 300 — 240 — 60 men, number who left.

Note.—It is of the utmost importance that the question bekept well in view in reasoning out for the statement of the First and Second terms ; otherwise, we may reason correctly and state incorrectly from a misconception of what is required.

Young students generally err in not keeping the question well in mind.

The method above described is that usually adopted, and is, perhaps, the most suitable for beginners ; but a knowledge of direct and inverse proportion, with a distinct rule for working each, will give a clearer perception of the nature of proportion, and afford a readier plan of stating at sight, than can possibly be attained by adhering to the above rule only.

Proportion—Direct and Inverse.—We therefore propose to give a full explanation of Direct and Inverse Proportion, with rules for stating.

Proportion is either Direct or Inverse.

In Direct Proportion the answer is larger or smaller as the term in the question is larger or smaller than the like term in the Data. That is, when the answer increases or diminishes, as the term in the question increases or diminishes, the proportion is Direct.

Thus, all other things being equal—

The cost of things increases as the number bought.

The work done increases as the number of workers.

The weight of anything increases as its bulk.

The number of things bought is in proportion to the money spent on them.

Work performed varies as the time occupied in doing it.

Interest varies as the amount lent and the duration of loan.

Note.—When one thing varies as two or more others (as, for instance, when Interest varies as the principal, the time, and the rate p. c.) the proportion is compounded of the result of these two or more ratios, and is termed Compound Proportion—explained hereafter.

In Inverse Proportion, the answer is larger or smaller in inverse ratio to the term in the question. That is, the answer increases as the term in the question becomes less; and vice versd, diminishes as the term in the question increases.

Thus, all other things being equal—

More workers are required as the time is less.

The time required will be less as there are more workers.

The less in size a thing is, the more required to fill a given space.

The less the value of a coin, the more required to make up a certain sum of money.

The higher a man’s wages, the less time he will take to save a given amount.

The more work done per day, the less days taken to do the work.

The less work done per day, the more days taken to do the work.

It is thus seen, that—

In Direct Proportion, more gives more, and less gives less.

In Inverse Proportion, more gives less, and less gives more.

Or in Direct Proportion, the answer varies directly as the term in the question.

In Inverse Proportion, the answer varies inversely as the term in the question.

Pules for Stating Direct and Inverse Proportion :—

Direct Proportion.—Pule.—State, As any term ill the Data : the similar term in the Question.

Inverse Proportion.—Pule.—State, As any term in the Question : the similar term in the Data,

Or, in short,

Direct, As Data I Question.

Inverse, „ Question ; Data.

Note.—These rules apply to the statement, at sight, of tho 1st and 2nd terms. The third terln is stated in the usual way.

When the Data comes before the Question, as it usually does, this method of stating amounts to this :—State straight on as the terms occur, in the case of Direct Proportion ; and backwards, in the case of Inverse Proportion.

Of course, the reverse of this should be done when the Question comes first.

Example I.—Direct Proportion.—If 16 lbs. cost 5s. 4d., what will 48 lbs. cost 1

Here it is evident that the proportion is direct; hence, after placing 5s. 4d. hi the third term we state straight on as the terms occur :—

As 16 lbs. ; 48 lbs. * * 5s. 4d. ; 16s. Answer.

Example II.—Inverse Proportion.—If 16 men take 57 days to do a piece of work, how long would 48 men take 1

Here it is evident that the proportion is Inverse (since more men would require less time, and less men would require more time). Hence, state, backwards—

As 48 men ; 16 men ; \ 57 days I 19 days. Answer.

The advantage of this method of working will be manifest in a question where it is not apparent whether the term of the data or that of the question is the greater ; for instance, in the sum,

Example III.—If 2 cwt. 1 qr. 14 lbs. of salt cost £1 18s., what will '125 of a ton cost 1

Here it is difficult to distinguish which is the greater quantity of salt, yet there is no difficulty in stating the terms; since, if the latter quantity be greater, the answer will be greater than £1 18s., and vice versa. Hence state, by the rule for direct proportion, as data ; question.

As 2 cwt. 1 qr. 14 lbs. : *125 of a ton ;; 38s. Answer.

Then reducing the first and second terms to lbs., we get—

As 266 lbs. : 280 lbs. :38s. : 40s. Answer.

Example IV.—If 3 men, 5 women, and 16 boys can do a piece of work in 216 days, how many days would 5 men, 7

women, and 7 boys take to do tlie same piece of work—a man doing 3 parts, while a woman does 2, and a boy 1 'i

Here time is required ; hence 216 days is the third term. Now, it is evident, as in Example II., that the proportion is inverse, since the time required will be in the opposite proportion to the number of workers; and, treating the workers as a whole, we must therefore state, As the workers in the Question—viz., 5 men, 7 women, and 7 hoys—is to the workers in the Data, viz., 3 men, 5 women, and 16 boys. Hence, the statement—

men    women boys men women boys    days

As 5    + 7    +    7    :    3    +    5    +    16    :: 216

3    2    1    3    2    1

15    + 14    +    7    :    9    +    10    +    16

Or,    days days

36 parts of work I 35 such parts 11 216 ! 210. Ans.

The statement being made, we now multiply the numbers representing the men, women and children by 3, 2 and 1, respectively, to obtain the number of parts of work performed respectively by the workers in the first and second terms ; when, as 36 parts represents the work done by the workers in the first term, and 35 parts that done by the workers in the second term, the statement amounts to—

As 36 parts : 35 such parts :216 days ; 210 days.

Problems in Proportion frequently assume a form which renders it necessary to employ Audition or Subtraction either before, or after the application of the Rule of Three.

Example V.—A bankrupt pays 13s. 4d. in the £, what will a creditor lose on a debt of <£42 1

First Plan.—Here we find the loss on every £1 by subtracting the 13s. 4d. which the creditor receives. Thus, £1 — 13s. 4d. = 6s. 8d. Then we reason, if 6s. 8d. be lost on £1, more will be lost on £42.

We therefore state, As £1 : £42 :: 6s. 8d. : £14 Answer.

Second Plan.—We might first find the amount the creditor will receive, and then subtract this from the amount owed him. Thus, we reason, if he receive 13s. 4d. on £1, ho will receive more on £42.

Therefore state, As ¿£1 : ¿£42    13s. 4d. : ¿£28.

Then, since £28 is the amount received, the amount he loses is £42 — £28 = £14 Answer.

Example VI.—The purchaser wished to have the perch reckoned as 7 yards in length : what would have been his estimate of the area of a farm containing 207 acres 2 roods 11 poles 3 square yards 1

In this problem we first require to find how many square yards are contained in a square whose side is a perch of 7 yards. It will evidently contain 7 times 7, or 49, instead of 30^. We, then, reason thus If the purchaser requires 49 square yards to the square perch instead of 301, his estimate of an area would be less than its real extent in the proportion of 49 to 301. We, therefore, state—

As 49 : 30|;; 207 ac. 2 ro. 11 per. 3 sq. yds. ; required area.

Then, 207 ac. 2 ro. 11 per. 3 sq. yds. X 301 -f- 49 = 128 acres 22 PH square perches. Answer.

Example VII.—At what exact times between 12 noon and midnight are the two hands of a clock together; and when are they exactly opposite each other 1

It is evident that the two hands will be together every time the large hand has gained a full round ; and that they will be opposite every time the large hand has gained an odd number of half rounds. Now, the large hand revolves 12 times while the small hand revolves once. The large hand, therefore, gains upon the small hand 11 rounds in every 12 hours. We now reason that it will take less time to gain 1 round, and hence state—-

As 11 rounds : 1 round :12 hrs. : 1 hr. 5T5T min.

The hands will therefore be together every 1 hr. 5pT min., and opposite, at £ of 1 hr. 5pT min. ; or, 32T\ min. past 12, and every 1 hr. 5pT min. after that. Hence, the answer is obtained by multiplying 1 hr. 5y\ min. by 1, 2, 3, 4, 5, 0, 7, 8, 9, 10, to find when the hands will be together ; and by subtracting 32TST min. from each of these answers to find when they will be opposite.

Note.—If the hour be clearly seen, it will only be necessary to multiply 55 min. by 1, 2, 3, &c., to find the time.

1.    Together at 1 hr. 5j\ min., or 5fT min. past 1 5 2hrs. 10p^, or 10-^ min. past 2 ; 3 hrs. 16T4T min., or 16^ min. past 3; 4 hrs. 21t9t min., or 21T9T min. past 4 ; &c., &e.

2.    Opposite at 32TST min. past 12, or 27-rr min- to 1 ; and at every subsequent 1 hr. 5j\ min.

Example VIII.—What should be the price of a bag of flour when the fourpenny loaf weighs 4 lbs., if it was £2 16s. a bag when 3| lbs. of bread cost 4d. 1

In this problem, the price of the loaf is the same in both Data and Question ; and when this is the case, ice neglect to state such terms, since they do not affect the answer. We, however, make use of such a term in reasoning to obtain the statement of the other terms.

We reason thus :—If we get more bread for the same money, flour must be cheaper. Hence state,

As 4 lbs. •' 3|- lbs. :: 56s. : 49s. Answer.

Example IX.—If a map is drawn upon the scale of 3 d inches to a mile, how many acres will be represented by 1 sq. ft. 3 sq. in. on the map 1

Here, 3^ in. represents a mile ; consequently 3£ times 3|, or 12J sq. in., will represent a square mile, or 640 acres. Hence state,

As 12^ sq. in. : 147 sq. in. ;; 640 acres : 7680 acres. Ans.

Example X.—126 masons contract to build a wall in 36 days; they complete one-half of it in 22 days : how many additional hands must then be employed that the work may be finished within the contract time ?

Here, 126 masons do one-half the work in 22 days, thus leaving (36-22) or 14 days for the other half of the work to be done in. Now, we reason, more men will be required to do the same work in less time. Hence, we state,

As 14 days : 22 days :: 126 men : 198 men.

But 198 is the number required to do the work, and 126 are at present engaged ; so that the additional number required will be 198 - 126 ~ 72 men. Answer.

Exercise    XXIII.—SIMPLE    PROPORTION.—

FROM TEACHERS’ EXAMINATION PAPERS.

1.    If 12 books cost £2 14s., what will 20 books cost ?

2.    If 34 acres cost £40 16s., what will 68 acres cost ?

3.    If 3 lbs. of butter cost 4s. 9cL, what quantity of butter may be bought for 11s. Id. ?

4.    If 17 yards of cloth cost £2 6s. 9d., how many yards maybe bought for £2 12s. 3d. ?

5.    If 880 articles cost 10 guineas, how many articles could be bought for 7 guineas ?

6.    If 792 articles cost 9 guineas, what will 88 articles cost?

7.    If 12 men could do a piece of work in 35 days, how many men will be required to do the same piece of work in 30 days ?

8.    If 3 cwt. 3 qrs. 9 lbs. cost £7 11s. 6d., what will 5 cwt. 0 qrs. 12 lbs. cost ?

9.    If 21 yards of carpet, 2 ft. 8 in. wide, will cover the floor of a room, how many yards will it take of carpet only 2 ft. 4 in. wide ?

10.    If 24 planks, 15 feet long, will reach across a river, how many planks, 18 feet long, will span the river ?

11.    If it cost £5 12s. 6d. to carry 10 tons 60 miles, what will it cost to have the same weight conveyed 36 miles ?

12.    If 625 potato cuttings will plant a furrow when placed 13 inches apart, how many cuttings will be required when they are planted 12f inches apart ?

13.    If 512 trees will plant a street, when planted 15 yards apart, how many would plant it if planted 24 yards apart ?

14.    If 42 coins, each worth 6s. 8d., will pay a certain debt, how many half-crowns will pay the same debt ?

15.    If 25 coins, each worth 7s., will pay a certain debt, what must be the value of another coin when 35 of them will pay the same debt?

16.    If 12 months’ rent be £81, what is the rent for 1 year S months?

17.    If the velocity of a body moving through space varies inversely as the time occupied in its passage, and a cannon ball whose velocity was 2,000 yards per minute reached a target, 2000 yards off, at the same moment as another cannon ball fired from the same place 12 seconds later ; what was the velocity per minute of the second ball ?

18.    If a labourer receive £70 for 9 months’ work, what should he receive for working 1 month 1 week 1 day ?

19.    If 3 cwt. 1 qr. 12 lbs. be carried 120 miles for ISs. 4d., how much should be paid for the carriage of 5 cwt. 2 qrs. 4-g lbs. the same distance ?

20.    If 16 men take 108 days to do a piece of work, how long will 27 men take ?

21.    If the tax upon a house assessed at £65 per annum be £4 17s. <xL, what is the tax upon property assessed at £117 per annum?

22.    If 11 men can dig 132 cubic yards of earth out of a dam in 13 weeks, how much can 9 men dig in the same time ?

23.    The quotients obtained by dividing different dividends by the same divisor are to each other as the numbers divided. If dividing 11011 by 143 produces 77, find, by Proportion, the quotient of 44044 by 143.

24.    The quotients obtained by dividing the same number by different divisors are to each other in inverse proportion to the divisors ; that is, the quotient increases as the divisor diminishes. If 22022 divided by 26 produces 847, find, by Proportion, the quotient of the same number by 91.

25.    One paddock is 1880 yards long and 820 yards broad ; what is the breadth of another paddock 1640 yards long which contains the same area ?

26.    12 men can earn £17 10s. 9fd. in 20 days ; how many days should 8 men work to earn the same amount ?

27.    14 horses could plough a field of 30 acres 2 roods in 35 days ; how many horses could plough it in 49 days ?

28.    If 2 qrs. 2 nls. 2 in. is the ninth part of 6 yds. 0 qrs. 2 nls., find, by Proportion, the ninth part of 18 yds. 1 qr. 2 nls.

29.    A gentleman wishing to find the height of a tree which cast a shadow 135ft., observed that the shadow of his walking-stick, which was 2 ft. 9 in. long, measured 4 ft. 7 in. ; what was the height of the tree ?

30.    A tree growing on the edge of a pond cast a shadow within 2 ft. of the opposite edge. If the width of the pond was known to be 74ft., and a reed 3 ft. above the water cast a shadow 3 ft. 4£ in., what was the height of the tree ?

31.    The property in a shire is assessed at £37548 ; what rate

must be struck in order to raise £3129 ?    ,

32.    A bankrupt’s assets are £75 16s. 2d., and he pays 3s. 8d. in the £ ; what are his debts ?

33.    If the quartern loaf cost 9d. when wheat is at 72s. a quarter, what should it cost when wheat is at 68s. a quarter ?

34.    If the sixpenny loaf weigh 4 lbs. when flour is at £2 a bag, what should it weigh when flour is £ 3 4s. per bag ?

35.    A gentleman whose income, above £500, is taxed at the rate of 8d. in the £, pays a tax of £17 17s. 4d. ; what is his total income ?

36.    If a train, running 37 miles per hour, runs the distance between two towns in 1 hr. 45 min., how long will another train take, running 35 miles per hour ?

37.    A tradesman marks his goods at Is. 4d. in the £ above the cost price ; what does he gain on sales amounting to £18 2s. 8d. ?

38.    The large hand of a watch revolves 12 times, while the hour hand revolves once. If the minute hand thus gains 11 rounds while it goes 12, how far has it to go in order to gain 1 round only ?

39.    One of two engines consumes as much water in 5 minutes as the other does in 7 minutes. If a tank will supply the former with water for 21 hours, how long will it supply the latter ?

40.    A man walked a certain distance in 37J days by walking 11 hours per day ; how many days would he have taken had he walked but 7| hours per day ?

41.    A can walk the same distance in 36 minutes as B can in 35 minutes ; how long will B take to walk a distance which A walks in 5 hours 24 minutes ?

42.    A can walk the same distance in 26 minutes as B can in 27 minutes ; how far will B walk while A walks 135 yards ?

43.    A ship’s provisions will last 57 weeks when the allowance is 17 ozs. per man ; how long will they last when the allowance is reduced to 12f ozs. ?

44.    If I can have 36 tons carried 10 miles for £16 7s. ; for how much may I have the same weight carried 75 miles?

45.    If I can have 36 tons carried 10 miles for £16 7s., what weight may I have carried the same distance for £5 9s. ?

46.    If I can have 36 tons carried 10 miles for £16 7s., how far may -• I have the same weight carried for £49 Is. ?

47.    A exchanged 39 yards of cloth worth 5s. 6d. a yard with B, for doth worth 6s. 6d. per yard ; how many yards should B give A ? *

48.    How many yards of carpet 2 feet 3 inches wide will cover 45 yards of oilcloth 1 foot 9 inches wide ?

49.    After paying an income-tax of 6d. in the £ a person has £258 . 17s. 3d. remaining ; what was his total income?

50.    How much land may be rented for £70 10s. 6d., if 5 acres be rented for £4 13s. 4d. ?

51.    If 97 gals. 2 qts. of wine cost £55 10s. 9d., how much can be bought for £9 5s. lgd. ?

52.    If 27 men do a piece of work in 11 days, working 9 hours a day, in what time would they have done it had they wrought but 7 hours daily?

53.    If 37 acres 3 roods 6 poles cost £115 16s. Sd, what will 56 acres 2 roods 29 poles cost ?

54.    If the carriage of 17 cwt. 2 qrs. 7 lbs. cost £110 12s. 10|d., what weight could be carried for 6 guineas ?

55.    If 5 tons 12 cwt. of goods can be carried 80 miles for £16 12s. 6d., what weight could be carried the same distance for 20 guineas?

56.    If 8 tons 14 cwt. of goods can be carried 76 miles for £24 15s., how far could 3 tons 80 lbs. be carried for the same price ?

57.    A bankrupt paid 7s. 9^d. in the pound ; his assets were £2898 10s. What were his debts ?

58.    One map is drawn upon a scale of 2 inches to a mile ; another, upon a scale of 3 inches to a mile. How many square inches of the former will represent an area represented by 117 square inches of the latter ?

59.    In what time will £472 10s., lent at 4£ per cent, per annum, simple interest, amount to £567 ?

60 A person finds that his income for the 6 years ending 31st December, 1880, will be at the rate of £411 Is. Sd. per annum and he wishes to have saved by that time 144 guineas ; what should be his daily expenditure ?

61.    A farm, containing 37 5 acres, was let for £75; what rent should be paid for another farm, containing 262 acres 2 roods ?

62.    If 43 acres 3 roods cost £215 14s. 10d., how much can be bought for £323 12s. 3d. ?

63.    A baker, having 200 customers, bakes 180 four-pound loaves daily ; how many additional customers should he have that his consumption of hour may be 12 bags per week, each bag yielding 135 four-pound loaves ?

64.    A garrison, provisioned for 8 weeks, was reinforced by 100 men, and the provisions were then consumed in 6 weeks ; how many men were in the garrison at first ?

65.    An army of 2,000 men had 17 weeks’ supplies, but one-eighth of the men were slain and a number of prisoners taken ; if the food was now consumed in 16 weeks, how many prisoners were taken ?

66.    A contractor employing 20 men engages 5 other labourers, and by this means finishes his contract two days earlier ; how many days’ work had the 20 men to do when the 5 men joined them ?

67.    The crew of a ship provisioned for 8 weeks was reduced by fever to two-thirds of the original number. Meeting with a wreck, 24 men were taken aboard, and the provisions were exhausted in 4 weeks. Of how many men did the crew at first consist ?

68.    A man-of-war with 1200 troops on board had sufficient provisions to last 17 weeks. The survivors of a wreck having been taken aboard, the provisions were consumed in 15 weeks. How many were taken aboard ?

69.    72 men had provisions for 44 days, but after 5 days 20 men went away. How long did the provisions last the remaining men ?

70.    A baker having 160 customers makes 180 loaves daily. How many additional customers would cause him to bake 270 such loaves daily ?

71.    If 24 men could have done a piece of work in 30 days, in what time was the work completed when the contractor put on 6 extra

men ?

72.    A farmer could plough his paddock of 76 acres 2 roods, 26 poles in 45 days, but hiring 5 extra teams he had his land ploughed in 37^ days. How many teams were at first employed ?

73.    A garrison having provisions for 60 days, allowing 12 ozs. per man per day, was re-inforced by three times as many men as it already contained. How long should the provisions last if no reduction be made in the daily allowance ?

74.    A fortress commanded by 100 soldiers had 8 weeks’ provisions, but having been besieged, the daily allowance per man was reduced from 18 ozs. to 12 ozs. How long did the provisions last ?

75.    A ship leaves port with sufficient provisions to last 14 weeks ; 6 of the crew absconded upon setting sail, and the voyage lasted 16 weeks, at the end of which time it was found that the provisions were just exhausted. Of what number did the full crew consist ?

76.    A tax of 6d. in the £ is proposed to be levied upon the excess of all incomes over and above the first £200 ; what tax will gentleman pay whose annual income is 500 guineas ?

77.    If the Victorian Legislature pass an Act levying a tax of 6d. in the £ upon the excess of incomes above £240 per annum ; what will be the net yearly income of a gentleman whose monthly salary is 40 guineas ?

78. If 300 men could do a piece of work in 40 days, working 8 hours a-day ; how many men must be sent away so that the remainder shall finish the work in 50 days, working the same number of hours a-day ?

79.    A piece of work has to be completed in 48 days, and 35 men are employed. It is found at the end of 28 days that one-half only of the work is done ; how many additional men must be engaged in order to complete the work within the proposed time ?

80.    A dishonest trader used a weight which he called 1 lb. weight, whereas, in reality, twenty-four of such weights weighed 22^ lbs. only. What was the loss to the purchaser on 108 lbs. ?

81.    A man sold me a bag of onions as weighing 1 cwt., but I found there was a deficiency of 7 lbs. in the weight. What weight was sold me for each 1 lb. ?

82.    A retail dealer bought 4 cwt. 3 qrs. 8 lbs. of coffee for £33 19s., and sold at such a price that he gained £5 8s. 6d. What was his selling price per quarter ?

83.    A grocer bought 3 cwt. 1 qr. 16 lbs. of coffee for £22 14s. 9d., and sold at such a price that he gained £3 7s. 9d. What was his selling price per lb. ?

84 Suppose there be levied a house-tax of 6d. for every £1 of annual rental above £80, what tax will be payable for a house whose rental is 175 guineas ?

85.    Twenty-seven men engaged to complete a piece of work in 10 days of 8 hours each, but three of them were unable to attend. How many hours daily would the remainder have to labour in order to finish the work in the given time ?

86.    Sixteen men engaged to do a piece of work in 12 days of 10 hours each ; but, thinking the day too long, they agreed to work but 8 hours daily, and to take on extra hands to complete the work in the stipulated number of days. How many extra men were taken on ?

87.    Suppose there should be levied a tax of 6d. in the £ upon the excess of incomes above £500 per annum ; what sum will be payable upon an income of SI5 guineas per annum ?

88.    The rateable property in a borough is estimated at £6435 7s. 6d., on which a rate is to be levied of £214 10s. 3d. ; how much must be paid by an estate valued at £137 12s. 6d. ?

89.    If the cost of making an English mile of road be £1265 12s. 6d., \yhat would be the cost of making an Irish mile, 14 English miles, being equal to 11 Irish miles ?

90.    An apothecary uses, instead of a pound, a weight deficient by 3. drams ; how many of such weights will be required to make up 116-4-lbs. of true weight ?

91.    Find by Proportion the value of 17 lbs. 6oz. 10 dwt. of gold at £3 17s. 10£d. per oz., and prove the correctness of the answer by Practice.

92. A bankrupt’s estate can only pay 13s. 5^d in the £ ; what shall I receive for a debt he owes me of £5, and how much shall I lose by it ?

/    93. If the rent of 42 acres of land be £66, what will be the rent of 91

v acres of the same land, and how manj^ acres of it may be had for £121 ?

94.    If 36 cwt. 3 qrs. 16 lbs. of hay last 35 horses for 16 days, how many horses will the same last for 28 days ?

95.    A servant engages herself on the 27th June, 1870, at the rate of £36 a year ; she leaves her place on the 13th of December, 1870. What wages are due to her ?

96.    Four hogsheads of sugar weighed, in the gross, 6 cwt. 3 qrs.

2 lbs. ; each hogshead, without the sugar, weighed 60 lbs. The whole cost £24 5s. 7^d. ; what was the price of 8 lbs. of sugar ?

97.    A bankrupt’s debts amount to £3500, his assets to ¡£1370 16s. 8d. ; what will a creditor lose on a debt of £650 ?

98.    A bankrupt fails for £4000. Find the amount of his assets when a creditor lo^es £214 17s. 6d. on a debt of £350.

99.    I bought 13 acres 2 roods 36 poles for £70 6s. 9d., and paid for the conveyance of the land £15 10s. I want to cut up and sell my purchase so as to cover expenses and gain £20 on my bargain. What must I charge for 9 acres 3 roods 6 poles ?

100.    Bought 4 lbs. 6 ozs. 12 dwt. of gold for £211 11s. 6d. ; paid £1 7s. carriage ; 1 wish to gain £10 on my purchase. What must I charge for 1 lb. 2 ozs. 10 dwt. ?

101.    Bought 750 yards of cloth for £126 11s. 3d. ; at what must I sell it per ell English so as to gain £29 13s. 9d. on the whole?

102.    A bankrupt’s debts amount to £13488, and his assets to £5915 Is. Od. ; what will his creditors receive in the £?

103.    If a pillar of stone weigh 27 cwt. 3 qrs. 20 lbs., what will be the weight of a pillar of cedar of the same dimensions, supposing that the weights of equal bulks of cedar and the stone be as 561 to 2496 ?

104.    A wine merchant uses, instead of a gallon, a measure deficient by three-quarters of a pint. How many of such measure will be required to make up 130§ gallons of true measure ?

105.    A block of granite weighs 45 cwt. 2 qrs. 18 lbs. : what «will

be the weight of a block of mahogany of the same dimensions, supposing that the weights of equal bulks of mahogany and granite be as 1063 to 2784?    .

106 A plan of an estate is drawn on a scale of 4 feet to a mile : how many square inches of the plan will represent 3 acres 3 roods ?

107.    A bankrupt’s assets amount to £695 12s. ; his debts are £5768 10s. What will a creditor lose on a debt of £57 12s. 6d. ?

108.    The rateable property in a parish amounts to £7260, and a rate of £225 10s. is to be raised. What will be the net income from a pi'operty producing a gross rental of £875 12s. 6d. ?

109.    I rent a house at the rate of £64 a year, and occupy it from llt’n February to 15th September, 1870. What rent is due ?

110.    Four lbs. of sugar cost 3s. 9d. ; 4 hogsheads of sugar weighed in the gross 6 cwt. 3 qrs. 2 lbs. : each hogshead, without the sugar, weighed 60 lbs. What was the value of the sugar ?

COMPOUND PROPORTION.

Compound Proportion.—"When the mutual relationship of two quantities is affected by several different circumstances, the proportion is compound | and the ratio of the third and fourth terms is compounded of the several ratios expressing the alterations to be made.

Note.—Ratios are compounded by multiplication.

Here are given two or more complete ratios, and the antecedent of an incomplete compound ratio, of which the consequent is the fourth proportional required.

Definition.—When one thing varies as several other things, the proportion is Compound.

Thus, for example :—

Interest varies with the principal, time, and rate per cent.

Work done varies with the number of workers, and the time.

The cost varies with the number bought, and the value of one.

The weight of a thing is as Us bulk and specific gravity.

The force, or momentum, of a thing is as its weight and velocity.

The number of labourers required varies directly as the work to be done ; and inversely as the time allowed to perform the work, and the number of hours they labour each day.

When several circumstances cause a number to vary, the effect of each may be found separately by means of successive simple proportion statements.

Thus, since the cost of carriage varies with the weight and the distance, if we require to find the cost of the carriage of 2 tons for 150 miles when that of 10 cwt.for 50 miles is £1 3s. ltd., we may first find the cost of carriage of the greater weight—2 tons—for the same distance ; and then, making this answer the third term, find, by another statement, what the cost will be for carrying this weight the greater distance.

We reason :—A greater weight will cost more.

Hence—As 10 cwt. ; 40 cwt. ; 1 £1 3s. 4d.    £4 3s. 4d.

Then—The greater distance will increase the cost.

Hence—As 50 miles ; 150 miles ; I £4 13s. 4d. ; £14. Answer.

By Compound Proportion the same result is arrived at by one simple process, the statements being first made in precisely the same way, save that the first and second terms of the second statement are placed under the first and second terms of the first statement; thus—

As 10 cwt. ;    40 cwt. ; I £1 3s. 4d. \ £14. Answer.

As 50 miles ; 150 miles

The process now employed is that of multiplying the third term by the product of the second terms, and dividing the result by the product of the first terms.

Thus, <£1 3s. 4d. x 40 times 150 -j- 10 times 50 = £14. Answer.

In Compound Proportion, the ratio of the third term to the fourth term is that compounded of the ratios of the several first and second terms.

Definition.—A Compound Ratio is the product of two or more ratios.

It has been explained, that Ratio is the number of times one number contains, or is contained, in another; and that the ratio of one number to another is expressed fractionally by placing the number to be compared as numerator, and the number with which it is compared as the denominator. Thus, the ratio of 7| to 10 is 7^-lOths or |.

To obtain a compound ratio, we must therefore multiply together the fractions expressing the given ratios.

Rule.Multiply the numerators together for the nev) numerator and the denominators together for the new denominator.

Thus, the Compound Ratio of 2 : 3 and 4 : 5, is f of \ = f ; that of 1 to 2 and 3 to 4, is £ of f = £ ; and that of 2 : 3 and 3 : 4, is $

1 = A — i-

Now, the fourth term is to the third as the compound ratio of the several second terms to the first terms. Hence, in the. above example, where 40 cwt. is 4 times 10 cwt., and 150 miles is 3 times 50 miles, the answer is 4 times 3 times the 3rd term ; that is, 12 times £1 3s. 4d. = £14. In every example of Compound Proportion, the answer may be obtained by thus multiplying the third term by the compound ratio of the second and first terms; but, for the sake of convenience, the rule is stated thus :—    ,

Rule.—Divide the continued product of the second and third terms by that of the first terms.

As preparatory to making the statement, the following plan of placing the similar terms is not unfrequently adopted.

Rule.—Write down the terms of the Data in a horizontal line, and place directly under them the corresponding terms of the Question, with a note of interrogation under the odd term.

This plan of arranging the like terms enables us to see, at a glance, which two terms should be taken together in considering them with reference to the answer.

Example.If the sixpenny loaf weigh 4 lbs. when wheat is at 8s. Gd. a bushel, what weight of bread should be purchased for 7s. 6d. when wheat is at 6s. 6d. a bushel ?

Place thus :—

Supposition

Demand


Cost of Weight of Price of Bread. Bread. Wheat. 6d.    4 lbs.    8s. 6d.

7s. 6d.    Ì    6s. 6d.

In considering which are like terms, it must be remembered that they are not necessarily of the same name, but that they must be of the same nature.

Example.If 15 horses in 16 days draw 22 tons for 100 miles, IN HOW many days could 40 OXEN DRAW 35 TONS A DISTANCE of 70 MILES, SUPPOSING THAT 2 HORSES CAN DO AS MUCH AS 5 OXEN ?

Here, 15 houses and 40 oxen are similar terms, the comparison being made between the work done by them.

Again, since 2 horses can do as much work as 5 oxen, it is clear that a horse does more work than an ox in the proportion of 5 to 2 . so that the workers named in the supposition do 5 parts each while those in the demand do 2 parts each.

Parts of

Workers. Time. Weight. Distance. Work. Supposition    15 horses    16 days    22 tons    100 miles    A horse, 5

Demand    40 oxen    ?    35 tons    70 miles    An ox, 2

Statement.

As 40 : 15 :: 16 days.    Working.

„    22:35    16x 15x35x70x5 Answer.

„ 100 : 70    -= 16|Tdays.

„    2:5    40 x 22 x 100 x 2

A very common practice obtains of finding, by simple proportion, how many horses the oxen are equal to ; thus, if 5 oxen are equivalent to 2 horses, then 40 oxen are equivalent to 16 liorses. The statement is then made, As 16 horses is to 15 horses. Since, however, the equal parts done by a horse and an ox respectively in the same time may be at once obtained by changing the numbers which express the equivalent of each other, the above plan is preferable. It has also the advantage of showing the full statement at one view.

Eule for Stating—

1.    State the 3rd term, as in simple proportion.    ,

2.    Select two like terms, consider their effect upon the answer, and state as in simple proportion.

3.    Select two other similar terms ; reason, and state in the same way, placing them under the terms first stated.

4.    Continue to select and state similar quantities in this way until all the terms have been dealt with.

In reasoning out the effect upon the answer of any term in the question being greater or less than the corresponding term in the data, carefully leave out of present

consideration all other circumstances affecting the answer, and

1.    Keep the question well in view.

2.    Read the question with the 3rd term each time.

3.    Consider all other terms as equal in data and question.

f. Substitute the words “ the same ” for each quantity in

the question other than the one under consideration.

Example I.—If 48 persons eat 35s. worth of bread in 8 weeks, when wheat is at 56s. a quarter ; how many persons could be kept 6 weeks in bread for 90s., when wheat is at 64s. a quarter ?

1.    Place 48 persons in the 3rd term, because the number of persons is required.

2.    Ask—If 48 persons eat 35s. worth of bread in 8 weeks how many may be kept same time for 90s., when wheat is at the same price ? Ans.—More. Then state—As 35s. I 90s.

3.    Ask—If 48 persons eat 35s. worth of bread in 8 weeks, how many persons may be kept 6 weeks for the same money when wheat is at the same price? Ans.—More. Then state—As 6 weeks ; 8 weeks.

4.    Ask—If 48 persons eat 35s. worth of bread in so many weeks, when wheat is at 56s. a quarter, how many persons may be kept the same time for the same money when wheat is at 64s. a quarter ? Ans. —Less. Then state—As 64s. 1 56s.

Statement.—As 35s. I 90s. t48 persons ,, 6 wks. 1 8 wks.

,, 64s.    : 56s.    144 persons. Ans.

Working.—Multiplying together the 2nd terms and 3rd term, gives a continued product of (90 X 8 X 56 X 48), 1935360, which, divided by 13440 (the continued product of the 1st. terms), gives 14persons answer.

The very tedious labour of multiplication and division may often be very much shortened by cancelling the 1st terms with the 2nd and 3rd (see page 32); thus, the above working might have been shortened very much by drawing a lin between the 1st and 2nd terms and cancelling any number

on the one side of the line by any number (or factor of a number) on the other side; thus,

As


18::^ 0

£    18 x 8=144 Ans.

8

Remember.—The first term may be cancelled with the 2nd or 3rd; but the 2nd term must never be cancelled with the 3rd.

Then, the numbers on the right of the line being the factors of a dividend to which the numbers on the left form the factors of the divisor, we have no divisor remaining, but have merely to multiply together the remaining factors (on the right), 18 and 8, in order to obtain the answer. Hence, the answer =18x8 = 144.

Note.—When the 1st and 2nd terms of the same kind are of different denominations, they must be reduced to the same denomination before working.

Complete Rule for Working,

1.    Bring eacli pair of terms to the same name.

2.    Cancel (where possible) the 1st terms with the

other terms.

3.    Multiply together the remaining factors of the

2ml and 3rd terms, and divide the product by that of the factors left in the 1st terms.

Note.—In order to facilitate cancelling, the 2nd and 3rd terms are often placed as the numerator of a fraction (with the sign» (into) between them), and the first terms (with the same sign between them) as the denominator. Thus, the above statement would be placed as under :—

3

4^    X

X ^ X ^ X 48

----= 144 Ans.

\    $

jfoTE._The answer is of the same sort and name as the 3rd term.

That concrete numbers, when once stated, must be regarded as abstract applies also to Compound Proportion.

Example II.—If 15 men eat 28s. worth of bread in 14 days WHEN WHEAT IS AT 52s. PER QR. ; WHAT MUST BE THE PRICE OF WHEAT THAT 1SS. MAY PROVIDE BREAD FOR 13 MEN FOR 10 DAYS ?

The price of wheat is required : hence place 52s. in the third term. There are here three things to be considered in reference to the answer :—

1.    The number of men to be kept for a given time on a certain sum.

2.    The money allotted to keep these men in bread for the given time.

3.    The number of days the money must keep these men in bread. These three things do not regulate the price of wheat, but are

rather themselves affected by the price varying : hence, the stating of this sum presents greater difficulty than that of Example I.

To state the 1st and 2nd terms—

1.    Ask—Must the price of wheat be more or less than 52s. that the same money may keep 13 men only, instead of 15, for the same

time ?

A ns. More. Hence, state, as 13 men : 15 men.

Or, reasoning by the rules for direct and inverse proportion, must the wheat be dearer or cheaper when the same mopey will keep less men in bread? Ans. Dearer. Then, since less gives more, the proportion is inverse. Hence, state, as Question : Data, or, As 13 men : 15 men.

2.    Ask—Must the price of wheat be greater or less than 52s., that 18s. may keep the same number of men in bread for the same time instead of 28s. ? Since less money is to keep the men, the wheat must be cheaper.

Hence, Answer. Less. State, As 28s. : 18s.

0 r, by direct and inverse method. If less money keeping the men shows that the wheat is cheaper, less gives less, and the proportion is direct. Hence, state as Data : Question, or, As 28s. : 18s.

3_ Ask_Must the price of wheat be greater or less than 52s., that

the same money may provide bread for the same number of men for 10 days, instead of 14 days? If the same money lasts for a shorter time, the wheat must be dearer. Ans. More.

Hence state, As 10 days I 14 days. Working.

Statement.    3    -    ^

= 54s. Ans.


As 13 • 15 11 52s.    ^ X 18 X \\ X

io : 14


„ 28 : 18

Exercise XXIV.—COMPOUND PROPORTION.— FROM TEACHERS’ EXAMINATION PAPERS.

1.    One gas burner consumes 150 cubic feet in the same time as another consumes 115 feet; if the former be charged 17s. 6d. for 36 days, when gas is 13s. per thousand feet, what should the latter be charged for 49 days when gas is 10s. per thousand feet ?

2.    If £545 15s. 4d. gain £77 19s. 4d. in 2| years, what sum will gain £165 13s. 8d. in 3£ years?

3.    If 56 men dig a trench 180 yards long, 4 feet wide, 5 feet deep, in 9 days, what depth of trench 144 yards long and 2 yards wide could 80 men dig in 7 days ?

4.    If 14 men eat 11s. 6d. worth of bread in 3 days when wheat is 52s. per quarter, what must be the price of wheat that £2 5s. may provide bread for 26 men for a week ?

5.    If 16 men eat 17s. worth of "bread in 4 days when wheat is 51s. per quarter, what will it cost to provide 25 men with bread for a week when wheat is 4Ss. per quarter ?

6.    10 men dug a trench 40 yards long, 4 feet deep, in 3 days, and 6 men dug a ditch 20 yards long in 5 days ; how deep was the latter ditch ?

7.    If 8 men mow 7 acres in 7 days, working 8 hours a day, how many men will mow 15 acres in 6 days, working 10 hours a day ?

8.    A garrison of 800 men provisioned for 35 days at the rate of 18 oz. per man per day, receives an augmentation of 250 men ; by how much must the daily allowance be diminished, that the provisions may serve the whole of them for 44 days ?

9.    If 5 men earn £7 7s. 6d. in 10J days, how many days must 21 men work to earn £81 2s. 6d ?

10.    If 7 men earn £9 10s. 6d. in 101 days, what will 25 men earn in 31£ days ?

11.    If 24 men earn £153 12s. in 16 days, for how long would £183 15s. pay the wages of 15 men, the latter being paid J less per day than the former ?

12.    If a tank 30 feet long, 8 feet deep, 24 feet wide, be filled in 14 hours by a tap which runs 40 gallons per minute, how long would it take to fill a tank 6 feet larger every way with a tap which runs 50 gallons per minute ?

13.    A tank 30 feet long, 8 feet deep, 24 feet wide, was filled in 14 hours by a tap running 40 gallons per minute; how many gallons per minute must a tap run which will fill a tank 6 feet larger every way in 15 hours ?

14.    A person with a capital of £326 '03 gained £6 '05 in 3 '45 months ; how long will it take, at the same rate, for a capital of £659 ’62 to gain £34-48?

15.    A person with a capital of £659-62 gained £34-48 in 972

months; how long will it take, at the same rate, for a capital of £326 "03 to gain £6 "05 ?    '

16. If the cartage of 5f cwt. for 150 miles cost £3 7s. 4d., what must be paid for the cartage of 7 cwt. 2 qrs. 25 lbs. for 64 miles at -the same rate ?

17.    If 22 loaves last 9 persons 7 days, when the loaf costs 20d., how many persons may be fed 12 days at the same cost when the loaf is 7d., the rate of consumption being the same?

18.    If 35 persons consume 52 quarters of wheat in 7 months, when wheat is 52s. per quarter, what quantity should they consume in 9 months, when wheat is at 60s., so as not to increase the rate of expense ?

19.    A vessel with a crew of 32 men was provisioned for a voyage of 84 days, at the rate of 19^ ozs. per man per day, but at the end of 57 days 13 shipwrecked men were picked up, and the voyage delayed : to what must each man’s allowance be reduced to make the provisions last for 96 days ?

' 20. If the carriage of 54 cwt. 2 qrs. 7 lbs. for 46 miles be £1 15s., what distance may 23 cwt. 1 qr. 15 lbs. be carried for £2 5s. 6d. ?

21.    If 5'63 cubic inches of water weigh 3254 ozs. avoir., what will be the weight of 7 '9 cubic inches of nitric acid having a specific gravity of 1 ’220 ?

22.    If 69 men engaged at excavations remove 6932 cubic yards in . 35 days of 8 hours each, how long will it take 27 men, working 9” hours daily, to remove 10398 yards ?

23.    If 21 men take 4 days to paint 45,516 square feet of wall, working 6 hours daily, in what time will 19 men, working 8 hours daily, paint 72,067 square feet ?

24.    If the earnings of 24 men for 15 days’ work of 8 hours amount

1 to £150 13s. 4d. ; for how many days will £200 15s. pay 16 men -when wages have fallen supposing the men to work 10 hours a day, and to be paid proportionally for the overtime ?

25.    If 5 compositors in 16 days of 11 hours each can compose 25 sheets of 24 pages in each sheet, 44 lines in each page, and 40 letters in each line ; in how many days of 10 hours each may 9 compositors compose a book containing 36 sheets of 16 pages in each sheet, 50 lines in each page, 45 letters in each line ?

26.    If 5 compositors in 16 days of 14 hours each can compose 20 sheets of 24 pages in each sheet, 50 lines in a page, 40 letters in a line ; in how many days, 7 hours long, may 10 compositors cordpose a volume containing 40 sheets of 16 pages to the sheet, 60 lines in a page, and 50 letters in a line ?

27.    If 17 men in 54 days, by working 8 hours a day, made an

excavation 121 feet 6 inches long, 25 feet 6 inches broad, 24 feet deep, for what length of time daily must 18 men work during 51 days m order that they may complete an excavation whose length and breadth are 1 foot 6 inches less, and its depth 1 foot 6 inches greater, than the preceding one, supposing that 9 men of the latter do as much as 10 men of the former ?    I

28 If 45 men, by working 9 hours a day, make an excavation 180 feet long, 43 ft. 6 in. broad, and 40 ft. 6 in. deep, in 87 days, how long will it take 58 men, who work 8 hours a day, to make an exca-

vation whose length is 2 yards shorter, and whose breadth and depth are each 1 yard shorter than the preceding one, supposing that 6 1 men of the former gang do as much work as 7 of the latter gang ?

29.    If 95 men can form a trench 150 yards long, 15 yards wide, and 3 feet deep, in 7 days of 8 hours each, in how many days of 7 hours each can 135 men dig a trench 450 yards long, 27 yards wide, and 5 feet deep ?

30.    If when gas is 7s. 6d. per 1000 cubic feet, the cost of the gas consumed by 10 burners, lighted 3 hours daily, amounts to £4 13s. 9d.: for 75 days, what will be the cost of the gas consumed by 8 burners, lighted 5 hours daily, for 60 days, when gas is 10s. per 1000 cubic feet ?

31.    If-4 men, or 5 women, can gather 4 ac. 3 ro. 8 per. of potatoes in 3 days of 8 hours eacl\, how many hours per day should 3 men and 5 women be employed in order to gather 35 ac. 2 r. 18 per. in 11 days?

32.    If 6 women and 5 men ci n gather 48 ac. 2 ro. 16 p. of potatoes in 12 days of 9 hours each, an l 2 men' can do as muclf work as 3 women, how many hours per .ay. must 9 women be employed in order to gather 31 ac. 1 r. 16 pe:. in 11 days ?

33.    The paving of a yard w th tiles 9 in. square cost £8 Is. 4d. What would have been the co, t had the tiles been 9 in. long and 6 in. broad, their price being to that of the former as 5 : 8, labour disregarded ?

34 If 3 horses eat as much as 5 oxen, and 4 oxen as much as 9 sheep, how many horses, in addition to 20 sheep, may be put for 12 days into a paddock cracaining 42 acres, when 24 acres will feed 12 oxe> id 30 sheep for 19 days ?

I its men build a wall 40 ft. long, 12 ft. high, and 18 inches tAick in 5 days ; what length of wall, 10 ft. high and 2 ft. thick, could 25 men build in 4 days ?

36. A vessel with a crew of 27 men, provisioned for 90 days at the rate of 22 ozs. per man per day, was, after 27 days, forced by stress of weather to lie at anchor for a fortnight, at the end of which time 3 men died; how must the provisions be now apportioned, that they may hold out the extra time, i.e., for 104 days in all?

37 If 150 men can make an embankment 400 yds. long, 12 ft. broad, and 8 ft. high, in 18 days of 9 hours each ; in how many days of 8 hours each would 225 men make an embankment 650 yards long, 10 ft broad, and 1£ yds. high ?

38.    If 24 men build a wall 13 ft. high in 45 days ; how many men would be required to build a wall half as long again, but only 11 ft. high, in 56 days ?

39.    If a tenpenny loaf weigh 3| lbs. when wheat is at 9s. 6d. per bushel ; what ought to be paid for 1 cwt. when wheat is at 8s. 3d. per bushel ?

40.    If 2 men can do as much work in a given time as 3 women,

and 4 women as much as 5 boys, how many men must be employed to assist 12 boys to reap 126 acres in 18 days, when 8 women and 10 boys together can reap 140 acres in 25 days ?    .

11

41.    If 25 chests of tea, each containing 168 lbs., cost £306 5s., what should be given for 18 chests, each containing 150lbs., 11 lbs. of the latter being worth 12 lbs. of the former ?

42.    The paving of a court-yard, with tiles 6 inches square, cost £9 Is. 6d. : what would it have cost had the tiles been 12 inches long and 9 inches broad, their price being to that of the former as 23 I 8,—labour disregarded?

43.    If 40'32 tons are conveyed 73542 miles for 19'S1 pounds sterling, how far ought 1287 tons to' be conveyed for 45'64 pounds sterling ?

44.    A farm containing 37 '5 acres was rented for 36 months for £56’25 : for how many months should another farm containing 56 ‘25 acres be rented for £50 '625, the rent of 5 acres of the latter being equal to the rent of 4 acres of the forri er ?

45.    If 41-36 tons be carried 50’24 n iles for 30 ’806 pounds sterling, what weight should be carried 13‘J'32 miles for 46'209 pounds sterling ?

46.    If 18’27 cwt. be carried 63‘J miles for £3'.75, how much should be charged for the carriage of 2'34 tonp for 47*74 miles ?

47.    If the carriage of 1'75 tons fir 4800 miles came'to £70, what would be the proportionate chargB for carrying 5'125 cwt. 246 miles ?

48.    If £225 gain £6 11s. 3d. in f months, what sum should gain

£21 18s. 9d. in 5§ months ?    .

49.    If £150 gain £2 16s. 3d. in 4$ months, what sum should gain £13 in 3f months ?

50.    If 84 hdds. of wine cost £1732'5, what will be the cost of 54 hdds. of another Lind, 9'75 gallons of which are wortfn 14'625 <ra’6ms

of the former ?

51.    If the rent of a farm of 17 *7625 acres be £39 4s., what wi\\ be the rent of another farm containing 28'42 acres, if 6 acres of the former be equal in value to 7 acres of the latter ?

52.    If 7'08 tons are carried 52'08 miles for £14'46, what distance

should 7'947 tons be carried for £20 ?    . ,

53.    If the carriage of 48*28 tons for loO miles cost £36 21, what distance would 5'25 cwt. be conveyed for 5s. 3d ?

54.    If 28 loads of stone, of 15 cwt. each, build a wall ~0 feet long and 7 feet high ; how many loads of 19 cwt. will build one 323

feet long and 9 feet high ?    „    . D ,

55 Three men, working 8 hours a day. can reap 15 acres m 8 days; how'many men, working 12 hours a day, can reap 30 acres m 4

da?6 The rent of a farm containing 41'25 acres for 39 months was £89-375- what would be the extent of another farm whose rent for 33 months was £103'125, 4 acres of the latter being worth as

much 3iS 3 of the former ?    i •    1

57. If 7 men, or 9 women, can do a piece of work in -4 days, in

what time could 9 men and 7 women together do the same?

58 If 3 horses, or 5 oken, eat a ton of hay m days, in what time would 5 horses and 3 oxen together consume the same?


CHAPTER IV.

INTEREST, VULGAR AND DECIMAL FRACTIONS. INTEREST.

INTEREST is money charged upon a loan or debt, the charge being usually made at a given rate upon each <£100 for a year.    •

The Principal is the money lent.

The Interest of £100 for a year is called The Rate per Cent, per annum, and is usually marked p.c., or %.    .

The Amount is the principal plus the interest.

It is so called as being that which the debt amounts to; and, consequently, the amount to be repaid by the borrower.

Interest may be either Simple or Compound.

In Simple Interest usury is charged upon the original loan or debt only, and is payable at stated periods, generally yearly or half-yearly.

In Compound Interest the charge is made upon the sum of money owing at the beginning of each year ; that is, upon the prime debt with its interest for previous years. In Compound Interest, therefore, interest is charged upon interest.

The interest of any sum of money is in direct proportion to the money lent, the duration of the loan, and the rate of interest charged ; i.e., to the principal, time, and rate per cent. Hence, Questions on Interest are Questions on Proportion, and may all be solved by that rule.

The Principal and Pate are in inverse ratio to the Time.

Whenever, therefore, proportion is employed in order to find the Principal or Pate, the Time must be stated inversely; while, in order to find the Time, the Principal and Rate must be stated inversely, i.e., with the given rate or principal in the 1st term.

On the other hand, when the proportion exists between the Time and Interest, or the Time and Amount, the statements are direct; since these are in direct ratio to each other.

Whenever it is required to find the simple interest of any sum—

The Rate per cent., or the interest of £100, is the 3rd Term.

The given Principal and Time form the 2nd Term.

£100 and 1 yr. (the standards of reference )form the lsi Term.

For example, in finding the interest of £560 for 5 years at 6 p. c. per annum, we should state—

As £100 : £560 :1 £6 interest : required interest.

As 1 year : 5*years.

Then (£560 X 6 X 5) -f- (100 X 1) = £168. The required interest.

Since the first term is always composed of the numbers 100 and 1, a proportion statement is rendered unnecessary by the following rule.

To find the Simple Interest on any sum :—

Pule.—Multiply the principal by the rate per cent, and number cf years, and divide by 100.

Note.—The division by 100 is usually effected thus

Rule._Cut off the two last figures of the pounds and reduce them to

shillings, adding in the shillings of the sum of money to be divided. Cut off the two last figures and reduce to pence, adding in the pence. Again cut off the two last figures and reduce them to farthings, cutting off the two last figures as before.

Example.—Find the Simple Interest on £265 13s. 6|d. for THREE YEARS AT 12 PER CENT. PER ANNUM ; AND THAT ON £354 7s. llD. FOR FIVE YEARS AT 5 PER CENT. PER ANNUM.

Example I.

Example II.

£

s.

d.

£

s. d.

265

13

64

354

7 11

12

5

31S8

2

6

1771

19 7

3

5

95,64

7

6

88,59

17 11

20

20

12,87

11,97

12

12

10,50

11,75

4

4

2,00

3,00

Answer £95

12s. 10'|d.

Answer £88

11s.

In Interest there are four things to he considered, the Principal, Time, Rate per cent., and Interest; and when any three of these are known, the fourth may he found hy proportion. The above rule, however, applies to that case only where the amount of interest is required. In all other cases—namely, where the principal, time, or rate is required —it becomes necessary to employ proportion.

When the Interest is not stated, the amount of £100 is often a necessary term. This, however, can scarcely be regarded as a new term, since the Amount is the sum of Principal and Interest.

The principles of proportion being understood, no further explanation is necessary nevertheless, as a guide to beginners, we here give rules for dealing with the three cases where the principal, time, and rate are required.

Special Rules for Stating Questions on Interest.

Whenever the question takes the form of:—

I. —What sum of money lent at a certain rate per cent, will amount to so much in a given time ?

Rule I.—First find the amount of £100 for the given time at the given rate per cent, and then state:—

As the amount of ¿£100 the given amount ;; ¿£100 I Required sum.

Example I.—a sum of money at 5 per cent, amounted in 4f YEARS TO £618 15s.: FIND THE PRINCIPAL.

Interest on £100 for 4f years at 5 per cent. = 4f X 5 = £23 15s.

Hence, the amount of £100 for 4| years at 5 per cent. = £123 15s.

As, £123 15s I £618 15s. ; *. £100 : £500 Answer.

II. —At what rate per cent, will a certain sum amount to so much in a given time 1

Rule II.—Subtract the sum lent from the sum it amounted to in order to obtain the interest, and then state,—

As, The money lent : ¿£100 :: This interest : Ratep. c.

As, The given time : 1 year.

Note.—By comparison with the rule given to find the interest by means of a proportion statement, it will be seen that the first and second terms have here changed places.

Example IT.—at what rate per cent., simple interest, will £450 amount to £516 in 3 years 8 months ?

£516 - £450 = £66, Interest of £450 for 3 years 8 months.

As £450; £100 I *. £66 : 4 per cent. Answer.

3§ yrs. :    1 year

In this case bear in mind that the question “At what rate per cent." means “what is the interest of £100 for 1 year ?” and the whole interest forms the 3rd. term, the reasoning being, if this is the interest on the given sum, that on £100 will he less ; and, if this is the interest ‘or the given time, the interest for 1 year will be less.

III.—In what time will a certain sum lent at a given rate per cent, amount to so much ?

Rule III.Find the total interest, as before, by subtracting the money lent from the sum it amounts to; and then placing 1 year in the third term, state :—

As the rate per cent. I Total int. :    1 year I Req. time.

As The Given Principal I ¿£100.

Example III.—in what time will £150 amount to £187 2s. 6d.

AT 5£ PER CENT. ?

£187 2s. 6d.—£150 = £37 2s. 6d., total interest of £150.

As £5£ per cent : £37 2s. 6d. \ \ 1 year = years Answer,

£150    : £100

Note.—If the Interest be reauired, the third term is the rate per cent. ; if the Principal be required, it is £100; if Time, it is 1 year; and if Rate per cent, it is the whole interest.

In Compound Interest, the interest for each year is calculated separately, not upon the original loan or debt, as in Simple Interest, but upon the amount that has accumulated by adding to the original principal the interest for the previous years. That is to say, the second year’s interest is charged upon the amount due at the end of the first year, the third year’s interest upon the amount due at the end of the second year, and so on, the principal for each year being obtained by adding the last found interest to the last used principal. '

To find the total amount of Compound Interest, we then,—

1.    Add together the interests fen' the several years ; or,

2.    Find the last year's amount, and subtract the principal.

To find the Compound Interest on any sum :—

Rule.Find the interest for each year, separately, by multiplying the principal due at the beginning of the year by the rate per cent., and dividing by 100.

The rule given for stating, in order to find the principal or money lent at Simple Interest, is also applicable to Compound Interest ; but here the number of years is not made part of the statement, the interest of £100 for the given time being first calculated.

It is also worthy of notice that it is not necessary to make the time part of the statement in finding the rate per cent, by Rule II. Simple Interest, if the total interest be first divided by the number of years.

Example 1.—Find the Compound Interest on £250for thiee years at 5 per cent.

FIRST PLAN:—

£

250 first year’s principal.

