ARRANGED TO SUIT THE REQUIREMENTS OF
ISSUED BY
BY
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.
* |
« • |
PAGE 1 | |
Definitions * • * |
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. |
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 |
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
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 - - -
Notation and Numeration of 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 - - -
Exercise on Proportional Parts - - -
The Reciprocal of Numbers ....
Exercise on Profit and Loss ....
Miscellaneous Questions, from Teachers’ Examination Papers
ANSWERS .....
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, y^{5}.
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 _{T}^{s}r 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, _{T}y 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, 10^{3} - 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.
10^{2}, 10^{3}, 10^{4}, (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 6 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.
376495 1234567
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.
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.
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.
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.—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
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.
Division by any number less than 13 is called Short Division.
Short Division is worked as under :—
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 ?
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) = 8^{2} - 3^{2} = 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} = 8^{2} + 3^{2} + (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 + 2^{2} = 10^{2} = 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 8^{2} 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 100^{2} - 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 1 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 1 What number must be added to this latter number that the above difference may be subtracted just once more^{1}?
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 number^{1}?
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 they^{1}?
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 7 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.
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.
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.
N_{0TE-}—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.
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
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.
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 |
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.
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 subtract^{1} 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.
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. |
^{4}pks. |
^{2}gall. |
^{4}qts. |
^{2}pt. |
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 ^{20}cwt. ^{4}qrs. “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. ^{20}cwt. ^{4}qrs. |
“lbs. tons. ^{20}cwt. ^{4}qrs. |
“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
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.
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.
20 6
46
12
54
4
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.
£ 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 |
Sì |
£ 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 |
7Î | |||
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 |
5Ì | |||
8252 12 |
8§ |
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.
tons cwt qrs. lbs.
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.
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.
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 |
6« | |||||
£ |
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. |
27^{u} |
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.
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.
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 |
^{b}y |
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 |
5è |
■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.
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.
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 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.
2 scr. 18 grs. to grains.
£^{20} s.^{12} d. 325 18 9f 20
6518 shillings.
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.
tons^{20} cwt. 182 15
20
qrs.
3
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.
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.
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
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
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
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.
14 lbs. ~ I of 2 qrs. 28s. cost of 0 2 0
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. | |
H« 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 |
^{2}i — 2 4 = »» |
10 |
0 | |
Is. 3d. = |
i |
43 |
9 |
3-j^{9}fT - 8 4 — 9 9 |
2 |
6 | |
5d. = |
4 |
21 |
14 |
7M-78= „ |
1 |
3 | |
7 |
4 |
10^=87= » |
5 |
£2332 12 3_{t}V = 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.
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
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.
¿ 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
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
44. Find the weight, in tons, of 798 fodders of lead, a fodder being
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 relation_{t} to each other as two other things, the four are said to be in
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. 5_{T}^{5}T min.
The hands will therefore be together every 1 hr. 5p_{T} min., and opposite, at £ of 1 hr. 5p_{T} min. ; or, 32_{T}\ min. past 12, and every 1 hr. 5p_{T} 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 32_{T}^{S}T 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 5f_{T} min. past 1 5 2hrs. 10p^, or 10-^ min. past 2 ; 3 hrs. 16_{T}^{4}T min., or 16^ min. past 3; 4 hrs. 21_{t}^{9}t min., or 21_{T}^{9}T min. past 4 ; &c., &e.
2. Opposite at 32_{T}^{S}T 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 a 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
As 40 : 15 :: 16 days. Working.
„ 22:35 16x 15x35x70x5 Answer.
„ 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 14k persons 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
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.
1. Bring eacli pair of terms to the same name.
2. Cancel (where possible) the 1st terms with the
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» x (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 _{i }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?
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.
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.
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.
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
16,803920
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.
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 p_{f} 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.
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 a 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 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.
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.—^{1}THE 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, 2016_{v}& 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 all, the 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.
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.
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 _{T}7J> 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, _{x}^{e}y, ^, and _{T}^{7}¥ 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 _{T}^{8}^- 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
24 24'
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; A« 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
----- 60 — n
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 by “ of” 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.
Example II.—Express 13 cwt. 3 qrs. as the fraction of 1 ton.
13 cwt. 3qrs. 1540 lbs. 1540 11
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.
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---{ ^ -_{T}v }•.
2. Take out of outer brackets by doing the work within.
2
if " n ^ + ff ^{2}f- 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.
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^^{3}0. Answer - _{T}uu^{2} .
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^iU^{the}ii^{divisi0n }hA ^{continued with} this remainder 66, the ngures 8461.,3 will again and again occur in the quotient. Such an
andT82^{1S & ClrCulatin}S ^{or re}P^{eati}ng 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.
In these examples one figure continually repeats. The decinlal is then called a Pure Repeater.
i = 6)5-00000 . 12)7-000000 .
-or -83; _{X}V = - 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. ^of_{T}\ofifof_{T}«éV IS.HoílO^oí^olUh 14. 31 of*of 14ofiff. 15. 8_{T}\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. t^{8}t°*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 •
5 6 3 12 7*
515 Q 3147 115' 5 17*
7 8 7 4 3 13 1*
5 5 4 4 19 5*
V.—Reduce the following to the lowest common denomination ; then add the numerators together, and reduce the results to mixed numbers:—
7 5 5 7 IQ & 31.
9 9» 99» 11» ^ ^{u}9*
2 9» 3 1» 3 7»
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 ’ 5» |
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»
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-