EXAMPLES.

Lent £275 at 6 per cent. interest, and received £5 15s., what was the time?

Lent £45 16s. 8d. at 8 per cent., and received £5 10s., what was the time? Answer, 11⁄2 years.

Lent £116 8s. at 5 per cent., and received £2 8s. 6d., what was the time? Answer, 5 months

Given the principal, interest, and time, to find the rate per cent.

Rule. Multiply the shillings in the gain by 60 as before; this divided by the product of the principal and time will be the rate per cent.

EXAMPLE.

Lent £116 8s. for 5 months, and received £2 8s. 6d., what was the rate per cent.?

Lent £45 16s. 8d. (45%) for 1 years, and received £5 10s., what was the rate per cent. ?

Answer, 8 per cent.

OF DIVIDING ONE FRACTION BY ANOTHER.

After the fractions are prepared, it may not be amiss (instead of inverting the divisor) to make the denominators of divisor and dividend alike by multiplying or dividing their terms; the numerator of the dividend will be the numerator of the quotient; and the numerator of the divisor will be the denominator of the quotient.

This is the same as divided by +, and therefore the quotient is, which is, perhaps, the readiest method.

RULES

FOR DERIVING FROM TWO OR THREE GIVEN

QUANTITIES A THIRD OR FOURTH

THAT IS WANTED.*

The principles of the Rule of Three are very plain and simple, If 6d. be given for 12lb. of any article, Id. will will purchase 2lb. ; thus, 6d. : 12lb.:: 1d.: 2lb; the answer is found by dividing the second term by the first. Again, if 1lb. cost 6d., 121b. will cost 72d., i. e. the answer will be the product of the second and third; and, consequently, if the first and third be neither of them unity, the answer will be the product of the second and third divided by the first; 7:54; upon the same grounds it is plain that 3X7:3X5: 4×3: X3, i. e. the proportion will not be altered, if the terms are trebled; nor would it if they were increased ten fold, fifteen fold, or any number of times...

It follows too, that the product of the first and last will be equal to the product of the second and third. And again, if two factors, 5×12, be equal to 6×10, 5 will be to 6:10 12.

The working of the following example will sufficiently illustrate what I would recommend.

NOTE.

* Before a person can become a clever mental calculator, the rules laid down must be well practised in order that the judgment and skill of the arithmetician may thereby be improved. This rule teaches a great deal; and will enable him to think properly and justly. But when he has attained sufficient knowledge by reasoning, he will not always have recourse to its statings. For when either a written or verbal question is put to him, he will quickly know the divisor (if any) from the remaining terms, and will measure either of them by it.

EXAMPLE.

Gave £24 16s. for 16 yards of cloth, what will 24 yards cost at the same rate?

Here it must instantly be seen that 16 is the divisor, and 1, it follows that if to £24 16s., its half, be added, the answer will be £37 4s.

as

EXAMPLE.

If of an ell of Holland cost 3s. 44d., what will 381

ells cost?

Here 38. 44d.=% of a £, ell ells £. £. s. d. and 38 ells='ells, J쿱: 3급 :: : 179101. Also by inverting the divisor we have it thus,

60

By the above it may easily be seen that the answer is 311 times £; now, the ro of a £. is 1d.; consequently, the r of a £. is 1s. 14d; therefore 311 times 1s. 1d. or 1 of £15 11s. (311 shillings) is £17 9s. 10 d.

The reader is advised to adopt generally the above method after cancelling, for by that means he may be led to give his answer mentally. It has this besides to recommend to him, that he will the better understand the value of aliquot and fractional parts of pounds, &c. Again,

If of a yard cost of a £, what will 31% of a yard cost? yard yard

To : 28.3 ::

£.

Here, by inverting the divisor, and cancelling, we have XX, the product of these denominators is 192 ; now the of a £. is 14d., and it must be plain that there are 253 ten times 14d., or, which is the same thing, 253 at 18. Od', or £13 3s. 6d. for the answer; and, by reversing the above, it must also be clear that 10 times £1 6s. 41d. (253 times 1 d.) will give both one and the same answer.