5

12‘50 first year’s interest.

250

262 "50 first year’s amount, or second year’s principal. 5

13T250 second year’s interest 262 50

275'6250 second year’s amount, or third year’s principal. 5

13’781250 third year’s interest. 13‘1250    second year’s interest.

12*50    first year’s,interest.

£39-406250 total interest for the three years.

20

8,125000

12

1,500000

4

2,000000

Answer :—£39 "40625 = £39 8s. l^d Ans.

SECOND PLAN:—

Rule.—Find the amount for the given time, and ** ibti Ctoc the original principal.

£

250 first year’s principal.

5

12"50 first year’s interest.

250

262'50 first year’s amount. 5

13'1250 second year’s interest. 262 50

275'6250 second year’s amount. 5

13-781250 third year’s interest.

275-6250

289 "406250 third year’s amount.

250‘    subtract original principal.

Answer, £39"406250 = £39 8s. l|d., compound interest.

Example II.—Find the amount and compound interest of £624for two years at 124 per cent.

£

124 = 4 °f 100 = h 624 principal.

78 interest for first year.

624

124 = § of 100    702 second year’s principal.

87 15s. interest for second year.

£789 15s. amount at end of second year. 624 deduct original principal.

£165 15s total interest for two years. Answer :—Amount, £789 15^ Interest, £165 15s.

Example 111.—Find the compound interest on £201(8 for three years at £>§ per cent, per annum.

£

2048 first year’s principal.

12288 x 256

£125 "44 first year’s interest. 2048 2173’44 first year’s amount, or second year’s principal.

6*

1304064

27168

£133'1232 second year’s interest. 2173-44 2306 ‘5632 second year’s amount, or third year's principal.

H

138393792

2883204

! £141-276996 third year’s interest. £133‘1232    second year’s interest.

£125 "44    first year’s interest.

£399-840196

20

16,803920

12

9,647040

4

2,5SS160

Answer:—£399 16s. 9|d., and -58816q.

Example IV.— What sum will amount to £289 8s. l\d. in three years at 5 per cent. Compound Interest l

By proceeding as in Examples 1, 2, and 3, we find the interest of £100 for three years, at 5 per cent., to be £15 15s. 3d. So that, in the given time, £100 amounts to £115 15s. 3d. | and since the

\

principal, which will amount to £289 8s. l\d., must bear the same proportion to £100, which this latter amount does to £115 15s. 3d., we state—

As £115 15s. 3d. I £2S9 8s. l|d. : '. £100 : £250. Required sum.

By reference to Example 1, it will be seen that £250 is the sum which produces £39 8s. lid. in three years at 5 per cent., and hence amounts to (£250 + £39 8s. l|d.), or £289 8s. l|d.

Exercise XXV.—INTEREST.

I. Find the Simple Interest on—

1. £250 for 3 years at 5 per cent. 3. £275 for 5 years at 12 p. c.

5. £140 15s. for 2 years at 8 p. c.

7. £728 17s. 6d. for 5 years at 8^ p. c.

9. £115 18s. 4d. for 3£ years at 71 p. c.

11. £960 for 292 days at 6§ p. c.

13. £342 10s. for 3-| yrs atll^pc.

15. £927 for 1 year 219 days at H p. c. ‘


2. £360 for 2 years at 7^ p. c.

4.' £288 for S years at 12^ p. c.

6. £170 12s. for 3 years at 8^p. c.

8. £8412s.6d.for4yrs.atl6§p.c.

10. £72 13s. 4d for7i yrs at4p c.

12. £576 for 2 years 146 days at

Qâ pf Q'

14. £273 15s. for 1 yr. 8 mos. at 12 p. c.

16. £625 12s. 6d. for 3 years 73 days at 7 \ p. c.


II.    Find the Amount at Simple Interest of—

1.    £225 10s. for 2 years at 3 p c.    2.    £177 for 5 years at 8 p. c.

3.    £114 18s. 4d. for 2j years at    |    4.    £S4S for 3 years at 6£ p. c.

p. c.    j    6.    350guineasfor5yrs.at33p.    c.

5.    500guineasfor 7    yrs. at 4| p.c.    !    8.    224 guineas for 3-^ at    3f    p. c.

7.    140guineasfor2^yrs. at4f p. c.    i    10.    £84 19s. 2d. for 3 yrs. at 12p‘c.

9.    £76 13s. 4d. for 4 yrs. at 5 p. c.    12.    £100 8s 4d. for 6yrs. at8£ p    c.

11. £8817s 6d.for2iyrs at2£p.e.

III.    Find the Simple Interest on—

1.    £4000 for 4 years at 12£ per cent, per annum.

2.    £325 17s. 6d. for 8 years at 6f per cent, per annum.

3.    £5064 10s. 8d. for 3 years    at    5^ per cent, per annum.

4.    £418 18s. 4d.    for 2} years    at    7J per cent, per annum.

5.    £240 12s. 6d. for 3^ years at2| per cent, per annum.

6.    £412 13s. 6d. for 2i years at 2^ per cent, per annum.

7.    £516 14s. 2$d. for 3 years at 6§ per cent per annum.

8.    £8019 19s. lOd. for 4 years at 4f per cent, per annum.

9. £444 4s. 4d. for 7 years at 8| per cent, per annum.

10.    £1806 14s. 3|d. for 9 years at 4f per cent, per annum.

11.    £989 19s. lid. for 3 years 292 days at 4^ per cent, per annum.

12.    £1000 10s. lOd. for 5 years 146 days at 15 per cent, per annum.

IV. Find the Compound Interest on—

^ 1. £56 16s. 8d. for 2 years at 5 per cent, per annum, t s 2. £128 10s. for 3 years at 6 per cent, per annum.

/ 3. £625 for 2 years at 1\ per cent, per annum.

/ 4. £482 12s. 6d. for 3 years at 12^ per cent, per annum.

5.    £100 10s. for 3 years at 5 per cent, per annum.

6.    £355 13s. 4d. for 3 years at 25 per cent, per annum.

7.    £144 for 3 years at 16§ per cent, per annum.

8.    £606 12s. 6d. for 2 years at 37£ per cent, per annum.

Exercise XXVI.— INTEREST.

FROM TEACHERS’ EXAMINATION PAPERS.

1.    What sum of money lent at 5^ per cent., Simple Interest, will amount to £25 in 8 months ?

2.    What sum of money lent at 5^ per cent., Simple Interest, will amount to ¿125 in 18 months?

3.    What sum of money lent at 4£ per cent., Simple Interest, will amount to £2S in 10 months?

4.    In what time will £375 amount to £450 at 4£ per cent, per annum, Simple Interest ?

5.    What sum of money lent at 4J per cent, per annum, Simple Interest, will amount to £75 in 3 years 5 months ?

6.    What sum of money lent at 4^ per cent., Simple Interest, will amount to £100 in 5 years 7 months ?

7.    £356 at 4 per cent, amounted in a given time to £409 8s.: find the time.

8.    In what time will £650 amount to £798 13s. 9d. at 4£ per cent.,

Simple Interest ?

9.    Find the interest on £670 from 1st January, 1867, to 5th April, 1870, both days inclusive, at 7£per cent, per annum.

10.    £656 in 2£ years amounted to £783 2s. What was the rate of interest ?

11.    At what rate per cent., Simple Interest, will £460 amount to £544 14s. 4d. in 31 years ?

12.    In what time will £350 amount to £408 3s. 9d. at 4f per cent, per annum ; and what would be the amount for the same time at Compound Interest ?

>    13. What principal will amount to £501 4s. in 4 years at per

cent., Simple Interest ?

EXERCISE FROM EXAMINATION PAPERS. 157

14.    What principal will produce £113 15s. as interest in 5 years at per cent. ?

15.    £150 in 4^ years amounted to £187 2s. 6d. at Simple Interest. What was the rate per cent. ?

1(3. In what time will £690 amount to £783 3s. at 4J per cent, per annum ?

17.    At what rate per cent, per annum will £284 10s. amount to £337 16s. 10^d. in 3f years ?

18.    What will the interest on £256 amount to from the 1st of January to the 2Sth June, 1872, both days inclusive, at 4| per cent, per annum ?

19.    What sum lent at 6i per cent, per annum, Simple Interest, will amount to £650 at the end of 2 years 10 months ?

20.    What sum will amount to £769 10s. in 2^ years at 6 per cent, per annum, Simple Interest ?

21.    At what rate of interest will £266 13s. 4d. amount to £291 8s. 4d. in years, Simple Interest ?

22.    If in 9 months £60 gain 36s., in what time will £590 gain £25 11s. 4d. ?

23.    At what rate per cent., Simple Interest, will £375 amount to £529 7s. 6d. in 4^ years ?

24.    At what rate per cent., Simple Interest, will £468 6s. 8d. amount to £578 13s. 8jd. in 4| years ?

25.    In what time, will £637 10s. amount to £774 15s. 6d. at 3-f per cent, per annum, Simple Interest ?

26.    In what time will £987 12s., invested at 4j per cent., Simple Interest, amount to £1197 9s. 3-Jd. ?

27.    If I lend a friend £275 at 5f per cent., Simple Interest, and tell him to keep it till the principal and interest amounted to £375 2s. lid., how long will he have it ?

28.    In what time will a sum of money double itself at 5 per cent., Simple Interest ?

29.    Find the Simple Interest upon £625 from 1st December, 1874, to 8th April, 1876, both days inclusive, at 8^ per cent, per annum.

30.    Find the Simple Interest upon £575 from 21st November, 1874, to 1st May, 1876, both days inclusive,/at 9§ per cent, per annum.

31.    In what time will £91 13s. 4d. amount to £105 6s. 0£d. at 4^ per cent., Simple Interest ?

32.    A mining company pays a dividend of 15s. per share, which is at the rate of 6 per cent. Find the amount of a share.

33.    On what sum will £154 10s, be the Compound Interest for two years, at 6 per cent. ?

34.    At what rate, Simple Interest, will a sum of money amount, in 2 years, to the same as at 6 per cent., Compound Interest ?

35j Find the amount of £560 15s. in 4 years 7 months, at 4Jper cent, per annum, Simple Interest?

36. In what time will £325 lent at 6i per cent, per annum, produce the same amount of interest, as £550 lent at 5 per cent, produces in 4^ years ?

37. What sum lent at 4| per cent., Simple Interest, will amount to £400 at the end of 3 years 7 months ?

3S. At what rate per cent, must £7000 be invested to produce quarterly income of 125 guineas ?

39.    What principal, lent at 6| per cent, per annum, will produce a yearly income of £487 10s ?

40.    At what rate per cent, per annum, Simple Interest, will £450 amount to £516 in 3 years and 8 months ?

41.    What is the Compound Interest on £750 for 2 years, at 7£ per cent, per annum ?

42.    What is the Compound Interest on £650 for 2 years, at 8| per cent, per annum ?

43'. In what time will £325 12s. 6d. amount to £397 5s. 3d., at 4 per cent, per annum ?

44.    What principal, in 3f years, at 2\ per cent., will produce the same interest that £650 does in 4 years at 3 per cent per annum ?

45.    At what rate per cent, will £780 17s. 6d. amount to £937 Is. in 3 years 4 months ?

46.    In what time will pounds become guineas, at 6£ per cent, per annum, Simple Interest ?

47.    In what time will the interest on £375 13s. 4d. amount to £105 13s. l^d. at per cent., Simple Interest?

48.    At what rate per cent, per annum, Simple Interest, will pounds become guineas in 16 months ?

49.    What principal will amount to £334 16s. in 7 years, at 5 per cent, per annum?

50.    Find the Compound Interest on £325 for 3 years at 7 per cent.

51.    At what rate per cent, per annum, will 500 guineas amount to £567 in 5 years, Simple Interest ?

52.    What sum will produce £70 as interest in 5^ years at 4£ per cent, per annum, Simple Interest ?

53.    In what time will £325 amount to £360 10s. ll|d. at 3£ per cent, per annum.

54.    Find the amount of £625 10s. in 3^ years at 4| per cent., Simple Interest?

55.    What sum of money must be lent at 7 per cent., Simple Interest, that the interest on it at the end of 6£ years may amount to £500 ?

56.    Find the Compound Interest on £614 for 2 years at 2| per cent, per annum ?

57.    What principal, lent for 3 years at per cent.. Compound Interest, will amount to £793 7s. 2§d ?

58.    At what rate per cent, will £375, lent at Compound Interest for 4 years, produce as interest £174 0s. 9d. ?

59.    In what time will £1280, lent at per cent., Compound Interest, amount to £1590 2s. 9-gd ?

60.    At what rate per cent., Compound Interest, will £640 amount to £1025 3s. l^d. in 4 years ?

61.    What sum of money lent for 3 years at 5 per cent., Compound Interest, will amount in 3 years to £577 16s. 4|d ?

THE GREATEST COMMON MEASURE.

The Greatest Common Measure of numbers is the greatest number which will divide each and all of them. See Definitions, page 5.

To find the Greatest Common Measure of two numbers:—

Rule.Divide the greater number by the less. If tl^ere be a remainder, divide the last divisor by it, and continue on this plan to divide the last divisor by the new remainder until there is no remainder. The last divisor is the Greatest Common Measure.

Ex. I.—G. C. M. of 126 and 990.

126)990(7

882

Ex. II.—G. C. M. of 23 and 220.

23)220(9

207

13)23(1

13


10)13(1

10


108)126(1

108

I.    18)108(6    II.

108

3)10(3

9 1)3(3

3

1.—G. C. M. 18.    II.—G. C. M. 1, or unity.

In I, the G. c. M. might have been seen by inspection.. See Hints on Cancelling (11), pages 32 and 33.

In II, the G. c. M. is unity; hence, the numbers 23 and 220 are prime to each other.

Note.—Numbers are prime to each other when their g. c. m. is unity.

Whenever, at any part of the work, any divisor and the remainder it produces are both prime numbers, or are prime to each other, the rest of the work need not be done; for the G. c. M. is then unity Thus, in II., the first divisor and first remainder are both prime numbers ; hence the G. c. M. is unity.

Observe that the first divisor is divided by the first remainder; the second divisor by the second remainder ; the third divisor by the third remainder, and so on.

When one number is contained in another an exact number of times, the less number is the G. c. M. of both.

Note.—The g. c. m. of two or more numbers cannot be greater than the least of the numbers, otherwise it will not be contained in that number.

In finding the Greatest Common Measure, the process will frequently be shortened by placing in the quotient, at each division, the number which will give a product nearest to the numbers divided, placing the product above the dividend if it be greater.

Example.—Find the G. C. Measure of 68154 and 258494. Ordinary Method.    Abridged Method.

6S154)25S494(3

204462


54032;68154(1 54032


14122)54032(3

42366


272616

68154)258494(4

- 70610

14122)68154(5

- 14736

2456)14122(6

614)2456(4

2456

11666)14122(1    —

11666    G. C. Measure 614.

2456)11666(4

9S24 1842)2456(1

1842

Greatest Common Measure 614.    614)1842(3

1842

To find the G. C. M. of three or more numbers :—

Rule I.Find the G. C. M. of the two first numbers; then, by the same method, find the G. G. M. of this measure and the third number ; and continue to find the G. G. M. of the measure last obtained and the next number, until all the numbers have been dealt with.

The last obtained G. C. M. is the G. G. M. of all the numbers.

Example—Find the G. c. M. of 126, 990, and 27.

Here, the G. c. M. of 126 and 990 has already been found to be 18, so that we have only to get the G. c. M. of 18 and 27, which is seen to be 9. Answer—G. C. M. 9.

Rule II.■( 1). Divide each number by any measure common to all. (2). Divide the numbers thus obtained by any measure common to all. (3). Again, divide the numbers thus obtained by any measure common to all, and continue doing this until the numbers obtained have no common measure.

The product of the divisors will be the G. C. Measure.

Example I.—Find the G. C. M. of 1008, 1152, 1296, and 1728.

8)1008 1152 1296 1728 1st line

6)126

144

162

216

2nd line

3)21

24

27

36

3rd line

7

8

9

12

4th line.

8 X 6 X 3 = 144. The Greatest Common Measure.

Note.—Any other common measures might have been selected to divide by.

Thus, by Hints on Cancelling, pages 32 and 33, it is seen that, all the numbers in the 1st line are divisible by 2, 3, 4, 6, 8, 9, 12, &c., while those of the 2nd line are all divisible by 2, 3, 6, 9 and 18. Hence, either of these numbers might have been selected as divisors.

Had the numbers in the 1st and 2nd lines been divided by 12, the same result would have been obtained by two divisions ; thus— 12)1008 1152 1296 1728

12)84    96    108    144

7    8    9    12

12 X 12 = 144. The Greatest Common Divisor, or the G. C. M.

12

The G. C. Measure is principally required in reducing fractions to their lowest terms.

When we cannot tell by inspection whether or not two numbers are prime, Rule I. must be used ; as also, when we are in doubt as to whether or not any two numbers, obtained by division in applying Rule II., are prime to each other.

Example 2.-—Find the G. C. M. of 2233, 2552, and 2871.

11)2233 2552 2871

29)203    232    261    11 X 29 = 319, the G. C. M.

7    8    9

Here, we observe, by inspection, that the given numbers are divisible by 11 ; but it is more difficult to tell whether or not the numbers 203, 232 and 261 are prime to each other. We therefore apply Rule I. to these numbers and find that 29 is their G. C. M.: and 29 multiplied by 11 will give the G. C. M. of the original numbers.    t

Exercise XXVII.—1THE GREATEST COMMON MEASURE.

Find the Greatest Common Measure of—

1 384 & 360. 2. 735 & 945. 3. 693 & 1001. 4. 1056 & 1248.

5    1071    & 1197.    6.    539 & 833. 7. 819 & 1183. 8.    783 & 957.

9    3264    & 3904.    10.    783 & 1218.    11. 1365 & 1404.    12.    1326 & 1482.

13    7425    & 7500.    14.    1S63&2592.    15. 2412&2736.    16.    2880&6480.

17    936 & 1755.    18.    1134 & 1458.    19. 1463 & 2261.    20.    2380 & 2499.

21 9450 & 1215. 22. 6930 & 8064. 23. 16758 & 15219.

24 5985 & 6042. 25. 85, 119 & 153. 26. 90, 126 & 162.

27. 95, 133, & 171.    28. 144, 112, 128, & 80. 29. 45, 75, 90, & 135.

30. 84, 112, 98, & 42. 31. 448, 504, 512, & 528.

32 145, 174, 232, & 290.    33. 432, 405, 810, 864.

34 665, 1045, 1235, & 1615. 35. 539, 833, 1071, & 1197.

36 1715, 3430, & 3528. 37. 2352, 3136, & 504.

38! 6328, 7232, & 7119. 39. 24157, 38773, & 184933.

40. 12280, 14736 & 18420.    41. 105105 & 114345.

42. 19968 & 29568. 43. 19968 & 34944.    44. 29568 & 34944.

45 19968, 29568, & 34944.    46. 1989 & 4641.    47. 4080 & 4560.

48 13703 & 14861. 49. 5391 & 6589. 50. 30987 & 221291.

51 8585 & 9595.    52.    9717 & 11931.    5'3. 3080, 3136, & 3192.

54 3930 & 4323.    55.    1904, 2016v& 2128.    56. 6767,6834.

57‘. 945, 2016, & 3CfS7. 58. 8721, 7803, & 9180.

59. 30051 & 30618.    60. 5936 & 17136.

THE LEAST COMMON MULTIPLE.

The Least Common Multiple of two or more numbers is the least number into which each will divide without leaving a remainder. See Definitions, pages 5 and 6.

If numbers be prime to each other, their product is their L. C. M.

To find the L. C. Multiple of two numbers :—

Rule.—Divide one of the numbers by their G. C. M., and multiply the other by the result.

Thus, to find the L. C. M. of 57 and 95, divide either the 57 or the 95 by 19 (the G. C. M. of 57 and 95), and multiply the other by the quotient.

57 —f— 19 = 3, and 95 x 3 = 285.

Or, 95 -f- 19 = 5, and 57 x 5 = 285.

The product of two numbers divided by their G. C. M. will also give their L. C. Multiple.

Thus, 57 x 95 = 5415 ; and 5415 •— 19 =• 285, the L. C. M. of 57 and 95.

To find the L. C. Multiple of three or more numbers.

Rule.Get the L. C. M. of the two first numbers ; then find the L. C. M. of this multiple and the third number; then the L. C. M. of the multiple last obtained and the next number; and so on, until all the numbers have been used. The last found L. C. M. is the Least Common Multiple of allthe numbers.

Thus, to obtain the L. C. M. of 57, 95, and 60, we first find the L. C. hi. of 5/ and 95, as before j we have then only to find the L. C. M. of 285 and 60. It is seen, by inspection, that the G. C. Measure of these is 15; hence, their L. C. Multiple is obtained by dividing 60 by 15, and multiplying 285 by the result.

60-5-15 = 4; and 285 x 4 = 1140, the L. C. Multiple.

Axiom.—A multiple of any number is also a multiple of any factor of that number. Hence, in finding the L. C. M. of several numbers, we may disregard any number which is contained in one of the others, and find the L. C. M. of the others.

The following is a better method of finding the L. C. M of several numbers, and is that generally used :—

Rule—

I.    Disregard, or strike out, any number contained in one of the other numbers.

II.    Select a number (either a prime number or the square of a prime number) which will divide two or more of the numbers left; divide these by it, and bring doom all numbers which are not divisible.

III.    Treat the second and subsequent lines in the same vianner until the resulting numbers are all prime to each other.

IV.    The L. C. Multiple will be the product of these last numbers and the various divisors.

Note.—In any line, any number contained in another may be struck out.

Example /. — Find the L. C. M. of 6, 8, 12, 15, 20, 24, 25, and 60

Here, we disregard the 6, 8, and 12, since they are factors of 24 ; and the 15 and 20, which are factors of 60.

We have therefore to find the L. C. M. of 24, 25, and 60 ; and this will also be the L. C. M. of 6, 8, 12, 15, 20, 24, 25, and 60.

4)    24    25    60

5)    6    25    15

6    5    §

Then, 4 x 5 x 6 x 5 = 600, the L. C. Multiple. Answer, 600.

Example II.—Find the L. C. M. of 12, 14, 16, 18, 24, 30, 35, and 36. Here the 12 and IS may be disregarded, as being contained in 36.

2) 14

16

24

30

35

36

4) X

8

12

15

35

18

3)

$

15

35

18

*    35    6

Then, 2x4x3x 35 x6 = 5040, The L. C. Multiple.

^0XEi—Care must be taken to select either a prime number or the square of a prime number to divide by; otherwise, the Common Multiple found may not be the Least Common Multiple.

For example, if 8 be selected as the first divisor in obtaining the L. C. M. of any three even numbers, two only of which are divisible by 8, the least Common Multiple will not be obtained.

Example.—    8)    12    16    40

12    ^    5

Here 8 X 12 X    5    =    480.    But    240    is the L.    C.    M.

There is a much    shorter method of    finding the    L. C. M. in which

a number which contains most factors common to the given numbers is selected as the divisor, care being taken to divide any number capable of division by the number selected or by any of its factors.

Thus, the L. C. M. of 14, 16, 18, 20, 24, 30, 35, and 36 may be got in one line by selecting 12 as the divisor, taking care to divide all numbers capable of division by 12 or by any factor of 12.    Thus,

12) 14 16 18 20 24 30 35 36    '

\ 35    3

Then, 12 X 4 X 35 X 3 = 5040, the L. C. Multiple.

Note.—The Least Common Multiple of two or more numbers cannot be less than the largest number; otherwise it will not contain that number.

The L. C. M. is mostly required for bringing vulgar fractions to equivalent ones having the same denominator.

Example III.—Find the least number divisible by 8, 14, 16, 20, 21, 24, 30, 35, 40, 56, and 60.

56) $ \\ 16 5$ 21 V& % 35 40 56 60 2    §    %    $    15

The L. C. Multiple — 56 x2x 15 = 1680, Answer.

The L. C. M. of the denominators of several fractions is called their Least Common Denominator.

The L. C. Multiple is also required for the solution of such questions as the following :—

Sum 1.—Find the least size cask that will contain an exact number of each of the following measures : 1 gallon, 3 gallons, 4 gallons, 6 gallons, and 8 gallons.

Solution.—The L. C. M. of 1, 3, 4-, 6, and 8, is 2If.. Hence, a cash of 2It- gallons will contain each—Answer, 24 gallons.

Sum 2.—A machine has eight hammers, the first falls every second ; the second, once in two seconds ; the others in 3, 4. 5, 6, 7, and 8 seconds respectively. If they be observed to fall together, how many minutes must elapse before they will again strike altogether ?

Solution.—The L. C. M. of 1, 2, 3, 4, 5, 6, 7, 8, is 840. Hence it will he in 840 seconds, or in 14 minutes. Answer, 14 min.

Sum    —Find the least sum of money which contains an exact

number of guineas, sovereigns, half-sovereigns, crowns, half-crowns, sixpences, and pence.

Solution.—Here ws may disregard all hut the sovereign and guinea in finding the L. O. Multiple j since all the others are exactly contained in a sovereign.

The L. C. Multiple of a guinea and a sovereign, when expressed in shillings, is 21 times 20, or 420 shillings. Hence the sum of money required is £21. Answer, £21.

Exercise XXVIII.—LEAST COMMON MULTIPLE.

Find the Least Common Multiple of—


16.    12, 33, 77, 84, and 132

17.    28, 32, 56, 63, and 72

18.    132, and 176


19.    9, 13, 17, and 19

20.    18, 45, 75, 90, 120, and 10O

21.    12, 36, 48, 63, and 64

22.    17, 85, 187, 55, and 170


1.    2, 4, 8, 12, and 16

2.    3, 6, 9, 12, and 18

3.    5, 10, 20, 25, and 50

4.    5, 6, 12, 15, 24, and 30

5.    8, 12, 16, 24, and 32

6.    4, 7, 14 16, 21, and 28

7.    9, 16, 18, 36, and 72

8.    11, 12, 22, 33, and 44

9.    7, 8, 14, 16, and 32

10.    16, 18, 32, 36, and 96

11.    9, 36, 35, 42, and 63

12.    48, 64, 72, and 144


13.    9, 15, 45, 75, and 135

14.    48, 84, 128, and 168

15.    117, 171, and 711

1C IO QQ T'T C/f otwI 1 ‘3


^    .    1 O Ot7) I O j j    li A I

23.    13, 14, 21. 52, and 91

24.    11, 12, 18, 16, and 99


FRACTIONS.

NOTATION AND NUMERATION OF FRACTIONS.

Fractions are either Vulgar or Decimal.

Vulgar Fractions are expressed by placing one number above another with a straight line between them.

The number above the line is called the Numerator.

The number below the line is called the Denominator.

The Numerator shows the number of parts represented.

The Denominator names these parts, and shows their value.

In the fraction f, the denominator 8 shows what part of the whole each of the seven parts is;—namely, the eighth.

Again, f denotes five parts of such a size that six of them would make up the whole.

■A. denotes five parts, each a twelfth of the whole.

f denotes eight such parts as that nine would make the whole denotes twelve out of a hundred equal parts.

To denotes that of ten equal parts which the whole is to be divided into, nine of these parts are to be taken.

The Denominator thus shows into how many parts of an equal size the whole is divided.

The Numerator shows how many of such parts are under consideration.

Notation.—To notate eleven sixteenths, we simply write 11 over 16 and draw a line between them ; thus, '

Numeration.—To Numerate a Fraction :—

is read seventeen one hundredths ; fifteen thirty-eighths, and so on, th or ths being added to all denominators except 2 and 3.

The words expressing the numerator and denominator are frequently connected by a hyphen ; especially when each is a single word, as five-billionths, seven-tenths.

Decimal Fractions are also expressed by ordinary figures; but here the numerator only is written ; the denominator being represented by a dot called a decimal point, thus,

. K 5 .    - 5

O T7J> UO Toll '

A decimal fraction has 10, 100, 1000, or some other power of 10, as its denominator, and the dot is so placed as to show which power of 10 is the denominator.

The decimal point is placed either between or before the figures of the numerator; thus, 3T05, *75, *125.

The denominator is that power of 10 indicated by the number of figures following the decimal point; thus, in • 625, the denominator is the third power of 10, or 1000, because 3 figures follow the decimal point; in • 25 it is the second power, or 100; in *0125, the fourth power, or 10,000.

To find the denominator we may, therefore, consider the dot to represent 1, followed by as many noughts as there are figures after it.

To Numerate a Decimal.Read the significant part of the decimal for the numerator ; and for the denominator supply 1 and as many cyphers as there are figures following the decimal point. Thus, ’017 is seventeen thousandths; ’000101 is one hundred and one millionths.

To Notate a Decimal.—Place the numerator as though it were a whole number ; and to express the denominator point off by the following rule:—

Rule for Pointing. —Beginning at the units' figure of the numerator and proceeding to the left, say tenths, hundredths, thousandths, tens of thousandths, dec., until you come to the name required ; then place the decimal point before the figure you have arrived at.

Remember.—A decimal of 1 place is tenths ; of 2, hundredths; of 3, thousandths; of 4, ten thousandths, &c.

Note.—If there be not sufficient figures in the numerator to carry us to the required place, prefix noughts to the number.

Thus, to express 975 thousandths, we write 975 as the numerator, and, placing our pen on the 5, say tenths; on the 7, say hundredths ; on the 9, say thousandths; and then place the point before the 9, thus -975.

975 millionths would be expressed, ’000975.

VULGAR FRACTIONS.

Although the fraction means two-thirds, yet it is often treated as meaning two divided into three equal parts, it being frequently more convenient to regard a fraction as expressing the division of the numerator by the denominator.

Thus <£-§- may be considered as representing £2 divided into three equal quarts, instead of £1 divided into three equal parts and two of these parts taken. The result will be the same—viz., 13s. 4d.

The convenience arising from regarding a fraction as expressing the division of the numerator by the denominator has caused the straight line between the numerator and denominator to equal the sign of division ; so that § is often read 2 upon 3, or 2 divided by 3.

Fractions apparently unlike each other, and read differently, may be but different expressions for the same value. Thus, xey, ^, and T7¥ are of equal value, each representing \ ; for in each case the numerator is one-half of the denominator, showing that as many parts, compared with the number into which the whole is divided, are taken in each case.

Therefore, if anything be divided into a greater number of parts and a correspondingly greater number of these parts be taken, the sam,e value will result.

Hence, fractions whose numerators are the same parts of each other as their denominators are of each other, are equal in value ; thus,    f, and T8^- are all equal.

From this fact we obtain the following :—

Axiom.—If both the numerator and denominator of a fraction be either multiplied or divided by the same number, the value is unaltered.

22x4    8    9    9 -f- 3    3    75    3

Thus, - = - = —;    — =---= -;    & — = -

3    3x4    12    12    12-r 3    4    200    8

Upon this axiom depend the rules for the addition and subtraction of Vulgar Fractions; as also do the rules for cancelling.

It has been shown that concrete numbers cannot be added or subtracted unless of the same name; in like manner, fractions can be added or subtracted only when bearing the same name; or, in other words, when they have the same denominator.

When fractions have the same denominator, they may be added or subtracted by simply adding or subtracting their numerators : thus, + tt ~ H"?or !•     tV = tz> or 1-

If fractions have not the same denominator, they must be brought to equivalent ones having the same denominator before they can be added or subtracted.

To bring Fractions to equal ones of the same name :—

Rule._Find the L. C. M. of their denominators ; divide

this separately by each denominator, and multiply the numerator's respectively by the respective results.

Thus, to bring £, f and § to equivalent fractions of the same denomination :—

12,    16,    18,    15

h l l I =----

24 L. C. M. of the denominators.

Say, 2 into 24, 12; 12 x 1 = 12; 3 into 24 = 8, and 8 x 2 = 16 ; 4 into 24 = 6, and 6 x 3 — 18 ; 8 into 24 = 3, and 3 x 5 = 15. Hence the equivalent fractions are, -if, ff, ff, and if.

These fractions can now be added or subtracted by adding or subtracting the numerators, as shown above.

Hence, to Add or Subtract Fractions :—

Rule.Bring them, to the same denomination, and then add or subtract the numerators according to the signs, retaining the common denominator.

Thus, since § and f, are equal to if and if respectively ; their sum is —+-— or 31 twenty-fourths = ff; while their difference is

16 - 15 _ I

24    24'

To Multiply Fractions :—

Rule.—Multiply the numerators together and the denominators together.

Thus, |xf x \ — if Am.

To Divide one Fraction by another :—

Rule.—Invert the divisor and multiply; or multiply by the reciprocal of the divisor.    ,

Thus, f divided by f = f X f = ff Ans.

Note.—The reciprocal of a fraction is the fraction formed by inverting its terms ; that is, by placing its numerator as denominator, and its denominator as numerator. See Definitions, page 3.

Fractions connected by the signs x and s- are factors of one and the same quantity; while those connected by + and - form distinct quantities. Hence, when fractions are connected by various signs, the work of multi, plication and division must precede that of addition and subtraction.

If a thing be cut up into 8 equal parts,—

1.    8 of such parts will make the v:hole.

2.    More than 8 of such parts cannot be taken from one alone.

3.    More than 8 of such parts would be more than the whole. Again, if a thing be divided into any number of equal

parts :—

1.    This number of such parts will make the whole.

2.    More than this number cannot be got from one whole.

3.    More than this number of such parts would be more than one whole.

Hence,—1. When numerator and denominator are alike, the fraction equals unity:—Thus, §, ^, f, f-, f and are each equal to 1.

2.    In treating of a single thing, the numerator cannot exceed the denominator ; for we cannot take more than the whole of that thing.

3.    When the numerator exceeds the denominator, more than unity is expressed; thus, § = 1^; f = 1^ ; § = 2|.

Definitions:—

A Proper Fraction represents any part or parts less

21    ' .

than the whole ; as, f, f, rf. Hence, its numerator is

less than its denominator.

An Improper Fraction represents a whole or more than

71    .

a whole; as, f, f, A Hence, its numerator is not less than its denominator.

A Mixed Humber is a whole number and a fraction; 91

AS, *J-£, O-g-.

A Compound Fraction is a fraction of a fraction; as,

§ of £ of ; or, \ of £ X | of \ -r* -7-.

Tif

A Complex Fraction has a fraction in the numerator or denominator, or in both ;

3f_3f    4

As, — > —*> 2 + ^ of 15 Qr 4$ 4j 2h 5 7| 4J 5-f of 20    9f-16^ + 53!

A Simple Fraction has a simple number both as numerator and denominator; thus, f|*, %5, and are simple fractions.    -    •

A fraction is in its lowest terms when its numerator and denominator are prime to each other.

In dealing with fractions, it is generally necessary to reduce complex and compound fractions to simple fractions; and to express these last in their lowest terms.

A compound fraction is reduced to a simple fraction by multiplication.

Thus, f of 4 °f ■§■ =


2x6x5

----- 60n

3x7x8 Tes — tí-


Thus, 13.


A fraction is reduced to its lowest terms by dividing both numerator and denominator by their greatest common measure. 12-f-3    „

21—3

To Reduce a Complex Fraction to a Simple Fraction :— Rule.Reduce its numerator and its denominator each to a simple fraction, and then divide the former by the latter.

Thus,


f °f 4 X f-U-k + i


2 0 v 6 3    _

6 3    * oT -


Since compound fractions are reduced to simple ones by multiplication, the word “of ” may be regarded as being equivalent to the sign x. It denotes, however, a closer relation than that sign, so that when several fractions are connected by “of,” X, arid -f-, theparts connected byof” must be worked before the other parts.

Thus, i -r | of T%- X 3 of f X § = f i X * X $ = £ X | X

Work indicated by the signs X and -f- must be done in order as these signs occur.

To Reduce one thing to the Fraction of another:—

Rule.—Bring both to the same name ; divide the one by the other, and express the result as a fraction in its lowest term.

In other words, the quantity to he reduced must be made the numerator of a fraction and the other quantity the denominator. Both are then reduced to the same name, and the numbers expressing them are considered as abstract.

Example I.—Express 11s. 8d. as the fraction of 1^ guineas.

11s. 8d.    140d.    140    10

1£ guin.    378d.    378    27

Example II.—Express 13 cwt. 3 qrs. as the fraction of 1 ton.

13 cwt. 3qrs.    1540 lbs.    1540    11

1 ton.    2240 lbs.    2240    16    ’

When it is required to reduce one quantity to the decimal of another, the same plan is followed; but the vulgar fraction is then reduced to a decimal, as explained hereafter.

When fractions are connected by the sign of multiplication, the numerator of one may be cancelled with the denominator of another.

Directions to be observed in working fractions.

1.    Clear away all brackets by doing the work, the inner ones first.

2.    Simplify all compound and complex terms.

3.    Work the parts connected by the signs x and -f- in order as the signs occur.

f Work the parts connected by + and - last.

5. Reduce the result to its simplest form.

Example:—

■{ (t ~ if ) + i 1" ~ « X it + Iff -r2 f x | of

< (5-2)-(i - m-


2


■y


9


1. Clear away the inner brackets by doing the work within.

i 2 + $ y - a x 2 + u -f- i x t of i---{ ^ -Tv }•.

2.    Take out of outer brackets by doing the work within.

2

if " n ^    + ff 2f- f X f of f - — - ff

*

2.

3.    Simplify f of f, and —

4.4    _ _2_ x -fi__h 4J- — -l    2 v 2    _ 3    _ .1 a

li 2i 2 2 1 V8 • i X H 2    T2"-

4.    Do the work of multiplication and division as the signs occur.

This gives    +

5.    It is frequently better to add the fractions preceded by the sign +, and then to subtract the sum of those preceded by -.

Thus, if + U = m, and i + # + ff - W-832    163    9984 - 11410    1426    713

’ 210    36    2520    2520    1260    AnSWer

Note.—Here we cannot take 11410 from 9984, so we take the positive numerator 9984 from the negative numerator 11410, and thus obtain the negative answer - -j^30. Answer - Tuu2 .

Note.—This answer shows how much must be added to before y%3 can be taken from it. Hence it is a negative fraction. See page 35.    ,

Remark.—The fractions within the brackets might have been taken out without performing the work indicated by the signs connecting them.

The rule is, “If the sign plus precede a bracket, take the various quantities oat as they occur, preserving the signs which precede them ;

if the sign minus precede the bracket, alter the sign of each quantity,_

changing + into -, and - into +.”

DECIMAL FRACTIONS.

It has been shown that multiplying or dividing both the numerator and the denominator of a fraction by the same number does not affect the value.

Now, adding one, two, three noughts to a decimal, has the effect of increasing both the numerator and the denominator tenfold, a hundredfold, a thousandfold.

Thus, •5 is 5 parts out of 10; '50 is 50 parts out of 100 ; *500 is 500

parts out of 1000 ; and '5000 is 5000 parts out of 10000.

Each of these is equal to or ‘5

Again, omitting a nought at the end of a decimal decreases both numerator and denominator tenfold.

Hence, 250 and ’25 are equal in value—both expressing J.

It is thus shown that

1.    Adding noughts to a decimal does not alter its value.

2.    Omitting noughts at the end of a decimal does not afect its value.

In Yulgar Fractions, it was shown to be necessary to reduce fractions to the same denomination before they could be added or subtracted. Decimals may be reduced to the same denomination, at once, by adding cyphers to make the number of decimal places equal.

Thus, "5, ’25, and *0020 may at once be expressed as thousandths. They equal '500, '250, and ‘002.

Again, 3'5, 17, and 18-0125 = 3‘5000, 17'0000, and 18-0125.

A decimal may be at once multiplied by 10, 100, 1000, dec., by removing the decimal point one, two, three, dec., places to the right; and divided by 10, 100, 1000, dec., by removing the point one, two, three, dec., places to the left.

Thus, 3775 x 10 = 37-75; 337515 x 100 = 3375-15.

And, 62 5 -MO = 6 25 ; 62 5 -r 1000 = -0625.

Again, 5 X 10 = 5 ; "56 X 100 == 56 ; and *5 —j— 10 = 05.

From these examples, it is seen,

1.    That multiplying a decimal of 1, 2, or 3 places by the 1st, 2nd, or 3rd power of 10, produces a whole number.

2.    That placing a cypher between the point and the figures of the decimal has the effect of dividing by 10 ; placing two cyphers, of dividing by 100, Ac., dec.

To Add or Subtract Decimals :—

Rule.—1. Make the number of decimal places of the addends equal by adding noughts. 2. Arrange the numbers as in Simple Addition and Subtraction. 3. Add or subtract, as in Simple Addition or Subtraction. 4. Place a decimal point in the answer under the other decimal points.

Note.—Care must be taken to keep the decimal points directly under one another, so that the tenths shall be under the tenths, hundredths under hundredths, &e. If this be duly attended to, the first part of the above rule will become unnecessary.

Example.—Add together 30'6025, ‘00375, 6-0265, 300, -2, -175, and 40-0125.

First Plan. 30-60250


Second Plan. 30-6025


•00375 6 02650


•00375

6-0265


300-00000


•20000

•17500


40 01250


300-

•2

•175

40-0125


377-02025 Answer.


377"02025 Answer.

13


Examples.Take *6005 from 2'5; and

"225 from "65.

First Plan.

Second Plan. .

2-5000

•65

•6005

•225

1-8995

•425

Note.—It will be observed that the second plan in each of the above examples differs from the first, merely in neglecting to supply the cyphers necessary to bring the decimals to the same denomination.

To Multiply Decimals:—

Rule.Multiply as in whole numbers, and then mark off

as many places for the decimal as

there are figures in the

decimal parts of the multiplicand and multiplier.

Example.Multiply 17 "972 by 10"05 ;

and 8'225 by ‘0046.

I.

II.

17972

8225

1005

46

89860

49350

1797200

32900

180'61860 Answer.    ‘0378350 Answer.

In II., the product contained six figures only, and since seven figures had to be marked off, the seventh place is made by prefixing a cypher. In both answers, the final 0 might be omitted after pointing.

Rule for the Division of Decimals :—

1.    Make the number of decimal places in divisor and dividend equal, by adding noughts to that which has the less number of decimal places ; and divide as in whole numbers.

Note.—1. If the divisor is then contained in the dividend an exact number of times, the answer is a whole number.

2.    If there be a remainder, place a decimal point in the quotient, and add a cypher to the remainder ; if the divisor is not contained in this, place a nought in the quotient and add another cypher to the remainder. Continue to add cyphers and carry on the division until no remainder is left, or until some remainder occurs ivhich has previously been obtained.

3.    When the divisor is not contained in the dividend, place a decimal point in the quotient at once, and proceed as described in 2. In this case, the answer will be a decimal without a whole number.

Example I. Divide 69 ‘96 by 1 '749,

Here adding a nought to the decimal in the dividend to make the number of decimal places equal.

1749)69960(40 Answer, a whole number.

6996


0

Example II. Divide 1179‘0084 by 8"6 Here add 3 noughts to the decimal in the Divisor.

86000)11790084(137-094 Answer. 86000


319008

258000


610084

602000


808400

774000


344000

344000


Example III. 78)690(-8846153, &c. 624


660

624

360

312


Divide 6-9 by 7'8

Answer, a decimal only. Where we find a remainder 66 occurring which has been obtained previously.


480

468

120

78


Ans. *8846153846153846153, &c. Expressed thus, -8846153.


420

390

300

234

66


fi^r°p?8^iUtheiidivisi0n hA continued with this remainder 66, the ngures 8461.,3 will again and again occur in the quotient. Such an

andT821S & ClrCulatinS or rePeating decimal. See pages 181


To Reduce a Yulgar Fraction to a Decimal:—

Rule.—Divide the numerator by the denominator, as described in division of decimals.

8)5-000

Thus, | =-= -625

•625

The 'prime factors of a number are the only prime factors of any power of that number.

Thus, 3 is the only prime factor of 9; hence 3 is the sole prime factor of 81, 729, or any power of 9.

Again, 5 and 2 are the only prime factors of 10; hence 5 and 2 are the only prime factors of any power of 10, as 100, 1000, 10000, &c.

One number is contained an exact number of times in another, only when the former contains no prime factor which is not also a factor of the latter.

Thus, 6 is not contained an exact number of times in 64, because it contains the factor 3, which is not a factor of 64.

Hence, if a divisor have any prime factor which is not also a factor of the dividend, the divisor is not exactly contained in the dividend ; hence, there must be a remainder.

From the above, it will be seen that—

In division of decimals, an exact answer will be obtained only when all prime factors of the divisor, other than 2 and 5, are also factors of the dividendl.

2 and 5 are necessarily made factors of the dividend by the addition of cyphers, which has the effect of multiplying by 10, 100, &c.

And, since a fraction in its lowest terms has had all factors common to its numerator and denominator expunged, such a fraction can be reduced to a terminating decimal only when its denominator is composed solely of the factors 2 and 5.

Thus, £, §, I, 4, y, -y, y. can all be exactly expressed as finite decimals ; for 2 4, S, 5, 10, 16, 32 may be reduced to unity by division by 2 and 5.

If a fraction, when in its lowest terms, have a denominator which • contains other prime factors than 2 and 5, the division of the numerator by the denominator (as described in division of decimals) will never terminate, but will produce a quotient which, however far it may be carried out, will still leave a remainder.

The quotient, however, will consist of a set of figures the whole or a certain part of which will continually recur in a regular order ; whilst the number of decimal places of which it may possibly consist before the end of this recurring period is reached, is limited to the number of units composing the denominator. For, since the remainders which occur in working division must each be less than the divisor, it is clear that the number of different remainders that can possibly occur must be less than the number of units in the divisor ; and whenever a remainder is obtained that has occurred at a previous stage of the work, it is plain that again adding cyphers and continuing the division will produce the same result in the quotient and leave tins remainder again ; that is, the paid of the quotient last obtained will again and again repeat and leave the same remainder so long as we may continue to divide.

In illustration, let it be required to reduce 4 to a decimal. In dividing by 7 the various remainders which may occur are 1, 2, 3, 4, 5, and 6, and when each of these has occurred once, the division, if continued, will reproduce one of them, causing the quotient to repeat itself also.

Thus, ¿ = 7 \_jf_

• 714285714285 and 5 over = • 714285

In this case the whole of the quotient repeats, and the decimal is called a Pure Circulator.

§ = 3)2-0000    .

-or 6

'6666, &c.


= -5; f = -6; % = -7; f = “8.

In these examples one figure continually repeats. The decinlal is then called a Pure Repeater.

i = 6)5-00000    .    12)7-000000    .

-or -83; XV =    - or -583.

•83333 &c.    -583333 &c.

In these examples the last figure, 3, only recurs ; for, dividing 5-00, &c., by 6, we obtain 8 and a remainder 4, which, when augmented by a cypher and divided by 6, gives 3 and 4 over, which, augmented and divided, will repeatedly give 3 and 4 over.

When the last part only repeats or circulates, the decimal is called a Mixed Repeater or Circulator.

We now proceed to show what fractions will (when in their lowest terms) produce pure, and which will produce mixed, circulating decimals.

In doing so, we shall use the terms repeating and circulating as synonymous, since no advantage is to be gained by preserving the distinction.

Whenever the denominator is a prime number, or a composite number not containing 2 or 5 as a factor, the decimal will be a pure repeating or circulating decimal. When, on the other hand, 2 or 5 enters as a fraction of the denominator in connection with some other factor, the result will be a mixed circulator. The number of figures which will compose the circulating period will, in each case, depend upon the least number of 9’s which are required to indicate a number divisible by the prime factors (other than 2 and 5) composing the denominator. Thus, because 3 and 9 are contained in one 9, any proper fraction having 3 or 9 for its denominator will repeat

on the first figure ; thus, f = *2, J = '3. Again, because 11 is contained in 99 any fraction with 11 as a denominator will repeat after the second figure. Again, because 7, 13 and 39 are measures of 999999 (that is, of six nines), any proper fraction with 7, 13, or 39 as a denominator will produce a pure circulating decimal with six figures

in the repeating period ; thus, % = ‘"571428 ;    = ‘384615. Again,

37 is contained in 999 (i.e., in three 9’s), hence, will yield a pure

circulating decimal of three places, = ‘432 ; -J-f = ‘459.

Note.—All prime numbers, except 2 and 5, are contained in some number composed entirely of 9’s ; hence it is that all vulgar fractions may be expressed as repeating decimals, if they cannot be expressed as finite decimals.

Vulgar fractions, in their lowest terms, produce :—

1.    Finite decimals when the only prime factors of the denominator are 2 and 5.

2.    Pure repeating or pure circulating decimals when neither 2 nor 5 is a factor of the denominator.

3.    Mixed repeaters or mixed circulating decimals when the denominator has other prime factors in addition to 2 or 5.

In reducing fractions, expressed in their lowest terms, to decimals :

X._The number of decimal places in a finite decimal coincides with

the number of cyphers in the least power of 10 which contains the denominator an exact number of times.

II.—The number of decimal places in a pure repeating or circulating decimal coincides with the number of 9’s which express the least number (composed solely of the figure 9) which contains the denominator an exact number of times.

In like manner, the number of figures in the circulating and noncirculating parts of a mixed circulator depends upon the prime factors of the denominator.

Exercise XXIX.—VULGAR FRACTIONS.

I.—Reduce the following compound Fractions to simple ones:—

I.    § of | of l 2. | of i of f.    3. | of | of |.    4. | of f of f.

5. 4 °f II °f o    6. g of T\ of of ■j'Sf. 7. TT II II °f H•

8* 9 tt    L 9- tI I! §!• 10.    of of U of 1¿.

11. fof§offoff. 12. ^ofT\ofifofT«éV IS.HoílO^oí^olUh 14. 31 of*of 14ofiff. 15. 8T\j°fI°f21 of8f. 16. foff of f of fof*. 17. i of § of | of i of f of $ of l 18. if of 8i of 15f of fgg of f. 19. & of U of ff of f§ of iff. 20. f of * of * of ff of ff of 21.

II. —Reduce the following Fractions to their lowest terms:—

1. if. 2. if. 3. ff. 4. fif. 5. II§. 6. ffg. 7. fff.

8. TBTjV 9. fiif. 10. ffff. 11. ifUir- 12- iffi" 18- IliTlI-

14. Hff. 15. ffff. 16. »    17- Ill- 18. ffff.

19. t8t°*Vt- 20. ffilif. 21. iHf|. 22. ff§f. 23. ffgfffif-

24. fggf. 25.    26. iff. 27.    28. ffff.

29. fffff. 30. ffffff. 31. ffff. 32. tWVW- 33. fffff.

34. mh 35. ffff. 36. ffffff«. 37. f?f. 38. fffff.

39. fffff. 40. ffffff.

III.—Reduce the following mixed numbers to improper Fractions:—

1. 3f. 2. 15f. 3. 174- 4. 201 g. 5. 101fg. 6. 75f.

7. 100f. 8. 33 J. 9. 5ff. 10. S9f 11. 98£.    12. 23|g.

IV.—Reduce the following improper Fractions to whole or mixed numbers:—    -

1.    2. W-


3.


5J7J5 7 9 *


4.


5.


£    4 .»i    4

99    •


7.


5 6 3 12 7*


515 Q 3147 115'    5 17*


10.


7 8 7 4 3 13 1*


11.


5 5 4 4 19 5*


12.


V.—Reduce the following to the lowest common denomination ; then add the numerators together, and reduce the results to mixed numbers:—

I.

4.'

7.

10.

13.

16.

19.


7 5    5 7 IQ & 31.

9 9» 99» 11» ^ u9*


2 9» 3 1» 3 7»


& iff.


b'b & l 2. f, I, I, b b & H- 5.

4    1_5_    1 9 JL± Rx, _7_

19» 3 8» TU’ 5 7» ^ 10*

3 _5_ _8_ AA Rr 4-X

7» 14» 21» 42»    ^6*

1JL 5_8    05 5 .Or 1 1 A

2 3» 6 9» ^46» 00 A 6

3.

b

I, b i

T2

4

1 3 1 5»

2 2

2 5»

6.

7 4 2 5 11? 13» 20»

, 8.

1 1 16»

§,

f, i & 4.

Q 7. Z -J-9» 5» 12’

7 __7_ 2 4» 00 3 6*

11.

3

14»

_5__7_ Rr

12» 24»

1 9 4 2’

12 -5- a

11» 9»

1, b & è-

14.

5

3 2»

3_9 2_5 ¿r 04» 32» ^

3    7

4    0*

1 K 4 5 6 ±tJa 5» 0» 7»

TF» & AT'

IS 18 XL    IS    19 5 9 3 5 fir

TT» T9» 00 2 3*    -LO’ 24» 7 2» 36»

3 3    37    43 Rr 4 7    9]    15    15    15

43» TT» TT» 00 57*    29» 3 1» 4 1»


17.

20.


Exercise XXX.—VULGAR FRACTIONS.

Addition and Subtraction.

I.—Add together-

1    5    _7_ ü 1 1    ' O

12» 12) 12") 12» 1 12*

A» A» 3» 4:» 8» 8*


1.


a


q 1 i a a a

°' 3» 4» 3» 4» 6*

113 7    /C _3_ X8. Ai 2-8 m

2» 4» 7» 8’    u' 19» 19» 5 7» 5 7’ 57'


n la 5 il 3 17    Q A _5 _7_ _9_ _5_ XA    Q 2 A A A_o li A®

/• 6» 4» TA AT’ 8» AT' /    9» lA> 16» 16» 18» AT'    3» 7» 9» 2 1» AT» 63'


in 5    17    11    19 _8_ AA 11    2-    A %. A _2_ O IO 1 a A X AA

IO. g-j AT» Tfí’ AT» 2T» 7 2' Ai' 3» 5» 7» 9» 11»    L¿á% 2» 4» 6» 5» 10» 12'

13. f» y, fj) tÍ» *' TA» TA> AT» "e^T» AT' 1®* A> "To» TT> As» 5&. 16. 5L3i,4L5i,6i- 17.2¿,41f,20*,5*. 18. 3*, 15*, 30¿, 100§. 19. 50y, 30§, 14*, 3*.    20. *, *, ff, *, 1*.

91 1 5 Q 5 1 9 IOTA 1 flg-^1.    ^68» 85» AV/34» 17 0*


II.—Find the difference between—

1. * and *. 2. * and *. 3. f and *. 4. * and *. 5. f and f.

6. * and *.    7. 4-f and J. 8. 3^ and 2£.    9. 14f| and 5*.

10. 8* and *.    11. 9* and 9f.    12. 17* and 15£.

13. £ of * and f of *.    14. g of * and £ of 1*.    15. 6* and Iff

16. x* and T*. 17. 7* and 9£. 18. 12 and *. 19. 5* and *. 20. * and 100.

III. —Find the remainders (1) by successive subtractions, and (2) by subtracting the sum of the fractions preceded by the sign - , and show that the results coincide.

1    1    3 — 1    / 9    23-1    — 1    A£__3__A£ — 1

1. g _ g £•    2T    4 3*    10    10    32 8-

A 03 — 5 — 1_7    K    AA — 1 —    A__A_    ~6    11 — %■ \--A

4- eT 10    32'    "•    21) 4    5    10"    U-2S 7    4    50

17    99 _ 29 - 1 of 1    ft    JA _ Jj_ _ A3 _ _4_ O    AA — AJL — AA —

/• Hi) HO 8 01 4'    7 5 2 5 7 5 151 -''1    7 2    3 0    4 8    Ti'

10.    5f-3f-2f    11. 100 - 40f-29f. -12. 10-5|-3f-li.

IV. —Simplify the following by subtracting the sum of the negative from the sum of the positive.

1    + _ 5 _l 4 _*£ 4- A _ A    -9 A — JL-Lfi — A 4. £ _ S A- AA.

1 * "7 V ‘ "5"    5 *    5 9^4 5~9 4 ‘    30*

- 3. -H-I-I+I+*. 4-

5. 3±-4k + 3k-2i + lU-^6. A_|_5_li-AA + 7.

„ 7. 3-*-5£-24 + 4f-§. 8. 80H-3^-*-27t23-^.

9. 18^-21+11-4^-11-1.    10. 6f-5A + 4y-3f + 2|-li.

11.    llii-10H + 9T%-8|-6f+7|. 12. If-7g-6-? + 5f+4±+3f.

