itself, is=1000, consequently any number of farthings, increased by their part, will be an exact decimal expression for them: Whence, if the number of farthings he more than 12, 4 part is greater than qr. and, therefore, 1 must be added; and when the number of farthings is more than 36,7 part is greater than 14qr. for which 2 must be added.

EXAMPLES.

1. Find the decimal of 13s. 9ąd. by inspection.

Add

·6..

of 12s.

•691 =decimal required.

2. Find by inspection the decimal 17s. 81d.

CASE IV.

expressions of 18s. 31d. and Ans. £914 and £885.

To find the value of any given decimal in the terms of the integer.

RULE. 1. Multiply the decimal by the number of parts in the next less denomination, and cut off so many places for a remainder, to the right hand as there are places in the given decimal. 2. Multiply the remainder by the next inferior denomination, and cut off a remainder as before

3. Proceed in this manner through all the parts of the integer -and the several denominations, standing on the left hand, make the answer.

185. What is the rule for finding the value of any given decimal in the terms of"

the integer ?

COMPOUND MULTIPLICATION is the multiplying of sums of different denominations; it is useful in finding the value of Goods, &c. And, as in Compound Addition, we carry from the lowest denomination to the next higher, so we begin and carry in Compound Multiplication; one general rule being to multiply the price by the quantity.

NOTE. The product of a number, consisting of several parts or denominations, by any simple number whatever, will be expressed by taking the product of that simple number, and each part by itself, as so many distinct questions: Thus, 331. 15s. 9d. multiplied by 5, will be 1651. 758. 45d. (by taking the shillings from the pence, and the pounds from the shillings, and placing them in the shillings and pounds respectively,) 1681. 18s. 9d. and this will be true when the multiplicand is any compound number whatever.

CASE I.

When the multiplier or quantity does not exceed 12.

Multiply the price of one yard, pound, &c. by the whole quantity or number of yards, pounds, &c.—the product will be the

answer .

186. What is Compound Multiplication ?-187. What is the rule when the muitiplier does not exceed 12 ?

EXAMPLES.

1. What will 5 yards of broadcloth amount to, at 1£. 12s. 44d. a yard ?

£.

S.

d.

12

41

1

£8

5

1 101

In this example, we write down 1£. 12s. 4 d. the price of one yard, and then write 5, the number of yards, under the least denomination. We multiply 2 farthings by 5, and the product is 10 farthings, which we bring into pence by dividing them by 4; we write down the remaining 2 farthings, and reserve the quotient, 2 pence, to be added to the product of the pence. We then multiply 4 pence by 5 and the product is 20 pence, and 2 pence which we reserved are 22 pence, which we bring into shillings by dividing them by 12; we write down the remainder 10 pence, and reserve the quotient, 1 shilling, to be added to the product of the shillings. We then multiply 12 shillings by 5, and the product is 60 shillings, and 1 shilling that we reserved are 61 sbillings, which we bring into pounds by dividing them by 20; we write down the remainder 1 shilling, and reserve the quotient, 3 pounds, to be added to the product of the pounds. We then multiply 1 pound by 5, and the product is 5 pounds, and 3 pounds which we reserved are 8 pounds; this being the highest denomination, we write down the whole amount 8 pounds and find the product or answer to be 8£. 1s. 10d.

2. Multiply £4 13s, 43d, by 10,

3. Multiply £8 15s. 112d. by 11.

4. Multiply £13 12s. 11d. by 7.

5. Multiply £14.17s. 84d, by 9.

Ans. 133£. 19s. 24d.

REMARK.-The facility of reckoning in Federal Money, compared with pounds, shillings, &c. may be seen from the examples given in this and the following cases.

The general rule is

Multiply as in simple multiplication, and from the product point off so many places for cents and mills, as there are places of cents. and mills in the price.

Where the multiplier, that is, the quantity, is above 12. Multiply by two such numbers, as, when multiplied together, will produce the given quantity, or multiplier.

CASE III.

When the quantity is such a number, that no two numbers in the table will produce it exactly.

Multiply by two such numbers as come the nearest to it ; and for the number wanting, multiply the given price of 1 yard by the said number of yards wanting, and add the products together for the answer; but if the product of the two numbers exceed the given quantity, then find the value of the overplus, which subtract from the last product, and the remainder will be the an

NOTE. This may be performed by first finding the value of 48 yards, from which if you subtract the price of 1, the remainder will be the answer as above.

189. How do you proceed when the multiplier or quantity is such a number as no twe numbers in the multiplication table will produce exactly?