The statings in this rule may always be brought to whole numbers, taking the last for example.

Here, by multiplying the terms of the first fraction by 4, and that of the third by 5, we have their denominators the same, and therefore they may be rejected, and they stand thus, yards £.

yards

36 : 283 : : 15

There being now only one denominator, multiply the 1st term by it, and we have it thus,

yards

288

yards

£. £. s. d. 253 15 : 13 3 6

as before, and thus we have all the terms cleared of fractions.

Here it may be observed that, by the above process, making the denominator of the divisor and one of the other terms the same by multiplication, and then destroying their

denominators, is the same as multiplying those terms by one common number; for instance, in destroying the denominators of the 1st and 3d terms, they are each multiplied by 40; consequently, the terms being muliplied by the same, the correctness of the statings is not altered.

2dly. The remaining term having a denominator, which in its nature is a divisor, being multiplied into the first term, which is also a divisor, the whole is cleared of fractions; and in this manner may fractions always be removed. The above has been introduced merely as a novelty, and the reader, by adopting such methods, may become more convinced of the utility of fractions.

EXAMPLES.

If of a lb. cost 7s. 9d., what will 15 lb. cost? Answer, £15 15s. 2d. If of of a yard cost of of a £, what will of 16 yards cost? Answer, £5 14s. 3 d. Suppose I have by me 300 yards of Cambric, which cost me £100 13s. 4d., but some damage having happened to it, I am willing to lose £17 10s. by the whole, at what rate must I sell it per ell English?

Answer, 6s. 11¿d.

The above question being admirably fitted for mental calculation, it may be proper here to explain the method.

First. From 300 yards deduct its, i. e. 60, to reduce them to ells English, and we have 240, which answers to the number of pence in a £.; therefore, by deducting one sum from the other in the above question, we have £83 3s. 4d., or £83, which being considered, as so many pence, we have 6s. 114d. the answer.

It must be here evident that any measure or multiple of the above 300 yards may be brought to ells English, and the answer found mentally; thus, for instance, allowing the money to be the same, suppose it had been 60 yards.

First, 60 yards=48 ells English, and this 48 is again equal to of the number of pence in a £., consequently, 83+ X5 =415% pence, or £1 14s. 7 d. which is the answer.

AGAIN.

Suppose it had been 900 yards, which is equal to 720 ells English, or 3 times the number of pence in a £, then 83=£27+, this also considered as pence, is 2s. 3 d. for

3

the answer.

Thus, it must be observed, that when a divisor of any

arithmetical operation is a measure or multiple of 240, the same means may be used; the rule is, therefore, well worth the best attention.

When the product of the denominator of the 2d and 3d terms are equal to the denominator of the 1st term, the denominators of the whole of them may be rejected.

EXAMPLES.

If of a lb. cost £, what will of a lb. cost?

Here 8×5=40, the denominator of the 1st term, and we have it thus stated.

If of a yard cost of a £, what will of a yard cost? Answer, £2.

If of a yard cost of a £, what will a yard cost? Answer, £3 6s. 8d Here, by substituting for the half yard, all the denominators may be rejected as before.

How many yards of cloth at 5s. 8d. per yard may I give for 57 yards at 4s. 3d. per yard?

Answer, 43 yards.

This question belongs to inverse proportion; but it is of little moment to what it belongs; common sense tells us that 5s. 8d. is the divisor; therefore we put its equal in pence as a denominator, and 4s. 3d., or its equal in pence, as a numerator; and we have, which shews that by deducting from 57%, its part, we shall have 43 yards, which is the answer.

A right understanding of the rule of three is one of the great principles of arithmetic; it is not only necessary that the reader should understand his rules, but that he should think properly and justly; this, indeed, will enable him to become a good mental calculator.

CONCISE METHODS

FOR WORKING SUMS IN THE RULE OF THREE.

When the first term does not measure either the second or third term, it may by a little observation be readily