13. 106 - 3£ - 75Th - 13+f.    14. 64|-80| + 15H-12^ + 30f

15. 14A+18|-15|-17f+l.    16. 72 + .|_§_f-AA-AA

17. 3f-7^-5*-8f4 + 20.    18. 12i-|-f-|-§-i-8.

19. 137-51f-24-71-43. 20. 35-304-2^-131 + 12.

Exercise XXXI.—VULGAR FRACTIONS. Multiplication and Division.

Multiply—

1- * by f 2. TST by if 3. £ by T\.    4. f by f 5. £ by 1TV-

6. £ of f by § of 4|.    7. f of 4 of §£ by 4 of fi of 5f ,

8. 5f of i by 4 of 4 of 6.    9. -£ of 1| of £ by 1£ of | of 14.

10.    4 of of T5* of § of 44 by i of of 54 of fT of 28.

11. —Divide—

1. g by £-.    2. Ü by g^V 3. T7T of 3£ by of 4V*

4. 2£ by 4fg. 5. ft of 3£ of 7£ by 144 of 6£ of £.

6. § + £xi + fxii-+§xfxf + i-7-U.

III.—Simplify—

Te^TT °f A'x    !•

9    3_!_iv ^ A V A -1- 7 y 5 JL 3 y 1 1 1 I I 02

7,9XTT*7X7,9X8,TX 1 2 O 2 6 ’ ^3*

3.    § of £ of 4 of f of f of £ of of 4t of 44 °f 12.

4.    24 of f4 + -g-g"^ tV ~ 24 -r 1M + 4 "V 34 -r J-4^.

5.    § of f - § of f+ § of 4 - § of 1 + i of 4k - TVo.

6.    4of 4of£+§off <A-f--!-ri4xg-4-f5.

7.    §+6' + f+fI~f-54-5 + f + T^> + 1T)_ 4 7 7 2 Ó ~ TTO-

3. (2J-li)-r(3-§of 7 + 4 of 34) X (4 of | £ of

(34 of 4 -14) - 6§ x 44 of 4 + 5f x 4^ + |f.

10.    (24 + s -£) 4- (4 Of 44) + (i of 4 Of 44 of 124-14)

11.    4 (S+!-*)*(*-21 Of i) J-4- 4 (74-3§) + (*ofM-5-«ofif)}..

12.    (44 - 2*) -V- 4 (1 + i of 8) - (I of 24) K (f of fa+ 4).

Exercise XXXII.— VULGAR FRACTIONS. FROM TEACHERS’ EXAMINATION PAPERS.

l-L    24 _ 1-5.    81

1.    By what fraction must of 1 4    A —be divided in

.    . AT2    ? + ]6 fa

order to give 4f -os quotient ?

2.    From the sum of 3f and 5-fa, subtract the difference between 2|

91

and 3 A-; and divide the remainder by 6 A of -3-

•■5

3.    Multiply the sum of 44 and 5fa by their difference ; and divide

24

the product by fa °f ~io Ta*

4.    Find the product of the sum and difference of 4 and 24 ; and divide it by 4^5 of 3 of f -r- 4.

5.    Multiply the sum of § and 34 by their difference, and divide the product by 24 of 6 of


6. To the difference between 2| and 3t9<j add the sum of 44 and

33

4-| ; and divide the total by 2f of

7.    From the sum of 2| and -J, subtract the difference between 1| and 3t3o ; and divide the remainder by the product of 52 and

m

u*

8.    Multiply 3f by 4g ; subtract § from the product, and divide the difference by of 8.

9.    Divide 3b by 4J ; from the quotient subtract f, and multiply the remainder by ^7 of g .

10.    Add together 3£, 5§, i of £; subtract from the sum and divide the difference by yy .

11.    Add together 6f, 4§, and ; subtract 3J from their sum, and divide the difference by yy ,

A9

11 _ A of % + &

12.    Reduce q4 0 ,z ,—yd — (-¡% of 2-||) to its simplest form.

¿9 \o '    12/

Exercise XXXIII.—REDUCTION OF VULGAR FRACTIONS.

I.    Reduce to the fraction of ¿£1—

I.    16s. 8d.    2.    18s.    4d.    3.    17s.    6d.    4.    6s. 3d. 5. 9s. 4d.

6. 14s. 8d.    7.    13s.    9d.    8.    19s.    2d.    9.    17s. 4|d. 10. 13s. 5fd.

II.    7s. HJd. 12. 19s. 7|d.    13. 17s. 4|d.

II.    Reduce to the fraction of an acre—

1. 2ro. 16po. 2. 3rds 14po. 3. lro 5po. 7T%yds. 4 3ro. 11 yds. 5. 12 yds. 8sq. ft. 6. 35 sq. yds. 7. lro. 20 po. 15isq. yds.

8. 2 ro. 30 po. 9. 14 yds. 3 ft. 100 sq. in. 10. 15 per. 9 sq. yds.

II.    18per. 8 sq. ft. 12. lro. 11 per. 15sq. yds.

III.    Express as the fraction of 1 cwt.—

1. 3qrs. 15 lbs. 2.    2qrs. 4    lbs. 8 ozs.    3. 1 qr. 14 lbs. 14 ozs.

4. 3qrs. 18 lbs. 5.    1 qr.    20    lbs. 15 ozs.    6. 3qrs. 26 lbs. 14 ozs.

2. 2qrs. 6shts. asthefr. ofaream. 4. 30 c. ft. as the fr. of 8 c. yds. 6. 10 dwt. 5 grs. as thefr. of 5 ozs. 8. 5 gals. 2 qts. as the fr. of 1 pipe. i0. 1 pt. 3 gills as the fr of 2 qts. 12. 1 oz. Av. as thefr. of 1 lb Troy.


IV.    Reduce—

1. 5yds. 2ft. to thefr. of 1 mile.

3. 17s. 6d. to the fr. of a guinea.

5. 5G0 ac. to the fr. of a sq. mile.

7- 3dys.20hrs. to thefr.of 1 wk.

9. 2stones 121hs.tothefr.of 1 cwt. 11. 8 sq. ft. to the fr. of 3 sq. yds 2 yrs. 219 days.

V.    Express—

1. 16 gals, as the fr. of 1 barrel.

3. £2 16s. as the fr. of 3 guineas.

5. 1 ell 2§ na. as the fr. of 1 yd.

7. 2scr. 12grs as thefr. of 2ozs

9. 15dwt 20 grs. as thefr. of lib.

11. Gdrs. 21 sc. as the fr. of 1 oz.


2. 12dwt 20grs to thefr. of alb.

4. 5 cwt 3 qrs to the fr of 1 tori.

6. 2 qts. 1 pt. to the fr. of a gal.

8. 7 fur. 80 yds. to the fr. of 1 mile. 10 3 bus 2 qts. to the fr. of 1 load. 5 ft. 12. 73 days to the fr. of


Exercise XXXIV.—DECIMAL FRACTIONS.

I.    Express as Decimals—

I.    57 hundreths. 2. 17 ten thousandths.

3. 3008 hundreths. 4. 61 ten thousandths.

5. 9 thousandths. 6. 127 millionths. 7. 7 millionths.

8. 3 thousandths. 9. 51 millionths. 10. 36 thousandths.

II.    §.    12. ^0.    13.    14. 57 nine-hundredths.

15. 116 ten-millionths. 16. jg. 17. fff. 18. f.

19. 205 tenths. 20. 5 hundreds of thousands.

II.    Express as Vulgar Fractions in their lowest terms— 1. -085. 2. 3-075. 2. 2 06. 4. -0625.

5. -0588235294117647.    6. -05263157894736842L

7. T42S57.    8. -285714.    9. '428571. 10. -571428.

II.    -714285.    12. -857142    13. -583.    14. -4083. _    _

15. -384618.    16. -432. 17. 5‘08. 18. 3’07.    19. -325.    20. -1903.

III.    Find the sum of :—

1 -025, 2*005, -0006, ‘5, and -05. 2. 5*75, -25, -225, and -00625. 3 8-225, 6-725. 70 5, and '0025.    4. 12-0125, -006, -8275, and -5.

5. 7-213, 80, -08, 5-5, and 11T125.    6. -0001, ‘Oil, -11, IT, and

no.

7. 57-375, 15-25, -07, and 1250-3. 8. 11T25, 9‘02, -006, and 'll.

9. 5-3, 8-5, 4-45, and 12;54.    10. 10-83, 5-405, -6, and -25.    # >

11. 3-75, 1-4, -345, -01236, and *12345. 12. -0132, 5'045, and 23T56.

IV. Find the difference between—

1. -0061 and *000129. 2. *009 and *000007.

3. 3 thousandths and 51 millionths. 4. *6 and *036.

5. 15 and-015. 6. 8*175 and 8*16.    7. *109375 and 0875.

8. 2-59375 and *54140625., 9. 6-65625 and 2-7182818._    _

10. -06 and-021484375. /il. *875 and *859375.    12. *791 and *145

Exercise XXXV.—DECIMAL FRACTIONS. Multiplication and Division.

T. Multiply—

,. 7-04 by -25.    2. *0012 by *07.    3. *0017 by 30-08.

4. *000635 by -0173. 5. *00625 by -0125. 6. *32 by *0005.

7. 0061 by *000129. 8. ‘009 by "000007. 9. '003 by -000051.

10. -0002 by 8.* 11. -073 by -000085. 12. 42*7 by 9-7123.

13. 3-4Ô9 by 1-925. 14. (1*07-r '007) by -07.

15.    10-5 by -0075.    16 37-805 by -0012.    17. -007 by 2-04.

18.    11-6125 by 10-75.    19. 1*3 by 1*3 by 1*3 by 1*3.

11.    Divide—

1. 32 by -004.    2. 36 by -009.    3. -0017 by 30-08.

4. 30-08by-0017. 5. (-075 - -0075)by -004. 6. ( 0035wr-035) by *008

7. 9-984 by -001953125.    8. 24 64 by -09375.

9. *405 by-0625. 10. 1*07 by *007.

11. (-0021 X 7*31) by 1 oi. 12. *1875 by 321.

13. -4 by -002.    14. 3 by '006.    15. -04 by 0008.

16. -03 by -0006 17. *0061 by 000129. 18. 009 by *000007.

19. 018 by 0005. 20. *003 by -000051. 21. -6 by -036.

22. *036 by ‘6. 23. *05 by -0025.    24. *0025 by 05.

25. *04 by -0018. 26. 0018 by -04.    27. 0002 by 8.

28. 8 by -0002. 29. 9 by -003.    30. ’073 by 000085.

Exercise XXXVI.—REDUCTION OF DECIMALS, FROM TEACHERS’ EXAMINATION PAPERS.

Express—

1.    7 drams and ’0624 of a ton as decimals of a stone.

2.    15 perches and "0385 of a sq. mile as decimals of an acre.

3.    5 yards 1 ft. 6 in. as the decimal of a mile.

4.    '3257 as a vulgar fraction, and -fas. as the dec. of a half-guinea.

5.    ’02025 of £5, and 3|d as decimals of a crown.

6.    16 yards 1 ft. 6 in. as the decimal of a mile.

7.    T9 of a rood —5’2 perches as the dec. of an acre.

8.    ’93 acres—1 ’21 roods as the dec. of an acre.

9.    (| crown ^ £T9g) as the dec. of half a guinea.

10.    (’015 wr ’015) as a vulgar fraction in its lowest terms.

11.    ’013375 of a day and 6f sec. as decimals of 1 hour.

12.    14 dwt. 5 grains as the decimal of a pound.

13.    6 dwts. 4 grains as the fraction of a pound and a quarter.

14.    13s. 4£d. as the decimal of a pound.

15.    (| half-crown ^ f guinea) as the decimal of £1 6s. 8d.

16.    (1’03 half guineas ^ 2T25 crown) as the decimal of 6s. 8d.

17.    (3’01 dwts. v/-' ’025 lb.) as the dec. of a pound avoirdupois.

18.    ’59375 of an acre in roods and perches.

19.    6 ozs. 3 dwts. 4 grs. as the decimal of a lb.

20.    | lb. — f- oz. avoir, as the decimal of 1 cwt.

21.    3’09 oz. avoirdupois as the decimal of 1 cwt.

22.    ’14285*7 of an acre as perches, yards, &c.

23.    (£1 ’04 -j- ’34 of a guinea) as the dec. of a crown.

24.    (|. guinea § crown) as the decimal of £1 13s 4d.

25.    T6 yd. — 2’05 in. as the decimal of a furlong.

26.    (’41 of a day ^ T6 of an hour) as the dec. of a week.

27.    2s. l£d. as the decimal of a crown.

28.    3 lbs. 3 ozs. Avoirdupois to the decimal of 2 lbs., Troy.

29.    ’75 of 3 galls. 2 qts. to the decimal of U barrels.

30.    5 hrs. 15 min. 15 sec. to the decimal of 1 week, 1 day, 1 hour.

31.    3 miles 44 yds. to the decimal of a league.

32.    fa of 7 crowns to the decimal of 2£ guineas.

33.    •] 428 of a shilling to the decimal of 3 florins.

34.    60 acres 3 roods 16 perches to the decimal of a square mile.

35.    18 lbs. 6 ozs. 8 drs. to the decimal of 1 ton 15 cwt.

36.    1 English ell, 3qrs. to the decimal of 11 French ells.

37.    11 sq. per. 24 yds. 6| sq. ft. to the decimal of 1 acre.

38.    5 ozs. 3 dwt. 7 grs. to the decimal of 1 lb. _

39.    10 cubic ft. 720 cubic in. to the decimal of 1 cubic yard.

40. " 1 gall. 3 qts. 1 pt. (wine) to the decimal of 1 hogshead.

Exercise XXXVII.—COMPOUND AND COMPLEX FRACTIONS.

FROM TEACHERS’ EXAMINATION PAPERS.


Reduce to their simplest form-1. (2i - H)-r(3 - § of 7 + | of 3*). 2. (3f of 44 - *)-H4* of 16 J). 7i + 2f . 5Ty-6of 4 T 41M-3I    31 + J


3.


Hof 7-34 •    64-3f


4.


61-24 of 24 • If of & of §


_ .    6§ - 3f . 2* of (* + 4)    » 3f -r*.T of 24

—-    a —-


H of 2 + x®5 * 4f-2^of


6.


ti of 6|


24 + Hi ■ 5744 -Hf 9. 3f -H (§ - * + i).

1-2


8.


34 of 7| -r- 3f 4f - 2§ of If


10. 4 Of 4 -h^-S-of H A of 104.


11.


13.


i-(H of I) ' 31 + 2| of


H-H + 2A . 3f-(tQfH)

3§ of 54    “ J_

24    • 34


42 - 34tt 54 -r- 64


of 7.    14.


24 + 34

4_L _    6.


1

34


4 of 4 of 9


15.


3|-2f+ I* . 34-f- 24    .. 5?1 °f 2i of 3


17.


34-24 + 3 j\

4f + 3§ ^

- 1-I7T *


5f ~ 1A’

24 - 14


16.


335

_ 2


2To + 1A -    ' 34 -r 4


of


3f + 2f


18.


3? - 24 + 5 A

24 of A -f 4f


19.


34 -24 + 5A    9n 3| - 24 -t- 3A f K1 . 24 X 4

24 of|-r- 24 * ZU> 14of 34    01 ^ “ 31


21.


6f-i- (34 of    yg j

3| - (4 +-f of 9) X 14 of 5 8Ó.


171


(%+-i_)

V 44    8-5* /


4f 8-5f ' 6fof7


74 of 4-X


54


-2A


24.    -7- (if X If- ff).

.,25.

5 • 7

2f + 34 3§ 4i + 5| + 10i

0 4 9 7

3| 26. 4f

3f , * H + 24-

^11 . ^11 2f • 8to

19f-

H*

+

Jr

rH

1


4j of (3j-j)    _ji±ILzVL .

7‘    § -h §    0    9,> §of*H-fof2^.

of 6i + 3i.


4 - jPf t-Hg Of T9o

9Q    4

8à - (t-7 of 2i-j~t of 12)

A “ 5

è of t of 19.


on - (§Qff) + l    2$-jof •&-§

2y^ -r 3| •    6l- It-(I-hi)

5i_ZÌs±l ,

32. 3 a “h rr


3f


2|-


5    24 ■+■ 7tV


5f - 3g of § . f« _4±3p

33.    „ 24

7-‘9


(2f+*-*)-* + * . f 3f °f 11 A-lì 34‘    7F+1    “^ + 1 A-hiJ

Note.—This fraction — ^39q . (56 ittìì) *““* 439^ • (w +

R40 1- 478 128

134Ì/ - 6 17 308?-    -

(I-*) + m- ^(2-H)-(l-i)K

35.


0*_ 8-24- 11 of F5 4    2-1 °    2Jr

U


l^ + 2|of 3i-


36.


B -f- (6i - -| of 5è).


1 -


4-S-r-li

* To subtract a negative quantity, change the sign into + and add.

COMPOUND AND COMPLEX FRACTIONS. 193 1

5 +


3


1 -


44


v. è of (si of 7 ] X * of


12 + (1-1)


7* (14 of*)

* Where no sign occurs, as here, the sign of multiplication is understood.


38. 84 of 5* ~

7 Q3 .    9

°4    20


5i-lot

4

5

3

n «ft

12 1 T


of a H- -A) 3    ■

34 + 24


39.

34 of f -h 14 of |

5

40 YT"Ui—y- + T ~ I °f 6. This fraction, given at the • 14 o °    Teachers’ Annual Examination,

I ~ j    December, 1876, yields a negative

4    answer. See pages 175 and 35.

Simplify the following fractions, and reduce the result in

each case to a decimal.

4g + 34 - (§ of 2*)

41. K1 . n, f 1    •    42.

54 — (H of-wn)

^To


43. 3g — 24 .    24    .

44 + f * i of I of 7


(24 -f) + W+|j)

4 of 31 of 6*    2 '

84 - 5f a of f of 7


44.


71+84


45


,    - i1-)15—• «• 2* ^ J

- S|


1*

# of 64 of 1


n


5-4 * 34-23 + 5*’

5f - 24 - 24-


47. 54 f 2| + (hf of 5| of - ).


48.


1


2-8 of 2-27

49. -r.— X

1T36


4-4-2-83 1-6 + 2 629


of


34 of 5 - f of 11 + 12*

6-8 of 3 2-25 *


50.


1


+


1


1X2 1 1x2x3    1 X 2 X 3 X 4


etc., correct to 7 places.


Note.—In this example a new factor is introduced into the quotient each time.


+ etc<wlUbe Ix2x3x4x5 +


1

1 X 2x 3 X 4 X 5 x 6>


and so on, and may be worked by reducing the first fraction to a decimal, and dividing this decimal by the new factor, thus


1X2    — h — -5 - -5

x x 2 x~3 = £ of £ = £ of -5 = T6666666.


1 X 2 x 3"x4 = i of 1X2X3 = i of '16' = -°4166666.

Continue on this plan until the first 7 figures of the decimals produced are cyphers, then add up the results.

51. Find the value of 3-27 of '5 of of 10£f lbs. Troy, + '24 x *227 of 2^ ozs., and reduce the result to the decimal of 2 lbs. 0|- ozs-Troy.


52. Express in drams, scruples, and grains, Apoth., the difference between 10 ozs. 4 drs., Avoir., and 10 ozs. 2 dwt., Troy.

<*»

Note.—The grain in Troy weight is the same as that in Apoth. weight.


53. -{(*-«)!+!}• 1 A

-{ (2 — ££) —(£_ i)     a52§-

ka n - if + 2$ ^ 12* 3f 4- 2i of 3i • 9££*


56.


13-


4 - -f of 6


x

x

8__ 4-TT +

T Ì X § of

3 _ 2 JL 4 5 3 • 9

.55.

7f -

di + 2J) .

10 7 A ~ 1 7

m-

•b 2D x 3| •

iUJL

^12

f of

3 _

5

-Ì v | of f +

. J.1 -L- 1JL

13 • A0*

_9_ x 10

(if-

t)


57.


3-


fof b 1-è -


i


58.


- è


^0TE_Jn working the three last, commence at the lowest part.

CHAPTER V.

DISCOUNT, EXCHANGE, AND OTHER RULES. DISCOUNT AND PRESENT VALUE.

DISCOUNT is money taken off a debt for payment before it is due.

Discount, like Interest, is usually calculated at a given rate upon each ¿£100. Not, however, upon each ¿£100 of the debt; but upon each ¿£100 of that sum which, put out at interest at the given rate per cent., would amount to the debt in the time that must elapse before the debt falls due. This sum of money is called the Present Value of the debt.

Discount is therefore the interest of the Present Value. <£100 plus its interest for the period of prepayment, represents the debt on which £100 is the present value, and the interest of £100 the discount. That is to say, the amount of £100, at Simple Interest, furnishes us with a debt which is to the given debt as the interest of £100 is to the required discount.

Rule for Present Value and Discount :—

Rule.Find the interest and amount of £100, and state

As the Amnt. . I The given )..(    £100    ; Present Value.

of £100    * \ sum or debt \ ” \ Orlnt. of £100 ; Discount.

Example I.—Find the Present Value and Discount of £261, due two years hence, at 8 per cent per annum.

Int of £100 for 2 yrs = £8 X 2 = £16 Amt of £100 — £116.

As £116 : £261 :: £100 : £225, the Present Value.

4s £116 ; $261 11 £16 : £36, the Discount.

Note.—The Discount may be more easily obtained by subtracting the Present Value from the Debt ; so, also, when the Discount has been first obtained, the Debt — the Discount = the Present Value.

This is evident from the fact that—

The Present Value + The Discount = The Debt.

Having given tlie rule for Discount in conjunction with that for Present Value, in order to draw attention to the fact that they differ only in regard to that which forms the third term, we now give the rules for stating separately.

To find the Discount on any given sum or debt :

state, As Amount of £100 : Given Sum :: Int. of £100.

To find the Present Value of any given sum or debt:

State, As Amount of £100 : Given Sum :: £100.

It is thus seen that the rule for finding the Present Value is precisely the same as that given to find what sum of money will produce a given amount. See Rule I., Page 150.

Practically, Discount is a rule seldom or never employed in business transactions, it being usual to charge interest upon the debt instead of discount.

This allowance is termed commercial dis^^stt.

However, the principles involved in what is termed true discount so frequently occur in questions coming under the head of “ profit and loss,” that it would be unwise to omit their explanation here.

]y0XE__The per centage which a tradesman takes off the amount

of his account for immediate, or cash, payment, is also called rebate or DISCOUNT.

Remark.—Discount, like all other rules depending upon the principles of proportion, may he solved by first principles.

Example.Find the Discount on £290 for 3 years at 15 p. c.

The interest on £100 for 3 years at 15 p. c. == £45.

The discount of £145 is £45.

The discount of £290 is o{ £45 = f of £4&> or £90-

Exercise XXXVIII.— DISCOUNT AND PRESENT WORTH.

I. Find the Discount upon—

1.    £788 2s 7^d. for 3^ years at 5 per cent., Simple Interest./

2.    £481 18s. 3d. for 3£ years at 3£ per cent., Simple Interest.

3.    £4279 5s. 6d. for 5 years at 6J per cent , Simple Interest.

4.    £945 14s. Ofd. for 3^ years at 3f per cent., Simple Interest.

5.    £1858 3s. 3^d. for 9 months at 3£ per cent., Simple Interest.

6.    £769 18s. Rd. for 2 years 94 days at 3£ p. c., Simple Interest.

7.    £2406 6s. 2Jd. for 5 years 3 months at per cent., Simple Interest.

8.    £768 4s. 4£d. for 1 year at £2 8s. 7d. per each £100.

II. Find tlie Present Worth or Present Value of—

1.    £1011 9s. 9d. due 2 years hence, at 4per cent., Simple Interest.

2.    £442 19s. 10|d. due 5 years hence, at 5 p. c., Simple Interest.

3.    £187 2s. 6d. due 4^ years hence, at 5.^ p. c , Simple Interest.

4.    £546 5s. due 3 years hence, at 5 p. c , Simple Interest.

5.    £2132 Is. 3d due 7g years hence, at 9£ p. c , Simple Interest.

6.    £744 16s. l£d. due 8f years hence, at 8| p c . Simple Interest.

7.    £361 6s 5d. due 3 years hence, at 12 p. c., Simple Interest.

8.    £457 12- 6d due 6£ years hence, at 4| p. c, Simple Interest.

9.    £147 due 2 years hence, at 5 p. c., Compound Interest. \

10.    £289 8s.- l£d due 3 years hence, at 5 p. c., Compound Interest.

11.    £281 7s. 6^d. due two years hence, at 6 p. c. Compound Interest, the interest being reckoned half-yearly.

12.    £2807 14s. 4fd. due R years hence, at 4 p. c., Compound Interest, the interest being payable quarterly. /

III. Find the Commercial Discount upon—

1.    £158 6s. 8d. for 3 years at 5 per cent., Simple Interest.

2.    £173 17s. 4|d. for 4 years at 6 per cent., Simple Interest.

3.    £333 6s. 8d. for 3 years, at 8R per cent., Simple Interest.

4.    £715 6s. 3d. from May 9th to October 2nd, at 2\ per cent, Simple Interest.

5.    £935 2s. 7id. from September 19th, 1877, to July 8th, 1878, at 5 per cent.

6.    £1999 17s. 2§d. from July 22nd to October 2nd, at 15 per cent.

7.    £752 5s. 5fd. from August 9th, 1879, to May 27th, 1880, at 7| per cent.

8.    £1171 15s. from August 2nd, 1877, to March 9th, 1878, at 12£ per cent.

EXCHANGE.—THE CHAIN RULE.

Exchange is the name given to the process by which we reduce a sum of money of one country to its equivalent in the currency of another country.

In Exchange and other rules where the equal values or the ratios of the quantities are given, a series of successive proportion statements is saved by the employment of the CHAIN RULE.

CHAIN RI LE.

1, Place to the right of the word “ Answer ” the quantity which it is to he the equivalent of. 2. Select the quantity which is of the same kind as this and put under the word “Answer,” placing its equivalent opposite it. 3. Continue, on this plan, to place the quantity of the same kind as that last stated on the left with its equivalent opposite it until that term is stated which is of the same kind as the answer. 4. Divide the product of the terms on the right hy that of the terms on the left.3

In short—

Place opposite each other equal values, and select the quantity of the same hind as the last to state next.

Example I.—If £1 = 25 francs (French); 50 francs = 22 75 florins (Dutch) ; \\ florins = 240 rees (Portuguese) ; and 800 rees = 7s. Flemish; find how many Flemish shillings are equal in value to £40.

By Chain Rule—

Ans. = £40. Then—

= 637s. Ans.


£1 = 25 francs 50 francs = 2275 florins florins = 240 rees 800 rees = 7s Flemish


40 x 25 x 22 75 x 240 X 1


1 x 50 x 1-5 x 800 40 X 25 x 2275 x 240 x 7 50 x 150 x 800


PROPORTIONAL PARTS.

Proportional Parts are such as bear to each other

given ratios.

To divide anything into proportional parts.

Rule.—!>ivide the whole hy the sum of the equal parts, and multiply the result, respectively, by the number of such equal parts in each of the given parts.

Example I.—The sum of ¿£60 is set apart to be divided among the five directors of a company for their services, in proportion to their number of attendances. A attends 7 times, B 8 times, C 9 times, D 10 times, and Ell times. What will each receive 1

The sum of the equal parts =z7 + 8 + 9+10 + lln: 45.

£60 divided by 45 iz £1 6s. 8d., the equal part of which A is to get 7 such, B 8, C 9, D 10, and Ell.

Hence A gets 7 times £1 6s. 8d., or £9 6s. 8d. ; B, 8 times £1 6s. 8d., or £10 13s. 4d. ; C, 9 times £1 6s. 8d., or £12; D, 10 times £1 6s. 8d., or £13 6s. 8d. ; and E, 11 times £1 6s. 8d., or £14 13s. 4d.

Example II.—Eighty-seven head of live stock, consisting of an equal number of horses, oxen, and sheep, were bought for ¿£645 19s. 6d.; an ox cost eight times as much as a sheep, and a horse twice as much as an ox and a sheep together : find the price of each.

Here, we consider the cost of a sheep to be one equal part, such that an ox cost eight, and a horse twice (8 + 1), or eighteen. 87 divided by 3 zz 29, the number of animals of each kind.

29 horses costing 18 parts each zz 522 )    £645    19s. 6d. -f 783 =    \

29 oxen costing 8 parts each zr 232 > £0 16s. 6d., cost of a sheep. ( 29 sheep costing 1 part each ~ 29 ) £6 12s. 0d., cost of an ox. i

- £14 17s. 0d., cost of ahorse. )

Total number of equal parts z: 783

Example III.—Divide 234 into 3 parts, which shall he

as l 1 and 1.

Here, in order to enable us to compare the fractions, we reduce them to the same name, obtaining T\, and T\.

The parts must therefore be as 6, 4, and 3. Sum of parts, 13.

234 -f- 13 = 18, the equal part.

18 X 6 = 108, the first part; 18 X 4 ~ 72, the second part; and 18 X 3 ~ 54, the third part.

Example IV.—A merchant ordered 36 pieces of silk, each containing 28 yds. 1§ nls., to be made into dresses of three different sizes ; those of the largest size contained 15| yds. each, those of the next size 12 yds. Oqrs. 1 nl, while the smallest contained 9 yds 2 qrs. 2j nls. There was to be an equal number of each size. Find the number made, and the value of the whole silk, at 16s. 8d per ell, French

Whole cloth =: 36 times 28yds. 1§ nls. = 1,011 yds. 2 qrs 2\ nls.

Sum of Parts = 15| yds + 12 yds. 1 nl. + 9 yds. 2 qrs. 2\ nls. 37 yds. 1 qr. 3£ nls.

Whole cloth divided by the cloth required for 1 dress of each size = 1,011 yds. 2 qrs. 2£ nls. -f- 37 yds. 1 qr. 3^ nls. = 27.

Hence there are 27 dresses of each size, or 81 in all.

The cost may be found by several different plans. The following, however, is one of the best :—

1,011

yds.

2 qrs. 2\ nls.

= 674 F. ells 2 qrs. 2-:

nls.

10s.

= h

674

F. ells

1 qr. 2 nls. |

16s.

8d.

6s. 8d.

= 1

337

3 nls. ^

4s.

2d.

224

13 4

l^nls.|

2s.

Id.

0

7 3 h

Is.

0'd.

Answer. £562 0 7\    Cost of 2 qrs. 2\ nls. 7s. 3£d.

Exercise XXXIX. — PROPORTIONAL PARTS. FROM TEACHERS’ EXAMINATION PAPERS.

1. A farmer bought pigs at 12s. 4d , turkeys at4s. 9d., geese at 3s. 6d.,' and fowls at Is. lOd. each, in the proportion of three geese, four pigs, and five turkeys to each fowl. Required, the number bought, the sum laid out being £38 8s. 9d.

2.    The sum of £1148 2s. Cfd. is to be divided amongst an equal number of men, women, and boys—in all 87. Each woman is to receive three times as much as a boy, and each man as much as a woman and two boys together. How much did each receive ?

3.    A father divides his estate, consisting of 1817 acres 2^ roods, among his three sons. For every 7 acres the eldest one receives, the second has 5 acres ; and for every 2 acres the second receives the third gets 3 roods. How much land did each receive ?

4.    A gentleman left his estate to be divided among his three children in such proportions that, for every 5 acres the eldest received, the second should receive 3 acres, and for every 2 acres the second received the youngest should receive 1J acres. W hat number of acres fell to the share of each—the land being worth £10,974 6s. 8d., valuing it at £3 13s. 4d. an acre ?

5.    Ten square miles of land are to be divided amongst 126 persons ; 57 of them receive ea'ch 38 acres 3 roods 35 perches, and the remainder is equally divided amongst the others. How much does each of the latter receive ?

6.    A person leaves £3000 to be divided amongst his 4 sons : the eldest is to receive £1 for every 15s. the second receives ; the latter, 18s. for every 10s. the third receives ; and the youngest is to receive £400. How much did the other sons receive respectively ?

7.    Divide 336 yds. 3 qrs. 3f nls. of broad cloth between A, B, and C, so that for every 5 yards A has, B shall have 3 and C If yds. Find the value of the cloth at 17s. 9d. per ell English.

8.    In a certain State School 5 the scholars were in the 1st class, -i-in the 2nd, fifteen in a class between the 1st and 2nd, J- in the 3rd class, -Jg in the 4th class, and in the 5th and 6th classes. Required, the number of children in the school.

9.    In a certain State School \ of the children were in the 1st class, three-fourths of this number in the 2nd class, ^ of the whole in the 3rd class, £ in the 4th class, -JT in the 5th, and 2 in the 6th class. Required, the number in the school.

10.    The proportions used in making sovereigns are 21 parts of pure gold and 2 parts of alloy. Find the weight of pure gold in 500 sovereigns—supposing a sovereign to weigh 5 dwts. 3f grs.

11.    A landowner wishes to cut up a large paddock of 620 acres

10 perches into 30 smaller paddocks—15 of each size. The 15 larger ones are to contain each 29 acres 2 roods 15 perches. What will each of the 15 smaller ones contain ?    '

12.    A sum of £1,031 19s Hd. is to be divided amongst an equal number of men, women, and boys—in all, 78. Each woman is to receive 5 times as much as a boy, and each man as much as a woman and two boys together. How much wiM each receive ?

13.    A person leaves £3,000 to be divided among his four sons—the eldest is to receive £1 for every 15s. the second receives, the latter is to receive 18s. for every 10s. the third receives, and the youngest receives £400. How much did the other sons receive respectively ?

THE RECIPROCAL OF NUMBERS.

The Reciprocal of a number has been defined as that which the number must be multiplied by to produce unity; or, as the fraction formed by placing the number in question as the denominator, and 1 as the numerator. Thus, the

1 2

of 8 is -g- j of 171 is —- , or

Since a fraction multiplied by itself inverted produces unity, the reciprocal of a fraction is obtained by inverting the fraction.

Thus, the reciprocal of f is ■§, of f- is f, of f- is — of 3| is -fe.

If one thing be a certain part of another, the second will be that part inverted of the first. That is,

Whatever fraction one thing is of another* the reciprocal of that fraction expresses the relation of the latter thing to the former.

Thus, if one thing be \ or £ of another, the latter will be twice or thrice the former.

If 7s. 6d. be | of £1, then £1 will be f of 7s. 6d.

ki    12

If A’s money be of B’s, then B’s money is of A’s.

12    5$

If A can do j- as much work as B, then B can do \ as much as A.

If 12^ be 5 times 2£, then 2^ is A of 12|.

If the cost price be T8T of the selling price, the selling price is y 0f the cost price.

If A be to B as 7 : 15, then B is to A as 15 I 7.

If the gain be 15 per cent, on cost, then the cost was *££ of the gain.

The application of the above principle will often save much labour in solving questions on Proportion.

Thus—

1.    When unequal workmen can do the same work in different times, the relative amount of work which they can do in the same time will be expressed by the reciprocals of the numbers denoting the different times. Thus, if a man can do as much work in 5 hours as a woman can do in 6, or a boy in 7, the relative amounts of work which a man, woman, and boy can respectively do in the same time are +, i, and i, or 42, 35, and 30 ; or, a man will do 6 parts to a woman’s 5, and 7 parts to a boy’s 5.

2.    When different numbers of things can be bought for the same

money, the relative costs of the same number of each are as the reciprocals of the numbers bought. Thus, if 12 sheep can be bought for cash for the same money as will purchase 11 on credit, the relative prices for cash and credit of the same number of sheep will be    and -Jy,

or 11 units of money as the cash price of one for every 12 units of money in the credit price.

Example I.—A can walk the same distance in 37 min. as B can in 385 min. If, in a race of 2 miles, A gives B 159 yards start, who will win, and by how much ?

Solution.—The comparative distances walked by A and B in the same time will be to each other as the reciprocals of the different times in which they walk the same distance. Therefore A can walk 385 measures of a distance (be it mile or yard,) while B walks 37 6uch measures.

A walks the same distance in 37 min. that B walks in 38J min.

Then A can walk 381 yds. while B walks 37 yds.; or,

A gains 1J yds. upon every 37 yds. that B walks.

And as B has to walk (2 miles — 150 yds.), or 3370 yds.,

A can give him a start of ^ of 3370 yds. = 136ff yds.

But, since A gave B 150 yds. start, B wins by (150 — 136ff) yds.

Hence, Answer, B wins by 13|f yds.

The solution of the same by Proportion :—

First.—Let unity represent the distance that B walks in 38| min. Then, since A walks this distance in 37 min., he will walk farther in the time that B takes—i.e., in 38^min. Hence, to get the comparative distance which A will walk in 38£ min., state—

As 37 min. : 38^ min. I1 distance ly\ such distances.

Secondly.—Since A can walk 1^ of any distance walked in the same time by B, A gains upon B ^ of any distance that B may have to walk ; or, in other words, A gains 3 yards upon every 74 B walks. And since B has to walk 3370 yds., we state,—

As, 74 yds. : 3370 yds. ; 3 yds. gain : 136ff yds. gain.

A gave B 150 yds. start, and gains 136f^ yds. only, so that the latter wins by the difference ; or A loses the race by the difference between 150 yds. and 136ff yds.

That is,—B wins by 13-|f yds. Answer.

Example II.—“ A flock of sheep were sold, some for cash

AND SOME ON SIX MONTHS’ CREDIT. 12 OF THOSE SOLD FOR CASH BROUGHT THE SAME MONEY THAT 11 OF THE OTHERS WERE SOLD

for. What was the true discount charged ?

The comparative cash and credit prices of a sheep are to each other as the reciprocals of the numbers which express the different number of sheep which could be bought for the same money.

Hence the cash price was 11 units of money (whether shillings or pounds,) for every 12 in the credit price.

For credit, therefore, 1 additional unit was charged upon every 11 of the cash price. Hence, the true discount (which, it must be remembered, is the interest on the present worth or cash price,) is 1 on 11, or 9yT per cent. But as 9TlT p. c. is the interest for six months, the rate p. c. per annum is 9-^ x 2 = lSyU per cent. Answer.

Example III.—By selling an Arithmetic at 3s. 9d., a bookseller gains f of the prime cost. What per cent, would he gain by selling it at 4s. Id.

Solution.—The selling price = cost + of cost, = f of the cost.

Prime cost = £ of the selling price ; | of 3s. 9d. = 2s. lid.

4s. Id. - 2s lid. = Is 2d., the gain by selling for 4s. Id.

And Is. 2d. gained upon 2s lid = -§-f of 100; or 40p.c. Answer.

Example IV.—A can do a piece of work in f of the time B takes; B can do it in f of the time C takes, and C requires J as long again as D. Compare the time D would take with the time A takes.

A requires § of f of -§• of the time which D takes.

D requires | of f of -f = f of the time that A requires. Answer, 1| of the time.

PROFIT AND LOSS.

Questions coming under the head of Profit and Loss involve no new principle. There are, however, a few things which should he kept well in view.

I. —In the case of gain :—

The selling price = the cost price + the gain.

II. —In the case of Loss :—

The selling price = the cost price — the loss.

Example /.—“By Selling a Book at 3s. 6d., a Bookseller

GAINS 31| PER CENT. OF THE PRIME COST. WHAT IS HIS GAIN IF HE SELL IT AT 4S ? ”

Solution :—

(1.) The selling price = the cost + the gain.

311

(2.) The gain is jqq of the prime cost.

31£    isix

;he selling price = the cost —J—Jqq    cos^ ~ TÖ0


By the reciprocal principle explained at page 202.

The cost is of the selling price, or of 3s. 6d. = 2s. 8d. When selling at 4s., the selling price - the cost = the gain.

Hence, the gain = 4s. - 2s. 8d. = Is 4d.; that is, the gain is 16 pence upon every outlay of 2s. 8d. = 50 per cent. Ans.

Example II.—“B-y Selling Quills at 16s. 4.n. per 1000 a Stationer Loses 6f per cent, of his Money. How much per CENT. WILL HE GAIN BY SELLING THEM AT 19s. 3d. PER 1000 ? ”

Solution :—

(1.) The selling price == the cost price — the loss.

(2.) The loss is 100 of the cost.

Hence, the selling price = the cost — |qq of the cost = °f the cost.

By the reciprocal principle explained at page 202.

The costisggx of the selling price, or^p of 16s. 4d. = 17s. 6d. When selling at 19s. 3d. per 1000, the selling price - the cost = the gain.

Hence, the gain — 19s. 3d - 17s. 6d. = Is. 9d. or 21 pence. And Is. 9d. gain upon 17s. 6d. is ^Vo x 100 = 10 per cent. Ans.

From I. we obtain,

In the case of gain,—

III.    The cost price = the selling — the gain.

IV.    The gain = the selling price — the cost.

From II. we obtain,

In the case of loss,—

V. The cost price = the selling price + the loss.

VI. The loss = the cost price — the selling price.

Example III.—By the sale oe goods which cost me £9 2s., I lost

A SUM EQUAL TO 8J PER CENT OF THE MONEY I RECEIVED ; AND BY THE SALE OF OTHER GOODS WHICH COST ME £27 18s., I GAINED 12£ PEE CENT OF THE MONEY I RECEIVED. WHAT PER CENT DID I GAIN ON THE COMBINED TRANSACTIONS ?

That is, the cost pricey £9 2s., is Hence, taking the reciprocal—

The selling price is    of £9 2s. = £8 8s., and

The^oss is £9 2s. - £8 8s. = 14s.

In the second transaction a gain is made ; hence, by III.,

The cost price = the selling price - the gain.

121    .

The gain — of the selling price. Hence,

■X


In the first transaction a loss is sustained ; hence, by V.,

The cost price = the selling price + the loss.

. 84

The loss is of the selling price. Hence,

81 , -,

The cost price = the selling price + —t of the selling price.

108^ r , ... .

„ of the selling price ;


100


874

The cost price, £27 18s., is of the selling price.

Hence, taking the reciprocal,

The selling price is gyy of £27 18s. = £31 17s. S-fd., and

The gain is £31 17s. 8*<L -£27 18s. = £3 19s. 8*d.

The gain, £3 19s. 8fd., exceeds 14s., the loss on the first transaction, by £3 5s. 84d.    .    ,

Hence, in the combined transactions, £3 5s. 8-^d. has been gamed

upon (£9 2s. + £27 18s.), or upon £37. Hence, to get the p. c. gained, State, As £37 : £100 ; I £3 5s. 8|d. : 8fff P- c- Answer.


EXERCISE ON PROFIT AND LOSS.

207


Exercise XL.—PROFIT AXD LOSS.

1.    A draper sold a piece of cloth containing 1000 Flemish ells for 850 guineas, and gained upon every yard | of the prime cost of an English ell. What did the whole piece stand him in ?

2.    By selling wine at 2s. 6d. per gallon a grocer would lose 6j per cent. He sells at £4 10s. per quarter-cask of 30 gallons. What was his gain per cent ?

3.    A grocer bought 2 tons 3 cwt. of sugar for £120, and paid £2 10s for expenses. He wished to gain 50 per cent, on his outlay. What must he charge for 3£ cwt. ?

4.    At what rate must I sell oats which cost me 12s. 6d. per bag, in order that I may gain on every £25 of outlay the selling price of 8 bags ?

5.    At what rate must I sell oats which cost me 15s. a bag, in order that I may gain on every £22 10s. of outlay the selling price of 6 bags ?

6.    A merchant imported a cargo of tea and expected to realize a

profit of 20 per cent., but the tea proved to be damaged and was sold for £432 10s., which was 12^ per cent, less than the cost price. What was the difference between the sum realized and that expected ?    #

7.    By selling oats at 5s. 8d. per bushel, a corn merchant gains _2- of the prime cost. What would be his gain per cent, if he sold them at 5s. 9d. per bushel ?

8.    By the sale of goods which cost me £7 16s. I lose a sum equal to 15f per cent, of the money I receive ; and by the sale of other goods which cost me £27, I gained a sum equal to 12£ per cent, of the money I received. What per cent, did I gain on the whole ?

9.    A merchant sells goods at £2 15s. 3d. per cwt., and by so doing loses 15 per cent. What will be his gain per cent, by selling at £3 8s. 3d. per cwt. ?

10.    A man bought 9 acres 3 roods of land for £130. The expenses of conveyance amounted to £22 10s. He wishes to sell it at a gain of 25 per cent on his outlay. What must he charge for 4£ acres ?

11.    By selling sugar at £32 13s. 4d. per ton, a merchant lost 6f per cent. What would he have gained on a cargo of 256 tons had he sold at 4|d. per lb ?

12.    A grocer sells sugar at £2 3s. 2d. per cwt., and by doing so loses 12^ per cent. What was the prime value ?

13.    By selling corn at 4s. 10^d. per bushel, a grocer lost 2^ per cent. What would have been his gain per cent, had he sold it at £2 3s. 4d, per qr. ?

14.    I sell cloth at 7s. 6d. per Flemish ell, and gain 20 p. c. of the cost price of an English ell. What did it cost per French ell ?

15.    I bought 10 dozen oranges for 7s., some at 4 for 3d., and the rest at 6 for 4d. How many did I buy at each rate ?

16.    A merchant, by selling tea at £8 17s. 3^d per chest of 74 lbs,, gained 25 per cent. What was the prime cost of 9 lbs. ?

17.    A draper bought 360 yards of cloth at 5s. 4d. a yard. He sold 210 yds. at 7s. 2d. per yard, and the remainder at 9s. a yard. What did he gain per cent ?

18.    A grocer, by selling sugar at 3|d per lb., lost 16 per cent. What was the prime cost of 1 cwt ?

19 A wine merchant mixes together 12 gallons of spirits at 18s. lid. per gallon, 15 quarts at 16s. per gallon, and 4£ gallons at 20s. per gallon. How much water should he mix with the spirits in order that, by the sale of the mixture at 16s. 9|d. per gallon, he may gain Is. 8d. in the £ ?

20.    A publican mixes together 10 gallons of spirits at 12s. 4d. per gallon ; 12 gallons, at 13s. 8d. per gallon ; and 6 gallons at 16s. 9d. per gallon. What will he gain or lose per gallon by retailing the mixture at 3s. 6d per quart ?

21.    A merchant imports 150 chests of tea ; each chest measures 2 cubic ft., and weighs 50 lbs. ; he pays for freight, 5s. 3d. per cubic yard. A wharfage fee of 2s. 4d. per cwt, and an ad valorem duty of 10 per cent on the invoice price, 2s. 6d per lb. For how much must each chest be sold so as to just clear expenses ?

22.    A grocer mixes together l|cwt. of tea at Is. 5d. per lb. ; 70 lbs. at Is. 7d. ; and 96 lbs at Is 10d., and sells the mixture in 21b packets. What must be the price per packet, that he may gain 6 guineas on the transaction ?

23.    By selling The State School Grammar at Is 6d., a bookseller gains of the prime cost. How much per cent, will he gain if he sell it at Is. 3d ?

24.    A grocer sold 12 bags of sugar at a profit of 12^p c , and 16 ba»s at a&profit of 10 p.c If he had sold the whole at a uniform profit of 12 pc., he would have received £1 6s more than he did. What was the cost of the sugar per bag ?

25 A tailor buys cloth at a wholesale rate of 16-jpc below the retail price, and then charges 33^ p c upon the retail price for making up into suits. What profit does he gam on a suit for which he charges £4 16s. ?

in the same name before


ARITHMETICAL EQUATIONS.

Definitions.—An Arithmetical Equation consists of two equivalent expressions for the same amount: As, 6x4+1 = 100 -+ 4.

The equal amounts are generally placed opposite each other with the sign = between them, the whole being called an equation or identity. Thus, 5s. + Is. 8d. = ¿61 -f- 3, means that the result of the addition of the two sums of money on the left of the sign =, is identical with the result of the division indicated on the right of that sign. In other words, that 6 s. 8d. equals one-third of a pound.

The principles of equations are both simple and easy of comprehension j yet they are of such universal importance that upon them most of the operations of arithmetic hinge. A knowledge of these principles is quite indispensable to the intelligible solution of many questions to be found in the most popular treatises on arithmetic. It is, therefore, proposed to state the most useful of them here in a form suitable to the student who is not also an Algebraist.

Axiom 1—If equals be added to equals the wholes (or sums) are equal.

If 9 + 4 = 13 ; then 9 + 4 + 7= 13+7.

Axiom 2.-if equals be taken from equals the remainders are equal.

If 12 = 7 + 5 ; then 12 — 8 = 7 + 5 — 8.

Axiom 3 If equals be multiplied by the satne number the products will be equal.

If 12 = 9 + 3; then 12 x 5 = (9 + 3) x 5.

Axiom 4. if equals be divided by the same number? the quotients will be equal.

If 12 = 9 + 3 ; then 12 + 4 = (9 + 3) + 4.

Axiom 5.—If two expressions be equal, similar multiples, powers, fractions, or roots of them will also be equal.

Thus 122 = (9 + 3)2 ; ^12 = ^(9 + 3); | of 12 = | of (9 + 3).

We thus see that we may add equal amounts to, or take equal amounts from, both sides of an equation, or multiply or divide both sides by the same number, without destroying the equality of the identity.

From the above axioms, we get the following modes of working in order to solve a problem or to find the unknown value of any term it may contain. We may here remark that all the operations described are, in effect, equal additions, equal subtractions, equal multiplications, or equal divisions.

Terms of an Equation :—

Each number preceded by a plus or minus sign is a distinct term of an equation, unless several are enclosed within brackets, in which case the whole bracket is a term. Numbers connected by the signs of multiplication and division form part of one term. Thus in 8 + 4- 7 + 8-i-2-(6-2)x3, the terms are 8, 4, — 7, 8 -f 2, and - (6 — 2) X 3 ; while f of -f f 0f 9 X (7 — 5) is one term only. This principle is sometimes expressed thus—Numbers connected by X and -r form parts of one quantity ; numbers connected by + and - form separate quantities. Hence the rule—The work indicated by the signs X and -f must always be done before that indicated by + and —, except where brackets occur.

Modes of operation—

1 TRANSFERRING.—Any term may be moved to the opposite side by changing the sign + into -, or vice versa, minus into plus.

If 9 + 3 = 12, then 9 = 12 - 3.

3 TRANSPOSING.—The two sides of an equation may change places without any alteration of the signs.

If 7 + 2 -6 = 8 — 5, then S-5 = 7 + 2-6.

3.    CHANGING SIGNS,—All the plus and minus signs may be exchanged without affecting the equality.

If — 8 + 5 — 3J == — 5J — 1, then 8 — 5 + 3^ = 5^ + 1.

Note.—Signs within brackets must not be changed.

If_8 + 5-3J=- (5* + 1), then 8 - 5 + 34 = (5* + 1).

4.    Any MULTIPLIE It of the whole of one side may he placed as a DIVISOR of the whole of the other side, and vice versa, any DIVISOR affecting the whole of one side may be placed as a MULTIPLIER to affect the whole of the other side.

If (8 -I- 7 + 5) X 6 = 1200 -h 10, then (8 + 7 + 5) X 10 = 1200 -r- 0.

5.    A fractional factor of the whole side may be transferred to the opposite side by inverting it.

If I of (9 + 7) = 12, then (9 + 7) = $ of 12.

6 A numerator of a fraction on one side may be placed as a denominator on the other side; and vice versa.

If I of (9 + 7) = Vi then 3 X (9 + 7) X 2 = 24 X 4.

The first three of these operations depend upon the facts stated in Axioms 1 and 2—viz., that the equality of an equation is not destroyed by the addition to or the subtraction of equal amounts from both sides.

The three last are particular applications of the axiom that multiplying or dividing every term of both sides of an equation by the same number does not destroy the equality.

To Solve a Problem :

1.    Clear it of fractions by equal multiplications.

2.    Collect all the terms of the unknown value on the left, and the known ferms on the right, by transferring, transposing, &c.

3.    Add together separately the unknown terms and the known terms, and then find the value of the unknown term by the ordinary rules of Arithmetic.

4.    When any power of a quantity is given, the number required is found by extracting the root corresponding to that power.

5.    When two distinct values are shown for two unknown numbers, by simultaneous equations,4 make the number of unknown things of one kind equal in both equations, by multiplication, and then subtract one equation from the other before solving it.

TO TAKE NUMBERS OUT OF BRACKETS :—

(a)    If no sign, or the, sign plus, precede the brackets, they are taken out without altering any of the signs.

(3 + 7 - 5) + (3 - 9+ 7) = 3 + 7- 5 + 3- 9 + 7.

(b)    If the sign - precede the bracket, change the sign of each term as it is taken out.

(3 + 7 - 5)-(3-9 + 7) = 3 + 7- 5- 3+ 9-7.

(c)    When a number is placed before or after a bracket and connected by the sign x or -+, the multiplication or division affects the whole of the terms within the bracket. That is, 2 x (3 + 7 - 5), + 3 — (3 x 2-+3) + (7 x 2+-3) - (5 x 2 + 3) = 2 + 4§ - 3£ = 3J; or it may be considered as the result of (3 + 7 - 5), i e. 5, multiplied by 2 and divided by 3. Now (3 + 7 - 5) = 5, and 5 x 2 +- 3 = 3^.

(d)    If both outer and inner brackets occur, take the terms out of the inner brackets first and then out of the outer.

-■{ (3 +7- 5) -(3 + 9 +7) +8 ¡4 = -,¡3 + 7 -5-3-9-7+81-= - 3 - 7+5+3+9+7 - 8=6.

Note —When no sign stands before a number, the sign +- is understood. When no sign comes between numbers, the sign x is understood.

Arithmetical Equations :—

Problem I.—Two tanks of equal dimensions are full of

WATER, AND HAVE TAPS WHICH WILL EMPTY THEM RESPECTIVELY IN 3 AND 4 HOURS. If BOTH TAPS BE OPENED TOGETHER, WHEN WILL THE FIRST CONTAIN JUST ONE-HALF AS MUCH AS THE SECOND ?

In every hour £ of the water runs out of the first, but | only out of the second.

Now, ^ of the water X the required number of hours will be the amount run out of the first tank, and \ of the water X the required number of hours will be the amount run out of the second tank.

Again, by question, the whole contents - the amount run out of the frst, = one-half of the whole contents - the amount run out of the second. Or—

Contents - ¿Cont. X No. of hrs. = |(Cont. - |Cont. X No.ofhrs.)

Taking out of brackets

Contents - J Cont. X No. of hrs. ~ £ Cont. — | Cont. X No. of hrs.

Transferring

Contents — \ Cont. zz J Cont. X No. of hrs. — £ Cont. X N®. of hrs.

Collecting

£ Contents = (£-£) Cont. X No. of hours.

Taking out of brackets

\ Contents — Contents X No. of hours.

Removing denominators to opposite sides

24 Contents z2 X 5 Contents x No. of hours.

Changing sides

No. of hours x 10 Contents 24 Contents.

Transferring a factor

w t i    24 Contents.

No. of hours — ——-

10 Contents.

Answer.


Hence, No. of hours required ~    ~ 2f hours.

Note.—The above work might have been shortened very much, but it is here given in full in order to show how to transfer, collect, remove brackets, &c.

Note 2.—Any one of the above operations might have taken precedence of those before it without affecting the result.

Problem II.—Find two numbers such that J of the one added to ^ of the other shall equal 16 ; and that £ of the one added to \ of the other shall equal 18.

By question.    £ of the one No. + J of the other No. = 16 (i.)

Multiply i. by 2.

f of the one

+ J of the other = 32

(ii.)

By question.

5 of the one

+ \ of the other = 18

(iii.)

Subtract iii. from ii

of the one

= 14

(iv.)

From iv. fg of the first No. = 14 ; first No. = \5 of 14 = 30.

By i. £ of 30 + \ of the second No. = 16 ; the second No. = 24.

AVERAGE.

Definition.—The Average of several things is the result of taking one with another to get the general or mean result.

I —The average thus equals the sum divided by the number of things.

The average age of a class of six children, whose ages are six, five, seven, ten, eight, and nine years respectively, is, (6 + 5 + 7 + 10 + 8 + 9) -r 6 = V = 74 years.

II. —The average of two numbers + half their difference = the greater number.

The average of two numbers - half their difference = the less number.

Hen«e, of two numbers,

£ sum + \ diff. = the larger.

£ sum - ^ diff. = the smaller.

Thus, if the average of two numbers is 12, and their difference 6, the greater is 12 + 3, and the less 12 - 3 ; or the numbers are 15 and 9.

III. —To get the average rate of various things at different rates.

Rule.—Multiply each number by its particular rate, add the products together, and divide by the sum of the numbers.

Thus, the average cost of 20 lbs. of butter at Is. 6d., 10 lbs. at Is. 9d., and 30lbs. at 2s. 5d. = \ (Is. 6d. x 20) +- (Is. 9d. x 10) + (2s. 5d. x 30) 4- (20 -+- 10 +- 30) = 120s. 4- 60 = 2s. ; or, as generally shown,

20 lbs. at Is. 6d. = 30s. Od.

10 ,,    Is. 9d. = 17s. 6d. And£^ =2s., average cost.

30 „    2s 5d. = 72s. 6d.

Total, 60 Total cost £6 0s Od.

IV._-©ne of two quantities x tlie difference between its rate and tlie average rate = the other quantity x the difference between its rate and the average rate.

Thus, if the average cost of two lots of tea is 36§d., and one lot is 56 lbs. at 50d. per lb., and the other 448 lbs. at 35d. per lb., the deviationfrom the average is (50d. - 36§d.), or 13|d., in the first case, and (36§d. — 35d.) = l§d. in the second case.

Then 56 lbs. x 13J must equal 448 X 1§.

This principle helps us to solve many difficult questions ; for instance, such sums as Sum 9, page 290, B. Smith, or Sum 14, page 260.

Sum 1.—I bought 10 Dozen Oranges for 20s. Some at 2 for 3d., and the rest at 4 for 11d. How many did I buy at each

RATE ’

120 for 240d. = 2d. each. The average cost.

2 for 3d. = l*d. each, or |d. below the av. cost.

4 for lid. = 2fd., or fd. above the av. cost.

In what proportion must I buy oranges at two farthings below and three farthings above the average price respectively ?

The greater the deviation from the average of the one rate, the less in number will be required at this rate to counteract the effect of the less deviation of the other rate. Hence, there must be less at the second rate in proportion as 3 farthings is greater than 2 farthings. For, every 2, at Jd. above, will make up for 3 at Jd. below.

The comparative number bought at the rates named must, therefore, be 3 at the first rate for every 2 at the second. Or, of every 5 bought, 3 will be at the first rate and 2 at the second. Hence—

No. bought at first rate = f of 120 = 72    \ Ar|R

No. bought at second rate = f of 120 = 48    )    '

Rule.—Find the deviation of the several rates from the average Then the comparative numbers at these rates will be in inverse proportion to the deviations-

Sum 2.—A settler bought 20 sheep for 20s. Rams at 5s. each, Lambs at 4d. each, and Ewes at Is. each. How many

OF EACH DID HE BUY ?

20 for 20s. = Is. each, the average cost.

Rams at 5s. each = 48d. each above the average.

Lambs at 4d. each — 8d. each below the average.

Ewes at Is. each = just the average cost.

Since the ewes are at the average, we have only to make the cost of the rams and the lambs, in shillings, coincide with the number of them bought; that is, we must find how many lambs at 8d. below the average cost must be bought for every ram at 48d. above the average cost. By the above rule more lambs will be required in the proportion of 48 to 8 (the reverse of the deviations), and these numbers are as 6 to 1.

Hence— Ans. 1 Ram, 6 Lambs, and 13 Ewes ; or 1 2 Rams, 12 Lambs, and 6 Ewes )

Sum 3.—A Grazier bought 100 Animals for £100, Cows at £5, Sheep at £1, and Goats at Is. How many of each sort did he

BUY?

100 head for £100 = £1 each, the average cost.

Cows at £5 each = 80s. above the average cost.

Sheep at £1 each = 20s., just the average cost.

Goats at Is. each = 19s. below the average cost.

Since a sheep costs the average, we have simply to make the price, in pounds, of the cows and goats, together, coincide with the number bought,—that is, to buy as many cows and goats as will cost an average of £1 each.

The cost of the cows differs from the average times as much as the price of the goats does; and, since as many goats must be bought at 19s. below as will reduce the cost of the cows by 80s, it is evident that a greater number of goats than cows must be bought in inverse proportion to the deviations from the average, or that 80 goats must be bought for every 19 cows. Hence, ans. 19 cows, 80 goats, and (100 - 99) or 1 sheep.

Again (by iv.) the number of goats x the deviation of their cost from the average — the number of cows x the deviation of their cost from the average. Hence, the number of goats x 19 = number of cows x 80 ; therefore, there must be 80 goats for every 19 cows.

Sum 4.—A man buys 100 Head for £100, Horses at £5, Cows at £1, Sheep at 5s. How many of each sort does he buy ?

£100 for 100 head is an average of £1 per head.

The price of the gows here coincides with the average cost.

Horses at £5 — 80s. above the average; while sheep at 5s. = 15s. below the average. Hence there must be 80 sheep for every 15 horses, or 16 for every 3.

Hence, Ans. 3 horses, 16 sheep, and

(100 - 19) or 81 cows.

or 6 ,,

32

99

(100 - 38) or 62 cows.

or 9 „

48

99

(100-57) or 43 cows.

or 12 „

64

99

(100 - 76) or 24 cows.

or 15 ,,

80

99

(100-95) or 5 cows.

It is thus seen that many questions of

this kind admit of several

answers.

Sum 5. —The attendance at a certain State School has increased during the year from 400 to 420. The average of the boys alone has decreased 8^ per cent. ; while that of the girls has increased 25 per cent. Find how many boys and girls respectively were in average attendance at the beginning of the year.

The average increas^jin the attendance is 20, or 5 per cent.

The difference between 5 p.c. increase and 8-| p.c. decrease is lSJp.c.

The difference between 5 p. c. increase and 25 p c. increase is 20 p. c.

The comparative number of boys at 13 j p.c. below the average and girls at 20 p.c. above the average, must be 20 boys for every 13^-girls, or, 3 boys for every 2 girls ; for the numbers required to produce a given average are to each other as the reverse of the deviations from the average.

Hence, there were f of 400, or 240 boys; and -§ of 400, or 160 girls.

Sum 6.—By selling a mixture of equal quantities of black and green tea at 2s. 9d. per lb. a grocer gains 10 per cent. If he sold the black alone at 2s. 4d. he would gain 12 per cent, on its prime cost. What was the prime cost per lb. of the green tea ?

Solution.—

The average cost per lb. of the mixture =    of 2s. 9d. = 30d.

The prime cost per lb. of the black tea —    0f 2s. 4d. — 25d.

2 lbs. of the mixture cost 60d., and 1 lb. of the 2 lb. cost 25d.

Hence, the other must have cost 60d. -25d. = 35d., or 2s. lid. Ahs.

EXAMPLES WORKED OUT.

1.    The pasturage of a certain field will feed 1125 sheep for 10$ weeks, whereas a flock of 1250 sheep consumed it in 9 weeks. How many sheep would the field graze regularly—the grass growing uniformly ?

Solution.—If 1250 sheep consumed the grass in 9 weeks, we find by Proportion that 1125 sheep should consume it in 10 weeks. But, by the question, the grass lasts 10$ weeks, or half-a-week longer.

The reason of this is, that the grass has been growing for 10^ weeks instead of 9 weeks. The growth of the extra week and a half has, thei'efore, supplied pasturage for the 1125 sheep for half-a-week. Hence, 3 weeks’ growth would last them for 1 week, or 1 week’s growth would last $ of 1125 for one week—i. e., the weekly growth will feed $ of 1125 sheep, or 375 sheep. Ans. 375 sheep.

2.    By selling a horse for £16 7s. which I had just bought for £13 10s., with 8 months’ credit, I gained 20 per cent. How long credit did I give, reckoning interest at 12 per cent, per annum ?

Solution.—The present value of £13 10s. due 8 months’ hence, at 12 per cent. =£12 10s. A profit of 20 per cent, upon this makes my cash selling price of this, and Lru of £12 10s. =£15.

£16 7s. - £15 = £1 7s., the interest charged upon £15.

We now require to find for how long £1 7s. is the interest upon £15 at 12 per cent, per annum, and therefore state—

As £12 : £1 7s. 11 1 year : 9 months. Answer.

£15 : £100.

3. If a rate of 2s. in the £ be struck upon 75 per cent, of the rent, and the rent and rates together form what is termed the estimated annual value ; find the ratable value of property of which the estimated annual value is £172.

Solution.—The ratable value is that upon which the rates are calculated—i.e., 75 per cent, of the rent.

The rates are ^ of ^ of the rent; = ^ of the rent.

The estimated annual value is the rent + the rates — rent + *o

of rent.    .

That is, the estimated annual value is 0f the rent.

Hence, the rent is £§ of the estimated annual value — of £172

And°the ratable value is 75 per cent, of this = TVo of £160 “


£120. Ans.

4. A person bought at Sydney 4 boxes of oranges, and sold them in Melbourne, after paying a duty of 20 per cent., at a loss of 15 per cent. Had he received 16s. more he would have gained 5 per cent, on his outlay. What did he pay for the oranges ?

Solution.—16s. makes a difference of 5 per cent, gain instead of 15 per cent, loss

Hence, 16s. = 20 per cent of total cost; V cost was 80s.; and the total cost being 120 per cent, of the first cost, the original cost was of 80s. = £3 6s. 8d. Answer.

5.    A farm is let for £90, and a certain number of tons of potatoes. When potatoes are at £5 a ton, the whole rent is 20 p. c. more than when they are at £4 a ton. Find the number of tons that formed part of the rent.

Solution.—£1 per ton increases the whole rent 20 p. c.

The whole rent = value of the potatoes at £4 per ton + £90.

Value at £1 per ton = of (value at £4 a ton + £90); or, value at £1 per ton = value at 16s. a ton + £18.

It takes 16s. of this extra £1 -per ton to increase the value of the potatoes by 20 p. c., thus leaving 4s. per ton to increase the fixed part of the rent—i. e., £90 by 20 p. c. Now, how many tons at 4s. will produce ffc of £90, or 360s ?

360 -r- 4 = 90 tons. Answer

Or, by an equation—

The rent in the first case = £5 X number of tons + £90.

The rent in the second case = £4 X number of tons + £90.

The difference, or £1X No. of tons = -5^ of (£4 X No. of tons + £90) ; i. e., £1 X number of tons = | of £4 X number of tons + | of £90 ; or, 20s. X number of tons = 16s. x number of tons + 360s.

Subtracting 16s. x number of tons from both sides of the equation, 4s. x number of tons = 360s. Hence number of tons = 360-f 4 = 90. Ans.

6.    A square paddock enclosed with a three-rail fence contains as many acres as there are rails in the fence. Supposing the rails to be 9 feet long, how many acres does the paddock contain ?

Solution.—Since the 3 rails in each panel extend 3 yards, each rail will represent 1 yard of fencing. For each yard in the fence of one side, there are four yards in the whole fence ; so that for each yard in the whole fence to represent an acre there must be four acres for each yard in the fence on one side. An acre of 1 yard in width must be 4840 yards in length; therefore, for every yard to represent 4 acres the length must be four times 4$40 yards = 19360 yards. Ana that the paddock may be square, the fence on each side must be 19360 yards long. Hence, the required area is 19360 yards X 19360 yards = 77440 acres. Ans.

7.    A father and his son, by working together, can accomplish a piece of work in 48 days. The father works 8 hours a day, the son 9 hours. The father can do in 5 hours as much as the son can do in 6 hours. In what time could the father alone do the work ?

Solution.—The father can do 6 parts in the same time that the son takes to do 5 such parts ; and since the father works 8 hours a day, and the son 9 hours, the father will do 8 times 6, or 48 parts a day, to the son’s 5 times 9, or 45 such parts. Therefore, of every (48 + 45), or 93 parts done by both, the father does 48. And since they together do fg 0f the work per day, the father alone can do of Jg., or J-g- of the work per day. Hence, he will take 93 days. Ans.

8. Two equal tanks are full of water. No. 1 has a tap which will empty it in 11 hrs. ; and No, 2, a tap which will empty it in 12 hrs. If both taps be opened together, when will No. 2 have four times as much water left in it as there is in No. 1 ?

Solution.—In the required time, 12 parts run out of No. 1 for 11 such parts that run out of No. 2 ; and, by question, there are four parts remaining in No. 2 for one such part left in No. 1. That is, taking the water left in No. 1 as one part, there are 3 more such parts left in No. 2 than in No. 1. Now, these 3 extra parts must fill the space representing the one part less that was emptied from No. 2. The part remaining in No. 1 is, therefore, one-third as large as one of the 11 parts which have been emptied out of No. 2, or one of the 12 parts emptied from No. 1. Hence, if the whole contents of either tank be divided into 12£ equal parts, the required time is that in which 12 such parts will run out of No. 1, or 11 out of No. 2.


Hence, Ans. = ^ of 11 hrs., or ^ 0f 12 hrs. = 10ff hours.

By an equation—

Let C represent the whole contents of either tank.

Then, T*r C x required number hours = quantity which runs out of No. 1.    .

And, iC x required number hours = quantity which runs out of No. 2.

C - water run out of No. 1 = \ (C - water run out of ISo. 2).

Or C - xt C x number of hours = | of (C - T\ C x number of hours).    '

C - -j-L- C x number of hours = 4 C — ^ C x number of hours, f C = (xt - C x numher of hours = ffg c x number of hours.

Hence, required number of hours = f of    = 10ff hours.

Ans.

9. In Lake Terang I observed a bulrush, the head of which, seven inches in length, was just above the surface of the water. Reaching it with my walking-stick, I drew it towards me, and found that it submerged at a distance of 35 inches from the place where it grew. What was the depth of the lake in that part ?

Solution —Suppose a circle drawn, having for its radius the entire length of the rush. Then, twice the length of the rush gives us the diameter of this circle ; while the distance between the two points of submersion (if the rush were first pulled in one direction and then in the opposite) is a chord of the same circle.

Now, by Euclid, Book III., Prop. XXXV.—If two chords of a circle cut one another, the 'product of the two parts of the one is equal to the product of the tivo parts of the other.

So that the 'product of the known part of the diameter, 7 inches, by the unknown part, is equal to the product of the two parts of the other chordi. e., to (35 x 35) in. Hence, the unknown part of the diameter is (35 x 35 -f 7) in. = 175 in., to which, if 7 in. be added, the sum will be the whole diameter, one-half of which is the radius— i. e., the length of the rush. Hence, the length of the rush is (175 + 7) ■— 2 = 91 in. But the depth of the lake is 7 in. less than the length of the rush. Hence, Depth of lake == 91 in. - 7 in. = 84 in. = 7 ft. Ans.

10.    A block of land was sold to A at a certain loss per cent, upon the prime cost. A sold it to B at the same losing rate ; but B, by selling it to C at 21 per cent profit, received the original cost of the land. At what losing rate per cent, was the land sold to A and B respectively ?

Solution.—A gave for the land a certain fraction of the prime cost (say, for short, fraction of cost). B gave the same fraction of this for it (i.e., fraction of fraction of first cost); and he sold it to for of this, and, by so doing, received the original cost.

Hence, huu °f fraction of fraction of first cost = the prime cost fraction x fraction of prime cost =    of prime cost—i.e., the square

of this fraction = of prime cost. Hence, the fraction of the prime cost at which it was sold to A = v/lf or tt > so that was lost; and = 9-jh: per cent. Ans.

11.    A sum of £204 is divided amongst A, B, C, andD. A gets 5s for 4s. B gets ; B 3s. for 4s. that C. gets, and C gets 5s. for 2s. 6d. that D gets : how much does each receive ?

Solution. A    B    C    D

5s.    4s.    :    :

3s.    4s.    :

5s.    2^s.

A gets 5s. to B’s 4s., or to C’s | of 4s., or to D’s ^ of f of 2^s. ; or A gets 5s. to B’s 4s., or C’s 5s. 4d., or D’s 2s. Sd.

Hence, sum of parts = 5s. + 4s. -f 5s. 4d. + 2s. 8d. = 17s.

And A gets -fr of £204 = £60 ; B, of £204 = £48 ; C, 5s. 4d. of every 1/s , or £64 ; and D, 2s. 8d. of every 17s., or £32. Ans.

12.    There are three merchants in partnership, and their stock is worth £344. A’s money has been in the business 5 months, B’s 6 months, and C’s 9 months. The gains have been distributed equally. What stock did each invest in the business ?

Solution.—Since each received an equal share of the profits, the comparative amounts of the sums invested must be to each other in inverse ratio to the several times.

Hence, since A’s money has been in use 5, B’s 6, and C’s 9 months, the sums they put in must be to each other as }, |, and |; or as 18, 15, and 10.

Sum of Parts = 18 + 15 + 10 = 43.

A put in of £344 = £144 ; B, of £344 = £120 : and C of £344 = £80. Answer.

Exercise XLI.—MISCELLANEOUS QUESTIONS.— FROM TEACHERS’ EXAMINATION PAPERS.

1.    The sum of two numbers is one hundred and twenty thousand and ninety, and one of them is half as large again as the other. Find their product.

2.    Find the amount of 690087905 repeated eight hundred and ninety thousand and six hundred times.

3.    Find by practice the value of 109ozs. lldwt. 13J grs. of gold at £3 15s. 10|d. per oz.

4.    A and B rent a field between them, A paying half as much again

as B. A keeps in it 16 oxen for 25 days : how many horses may B keep in it for 20 days, supposing 3 horses to consume as much as 5 oxen ?    '

5.    In an infant school, where the ages range from 3 to 7 years, £ of the children are 3 years old, \ are 4 years old, £ are 5 years old, § of the remainder are 6 years old, and there are 13 children 7 years old. Find the number at each age.

6.    Seventy-two men had provisions for 35 days, but after 5 days 20 men left : how long did the provisions then last the number left ?

7.    A fixed rent of £561 is converted into a com rent of two-thirds

wheat and one-third oats, at the average price of 8s. 6d. per bushel for the wheat and 5s. 6d. per bushel for the oats. To what does the rent amount when the price of wheat has fallen to 7s. 6d. per bushel and that of oats has risen to 6s. 6d. per bushel ?    i

8.    A rate of Is. 8d. in the £ is struck upon 90 per cent, of the •rent, and the rent and rates together form what is called the estimated annual value. Find the ratable value of property, the estimated annual value of which is £516.

9.    An estate is bought at 13£ years’ purchase,, what interest does the purchaser gain upon his outlay ?

10.    How many years’ rental should be paid for a house that die investment may yield 15 per cent, interest upon the outlay?

] 1 For what sum should a cargo worth £3412 be insured at of per cent., so that the owner, in case of loss, may recover the value

of both cargo and premium ?    , u .

12 What ready money ought to be paid for a debt oi £1404 10s, due two years hence, allowing 6 per cent, per annum comp. mi. ?

13. There are three merchants in partnership, and their stock is worth £17200. A’s money has been invested in the business 5 months, B’s 6 months, and C’s 9 months. The gams have amounted to £3600, and these have been distributed equally. What stock did each invest in the business? Note. See hum 12,

^a?4 *After a certain number has been repeated as an addend 10013 times, and the sum subtracted from 11090301, the remainder is 7015010. Find the number.

15.    A flock of sheep increases at the rate of 75 per cent, annually. If a grazier buys a selection stocked with 1600 sheep : how many will his flocks number at the end of three years, supposing him to sell 160 sheep at the end of each year ?

16.    A ship leaves port with sufficient provisions for a voyage of 14 weeks’ duration, six sailors having absconded when the anchor was heaved, the supplies held out for 16 weeks. Of how many did the full crew consist ?

17.    47 qrs. 3 bus. 3£ pks. of wheat at £2 16s. 4d. per quarter.

18.    A sum of £68 is divided amongst A, B, C, and D. A gets 5s. for 4s. B gets, B gets 3s. for 4s. that C gets, and C gets 5s. for 2s. 6d. that D gets. How much does each receive ?

19.    Find the value of 207 ozs. 9 dwts. 10 grs. of a metal composed of equal weights of gold and silver, the gold being worth £3 16s. 9d. per oz., and the silver £3 6s. per lb.

20.    Six square miles of land are to be divided among 98 persons, 40 of them are to receive the same quantity, and the remaining 58 are each to have 2 a. 1 r. 24 p. less than each of the others. How much will each receive ?

21.    Find the difference between the simple and compound interest on £750 for 2^ yrs. at 4 per cent.

22.    A grocer bought 2 cwt. of tea. He sold 80 lbs. at 3s. per lb., and found that he had gained 12^ per cent. At what rate must he sell the remainder so as to clear 15 per cent, on the whole ?

23.    Find the square of 9058070.

24.    Find the product of the sum and difference of 40008072, and 12067100, and show that it equals the difference ©f their squares.

25.    A school with an attendance of 529 has increased during each of the last two years at the rate of 15 per cent. : what was the attendance two years ago ?

26.    If 3 men can reap 25 acres in 10 days, and the work of 2 men

be equal to that of 5 boys, how many boys must be employed to assist 4 men in reaping 52 acres, so that the work may be completed in 12 days ?    .

27.    If the lining of a room with paper If ft. wide at 5|d. per yd. cost £2 9s. 4d. ; what would it have cost to line it with paper £ yd. wide at 7|d. per yd. ?

28.    One map is drawn upon a scale of 5 in. to a mile, and another upon the scale of 3 in. to a mile. How many sq. in. of the former will represent an area represented by 135 sq. in. of the latter ?

29.    A possessed f- of a vessel and sold f of £ of his share for £360. What was the value of the vessel ?

' 30. A room is 16 ft. long, 14 ft. wide, and 9^ ft. high. Whart will it cost to cover the walls with paper at Is. 9d. a piece, each piece of paper being 20 ft. long and 21 in. wide ?

31.    Find the value of 2‘ 145 of 5s. 8|d. ?

32. Find the value of 17 lbs. 13 dwt. 14 grs. at £3 18s. 2d. per oz.

33. A gallon contains 277‘274 c. in. How many gallons will a tank hold whose top is 8 ft. 7 in. square, and its depth 10 ft. 4 in. ?

34.    A rectangular plot of ground contains 11 ac. 2 ro. 32 po., and its breadth is found to be 10 chains 40 links. What is its length ?

35.    A vessel from which guns are being fired at intervals of 1 min. is approaching the land at such a rate that the reports of the guns are heard by the people on shore at intervals of 59 seconds. If sound travels 1093 ft. per sec., find at what rate per hr. the vessel is approaching the land.

36.    Find the value of 299 qrs. 5 bus. at £20 14s. per qr.

37.    A lends B £283 10s from 1st June to 30th September: how long should B lend A 330 guineas to requite the loan ?

38.    69 cwt. 151bs. 10 oz. at £3 14s. lid. per cwt.

39.    If when malt is at £3 10s. per qr. beer brewed with 3^ bus. to the barrel be sold at Is. 9d. per gal. : how much per gal. should be paid for beer brewed with 5 bus. of malt to the barrel, when malt is at £2 12s. 6d. per qr. ?

40.    Find the continued product of seven millions eight hundred thousand and ninety, fifty-four thousand and sixty-nine, and nine hundred and thirty thousand seven hundred.

41.    What number divided by 709530 gives a quotient 906007 and a remainder 503298 ?

42.    If a crop of wheat raised from one mile square of land gives a profit of £2 3s. 6d. per acre after paying the following expenses, viz. :—fencing, £1 5s. per chain ; grubbing, £2 Is. 3d. per acre; ploughing, 7s. 6d. per rood ; all other expenses, £75 12s. 9d. For how much did the whole crop sell ?

43.    317 sq. yds. 6 sq. ft. 54 sq. in. at £7 11s. 7^d. per sq. yd.

44.    If it takes 21 men 5 weeks to clear 350 acres working 5 days a week for 8 hours a day ; how many weeks will one-third of them be in clearing 245 acres, if they work 7 hours a day and 6 days a week?

45.    A person who owned -f- of a ship sold § of § of his share for u. £2083^. Find the value of 24 4- 3^ of his remaining interest at the

same rate.

46.    Find the cost of post and rails necessary for the erection of a two-rail fence around an allotment having a frontage of 33 ch. 25 Iks. and a depth of 42 ch. 75 Iks. The rails are 8 ft. 6 in. long, but wheu erected they overlap one another to such an extent that the length of one rail out of every 34 is lost. The posts and rails cost £3 0s. 5d. per 100 with the exception of the corner posts, each of which costs as much as two other posts.

47.    Find the cost of post and rails necessary for the erection around an allotment having a frontage of 38 ch. 25 Iks. and a depth of 51 ch. 75 Iks. The rails are 8 ft. 6 in long, but when erected they overlap one another to such an extent that the length of one rail out of every 34 is lost. The post and rails cost £2 19s. 7d. per lOOj with the exception of the corner posts, each of which costs as much as two other posts.

48.    I bought a house, and agreed to pay for it the sum of £504 at the end of six months. Instead of carrying out his agreement, the seller asked me to pay £360 cash. What length of time should he allow me to pay the balance, supposing money worth 10 p.c. per ann.?

ANSWERS

ANSWERS TO EXERCISE I.—SIMPLE RULES.

Page 43.

1. Units, hundreds, millions, tens of millions, billions.

2. 1000067000509.    3. 10,000,100 and 1010000; Product

98981900010000.    4. 4897478703226684 (Numerate). 5. Sum

865436009 (Numerate) ; Difference 745364009 (Numerate) ; Prod. 48352994940324000 (Numerate). 6. 500890.    7. 10111021110000

(Numerate). 8. 36090050780 (Numerate). 9. 7600940.    10. Words.

11.    7280 and 1284 Remainder.    12. 6370 and 23529 Rem.

13. 9568 and 700690-    14. 10110000001021110000 (Numerate).

15. Prod. 324199298131263 (Numerate); Sum 65436816 ; Diff. 54636798.    16. Quotient 59600077 (Numerate) ; Rem. 3997307.

17.    284750605 (Numerate). 18. Quot. 202047 (Numerate) ; Rem.

43396 (Numerate). 19. Words. 20. 4005.    21. 1346 and 29 Rem.

22. 349 and 46 Rem. 23. 5078; 1503.    24. 441719913985918200

(Numerate). 25. Words. 26. 556513698890244460 (Numerate). 27. 7861 times. 28. 34960078 (Numerate). 29. 50304007.

30. 78501.    31. 9700980.    32. 48665686179840000 (Numerate).

33. 1040884057650 (Numerate). 34. 96079999999. 35. 7042 times; 8780S2.    36. 186543480344283 (Numerate), 37. 5468438481300.

38. 96079.    39. 601305370 (Numerate). 40. 205190049 (Numerate).

41.    2406328165935.    42. The numbers 2090830 and 896070 ; Prod.

1873530038100.    43. The numbers 769 and 61589210 ; Quot. 80090.

44. 6800790.    45. 59201284846176.    46. 7009.    47. 46094300.

48. 1129.    49. 397484003746420 (Numerate). 50. Sum 11001010 ;

Diff. 9000990 ; Prod, of Sum and Diff. 99019980999900. 51. 5352885398597 (Numerate). 52. 601058718461334 (Numerate).

ANSWERS TO EXERCISE II.—SIMPLE RULES.

Page 50.

1. 38483.    2.    111436.    3. 900970.    4. 15054.    5.    5484700.

6. 62391.    7. 899095.    8. 700991.    9. 3490.    10. 390.    11. 8100.

12.    3000410. 13. 203612. 14. 5970. 15. 607. 16. 11091. 17. 54.

18. 89.    19. 100.    20. 50.    21. 20209.    22. 4616.    23. 56796

24. 16630. 25. 161928. 26. 16180. 27. 5394. 28. 148. 29. 365 30. 35210.    31. 1162.    32. 46712.    33. 141968.    34. 200990

35. 123432.    36. 3332902607300 (Numerate).    37. 745833

38. 468441.    39.    23316.    40. 500085 (Numerate).    41.    5080007.

42.    310094.    43.    4800005.    44. 9070006.    45. 10128    and    55 Rem.

46. 54, 35.    47.    3030824947990.    48. 17.    49. 331.    50. 2203.

51. 1409 and 1239.    52. 217789.    53. 268608.    54. 1189

55. 496217 and 491436. 56. 1588065. 57. 987. 58. 2908 and 546. 59. 5760 and 1760. 60. 16, 48, 144. 61. 17, 26, and 35. 62. 6132.

lft

63. 569608888896009 (Numerate). 64. 2352.    65. 5280.    66. 19.

67. 804 and 742609 Rem. 68. 19169.    69. 1061.    70. 3407.

71. 578340.    72. 540500.    73. 99.    74. 229500.    75. 78921.

76. 614.    77. 981.    78. 132.    79. 648.    80. 112.    81. 345.

82.116.    83. 1403243043040000 (Numerate).    84. Five.

85. 24400489920.    86. 2985984.    87. 48627125.    88. 2800733.

89. 100.    90. 86400.    91. 119 ; and by 7.    92. 3509 and 63 Rem.

93. 261 and 36 Rem. 94. 90060.    95. 10010961 (Numerate).

96. 35721. 97. 7000 & 5760.    98. 108 & 16. 99. 43 yrs. and 8 yrs.

100.    Sum 450723635 (Numerate) ; Diff. 450722365 (Numerate) ; Prod. 286209105000 (Numerate) ; Quot. 709800 (Numerate).

101.    4999999995.    102. 222222222.    103. 1117711177.

104. 79 and 7663.    105. 19 and 95.    106. 97 and 679.

ANSWERS TO EXERCISE III.—COMPOUND ADDITION. Page 84.

1. £154 8s. 9d.

4. £306 Is. Ofd.

7. £2631 18s. 3£d.

10. £3692 16s. lfd.

13. £3673 18s. 4fd.

16. £3223 Is. 9d.

19. £29086 11s.

22. £27440 15s. 5^d. 25. £35135 17s. 9fd.

28. £2809 18s. 8èd.


2. £85 4s. lid.

5. £353 11s. 7id.

8. £3151 16s. 9fd.

11. £32093 14s. 2fd.

14. £3117 11s. 3d.

17. £2988 7s. 6d.

20. £322 7s. 10fd.

23. £3634 11s. 8£d.

26. £26577 10s. 6fd.


3. £346 3s. 7id.

6. £260 Ils. 5±d.

9. £3234 10s. O^d.

12. £31622 19s. 2d.

15. £3467 Is. lfd.

18. £2088 15s. 11 ‘d.

21. £20513 14s. Rd.

24. £3637 19s. 8d.

27. £37096 16s. 3fd.


ANSWERS TO EXERCISE IY.—WEIGHTS AND MEASURES. Page 85.

1.

174 lbs. 5 oz. 3 drs.

2.

71 cwt. 1 qr. 1 lb.

3.

151 tons 1 cwt. 1 qr.

4.

3 cwt. 1 lb. 11 oz.

5.

65 cwt. 3 qrs. 19 lbs.

6.

119 cwt. 3 qrs. 1 lb.

7.

98 lbs. 2 ozs. 3 drs.

8.

185 tons 6 cwt. 57 lbs.

9.

84 lbs. 1 oz. 0 dwts.

10.

78 lbs. 9 oz. 14 dwts.

11.

33 oz. 3 dwts. 10 grs.

12.

30 oz. 3 dwts. 21 grs.

13.

21 drs. 1 scr. 4 grs.

14.

26 drs. 1 scr. 14 grs.

15.

21 oz. 4 drs. 2 scr.

16.

37 oz. 4 drs. 2 scr.

17.

64 lbs. 0 oz. 4 dwts.

18.

37 lbs. 9 oz. 6 dwts.

19.

129 lbs. 10 oz. 3 dwts.

20.

79 oz. 3 dwts. 13 grs.

21.

70 cwt. 0 qrs. 19 lbs.

22.

59 cwt. 0 qrs. 11 lbs.

23.

54 cwt. 1 qr. 3 lbs.

24.

212 tons 7 cwt. 0 qrs.

25.

92 yds. 1 ft. 3 in.

26.

41 yds. 1 ft. 11 in.

27.

56 yds. 2 ft. 11 in.

28.

74 yds. 1 ft. 5 in.

29.

2526 tons 1 cwt. 0 qrs. 8

lbs.

30.

2978 tons 16 cwt. 0 qrs. 8 lbs.

31.

2211 tons 6 cwt. 3 qrs. 21

lbs.

32.

3052 lbs. 3 oz. 5 dwts. 20 grs.

33.

2491 lbs. 4 oz. 6 dwts. 12

grs.

34.

2867 lbs. 10 oz. 4 dwts. 9 grs,

35.

161 lbs. 10 oz. 6 dwts. 16

grs.

36.

235 lbs. 7 oz. 12 dwts. 12 grs.

37. 279 lbs. 8 oz. 4 dwts. 8 grs.

39. 29 fur. 3 per. 0 yds. 1 ft. 6 in.

41. 2185 ac. 2 r. 7 per. 22 sq.yds.

43. 103 per. 20| s. yds. 6 ft. 128 in.

45. 154 qrs. 3 bus. 3 pks. 0 gals. 47. 169 c. yds. 24 c. ft. 500 c. in.

49. 268 c. yds. 24 c. ft. 1579 c. in.

51. 59 wks. 4 dys. 21 hrs. 49 m.


38. 167 mis. 3 fur. 32 per. 2f yds.

40. 133 per. If yds. 1 ft. 8 in.

42. 18S3ac. Or. 18p. 13isq.yds.

44. 101 gal. 3 qts. 0 pts. 0 gills.

46. 123 bus. 0 pks. 1 gal. 1 qt.

48. 210 c. yds. 17 c. ft. 197 c. in.

50. 112 wks. 3 dys. 20 hrs. 15 m.

52. 75 dys. 21 hrs. 13 m. 47 sec.


ANSWERS TO EXERCISE V.—COMPOUND SUBTRACTION. Page 87.

1. £1131 2s. 3d.

4. £2 10s. 4fd.

7. £170 12s. 4fd.

10. £1798 15s. lfd.

13. £324 Os. Od.

16. £155 14s.

19. £201 13s. Ofd.

22. £81 16s. Ofd.

25. £349 19s. lfd.

28. £52 18s. lOfd.

31. £28 3s. Ofd.

34. £509 19s. llfd. 37. £258 17s. 2fd.

40. £61 18s. 9fd.

43. £91 Is. 7d.

46. £491 19s. llfd.

49. £175 Os. 3d.

52. £12 6s. 3fd.


2.

£46 Is. Ofd.

5.

£233 Is. llfd.

8.

£28 Os. Od.

11.

£36 19s. lOfd.

14.

£10 10s. Ofd.

17.

£30 13s. Ofd.

20.

14s. lfd.

23.

£74 19s. 5|d.

26.

£153 18s. llfd.

29.

£602 Is. 8fd.

32.

£83 18s. lOfd.

35.

£5 2s. Ofd.

38.

£118 16s'. 9|d.

4L

£75 19s. Ofd.

44.

£503 18s. Ofd.

47.

£489 18s. 11RL.

50.

£134 2s. lfd.


3.

£152S Os. 5d.

6.

£5484 14s. 2fd.

9.

£1766 5s. 3fd.

12.

£73 13s. 7d.

15.

£1293 11s. lOfd.

18.

£2 3s. 7fd.

21.

£100 4s. 8Id.

24.

£511 18s. llid.

27.

£419 11s. lOfd.

30.

£19 18s. lfd.

33.

£112 17s. lOfd.

36.

£333 2s. llfd.

39.

£3S2 7s. lOfd.

42.

£1385 3s. lOd.

45.

£150 Os. llfd.

48.

£111 15s. 9fd.

51.

£28 8s. 0|d.


ANSWERS TO EXERCISE YI.—WEIGHTS AND MEASURES. Page 89.

1. 1 lb. 15 ozs. 3 drs. 2. 1 cwt. 3 qrs. 3 lbs. 3. 2 lbs. 3 ozs.

4. 3 lbs. 14 ozs. 12 drs. 5. 1 cwt. 1 qr. 16 lbs. 6. 56tonsl9cwt. 2qrs. 7. 25 lbs. 14 ozs. S. 17 tons 16 cwt. 2 lbs. 9. 12 lbs. 10 ozs. 19 dwts.

10. 9 ozs. 3 dwts. 21 grs. II. 14 lbs. 10 ozs. 14 dwts. 12. 3 ozs. 18 dwts. 23grs. 13. 4 drs. 0 scrs. 15grs. 14. 1 lb. 6 ozs. 5 drs. 15. 1 lb. 3 drs. 2 scrs. 16. 4 drs. 1 scr. 16 grs. 17. 1 yard 1 foot 3 in.

18. 17 miles 1 fur. 35 poles. 19. 3 fur. 31 per. 4f yds. 20. 7 per. 5 yds. 0 feet. 21. 18 acres 0 roods 32 per. 22. 4acres 2 roods 27 per.

23. 1 rood 29 per. 271 sq. yards. 24. 9 per. 29f yards 7 feet.

25. 14 yards 6 feet 136 sq. in.    26. 2 per. 29f yards 8 feet.

27. 5 yards 3 feet 97 sq. in. 28. 20 acres 0 roods 39 per. 271 yards.

29. 20 cub. yds. 2 ft. 1635 in. 30. 310 cub. yds. 2 cub. ft. 851 cub. in.

31. 18 cub. yards 22 cub. feet 11 in. 32. 42 cub. yards 22 cub. feet 1727 cub. in. 33. 1 gall. 3 quarts 0 pints. 34. 1 pint 3 gills.

35. 11    bus. 2 pks. 1 gall.    36.    6 pks. 0 galls. 3 qts. 37. 1 qr.

7 bus.    1 pk.    3S. 9 bus. 3    pks. 1    gall. 39. 9 bars. 27 galls. 3 qts.

40. 4 qrs. 5 bush. 3 pks. 41.    87 days 19 hours 58 min.

42. 4 hours 49 min. 50 secs. 43. 4 days 1 hour 59 min. 56 sec.

44. 1 week 1 day 22 hours. 45. 10° 14' 54".    46. 7° 45' 37".

47. 3 loads 4 qrs. 7 bus. 2 galls. 48. 4 qrs. 2 bus. 3 galls.

49. 17    miles    74 chains 75    links.    50. 29 chains 20 yards 2 feet.

51. 39    acres    8 chains 1287    links.    52. 236 acres 3 chains 9881 links.

ANSWERS TO EXERCISE YII.—COMPOUND MULTIPLICATION. Page 90.

1. £24 13s 4d, £49 6s 8d., £74, £98 13s. 4d., £123 6s. 8d., £148

2. £251 13s. 9fd., £419 9s.    8Jd.,    £587    5s.    6fd.,    £755    Is. 5£d.,

£922 17s. 3|d. 3.    £2923 18s.    4d.,    £3654    17s.    lid.,    £4385    17s.    6d.

£4020 7s. 8^d. 4. £172 10s. 5d., £197 3s. 4d., £221 16s. 3d., £246 9s. 2d.    5.    £427 13s.    6d.,    £481    2s.    8^d.,    £498    19s.    Id.,

£534 11s. 10^d.    6. £760    14s.,    £832    Os.    3Jd.,    £855    15s.    9d.,

£950 17s. 6d.    7. £7389 5s. 9d., £7741 3s. 2d., £7917 Is. 10id.,

£8620 16s. 8M. 8. £13030 10s. 5d., £14072 19s. 3d., £14333 11s 5U, £14594 3s. 8d.    9. £49659 18s. 9d., £52142 18s. 8pl., £52970 12s.,

£54625 18s. 7£d.    10. £10555 11s. 3d., £10857 3s., £11611 2s. 4^d.,

£12063 10s.    11. £24371 5s. lljd., £25273 18s. 9d., £26477 9s. 2d.,

£27079 4s. 4£d.    12. £1919 18s., £1959 17s. ll£d., £1979 17s. ll|d.,

£1999 17s. lid. 13. £179 4s. 10£d., £200 6s. 7£d., £305 15s. 4kl., £326 17s. ljd. 14. £426 17s. 5fd., £544 12s. 7fd., £839 0s. 6fd., £1427 16s. 4|d.    15. £410 Is. Ud., £599 6s. 7*d., £914 15s. 4|d.,

£1545 12s. lO^d. 16. £2845 Is. 0|d., £4824 4s. 44d., £7298 3s. 6fd„ £9772    2s. 8£d.    17. £307041 17s. 9fd., £333576 7s. 6d.,

£125091 2s. 9|d.    18. £20492 3s., £30738 4s. 6d., £43911 15s.

19. £401924 9s. 10^d., £320004 3s. 4kl. 20. £2896994 19s. 2d., £1593934 2s. 9|d.    21. £2745243 17s. 6d., £40423442 2s. Cd.

22. £27854228 3s. lO^d., £181317192 7s. ljd. 23. £110257741 2s. 3pl, £256165^35 17s. 6£d. 24. £3037779382 7s. 2|d., £2745850893 17s.9|d.

25. £33587736, £31754486 8s. 26. £33059 15s. 2d., £33377 12s. 9|d., £34331 5s. 9d. 27. £78319 15s. 5d., £79743 15s. 4d , £85439 15s.

28.    £119540    11s.    3Jd.,    £122504    7s.    7d.,    £123492    6s.    4*d.

29.    £240350    10s.    3d.,    £244165    12s.,    £247980    13s.    9d.

30.    £357224 17s. 3d., £378874 17s. Id., £389699 17s.

ANSWERS TO EXERCISE VIII.— MULTIPLICATION, USING THREE FACTORS. Page 91.

1    £26412    Is.    10^d.,    £27166    14s.    6d.,    £27669    16s.    3d.

2    £200569    8s.    6d.,    £210120    7s.,    £222854    18s.    4d.

3    £354385    15s.    10d.,    £389824    7s.    5d.,    £398684    Os.    3fd.

4    £2142930 7s. 5^d., £2361596 14s. 9d., £2474053 14s. 6d.

5    £3395491    13s.,    £3466231 Is. 2RL, £3631289 13s. 7U

6    £1207295    9s.    4d.,    £1188431    9s.    6d.,    £1358207    8s.

7. £5293831 9s. 10d., £5568835 14s. 6d., £5775088 18s.

8.    £5486977 19s. 4Jd., £5780520, £5961161 5s. 9. £2873573 2s.

£2817032    17s.    9£l.,    £2963372    5s.    2Jd.    10.    £9126195    18s. 6d.,

£9175580    2s.    OJd.,    £9600284    Os.    6d.    11.    £2622727    Is. 6fd.,

£2997402 7s. 6d., £3179063 2s. 6d.    12. £6249992 16s. 9^d.,

£9679988 18s. 2d., £9899988 13s. l^d. 13. £4339908 7s. 3|d., £3429063    8s.,    £6199869 Is.    lOJd.    14.    £6749215    4s.    7d.,

£7713388    16s.    8d.,    £5153946    3s.    6d.    15.    £1769250    5s.    3d.,

£2471333 14s., £3089167 2s. 6d.    16.    £6273663 4s. lid.,

£7667810 11s. 8£d., £9954903 12s.

ANSWERS TO EXERCISE IX.—ANSWERS TO MULTIPLICATION BY FACTORS. Page 91.

Extended Multiplication Tables.

1. £3065 0s. 5d., £4597 10s. 7£d., £6130 0s. 10d., £7662 11s. Oid.

2.    £37113 2s. 74d., £43298 13s. Ofd., £49484 3s. 6d., £55669 13s. llld.

3.    £9176 Is. lid., £13764 2s. 10£d., £18352 3s. 10d., £22940 4s. 9£d.

4.    £22929 14s. 6d., £26751 6s. lid., £30572 19s. 4d., £34394 11s. 9d.

5.    £21778 Is. lOgd., £32667 2s. 9fd., £43556 3s. 9d., £54445 4s. 8fd.

6.    £21215 Is. 3d., £24750 18s. l±d., £28286 15s., £31822 11s. 10£d.

7.    £20031 11s. 4d., £30047 7s., £40063 2s. 8d., £50078 18s. 4d.

8.    £68916 10s., £80402 11s. 8d., £91888 13s. 4d., £103374 15s.

9.    £31580 13s. 9d., £47371 Os. 7^d., £63161 7s. 6d., £78951 14s. 4\d.

10.    £44639 6s. 10^d., £52079 4s. 8RL, £59519 2s. 6d., £66959 Os. 3fd.

11.    £22565 3s. 6d., £33847 15s. 3d., £45130 7s., £56412 18s. 9d.

12.    £44834 19s. 3d., £52307 9s. IJd., £59779 19s., £67252 8s. 10$d.

13.    £237799 8s. 2Jd., £35669 2s. 3fd., £47558 16s. 5d., £59448 10s. 6|d.

14.    £26554 3s. 3d., £30979 17s. ljd., £35405 11s., £39831 4s. 104d.

15.    £115560 13s. ljd., £151117 15s. 7£d., £168896 6s. lOAd., £176007 15s. 4gd.

ANSWERS TO EXERCISE X.—COMPOUND MULTIPLICATION. Page 92.

Weights and Measures.

1. 190 cwt. 1 qr. 4lbs. 2. 31 tons 15 cwt. lqr. 3. 206lbs. 15ozs.

4. 22 tons 1 cwt. 2 qrs. 2 lbs. 5. 1 ton 9 cwt. 3 qrs. 2 lbs. 3 ozs. 8 drs.

6. 388 tons 15 cwt. 1 qr. 7. 700 lbs. 10 ozs. 8dwts. 8. 391 lbs. 3 ozs. 6dwt. 9. 590 lbs. 11 ozs. 9 dwt. 4grs. 10. 110lbs. 10 ozs. 14dwt. 23grs. 11. 538 lbs. 12. 1040 lbs. 9 ozs. 4 drs.' 2 scr.

13. 3876 yards. 14. 29 mi. 7 fur. 9 per. 1yd. 2 ft. 10 in. 15. 14200 miles. 16. 673 mi. 7 fur. 11 per. 2^ yds. 17. 20968 acres 3 roods. 18. 13409ac. 3ro. 12per. 19. 41 ac. 2ro. 22per. 26£sq. yds.

20. 1 ac. 2ro. 39per. 26jsq. yds. 5sq. ft. 119sq. in. 21. 1226 c. yards 1 c. ft. 1296 c. in. 22. 3753 c. yds. 3 c. ft. 163 c. in. 23. 2964 Eng. ells 3 qrs. 3 na. 24. 3528 Fr. ells 4 qrs. 2 na. £ in.

25. 4213 FI. ells 2 na. 26. 1009S Eng. ells. 27. 17856 ac. 6 sq. ch. 28. 14632 ac. 3sq. ch. 5200 sq. links. 29. 259051eagues.    30. 104809

sq. miles 240 acres. 31. 13600 gallons 2 qts. 1 pt. 32. 1007 qrs. 7 bus. 2 gals. 33. 37893 qrs. 7 bus. 2pks. 34. 4663 bus. 1 pk. 1 qt.

35. 8200 yds. 2 qrs. 36. 3118 gals. 2 qts. 37. 34 loads 5 c. feet.

38. 38loads24c. ft. 39. llSlbs. 11 ozs. 6 drs. 0 scr. 1 gr. 40. 359lbs. 3 ozs. 3 drs. 1 scr. 41. 9623 hdds. 31 gals. 2 qts. 42. 3131 hdds. 53 gallons. 43. 1489 bar. 17 gals. 3 qts. 44. 1990 pipes 44 gallons.

45. £15779 1 A. 6 c. 9 mils. 46. £29815 5 fl. 2c. lmil. 47. 399 days 18 hours. 48. Ill days 10 hrs. 33 min. 36 sec. 49. 80° 56'15".

50. 189° 6'.    51. 802 gals. 2 pts. 12 fl. ozs. 52. 736 fl. ozs. 5 fl. drs. 20

minims. 53. 1203 ac. 3 ro. 24 per. 6 yds. 54. 1236 ac. 36 per. 12 sq. yds. 55. 1362 ac. 2ro. 56. 14 ac. 3ro. 23 sq. per. lOf sq. yds.

57. 592 miles 7 fur. H chains. 58. 753 miles 1 fur. 2 chains. 59. 101 tuns 1 hdd. 60. 140 tuns lbuttO hdd. 4 gals. 61. 2417 pks. 1 gal. 1 qt. 62. 13672 bus. 2 pks. 1 gal. 2 qts. 63. 1117 lunar months 8 days. 64. 3400 years 310 days. 65. 11830 chains 20 yds. 66. 492 miles 60 ch. 16 yds. 67. 350 miles 31 ch. I per. 68. 5060 chains. 69. 3305 acres 8 sq. ch. 12 sq. per. 70. 231 acres 6 sq. ch. 7 sq. per. 28f sq. yds. 71. 1775 Eng. ells 2 qrs. 3na. 72. 95 yds. 3qrs. 3nls. Of in.

73. 61 tons 6cwt. 3 qrs. 14lbs.    74. 7 tons 2cwt. 2qrs. 61bs.

75. 2209 lbs. lOozs. 1 dwt.    76. 203 lbs. 5 ozs. lOdwt. lSgrs.

77. 1349fathoms2feet. 78. 633mi.5fur. 33per. lyd. 79. 1190c.yds. 10 c. ft. 1608 c. in. 80. 624 c. yds. 2 c. ft. 700 c. in. 81. 48 Ids. 5 c. ft. 82. 453 Ids. 10 c. ft. 83. 509 bales 4 reams. 84. 761 reams 17 qrs. 12sheets.

ANSWERS TO EXERCISE    XI.—COMPOUND

DIVISION. Page 93.

1. £29 3s. 9d., £19 9s. 2d., £14 11s. 10£d., £11 13s. 6d.,£9 14s. 7d., £7 5s. lUd., £5 16s. 9d, £4 17s. 3£d.    2. £3 12s. 2fd., £2 3s. 3fd.,

£1 10s. Hid., £1 4s. Ofd., £0 19s. 8fd.,    £0 18s. Old., f q.

3. £37 13s. 5fd. 4. £2 9s. 3fd. 5. £48 17s. 7|d. 6. £34 17s. bid.

7. £78 2s. 8d.    8. £17 8s. 3d. 9. £75 14s. 2fd. 10. £235 6s. 3{d.

11. £134 6s. 9d. 12. £7 19s. 7|d. 13. £173 lbs. 5fd. 14. £3 16s. Ofd. 15. £124 5s. 4d.    16. £11 11s. llfd. 17. £183 12s. lOfd.

18. £9 9s. 9fd. 19. £128 17s. 3d. 20. £7 7s. 7id. 21. £333 12s. 3fd. 22. £654 3s. 2fd.    23. £17 17s. 7fd.    24. £19 19s. llfd.

25. £54 10s. lid.    26. £50 11s. 9|d.    27. £73 19s. 9£d.

28. £299 9s. 9fd. 29. £711 11s. llfd. 30. £89 19s. Sfd.

31. £99 19s. llfd. 32. £88 18s. 84d.

ANSWERS TO EXERCISE XII.—DIVISION BY FACTORS. Page 93.

1. £3 7s. 6d., £2 19s. Ofd., £2 12s. 6d., £2 7s. 3d.    2. £2 5s. 10d.,

£2 3s. 9d., £2 Os. lfd., £1 18s. 6d.    3. £17 6s. 6d., £16 Os. lOd.

£15 9s. 4|d., £14 8s. 9d. 4. £14 2s. 4d., £13 4s. 8fd., £12 16s. 8d.,

£12 2s. 5. £10 8s. 6UU £9 7s. 8id., £8 18s. 9d., £S 10s. 7èd-6. £8 3s. 4d., £7 13s. Hd., £7 10s., £7 7s.    7. £3 4s. 2d., £3 3s.,

£3 Is. 10^d., £2 17s. 9d. 8. £25 13s. 4d., £25 5s. 3|d., £24 10s., £23 2s. ~    9. £20 Is. Old., £19 5s., £18 15s., £18 Os. lljd.

10. £6 8s. 4d., £6 3s. 9d., £5 18s. E>d., £5 15s. 6d. 11. £42 2s. 2±d., £41 5s., £40 16s.    8d., £40 8s. 6d.    12. £10    16s.    6fd., £10 14s. 6d,

£10 8s. Old., £10    4s. 9d. 13. £117    4s. 4R1.,    £109    8s. Id., £108 10s ,

£99 9s. 2d.    14.    £109 4s., £108    6s. 8d,    £106    12s. 9|d, £105.

15. £1 6s. 3d , £1    5s. 8d , £1 4s. 9d ,    £1 4s. Ofd. 16    £27 10s , £26 19s.,

£26 5s , £25 5s. 3|d.

ANSWERS TO EXERCISE XIII.—COMPOUND DIVISION. Page 94.

£

s.

d.

£

s.

d.

£

s.

d.

£

s.

d.

1.

13

15

2

2.

15

7

6

3.

17

12

5

4.

21

4

44

5.

23

10

m

6.

30

10

os

7.

38

14

74

8.

105

11

ID

9.

26

17

3*

10.

222

2

24

11.

37

15

54

12.

59

19

6|

13.

23

13

34

14.

701

11

71

15.

19

10

85

16.

96

16

4

17.

42

6

6

18.

300

10

10i

19.

50

18

9|

20.

980

7

6|

21.

47

11

HI

22.

204

18

34

23.

36

16

H+W

24.

199

19

9f

25.

11

4

8

26.

70

0

84

27.

5S

17

34

28.

26

18

8f

29.

60

0

54

30.

39

10

104

31.

400

0

04

32.

315

19

HI

33.

67

17

U

34.

299

19

9f

35.

55

15

54

36.

99

19

94

37.

188

8

84

38.

777

7

7

39.

266

6

6¿

40.

614

16

4|

ANSWERS TO EXERCISE XIV.—COMPOUND DIVISION. Page 94.

1. 13 and 1 far. over. 2. 19. 3. 509. 4. 799. 5. 78. 6. 220. 7. 176 and 8189 far. over. 8. 174-    9. 99.    10. 437.

11. 1096 and 534 far. over. 12. 960. 13. 1000. 14. 909.    15. 789.

16. 614.

ANSWERS TO EXERCISE XV.—COMPOUND DIVISION. Page 95.

1. 105. 2. 12. 3. 37. 4. 58. 5. 97.    6. 79.    7. 243. 8. 209.

9 103. 10. 1008. 11. 198. 12. 491. 13. 16 and 170190d. over.

14. 72. 15. 280. 16. 28S. 17. 1111. 18. 1728. 19. 5760. 20. 4840.

ANSWERS TO EXERCISE XVI.—COMPOUND DIVISION. Page 95.

Weights and Measures.

1. 2 tons 19 cwt. 1 qr. 25£ lbs. 2. 1 qr. 18 lbs. 11 ozs.

3. 4 cwt. 1 qr. 24 lbs. 6 ozs. 4. 7 tons 14 cwt. 2 qrs. 14 lbs.

5. 17 acres 3 ro. 32per. 11 sq. yds. 6. 4 acres 1 ro. 39 po. 15 yds. 1 sq. ft. 3 sq. in. 7. 2.ac. 3 ro. 25 per. 7 sq. yds. 8. 12ac.3 ro. 12 per. 4 sq. yds!

9. 7 miles 7 fur. 37 per. 4 yds. 10. 3 ro. 25 per. 20 sq. yds. 1| sq ft!

11. 3 fur. 25 per. 3 yards 2 ft. 12. 5 miles 55 chains 55 links!

13. 26 acres 8 sq. chains 7965 sq. links. 14. 8 acres 2 roods 15 per.

15. 17 qrs. 7 bushels 3 pecks. 16. 28 qrs. 3 bushels 2 pks. 1 gall!

17. 11 lbs. 11 ozs. 11 dwts. 11 grs. 18. 8 ozs. 18 dwts. 18 grs!

19. 13 weeks 5 days 13 hours. 20. 23 hours 34 min. 34 secs!

21. 19 loads 4 qrs. 5 bushels. 22. 4 barrels 13 galls. 3 quarts 1 pint

23. 51 yards 2 qrs. 2 na. 1J inches. 24. 15 E. ells 3 qrs. 3 na. 1 in! 25. 1 hogshead 60 gallons. 26. 40 gallons. 27 11 tierces 22 gallons! 28. 9 barrels 6 gall. 29. 1 sq. mile 320 ac. 30. 11 oz. 17 dwt. 18 grs! 31. 2 roods 15 per. 23 sq. yards. 32. 2 galls. 2 pts. 8 fl. ozs. 7 fl. drs!

ANSWERS TO EXERCISE XVII.—COMPOUND DIVISION. Page 96.

Weights and Measures.

1. 12.    2. 57.    3. 767AV 4. 650.    5. 75.    6. 804.    7. 100.

8. 960. 9. 114. 10. 854ia±f 11. 42. 12. 72.

ANSWERS TO EXERCISE XVIII. Page 96.

1. £347 18s. 6|d.    2. £9061 Is. ll|d. 3. £15585 11s. 2d.

4. 69 cwt. 3 qrs. 15 lbs. 8 oz. 5. 88 lbs. 10 ozs. 10 dwts. 14 grs.

6. 2639 lbs. 9 ozs. 2 scr. 7. 8576 ac. 3ro. 30 per. 8. 782mi. 0fur. 24 per.

9. 20 mi. 4 fur. 17 per. 4,^ yds. 10. 95 mi. 1 fur. 34 per. 1 yd. 11. 331 lbs. 9 ozs. 2 drs. 12. 13 times. 13. 3 times. 14. £3269017s. 11 ¿d. 15. 57 yds. 3 qrs. Iff na.    16. 459 mi. 2 fur. 19 po. f yd.

17. 31250 sq. links. 18. £1 11s. 3d.    19. £50103 14s. 7-Bd.

20. £33964 15s. 9Id. 21. lcwt. 3qrs. Olbs. 13-f-ozs. 22. 371589T2c.in.

23. 125. 24. 147. 25. £8255 2s. 8d.    26. £30.    27. £1 6s. Offfd.

28. 6^d.    29. 15040 gills'3008 quarts, and 15040 is five times 3008.

30. Is. 9d. 31. £1 3s. 6d.    32. £15 Us. 0TyTd. 33. 207360 c. inches.

34. 69120 c. inches. 35. 9^\.    36. lOjf.    37. 8 min. 20|| sec.

38. 9 mi. 1 fur. 24 per. 2f yds. 39. Ans. 1491 yds. 2 qrs. 1 na.

ANSWERS TO EXERCISE XIX.—REDUCTION OF MONEY. Page 104.

I. —1. 15504.    2. 453984.    3. 180250.    4. 359903.    5. 1982369.

6. 927302.    7. 360882.    8. 614339.    9. 8641534.    10. 959510.

11. 8259983. 12. 6146879. 13. 126000. 14. 3689784. 15. 8461080.

16. 1983600. 17. 9182094. 18. 3134181. 19. 2133706. 20. 3200319. 5

8. £92o 18s. 6cl. 9. £42982 18s.    10. £21329 18s.    11. £14781 7s.

12. £29841 4s.    13. £819.    14. £953 2s. 3d.    15. £937 19s. 4d.

16. £11607 2s. 6d. 17. £7656 5s. 18. £34466. 19. £199 14s. 3d.

20. £1470 16s. 8d.

III.—1. 302400 far. 2. 3009989 guineas and 18s. 3. 1652616 halfpence. 4. £226906.    5. 69216 half-crowns. 6. 863520 sixpences.

7. 4052f crowns. 8. 93550 half-crowns. 9. £128600 16s. 5Jd.

10. 979815 guineas and 16s. 8^d.    11. £1361 12s.    12. £861 3s. 7fd.

13.    £75 12s.    14. £198 Is. 9d.    15. 360 sh. and 3d.    16. 5241

crowns 3s.    17. 17920 threepences.    18. 22771 guineas 9s.

19.    4820 guineas and 20s.    20. £800719 10s.    21. 288 crowns.

22.    8750 florins. 23. 1829 guineas and 11s.    24. 3226 sixpences

and 4cl. 25. 838 guineas 2s.    26. 8316 sovereigns. 27. 10481

halfpence. 28. £286 12s. 6d.    29. £145274.    30. £39 7s. 6d.

31. 18900 florins. 32. 140 half-crowns. 33. 275 guineas Is. 6|d. 34. 48 marks 8s. 6d.    35. 1308 crowns 4s. llfd. 36. 302772

florins Is. 37. £104 9s. 6d.    38. 8639 halfpence. 39. £1050.

40.    952 guineas 8s.

ANSWERS TO EXERCISE XX.—REDUCTION OF WEIGHTS AND MEASURES. Page 105.

1.    40992 lbs. 2. 132160 lbs. 3. 8400 lbs. 4. 23168 ozs.

5. 183 cwt. 3 qrs. 20 lbs. 6. 23 tons 6 cwt. 3 qrs. 19 lbs.

7. 40 cwt. 3 qrs. 11 lbs. 6 ozs. 8. 145 lbs. 9 ozs. 4drs. 9. 3343 drams.

10.    41788    lbs.    11.    12 cwt. 2 qrs. 9 lbs. 6 ozs., 4 drs.

12. 426 cwt. 1 qr. 16 lbs. 2 ozs. 13. 6 lbs. 5 ozs. 12dwt. 16grs.

14.    3170 dwt. 14 grains. 15. 13 lbs. 2 ozs. 17 dwt. 11 grs. 16.80616grains. 17. 73463grains. 18. 96686grains. 19. 17220grains.

20.    68lbs. loz. 4drs. 21. 485760yards. 22. 7miles lfur. 4per. 4yds.

23.    152020 yards. 24. 2130974 feet. 25. 7 miles 1 fur. 4 per. 1 yd.

26. 1014 miles 6 fur. 24 per. 3 yds.    27. 20168 yds. 2 ft. 10 in.

28. 13 miles 7 fur. 36 per. 1 yd. 10 in.    29. 118 fur. 37 per. \ yard.

30. 238 miles 4 fur. 30 per. 31. 4895 perches. 32. 18335734 yds. 33. 778 ac. 2 ro. 36 per. 3 yds. 34. 476 ac. 2 ro. 35. 17424000 sq. yds.

36. 1550463| sq. ft. 37. 39732 square inches. 38. 5615 sq. feet.

39. 4ac. 24per. 12sq. ydsi 5ft. 72in.    40. 864ac. 3ro. 38per. 5|sq. yds.

41. 3566ac. lro. lOper. 274sq. yds. 42. 1087sq. miles 124 acres.

43. 39 E. ells 2 nails 4 in. 44. 675 Fr. ells 3 qrs. 1 nail f in.

45. 1344 yds. 1 qr. 3 nails J in. 46. 6S4 nails. 47. 1634 Eng. ells.

48. 1440 nails. 49. 442 E. ells 4 qrs. 50. 471 E. ells 3 qrs.

51. 1353024 c. in. 52. 2564 c. ft. 53. 7 c. yds. 23 ft. 95,1 in. 54. lyd. 18c. ft. 882c. in. 55. 160 loads 8c. ft. 56. 744loads 48c. ft. 57. 1728 tons 28 c. ft. 58. 92 loads 2 c. ft. 59. 85100544 c. inches.

60. 1418 loads. 61. 5 days 11 hrs. 50 min. 24 sec. 62. 1396S0 min. 63. 1489800 seconds. 64. 23940 minutes. 65. 25 days 1 hr. 12 min. 66. 211 hrs. 54 min. 67. 108 yrs. 4mos. 2 wks. 68. 5928 days. 69. 455 cal. months. 70. 6404 years 6 months. 71. 4600 gallons. 72. 677 pints. 73. 19515 bush. 2 pks. 74. 1024 qts. 75. 7488 pints.

76. 16380 quarts. 77. 910 gals. 1 qt. 78. 405 pecks. 79. 4352 qts. 8Ô. 501 quarters. 81. 1442 bush. 22 lbs. 82. 1612 bushels 20 lbs. 83X3640 lbs. 84. 680 lbs. 85. 1608 feet. 86. 26784 pints. 87. 100800 pints. 88. 1185 chains 14 yds. 89. 15967 qrs. 2 nails. 90.    23 lbs. 3872 grains. 91. 6192 lbs. 92.    102965 lbs.

7 ozs. 10 dwt. 93. 4990 stones. 93. 6782 cwt. 2 qrs. 95. 100 acres. 96. 12560 chains. 97. 3301 chains 18 yds.

98.    882 miles 68 chains. 99.    8648 acres S chains.

100. 7629000000 sq. links. 101.402362400 ft. 102. 432 miles 1680 yds.

103. 207333^ sq. yds. 104. 72 ro. 7 per. 1J yds. 3 ft. 105. 322560 grains. 106. 322560 grains. 107. 2271 sq. per. 27| sq. yds. 108. 25962000 nobles. 109. 964 marks. 110. 3192 sovereigns.

111. 1328 months20days. 112. 3170lbs. 3 ozs. 10dwt. 113. 194560 pints. 114. 23 qrs. 6 bus. 2 pks. 115. 46200 ft. 116. 2093 years 80 days. 117. 248156480 sq. yds. 118. 587827200 minims,

119. 4414080 yards.    120. 4033 Fr. ells 2 qrs. 121. 9 pints 4oz.

5 drs. 28 minims. 122. 180 cwt. 1 qr. 21 lbs. 123. 6080 miles, i 124. 1568 sq. miles Eng. 125. 189738 sheets. 126. 164 rms. 6 quires

4    sheets. 127. 35876 seconds. 128. 11 deg. 40' 35".    129. 6600.

130. 102 days 21 hrs. 8 min. 131. 10018008 half-pence. 132. 31 loads 1 qr. 2 bus. 133. 22 stones 9 lbs. 4 ozs. 4 drs. 134. 644 tons. 135. 2281284 pence. 136. 1 ac. 24 sq. perches, 21 sq. yds. 21 sq. in. •137. 94256 nails.    138. 9287960 sq. yds.    139. 101 acres.

140. 1574 years 6 wks. 141. 14 miles 7 fur. Oper. 5 yds. 2 ft. 6 in. 142. 306240 minims. 143. 3156 perches. 144. 240 qrs. 145. 8704 chains. 146. 188 quarts. 147. 357 sq- chains. 148. 161 scruples. 149. 9275168 ozs. 150. 181920 sheets. 151. 5061 lbs. 11 oz. 15 dwt. 152. 9990 florins.    153. 66816 lbs.    154. 132 threepences.

155. 49000 lbs.    156. 1323 nails. 157. 77500000 sq. links,

158. 18144 pecks. 159. 726^ acres. 160. 28512 pints. 161. 23058 paces. 162. 52hdds. 4gals. 2qts. 163. 40years. 164. 39488pecks. 165. 4053600 lbs. 166. 1531020 sec. 167. 1920lbs. 168. 176197 sec. 169. 8781 bushels. 170. 85 deg. 16 min. 48 sec. 171. 3840 lbs. 172. 30 Eng. ells.    173. 1800 lbs. 174. 69 yds. 2 ft. 9f in.

175. 20431 bushels lOlbs. 176. 178Flem. ells. 177. 3 sq. mi. 166ac. 178. 483 gallons. 179. 12 Eng. ells 2 qrs. 2na. l^in. ISO. 2640 quarts. 181. 271814400 sq. yds. 182. Smiles. 183. 6lbs. 11 ozs.

18 dwt. 15grs. 184. 63437 sq. yards. 185. 175^ inches. 186. 7840 acres.

ANSWERS TO EXERCISE XXL—PRACTICE (Simple). Page 114.

1. £303 12s., £624.    2. £4722 13s. 4d., £4266 13s. 4d.

3 £23149 10s., £20943 6s. 8d.    4. £88406 12s. 8d., £44791.

5    £2977 9s. 4d., £4123 18s. 6. £9720., £45540. 7. £93434 7s. 6d.,

£140770. 16s. 8d.    8. £6973 10s., £6420.    9. £761846 5s.,

£507669 7s. 6d. 10. £9023 Is. 3d., £27507 18s. 9d. 11. £17204.,£72653 12 £114046 4s., £110242 16s.    13. £1340 2s. 8d., £7146 2s.

14. £696 17s. 2d., £853 6s. 7d. 15. £2069 7s. 6d., £2300 18s. ll|d.

16. £1262 2s. 6gd., £1583 13s. 3|d.    17. £3294 16s. 2£d.,

£3121 4s. 3|d.    18. £6314 10s. 9Jd., £3009 2s. 6yf.

19. £1697 9s. 3d., £6104 3s. lOjd. 20. £509 6s. 6|d., £1062 Os. 3|d. 21.    £3078 Is., £5620 16s.    4d.    22.    2595 10s.    lx\.,

£3886 16s. 7x6d-    23. £10575 12s. l^d., £19921 18s. Ilfid.

24. £2791 14s. 8#d., £6405 9s. 8fd.    25. £4906 9s. 9-id.,

£6322 10s. 10T\jd. 26. £329 6s. 6|d., £987 19s. 8fd. 27. £1078 13s., £692 6s. 6d. 28. £716 13s.9££d.,£8395s. lOffd. 29. £13965 14s. 11-ftd., £13547 4s. 4gVd. 30.    £6793 6s. 2$£fcd., £8890 5s. 0^1.

31. £1073 6s. 0£d. 32. 412051 ac. 2ro. 24 per. 33. 2378 galls. 2 qts. 34. 110376 cwt., or 5518 tons 16 cwt.    35. £29275 12s. 6d.

36. 98 tons 5 cwt. 3 qrs. 12 lbs. 37. £4824 19s.    38. 876864 En. ells.

39. 3526 miles 4 fur. 6 per. 40.    212 lbs. 2 oz. 12 dwts.

41. 685924 days 21 hours 26 min. 24 secs. 42. 79200 inches. 43. 81675 gallons. 44. 778 tons 1 cwt. 45. £1338. 46. 2090 lbs. 3 oz. 6 dwt. 16 grs. 47. 116 tons 1 cwt. 3 qrs. 4 lbs.


1.

3. 5. 7. 9. 1113. 15. 17. 19. 21•' 23. 25. 27. 29. 31. 33.


ANSWERS TO EXERCISE XXII.—PRACTICE (COMPOUND). Page 116.

£1 8s. OgVod*, or -065625d.

£6 18s. 9-fJd., or -53125d.

£3332 16s. 3d.

£50 2s. 10fd., or -4d.

£1464 8s. 24pU or -7083d.

£4520 14s. If¿d., or -5125d.

£2332 12s. 3xVh, or -1875d.


2. £96 18s. 2ffd., or -390625d. 4. £33 17s. 9fd., or -8d.

6. £676 0s. 9ffd., or -85d. 8. £93 12s. 2ffd., or -517857l42d.

10. £402 Is. 3d. 12. £2889 4s. 5^d., or -15d. 14. £253 2s. 2fd., or -625d. £2141 12s. 2f|d., or -859375d. 16. 48 allot. £1095 2s. 2xVl, or -Id £106 12s. Id.    18. £19 11s. 8ffd., or -825d.

£23984 9s 6iffd., or -69375d.    20. £245 3s. 7ffd., or -96875d.

£371 13s. 5yLd., or -4375d.    22. £615 Is. 9fd., -375d.

£1771 2s. 3ff4-id,, or -9134765625d. 24. £4185 19s. 7fd., or -75d £1440 16s. llffd., or ’90625d.    26. £286 14s. Ofd., or -25d.

£125 6s. 2fd., or -375d.    28. 159 7s. 9fd., or -375d.

£283 15s. 5fd., or -25d.    30. £227 2s. l^d., or -5d

£2226 14s. llfffd., or '559375d. 32. £2322 18s. Offd., or -S4375d. £348 7s. 7HId., or -54017S57l42d.    34. £19 9s. 6ffd., -725d.


35. £165 0s. 3fid., or -4772d.    36. £103 10s. 10ffd., or -8125d.

37. £143 11s. 71 Ad, or -775d.    38. £45 19s. 6AWd., or -2649 &c. d


39. £2749 9s. 0Z% or -075d. 41. £133 8s. 6fd., or -25d.

/ 43. £89 9s. 2ffd., or -2125d.

45. £52 0s. 8jfd., or -9375d. >

♦ 47. £575 0s. 5jfid., or -94375d.

49. £114 7s. 9fd., or '125d.

\ 51. £52 7s. 8TVb, or -5625d. 53. £62 16s. 6gd., or -125d.


40. £2236 3s. 8xVd., or -0625d. 42. £79 Is. Offd., or -421875d. 44. £162 12s. 3xVod., or -5'083d. 46. £93 16s. 5ffd., or -96875d. 48. £223 10s. t^d., or -4375d. 50. £150 4s. 2fd., or -625d. \ 52. £57 Is. l|d., or '75d. \ 54. £59 5s. 9xd., or ’5d.


55. £418 3s. Ofd., or -42S57ld.    56. £2880 16s. 6A7od., or -19583d.


57. £113 19s. 6&cL, or -05d.    58. £415 14s. 2ftfd., or -8828125(1.

59. 27 of each kind, or 81 in all, value £562 Os. 7id., or '5d.

60. A 172 yds. 3 qrs. 1 nl., B 103 yds. 2 qrs. 3 nls., C 60 yds. 1 qr. 3| nls. Value £239 5s. 2^id., or '1375d.    61. 120 ac. 2 ro. 33 per.

62 £3867 10s. Od.    63. 444600 bushels.

ANSWERS TO EXERCISE    XXIII.—SIMPLE

PROPORTION. Page 130.

1. £4 10s. 2. £81 12s. 3. 7lbs. 4. 19yds. 5. 616. 6. £1 Is.

7. 14 men. 8. £10 2s. 9.24 yds. 10.20.    11. £3 7s. 6d.    12. 650.

13. 320.    14. 112.    15. 5 shillings. 16. £135.    17. 2500 yds. per

min. IS. £10.    19. £1 10s. 3d.    20. 64 days. 21. £8 15s. 6d.

22. 108 c. yds. 23. 308.    -24. 242. / 25. 940 yds. 26. 30 days.

27. 10. \ 28. 2 yds. Oqrs. Ona. 1* in. 29. 81 ft* 30. 64 ft. 31. Is. 8d.

• 32. £413 10s.    33. 8$. 34. 2|lbs. 35. £1036. 36. 1 hr. 51 min A

37. £1 2s. 8d.    38. 65T5Tmin. 39. 29fhrs. 40. 55 days.

41. 5hrs. 15 min. 42. 130 yds. 43. 76 weeks. 44. £122 12s. 6d. 45. 12 tons. 46. 30 miles. 47. 33 yards. 48. 35 yards. 49. £265 10s.

50. 75 ac. 2 ro. 10 po. 51.16 gals. 1 qt. 52. 144 days, or 14 days 1 hour.

53. £173 15s.    54. 1 cwt. 55. 7 tons 1T% cwt. 56. 217f£jj-

57.    £7440.    58. 52 sq. in. 59. 4^ years. 60. £1 Is. Hcl.

61. £525.    62. 65 ac. 2ro. 20 per. 63. 100.    64. 300.    65. 375.

66. 10. 67. 18.    68. 160.    69. 54 days.    70. 80.    71. 24 days.

72. 25 teams. 73. 15 days. 74. 12 weeks. 75. 48. 76. £8 2s. 6d.

77. £497 8s.    78. 60.    79. 14.    80. 7 lbs. 14 ozs. 81. 15ozs.

82. £2 0s. lOd. 83. Is. 4^d.    84. £2 11s. 10id.    85. 9 hours.

86. 4 extra men. 87. £8 17s. 10|d.    88. £4 11s. 9d.

89. £1610 15s. 10fd. T7T q. 90. 120. 91>£819 12s. 8^d. 92. £1 12s. 8|d. * 93. £143, 77 acres. 4)4. 20liorses. v95. £16 15s. 4^^., £16 13s. 4^fd. including one day only. 96. 7s. 6d. 97. £395 8s. 4d. 98. £1544 5s. 8-i-d. \

99. £75 9s. 5|fid.' 100. £59 4s. O^d. 101. 5s. 2|d.    102." 8s. 9-]d.

103. 6cAvt. 1 qr. 3^^ lbs. 104. 144.    105. 17 cwt. 1 qr. 20^1^ lbs.

106. 13i sq. in 107. £50 13s. 6Tf|^d.    108. £848 8s. 6ftd.

109. £37 17s. 5fd. ^q. 110. £24 5s. 7$d.

ANSWERS TO EXERCISE XXIY. — COMPOUND PROPORTION. Page 143.

1. 14s.04||d. 2. £828 8s. 4d. 3. 4ft. 7|in. 4. 46|fs. 5. £2 3s.9d.

6. 8 feet. 7. 16 men. 8. By 7^ ozs. 9. 27^ days. *10. £102 Is. Ofd.

11. 35 days. 12. 29-f hours. 13. 98 gallons. 14. 9'72 months nearly.'.

15. 3‘45 months nearly. 16. £1 18s. 7-^Vsd.    17. 15 persons.

18. 57fjl quarters. 19. 9fozs. 20. 139^gmiles. 21. 5‘570524333etc.ozs. 22. 1194Tdays = 119days2lKmrs20min.\ 23. 51days = 5days 2hours. X 24. 27^44 days. 25. 12 days. 26. 32 days. 27. 71 hours.

28.    73 days, 6 hours, 37? min. 29. 50§ days. 30. £6 13s. 4d.

31. 91 hours. 32. 9J hours. 33. £7 11s. 3d.    34. 36U arithmetical

answer; 36 horses, sensible answer. 35. 45 feet. 36. 191 ozs.

37. lOf f £ days. 38. 24^- men, arithmetical answer ; 25 men, logical answer. 39. £1 Is. 7i^d.    40. 6ff arithmetical answer; 7

men, sensible answer. 41. £214 15s.    42. £8 13s. ll|d.

43.    53 miles 0 furlongs 25‘899, &c., perches. 44.    27 months.

45. 23'5556, &c., tons; or 23 tons 11 cwt. 12 lbs., &c. 46. £7 3s. 9gyTd. 47. £0 10s. 6^d.    48. £936.    49. £832.    50. £1670 12s. 6d.

51. £53 15s. 2|d. 52. 64i£ff§£miles = 644745, &c.    53. 200 miles.

54.    459 loads. 55. 8 men. 56. 75 acres. 57. llff. 58. ll-A-

EXERCISE XXY.—INTEREST.    Page 155.

I. —1. £37 10s.    2. £54. 3. £165.    4. £288.    5. £22 10s. 4fd.

6.    £42 13s.    7. £295 11s. llfd. 8. £56 8s. 4d.    9. £28 19s. 7d.

10.    £21 16s.    11. £51n4s.    12. £129 12s.    13. £137.    14.'£54 15s.

15. £16 9s.' 7-g-d.    16. £150 3s.

II. —1. £239 0s. 7fd. 2. £247 16s. 3. £136 9s. 3Jd. 4. £1007.

5. £690 7s. 6d.    6. £43116s. 3d.    7. £163 3s. 4id. 8. £263 8s. 5f| d.

9. £92.    10. £115 10s. lOfd. 11. £93 6s. 4£d. ° 12. £150 12s. 6di"

mi-1. £2000.    2. £162 18s. 9d.    3. £835 12s. llffd.

4. £73 6s. 2£d. 5. £18 0s. llfd. 6. £25 15s. 10^d. 7. £103 6s. 107-d.

8. £1374 17s. lffd- -    9. £264 6s. 2^d.    10. £722 13s. 84d. |q.

11.    £169 5s. 9‘429d. 12. £810 8s. 9T%d.    51

IV.—1. £5 16s. e^d. 2. £24 10s. lOfd. 73376 q. 3. £97 5s. 3fd.

4. £204 11s. O^fod. 5. £15 16s. 9d. ‘915 d. 6. £27 6s. lid. -38375 d

7.    £84 13s. 4d. 8. £540 5s. 6fWd.

ANSWERS TO EXERCISE    XXYI.—INTEREST.

FROM TEACHER’S EXAMINATION PAPERS. Page 156.

1. £24 2s. 3fffd. 2. £115 9s. 5£ffd. 3. £26 19s. 9/M.

4. 4£ years. 5. £65 9s. 9ff£fd. 6. £79 18s. 4T%2f>Td.    7. 3| years.

8.    5TV years. 9. £163 19s. 3ffd., or £163 16s. 6ffd., by calculating

for 3 years 95 days. 10. 7f per cent. 11. 5§ per cent. 12. 34 yrs., £411 16s. 8-5S2401640625d. 13. £410 16s. 4£fd. 14. £350. 15. 5ff percent. 16. 3 years. 17.'5 per cent. ' 18. £5 13s. 7£ffd.v, or £5 13s. 3fd. if 366 days be made the 1st term. 19. £548 18s. 2-2^L2-d.

20. £675.    21. 4f per cent. 22. 13 months. 23. 9J per cent.

24.    4£ per cent. 25. 5§ years. 26. 5 years. 27. 64 years.

25. 20 years. 29. £68 15s. 30. £75 18s. 31. 3£ years. 32. £12 10s. 33. £1250.    34. 6^ per cent. '35. £676 Ss.O^d. 36. 5^^ yrs.

37. £344 9s. l^d- iff 9-    38. 7£ per cent. 39. £7500 0s. Od.

40. 4 per cent. 41. £110 13s. 7|d. 4 q. 42. £109 18s. 3f d. f q. 43. 54 years. 44. £832.    45. 6 per cent. 46. 9T33 months.

47. 3 years 9 months. 48. 3£ per cent. 49. £248. 50. £73 2s. 9J-”-d. 51. If per cent. 52. £312 10s.    53. 3f years. 54. £716 19s 7-J-d"

55. £1098 18s. 0£d. rs3q. 56. £31 Is. 8y\,d. 57. £624. 58. lo'per cent. 59. 3 years. 60. 124 per cent. 61. £499 2s. 9715d.

ANSWERS TO EXERCISE XXVII.—GREATEST COMMON MEASURE. Page 162.

1. 24.    2. 105.    3. 77.    4.    96.    5.    63.    6.    49. 7. 91. 8. 87

9. 64. 10. 87.    11. 39.    12. 78.    13. 75. 14. 81. 15. 36. 16. 720

17. 117. 18. 162. 19. 133. 20. 119. 21. 135. 22. 126. 23. 171. 24. 57.    25. 17. 26. 18. 27.    19.    28.    16.    29.    15. 30. 14. 31. 8

32.29.    33.27.    34.95.    35.7.    36.49.    37.56.    38.113.

39. 203.    40. 1228.    41. 1155.    42. 384.    43. 4992.    44. 2688.

45. 384.    46. 663.    47. 240.    48.    193.    49.    599.    50. 313. 51. 505.

52. 123.    53. 56.    54. 393.    55.    112.    56. 67.    57. 63. 58. 459.

59. 567. 60. 112.

ANSWERS TO EXERCISE XXVIII.—LOWEST COMMON MULTIPLE. Page 166.

1. 48. 2. 36. 3. 100. 4. 120. 5. 96. 6. 336.    7. 144. 8. 132.

9. 224.    10. 288.    11. 1260.    12. 576.    13. 675.    14. 2688.

15. 175617. 16. 924. 17. 2016. 18. 528. 19. 37791. 20. 1800. 21. 4032.    1870. 23. 1092. 24. 1584.

EXERCISE XXIX.—VULGAR FRACTIONS. Page 183.

f.—1.

b 2. f

3. J. 4

. W 5.

11

3 0"

6.

JL

3 6*

7. 4

• 8. *.

9. f. 10. f. 11

_i_ io . 12.

tV- 13.

1.

14.

4 4 0 3 2 3 2 7 2 5*

15. 92f|.

16 -S-3-

2 5 0*

17. b

18. 12.

10- AV

20.

l-f

IL—1

• §• 2. ^

- 3 -7-

4. if 5.

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6.

3    6 1

4    3*

7 5 5 ’ 6 4

• 8. ff

q es

V. 7 3.

io. H-

11

-1 A- 64 3'

12. \b

13. 4

3 9 7 7 1 2 *

14.

1 5

'2 9*

15.

i6. m-

17 -31 1 * 14*

18. if

IQ 6 5 8 89 7*

20

2

* 7*

21.

19 3 4*

22. if.

23. «.

94 91

25. If

26. f.

27.

1 3 7 7*

28.

13 1 7 •

29. t5t.

30. m

Q1 8 5 o±. i07*

32- tW-

33. g.

34.

9

1 r*

35.

2 3 29

36 ~6-i-7-

28 39*

37. f

38 uojl

OO. U1.

39. §. 40

1 200 3 • 2 5 04*

III.—

1. y. 2.

78 Q 124 5 ' 1

A 4 19 • 20 ’

5.

1 3 25 fi 13* U*

68 3 9 *

7 8 0S *8

s. ir-

9. W0.

10. eu-

11 1 AX. g .

12.

5 5 1 2 3

IV.

1. 331. 2.

7W 3.7|

f. 4. 10*. 5. 16

ttV- 0.46.

7 4-5JL-

8. Mb

9. 6tW-

10. 2MV

11. 28.

12

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V.—1

1* 2.1

3 7 3 11 TUT* 16'

■ 4.4*.

5. 3

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1 5 0*

6. 2

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7 9571

8. 6*.

9. 2if.

10. 5||

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12.

1 4 0 2 7 .5544

13. 16*.

M. 2ff£

. 15.4M-

16. 5ff.

17.

3

r 7 •

IS.

2 H-

IQ 01116146 ^ 1 3 0 3 7 83*

20. 3fM

mi- 21

11279570 • 11679BU9'

ANSWERS TO EXERCISE XXX.—VULGAR FRACTIONS. Page 184.

I.—Addition—1.

3*. 2.

2*.

3. 2|.

4. '

5.

3f

6.

2§.

7. 2|.

8- 2£f

Q 9 50 47 • °189

. 10.

0 1 3

üios*

11

92621 * °3T65-

12.

4ff*

13. 4*.

14.

1_A1_

. 15. 7

if. 16

• 24*.

17

. 70|-.

18.

149

1 7 9 14Ï*

19. 98ff.

20,

, 13**. 21.

91 3 2 9 Z,A 34 0*

II.—Subtraction-

-1 -3-

2- *•

3- *

4.

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5.

1

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6.

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7 -JL fi

1 1

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9.

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10. 8*

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2

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12.

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13.

2 1 1 5 4 0*

14. -h-

15.

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1 G 5

17. Iff.

18.

1GV

19.

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20. 99jf.

III.—1.

i.

2. §.

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4 -9** 6 4*

5. f.

6.

3

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7.

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1 9

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9. JAr. 10. 0. 11. 30. 12. 0.

IV.—1. *.    2. f|.    3. è. 4. 0.    5. 1.    6. 4f.    7. f.

fi /j Q 3 O O 2    q    015 4 9 IA Q10 1    11 Q 1013 IO 38 8 1    IQ 19 9

°* ^3003'    ^3 46 5*    A^* ^4 2 0*    AA* ^27 7 20*    A^*    2520*    AO* A°25*

14. 17-fè-    15.    fi 16. 70*. 17.    2*5. 18. *. 19. 32f.    20. 1*.

ANSWERS TO EXERCISE XXXI. Page 185.

I.-1. *. 2.

4 3 1

•j. 0. 2.

4.

1

5. f.

6. f

7.

i 8. 1.

9.

1. 10. 5ff.

«rx

II.-1. If. 2.

20§. 3.

40ff.

4.

7 6

TTT*

p; i28 °* T3T*

6.

1.

III. 1. 3?*.

2. f.

3. 2f

4. 2f.

5.

0.

6. tV*

7.

14*. * 8. 0.

9. f

10.

3|.

11.

012 32 11*

12.

1

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Page 186.

3 09    0 9

TÔOOO* V. O 2 2

12. 0.


ANSWERS TO EXERCISE XXXII.

1- sVo- 2. 1*V    3. 8H-    4. UH- 5.

7. *.    8. 12*.    9. 3*.    10. Ilf 11. lltff.

ANSWERS TO EXERCISE XXXIII. Page 187.

I.-l. 4-


7.    -2-i 8 H.

' "    1^>* °* 2 4*


O 11    Q    7    4.    _5_    K

^* 12.    o.    §.    16.    o.

6 4 7    11    1    1    O    1 5 :

'    900* J-A' ÏÏ2Ü*    16 C

II.—1. f

7- 414.    8.

0 6 7    O 1

¿f. 8 0*    °* 6

Q 4 6 6 9 Um 1568 160*

4 AAA •    ^*    440*

10 -lAAi-

5.

10 8 90* . 81 7 8 7 12ÏÏ

III.—1.

99

112*

2,

J2Ì • 2'24*

0 4 9 °* 12 8*

4.

5 1 K 5 6* ^*

7 8 3 1 7 9 2*

6.

8 8 7 89?*

IV.— 1.

Tf f fo".

2.

7 7

14 4(5'

Q 5

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A 2 3 rr* 80*

5. f.

6- f

7.

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8- fi.

9.

5

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10 -ti>-

AV/* 040*

11. f.

12.

8.

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f 2.

9

8 0

O 8 O. g.

4 -5-

rt* 3 6*

5.

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4 9

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7.

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2 5 2*

9. f*.

10.

11 ^

LA* 32*

12.

17 5 2 A 0 4 *

ANSWERS TO EXERCISE XXXIV. Page 188.

L—1. -57. 2. 0017. 3. 30-08. 4. -0061. 5. -009. 6. -000127 7. -000007. 8. -003. 9. -000051. 10. -036. 11. -375. 12. -312o!

13. 0025.    14. -063.    15. -0000116. 16. -296875.    17. ‘826171875^

18. -714285.    19. 20'5. 20. '500000.

7. -1

• •    7 •


___    2. \%3.    3.    4.    5. -Jjr. 6. -fa.

8. f 9 f    10. «. 11. f 12. #. 13.    14. tVò- 15.

16. U. 17. W- 18-    19. Mè- 20.

III.—1. 2-5806.    2/ 6-23125.    3. 85-4525.    4. 13-346.

5. 103-9055.    6. 111*2211.    7. 1322995.    8. 20 261.    9. 30’8.

10. 17-1580.    11. 5*68127. 12. 28-2Ì48239.

IV.—1. -005971.    2. -008993. 3. -002949.    4. -564.    5. 14-985.

6. -0083.    7. -0218751.    8. 2-05234375.    9. 3-9379682.

10. -0451822916.    11. -015625.    12. -6463372.

ANSWERS TO EXERCISE XXXV. Page 189.

I. —1. 1-76.    2. -000084.    3. -051136.    7 4. -0000109855.

5. -000078125.    6. -00016.    7. ’0000007869.    8. -000000063.

9. -000000153.    10. -0016.    11. -000006205.    12. 415-47365, etc.

13 6-5625.    14. 11 *963.    15. 07875.    16. -045366.    17. -01428.

18.    124-834375. 19. 3-Ì 60493827.

II. —1. 8000. 2.4000. 3. 0000565159, etc. 4: 17694 11764, etc.

5. 16-875.    6. 3-9375.    7. 511U808.    8. 262-826.    9. 6‘48.

10.    152-428571.    11.    *01526917    12 -000584112, etc. 13. 200.

14. 500.    15. 50.    16; 50. 17. 47;2868217, etc. 18. 12S5;7142S5.

19.    36.    20. 58-8235294117647058.    21. 16*6.    22. -06    23. 20.

24. -05. 25. 22-2    26. 045. 27. 000025. 28. 40000.    29. 3000.

30. 858-823529411764705S.

ANSWERS TO EXERCISE XXXVI. Page 190.

1. 001953125 and 9 984.    2    09375 and 24 64.    3. ‘003125.

4. and “0857142. 5. ’405 and ‘0625.    6. '009375.    7. 017361.

8. -6305.    9. '65476190.    10. ^.    11. -321 and -001875.

12. -05920138.    13. t|^.    14. -66875.    15. -1046875.    16. -03375.

17. -010247619Ó.    18. 2 ro. 15 per. 19. -513194. 20.-0064732Ì42857.

21. 0017248376623.    *    22. 22 per. 25 yds. 8 ft. 51f sq. in.

23. 5-588.    24. -278.    25. -00046843.    26. -05761904.    27. -425.

28. 1-9368489583.    29. -0486L    30. ’02722366, etc. 31. 1*0083.

32. -27.    33. -0238.    34. -095078125.    35.    00469547, etc.

36. -12.    37. -073863.    38. -43038194.    39. -38580246913.

40. -029761904.

ANSWERS TO EXERCISE XXXWII.—COMPLEX FRACTIONS. Page 191.

1. 1. 2. m 3. 1. 4. 3f. 5. 1^. 6- m- 7. 13f. 8. 5ff. 9.    10.^11.141- 12.^8-^13.19^0. 14. 35^_ = 35 07478632.

^15. im ,16. ^. 17. 71F&. *8. 13H- 19. 6H- 20. ff 21.

23. 1044.    24.    25. i. 9«    97 ai? or a <lt


22.

29.


_o_u_

1 7 5* '

M:' 30.


..    _    _ 26. tH- 27.41^,28.8^.

„„    13MH- 31. tV* 32^Vv 33.    34.

35. f.    36. 4.    37. Lj? 38. U4T. 39. 26f.    40. Neg. fr. - 1

41. *7714285.1    42. -03853649, etc. 43. 2T5^2-3897, etc.

44. xffie = -0302654, etc. 45. -678558721, etc. 46. $*£ = -355519480. 47. 3!^=-005242, etc.    48. 26-02285714.    49. 19 04.

50. •7182818, etc. 51. ’75.    52. 6 drs. 0 scr. 3f grs. 53. 0.

54. 14H-    55. | um- 56. 15T9^. 57. 2T4^. 58. 4. 59. 173.

ANSWERS TO EXERCISE XXXVIII.—DISCOUNT, &c. Page 197.

I.—1. £H7 7s. 7&d.    2. £52 11s. lO^d. 3. £1003 5s. 6d.

5-*-^8-7-d

°*3SQ. 1U-


4. £105 Is. 6|d.    5. £44 4s. 3Jd.    6. £51 17s.

7. £679 16s. 2|d.    8. £18 4s. 4Jd *

II.—1. £936 11s. 3d.    2. £354 7s. lid. 3. £150.    4. £475.

5. £1245.    6. £429 16s. 5|d.    7. £265 13s. 6|d. 8. £350.

9. £133 6s. 8d. 10. £250. 11. £250 nearly. 12. £2645 nearly.

HI.—1. £23 15s.    2. £41 14s. 6|d.    3. £86 17s. 6d.

4. £7 3s. Ofd. 5. £37 8s. Rd. 6. £59 3s. 54fd. 7. £45 2s. 8|d.

8. £87 17s. lid.

ANSWERS TO EXERCISE XXXIX.—PROPORTIONAL PARTS. Page 200.

1. 9 fowls, 27 geese, 36 pigs, and 45 turkeys. 2. £4 ,7s. llid. • £13 3s. ll|d. ; £21 19s. 10|d. 3. 917 ac.; 655 ac. ; and 245 ac. 24 ro. 4. 1460 ac.; 876 ac.; -and 657 ac. 5. 60 ac. 2 ro. 9£± per. 6. £1200 • £900; and £500.    7. 172 yds. 3 cjrs. 1 na. ; 103 yds. 2 qrs. 3 na •

60 yds. 1 qr. 3f na.—Cost £239 5s. 2-^d.    8. 450.    9. 840,’

10.    9 lbs. 9 ozs. 4dwt. ‘ lO-^sgrs. 11. 11 ac. 2ro. 39 per.

12.    A man, £21 7s. 5%d. ; a woman, £15 5s. 3|d. ; a boy, £3 Is. Qfd.

13.    '£1200 ; £900 ; £500.

ANSWERS TO EXERCISE XL.—PROFIT AND LOSS. Page 207.

1. -§-| of 850 guineas ; or, £771 17s. 10^d.    2. 12| p. c.

3. £14 19s. lf-fd. 4. 15s. 7R 5. 18s. 9d. ' 6. £160 12s. lOfd. 7. 15 p. c. 8. 8TVg P- c-    9- 5 p. c. 10. £87 19s. 7Ad.

11. £1194 13s. 4d.    12. £2 9s. 4d.    13. 8£ p.c. 14. 11s. 3d.

15. 72 and 48. 16. 17s. 3d. 17. 48f£. 18. £1 18s. 10§d. 19. 4 gal.

20. 1-Hd.    21. £6 18s. ll^d. 22. 3s. llif|d. 23. 15tV 24. £5.

25. £1 16s.

ANSWERS TO EXERCISE XLI. Page 222.

1. 3461185944.    2. 614592288193000.    3. £415 14s. 2j£fd.

4. 8 horses. 5. 60, 45, 36, 26, and 13.    6. 41T7g days. 7. £551.

8.    £432.    9.    74 per cent. 10.    6§ years. 11.    £3600.

12.    £1250.    13. A,    £7200;    B, £6000;    C, £4000. 14. 407.

15.    7645.    16. 48.    17. £133    14s. llT^d.    18. A, £20; B, £16 ;

C, £21 6s. 8d.; D, £10 13s. 4d.    19. £426 12s. 2J^d.    20. 40 receive

each 40 acres 2 roods 16ff perches ; 58 receive each 38 acres 0 roods 32ff perches. 21. £2 8s. 5’76d. 22. £0 3s.lHd. 23.82048632124900.

24. 1455030922747184.    25. 400.    26. 3 boys. 27. £2 9s. 4d.

28. 375.    29. £5600.    30. £1 8s. 6d.    31. £0 12s. 3|d.

32. £799 19s. lxihjd- 33. 4744-46215, etc. 34. 11 chains 25 links. 35. 12 miles 3 furlongs 14 perches 3 yards.    36. £6202 4s. 9d.

37. 99 days. 38. £258 19s. 8f£fd. 39. £0 Is. lO^d-40. 392516271721647000.    41. 642S39650008.    42. £4147 12s. 9d.

43.    £2408 12s. 6Ad.    44.    10 weeks.    45. £2333 6s. 8d.

46    £110 6s. 5d.    47.    £128 16s.    4|d.    48.    24 months; or 2 years.

APPENDIX

SOLUTIONS OF DIFFICULT QUESTIONS.

1.—Sum 14* Exercise XL. Page 208.

The gain upon 3 qrs. is -jffo of 5 qrs. = a of 5 cjrs. = 1 qr.

The gain is therefore f of the cost.

Cost + gain = sell, price = 7s. 6d. per Flem. ell.

i.e., (cost + -3- cost) or of the cost per Flem. ell = 7s. 6d.

Hence, cost per Flem. ell = f of 7s. 6d. = 5s. 7id. ; and,

Cost per French ell = 5s. 7|d. x 2 = 11s. 3d. Answer.

2.—Sum 1. Exercise XL. Page 207.

The gain upon 1 yd. is f of 5 qrs. = § qr. = ^ yd.

The gain is therefore of the cost.

Cost + gain = selling price = 850 guineas.

i.e., (cost + 3^- cost) or f a of the cost, = 850 guineas.

.*. the cost = ff of 850 guineas = £771 17s. lOffd. Answer.

3. —Sum 5. Exercise XL. Page 207.

At 15s. per bag, 30 bags are bought for £22 10s.

Put 6 bags away, and sell the other 24 for £22 10s.

£22 10s. for 24 bags is at the rate of 18s. 9d. per bag.

Then, by selling the 6 bags at this rate, their price is gained.

4. —Sum 12. Exercise XL. Page 207.

£2 3s. 2d. is 87| per cent, or | of the prime cost.

the prime cost was f of £2 3s. 2d. = £2 9s. 4d. Answer.

5—Sum 6. Exercise XL. Page 207.

What the merchant intended to sell for £120, he did sell for £87|;

i.e., for £32f less. Hence, the difference between the price expected and that realized was £32£ for every £87f he received Hence state As £S7| ! £432 10s. : I £32£ ; £160 12s. lOfd. Answer ’

6\—Sum 8. Exercise XL. Page 207.

First Transaction :—

The loss = 15£ p. c. of proceeds = ■£-,rof the proceeds.

In the case of loss . Cost - selling price + the loss.

Hence, cost = (proceeds -t- of proceeds) = f-f of proceeds. .*. the proceeds were of cost ; and -if of £7 16s. = £6 15s, Hence, the loss was £7 16s. - £6 15s. = £1 Is. Loss.

Second Transaction :—

The gain = 121 p. c. of proceeds = 1 of the proceeds.

In the case of gain : Cost = selling price - the gain.

Hence, cost = (proceeds - 1 of proceeds) = | of proceeds.

the proceeds were f of the cost ; and f of £27 = £30 17-,-s. Hence, the gain was £30 17-yS. - £27 = £3 171s. Gain.

Gain, £3 17^-s. - loss, £1 Is. = £2 161s., the net gain.

Total outlay = £7 16s. + £27 = £34 16s.

As £34 16s. I £100 ; *. £2 16|s. I 8-^ p. c. Answer.

second method.

Proceeds, 100 + loss, 15f = cost, 115f.

Hence, the loss was 15£ for every 115f in the cost.

As £115f : £7 16s. :: £15f : £1 Is. The loss.

Proceeds, 100, - gain, 121, = cost, 87-j.

Hence, the gain was 121 for every 871 in the cost.

As £871 I £27 :£12i : £3 17|S. The gain.

Then, as before, £2 16+s. upon £34 16s. is 8^%% per cent. Note.—See similar sum worked on page 206.

7.—Sum, 66. Page 133. Exercise XXIII. Simple Proportion.

Five men save 20 men two days’ work.    _

To save 20 men two days’ work, 5 men must work eight days. But the'20 men would have taken two days more, or ten days. By Proportion, As 5 men | 25 men I'. 2 days : 10 days. Answer.

S.—Sum 72. Page 133. Simple Proportion.

The 5 extra teams, by working 371 days, save the teams already employed (45 - 371) or 71 days’ work; i.e., the teams at first employed would have done in 71 days the work which has occupied 5 teams for 371 days. Hence, there were more than 5 teams in the ratio of 371to 7^> or 5 to 1- Answer, 5x5 = 25 teams.

9.—Sum 58. Exercise XXVI. Page 158. Interest.

£174 Os. 9d. 4- 375 =    = £‘4641. Interest of £1.

£1 + £'4641, or £1’4641 is the amount of £1 for four years.

But, the amount of £1 for four years at compound interest is the fourth power of (£1 + interest of £1 for 1 year) ;* the fourth root of £1'464I gives us £1 + its interest for 1 year.

The fourth root of £1'4641 = sq. root of the sq. root of £1-4641.

V l'464l = 1’21 ; and 1*21 = £1'1 amount of £1 for 1 year.

Therefore, £T is the interest of £1 for 1 year.

And, £T x 100, or £10 is the interest of £100 for 1 year, or the rate per cent. Hence, Ans. 10 per cent.

10.Sum 97. Exercise II. Page 55. (For rides seepages 3S and 3^.)

4 times 40320000 + 1240 times 1240 = 162817600.

-■J 162817600 = 12760, the sum of the two numbers.

Then, to find two numbers whose sum is 12760 and difF. 1240.

(12760 + 1240) 4- 2 = 7000, the greater of the two numbers.

(12760 - 1240) 4- 2 = 5760, the smaller of the two numbers.

77.—Sum 104. Exercise II. Page 55. (For rule see page 36.)

~s/ (605377 x 97) = 2>/ 58721569 — 7663, the greater number.

■>/ (605377 4- 97) = 2\/ 6241 = 79, the smaller number.

12.—Sum 59. Exercise XX VI. Interest. Page 158.

£1590 2s. iUd. - £1280 = £310 2s. 9|d., Compound Interest of £1280.

£310 2s. 9-ikl. 4- 1280 = £’242296875, Compound Interest of £1 for required time.

.". £1’242296875 = amount of £1 at Compound Interest for the required time.

£107^ = 100 = £1 ’075, which is £1 + its interest for 1 year. Now, (l-075)3 = (1’075 x 1-075 x 1’075) = £1’242296875; i.e., £1’075 must be raised to the third power to give the amount of £1 at Compound Interest for the required time. Hence, the time required is 3 years ; answer. See note to No. 9.

* NOTE.—The amount at compound interest upon any sum maybe obtained by raising (£1 + its interest) to the power denoted by the number of years, and then multiplying the result by the number of pounds in the principal.

/


SUPPLEMENT

TO

THE STATE SCHOOL ARITHMETIC.





( '





SUPPLEMENT

TO THE

STATE SCHOOL ARITHMETIC

ARRANGED TO SUIT THE REQUIREMENTS OF

CIj£ ^programme of Instruction

ISSUED BY

THE EDUCATION DEPARTMENT

BY

JOHN J. BURSTON

AUTHOR OF “ THE STATE SCHOOL GRAMMAR

GEORGE ROBERTSON & CO., LIMITED, MELBOURNE, SYDNEY, AND ADELAIDE

1884

SUPPLEMENT.]

INVOLUTION AND EVOLUTION.

A Power of a number is a product obtained by using that number alone as factor any number of times.

Thus, since 5x5 = 25; 5x5x5 = 125; and 5x5x5x5 = 625, then 25, 125, and 625 are each powers of the number 5.

If the number enter as factor twice, the product is called the square ; if thrice, the cube or third power; if four times, the fourth power; if five times, the fifth power ; and so on.

Tlias, 25 is the square of five ; 125, the cube of 5 ; and 625, the fourth power ; while 5 itself is called the root, or first power of 5, the square root of 25, and the cube root of 125.

A Poot of a number is therefore the number which must be multiplied by itself twice, thrice, &c., to produce the given number.

If the root must be used as a factor twice, it is called the square root ; if thrice, the cube root, and so on. Thus, 10 is the square root of 100 ; and 12 is the cube root of 1728.

Note.—The power of a number is frequently expressed by a small figure called an index—thus, 12- signifies the square of 12 ; and 12!, the cube of 12 ; while the product of five threes is denoted 35, and that of seven tens, 107.

The various roots are also indicated by the signs 3f, 4N/, &c.; read the square root; cube root; fourth root, &c. Thus 25 ; or ‘•V 25 means the square root of 25, which is 5 ; and 3V/1728 means the cube root of 1728, which is 12.

The process of raising a number to a given power is termed Involution ; while that of extracting any required Root is called Evolution.

Involution is, therefore, simply multiplication in which all the factors are the same.

Rule.—Multiply, using the given number as a factor twice for the square, thrice for the cube, and so on.

Examples.—Square the numbers 20, and 5|; and cube the number 12.

20 x 20 = 400 ; 5| x 5£ = 30|; and 12 x 12 x 12 = 1728.

Evolution is not so readily accomplished; while the methods employed depend upon the following results of Involution:—

If a number consist of two parts, then—

A. The square of the number itself — the square of each part + tiuice the product of the two parts.

Thus, 252 = (20 + 5)2 = 202 + 2 (20x5) + 52

= 400 + 200 + 25 = 6 25 = 252.

B. The cube of the number itself — the cube of each part + triple the products obtained by multiplying the square of each part by the other part

Thus, 253 = 203 + 3 (202x 5) + 3 (20x52) + 53 or, 253 = -j 8000 + 6000+1500 + 125 f = 15625 ; which = (25 x 25 x 25), or the cube of 25.

The method of obtaining these results may be exhibited thus:—

25

— 20+ 5

25

20+ 5

125

100-t- 25

50

400+ 100

: 625 ~

(a) 400+ 200+ 25

25

20+ 5

3125

2000+1000+125


20+5 20 +5


Multiply by,

202+ (20x5)

(20x5) + 52

202+2 (20x5) + 52 square of i 20+5 Multiply by,

. 2034-2 (202 X 5) +    (20 X52)

(202x5) + 2 (20x52) +53    1250    8000+4000+ 500

203+3 (202x5) + 3 (20x52) +53 zr 15625 zz 8000+6000+1500+125

Therefore, 252 = (20+5)2 = 202+2 (20x5) -|-52 = 400+200+25 = 625 : and, 253 = 203+3 (202x5)+3 (20 x52)+53    8000+6000+1500

+125 = 15625.

From A, we shall proceed to deduce the rule for finding the Square root of a number; and, subsequently, the rule for Cube root from B.

We must first, however, point out that—

Since, 92 = 81 ; 992 = 9801 ; 9992 = 998001, and 93 = 729 ; 993 = 970299 ; 9993 = 997002999,

it is evident that the square of any number cannot contain more than twice as many figures as the number squared; nor the cube of a number more than three times as many as the number itself.

For 9, 99, and 999, are the largest numbers of one, two, and three figures, and their squares have but two, four, and six figures respectively ; while their cubes consist of three, six, and nine figures respectively.

Again—

Since, l2 = 1 ; 102 = 100 ; 1002 = 10000 ; 10002 = 1000000, and, 18 = 1 ; 103=1000 ; 1003 = 1000000 ; 10003 = 1000000000, then, the square cannot contain less figures than twice as many minus one; nor the cube less than thrice as many minus two.

For 1, 10, 100, 1000 are the smallest numbers of one, two, three, and four places respectively, and their squares contain one, three, Jive, and seven figures, while their cubes contain one, four, seven, and ten figures.

Furthermore—

Since T2 = ‘01 ; ‘012 = *0001 ; ‘0012 = ‘000001 ; and T3 = •001 ; ‘OF = ‘000001 ; ‘0013 = ‘000000001, then the square of any decimal contains twice as many decimal places as the decimal itself, and the cube contains three times as many decimal places.

If any number, either integral or decimal, have an exact root, we can at once determine the number of figures of which the root consists, by the simple process of marking off into periods of two for square root, three for cube root, four for the fourth root, &e.

Thus, the square roots of 21025 ; 173056 ; 99‘800i and 1‘0*60*9 each contain three figures, being 145, 416, 9‘99 and 1‘03.

While the cube roots of 970299; 9T125; i728 and ‘001728 each contain two figures, being 99, 4‘5, 12 and ’12.

If, therefore, we first place a dot over the units figure, and thence to the right and left, one over every second figure, in

finding the square root, and over every third figure in finding the cube root, the number of dots or points at once indicates the number of figures in the root.

Now, this marking off into periods of twos or threes, is the first step in the process employed in order to find the square root or cube root of a number.

Hence, to divide into periods for square root,—

Rule.Marh the number off into groups of twos by placing a dot over the units figure, and thence over every second figure in both directions.

We next inspect the period farthest on the left to find the largest digit whose square it contains. This digit is the first figure of the root.

We now regard the root as consisting of two parts,—of which this first figure, considered as tens, is the first part, and the second figure, considered as units, is the second part.

Now, by A, it is seen that : If the square root of a number consist of two parts, then the number itself contains the square of the first part + twice the product of the two quarts + the square of the second part.

Thus, since the square root of 625 = 25 = 20 + 5, then, 625 = 202 + 2 (20 x 5) + 52, i.e., the square of (20 + 5).

But, 202 + 2 (20x 5) + 52 = 202+ (40 + 5) 5.

Hence—

A number contains the square of the first part of its root + the product of twice the first part with the second part added in, multiplied by the second part of the root.

of 625.


Let us now apply this principle to the extraction of the square root

b

625 (25


4


15    225

225


a

625 (20 + 5 = 25 400

(40 + 5) 5 225 = 45x5 = 225

Having divided the number into two periods, 25 and 6 (i.e., 600), we find that 2 is the greatest number whose square is contained in 6, and therefore set this as the first figure of the root, considering it as 2 tens, or 20. We then place twice the first part of the root { i.e., 40), to the left of the number remaining after subtracting the square of the first part of the root, and find that it is contained in 225 about 5 times ; this 5 we'place as the second figure of the root, and also after the Trial Divisor, fi), to complete the True Divisor, Ifi>, which is then multiplied by the second part of the root to obtain the subtrahend.

NOTE.-r—At b., which exhibits the common plan of working, the cyphers showing the local value of the figures 2 and 4 are omitted.

In order to exhibit the analogy between the above process and the method employed in finding the square root of a compound algebraic expression, we take the following from John P. Wilson’s admirable “ Outlines op Algebra ” :

Since, the square of (a b) is a2 + 2ab •+- b2, and or -f- ‘lab -f- b~ — a2 + (2a -j- b) b,

it appears that a, the first term of the root, may be obtained by taking the square root of the first term of the given expression ; and that b, the second term of the root, is contained 2a + b times in the remaining terms, and therefore may be determined by dividing the second term of the expression by 2a.”

We may now give the rule in full, merely remarking, first, that at each stage the root is regarded as consisting of two parts—the part of the root already found and the figure to be next obtained.

TO FIND THE SQUARE ROOT.

Rule.—1. Place as the first figure of the root the greatest number whose square is subtractable from the first period. Subtract and bring down the next period.

2.    Double the part of the root already found, and place to the left of the number so formed as the Trial Divisor.

3.    Divide all but the last figure, of the new dividend, by the trial divisor, placing the result after the trial divisor and also in the quotient.

4.    Call the divisor, now, the True Divisor, multiply it by the figure just placed in the root, subtract the product and bring down the next period.

5. Repeat the process described in 2, 3, and 4, for each of the other figures of the root.

Ex. 1, 2. Find the square root of 3526884 and of 2741 ‘5696.

(1).

3526884 (1878

(2).

2741-5696 (52'36

1

25

28

252

102

241

224

204

367

2868

1043

3756

2569

3129

3748

29984

10466

62796

29984

62796

Ex. 3, 4. Find the square root of 92446'4025, and of '00010816.

(3).

92446'4025 (304-05

(4).

•00010816 ('0104

9

00

604

2446

01

2416

1

60805

304025

204

0816

304025

816

Note 1.—When dividing by 28, in Ex. 1, we have 28 as a remainder ; yet 29 would not have been contained 9 times in 252. It here frequently occurs that the remainder is even greater than the True Divisor.

Note 2._In (3), it was found that the trial divisor 6 (considered

as tens) was not contained in 24, we then put 0 in the quotient and after the 6, and bring down another period 46 ; when we find it will go 4 times. We have to act similarly when we find that 608 (as tens) is not contained in 3040.

THE THEORY OF PERFECT SQUARES.

Def. A perfect square is a number which has an exact square root.

The squares of the numbers 1 to 10, are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Hence, the squares of 1 and 9 end in 1 ; of 2 and 8 in 4 • of 3 and 7 in 9 ; of 4 and 6 in 6 ; that of any number ending in 5 in 25 ; and that of any number ending in cyphers ends in twice as many cyphers.

i.e., (a) All perfect squares end in 1, 4, 9, 6, 25, and 00.

Again, (b) All perfect squares are either divisible by 4 or become so when lessened by unity. 6

From (a) and (b) it follows that no number has an exact square root, unless its two last figures are 01, 0i, 09, 16, 21, 24, 25, 29, 36, 41, 44. 49, 61, 64, 69, 76, Si, 89, 96 ; or an even number of cyphers preceded by one of these, as 4900, 5760000.

(c.) The square root of any proper fraction is itself a fraction.

For a whole number multiplied by itself cannot yield a fractional product. Neither can a fraction squared produce a whole number.

Note. —There is an apparent exception to (c) for the square root of

*9 is 1 ; but then ’9 is not, strictly speaking, a proper fraction, since it represents unity.

(d.) Again, The square of any proper fraction is itself a fraction of less magnitude than the fraction squared, because the product of two proper fractions is always less than either of them.

From (e) and (d) it follows that if the exact root of an integer be not itself an integer, an approximation only can be obtained by continuing the extraction decimally.

(e.) No decimal of an odd number of figures is a perfect square.

For all decimals, when squared, produce products containing an even number of figures. Thus, -32 = ’09 ; ’122 = '0144.

(f.) All roots and powers of unity are expressed by unity itself. Thus, l3 = (1 X 1 X 1) = 1; V 1 = 1 ! and V :9 = V 1 = 1.

(g.) A surd is a number which is not an exact power of any integer or fractioni.e., a number of which the exact root required cannot be found ; as \/ 3, V 12, SN/ 100. See page 7.

Thus, all numbers ending in 2, 3, 7, 8, or an odd number of noughts or decimal places, are surds when their square root is required.

(h.) The square root of any number may be positive or negative ; for the square of any negative number is the same as that of the corresponding positive number. See page 36.

Thus, since both (- 5 multiplied by - 5), and (5x5) — 25 ; the square root of 25 may be considered as - 5 ; or 5. Hence it is that many problems involving square root admit of two answers.

SURDS OR INCOMMENSURABLE NUMBERS.

"When the exact root of a number cannot be obtained, it is called a Surd or Irrational number.

Thus, the square roots of all numbers ending in 2, 3, 7, 8, or an odd number of cyphers, are Surds ; because no perfect square can terminate in these figures.

Again,—If the root of any whole number be not an integer, it is a Surd ; for every power of a mixed number is itself a mixed number.

To find approximately the square root of a Surd :—

Rule.Find the nearest square root of the integral part; then place a decimal point in the quotient and continue the extraction by bringing down, after the remainder, two decimal places at a time until the required, number of root figures is reached. Should there be no decimal, ajfix two cyphers at each step.

To extract the square root of a vulgar fraction :—

Case 1.—If the numerator and denominator be both perfect squares—Extract the square root of each for the fractional root.

Case 2.—If the denominator be not a perfect square,— Make it such by multiplying both numerator and denominator by all the prime factors of the denominator which occur an odd number of times. Find, then, the approximate root of the numerator to the required number of decimal places, and divide the result by the square root of the denominator.

rri -    7    _    7x2    —    . nr ii .

Thus :    - 3X2x3 ' 3x2x3x2    6X6 ’ 0r ” ’

then,    ^ =('/U + 6) = l3'7416- + 6)- 6236...Ans

8 _ 8


Again, _ =


’45    3x5x3


8x5    _    40


3x5x3x5    15X15


40

2 2 5 ?


then, x/TT = \/WV=(\/40-f 15) = (6-324 &c.-f-15) = *4216...Ans.

The more common plan of applying the principle involved in the above is to, Multiply the numerator by the denominator, and then divide the square root of the product by the denominator.

Thus, \/-f = \/40 -f- 8 = (6'324 &c. 8) = *7905 &c. Answer.

Exercise XLII.—SQUARE ROOT.

I.    Find the Square roots of :—

(1.) 1225.    (2.) 3025.    (3.) 164025.    (4.) 1296. (5.) 962361. (6.)

56169. (7.) 197136.    (8.) 19321.    (9.) 627264.    (10.) 419904.    (11.)

9-86965056.    (12.) 33177600.    (130 36966400.    (14.) 768S0'871076.

(15.) -000060559524. (16.) 133401-877809559524. (17.) 30103-291009. (18.) 90619U609.    (19.) -0906190609. (20.) -96196864.    (21.) 1 -4641.

(22.) 100-81766464. (23.) 9802-782081. (240 2571 *402681.    (25.) U.

(26.) Iff. (27.) i|f. (28.) Iff. (29.) |fff. (30.) 19ff. (31.)-ft

II.    Apply Square root to the solution of the following

1.    Find the two numbers whose sum is 8058 and product 7106400. See Rule, page 38.

2.    Fjnd the two numbers whose product is 200'15503, and difference 35 "013. See Rule, page 38.

3.    The product of two numbers is 64089320. and the quotient of the larger divided by the smaller is 1*7. Find the numbers. Rule, page 36.

4.    The product of two numbers is -5050125, and the quotient of one divided by the other is '02. Find the numbers. Rule, page 36.

5.    The product of two numbers is 99825, and the quotient obtained by dividing the larger by the smaller is 33. Find the numbers. Rule, page 38.

6.    Three numbers are proportional—i.e., the first is the same part of the second as the second is of the third ; if the product of the first and third be 28224, what is the second number ?

7.    Three numbers are proportional; if the first be 48, and the third be 4107, what is the second number ?

III.    1. A team of cricketers went into the country to play a match and agreed to share the expenses equally. They paid between theirf ill os., and then found that each had paid as many shillings as there were players. Of how many did the team consist ?

pence per cubic yard as there are cubic yards excavated. What is the depth of the mine ?

3.    A pleasure party, who engaged a boat for the day, and agreed to share the expenses equally, found that each had paid as many pence as there were persons in the party. If the total amount paid was £17 12s. Id., how many persons were there, and what had each person to pay ?

4.    A glazier put in a number of panes of glass at as many pence per pane as there were panes put in. If he received £3 10s. Id., how many panes did he put in ?

5.    A plasterer colours the walls and ceiling of a square room, whose floor contains 182J sq. ft. If the walls are 12 ft. high, how many sq. yds. of colouring does he do ; and what will he receive for the work if he gets as many farthings per yard as there are yards coloured ?

6.    The side of a square is found by extracting the square root of its area. Find the length of fence required to surround a square paddock containing 490 acres.

7.    Find the length in yards of the side of a square field containing 86 acres 2 roods 13'29 perches.

8.    A horse-dealer bought a horse for £16, and sold it to me at a certain rate per cent, profit. If by selling it at £25 I should gain at the same rate per cent., what did I pay for the horse, and what was the rate per cent, profit at which he was sold ?

9.    At what rate per cent., compound interest, will £156 5s. amount in two years to £182 5s. ?

Note.—Divide the amount by the principal, extract the square root, and multiply the result by 100 for the amount of £100 for 1 year. This, minus £100, is the rate per cent.

10.    A body falling through space increases in velocity as the square of the time occupied in its descent. If a stone fall 16 ft. the first second, how long will it take to reach the bottom of a vertical mining shaft 729 ft. in depth ?

CUBE ROOT.

At B, page 9J+8, it is explained that if any number consist of two parts, then—

The cube of the whole = the cube of each part + triple the products obtained by multiplying the square of each part by the other part.

Thus, since 25 is composed of 20 and 5,

then, 253 = 203 + 3 (202 x 5) + 3 (20 x 52) + 53

= 8000 + 6000    + 1500 + 125 = 15625.

For the method of obtaining this result, See page 248.

Any number may thus be separated into two parts and cubed without employing the ordinary process, thus

Rule.To the sum of the cubes of its two parts, add triple the products obtained by multiplying the square of each part by the other part.

Thus, to cube 112, which is (100 + 12), we have,

1123 = (100 + 12)3 = 1003 + 3 (1002 x 12) + 3 (100 x 122) + 123.

= 1000000 + 360000 + 43200 + 1728 = 1404928.

Now, in order to frame a rule for extracting the cube root of a number, we collect the three last terms of which a cube consists into one compound expression, so as to make the whole consist of but two terms—the cube of the first part of the root being the first term.

Thus—

1123 = 1003 + 3 (1002 x 12) + 3 (100 x 122) + 123.

= 1003 + \ (3 X 1002) + (300 + 12) X 12 j. x 12 = 1003 + \ 30000 + 3744 [ x 12.

= 1000000 + (12 times 33744) = 1404928.

These three terms may be collected in a variety of ways, of which we exhibit two below ; and since the rule for cube root depends upon the method employed to collect them into one expression, there are several modifications of the rule—each rule depending entirely upon the method here adopted.

A.    253    =    203    +    (3 x    202 x    5)    + (3 x 20    x    52) + 53    (a.)

=    203    +    -j (3    x 202)    +    3 (20x5)    +    52)x5    (b.)

=    8000    +    (1200 + 300    +    25) x 5    (c.)

=    8000    +    5 times 1525    =    15625.    (d.)

B.    253    =    203    +    (3 x    202 x    5)    + (3 x 20    x    52) + 53    (a)

=    203    +    (3    x 202)    +    (60 + 5)    x 5 x 5    (b.)

=    8000    +    5 times (1200    +    325)    (c.)

=    8000    +    5 times 1525    =    15625.    (c/.)

first figure to form the number which is to be multiplied by the second figure to complete the divisor.

When the root consists of three or more figures, the other figures are obtained in the same way as the second figure, the part of the root already found being considered as a number of tens, and its square tripled for the Trial Divisor.

The triple of the square of the part of the root already found may,

however, be more easily obtained by adding the square of the last figure to the last two lines, and affixing two cyphers

thus, 325 + 1525 + 52 = 1875, and 187500 is triple the square of 250.

We shall now write the rule more fully, and, as it is unavoidably lengthy, we shall first explain separately how the trial divisors are obtained.

CUBE BOOT.

RULE IN FULL.

To mark off into periods of threes :—Place a dot over the units' figure and thence over every third figure to the left,—and to the right, also, when there is a decimal.

To FORM THE TRIAL DIVISORS :—

a.    Place to the left of each new dividendformed by adding the next period after the remainder—triple the part of the root already obtained, leaving sufficient space between for the trial divisor.

b.    First trial divisor.—Multiply the number on the left by the first figure of the root, and add two noughts to the product.

c.    All other trial divisors.Add the square of the last figure (of the root) to the last two lines and affix two noughts7

BULE FOB CUBE BOOT.

a.    From the first period subtract the greatest cube number possible, and place its root in the quotient as the first figure.

b.    Bring down the next period, and divide by the First Trial Divisor (formed as above described), and set the

result in the quotient and also after the number on the left of the trial divisor.

c.    JTLutiply the number now on the left by the last figure in the quotient, and add the result to the trial divisor in order to obtain the True Divisor.

d.    Multiply the True Divisor by the last figure in the quotient, subtract, and bring down the next period.

e.    Form the next trial divisor, and repeat the operations described in b, c, d, continuing on this plan until the last period has been brought down and dealt with.

Example 1.—Find the cube root of 231475544.

231475544 (614 216

181

10800

181

15475

10981

10981

1834

1116300

7336

4494544

1123636

4494544

Example 2.—Find the cube root of 1131366088.

113i366088 (1042 1

304

30000

1216

131366

31216

124864

3122

3244800

6244

6502088

3251044

6502088

NOTE 1.—In Ex. 2, the trial divisor, 300, is not contained in the new dividend, 131, so we place a nought in the quotient, another after the number on the left, and two after the trial divisor, and continue the division by bringing down the next period.

262 THE STATE SCHOOL ARITHMETIC. Example 3.—Extract the cube root of 80901828T25.

80901828-125 (432'5 64

123

4800

369

16901 *

5169

15507

1292

554700

2584

1394828

557284

1114568

12965    55987200 280260125

64825

56052025 280260125

NOTE 2.—In Ex. 3, the last figure of the root is a decimal, because a decimal period had to be brought down to obtain it. We may also here note that the cube of any decimal has three times as many figures as there are figures in the decimal; and vice versd, in finding a root there must he one figure in the decimal for every three decimal places in the cube number.

Exercise XLIIL—CUBE BOOT.

I. Find the cube roots of the following :—

(1.) 15625000.    (2.) 2299968.    (3.) 1131366088.    (4.) 231475544.

(5.) 191102976000.    (6.) 6634074439.    (7.) 31 006494199296.    (8.)

355030270-485129.    (9.) 13-824.    (10.) 25-050878776.    (11.) fff.

(12.) 5Hff (13.) m•    (14-) -244140625.

H. —

I.    At what rate per cent., compound interest, will £62 10s. amount to £108 in 3 years? See note to sum 9 of III., page 256. But here extract the cube root, as the time is 3 years.

2.    The edge of a cube is found by extracting the cube root of its contents in feet or inches. Find the length in inches of a cubical block of stone containing 34 cubic feet 567 cubic inches.

3.    I want to excavate a cubical tank which shall hold 1953 "125 cubic feet; what depth must I make it ?

4.    A cubical iron tank contains 1638'4 gallons of water. Reckoning 6£ gallons to the cubic foot, what is the height of the tank ? _

5.    A cubical box contains 225 smaller boxes each 3 in. long, 2§ in. broad, and lg in. deep ; what is the length of the edge of the larger box ?

6.    The dimensions of rectangular blocks of granite are 16 ft. 8 in., 16 ft., and 3 ft. 9 in. ; what must the edge of a cubical block be in order to contain the same solidity ?

7.    A cubic foot of water weighs about 1000 ozs. ; how deep is cubical tank which holds 23 tons 18cwt. 1 qr. 21 lbs. 15 ozs. ?

8.    Find the sixth root of 9109555799784049, and the ninth root of r 5159780352.

SQUARE MEASURE;

OR, MENSURATION OF RECTANGLES.

Square Measure is so named from being the measurement of four-sided surfaces having square corners, or right angles. (See page 66.) By this measure we compute the magnitude of any plane or level surface of which the boundaries are straight lines and the angles right angles. (See page 59.) All such surfaces have four square corners (or right angles), and four straight sides, each opposite two of which are of the same length and parallel to each other.

Such a surface is called a Rectangle or a Rectangular Parallelogram ; also a Quadrangle.

Note.-—A rectangle which has its four sides of the same length is called a Square.

The measurement in the direction of the longer side is called the Length, and that in the direction of the shorter side the Breadth. Every surface has, of necessity, both length and breadth ; and these are called its Dimensions. The magnitude of a rectangle depends upon its dimensions. The magnitude of any surface is called its Area.

By the Area we mean the number of square units of any given size the surface contains.

Note.—In measuring the surface of a wall, the dimensions are its Length and Height.

The area of a rectangle increases as either the length or the breadth is increased ; and, if both its length and its breadth be increased, the area is increased in the ratio of the product of the numbers which show how many times the length and the breadth, respectively, have been increased.    o

Thus, while doubling either the length or the breadth increases the area twofold, doubling both increases it fourfold. Again, if we double one and triple the other, we increase the area six times; and if we triple both we increase it nine times. Conversely, if we take but one-half the length and one-fourth of the breadth, we diminish the area eightfold. That is, we multiply the area by \ of \, or by ¿th; and, consequently, diminish it to one-eighth of its former size.

In every rectangle there are as many square units of any given name (whether sq. inches, sq. feet, sq. yards, &c.), as is expressed by the product of the linear units of the same name contained in the length and breadth. See page 69.

Fig. A.

Thus, in Fig. A. the length contains 5 units and the breadth 3 units; and, consequently, the area contains 3 times 5, or 15, square units, the side of each „ square being the length of one of the linear units in the length or breadth.


1

2

3

4

5

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15


If the units in the length and breadth be inches, feet, or yards, then the 15 sq. units will be sq. inches, sq. feet, or sq. yards. Or, if they each represent a chain or a mile, then is the area 15 sq. chains, or 15 sq. miles, as the case may be.

From the principles we have explained we obtain the following simple rule to find the area of any rectangular surface.

Rule.—Multiply the length by the breadth; or,— Multiply together its two dimensions.

Note.—It is usual first to express the length and breadth in terms of the same name ; and it must be carefully remembered that the product is square units of that particular name.

Thus:—Length    in    inches    x    breadth in inches = area in sq. inches.

Length    in    feet    x    breadth in feet == area in sq. feet.

Length    in    yards    x    breadth in yards == area in sq. yards.

Length    in    chains    x    breadth in chains = area in sq. chains.

Length    in    miles    x    breadth in miles = area in sq. miles.

Since the area of a rectangle is the product of its two dimensions, we have only to divide the area by either dimension to obtain the other.

Thus, if a room 15 feet long contains 180 sq. ft. of flooring, then its width is (180 -7- 15) ft. = 12 ft.

Hence,

Area —■ length produces Breadth or Width.

Area -j- breadth or width produces Length.

Since the area of a square is the product of the two equal numbers representing its sides, The side of a square is found by extracting the square root of its area.

Thus, the side of a square field containing 86 acres 3664 sq. yds. is 648 yds., the sq. root of 419904, the number of yds. in the area.

Since a cube has six equal surfaces, its area is found by multiplying the square of its edge by 6.

Thus, the surface of a cube whose edge is 5 feet = (52 X 6) ft. =■ 150 sq. ft.

Conversely, if the surface of a cube be known, its edge may be found by dividing by 6, and extracting the sq. root of the result.

Thus, the edge of a cube the area of whose six surfaces is 150 sq. ft., is n/(150    6) = x/25 = 5 ft.

While square feet divided by linear feet give linear feet, and square yards divided by linear yards give linear yards, it must be remembered that square measure divided by square measure, or linear measure by linear, gives an abstract number.

Thus, we can only divide 46 sq. yds. 6 sq. ft. by 3 sq. yds. 8 sq. ft., when we wish to find how often the latter quantity is contained in the former.

Examples Worked Out.

Ex. I.—A sheet of foolscap is 16^ in. long and 13£ in. broad : into how many square pieces, each an inch in length, can it be cut ? .

16J inches X 13| inches = 218£ sq. in. Answer, 218f pieces.

Ex. 2.—A room is 16 ft. in length and 12 ft. 6 in. in breadth : find the area of the floor.

16 feet X 12£ feet = 200 sq. ft. Ans.

Ex. 3.—What is the area of a rectangular court-yard 84 ft. long and 60 ft. 9 in. broad ?

84 ft. x 60f ft. = 5103 sq. ft. = 567 sq. yds. Ans.

Ex. 4.—The length and breadth of a rectangular paddock are 30 ch. 75 links and 28 ch. 80 links : how many acres does it contain ?

3075 ch. x 28‘80 ch. = 885’6 sq. ch. = 88-56 ac. Ans.

Note.—Here observe that 885‘6 sq. chains is at once reduced to acres by removing the dec. point one place tc the left. See page 60.

Ex. 5.—A rectangular enclosure 85 yds. long and 65 yds. broad is surrounded by a foot-path 2£ yds. wide : find the area of the footpath. b. Find also the area of a strip of land, as wide as the path, dug for a flower-bed all around the enclosure inside the fence.

Note 1.—We have here to find the area of a hollow rectangular surface (commonly termed a hollow square), and this is most easily done by finding the area of the outer rectangle, and then subtracting the area of the inner one.

Thus :—Length of outer = 85 yds. Breadth of outer = 65 yds.

Area of outer = 90 yds. Area of inner = 85 yds.


+ 2\ yds. + 2\ yds. = 90 yds. + 2J yds. + 2\ yds. = 70 yds,

x 70 yds. = 6300 sq. yds. x 65 yds. = 5525 sq. yds.


Difference of areas, or area of the foot-path = 775 sq. yds. Ans. Again—

b. Area of enclosure = 85 yds. x 65 yds. = 5525 sq. yds.

Area, less flower-bed = 80 yds. x 60 yds. = 4800 sq. yds.

Difference of areas, or area of the flower-bed = ^25 sq. yds. Ans.

Note 2.—It will be noticed that when the area is required of a strip of uniform width surrounding a rectangle, twice its width is added to get the length and breadth of the outer rectangle if the strip be on the outside, and the same subtracted if on the inner side.

Exercise XLIV.—SQUARE MEASURE.

1.    How many square inches can a boy draw on a slate 12 inches long and 8 inches broad ?

2.    How many square feet and inches in the top of a table 4 ft. 6 in. long, and 2 ft. 10 in. broad ?

3.    A wall is 9 ft. 6 in. high, and 16 ft. 6 in. long ; how many square yards of paper will cover it ?

4.    A room is 18 ft. long, 12 ft. broad, and 10 ft. high ; how many square yards of oil-cloth will cover the floor ? See page 60.

5.    The Age newspaper is printed on paper of the size termed double elephant (See page 67, Table) ; how many square feet and inches will a sheet cover ? Find also the least No. of acres which any exact No. of sheets will cover, and the No. of sheets required.

6.    A board, 12 ft. 10 in. long, is 1 ft. in. in breadth ; find its area.

7.    What is the superficial measurement of a pane of glass 4 ft. 6 in.

in length, and 2 ft. 3^ in. in breadth ?    .

8.    What surface does a target present which is 8 ft. 6 in. high and 6 ft. 9 in. broad ?

9 The fence on one side of a rectangular garden is 18 roasm length, and that on the other side 15 rods; find the area of the garden.

10.    A street, a chain and a-half in breadth, is 35 ch. in length ;

what is the area of the street without the two footpaths, if they are each 11 ft. wide ?    _

11.    How many square feet does a verandah 5 ft. wide cover which extends around a house having 40 ft. frontage by 30 ft. depth ?

12.    A rectangular play-ground is surrounded by a plantation of shady trees, which is 7 feet in width ; if the play-ground itself be 61 yards long, and 51 yards broad, find the area of the plantation, and express it as the decimal of an acre, Irish measure.

13.    A square play-ground is enclosed by a fence, the erection of which cost £35, at 8s. 9d. per rod. What is the area of the ground ?

14.    How many trees may be planted in a belt or strip 74 ft. wide around the inside edge of a playground 80 yds. in length and 75 yds. in breadth, allowing 135 sq. ft. for each tree ?

15.    How many slates, 7 in. by 10 in., will completely cover the top of a desk 15 ft. long and 1 ft. 2 in. broad ?

16.    How many children should be taught in a room 60 ft. long, 18 ft. broad, and 12 ft. 6 in. high, if each child should be allowed 15 sq. ft. of flooring ?

17.    Find the area or superficial measure of the six sides of a cube, whose length is 1 ft. 104 inches.

18.    Find the area of the four walls of a room, wdiose floor is 12 feet square, and height 104 feet.

19.    How much paper, 2 ft. 3 in. wide, is required for the walls of a room 16 ft. long, 15 ft. broad, 11 ft. high ?

20.    What length of paper, 1 ft. 9 in. wide, will cover the walls and ceiling of a room 15 ft. by 14 ft. by 10| ft ?

21.    How many flooring boards, 20 ft. long and 6 in. wide, will floor a room 60 ft. long, 18 ft. wide, and 12 ft. 74 in. high ?

22.    A room is 14 ft. long, 12 ft. wide, and 9g ft. high ; how many square yards of wainscotting will line it all round to the height of 9 in. more than half way up ?

23.    What is the area of a grass plot in an enclosure 74 yards square, if a gravel walk 6 ft. wide extends all around it inside the fence, and a circular fish pond containing 60 square yards occupies the centre ?

24.    How many bricks, 9 in. long and 4 in. broad, will pave a square court-yard containing two-fifths of a rood, allowing for a path 3 ft. wide all round ?

25.    A storm-water channel is 20 chains long, 8 ft. deep, and 33 yds. wide ; how many cubic yards of earth have been removed ? And what is the area of the surface of the water in it when full ?

26.    A room is 16 ft. long, 14 ft. wide, and 94 ft. high. What will it cost to cover the walls with paper at Is. 9d. a piece, each piece of paper being 20 ft. long and 21 in. wide ?

27.    What length of paper, § yd. wide, will be required to cover the walls and ceiling of a room 30 ft. 7 in. long, 15 ft. 2 in. broad, and 14 ft. 10 in. high, there being a fire-place 4 ft. high and 3 ft. lOf in. broad, and doors and windows whose combined area is 42 sq. ft. ?

28.    Find the area of the four walls and ceiling of a square room, the area of whose floor is 121 sq. ft., and whose height is 9ft. 6in. How much paper, 2 ft. 9 in. wide, will cover the walls ?

29.    Find the area of a foot-path, 2 yards wide, all around a block of houses 956 yds. in length, and 766 yds. in breadth.

30.    A rectangular enclosure is five times as long as it is broad, and contains 4£ acres. What is the area of a strip of plantation, 7 ft. wide, running all around inside the fence ?

31.    Find in chains and links the side of a square field which contains 406 ac. 1 rd. 25 per.

32.    Painting the outside of a cubical tank at 9d. per square yard cost £1 2s. 2fd. ; what was its depth ?

33.    A rectangular area is five times as long as it is broad, and contains 50 acres ; find its breadth.

34.    The width of an enclosure which contains 9126 sq. yds. is two-thirds of its length. Find the latter dimension.

35.    A squatter has a square mile of land, and encloses with it a two-chain road on the north and south, and a road one and a half chains wide on the east and west. How many acres does the enclosure contain ?

36.    The Argus newspaper, Melbourne, contains eight pages, each 25 in. long and 18 in. broad. If a subscriber begin on Thursday, the 1st day of January, 1880, to save up his daily copy to paper the walls and ceiling of his house, which contains four rooms, each 12^ ft. square and 10^ ft. high, and a passage 4 ft. 2 in. wide, when will he have sufficient paper for his purpose, allowing 18 copies less for the space taken up by the doors, windows, &c. ?

37.    In slating a roof fths of each slate is covered by the overlap of other slates, flow many slates, each 13^ in. long and 8 in. broad, will cover the gable end roof of a school which is 60 ft. long and 20 ft. wide at the eaves, the top of the roof being 7 ft. 6 in. higher than the walls ? And what must be the height above the walls of a roof to the same school that Ath as many more slates may be required ?

38.    The breadth of a room is 15 ft. : the cost of plastering the walls, at 3s. per sq. yard, is £11 11s., and that of carpeting the room, at 4s. a yard, is £5 10s. What is the height of the room ?

39.    How many deal planks, each 6 ft. long, 6 in. wide, and £ in. thick, will floor a kitchen 14 ft. 8 in. long and 11 ft. 3 in. wide?

40.    The edges of a block of granite are in the proportion of 1, 2, 3; the area of its six surfaces being 49| sq. yds.; find the actual length of its edges.

CUBIC MEASURE;

OR, MEASUREMENT OF SOLIDITY AND CAPACITY.

Every solid body lias three dimensions—length, breadth, and thickness (sometimes called the height, width, and depth).

The number of cubic inches, cubic feet, or cubic yards, &c., in a solid body is called its Contents. (Sometimes also the Volume, Solidity, or Solid Contents; and in the case of a hollow space or a vessel, the Volume or Capacity.)

Cubic Measure is employed to find the contents of any defined space or solid body bounded by six rectangular surfaces. See pages 60 and 61.

Such a body is called a Rectangular Parallelopiped ; or, if its three dimensions be the same (i.e., of equal length), it is a Cube. See page 61.

Examples of the figure termed a Rectangular Parallelopiped are very common; as, for instance, a common box, an ordinary room, a square tank, squared timber, &c.

To find the cubical contents of such a figure :—

Rule.Multiply together its length, breadth, and thickness;

Or, Find the continued product of its three dimensions.

Note.—It is customary to reduce these dimensions to the same name before multiplying; and, as in square measure, it must be carefully remembered that the product obtained is of that name to which the dimensions were reduced ; but here they are, of course, cubic inches, cubic feet, &c., instead of square inches, square feet, &c. See pages 69 and 70.

Beginners are very apt to confound cubic measure with square measure ; we shall, therefore, here state a few general principles.

1.    Ail dimensions—length, breadth, &c.—are lineal measure.

2.    Area, or superficial contents, is always square measure.

3.    Solidity, or cubical contents, is always cubic measure.

Also,    '

1.    The product of any two dimensions is square measure.

2.    The product of any three dimensions is cubic measure.

3.    Square measure divided by lineal gives lineal measure.

We may repeat these principles thus :—

1.    Length, breadth, depth, &c., are all Long measure.

2.    Surfaces, land, and all areas are Square measure.

3.    Solidity, volume, and capacity are in Cubic measure.

Again,—

a.    Length X length produces Area in Square measure.

b.    Lineal X lineal X lineal gives Cubic contents.

c.    Square units X lineal units gives Cubic units.

And, therefore,

d.    Area or square units -f- lineal units gives Lineal units.

e.    Cubic measure -j- square measure gives Long measure.

f.    Cubic measure lineal measure gives Square measure.

In other words—

a.    Long inches X long inches gives Square inches.

b.    Inches x inches x inches gives Cubic inches.

c.    Square inches x lineal inches gives Cubic inches.

d.    Square inches -f- lineal inches gives Lineal inches.

e.    Cubic inches -f- square inches gives Lineal inches.

f.    Cubic inches ■— lineal inches gives Square inches.

NOTE.—Substitute the word “feet,”yards,” &c., for the word “ inches” in each of the last six statements, and the assertions made will still be true.

Remark.—In repeating the foregoing principles so often, we may be charged with tautology ; but experience has convinced us that it is in grasping the distinctions they point out that the learner finds his greatest difficulties.

Example 1.—A rectangular tank is 5 ft. 6 in. long, 4 ft. wide, and 3 ft. 9 in. deep ; find its cubical capacity.

5^ ft. x 4 ft. x 3| ft. *= (V x t x V) c- ft- = 82? c- feet- Ans-

Example 2.—A box is 3 ft. 9 in. long, 2 ft. 6 in. wide, and 2 ft. 8 in. deep ; find its cubical contents.

45 in. x 30 in. x 32 in. = 43200 c. in. = 25 c. ft. Ans.

Or, 3f ft. x 2J ft. x 2f ft. = (V5 x ■§ x f) c. ft. = 25 c. ft. Ans.

Example 3.—A cube contains 9 c. ft. 73 c. in. ; find its edge.

9 ft. 73 in. = 15625 c. in.; and V8625c. in. = 25 in. = 2 ft. 1 in. Ans.

Example 4.—In making an ash-pit 7 ft. 9 in. long, and 5 ft. 3 in. wide, 325i c. ft. have been excavated; what is the depth of the pit 1

325^ c. in. -r (7| x 5J) sq. ft. = (6f1 x ^Vr) ft. = 8 ft. Ans.

Exercise XLV.—CUBIC MEASURE.

1.    Find the number of cubic inches in a cube whose edge is 1 ft. 8 in.

2.    Find the contents, in cubic feet and cubic inches, of a box whose dimensions are,—length 3 ft. 8 in., breadth 2 ft. 4 in., and depth 1 ft. 10 in.

3.    Find the solid contents of a block of Harcourt granite, which is 5 ft. 10 in. long, 4 ft. 6 in. broad, and 2 ft. 9 in. high.

4.    Find the number of cubic feet, &c., in a beam of squared timber 18 ft. 6 in. long, whose end contains 128 square inches.

5.    How many children should be taught in a school-room 60 ft. long, 18 ft. broad, and 12f ft. high, allowing 135 c. ft. of air for each child ?

6.    How many loads of stone of 1 cubic yard each have been taken out of a quadrangular quarry which is 20 yds. long, 12 ft. broad, and 7 ft. 6 in. deep?

7.    How many boxes of pencils, each 6f in. long, 2| in. broad, and 2£ in. deep, can be packed in a case 3 ft. lOf in. by 3 ft. 9h in. by 3 ft. 6f in. ?

8.    How many State School Arithmetics, each 7f in. long, 5 in. broad, and f in. thick, will fill a case 4 ft. 4£ in. by 2 ft. 3f in. by 1 ft. 74 in. ?

9.    How many packets of note-paper, each 8 in. by 5 in. by J in., will make up a parcel containing 3 c. ft. 1536 c. in. ?

10.    How many packets of Chinese crackers, 4 in. square and f in. in thickness, will fill a cubical box whose edge is just 1 foot?

11.    Find the contents of a cupboard whose front is 4 ft. square and depth 3 ft.

12.    Of two boxes of nibs, one is 2f in. by 1| in. by fin., and the other 2^ in. by If in. by £ in. How many more of the smaller size than of the larger can be packed in a box whose dimensions are 10^ in., 8 in., and 7 in ?

13.    A wall is built a brick and a-half (or 13f in.) in thickness ; how many bricks will it require if its length be 40 ft., and height 14 ft., supposing the dimensions of a brick to be 9 in. by 4f in. by 2f in. ; and allowing 1 brick less in every 8 on account of the space filled by the mortar, and 28 bricks for ventilators inserted ?

14.    Find the cubical contents of a block of stone, measuring. 5 ft. 7 in. by 3 ft. 5 in. by 1 ft. 3 in. 8

16.    The edge of a cube is 1 ft. 5 in. ; what is the length of the edge of another cube, which contains 27 times as much matter ?

17.    A stone wall is four times as high as it is broad, and sixteen times as long as it is high. If it contain 256 cubic yards, what is its length ?,

18.    The water in a channel, which is 60 ft. wide, is of an average depth of 5 ft., and flows at the rate of 4£ miles per hour. How many

c. yds. does it convey per minute ?

19.    How many planks of Kauri pine, each 20 ft. long, 15 in. wide, and 3 in. thick, can be stowed away in a ship’s hold, which is 40 ft. long, 15 ft. wide, and 13^ ft. deep ?

20. A ship’s hold is 96 ft. long, 46 ft. broad, and 15 ft. deep ; how — many cases, each 3 ft. long, 1 ft. 6 in. broad., and 1 ft. 3 in. deep,

can be stowed away in it ?

21.    A wall, 3 times as high as it is broad, and 32 times as long as high, contains 166| loads of stone, each 1 cubic yard; what is the length of the wall ?

22.    A luggage van is 20 ft. long, 12 ft. wide, and 10 ft. high ; how many bales of wool, each 3 ft. 4 in. long, 2 ft. broad, and 2 ft. 6 in. deep, will it hold ?

23.    One cube contains 42875 c. ft., and another 42’875 c. ft. ; how many inches is the edge of the first longer than that of the second ?

24.    The area of the six surfaces of a cube is 1944 sq. in., and the cube itself contains 5832 c. in. ; find, by the application of simple division only, the length of the cube.

DUODECIMALS OR CROSS MULTIPLICATION.

Ry Duodecimals we are enabled to multiply the dimensions feet, inches, &c., by feet, inches, &c., without first reducing them to a single denomination.


Here the unit of measure is the foot (whether linear, square, or cubic), and this is subdivided into primes, seconds, thirds, fourths, &c., marked ' "    Thus, 9 ft. 8' 5" T" = 9 feet 8

primes 5 seconds 7 thirds. These terms decrease in value in a twelvefold ratio, a prime being the twelfth part of a foot, a second the twelfth of a prime, a third the twelfth of a second,

and so on.

A foot is thus divided into twelfths, twelfths of twelfths, &c., and the fractional equivalents of a prime, second, third, &c., are therefore TV tU> ttVs> &c*


The products obtained by multiplying together any given dimensions are in like manner subdivided into feet, primes, seconds, &c., and it is important to remember that:—

In Long Measure the primes are linear inches.

In Square Measure the seconds are square inches.

In Cubic Measure the thirds are cubic inches.

To place quantities for Cross Multiplication :—

Place one under the other, feet under feet, primes under primes, seconds under seconds, &c.

RULE FOR CROSS MULTIPLICATION.

Rule.—Multiply each term of the multiplicand by the feet (if any) in the multiplier, dividing each product, except feet, by 12 ; put down each remainder under the term multiplied and carry to the next product. Multiply next by the primes, and then by the seconds, &c., placing the first remainder one place farther to the right in each case. Lastly, add up the partial products, dividing each sum by 12 except the sum of the feet column.

Examples.—Multiply 10ft. 7in. by Oft. 8in. ; and 18ft. 11 in. 10 parts by 10 ft. 9 in. 8 parts.

ft.

in.

ft.

sq. ft.

a. 10

7

b. 18

11'

10'

205 r 10"

' 4"' g""

9

8

10

9

8

12

12

95

3

189

10

4

22 in.

56-144ths.

7

0

8

14

2

10

6

1

0

7

10

8

tVV sq. in. =

: * l- in.

102

3'

8"

205

1'

10"

4'"

' 8'"'.

Ans.

= 102 sq.

. ft.

44 sq.

in. Ans.

a.

Ans. b. —

205 sq.

ft.

22*

sq. inches.

Observe that we “multiply every term of the multiplicand beginning with the lowest by each term of the multiplier beginning with the highest. ”

NOTE.-Instead of dividing by 12, we may more readily reduce each product to units of the next higher name by applying the pence table. Thus, in 6., we may say 10 times 10 = 100, 100 pence = 8s. 4d., put down 4 and carry 8 ; then 10 times 11 = 110, and 8 = 118, then 118 pence = 9s. 10d., put down 10 and carry 9.

The following points should be carefully noted :—    '

1. Before applying the rule for Cross Multiplication, yards must be reduced to feet, and all fractions of feet must be expressed as twelfths, twelfths of twelfths, &c.

2.    Feet must never be divided by 12 in working duodecimally.

3.    The product of two dimensions is square, and the continued product of three dimensions cubic measure.

4.    The primes and seconds are reduced to square inches by multiplying the former by 12 and adding in the latter.

5.    The inches in cubic measure are obtained by multiplying the primes by 12 and adding in the seconds, and then multiplying the results by 12 again and adding in the thirds.

6.    In Square Measure the primes are often called duodecimal inches ; each represents an area one foot in length by one inch in breadth. Seconds in long measure are sometimes called parts.

7.    Feet X feet give feet; Feet X 'primes give primes; dec.

Primes X feet give primes ; Primes X primes give seconds; due.

Sec. X feet give seconds ; Seconds X primes give thirds ; dec.

Thirds X feet give thirds; Thirds X primes give fourths; die.

Exercise.

1.    Multiply 8 ft. 11 in. by 7 ft. 9 in. Ans. 69 sq. ft. 1' 3"; or 69 sq. ft. 15 sq. in.

2.    Multiply, duodecimally, 8 ft. 3' 10" by 7 ft. 11' 9". Ans. 66 sq. ft. 55^j sq. in. ; or, 66 sq. ft 4' 7" O'" 6"*.

3.    Find, by cross multiplication, the area of the four walls of a room of which the length, breadth, and height are—15 ft. 10 in., 13 ft. 7£in., and 10 ft. lOin. respectively.* Ans. 70sq. yds. 8 sq. ft. 38 sq. in.

4.    How much paper, three-quarters of a yard in width, will cover the walls of a room 12 ft. 7 in. long, 11 ft. 9 in. wide, and 9 ft 11 in. high?* Ans. 71yds. 1ft. 5§f in.

5.    Find, by duodecimals, the cubical contents of a cubical block

of granite whose edge is 17 ft. 9 in. 5 parts. Ans. 5625 c. ft. 2 10 IT" 3""    5""" ; or, 208 c. yds. 9 c. ft. 419Ty£r c- in-

6. The dimensions of a stone pillar are—20 ft /in., 12 ft. 3 in., and 80 ft. 4XV in. ; how many cubic yards of stone does it contain ? Ans. 750 c. yds. 17 c. ft. 1680£ c. in.

* Note._To find the area of the four walls of a room    _

Rule—Multiply twice the sum of the length and breadth by the height.

The length of paper is then found by dividing the area by the width of the paper ; so that, in the case of the width being given m the fraction of a yard or foot, we have only to multiply the area m yards or feet by the fraction inverted, for the area of the walls is also the area of the paper required to cover them.

Thus ayd. = ^ft. And paper required in sum 4 = (12 ft. 7 in. + lift’. 9in.) X 2 X 9ft. 11 in. X fit. = 214ft. in.


RULES IN ELEMENTARY MENSURATION.

Some of the following rules are taken from “ Chambers’s Mathematics.

1.    To find the area of any parallelogram, whether it is rectangular or not :—

Rule.Multiply the length of one side by the breadth measured perpendicularly to that side.

2.    To find the area of a triangle :—

Rule A.—Multiply the length of its longest side by the breadth measured perpendicularly to that side from the opposite angle, and divide the result by 2.

Rule B.From half the sum of the three sides, subtract each side separately; multiply together the three remainders and the half sum of the sides, and extract the square root of the continued product.

3.    To find the hypothenuse of a right-angled triangle :—

Rule.—Add together the squares of the other two sides, and extract the square root of the sum.

4.    To find one of the other sides of a right-angled triangle :—

Rule.—From the square of the hypothenuse, subtract the square of one of the sides, and the square root of the remainder is the other side.

Note.—Rules (3) and (4) depend upon the principles demonstrated by the 47 prop, of the 1st Book of Euclid, viz.—The. sum of the squares upon the shorter sides of a right-angled triangle is equal to the square of the, greatest side, that is, of the hypothenuse.

5.    To find the area of any plane surface, bounded by any number of straight lines :—

Rule.—Divide the surface into triangles, and add together the areas of these.

Or, Divide the whole into triangles a'nd quadrangles, and find the combined area of these.

6.    To find one of the dimensions of a rectangle :—

Rule.—Divide the area by the other dimension.

7.    To find the side of a given square :—

Rule.—Extract the square root of its area.

8. To find the circumference of a circle :—    •

Rule.Multiply the diameter by S'1416.

Note.—Or by 3|, when no great accuracy is required ; and by 3T415926536, when very great accuracy is required.

9.    To find the diameter of a circle :—

Rule.Divide the circumference by 3'14-16.

10.    To find the area of a circle :—

Rule.—a. Mult, the square of the radius by 3'1416.

b.    Mult, the circum. by one-fourth the diam.

c.    Mult, the square of the circum. by '0796.

d.    Mult, the square of the diam. by7854

e.    Mult, circum. by diam., and divide by 4

11. To find the edge of a cube :—

Rule.—Extract the cube root of its solidity; or the square root of the area of one of its surfaces.

12.    To find the surface of a cylinder :—

Rule.—Multiply together the circumference and height, and add in the area of the circles forming its ends.

13.    To find the surface or area of a sphere :—

Rule.—a. Mult, the square of the diam. by S'1416.

b.    Mult, the circum. by the diameter, f

c.    Take 4 times the area of its great circle. 9

14.    To find the solidity of a cone or a pyramid :—

Rule.—Multiply the area of the base by ^ of the height.

15.    To find the cubical contents of a sphere :—

Rule.—a. Multiply the area by \ of the radius.

b.    Multiply the cube of the diameter by ‘5236.

c.    Multiply the area of its great circle by two-thirds

of its diameter.9

16.    To find the solidity of a cylinder or a prism.

Rule.—Multiply the area of its end by its height.

17.    To find the capacity of a cylinder in gallons.

Rule.Divide its contents in cubic inches by 277'274.

When great accuracy is not required, the cubical contents of a cylinder and its capacity in gallons may be more readily found by Rules 18 and 19, following, where the dimensions must be taken in feet.

For rough calculations, reckon 6| gallons to the cubic foot.

18.    Contents of a cylindrical tank in cubic feet:—

Rule.Square diameter x depth x    (Correct to 1

in 1000.)

19.    Capacity in gallons of a cylindrical tank :—

Rule.a. Multiply the contents in c. ft. by 6 j-. (Rough.)

b. Square diameter x depth x ff. (Correct to 1 in 12000.)

20.    To find the cubic contents of a rect. prismoid f :—

Rule.—To the areas of the top and bottom add four times the area of the middle section, measured horizontally, and multiply the result by one-sixth of the depth or height.

Note.—The length of the middle section is found by taking half the sum of the length of the top and bottom, and the breadth by taking half the sum of the breadths. 10

21. To find the capacity of a vessel of the form of the frustum of a cone (as a zinc tub, or a common bucket).

Rule.To the product of the diameters of the two ends add the square of each diameter, then multiply by ‘2618 of the depth.

Note.—In place of Rules 20 and 21, it is customary, when roughly computing the solidity of a prismoid or the frustum of cone, to apply the following conventional rule Multiply the product

of the wean lenyth and breadth (in the one case), or the square of the mean diameter (in the other case), by *7854 and the result by the depth.

Exercise XLVL— MENSURATION.

1.    The length and breadth of a rectangle are 13 ft. 9 in. and 11 ft. 11 in., respectively, and another parallelogram of which the sides are also 13ft. 9 in. and 11 ft. 11 in., but which is not rectangular, has one of its longer sides but 7 ft. 8 in. distance from the other. Find the difference between the areas of the two figures.

2.    The length of a parallelogram is 21 ft. 9' 6", and its perpendicular height 12 ft. 6 in. ; find the area.

3.    A field 12 ch. 36 links long has its opposite corners similar, but they are not right angles; if the distance straight across the paddock in the direction of its width be 3 ch. 33| links, what is its area in acres ?

4.    The sides of a right-angled triangle are 51 ft. 3 in. and 11 ft. 7|

in. ; find its area.    *    -

5.    Find the area of a triangle whose height is 7 ft. 9^ in. and base

5 ft. 7 in.    '

6.    The hypothenuse of a right-angled triangle is 35 ft. and the base 21 ft. ; find the perpendicular.

7.    The two sides of a right-angled triangle measure 66 ft. and 494 ft. ; find the hypothenuse.

8.    I want a line to reach from the middle of a street which is

115 ft. 6 in. wide to a window 77 ft. from the ground; what length must the line be ?    .    _    •

9.    A tree growing on the edge of a river 72 ft. wide is 96 ft. high ; what length of wire rope will reach from the top of the tree to the edg§ of the bank opposite ?

10. If in the last question the tree were but 70 ft. high and the rope required 874 ft. long, what would then be the width of the river ?

11.    A ladder 56J ft. in length will reach 45 ft. on the wall of a

warehouse on one side of the street, and 33 ft. 9 in. high on a wall on the opposite side, the foot of the ladder being retained in the same position. What is the width of the street ?    _

12.    A circular fish-pond is 25 feet across; find the distance around it in yards.

13.    Upon measuring the circumference of a circular lake, I find it to be 5281 0296 yds.; what is the distance straight across it?

14.    If I tether a horse with a rope 17 yds. 1 ft. 1 in. in length, how much land can he graze ?

15.    What length of rope will enable a horse to graze 1 ac. 1 ro. T0624 sq. per. of pasture?

16.    How much pasture will 22 yds. of rope enable a horse to graze?

17.    The circumference of a circle is 57.12 ft., and the diameter 18 j2! ft. ; find the area.

18.    What is the distance around a square paddock which contains 818 ac. 980 sq. yds. ?

19.    A square enclosure contains 8 ac. 881 sq. yds. ; what is the length of the fence around it ?

20.    A circular race-course encloses 79 ac. 2 ro. 16 per.; find the length of the course.

21.    A. and B. start together from the same place at 6 o’clock a.m. A. travels due east, at a speed of 12 miles per hour, and B. due south, at a speed of 16 miles per hour. How far apart are they at 6 p.m. ?

22.    What is the distance, diagonally, across a square paddock

which contains as much land as another paddock whose length is 640 yds. and breadth 180 yds?--—— -—-—

Note.—One application of Square Root is sufficient to solve sums 22 and 24.

23.    A cubical mound of earth contains 254 c. yds. 1 c. ft. of matter. Find (1) the length of its edge; and (2) the area of the ground it covers.

24.    Find the distance from corner to corner, diagonally, across' a square field containing 112 ac. 802 sq. yds.

25.    A person undertakes the painting of a cylindrical boiler at 9d.

per sq. yard ; if the length be 25 ft. and the diam. of the end 8 ft. 4 in.*, what does he receive ?    ■

26.    How many sq. inches are there on the surface of a globe 15 inches in diameter ?

27.    A conical spire is 33 ft. high and 10 ft. broad at the base ; find its solidity.

28.    Supposing the earth to be a perfect sphere with a diameter of 8333J miles, express in words the number representing the cubic miles of matter it contains,

29.    A square water-hole of uniform depth of 8 ft. has vertical sides, and contains (when roughly estimated at 6| gals, per c. ft.) 387200 gallons ; what is the length of its side ?

30.    A rectangular tank, 15 feet deep and twice as long as it is broad, is roughly calculated to hold 108000 gals, of water; wha^Js its length ?

31.    The side of a square tank is 14 ft. 9 in. and the depth 9 ft. 3 in. ; how many gals, does it contain ?

32.    A farmer has a section of land containing 320 acres, the width being 40 chains. If he wish to mark out a square piece of 10 acres

for a homestead in the very centre of his block, how far irbm the front and side fences must he put in his pegs ?

33.    From the corner of a field of 280 ac. having 70 ch. frontage to one road and the rest to other roads, a farmer wishes to fence off a square paddock of 42 ac. 4 per. What will it cost to erect on the two sides a three-rail fence, at £2 5s. per 100 for the posts and railsj and 2s. 6d. per rod for the labour, supposing two panels to extend a rod, and allowing the cost of 4 extra posts for each of the three corner posts ? Also, how many chains of frontage has the other part of the field ?

34.    A circular racecourse, enclosing a meadow a mile in diameter, is 22 yds. wide, and occupies the centre of a square area. If the outer edge of the course be 9 chains from the centre of each side fence, find (1) the number of acres in the whole enclosure ; (2) the area occupied by the course itself ; (3) the distance of the corner posts from the nearest part of the outer edge of the course.

35.    A heap of road-metal is 11 ft. 8 in. long and 7 ft. 6 in. broad at the top, and 12 ft. 4 in. long and 8 ft. 6 in. broad at the base: how many cubic yards of metal does it contain, the height of the heap being 3 ft. ?

36.    Allowing 6\ gals, to the c. ft., find the capacity, in gals., of a cubical tank whose side is 4 ft. in length.

37.    A baker’s trough is 12 ft. long and 3 ft. 6 in. deep, the width at the top being 2 ft. 8 in., and at the bottom 2 ft. 4in; find its contents in cubic feet.

38.    A cask is found to contain 88 c. ft. 436-^ c. in.; find the exact number of gallons it will hold.

39.    A round zinc tub is 2 ft. 2 in. across the top and 1 ft. 10 in. across the bottom ; its depth is 13 in. How many gallons of water does it contain when full ? (Rule 21.)

40.    A cylindrical tin is 6 in. wide and in. deep ; how much

water will it hold ?    _

41.    How many c. in. of earth will a flower-pot contain which is 10 in. wide at the top, 10 in. deep, and 6^ in. wide at the bottom ? Note.—Apply Rule 21 for this and the following sum.

42.    A zinc bucket is 12| in. wide at the top, 6^ in. at the bottom, and 1(R in. deep ; how many gallons will it contain?

43.    A round well, at the school, is 10 ft. in diameter and 12 ft.

deep; if it is filled to within 2 ft. 6 in. of the top, how many gallons of water are in it ? (Reckon 6^ gals, to the c. ft.)    _

44.    Find the contents in c. ft. of a circular zinc tank whose diam.

is 4 ft. and depth 5 ft. 3 in. ; and find the number of gallons it will hold, allowing gals, to the c. ft.    ;

45.    A gable-end roof is to be put on a house of which the walls are 18 in. wide and 21 feet apart; if the ridge-pole be 9 ft. high, and the length of the roof 44 ft., find the length of the rafters required to allow 6 in. extension at the eaves. Find also the number of sheets of galvanized iron, each 2 ft. 3 in. wide and 5 ft. 9 in. long, that will cover the roof, allowing 9 in. off the length and 3 in. off the

breaiSlu of each sheet for the overlap and the extension at the

eaves.

46.    Find the capacity of a square tin dish whose side on the top is 15 in. in length and on the bottom 9 in., the depth being 3 inches.

47.    The straight trunk of a kauri pine is 50 ft. long, 2 feet in diam. at the butt end, and 1 ft. 8 in. at the smaller end. Compute roughly its contents in c. ft. See note to Rule 21, page 278.

48 A square paddock enclosed with a two-rail fence contains as many acres as there are rails in the fence. Supposing the rails to be 9 feet long and to overlap 9 inche^. how many acres does the paddock contain ? Note.—See worked exam. No. 6, page 219. and exam. No. 16, page 286.

49.    Compute roughly the No. of c. ft. m a Stack of hay 12 ft.

high, 20 ft. long at the base, 24 ft. at the eaves, and having a breadth of 14 ft. at the base and 16 ft. at the eaves—allowing 900 c. ft. for the top of the stack above the eaves.    Note.—Mult,

together the mean length and width, and the height.

50.    When fishing from a boat on Lake Wendouree, I observed a reed 5 in. above the surface of the water. On drawing it towards me, I found that it submerged at a distance of 25 in. What was the depth of the lake in that part ? See Ex. worked, No. 9., page 220.

THE UNITARY METHOD ;

OR, SOLUTION BY FIRST PRINCIPLES.

This is a method of working, by the first principles of multiplication and division, questions for which it is usual to employ the more advanced rules,—as proportion, etc.

It is called the UNITARY METHOD, because the plan adopted is to find first the value of a unit, and, from this, the value required.

Example 1.—If 20 men earn £76 13s. 4d. in 18 days, how many boys will earn £103 10s. in 27 days, the daily earnings of a boy being f those of a man ?

Solution by First Principles ; or, by the Unitary Method. If 20 men, in 18 days, earn £76 13s. 4d.; then, 20 men, in 1 day, earn £76 13s. 4d. £- 18 = £4 5s. 2fd. and, 1 man, in 1 day, earns £4 5s. 2fd -f- 20 = 4s 3^d.    9

and, 1 boy, in 1 day, earns f of 4s. 3^d — Is. 8±d.

1 boy, in 27 days, earns 27 times Is. Sfd. — £2 6s.

Now, £2 6s. is contained in £103 10s., 45 times.

Therefore, 45 boys will earn £103 10s. in 27 days. Ans. 45 boys.

Example 2.—A grocer bought 56 lbs. of tea and sold it at 2s. Id. per lb., by Which he lost as much money as he paid for 6 lbs.; at what rate per lb. should he have sold it at so as to gain the cost price of 6 lbs. ?

By the Unitary Method :—

56 lbs. at 2s. Id. per lb. ~ £5 16s. 8d., the proceeds.

By selling 56 lbs. for £5 16s. 8d., cost of 6 lbs. is lost.

Therefore, 50 lbs. cost £5 16s. 8d.

And 1 lb. cost £5 16s. 8d. -=-50 =z 2s. 4d.

,, 56 lbs. cost (2s. 4d. X 56) ~£6 10s. 8d.

,, 6 lbs. cost (2s. 4d. X 6) ~    14s. Od.

But 56 lbs. must be sold for the cost of (56 + 6) lbs.

i.e., 56 lbs. must be sold for £6 10s 8d. 4- 14s. or for £7 4s. 8d.

And 1 lb. must be sold for £7 4s. 8d.    56 2s. 7d. Answer.

By Proportion :—

By selling at 2s. Id. per lb. he regains the whole cost less that of 6 lbs.; that is, the cost of (56 - 6) or 50 lbs.; at what rate must he sell it per lb. in order to receive the cost of (56+6), or 62 lbs. ?

Now, as selling at 2s. Id. per lb. will produce the cost of 50 lbs. only, then it must be sold at a proportionally higher rate to produce the cost of 62 lbs.

Therefore, As 50 lbs. : 62 lbs.:! 2s. Id. I 2s. 7d. Answer.

Note.—On a fair comparison of the two methods upon which the above is worked, it will be seen that while the UNITARY METHOD is the simpler, as it proceeds step by step after the most easy process of reasoning, yet the latter process—viz., that of Proportion, entails by far the less figuring; and consequently presents less likelihood of errors occurring in the work. At the same time, Proportion requires a higher exercise of the reasoning faculties, notwithstanding that it affords less scope for reasoning out by successive steps.

For these reasons, we should recommend the Unitary Method to those who experience a difficulty in perceiving where the proportion really exists ; and “The Rule of Three” to those who excel in their perception of the principle involved, rather than in accuracy of working.

Example 3.—An insolvent pays 12s. 6d. in the £1 on one-half of his debts, and 13s. in the £1 on the other half ; find the amount of his debts if his total assets are £286 17s. 6d.

On £2 debt his assets are 12s. 6d. + 13s. ~25s. 6d.

#*# For every £1 5s. 6d. assets he has, he owes £2.

Or, for every 12s. 9d. he has, he owes £1.

Now, 12s. 9d. is contained in £286 17s. 6d., /fiO times.

Therefore, his total debts are £450. Answer.

Example 4.—A linen draper laid out £500 on a lot of muslin, but | proved to be damaged and was sold at 5s. per yard, by which £50 was lost. At what rate per English ell must the undamaged part be sold so as to clear £50 on the whole ?

Solution.

The damaged part cost 4 of £500 zz £166-f.

Loss upon the damaged part zz £50.

the damaged part was sold for £116f.

At 5s. per yard, 4 yards were sold for £1.

Hence, damaged part zz (116f X 4), or 466f yds.

Now, damaged part zz 4, and undamaged part ■§, of whole,

Hence, undamaged part ~ 466f yds. X 2 ~ 933£ yds.

And, 9334 yds. =: 9334 X 4> or 746f Eng. ells.

Now, the whole muslin cost £500 ; and to gain £50, it must be sold for £550.

But damaged part fetched £116|.

hence, the rest must be sold for (£550 - £116f), or for £4334.

i.e., 746f Eng. ells must be sold for £4334;

and, 1 Eng. ell for £4434 746f zz 11s. 7yd. Answer.

Example 5.—A can mow 5 ac. in 94 days, B 44 ac. in 7 days ; how long will it take them to mow 124 ac. %

Solution.

A mows 5 ac. in 94 days zz 5 ac. -f- 94, or -|4 ac. per day.

B mows 44 ac. in 7 days zz 44 ac. -r 7, or if ac. per day.

A and B mow (y-f + 4f) ac. = Iff ac. per day.

12| ac. contains l|f- acres 10f times.

Therefore, A and B can mow 12J ac. in 10£ days. Answer.

Example 6.—If a tradesman’s bad debts amount to 2\ per cent, of his sales, and he wish to gain 17 per cent, clear profit, find (a) at what rate per cent, of the cost he must mark his goods, and (b) what per cent, clear profit he would gain by marking them at 334 per cent, above cost.

Solution.

a.    To gain 17 per cent., he must receive 117 per cent, of the cost. Now, of every 100 he charges he receives but 974.

In order to receive 974 he must mark at 100.

And to receive at 117 he must mark at (117 X 100 -f 974) = 120. Hence, to gain 17 p. c., he must markatl20p. c. of cost. Answer a.

b.    Of every 100 he charges, he receives but 974-

He therefore receives (1334 x 974 4- 100) or 130 per cent, of cost, And consequently gains 30 per cent. Answer b.

Example 7.—For what amount must goods worth £390 he insured at 2-| per cent, so that, in case of loss, the owner shall recover both the value of the goods and the premium paid ?

Solution.

Of each £100 insurance, £2^ repays the premium, leaving (100 - 2|), or £97£, to repay part of the loss.

Hence each £97§ worth of goods must be insured as being worth £100 ; so that the insurance shall cover the value of the goods and premium.

And, goods worth £390 must be insured for (390 -f 97£ x 100) or for £400. Answer.

Example 8.—Two men, A and B, undertake to cut down 20 acres of wheat ; A can cut 5 ac. in 9| days, and B 4£ ac. in 7 days. How long will it take them working together ?

Solution.

A cuts down 5 ac. in 9| days or (5 ac. -f 9|) or if ac. per day.

A,    in 1 day, cuts down 5 ac. 4- 9-^ = yf acre.

B,    in 1 day, cuts down 4J ac. 4- 7 yf acre.

Both together cut down (-Lf +    ) ac. or 1-Lf ac. in 1 day.

And, since 20 ac. contains l-£f ac. 17 fy times,

They will cut 20 ac. in 17|y days. Answer.

Example 9.—Sold 60 bags at a profit of 6 per cent, and 30 bags at a profit of 8 per cent., but had I sold the whole at a uniform profit of 7 per cent. I should have gained 15s. more than I did ; what was the cost price per bag ]

Solution.

The average per cent, gained by selling 60 bags at 6 per cent, profit and 30 bags at 8 per cent, profit — (60 x 6 + 30 X 8) -h- 90 — 6f per cent.    _    .

But at 7 per cent, average profit 15s. more is gained ;

Therefore (7 per cent. - Of per cent.), or \ per cent, profit = 15s.

That is, 15s. is per cent, or ygqth part of the^cost.

Hence, cost of 90 bags was 300 times 15s. = 4500s. ; and,

Cost per bag - 4500s. 4- 90 = £2 10s. Answer.

Example 10.—A merchant finds his bad debts amount to 4 per cent, of his sales ; if he wish to clear 44 per cent, profit, at what percentage of the cost must he mark his goods l

Solution.

Of every 100 he charges, he receives but 96.

He wishes to receive 144 per cent, of the cost.

He must therefore charge yk0 144 P- c- = 1®® P* c- Answer.

Example 11.—A water-tank holds 147 gallons, and the pipe that supplies it brings in 14 gallons in 9 minutes, while the tap discharges 40 gallons in 31 minutes ; supposing the tank to have been empty, and the pipe and tap to have been both left running for 3 hours before the tap was shut, in what time afterwards would the tank have been filled ?

Solution.

l4 gal. x 180 = 280 gallons filled by pipe in 3 hours.

•|ngal. X 180 = 232g8T gallons emptied by tap in 3 hours.

Therefore, 47§f gallons are left in at end of 3 hours.

Now, (147 -47fi) gals. = 99^\ gallons still to be filled.

In 1 min. the supply pipe fills f of 14 gallons = If gals.

If gals, is contained in 99^ gals. 63ff f times ; hence,

99ff gals, will fill in 63fff minutes = 1 hr. 3-fff min. Answer.

Example 12.—If £67 15s. 9d. is the value of 20 lbs. 6 ozs. lOdwts. of silver, how much silver is there in a trophy of that metal worth £25 10s. 6d., the cost of workmanship in which is £6 11s. ?

Solution.

20 lbs. 6 ozs. 10 dwts. ; or, 246|ozs., cost £67 15s. 9d.

1 oz. silver cost £67 15s. 9d. 4- 246| == 5s. 6d.

Value of trophy, inclusive of workmanship, £25 10s. 6d.

Subtract cost of workmanship,    £6 11s. Od.

Value of silver contained in the trophy, £18 19s. 6d.

There must be 1 oz. of silver for every 5s. 6d. in £18 19s. 6d.

.*. No. of ozs. = £18 19s. 6d. -f- 5s. 6d. = 69ozs. — 51bs. 9ozs. Ans.

By Proportion.

As £67 15s. 9d. ; (£25 10s. 6d. — £6 11s. ) ; 1 246^ ozs. * 69 ozs. Ans.

Example 13.—Thirteen bags of potatoes, when newly dug, wreigh 1 ton lcwt. 14 lbs. If, in drying, each bag lose 14 lbs., how much per ton will a dealer gain who buys potatoes, newly dug, at £6 10s. per ton, and sells them, when dry, at an advance of 52s. per ton ?

Solution.

13 bags, when newly dug, weigh 21 cwt. 14 lbs., or 2366 lbs.

In drying, they lose in weight 13 times 14lbs., or 182lbs.

Hence, the loss in weight is 182 lbs. upon 2366 lbs.

As 2,366 -4- 182 = 13, the loss is ^ of the quantity bought.

Or, of 1 ton bought, there remains but ff of a ton to sell. '

1 ton, when dry, is sold for (£6 10s. + 52s.) = £9 2s.

If of a ton is sold for -f| of £9 2s., or for £8 8s.

And the gain per ton is (£8 8s. - £6 10s.) = £1 18s. Answer.

Example 14.—I buy bread deficient in weight to the extent of 8 oz. in the 4 lb. loaf, and in order to recover my loss I sell it out weighing it upon scales which represent 1 cwt. as being 1 cwt. 14 lbs. What shall I gain or lose per cent, upon my outlay?

Solution.

In buying 4lbs., I lose £ lb., or I lose § of my outlay.

In selling 112 lbs., I gain 14 lbs., or I gain £ of my sales.

Had I the whole I paid for, I should regain my loss.

But of every 8 lbs. I paid for I received but 7 lbs. ; and can therefore regain but g of my loss. Consequently, £ of my loss, i.e. (§ of or of my outlay, is still lost.

And, as ^ of 100 is 1^-, I lose lT9g p. c. on my outlay. Answer.

Again :—

Suppose 2d. per lb. to be the price of bread.

Then, I am charged 8d. for 3£ lbs.

I sell 112 lbs. as 126 lbs. ; that is, for 252d.

And 3| lbs. for (252d. x3|t 112), or for 7|d.

Hence, on 8d. outlay I lose |d. ; which is at the rate of lT9g p. c.

Example 15.—A and B do a piece of work in 12 days, A and C in 15 days, B and C in 16 days. Find the time they will together take to do the work.

Solution—

Each person works two days, one day with each of the two others—

A and    B do the whole work    in    12 days,    or    -j— of    it    per    day.

A and    C do the whole work    in    15 days,    or    tL- of    it    per    day.

B and    C do the whole work    in    16 days,    or    tL- of    it    per    day.

See reciprocal principle explained    on    page    202.

In 2 days A, B, and C do + tt + rg-) or of the work.

In one day they can do \ of or 0f the work.

They can, therefore, do the whole in days, or in 9^ days.

Example 16.—What is the area of a square field which contains an acre for each panel in the fence enclosing it, supposing each panel to be 8j ft. in length?

Solution.

Each panel in the whole fence is to represent 1 acre.

Then each panel in the fence on one side must represent 4 acres.

Now, each panel is 8J ft., or one-eighth of a chain in length.

An acre, if a chain in breadth, is 10 chains in length.

And, if -i- chain in breadth, an acre is 80 chains in length.

An area of 4 acres, if ^ chain broad, must be 320 chains in length.

That the field be square, the length and breadth must be equal

Hence the length and breadth of the field are each 320 chains.

And    the area is (320 ch. X 320 ch.) = 102400 sq. ch. = 10240 ac.

In the following Exercise it is intended that the questions shall he worked by the application of the Elementary Rules, and the principles explained under the heading of the Reciprocal of Numbers (page 202), and Average (page 214).

Exercise XLVII.—THE UNITARY METHOD ;

OR, THE APPLICATION OF FIRST PRINCIPLES.

1.    A can do a certain piece of work in 3 days, B can do it in 4 days, and C in 6 days. How long will they take, all working together ?

2.    How many oranges at 2s. a doz. must be mixed with 24 other oranges at Is. 3d. a doz. that they may be sold at 8 for Is. ?

3.    A and B can do a piece of work in 18 days, A and C in 15 days, and B and C in 12 days. How long will the three take working together ?

4.    A bankrupt, whose deficiency amounted to 3s. 5|d. in the £, had assets amounting to £454 17s. lid. ; what did his liabilities amo«nt to? and what will a creditor receive upon a debt of £100?

5.    How much in the £ will be paid by an insolvency estate in which the assets are £448 8s. 9d., and the deficiency £301 11s. 3d. ?

6.    A bankrupt can pay but half his debts, and a person to whom he owes half as much again as to all his other creditors receives £37 16s.; find the total amount of his debts.

7.    In buying turnips at |d. per lb. I am charged 4s. 8d.; what weight did I actually receive, if the greengrocer’s scales were 3j lbs. short in the qr. ?

8.    A fruiterer buys a case of oranges, containing 10 dozen, for 11s. 3d., and paid Is. 3d. freight; a dozen and a half go bad, and he sells the rest at 6 a shilling. Find his gain, and the number that must go bad in order that he may neither gain nor lose.

9.    If a tradesman marks his goods at 30 per cent, above cost, and then makes a reduction of 5 per cent, for cash, what ready money does he receive for goods marked at 26s., and what did such goods cost him ?

10.    A, B, and C receive £33 for a piece of work which they do in 36 days, and which A and C could do in 48 days. If A can do half as much again as C, what should be the daily wages of each ? Prove your answer by multiplying the sum of the daily wages by 36 to obtain the £33.

11.    An insolvent’s effects are worth £168 3s. 9d. If he pays 11s. 3d. in the £, what is the amount of his liabilities ?

12.    A bankrupt pays 3s. 6d. in the £, and Is. in the £ insolvency

expenses. To what do his debts amount, his effects being worth £475 17s. 6d. ?    '

13.    I bought 17 gallons of spirits for £12 15s., and after adding 3 gallons of water to reduce the strength, I wish to sell the whole so as to gain £4 5s. ; what must I charge for 14 gallons ?

14.    An insolvent’s assets are 5s. 2^d. in the £, and his total deficiency £739 11s. 8d.; how much was owed to a person who receives £36 19s. 7d., and how much to all the other creditors together ?

15.    After paying away 7^ per cent, of my money and spending J

of the rest, I find that I have £185 still left: how much monev had I ?    J

16.    Find the true discount on £813 9s. for 16 months at 4f per cent.

17.    A baker uses a lb. weight which is ozs. short ; what weight of bread does a customer receive on buying 27 4-lb. loaves ?

18.    A dealer bought a horse for £22 17s., and after keeping him for 6 weeks at a cost of 13s. 6d. per week, he sold him for £33 12s. 6d.; what rate per cent, profit did he make upon his outlay?

19.    A settler buys 18 head of cattle for £94 10s. and sells 10 of them at £4 15s. 6d. each ; at what price per head must he sell the others that he may gain £9 9s. on the whole ?

20.    If 25 men can do §rds of a piece of work in 11 days, in what time will fths of the same piece of work be done if 5 men leave after the first day ?

21.    A merchant bought 10 gals, of spirits for £7 15s. and after losing 2| gals, by leakage he sold the remainder at 18s. 4d. per gal., but his gallon measure was J pt. deficient. What did he gain or lose by the transaction ?

22.    A bankrupt pays 10s. 8d. in the £, and afterwards 5s. in the £ upon the remaining debt. If his creditors lose £249 4s., find the total amount of his liabilities.

23.    An insolvent pays his creditors three dividends of 11s. 3d., 3s. 6d., and Is. 9d. respectively in the £. If £234 10s. would pay one-half of the remaining debt, what were his total liabilities?

24.    The cost of supplying 13 mi. 5 fur. 40 yds. of telegraph wire was £400 6s. 8d. ; what was the cost per mile ?

25.    A shop cellar measures 20 ft. by 16 ft. by 10£ ft. deep, and is two-thirds full of water. If a pump throwing 7 gals, per minute be used to bail it, in how many hours will the cellar be empty, reckoning 6| gals, to the c. ft. ?

26.    A person bought a house for £125 12s. 6d., and paid £3 3s. expenses ; for what did he sell it again if he lost £8 8s. ?

27.    After paying an income-tax of 7£d. in the £, I have £823 8s. 9d.

left. What is my income?    _    _

28.    A person pays away £2 17s. 6d. poors’ rates, and gives his a^ent 2 3 per cent, of the remainder; what has he left clear out of £327 7s. 6d. ?

29.    Divide £387 among three persons so that the second shall receive as much more than the first as less than the third, and the first shall receive one-ninth of the whole.

30.    The contents of 108 packages of silk measure 4428 yds.; the average value of each package is £13 16s. 9d. Find the average price per yard.

31.    The hind-wheel of a coach is 15 ft. in circumference, and the

fore-wheel 11 ft., and it is found, in travelling from Taradale to Elphinstone, that the fore-wheel has made 288 revolutions more than the hind-wheel. What is the distance between the two townships ?

32.    Suppose incomes below £250 a year subject to an income-tax of 4d. in the £, and incomes exceeding £250 a year to a tax of 5d. in the £. A person whose gross income was less than £250 a year found that his net income was £245 11s. 9d. ; what income exceeding £250 a year would have left him 11s. O^d. poorer?

33.    A man and his son can together build a certain wall in 7^ days, and the man can do in 2J days as much as his son can do in 3f days. They work together for 4 days, and then the father leaves the son to finish the work. How long will he take ?

34.    The weekly receipts of a shipping company average £2450 7s. 9d. ; the current expenses average £6651 13s. 7d. per month ; and there is annually carried to a reserve fund £2600. If the capital of the company is half a million, what rate of interest should the shareholders receive ?

35.    A gives B 132 yds. start in a race of 3 miles, A running at the rate of 7 miles per hour, and B at the rate of 6| miles ; which won the race, and by how much ?

36.    A person who owes £1500, has cash £700, goods worth £500, and book-debts (on which he can recover 12s. 6d. in the £) amounting to £800. Find the amount of his surplus.

37.    A lb. of tea and 3 lbs. of coffee cost 7s. 8d.; but if tea were to-rise 12£ per cent, and coffee 20 per cent., they would cost 9s. Find the price per lb. of tea and coffee.

38.    If 2 men, or 3 women, or 5 boys can do a piece of work in 23 hours, how long will it take 3 men, 4 women, and 5 boys to do the same piece of work ?

39.    A grocer mixed together three kinds of tea,—54 lbs. of the first at Is. 5d. per lb., 81 lbs. of the second at Is. 7d., and 45 lbs. of the third at Is. 9d. At what rate per lb. must he sell it in order that he may gain 10 per cent, of the money he receives ?

40.    A surveyor measures a paddock with a Gunter’s chain from which 4 links each 7 '92 in. in length have been broken off ; if the chain should consist of 100 such links, what is the true area of a. parddock which appears to be 312 ac. 2 ro. by this measurement ?

41.    When A has travelled 76 days at the rate of 15 miles per day, B, who had travelled 60 days, overtook him. If both started from the same place, at what rate per day did B travel ?

42.    A tailor uses a yard measure which is 1 nail too short ; what was the true measure of a piece of cloth for which I paid him £4 2s. 6d. at 5s. 6d. per yard ?

43.    A bankrupt’s liabilities are such that £224 14s. 7d. will pay 11s. 8d. in the £ ; to what do they amount ?

44.    A bankrupt pays 4s. lid. in the £, and his total assets are £453 Is. 5d. ; find his liabilities.

45.    A teacher bought a house for £120 and occupied it 18 months»

thus saving rent at the rate of £15 per annum. If he expend £2 12s. 6d. on repairs, and then sell it for £110, what does he gain by purchasing a house instead of continuing to pay rent ?

46.    A settler bought 100 animals for £100—cows at £4 10s. each, pigs at 30s. each, and sheep at 10s. each. How many of each sort djd he buy ?

47.    By the sale of goods which cost me £14 3s. 6d., I lost a sum equal to 12£ per cent, of the money I received; and by the sale of other goods, which cost me £12 12s., I gained a sum equal to 14f per cent, of the money I received. What did I gain upon my outlay ?

48.    Out of a flock of 120 sheep and lambs, -i were ewes with two lambs each, and there were a fourth as many again with one lamb each. How many of the sheep were without lambs?

49.    For what sum should a cargo worth £5568 15s. be insured at 8^ per cent, so that the owner may, in case of loss, recover the value both of cargo and premium ?

50.    I bought a hind-quarter and a fore-quarter of beef, weighing together 252 lbs. ; I paid 7|d. per lb. for the hind-quarter and 5^d. per lb. for the fore-quarter, and found that I had paid 17|d. on the whole more than if I had bought both quarters at 6§d. per lb. Find the weight of each quarter.

51.    Three workmen, A, B, and C, did a certain piece of work, and were paid daily wages according to their several degrees of skill. A’s efficiency was to B’s as 4 to 3, and B’s to C’s as 6 to 5; A worked 5 days, B 6 days, and C 8 days, and the whole amount paid for the work was £8 14s. Find the rate of each man’s wages per day.

52.    A merchant bought 400 lbs. of tea and 1600 lbs. of sugar, the cost of the latter per lb. 16§ per cent, that of the former; he sold the tea at a profit of 33J per cent., and the sugar at a loss of 20 percent., gaining, however, on the whole £7 10s. Find his buying prices and his selling prices.

53.    A retail merchant bought a quantity of Ballarat tweed, and marked it at an advance of 25 per cent, on the cost, and in selling it used a yar<} measure which was f in. too short, his entire gain being £5 17s. Find the cost price of the cloth, and the amount the person gained by using the false measure.

54.    Divide £3 among A, B, 0, D, so that A may have twice as much as B, B one-fifth more than C, and D as much as A and C.

55.    I sell a horse for £20 8s., and lose 15 per cent, on what I gave for him, whereas I ought to have gained 25 per cent. How much did I sell him under the price I had fixed on ?

56.    If 30 men, working 8 hours a day, complete a certain amount of work in 24 days, how many men, working 10 hours a day, can do five times as much in 15 days?

57.    By selling a pair of boots at 18s.. a bootmaker gains fths of their prime cost. Find the cost.

58.    Fifteen cricketers have dinner at an hotel, but three of them left before paying their bill, and the others found that they had each

to pay Is. 10|d. more than his proper share. What was the amount of the bill ?    >

59.    A can do a piece of work in 18 days, and B in 10 days ; A

works at it alone for 8 days, B then works for 3J days, and C then finishes the work in 6 days. In what time could C have done the work by himself ?    _

60.    Two pipes, A and B, would fill a cistern in 20 and 30 min.

respectively; both are opened together for 5 min., and then A is turned off. If a tap, C, which would empty it when full in 1 hour, be now open, and pipe B allowed to continue running, when will the cistern be filled ?    _    _

61.    Four workmen, A, B, C, and D, can do a piece of work in 36 days, and the relative amounts of work they can do in the same time are as 5, 6, 7i, and 9. How long would each take to do the work single-handed ?

62.    A can do a piece of work in 24 days, B in 32 days, and C in 36 days. If all start together, and A and B work on till the whole is finished, and C leaves off 4 days before, what time does the work take ?

63.    A can do a piece of work in 48 days, B in 60 days, and C in 80 days. All begin together, but A leaves off 2 days and B 3 days before it is finished, C completing it alone. How long does the work take ?

64.    A, B, and C can do a piece of work in 20 days, A and B together in 26f days, and A and O together in 30 days. How long will it take B and C together to do it ?

65.    A, B, and C can together dig 3 acres in 20 days, A can dig an acre in 16 days, and B a rood in 5 days. Compare the times which they take to do the same amount of work, and find how long C will take to dig 3 acres.

66.    To do a certain piece of work, A by himself would take 16 hours, B 15 hours, C 12 hours. A and B work together for 5 hours, and then B and C for 1 hour ; how long will it take anew man (D) to finish the remainder of the work, if his rate of work is to B’s as 49 is to 48 ?

67.    How many oranges, at 2s. a doz., must be mixed with 24 other oranges, at 7^d. per doz., that they may be sold at 8 for Is.?

V

STOCKS AND SHAKES, j

"When the Government of a country wishes to raise money to defray the expenses of war, or to carry out any large work, as the construction of a canal or railway, the sum required for the purpose is often so great that it would

be a difficult matter to obtain sufficient funds from any of the ordinary sources of revenue—as by the sale of land, or by the imposition of duties, customs, and taxes.11

In such a case, the Government borrows money from persons who have it to lend, and pays a fixed rate of interest upon the loan.

This loan is frequently obtained for an indefinite period; sometimes upon the understanding that the loan itself will not be repaid, but that the interest only shall be paid regularly and constantly. As, however, the payment of this interest is always sure, and as a person may at any time part with his right and claim to another, capitalists are quite willing to lend their money to the Government in consideration of such interest, which is both fixed as to rate and constant in duration.

As the interest is paid at the end of each six months, it is called the half-yearly dividend.

To each person from whom it borrows, the Government issues, as security, a stamped document bearing the signature of its treasurer. These documents are called Bonds or Debentures, and as they entitle that person to receive the half-yearly dividend who holds these acknowledgements when the interest has become due, they are transferable or negotiable ;—that is to say, they may be bought and sold in the same way as other property, such as houses, land, livestock, &c.

These bonds or debentures represent the credit of the country, which engages to pay the debts incurred on its behalf by its parliament. They take the place of such legal securities as bills of sale, mortgages, acceptances, &c., usually given by a private individual who, wishing to employ more capital than he possesses, borrows money to enable him to do so.

The public debt is variously styled the National Debt, The Funds, Consols, &c., and all these are comprehended within the general term Stocks.

The paid up capital of banks, insurance and other public companies, is also termed Stock.

Stock may therefore be defined as money lent to Government or funded in Government securities ; and as shares in joint-stock companies, as hanks, insurances, &c.

When joint-stock companies are first formed, the capital is held in shares of a fixed and equal value, each ¿£5, £10, £25, or £100, &c., as the case may be, of which a part only is at first called up, and the remainder as it may be required afterwards. If all the capital has been called, or paid up by the shareholders, these shares are often converted into stock for the sake of convenience in trading, stock being saleable in any quantity, while in buying and selling shares each transaction is confined to an exact number of shares.

Stocks have two distinct values—a nominal value and a marketable or current value. The former is fixed and permanent, and represents the exact amount of debt held by a public creditor; while the latter is variable, and shows the amount of sterling money that can be obtained by selling such share of the debt. In order to illustrate these two values of each £100 stock, let us suppose that a loan of £2,000,000 is required to be raised, and that Government issues 20,000 3 per cent, bonds or debentures, each representing £100 stock. Each debenture represents £100 of the debt; and no matter what it may be sold for, in the course of exchange, it still bears in name the value of £100; and carries with it, i.e., entitles the holder to receive, the interest of that sum. This £100 is therefore the 'nominal value of £100 stock; and, as we have just said, this nominal value never suffers change. Let us now further suppose that these same stocks are subsequently selling in the market at 90; that is, at £90 for each £100 stock. The sterling or marketable value of each £100 stock would then be but £90 ; and although £100 stock would still bring in the holder £3 interest, yet he would not be able to sell at a higher price than £90 for each £100 stock.12

The real or marketable value of stock, therefore, like that of all other negotiable property, fluctuates; that is to say, the price rises or falls as there is greater or less demand for the stock, and as money is more or less abundant in the market; or as the affairs of a country are in a more or less flourishing condition.

When the market price of ¿£100 stock is—

(a)    £100, the price of stock is said to be at par.

(b)    Above £100, the price of stock is at a premium.

(c)    Below £100, the price of stock is at a discount.

The price even at which Government sells stock in the first instance, that is when raising a loan, is usually below the nominal value, i.e., below par, or at a discount; generally at prices ranging from £90 to £98 for each £100 stock.13

Stocks are named according to the interest they bear— the 3 per cents., the 4| per cents., ¿sc.

The purchase and sale of stock is usually effected by an agent, who charges per cent, upon each £100 stock he buys or sells; that is, he deducts 2s. 6d. from the price received when selling £100 stock, and adds 2s. 6d. to the price paid when charging a person in buying £100 stock for him.

This agent is called a broker ; and the charge he makes, brokerage.

We shall now explain the following question :—“A person who holds £3800 in the 3 per cents, at 80, sells out and invests the proceeds in the 4 per cents, at 95; what change takes place in his income 1 ”

1.    He has stocks bearing 3 per cent, interest to the nominal value of £3800 ; and as each £100 brings him in £3 interest, his income is (£3800-^100) x 3=£114.

2.    He sells each £100 stock for £80, and therefore receives (£3800-^100) x 80 = £3040 sterling.

3.    With this ¿£3040 he purchases stock bearing 4 per cent, interest, giving ¿£95 for each ¿£100 stock; he therefore buys (¿£3040 -i- 95) x 100, or £3200 stock.

4.    As each ¿£100 stock brings in ¿£4 interest, his altered income is (¿£3200-^100) x 4 = ¿£128.

5.    The change in his income is therefore the difference between ¿£128 and ¿£114. That is, he receives ¿£14 per annum more than he did.

Let it now be carefully remembered that in buying or selling stock, we are simply buying or selling the right to receive a debt of a fixed sum of money, together with the interest payable upon that debt; and a few other examples will suffice to show that most questions in Stocks are simple questions in direct proportion, and that they are all easily resolvable by the application of first principles, as illustrated on pages 281 and 282.

Examples worked out.

1.    Find the value of £500 stock in the 3 per cents., at 90.

£100 stock is worth £90 ;    £500 stock is worth £450.

2.    How much stock in the 3 per cents., at 90, can be'bought for £450 ?

£90 buys £100 stock; £450 will buy £500 stock.

3.    Find the yearly income derivable from £500 stock in the 3 per cent, consols, at 92£

£100 stock yields £3 interest;    £500 stock yields £15 interest.

4.    Find the annual income obtained by investing £450 in the 3 per cents., at 90.

£90 will buy £100 stock, and therefore entitle the buyer to receive the interest of £100 stock (viz., £3) ; and as £450 is five times £90, this sum will entitle the buyer to receive £15 interest. Ans., £15.

5.    Ho^ much stock at 92g can be bought for £930, allowing | per cent, for brokerage ?

Each £100 stock bought cbsts the purchaser (£92g + ££\ or £93.

Then, £93 buys £100 stock, and £930 will buy £1000 stock.

6.    How much money will a person receive for £500 stock at 92£, allowing £ per cent, for brokerage ?

£100 stock sells for £92J, but the broker keeps 2s. 6‘d., or £4; the seller receives £92 for each £100, or £460 for £500 stock ” ”

5 V

Exercise XLVIII.—STOCKS.

I.    Find the marketable or cash value of—

1.    £3750 stock in the 3 per cents., at 95.

2.    £1275 stock in the 3 per cents., at 96.

3.    £4780 stock in the 3 per cents., at 92|.

4.    £1225 stock in the 3^ per cents., at 92.

5.    £3200 stock in the 4 per cents., at 87§.

6.    £1700 stock in the 5 per cents., at 89|.

How much stock can be bought for—

7.    £784 cash in the 3 per cents., at 98 ?

8.    £492 5s. cash in the 4 per cents., at 89| ?

9.    £585 15s. cash in the 4§ per cents., at 97§ ?

10. £954 2s. 3d. in the 5 per cents., at 96§ ?

II.    £661 17s. 6d. in the 8 per cents., at 88J?

12.    £886 0s. 7|d. in the 6 per cents., at 90g ? f\

Find the half-yearly dividend payable upon—

13.    £3000 stock in the 3^ per cents., at 90.

14.    £4600 stock in the 3 per cents., at 93J.

15.    £5555 stock in the 3 per cents., at 98£.

16.    £2060 stock in the 4J per cent, bank stock.

17.    £6409 stock in the Victorian 5 per cents.

18.    £2050 stock in the Victorian 4\ per cents. A

Find the annual income obtained by investing— •

19.    £50976 cash in the 3 per cent, consols, at 88^.

20.    £5890 sterling in the 6 per cent, bank stock, at 76.

21.    £7056 sterling in the 4| per cent, bank stock, at 84,

22.    £7756 sterling in the Victorian 8 per cents., at 96.

23.    £8656 10s. cash in the Victorian 5 per cents., at 99£.

24.    £5307 10s. cash in the 12£ per cent, railway stocks, at 96J. 4

How much money must be spent in stocks to produce—-

25.    An annual income of £55 10s. in the 3 per cents., at 93 ?

26.    An annual income of £87 in the 5 per cents., at 95?

27.    A half-yearly dividend of £225 in the Victorian 4£ per

cents., at 98 ?

Allowing per cent, brokerage, how much sterling money—

28.    Must be paid for £1000 stock, at 93J ?

29.    Must be paid for £845 stock, at 85g ?

30.    Will the seller receive for £880 stock, at 73£ ?    ^

When the 3 per cent, consols are at 95—-

31.    What is the cash or marketable value of £475 stock ?

32.    How much stock can be purchased for £150 8s. 4d. ? ^

33.    What yearly income is. derivable from £2555 stock ?

34.    What annual income is obtained by investing £456 ?

35.    How much stock will be sold for £764 15s. ?

36.    How much cash will buy £750 stock, brokerage % p. c. ?

When the new 'Victorian 4| per cents, are at 99—

37.    How much stock can be purchased for £742 10s. ?

38.    How much sterling money will be received for £9255 stock ?

39.    What is the half-yearly dividend upon £5000 stock ?

40.    What sum must be invested to give a yearly income of £270?

41.    How much stock will £11181 6s. purchase, brokerage £ p. c. ?

42.    What shall I receive for £8560 stock, brokerage | p. c. ?

When the 8 per cent, railway stock is at 97|—

43.    What is the value in sterling money of £400 stock ?

44.    How much stock will £683 7s. 6d. purchase, brokerage ^ p. c. ?

45.    What is the annual payment upon £1262 10s. stock?

46.    What sum must be invested to produce £454 10s. per aim. ?

47.    What sum will purchase £1454T6T stock, brokerage J p. c. ?

48.    What sum shall I receive for £433^ stock, brokerage £ p. c. ?

When the London 5-|- per cent, bank stock is at 115|—

49.    What rate of interest does the investor receive ?

50.    How much stock will yield an income of £231 ?

51.    What money will a broker receive for £980 stock?

52.    What is the brokerage upon stock which cost £8200 10s. ?

53.    What is the brokerage upon sales amounting to £3773 ?

54.    What will £1040 stock fetch, brokerage J p. c. ?

.    55. What alteration takes place in a fundholder’s income by

selling out £1920 stock, and investing in Victorian 4^ per cent, debentures at 96 ?

56. To-day’s Age (March 15th, 1879) announces that the tenders for the Victorian new 4-|- per cent, loan were opened in London on the 13th March, and that the whole amount required (viz., ¿63,000,000) was allotted to tenderers at an average price of ¿699 for each ¿6100 debenture. Find—    \

(a)    The total amount of money raised by the loan.

(b)    The interest payable annually updn it.    »

(c)    The rate per cent, interest investors receive.

(cl) The amount of money a person must invest in order to receive a half-yearly dividend of £90, brokerage r per cent.

57.    A person holding 75 railway shares of £100 each, paying an annual dividend of £3 5s. per share, sells out when they are at°£79 8s. 6d., and invests in the new Victorian 4^ per cents, at 99. Find the increase in his yearly income, allowing brokerage | per cent.

58.    A man, by selling out of the Victorian 4| per cents, at 107£

realizes £49871 5s., and gains 8£ per cent, on his investment. (1). At what price did he buy ?    (2). What was his original outlay ? (3j!

What alteration will take place in his income by re-investing in the 4 per cent, bank stock at par ?

59.    Which is the better investment—bank stock paying 8^ per cent, at £170, or the Victorian 4^ per cents, at 99£ ? and what will a person gain per annum by investing £16915 in the better of the two ?

60.    I invested £7400 in the South Australian 5 per cents, at 92^, and sold out at 93|, paying commission } per cent. What was my net gain ? and how much commission did the broker receive ?

Exercise XLIX.—MISCELLANEOUS AND DIFFICULT SUMS.

1.    If a grocer’s scales give only 15 ozs. 4 drs, to the lb., out of how

much money is a customer cheated who buys' sugar to the amount of £4 16s. ?    '

2.    If a horse is bought at 15 per cent, less than £15, and is afterwards sold at 25 per cent, more than the cost, at what price is he sold ? Find also the seller’s gain.

3.    A and B can dig a cellar in 4 days, and with the assistance of C they can dig it in 3 days ; in how many days can C dig the cellar ?

4.    Find, in ac. ro. and per., the area of a field which is 76'28 rods long, and 1872 rods broad.

5.    Certain goods were marked for sale at an advance of 3,5 per cent, on the cost, but were sold at a reduction of 35^ per cent, on the price marked. What percentage was lost ?

6.    Bought oranges "at 5 for 4d., and sold them at the rate of 4 for 5d. Find the rate of gainu

7.    D does a piece of work in 12 days, D and E in 5 days. In how many days will E do it, working alone ?

8.    At what price shall I mark hats which cost me 6 shillings each, that I may allowT6§ per cefit. discoimt and still gain 25 per cent. ?

9.    What would be the cost of a stack .pf wood, 16 ft. long, 4 ft. wide, and 6 ft. high, at 13s. per cord of 128 c. ft. ?

10.    The expense of excavating a cellar, 18- ft. long and 15 ft. wide, at Is. 9d. per c. yard, is 5 guineas. Find its depth.

11.    How many rolls of paper, each 6 yds. long and 2 ft. wide, will

it take to cover the walls of a room 18 ft. iong, 15 ft. wide, and 12 ft. high?    .

12.    If 5 be added to each term of the fraction Ub will the value be increased or diminished, and by how much ?

13.    Find the side of a square field equal to a rectangular field 700 yds. wide and 2800 yds. long.

14.    A stack of wood, 24 ft. long, 18 ft. wide, was sold at 12s. 6d. per cord, and fetched £16 17s. 6d. What was the height of the stack ?

15.    A tree is partially broken off 25 ft. from the top, which strikes the ground 15 ft. from the foot of the tree. What was the full height of the tree ?

16.    How many square feet of boarding are there in a close fence surrounding a square allotment containing 40 acres, the fence to be 5 ft. high ?

17- Find the length in yards of the side of a square play-ground containing 4 ac. 3225 square links.

18.    Two rafters, each 22 ft. 6 in. long, meet at the ridge of a roof 13 ft. 6 in. above the attic floor. What is the width of the house?

19.    How many rails, each 85 ft. long, are required for a three-rail fence which shall enclose a square paddock containing as many acres as there are panels in the fence ? See page 219.

20.    A selection of 320 acres contains a rectangular scrub which is impenetrable to sheep, and is 16 ch. long and 12 ch. 50 links broad. What must be the grazing capabilities per acre of the rest of the land, that 600 sheep may be kept upon it ?

21.    The first edition of one thousand copies of an arithmetic cost £161 2s. 6d. to publish, and was sold by the publisher at 2s. 9d. per copy. Allowing £10 for gratuitous copies presented to the press, and £12 for advertising and other expenses, what loss was sustained on that edition ?

22.    How often can three million seven hundred thousand be subtracted from three thousand seven hundred millions ?

23. By what fraction must 1|_ q£ 2 + 2£ — If _


be divided


in order to give ?    1^    3    \ + If 7f

24.    How many times is the number expressing the quarts in 3 tuns 1 hhd. 50 gals, of beer contained in the number which expresses the gills contained in 3 pipes 1 hhd. 29 gals, of wine ?

25.    A grocer bought 1 cwt. of tea, at 3s. 2d. per lb., and sold it so as to lose the selling price of 21 lbs. What was the selling price per lb. ? and at what rate must he sell it to lose the cost price of 21 lbs. ?

26.    A grocer bought 1 cwt. of tea, at Is. 9d. per lb., and sold it so as to gain the selling price of 14 lbs. What was his selling price per lb. ?

27.    How much tea, at 2s. 8d. per lb., must I give for 352 lbs. of coffee, at Is. 9d. per lb., so as to gain 10 per cent, by the exchange?

28.    What number of lbs. is the same fraction of a ton that 8f lbs. is of the required number itself ?

29.    A bankrupt’s deficiency amounted to £117, and his estate paid 15s. in the £; what were his debts ?

30.    Assets £213 Os. 9d., deficiency 5s. 9d. in the £ ; find the amount of the liabilities.

31.    Assets 7s. in the £, deficiency £136 10s.; find the debts.

32.    Creditors receive 8s. in the £, and the whole assets amount to £630; find the total liabilities.

33.    An insolvent’s assets are 12s. 6d. in the £ ; how much does a creditor receive whose loss came to £37 10s. ?

34.    I buy a horse and cow for £25, and give half as much again for the horse as for the cow; if I wish to gain £9 on each, for what must I sell them ?

35.    I bought 21 bottles of colonial wine for £5, and sold them so as to lose the selling price of 4 bottles; what was my selling price per bottle ?

36.    A tradesman’s prices are 33^ per cent, above the cost; if he allow 8^ per cent, discount for cash payment, what profit does he make on such a transaction when he receives £3 6s. ?

37.    A grocer marks his goods at 25 per cent, above cost. If he take off 8 per cent, for cash payment, what rate of profit does he actually make ?

38.    A can do a piece of work in 6 days, B can do twice as much in 10 days, and C 3-| times as much as both pieces in 40 days. What time would it take the three together to do the work ? Compare, also, their rates of working.

39.    A bankrupt’s assets are £31 6s. 8d., and his deficiency £219 6s. 8d.; how much will a creditor receive whose claim is £100 ?

40.    A baker uses a 1-lb. weight which is 3^ ozs. too light; how

many nominally 4-lb. loaves does he make out of a bag of flour containing 200 lbs., supposing that 4 lbs. of flour will make 5 lbs. of bread, true weight ?    ,

41.    A farmer raised 780 bushels of corn, and sold it at 6s. 6d. per bushel; next year he raised but 507 bushels. At what price per bushel did he sell it if he received the same money ?

42.    I buy 12 cwt 3 qrs. 8 lbs. of loaf-sugar, at £58 6s, 8d. per ton ; at how much per lb. must I sell it so as to gain £13 9s. 3d. on the

whole?    ...

43.    A certain number of half-crowns, together with twice as many shillings, four times as many sixpences, and five times as many threepences, make up £4 13s. Find the number of half-crowns. _

44.    An insolvent’s assets are £62 5s. 9d., and his deficiency is £36 14s. 3d. How much will his creditors receive in the £ ?

45.    If a bankrupt pay 7s. 5d. in the £, to what must his debts amount that his deficiency may be £186 17s. 3d. ?

46.    If a gold buyer’s scales defraud to the extent of 4 grains on every oz., what will a miner receive who sells 6 ozs. of gold at £3 18s 4d. per oz. ?

47.    I sell 1436 lbs. of sugar for £38 6s. 4d. ; if it cost me £3I9s. 4d. per cwt , what do I lose by the transaction ? Find, also, how many lbs. must be sold at 2d. per lb. above the cost that the loss may be recovered.

48.    I buy 24 bottles of colonial wine for £5, and sell them so as to gain the selling price of 4 bottles. What is the selling price per bottle ?

49.    A tradesman’s prices are such that, after making a reduction of 5 per cent., he still makes 14 per cent, clear profit ; at what per cent, above cost does he mark his goods ?

50.    A person bought 74 gals, of wine for £37 ; if 14 gals, leak out, at what rate per gal. must he sell the rest so as to gain £3 on the whole ?

51.    A dealer buys 2 tons 13 cwt. 24£ lbs. of potatoes at £4 13s. 4d. per ton, and sells 3 cwt. 24£ lbs. at 4s. 8d. per cwt. ; at how many lbs. per shilling must he sell the rest so as to gain £11 13s. 4d. on the whole ?

52.    A farmer rents a farm, and is to pay a fixed sum of money and a fixed number of bushels of oats. If the whole rent is £91 12s. when oats are at 6s. per bushel, and £100 12s. when they are at 6s. 9d. per bushel, find the number of bushels that formed part of the rent.

53.    A farmer grew 586§ bushels of wheat, and sold at 6s. a bushel; next year he sold his wheat at 5s. 6d. a bushel, and received exactly the same amount of money. How much did he grow the second year ?

54.    A bankrupt can pay 7s. 6d. in the £, but were it not for the insolvency expenses (which amounted to £30), he could pay 9s. in the £; find the amount of his debts ?

55.    An insolvent who had sufficient assets to enable him to pay 12s. 6d. in the £, received a legacy of £290, and then paid 13s. 4d. in the £ ; what were his liabilities ?

56.    I buy goods deficient in weight to the extent of 16 lbs. in the cwt., and I sell them again, and weigh them on a scale which represents 1 cwt. as being 140 lbs. What do I gain per cent, upon my outlay? Find, also, my loss by selling 1 cwt. as 128 lbs. (See Sum worked, exam. 14, page 286.)

57.    Having nothing but sovereigns in m37 pocket, I wish to pay a friend 4s. 2d. I owe him. If my friend has nothing but dollars, worth 4s. 2d. each, to give me as change, how can I pay him ?

58.    A invests £480 in the 3 per cents., and B £405 in the 4^ per cents., and B gets per cent, more for his money than A gets. If their incomes together amount to £38, find the price of each kind of stock.

N.B.—The above sum, from page 76 of Colenso’s Algebra, can be worked arithmetically, by attention to the principles explained on page 215 of this Arithmetic.

59.    A spirit dealer mixes three kinds of brandy, worth respectively 12s., 15s., and 7s. per gallon. In what proportions must he mix them, in order to clear 25 per cent, profit by selling the mixture at 12s. 6d. per gallon ?

60.    112 areas of land contain altogether 4187 acres; the average value of each area is £52 6s. 9d. Find the average price of each acre.

61.    A and B contract to do a piece of work together, A agreeing to do half as much again as B. A employs on it 15 men for 27 days; how many boys should B employ for 5 weeks, supposing that 3 men can do as much work in a given time as 7 boys ?

62.    Find the height of a tree which casts a shadow 29£ yards long, at the same time that the shadow of a steeple, whose height is 124 ft. 6 in., measures 41^ yards.

63.    In a school where the ages range from 8 to 12 years, of the children are 8 years old, £ are 9 years, i are 10 years, § of the remainder are 11 years, and there are 28 children 12 years old. Find the number at each age.

64.    Find, by Practice, the value of 117 lbs. 10 dwts. 9 grs. of silver, at £3 17s. 4d. per lb.

65.    Find the forty-eighth part of five hundred and eight millions seventy thousand and ninety-six, repeated six hundred and seventy thousand eight hundred and nine times.

66.    What sum of money must be lent out at 6 per cent. Simple Interest, that the interest on it may amount to £330 in 5| years ?

67.    If I buy cloth for £5 5s. on 12 months’ credit, and sell it a-fc once for £6 3s., giving 6 months’ credit, what do I gain per cent., money being worth 5 per cent, per annum ?

68.    What do I gain per cent, by allowing a customer 10 p. c. upon the same price at which I bought, receiving 20 per cent, reduction ?

69.    Two trains—one 200 feet,'' the other 284 feet in length—pass each other. If the one be travelling at the rate of 25 miles and the other at the rate of 35 miles per hour, how long will they take to pass? and what distance did each travel? Ans. 201§ ft., 282^- ft.

70.    A train, 330 feet long, passes over the Taradale viaduct in 14 seconds; what is the length of the viaduct, if the speed of the train is 37| miles per hour ?

71.    A steamboat leaves Sandridge for Geelong, a distance of 39

miles, going 6£ miles per hour, and it arrives 10 minutes before a sailing vessel which started from Sandridge 20 minutes earlier. Compare the rates of sailing.    <

72.    A merchant mixes 12 gals, of brandy at 16s. 8d. a gal. with 13 gals, of a better kind worth 25s. a gal. How much water must be added to it in order to sell at 15s. per gal. and gain £3 on the whole ?

73.    A train from Sandhurst to Kyneton has to pass through two

tunnels. If it take 45 seconds to pass through the first when the tiviin is going 25 miles per hour, and 25 seconds to pass through the other when going 36 miles per hour, find how much longer one tunnel is than the other.    _    _

74.    A train passes through one tunnel, a quarter of a mile in length, at a speed of 24 miles per hour, and through another, a third of a mile in length, at a speed of 30 miles per hour. How much longer will it take to pass through the one tunnel than the other ?

75.    If teachers be divided into five classes or grades, with average annual salaries of £122, £176, £244, £318, and £405 respectively,

and a teacher has to pass 4, 5, 7, 6, and 6 years respectively in these grades, what will have been the average salary received by a teacher 50 years ef age, supposing him to have commenced at 22 years of age ?    .    .    .    .    .

76.    Having nothing but sovereigns in my pocket, I wish to pay a friend Is. 8d. that I owe him. If my friend has nothing but dollars worth 4s. 2d. each to give me as change, how can I pay the debt ?

77.    A ship sails from Liverpool to Melbourne, a distance of 16020 miles, in 60 days, there being a favourable wind that helps her on at the average raté of 57 miles per day,; if returning by the same route the wind retards her to the sam^extent daily as it assisted her in her outward passage, how long should she be in reaching London, which is 720 miles nearer Melbourne ?

78.    A ship with 15 weeks’ provisions on board has a full crew of 42 men, who are allowed 56 ozs. per man each day ; when at sea 5 weeks, 7 men are taken on board from a wreck. How many ozs. per day may each man now be allowed, if the voyage can be completed in 8 weeks ?

79.    If 8 men can do a piece of work in 12 days by working 9 hours a day, how many hours a day must 9 men work to do half the work in 6 days ?

80.    If 8 men can do a piece of work in 15 days 3 hours by working S hours a day, how many hours a day would 6 men have to work in order to do three times as much in 41 days ?

81.    During one half of the year the income tax is Sd. in the £, and during the other half it is but 6d. ; what income had a gentleman who has a net income left of £213 11s. 8d. ?

82.    If it cost £37 17s. 9d. to keep a flock of 245 sheep for 123 dayt> how much additional expense will it incur to keep the flock for 12 days longer when 35 other sheep have been added to it?

83.    If it cost £23 17s. 9d. to keep 49 sheep for 65 days, how much additional will it cost to feed 35 other sheep for 12 days ?

84.    A mail train leaves Echuca for Melbourne, a distance of 153 miles, at the same time that a goods train leaves Melbourne for Echuca. If the speed of the former be 11 miles per hour greater than that of the latter, and the two trains pass each other 3 hours after starting, at what speed per hour is each train travelling ?

85.    A farmer on the Murray travelled from his homestead to Melbourne in 5 days, and back again in 6 days. If on each day he travelled 1 mile less than on the preceding day, how far from Melbourne did he reside ?

86.    On 23 chests of green tea, a merchant pays a duty of £5 more than the value of 8 chests, and, paying at the same rate, he pays upon 15 other chests £15 less than the value of 8 chests. 'VV'hat is the value of the tea per chest ?

87.    A person sells as pure coffee, at 3s. per lb., a mixture of 2 lbs. coffee and 1 lb. chicory. If the chicory be worth only 9d. per lb., what does he gain per lb. by his dishonesty?

88. If a shopkeeper uses a false weight of 15 ozs. for 1 lb., how many lbs. will a 60-lb. box of raisins appear to weigh ?

S9. A spiritdealer uses a gallon measure which holds but 3 qts. pts. ; how much will a person receive who has ordered 4 gals of wine ? What quantity did another customer order, if he received 11 gals. 1 qt. ?

90.    A person measured a fence, which he is to be paid for erecting, with a Gunter’s chain which is 39’6 in. short; how much does he charge for 95 chains which he has actually erected, at 9s. 9d. per chain ? How much did he gain by his false measurement ?

91.    At what time next after 5 o’clock are the hour and minute hands of a clock directly opposite ?

92.    Two partners sell out their business, and divide the proceeds, A taking £237 16s. 9d., and B £163 7s. If A invested £49 13s. 2d. more than B, how much did each put into the business, and what profit per cent, did the business yield ?

93.    A trooper, in pursuit of a criminal, sees him half a mile off, riding at the rate of 10 miles per hour ; if the trooper follow him up, riding 11 miles per hour, when will he be within gunshot range of, say, 110 yds. from the criminal, and how far will the latter have then ridden after he was first sighted ?

94.    A policeman gave chase to a burglar, whom he observed escaping from the window of a house 100 yards off. While the robber ran 15 yards, the constable ran 16 yards; how far had the former gone when he was overtaken ?

95.    A prisoner, who was attempting to escape from Pentridge, was not observed by the warder until he was 220 yards off. If the warder instantly pursued him, and had run a mile before he retook the prisoner, compare the rates at which the warder and prisoner ran, and also their rates if the prisoner had run a mile before he was retaken.

96.    Gold is 14TS times as valuable as silver ; in what proportions must these metals be mixed that the compound may be three times as valuable as an alloy of 2 parts gold and 8 parts silver ?

97.    At what rate per cent., Simple Interest, will a sum of money amount in 2 years to the same as at 10 per cent. Compound Interest?

98.    A luggage train leaves Melbourne for Wodonga, a distance of 187 miles, 2 hours before the mail train. If the former travel at the speed of 18^ miles j)er hour, and the latter 23J miles per hour, at what distance from Wodonga will the mail train overtake the other train ?

99.    A butcher bought a hind-quarter of beef, which weighed 180 lbs., for 2^ guineas, and sells § of it for 33s. 4d.; at what rate per lb. must he sell the rest, in order to gain 10s. lOd. on the whole?

100.    An insolvent pays 12s. 6d. in the £ on one half of his debts* and 13s. in the £ on the other half ; tc what do his debts amount, if his total assets are £286 17s. 6d. ?

101.    A settler spends £792 of his money in purchasing land, and ^ of the remainder in buying cattle to stock it with; if he has £360 left, how much had he at first ?

102.    Two travellers start from the same place to walk round an island ; the first walks at the rate of 20 miles per day, and the second walks 12 miles the first day, 14 miles the second, 16 miles the third, and so on. If they meet half-way round, what is the circuit of the island ?

103.    25 cattle consume 10 ac. in 13J days, and 45 cattle consume 14^f ac. in 9 days; how many acres will 18 cattle consume in 12 days, the grass growing uniformly ?

104.    If 3 sheep be equal in value with 17 geese, and 5 geese cost 16s. 3d., find, by Compound Proportion, the value of 19 sheep.

105.    A draper bought 64 yds. of flannel, at 2s. 4d. per yd., and sold it so as to gain 12^ per cent, upon the whole; find, by a Simple Proportion statement, his selling price per yard.

106.    I bought 150 animals for £100, comprising sheep at 15s. each, calves at 7s. 6d. each, and turkeys at 3s. 4d. each. How many of each sort did I buy ?

107.    A stone, let fall from the top of the Taradale viaduct, takes 21-

seconds to reach the ground. If a body fall through the air 16 ft. in the first second of its descent, and with a velocity increasing at a rate shell that the distance it falls through in any given time is obtained by multiplying 16 ft. by the square of the time, in seconds, it occupies in falling, what is the height of the viaduct ?    (b.) How long

would the stone take to reach the ground if the viaduct were 100 ft. high ?

108.    A body falling through space increases in speed with the accelerated velocity indicated by the .series of odd numbers 1, 3, 5, 7, 9, &c. If a stone fall 16 ft. the first second, say how far it will fall during the fourth second, and show that its descent during the four seconds is coincident with that obtained by multiplying 16 ft. by 4a, and give the depth of a mine which a stone will take 4 secs, to reach the bottom of.

109.    A wholesale merchant supplied a storekeeper with goods for which he charged him £348, his profit being 33^ per cent. If the storekeeper fail, and pay but 15s. in the £, what does the merchant gain or lose upon his prime cost ?

110.    A merchant finds from experience that his bad debts amount to 3 per cent, of his sales. If he wish to gain 15 per cent, clear profit, at what percentage of the cost must he mark his goods ; and for what ready money could he sell goods for which he now charges £400 by doing a cash trade only ? See Ex. 6, page 283.

111.    Two kinds of wine are mixed, 18 gals, of the one with 8 gals, of the other. If the price of the latter be 12s. l|d. and that of the mixture be 11s. per gallon, what is the price of the first kind ?

112.    A vessel which holds 260 gals, is filled by 2 pipes, running successively, in 20 minutes. One runs 15 gals, per minute, the other 10 gftofc-per minute. How long did each run ?

\

'113 In a division sum the dividend is 93100, the divisor and quotient are alike, and the remainder is 75. Find the divisor.

114.    If 17 hags of onions, newly pulled, weigh 1 ton 10 ewt.

3 qrs. 7 lbs., and each bag lose 21 lb. in drying ; how much per ton will a dealer gain who buys onions newly pulled at £8 8s. per ton and sells them when dry, at £15 4s. 6d. per ton ? See Ex. 13, page 285.

115.    At what rate per cent, will the discount upon £356 Is. 4d. amount in 5 years to £65 8s. ?

116.    A person finds that his bad debts amount to 4 per cent, of his sales. If he wish to make 8 per cent, clear profit, at what percentage above cost must he mark his goods ?

117.    Divide £17 18s. 9^d. f q. by ■§.

118.    By selling wine at 5s. 5d. per gal., a grocer gains 8| per cent, of the prime cost. Buying at the same rate, what imist he sell at per gal. in order to gain 20 per cent of the selling price ?

119.    Two pipes, running successively, fill a cistern which holds 282 gals, in 16 minutes. If one run 15 gals, per minute and the other 21 gals, per minute, how long did each run ?

Note.Work upon the principles of average explained on page 215.

120.    A tank of 1440 gals, has 3 pipes which will together fill it in half an hour. One pipe runs 6 gals, more, and another 3 gals, less per minute than the third. How many gals, per minute did each pipe run ?

121.    The sum of two numbers is 5840, and the quotient of the larger divided by the smaller 72. Find the numbers.

122.    The length of a room is 18 ft. and its height 11 ft. The area of the floor is that of the four walls. Find the breadth of the room.

123.    What is the area of a square paddock which contains an acre

for every eight panels in the fence enclosing it, supposing each panel to be 8| ft. long? See Ex. 16, page 286. _    _

124.    What is the area of a square paddock which contains an acre for every four panels in the fence enclosing it, supposing the panels to be 8£ ft. long ?

125.    A traveller set out on a journey, walking at the rate of 3s miles per hour. After proceeding a certain distance he rested for half an hour, and on resuming his journey walked at the rate of 4 miles per hour. On reaching his destination he found that he had arrived £ hour earlier than if he had walked the whole distance at 3i miles per hour without resting, but U hours later than if he had walked the whole distance at the rate of 4 miles per hour without resting. Find the whole length of his journey, and how far he had

travelled before he stopped to rest.

126.    At what per cent, above cost must a tradesman mark his

goods so that, after allowing a reduction of 5 per cent., he may still clear 33 per cent, profit ?    .    ,,    „    . ,    ...

127.    In slating a roof, f of each slate is covered by the slates which overlap it. How many slates, each 13£ in. long and 8 in. broad, will cover the gable-end roof of a church of which the walls are 60 ft. and

17 ft. long respectively, the top of the roof being 12 ft. higher than the walls, and the projection at the eaves 6 in. ?

128.    If a stone let fall from the Studley Park bridge takes If seconds to reach the water, what is the height of the bridge above the Yarra ?

129.    The income tax being 6d. in the £, a person had to pay £2 6s. 8d. less than when the tax was 8d. Find his income.

130.    I sell cloth at Ss. 4d. per Flemish ell, and gain 20 per cent, of the cost of an English ell on every Flemish ell sold. What did it cost per French ell ?

131.    Having engaged a boat at Edwards’s boat-house (Prince’s Bridge), I rowed up the Yarra at the speed of 3| miles an hour, and returned at the speed of 5^ miles an hour, having been rowing 5 hours altogether. How far up the river did I go?

132.    Apiece of cloth appears to measure 12 yds., but \ in. has been worn off one end and ^ in. off the other end of the yard measure used. What is the true length of the piece ?

133.    A policeman runs after a thief who is 55 yds. ahead ; four of the thief’s steps are equal to five of the policeman’s, but the policeman takes four steps while the thief takes three. How far must the,policeman run before he catches the thief?

134.    Find the price of the 4J per cent, stock when the income obtained by investing £1050 is £52 10s.

135.    A man leaves Taradale forKyneton at the same time that his wife leaves for Castlemaine. The distance to both towns is the same, and they both wish to arrive at a certain time ; but the man walking

3    miles per hour arrives ten minutes before the time fixed upon, while his wife walking miles per hour arrives half an hour late. How far is Castlemaine from Kyneton ?

136.    Meeting several beggars, I gave 3 pence to each, and have

4    pence left. I should require 6 pence more to give them 5 pence each. How much money have I ?

137.    A man buys a case of oranges at 6d. a dozen ; he finds fifty spoiled, and selling the rest at IS for Is., makes 10s. 9d. profit. How many are there in the case ?

138.    Two blocks of land are sold for £120 each, one for 20 per cent, above cost, the other for 20 per cent, below cost. What is the loss or gain per cent. ?

139.    If ’busses start from Melbourne every 10 min. and travel six miles an hour, how long, after one passes a lady walking along the route at the rate of three miles per hour, will it be before the next will pass her ?

140.    Three first class tickets, and four second class tickets to a certain railway station cost 4s. 3d. ; two first and six second class to the same station cost 4s. 6d. Find the first and second class'fares.

- 141. I give £20 for 48 cwt. of goods ; what do I gain by selling them at 2s. 6d. for 28 lbs. ?

142. I sell goods at 2s. 3d. for 28 lbs., that cost me £10 per ton; what do I lose ?

143.    A sells a cargo to B, and gains 10 per cent, on his outlay, B sells it to C and gains 5 per cent., C sells it for £10395 and loses 10 per cent. Find A’s outlay.

144.    Two mounted troopers keep sentry, riding in opposite directions round a circular beat 30 miles in circumference ; one rides 10 miles an hour, the other only six. How far apart will they be six hours after starting from the same place ?

145.    If each of two numbers be divided by the other, the product of the quotients is always equal to unity. Find by division of decimals the quotient of a certain small number divided by a larger number, supposing the smaller number to be contained in the larger number 14'0625 times. What is the sum of the two numbers if the smaller number be 112 ?

146.    A gentleman left £200 to be divided equally among his servants. Each received as many half-crowns as there were servants. How many servants were there ?

147.    A selector paid £426 13s. 4d. to have his land cleared and fenced, and then found that he had paid as many pence per acre as there were acres in his selection. Of how many acres did it consist ?

148.    At what rate per cent. Compound Interest will £520 16s. 8d. amount to £620 6s. 5d. in 3 years?

149.    A wholesale merchant charges a shopkeeper 18f-f per cent, above his prime cost. If the latter receive a discount of 3 per cent, when paying his account, how much per cent, profit does the merchant clear upon his prime cost ?

150.    Two assistant teachers and 3 pupil teachers receive £140 10s. salary in 5 months, and 3 assistant teachers and 6 pupil teachers £178 12s. in4 months. How long will 1 assistant teacher and 1 pupil teacher, paid at the same rate, take to earn £158 12s. ?

151.    A grocer mixes 56 lbs. of tea at Is. 6d. per lb. with a quantity of an inferior quality worth Is. 2d. per lb. How much of the latter kind must he use that he may gain 33^ per cent, by selling the mixture at Is. 9d. per lb. ?

152.    In reducing vulgar fractions (expressed in their lowest terms) to equivalent decimal fractions, it has been stated that those having denominators containing no other factors than 2 and 5 produce finite decimals, while those having denominators containing neither of these factors produce pure repeating and pure circulating decimals. Of the latter, all prime denominators yield pure repeaters or circulators containing either as many figures minus one as there are units in the denominator or an aliquot part, as one-half, one-third, one-fourth, &c., of this number.

a.    How many denominators between 1 and 100 are included in

the last statement ?

b.    Of numbers less than 101, how many produce (1) finite

decimals, (2) pure repeaters and circulators, (3) mixed circulators ?

153.    Two zinc tanks of equal dimensions are filled with water.

One has a tap which will empty it in 3 hours, and the other a tap that will empty it in 4 hours. When will one tank have but half as much water in it as the other ? See Ex. 8, page 220.

154.    How often does the number 10'703S8 contain the number *0108, and what is the remainder?

155.    A lady observing a volunteer practising shooting at a target, took out her watch and observed that 33- seconds elapsed after she saw the flash before she heard the report of the rifle. Supposing sound to travel at the rate of 1116 feet per second, and that light travels so rapidly that sight may be considered instantaneous, how far off was the volunteer ?

156.    Express in grains the difference between a lb. of gold and a lb. of feathers.

157.    I have two 5-gal. casks, one full of wine and the other full of brandy ; and thrice I take from each cask one gallon and put it into the other. What are now the proportions of wine in the one cask and brandy in the other ?

158.    A mixture of 15 gals, of brandy and 6 gals, of rum is worth 30s. a gallon. If the value of the brandy is 36s. a gallon, what is the value of the rum per gal. ?

159.    A train leaves Sandhurst for Melbourne at the same time that another train leaves Melbourne for Sandhurst. If the speed of the first be 28 miles per hour, and that of the other 20 miles per hour, in what time and at what place will they pass each other, the two towns being 100 miles apart ?

160.    Two towers, 40 ft. and 50 ft. high respectively, are standing in the same horizontal plane 120 ft. apart: how far from each tower is that point in the line joining their bases which is equally distant from their summits ?

161.    I bought a hind-quarter and a fore-quarter of beef, weighing together 504 lbs. ; I paid 7£d. per lb. for the hind-quarter and 5^d. per lb. for the fore-quarter, and found that I had paid 10s. 6d. on the whole more than if I had bought both quarters at 6jd. per lb. Find the weight of each quarter.

162.    The mail train from Sandhurst is 130 ft. long, and it passes a goods train from Melbourne which is 200 ft. long in 5 seconds. If both were moving on parallel lines in the same direction, the mail train would take half a minute to pass the goods train. At what speed per hour is each train moving ?

163.    A meadow is let for £25 and a fixed number of tons of hay. When hay is at £3 12s. 8d. per ton, the whole rent is 20 per cent, higher than when it is at £2 15s. a ton. What number of tons formed part of the rent? See Ex. &, page 219.

164.    A grazier bought 100 head of cattle for £100, cows at £3 each, store cattle at £1 each, and calves at 3s. each. How many of each sort did he buy ? See page 216.

165.    I buy a house for £925 11s. 6d., upon terms half-cash and the balance in three equal instalments at 3, 6, and 9 months respectively. I sell forthwith for £990, and give such credit as

causes me to lose £3 14s. Id. by the transaction, allowing money to be worth 10 per cent. How long credit did I give ?

166.    At what price must the 4J per cent, stock be that the investor may receive 5 per cent, upon his outlay ?

167.    Find the contents in c. yards of a dam or tank 101 ft. long and 75 ft. wide at the top, with sides sloping in the ratio of 4 to 1 (i.e., 4 ft. horizontal to 1 ft. perpendicular), and 7 ft. 6 in. deep.

168.    What is the cubic contents in yards of a tank 36 ft. square at the bottom, having on three sides a slope of 1 to 1, and on the fourth side a slope of 5 to 1, the perpendicular depth being 9 ft. ?

169.    A tobacconist bought a drum of Barratt’s twist and paid 33| per cent, duty, and then sold it at a gain of 12£ per cent. Had he sold it for £3 less, he would have lost 1 per cent. What did he pay for it ?

170.    A shopkeeper buys a parcel of gold from a customer, who-agrees to take out the value in goods. If he defrauds to the extent of 5 per cent, in weighing the gold, and then defrauds again to the extent of 5 per cent, in weighing the goods, how much per cent, does the shopkeeper gain by cheating ?

171.    A person bought 3 cwt. 1 qr. 14 lbs. of salt at £2 12s. 6d. per cwt., and sold 1 cwt. 3 qrs. 21 lbs. at £2 15s. per cwt. At what, rate per cwt. must he sell the rest so as to gain 11s. 9|d. upon the whole ?

172.    Multiplying the circuit of a room by the height gives the area of the four walls ; and this in sq. yds., sq. ft., &c., divided by the width of paper expressed in terms of the same name gives the length of paper required to cover the walls. Show by a fractional expres- * sion for the paper 2 ft. 3 in. wide required to cover the walls of a. room 20 ft. 6 in. long, 20 ft. wide, and 10 ft. 3 in. high, that

2 (length + breadth) X height X reciprocal of fraction expressing the width of paper ~ the length of paper required.

173.    A can walk the same distance in 25 sec. that B can in 27£ sec. If in a race of miles A give B a quarter of a mile start, who-will win, and by how much ? See Ex. 1, page 203.

174.    Copies of a book were sold, some for cash and some on six

months’ credit. 13 of those sold for cash brought the same money that 12 of the others were sold for. What rate of interest was charged upon the cash price for the credit allowed ?    _ _    _

175.    In a town where the increase in the number of criminals is 1-6 p.c., the male criminals alone have decreased 2'4 p.c., while the female criminals have increased 5 6 p.c. Compare the number of male and female criminals.

VICTORIAN EXAMINATION PAPERS.

GENERAL EXAMINATION OF TEACHERS.— December, 1877.

ARITHMETIC.

Set No. I.

Three hours allowed for this 'paper.

1.    Explain and illustrate the use of the vinculum or bracket.

2.    The quotient of a sum is 515450, and the divisor is to the quotient as 9 : 65. Required the dividend.

•» 3. A gentleman left an estate of 735 acres to be sold, and the proceeds to be divided between his three sons in the proportion of 3, 4, and 5. What was each son’s share, A of the land having fetched £25 per acre ;    £20 per acre ; and the remainder £15 per acre?

4. Find by Practice the value of 367 qrs. 25 lbs. 13 ozs., avoirdupois weight, at £5 16s. 5d. per qr.

-    5. If 20 men earn £76 13s. 4d. in 18 days, how many boys will

earn £103 10s. in 27 days, the daily earnings of a boy being £ those of a man ?    & 5

6.    Sold 60 bags of sugar at a profit of 6 per cent., and 30 at profit of 8 per cent. ; but had I sold the whole at a uniform profit of 7 per cent., I should have gained 15s. more than I did. What was the prime cost per bag ?

7.    20000 loads of gravel were to be carted to a given place in 10 days ; but it was found that 54 horses had only carted 4,500 loads in 6 days. _ How many more horses would be required to cart the restin the time left ?

8.    State and prove the rules for determining mentally_

(a)    The value of 100 articles, the price of one being given.

(b)    The interest of any sum at 6 per cent, for months

9.    In what time will £158 6s. 8d. amount to £176 2s. lid. at 3 per

cent, per annum ?    ’’    ^

10.    A can mow 5 acres in 9^ days, B 4^ acres in 7 days How

long will it take them together to mow 12£ acres, and how many acres will each mow ?    J

11.    Simplify j (li + 2f)    j -r$.

12.    (a) Reduce 7'01236 to its equivalent vulgar fraction and

divide '001648019 by 907.    6    ’ and

(b) What decimal added to £ will make the total equal to 3 ^

6- 1

ARITHMETIC.

Set No. II.

Three hours allowed for this paper.

1.    Define Ratio: state how ratio may be generally expressed.

2.    What number multiplied by 3769 will give 2148669210 as a product ?

3.    What will be the cost of carpeting a room that is 18| ft. long and 16 ft. 6 in. wide, at 6s. 9|d. per lineal yard; the width of the carpet being 2 ft. 9 in., and 2J yards being allowed for the waste in fitting the pattern ?

4.    A linen draper laid out on a lot of muslin £500, but upon examination ^ proved to be damaged, and fetched but 5s. per yard, by which means he lost £50. At what rate per ell English must the undamaged part be sold so that he may clear £50 on the whole ?

\

5.    Find by Practice the value of 9 casks of sugar weighing each 6 cwt. 1 qr. 7^ lbs., allowing 25 lbs. per cask for waste, at £2 16s. 8d. per cwt.

6.    If 35 boys can earn £45 18s. 9d. in 9 days, how many women will earn £77 3s. 6d. in 13| days, the daily earnings of a woman being to those of a boy as 7 : 5 ?

7.    A body of 3500 men have provisions for 5 months, at the rate of 36 ozs. per man ; but at the end of 2 months 350 more men arrive. By how much must the allowance to each man be reduced that the provisions may serve the whole for 6 months ?

8.    State and prove the rules for determining mentally—

(a)    What any number of pence per day will amount to in a

year, omitting Sundays.

(b)    The interest on any sum at 5 per cent, per annum for

months.    .

9.    What sum will amount to £118 18s. 9d. in 7 years, at 4| per cent, per annum ?

10.    A water tank holds 147 gallons, and the pipe that supplies it brings in 14 gallons in 9 minutes, while the tap discharges 40 gallons in 31 minutes : now, supposing the tank to have been empty, and the pipe and tap to have been left both running for 3 hours before the tap was shut, in what time afterwards would the tank have been filled, the water continuing to flow from the supply-pipe ?

^ j

3| + 29 '


of


11. Simplify


2§ of 3f 4i of (f-f)

12. (a) Divide 41-9717 by 34-12, having first reduced each to its equivalent vulgar fraction.

(b) What fraction divided by 872 will give a quotient of TV-

313

JUPIL TEACHERS’ EXAMINATION.—Dec., 1877.

ARITHMETIC FOR FIRST CLASS.

Set No. I.

1.    If £250'5 gain £12’525 in 6’8 months, what will £400’8 gain in 44 months ?

„    2. Simplify    + -ft of

3.    At what rate per cent, per annum will £547 15s. amount to £629 18s. 3d. in 3 years ?

4.    State and prove the rules for determining mentally—

(a)    The interest on any sum at 5 p. c. per annum for months.

(b)    The value of a ton, the price per lb. being given.

Set No. II.

*    1. If the clearing of a paddock containing 29'325 acres cost

£75 072, what will the clearing of a similar paddock containing 35‘19 acres cost, wages having risen in the proportion of l-25 : 1 ?

„    3*-£of 34 q ci

2.    Simplify # ^-(Jof2£j “ § of *•

3.    In what time will £120 10s. amount to £134 16s. 2|d. at 4f per cent, per annum ?

4.    State and prove the rules for determining mentally—

(a)    The value of 100 articles, the price of one being given.

(b)    The value of a lb., the price per cwt. being given.

ARITHMETIC FOR SECOND CLASS.

Set No. I.

1.    Sold 9 casks of sugar weighing each 9 cwt. 1 qr. 14| lbs. Find the value of the same by Practice, at £2 Is. 7d. per cwt. ; an allowance of 19 lbs. per cask having been allowed for waste.

2.    If £625 gain £93‘75 in 12’625 months, what sum will gain £75

in 2 -525 months ?    _____

3.    Simplify (3f - 2|) -f (4f -f 24 - f). _ _

4.    State and prove the rules for determining mentally—

(a)    The value of a lb., the price per cwt. being given.

(b)    The interest on any sum at 6 per cent, for months.

Set No. II.

1.    Sold 17 casks of soap, each weighing 5 cwt. 1 qr. 154 lbs. Find by Practice their value at £3 5s. 4d. per cwt., an allowance of 3s. 9d. per cask having been made for waste.

2.    If the carriage of 1'9375 cwt. for 52-25 miles cost 17s. 5d., what will the carriage of ’625 of a ton come to for 27'9 miles?

3.    Simplify (f + § + -£?) (12S - 6J 4- |). #

4.    State and prove the rules for determining mentally—

(a)    The value of a ton, the price per lb. being given.

(b)    What any number of pence per day -will amount to in

a year, omitting Sundays.

ARITHMETIC FOR THIRD CLASS.

Set No. I.

1.    Find by Practice the value of 79 cwt. 2 qrs. 19 lbs. of sugar at £2 6s. lOd. per cwt.

2.    A bankrupt owes £1126 8s., and his assets amount to £985 12s. How much will his creditors receive in the pound ?

3.    Give the rules for calculating mentally—

(a)    The interest on any sum at 5 per cent, for any number of

days.

(b)    The value of a ton, the price per lb. being given.

Set No. II.

1.    Find by Practice the value of 69 ozs. 17 dwts. 19 grs. of gold at £3 18s. 4d. per oz.

2.    A bankrupt owes £1484 16s., and he pays 15s. 3fd. in the pound. Required his assets.

3. Give the rules for determining mentally—    _

(a)    What any number of pence per day will amount to in a

year.

(b)    The price of one article, the price per gross being given.

ADMISSION TO TRAINING.—June, 1878.

1.    (a) From what number must 790850 be subtracted 4900 times

to leave 90019 as remainder ?

(b) What is meant by saying that two or more numbers are prime to each other ?

2.    If A, in 1 hr. 28 minutes, can read aloud 14 pages of a book having 44 lines to the page and 16 words to the line, how many pages of a book having 36 lines to the page and 12 words to the line can B read in 2 hrs. 15 min., supposing that the latter reads 8 words while

the former reads 7?    n,

3.    Find by Practice the cost of 287 sq. yds. 8 ft 88 in. at £1 13s. Jd.

P<4 S What must be the length of a tank 20 ft. 10 in. wide and 12 ft deep to contain 642 tons 1920 lbs. of water, assuming that a gallon contains 277i cubic inches, and that a gallon of water weighs 10 lbs ?

5 If a person who bums gas 3 hrs. 45 min. daily when it is 6s. 9d. per 1000 cubic ft. finds that his gas bill amounts to 1£ Per cent- oi his income, for how long daily should he burn gas when it is at 7s. 6d. per 1000 cubic ft. that his gas bill may only amount to 1^ per cent, of his income ?

6.    A gun is being fired from a fort at intervals of 2 min. 10 sec. By the passengers in a train approaching the fort, the report of the first gun is heard 13 seconds after its discharge, and the report of the second gun 3 seconds after its discharge. Supposing sound to travel 1116 feet per second, find at what rate per hour the train is approaching the fort.

7.    (a) State and prove the rules for determining mentally—1. The

value of one article, the price per gross being given.

2. The value of a lb. avoirdupois, the price per oz. being given.

(b) How would you calculate mentally—1. The wages of 750 men for 8 hours at 7£d. per hour ? 2. The product of 7848 by 125 ?

8.    Find at what rate per cent. £471 6s. 8d. must be lent out at Simple Interest to amount to £542 0s. 8d. in 1-| years.

9.    Divide 5f of 3f -    of 8    10

8    5    2* - * - J . _i_.

9 Dy 31 - i of 41.

10.    The profits of a certain company are 8| per cent, of the market value of the shares. A paid £45 for the 1 of an eightieth share. Find the total profits of the company, and the amount of the dividend to which A is entitled.

11. (a) Divide the sum of (4^-3’46) and 1‘93 by 60i85, and express the result as a decimal.

(6) Reduce the difference between 13'0625 acres and 2‘25 roods to the decimal of a square mile.

EXAMINATION FOR EXHIBITIONS.—Dec., 1878.

Time allowed for this paper, from 12.30 to 2.30 p.m.llthDec., 1878.

1.    A rectangular enclosure is 123-A chains long and 80'5 chains wide. Find by Practice the cost of enclosing it with a post-and-rail fence at 5s. 3|d. per rod or perch.

2.    An estate which will carry two sheep to the acre is valued at £4 per acre, and taxed at the rate of 25s. for every £100 of its value. If the tax amounts to £288 2s. 6d., and is paid only on that portion of the value of the estate which exceeds £2500, find how many sheep the estate will carry.

3.    A traveller set out on a journey, walking at the rate of 3^ miles an hour. After proceeding a certain distance he rested for half an hour, and on resuming his journey walked at the rate of 4»miles an hour. On reaching his destination, he found that he had arrived £ hour earlier than if he had walked the whole distance at 3J miles an hour without resting, but 1J hours later than if he had walked the whole distance at 4 miles per hour without resting. Find the length of his journey and how far he had travelled before he stopped to rest.

4.    A bill drawn 21st December, 1875, at 4 months, was discounted on the 11th February, 1876. If the true discount at 6 per cent, per annum amounted to £5 8s., find the amount of the bill.

5.    A draper having bought 4000 yards of cloth, sold ■§ of it at a loss of 5 per cent., \ of the remainder at a gain of per cent., and the rest at a gain of 2| per cent. If he gained £2 Is. 8d. by the whole transaction, find the cost price per yard of the cloth.

6.    If, under similar conditions, 318 litres of nitrogen gas are of the same weight as 125 litres of chlorine gas, 18 litres of chlorine gas of the same weight as 79'5 litres of marsh gas, 15'2 litres of marsh gas of the same weight as 14'4 litres of ammonia gas, 18 litres of ammonia gas of the same weight as 152 litres of hydrogen gas, and if 900 litres of hydrogen gas weigh 1247'4 grains, find in lbs. avoirdupois the weight of 12000 litres of nitrogen gas.

• (Note.—Work by chain rule, page 198.)

7.    (a) Simplify—

7 x.8 -f U & °f 2to + H) 7§-fof l

6 (8 ^ 1|)


1* of

6 -f-


1-1

2 k

(b) In the preceding example point out an instance of a complex and of a compound fraction.

8. Find the value of—

•27    _

-of an acre + (7 "65 - 3'025) roods

■4629 •0425

- 1

•0025


perches - 214'84375 sq. yds. x T28)

GENERAL EXAMINATION OF TEACHERS.— December, 1878.

ARITHMETIC.

Set No. 1.

Three hours allowed for this paper.

1. (a) “Each digit has two values, an absolute and a Relative.’' Explain this.    . .    .    .    ..

(b) In a simple division question the divisor is six and a nail times greater than the remainder, and is one forty-

eighth of the quotient: find the latter, the remainder being 74068. Express your answer in words.    _

2. Supposing a postman travels 17 miles 1 furlong 141 perches in his daily rounds, how many yards does he travel in a year, allowing him two weeks’ holiday ?

(NOTE —Six days per week. He travels 301 days in the year.—J. J. B.)

3.    Three merchants, A, B, and C, purchased a consignment of tea consisting of 90 tons 16 cwt. 1 qr. 14| lbs. and divided it amongst them. A paid for his share £4669, B paid £6225 6s. 8d., and C paid £7781 13s. 4d. What part of the whole did each get?

4.    Pind by Practice how much B in the above sum would gain by selling at £294 per ton.

5.    If I gained 25 per cent, of the original price of an article by selling it at £3 10s. 71d., how much per cent, would I gain by selling it at £4 4s. 9d. ?

6.    In what time will 20 men set up 4 miles of fencing if 24 men require 18 days to do 7 miles, supposing the first set of men to work 9 hours daily as against 8 hours daily of the second set, but that 4 men of the second set can do as much in one hour as 5 men of the first set ?

7.    (a) To find the value of a single article, the price per score

being given.

(&) To find the discount on any sum at 5 per cent, per annum.

State and prove the rules for determining the above mentally.

8.    Find the principal which being lent for 2\ years at £6§ per cent, will yield £655 7s. 8|d. as Simple Interest.

9.    (a) Prove that f divided by f must be f.

(b) Three men together can finish a piece of work in 2 hours, but at the end of half an hour two of them leave, having done ^ and T\ of the work_ respectively. Find in what time the third man can finish it.

10.    Find the value of

24

2 -

2

2 +

2

2 +

24 of 4


21 -

of 26 acres 2 roods 24 perches.

11. (a) Find the value of    ^    ^    ^ ^s.

•03416


(b) Find the sum and difference of ‘026 and ‘00026; multiply them and divide the product by 16'9.

Set No. II.

Three hours allowed for this paper, 1. (a) Distinguish between quantity and number.

(&) Supposing there are one thousand two hundred and fifty-eight millions of people now living in the world, and that one thirty-fourth of them die off annually, if the number of deaths be the same in all parts of the world find by how many the deaths in the whole world exceed those of Europe in one year, the population of ,the latter being given at 301356314. Write your answer in words.

2.    Find the minutes which have elapsed since the beginning of the Christian era till the present hour—viz., 9.30 a.m. 19th December, 1878. Allow 365 days 5 hours 48 minutes to the year.

3.    Three men contracted to build a railway 59 miles 36 chains 32 links long, and agreed that each man should construct a length corresponding to the capital he could employ. A placed £6517 in the concern, B placed £7820 8s., and C placed £9123 16s. Find the length of line each had to make.

4.    What would be the cost of fencing the line of railway described above at £184 16s. 8d. per mile? Solve this by Practice.

5.    A merchant bought tea at Is. 3d. per lb., which he sold at a profit of 30 per cent, to a retail dealer, who again sold it at an advance of 50 per cent, on his price. Find the price to the consumer, and the total percentage of profit on the original cost.

6.    If 14 men can do a piece of work in 26 days, working 8 hours daily, in what time will 12 men, working 9 hours a day, do twice as much work if 9 of the latter set of workmen can do as much work in one hour as 13 of the former set ?

7.    (a) To find the price of one article, the price per gross being

given.

(6) To find the value of an article, the price per pound being given.

State and prove the rules for determining the above mentally.

8.    Find the principal which, being lent for 24 years at £3 15s. per cent, per annum Simple Interest, will amount to £3394 5s. 5d.

9.    Two men, A and B, undertake to cut down 20 acres of wheat. A can cut down 5 acres in 9^ days, and B 4£ acres in 7 days. Find how long it will take them together, and what proportion of the work each will do.

10.    Find the value of—

(5H + 2t3»), of If ~ H 0f i ton 9 cwt. 1 <p\ 24 lbs.

^    ^ 2 - f of 34

21

11.    (a) State the rules for division of decimals.

.x [b) Find the value of - .m jY~ of 22‘2 of ‘29 of ‘2*34 of 3 shillings.

(c) Find the sum, the difference, and the product of ‘0076 and ‘00076, and divide your last answer by 3‘61.

ARITHMETIC.—FIRST CLASS.

Set No. I.

1. If the weight of a cubical block of tin whose length, breadth, ,nd height are respectively 5 feet, 4'75 feet, and 3'5 feet, be 20*9 ons, what will be the length of a cubical block of silver weighing 7'18 tons whose breadth and height are 2'5 and 1*75 feet respectively ; the weight of silver to that of tin being as 1:436 to 1 ?

(NOTE.—I have slightly altered one number in the foregoing question.—J. J. B.)

2. Simplify


lid.


3& of li

' 3. At what rate per cent will £575 amount to £744 16s.

Sf years ?

4.    Reduce 9 inches to the decimal of 3£ yards.

Reduce *13 to a vulgar fraction.

5.    State and prove the rules for determining mentally—

(a)    The value of a cwt., the price per lb. being given.

(b)    The interest on any sum at 6 per cent, for a given number

of months.

Set No. II.

1. If the weight of water in a cubical tank 4*73 feet long, 3*5 feet broad, and 10 feet deep be 4*73 tons, what will be the height of a cubical block of ice whose length and breadth are 8*4 and 5'5 feet respectively, the weight of which is 5 *52 tons ; the weight of ice to that of water being as *92 to 1 ?

(NOTE.—I have slightly altered two numbers in the above question.—J. J. B.)

+ £


3fr x [l - | of |f ]    : 3V


2. Simplify'

/• 3. In how many years will £1250 amount to £1547 18s. 4d. at 6^ per cent. Simple Interest ?

4.    Reduce 7sd. to the decimal of 5 shillings. Reduce *27 to a vulgar fraction.

5.    State and prove the rule for determining mentally—

(a)    The value of an ounce avoirdupois, the price per lb. being

given.

(b)    The interest on any sum at 5 per cent, for a given number

of days.

ARITHMETIC.—SECOND CLASS.

Set No. I.

/ Find the cost of carpeting a room 22 feet long and 5 yards broad, the price of the caruet being 4s. 9d. per square yard.

/ 2. Simplify [| + 2^-f of -ft J -f [l + iV + l0* 3^) •

Reduce the result to its equivalent decimal fraction.

/ 3. If it takes me 5’5 days, each 10’2 hours long, to travel 182’325 miles, how many days of 8‘5 hours shall I take to travel 331'5 miles?

4.    Divide ’47377 by ’059.

Find the product of 5 ’75, ’125, and ’05.

5.    State and prove the rules for determining mentally—

(a)    The price of a pound troy, that of an ounce being given.

(b)    What any number of shillings per week will amount to in

a year.

Set No. II.

/    1. What will be ton cost of making a road 1 mile 30 chains long

and 10 perches wide, at 10s. 8|d. per square perch, a 9i 4- 1 _ AA

2. Simplify -3---3-3 2—

of


+ i -f 4 +


1 - t

Reduce the result to an equivalent decimal fraction.

,    3. If the cost of carriage of 4’75 cwt. for 160’8 miles be £3 8s.,

what will be the cost of carriage of 11’875 cwt. for 100 5 miles ?

4. Divide the sum of ‘00075, ’025, ’00065, and ’3726 by nineteen

thousandths.    .

5. State and prove the rules for determining mentally :—

(a)    The price of a ton, given that of a cwt.    _

(b)    What any number of pence per week will amount to in

a year.


ARITHMETIC.—THIRD CLASS.

Set No. I.

1.    Find by Practice the value of 57 acres 2 roods 17 perches at x £5 13s. 7d. per acre.

2.    If £67 15s. 9d. is the value of 20 lbs. 6 ozs. 10 dwts.^of silver, / how much silver is there in a trophy of that metal worth £25 10s. 6d.,

the cost of the workmanship i^k^wh^ch is £6 11s. ?

3.    State the rule for determining mentally—

(а)    The value of one article at a certain rate per score. _

(б)    The value of a gross, the price of one article being given.

Set No. II.

1.    Find by Practice the value of 69 miles 6 furlongs 17 rods of wire

at £315s. 9d. per mile.    .    .

2.    If the value of a gold cup, weighing 10 ozs. 6 dwts. 16 grams is £45* 14s. 8d., the cost of the workmanship being £5 10s., what weight of gold is worth £12 4s. 2d. ?

3.    State the rule for determining mentally—

(a)    The value of one article at a given rate per dozen.

(b)    The price of 100, that of one article being given.

ANSWERS TO EXERCISES IN SUPPLEMENT.

EXERCISE XLIL—SQUARE ROOT.

I.M 35.    2.    55.    3.    405.    4. 36.    5. 981.    6. 237.    7. 444.

8. 139.    9. 792.    10. 648.    11. 3T416.    12. 5760.    13. 6080

14. 277-274.    15. -007782.    16. 365-242218. 17. 173-503. 18. 30103

19. -30103.    20.    -9808.    21.    1 21.    22. 10-0408.    23.    99'009.

24. 50-709.    25.    |.    26.    27. If.    28. ff. 29.    30. 4f

31. -6236, etc.

IL—1. 7050 and 1008. 2. 40‘015 and 5-002. 3. 104S8 and 6140.

4. -1005 and 5‘025.    5. 1815 and 55.    6. 168.    7. 444.

III.—1. 15. 2. 57. 3. 65; 5s. 5d. 4. 29. 5. 92|; £8 17s. 3|d. ^q.

6. 3£ miles. 7. 647 "35 yards. 8. £20; 25 p. c. 9. 8 p. c. 10. 6f sec.

EXERCISE XLIIL—CUBE ROOT.

I. —1. 250.    2. 132.    3. 1042.    4. 614.    5. 5760.    6. 1879.

7. 3-1416. 8. 708 09. 9. 2‘4.    10. 2 926 11. f 12.    13. §.

14. -625.

II. —1. 20 p. c. 2. 39 in. 3. 12 ft. 6 in. 4. 6"4 ft; or, 6 ft. 4f in.

6. 1 ft. 3 in. 6. 10 ft. 7. 9 ft. 6 in. 8. (a) 457 and (b) 12.

EXERCISE XLIY.—SQUARE MEASURE.

1. 96. 2. 12 sq. ft. 108 in. 3. 17 sq. yds. 3 ft. 108 in. 4. 24 sq. yds. 5. 7 sq. ft. 62 sq. in. ; 107 acres ; 627264. 6. 20 sq. ft. 123 sq. in. 7. 10 sq. ft. 45 in. 8. 57 sq. ft. 54 in. 9. 1 ac. 2 ro. 3U per. 10. 4 ac. 403^ sq. yds. 11. 800 sq. yds. 12. 544 sq. yds. 4 sq. ft. = '0694 Ir. acres (plantation is outside playground). 13. acres.

14. 50. 15. 36. 16. 72.    17. 2 sq. yds. 3 ft. 13^ in. 18. 56 sq. yds.

19. 101 yds. 1J inches. V 20. 156 yds. 21. 108 boards. 22. 31£ sq. yds. 23. 1 acre. 24. 14400.    25. 38720 c. yds. ; 3 acres.

26. £1 8s. 6d.    27. 313fii sq. yds. ; or, 313 yds. 1 ft. 6f-f in.

28. 59 sq. yds. 8 ft. ; paper 50 yds. 2 ft. 29. 1 ac. 2064 sq. yds. 30. 1 ro. 20 per. 11 sq. yds. 2 ft. N 31. 63 ch. 75 links. 32. 6 ft. 8 in.

33. 1 fur. or 220 yds. 34. 117 yards. 35. 697 ac. 32 sq. per.

36. Oct. 27th, 1880. 37. 5000; 11 - 1S07-..ft. 38. Ilft.x39. 55planks. 40. 4 ft. 6 in.; 9 ft. ; and 13 ft. 6 in.

EXERCISE XLV.—CUBIC MEASURE.

1. 8000. 2. 15 c. ft. 1184 c. in. 3. 2 c. yds. 18 ft. 324 c. in.

4. 16 c. ft. 768 c. in. 5. 100. 6. 200. 7. 2261. 8. 1001. 9. 192. 10. 135. 11. 1 c. yd. 21 ft. 12. 97. 13. 9380 bricks. 14. 23 c. ft. 1461 c. in. 15. 39 c. ft. 1529 c. in. 16. 4 ft. 3 in. 17. 64 yds.

18. 4400 c. yds. 19. 1296. 20. 11776 bales. 21. 240 ft. 22. 144. 23. 378. 24. 18 in.

EXEECISE XLYI.—MENSURATION.

s,    .

1. 58 sq. ft. 63 in. 2. 272 sq. ft. 57 sq. in. 3. 4’12 acres. 4. 297 sq. ft. 10' 8" 3'" ; or, 33 sq. yds. 128^ sq. in. 5. 21 sq. ft. 108£ sq. in. 6. 28 ft. 7. 82^ ft. 8. 96 ft. 3 in. 9. 120 ft. 10. 52 ft. 6 in. 11. 78 ft. 9 in. 12. 26'18 yds.    13. 1681.

14. 946 sq. yds. 8 ft. 19^ in. 15. 44 yds. 16. 1 ro. 10 2656 per.

17. 28 sq. yds. 7T7T sq. ft.    18. 4 miles 920 yds. 19. 796 yds.

20. 14 miles. 21. 240 miles. 22. 480 yds. 23. 19 ft.; 40 sq. yds. 1 sq. ft. 24. 1042 yds. 25. £3 3s. 7-j^d. 26. 706-86. 27. 31 c. yds. 26'94 c. ft. 28. 303009259259-259.    29. 88 ft. 30. 48 ft.

31. 12541 "814 etc. gals. 32. 15 ch. ; 35 ch. 33. £50 6s. 3d; 179 chains. 34. (a) 1000 acres; (b)f25 ac. 1 ro. 31 per. 15-5364sq. yds.; (c) 29-710678 etc. chains. 35. 10|||; or, 10-8580246913.    36. 400 gals.

37. 105 c. ft. 38.    550 gals. 39. 21 2 etc. 40. 2 651 etc. qts.

41. 542-5805 c. in. 42. 2'773 etc. gals. 43. 589Q| gals. 44. 412 335.

45. 15 ft. 6 in. ; 132 sheets. 46. 441 c. in. 47. 131-99083 c. ft. 4S. 40960 acres. 49. 4860. 50. 5 ft.

EXERCISE XLVII.—THE UNITARY METHOD.

1. U days. 2. 12.    3. 9|| days. 4. £550 ; . £82 14s. 2d.

5. 11s. llid. 6. £126. 7. 98 lbs. 8. 4s. 6d. ; 45. 9. £1 4s. 8|d. ; £1    10 A 8s. 3d., B 4s. 7d., C 5s. 6d.    11. £299. 12. £2115.

13.’ £11 18s. 14. £142; £85S. 15. £300. 16. £48 9s. 17. 92 lbs. 4 ozs 18. 25 per cent. 19. £7 Os. 6d.    20. 8 days. 21. He loses

Ss 4d. 22. £712. 23. £2680.    24. £29 6s. 8d.    25. 33J hours.

26. £120 7s. 6d. 27. £850. 28. £316 7s. 9d. 29. £43; £129; £215. 30 6s 9d. 31. 2£ miles. 32. £250 5s. 33. 9| days. 34. 9 per cent ’35. A by 56f yds. 36. £200.    37. Tea 2s. 8d., coffee Is. 8d.

38 6 hrs    39. Is. 9d. 40. 2S8 acres. 41. 19 miles. 42. 14 yds.

0 qr 1 nl 43. £385 5s. 44. £1843. 45. £9 17s. 6d. 46. 10 cows, 10 mss 80 sheep ; or, 12 cows, 2 pigs, 86 sheep ; or, 5 cows, 30 pigs, andl kheep. & 10.. 64- 48. 10. «.«007«. 50. Hind 119 lbs., fore 133 lbs. 51. A 12s, B 9s., C 7s. 6d.    52. Buying prices, tea

Is. 10^d., sugar 3|d. ; selling prices, tea 2s 6d., sugar 3d. 53 £21 3s ; gain by false measure 11s. 3d.    54. A lbs., ±> ys.,

0 7s. 6d., D £1 5s. 6d.    55. £9 12s.    56. 192 men. 57. 14s.

58. £5 12s. 6d.    59. 27 days. 60. 35 min. 61. A 198, B 165,

C 132, and D 110 days. 62. ll^V days. 63. 21f days. 64. 34f days. 6o. As 12, 15, 20 ; 80 days. 66. 3 hours. 67. 42.

EXERCISE XLYIII.—STOCKS.

1. £3562 10s. 2. £1224. 3. £4421 10s. 4. £1127. 5. £2796.

6. £1523 12s. 6d.    7. £800.    8. £550.    9. £600.    10. £990.

11. £750.    12. £975. 13. £52 10s.    14. £69.    15. £83 6s. 6d.

16. £46 7s.    17. £160 4s. 6d.    18. £46 2s. 6d.    19. £1728.

20. £465. 21. £378. 22. £646 6s. 8d. 23. £435. 24. £687 10s.

25. £1720 10s.    26. £1653.    27.    £9S00.    28. £932 10s.

29. £726 14s.    30. £645 14s. 31. £451 5s.    32. £158 6s. 8d.

33. £76 13s. 34. £14 8s. 35. £805. 36. £713 8s. 9d. 37. £750.

38. £9162 9s.    39. £112 10s.    40. £5940.    41.    £11280.

42. £8463 14s. 43. £390. 44. £700. 45. £101. 46. £5539 4s. 44d. 47. £1420.    48. £421 19s. 2d.    49. 44-f p. c. 50. £4200.

51. £1131 18s. 52. £8 17s. 6d. 53. £4 Is". 8d.    54. £1199 18s.

55. He loses 33s.    56. (a) £2970000, (b) £135000, (c) 4T6Tp. c.,

(d) £3965.    57. £26 5s.    58. At 99 ; £46035 ; he loses £97 13s.

59. The Bank stock ; £80 15s.    60. £40 ; £20.

MISCELLANEOUS EXERCISE XLIX.

1. 4s. 6d. 2. £15 18s. 9d. ; £3 3s.. 9d. 3. 12 days. 4. 8 ac. 3 ro. 27-9616 per. 5. 124 p. c. 6. 564 p. c. 7. 8f. 8. 9s.    9. 39s.

10. 6 ft. 11. 22 pieces. 12. Increased by Tfy. 13. 1400 yds.

14. 8 ft. 15. 45 ft. 16. 26400.    17. 139 7 yds. 18. 36 ft.

19. 30720 rails. 20. 2 sheep to the acre. 21. £45 12s. 6d.

22. 1000 times.    23. ^.    24. 5 times. 25. 2s. 8d. ; 2s. 6^d.

26. 2s. 27. 210 lbs. 28. 140 lbs. 29. £468. 30. £299. 31. £210.

32. £1575.    33. £62 10s.    34. Horse £24, cow £19.    35. 4s.

36. 12s. 37. 15 p. c. 38. Iff days ; as 10, 12, 15.    39. £12 10s.

40. 80 loaves. 41. 10s. 42. 8^d. 43. 12. 44. 12s. 7d. 45 £297

46. £23 6s. Id. 47. £12 10s. lOd. ; 1505 lbs. 48. 5s. 49. 20 p. c! 50. 13s. 4d.    51. 12 lbs. 52. 240 bus. 53. 640 bus. 54. £400.

55. £6960. 56. 74 p. c. ; 2^ p. c. 57. I give him £5 ; he returns me 23 dollars 58. A £72 y B £101f.    59. 3 gals., 3 gals., 7 gals

60.    £1 8s. 61.    21 boys.    62.    87 ft.    9 in. 63.    40, 24, 20,    8,    and 28

64.    £452 Ils. 4g7ÿd. 65.    7100374854743. 66.    £1000. 67.    20 p.c!

68.    12J P- c.    69. 54    sec.    70.    440 feet.    71. As    13    to 12.

72.    14 gals. _    73. 110    yds.    74.    2i sec.    75. £264    15s. 8fd.

76. I give him £3 ; he returns me 14 dollars. 77. 100 days.

78. 60 ozs. 79. 8 hours. 80. 12 hours. 81. £220. 82. £9 9s. 54d! 83. £3 3s. 84. 31 miles; 20 miles. 85. 165 miles. 86. £6 11s. 3d!

87. 9d.    88. 64 lbs. 89. 3 gals. 3 qts. ; 12 gals. 90. £48 15s '

£2 8s. 9d. 91. At 6 o’clock. 92. A £158 Ils. 2d. ; B £108 18s;'

50 per cent. 93. 26J min.; 4 miles 3 fnr. 94. 1500 yds. 95. As 8 t 7; as 9 I 8. 96. 109 gold to 36 silver. 97. 10£ p. c. 98. 15f miles. 99. 4^d.    100. £450. 101. £1632.    102. 360 miles. 103. 6£Hf ac.

104. £1/ 9s. lid. 105. 2s. 7^d.    106. 120 sheep, 24 calves, and

6 turkeys. 107. (a) 121 ft., (b) 2£ sec. 108. 112 ft. ; 256 ft. 109. Nothing. 110. 118f^; £388.    111. 10s. 6d. 112. 12 min. and

8 min.    113.    305.    114. £5    5s.    115.    4J p. c.    116. 12£    p. c.

117. £20    3s.    7^d.    ^q. 118.    6s.    3d.    119. 9 min. and 7    min.

120. 21, 15, 12.    121. 5760 ; 80.    122. 16£ ft. 123. 160.    124. 640.

125. 35 miles ; 20 miles. 126.    40 p. c.    127. 6000. 128. 49 ft.

j£9. £280.    130.    12s. 6d. 131.    10* miles.    132. 11 yds. 2    ft. 2 in.

133.    4 mile. 134.    £90.    135. 20    miles.    136.    Is.    7d.    137.    974.

138.    4 p. c. loss.    139.    20 min.    140.    9d. ;    6d.    141. 20    p. c.

142. 10 p. c. 143. £10000.    144. 6 miles. 145.    07i ; 1687.

146. 40.    147. 320    ac. 148. 6    p. c.    149.    15 p. c.    150. 12 months.

151. 72 lbs.    152.    (a) 25 ; (6)    14 ;    40 ; 46.    153.    2 hrs. 24    min.

154. 991; -00108.    155.1302 yds. 156. 1240 grains. 157. 76 to 125;

or, 3-^5 gals. 158. 15s.    159. In 2 hrs. 5 min. ; 41§ miles from Mel

bourne. 160. 63| ft. from lower, 56£ ft. from higher. 161. Hind-quarter 288 lbs., fore-quarter 216 lbs. 162. Mail train 26| miles, goods train 18f miles. 163. 15 tons. 164. 17 cows, 43 store cattle, 40 calves.    165.    12 months.    166. At    90. 167. 970Jj-    c. yds.

168. 972 c.    yds.    169. £16 13s.    4d.    170.    9£p. c. 171. £2    17s. 4d.

172.    2 (204 + 20) x    10i    x    f = 369 ft.    = 123 yds.    173.    B    by 44    yds.

174.    16f p. c. 175.    As    1    to 1.

ANSWERS TO EXAMINATION PAPERS.

CERTIFICATE EXAMINATION, 1877.—SET I.

1. See page 8.    2. 36787666500.    3. £3430; £4573 6s. 8d.;

£5716 13s. 4d. 4. £2141 12s. 2ffd. 5. 45. 6. £2 10s. 7. 225.

9. 3|yrs. 10. 10£ days; A 5fac., B 6£ac. 11. 4. 12. or

; (a) -000001817 ; (6) 2-375.

CERTIFICATE EXAMINATION, 1877.—SET II.

1. Seepage 119 and Note on page 120. 2. 570090.    3. £13 9s. llfd.

4. 11s. 7-fd. 5. £155 7s. 9|d. 6. 28. 7. By 11^ 0zs. 9. £91 13s. 4d.

10.    1 hr. 3Mimin. 11. 8|.    12. (a) 34T2 = 34T234, and

x    = 123; (b) TV

PUPIL TEACHERS’ EXAMINATION, 1877.—FIRST CLASS.

Set I.—1. £13 5s. 2-^d. 2. 2. 3. 5 per cent.

Set IL—1. £112-608, or £112 12s. lffd. 2.    3. 2\ yrs.

PUPIL TEACHERS’ EXAMINATION, 1877.-SECOND CLASS.

Set. L—1. £172 6s. 9/^d.    2. £2500.    3. ff|£.

Set. IL—1. £296 Os. ll£d. 2. £3. 3. 5.

PUPIL TEACHERS’ EXAMINATION, 1877.—THIRD CLASS.

Set L—1. £186 Us. 2|£d. 2. 17s. 6d.

Set IL—1. £273 14s. 8^d. 2. £1136 16s.

TRAINING EXAMINATION.—JUNE, 1878.

1. (a) 3875255019 ; (b) That they have no common factor. 2. 40 pages. 3. £485 18s. 6èd. 4. 92 ft. 5 in. 5. 2hrs. 48 min. 45 sec.

6. 63 miles 720 yards per hour. 8. 12| p. c/^ 9. *21.    10. Total

profit, £2754 ; A’s share, £3 16s. 6d.    11. (a) *49846153 :

(b) -01953125.

EXHIBITION EXAMINATION.—DECEMBER, 1878.

1. £432 12s. 3£d. 2. 12775. 3. 30 miles ; 15 miles. 4. £455 8s.

5. 4s. 2d.    6. 33 lbs. 7. (a) ff6; (b) Complex, A. ; compound, f of

8. 1 ac. 2 ro. 24 per. 27J yards.

CERTIFICATE EXAMINATION, 1878.—SET I.

1. (a) See Local and Intrinsic Values, page 3; (6) 23109216. 2. 9096144| yds. 3. 22 tons 14 cwt. 10§ lbs. ; 30 ik>ns 5 cwt. 1 qr. 23£ lbs. ; 37 tons 16 cwt. 3 qrs. 8§ lbs. 4. £2674 18s. 61d. 5. 50 p. c. 6. 13f days. 8. £3932 6s. 3d.    9. (a) Because £x§=§;

(b) 3 hrs. 36 min. 10. 7 ac. 0 ro. 37 per. (fraction

11. (a) £2 2s. 5id. ; (6) Sum -02626 ; diff. -02574 ; prod. -00000676 : quot. -000039996.

CERTIFICATE EXAMINATION, 1878.—SET II.

1. (a) Quantity denotes how much ; number, how many ; (b) 28136579 2. 987712194. 3. 16 mi. 41 ch. 20 links ; 19 mi. 65 ch. 44 links • 23 mi. 9 ch. 68 links. 4. £10989 Is. 7Hd- (or £21978 3s. 2ffd. if double line of fence at £184 16s. 8d. per mile of fencing). 5. 2s. 54d. • 95 per cent. 6. 37^ days. 8. £3103 6s. 8d.    9. 17f-7- days : A does

9ff ac-> B 10R ac. *K). 1 cwt. 2 qrs. 16 lbs. (r\ x 0f 3300 lbs.) 11. (a) See page 178 ; (b) £170 2s. 8d ; (c) Sum -00836, diff. -00684 prod. -000005776, quot. '0000016.

PUPIL TEACHERS’ EXAMINATION, 1878.-FIRST CLASS.

Set. I.—1. 4^ ft., or 4-54 ft. 2. 2f. 3. 3§. 4. *076923;    ,

Set. II.—1. 4-jSj- ft., or 4'54 ft. 2. 1£. 3. 3§ yrs. 4. '125;

PUPIL TEACHERS’ EXAMINATION, 1878.—SECOND CLASS.

Set. I.—1. £8 14s. 2d.    2. 5 0784313725490196.    3. 12 days.

4. 8 03 ; -0359375.

Set II.—1. £2355 16s. 8d. 2.5 -0784313725490196. 3. £5 6s. 3d. 4. 21.

PUPIL TEACHERS’ EXAMINATION, 1878.—THIRD CLASS.

Set I.—1. £327 3s. 1-flfed. 2. 5 lbs. 9 ozs.

Set II.—1. £264 7s. 7rfftd. 2. 3 ozs. 2 dwts. 17^ grs.

Note 1.—In Sum 2, Certificate Examination, 1878, Set II., 987712181ff-|- min. is the exact answer, allowing a proportionate part of the extra 5 hrs. 48 min. for the year 1878 to belong to the= period 9.30 a.m. Dec. 19th to 12 p.m. Dec. 31st, 1878.

Note 2.The following admit of several answers : —

Ex. 47, Sum 46.—As 2 cows, 42 pigs, 56 sheep.

Ex. 49, Sum 59.—As 5, 1 and 5 ; 2, 1 and 4 ; 1, 2 and 3 ; 2* 7 and 8, &c., &c.

Ex 49, Sum 106.—As 125 sheep, 10 calves, 15 turkeys.

1

If we here first subtract the lower number from the equivalent of the unit changed into a lower name and then add the result to the upper number, the only difference between the two methods described will be that the unit changed to its equivalent in a lower name is in the 1st Method an addition to the minuend, while in the 2nd Method it is a removal from a higher part of the minuend. In other words, the 1st Method depends upon the principle that equal additions to minuend and subtrahend do not alter their difference ; while the 2nd Method is worked by changing part of the compound quantity forming the minuend into a lower name.

2

Here be careful to divide ff- by f before multiply by

3

Be careful to express the similar quantities obtaining the product of each side.

4

Simultaneous equations are those which show different values for the same unknown quantities. See Example II., next page.

5

—l.£3190 7s.5fd. 2. £3442 ls. 3. £7943 2s. 7R 4. £3395 5s.2d.

5. £763 8s. 10|d.    6. £1149 15s. 3d.    7. £1313 10s. 5id.

6

The reason of this is, that the square of any number is equal to the product of the squares of its factors. Now, all even numbers contain the factor 2, which when squared becomes 4. So that the square of an even number contains 4. Again, an odd number is an even number + unity, and the squure of a number consisting of two parts the square of each part plus twice the product of these parts. (See Ax. ix., page 37 ) So that, if an odd number be squared, it is equal to the square of an even number plus unity. And this is the square of the even number + (twice the product of the even number xl)+the square of 1 ; or (the even number squared + twice the even number + 1) ; and since the square of an even number contains 4, and twice any even number contains 4, it follows that we have only to subtract 1 from the square of an

odd number to leave a remainder exactly divisible bv 4. Thus ll2 — finj-n2 _

100 -+6 20 -f- 1 = 121, and (121 — 1) is divisible by 4.    ’    '

7

This, it has been said, is only a shorter plan of obtaining triple the square of the part of the root already found.

8

   Find, by Cross Multiplication, the number of cubic feet and inches in a cube, whose edge is 3 ft. 5 in.

9

The number 3-14159...expresses the number of times the circumference of a circle contains the diameter. The exact ratio of the circumference to the diameter cannot be expressed; but it has been found to be 3-1416 nearly.

f It is thus seen that the area of a sphere is equal to the convex surface of its CIRCUMSCRIBING cylinder ; that is. of a cylinder whose height and diameter are equal to the diameter of the sphere.

10

Note.—It will be observed that the solidity of a cone, sphere, and cylinder of equal diameters and height are to each other as the numbers 1, 2, and 3.

t As a heap of road-metal, a trough, or a water-dam having a rectangular top and bottom, and similar slanting sides.

11

NOTE.—It is only in newly settled countries that the sale of land can be regarded as an extensive source of revenue.

12

It will be well to remember that £100 stock simply means £100 of the loan or debt.

13

The Victorian new 4J per cent, loan has just been successfully floated on the London market at a little above £99 for each debenture of £100 stock.