COMPOUND DIVISION.

§ 158. Compound Division consists in resolving a compound number into several equal parts.

1. Divide £17 11s. 4d. by 8.

£ S. d.

PERFORMED. 8) 17 11 4

2 3 11

EXPLANATION. £17-8-2 and 1 remainder. This £1 remainder+11s.=31s., and 31s.÷8=3 and 7s. remainder, and 7s.+4d.=88d., and 88÷÷8=11d.

The same

It is obvious that the remainder of each division is of the same denomination as the number divided. is true of the quotient figures.

RULE.

Commence with the highest denomination of the dividend, and if, after dividing, there be a remainder, reduce it to the next lower denomination, and, adding whatever be given of the lower denomination, divide again ; and so continue to do through all the denominations.

may

If the divisor is a composite number, the division may be performed by factors, if preferred.

158. In what does compound division consist? With what denomination do we commence in dividing? What is done with each remainder? How divide if the divisor is composite?

REDUCTION.

§ 159. Reduction consists in changing the denominations of numbers without altering their value.

Since 1 penny 4 farthings, it is perfectly obvious that any value expressed by pence, is also expressed by a number of farthings, four times as great as the pence.

1. How many farthings are equal to 3 pence? Ans. 3× 4 12 farthings. How many farthings are equal to 7 pence? Any value in pence and farthings is expressed in farthings by reducing the pence to farthings and adding the given farthings.

2. How many farthings are there in 5 pence, 2 farthings? 5d.=20qr., and 20gr.+2qr.=22qr., Ans. How many farthings in 9d. 3qr.? In 8d. 1gr.? In 11d. 1qr.?

3. How many pence in 3s.? Since 1s. 12d., 3s.=3 times 12d. 36d., Ans. How many pence in 5s.? in 7s.? 4. How many pence are there in 4s. 3d.? In 4s. there are 48d., and 48+3=51d., Ans.

To find the pence in a given number of shillings and pence, multiply the shillings by 12, and add the given pence.

5. How many pence in 5s. 8d.? In 4s. 7d.? In 6s. 6d.? 6. How many shillings in £3? In £3 6s.? In £2 9s.?

§ 160. To obtain the value of any denomination in terms of higher value, divide by the number required of the given denomination to make one of the denomination of higher value.

1. In 16 farthings, how many pence?

It requires 4 farthings to make 1 penny; therefore, 16qr.÷4=4d., Ans. How many pence in 12 farthings? In 20 farthings? 2. In 36 pence how many shillings?

12d. 1s.; there

fore, 36d.12=3s. In 48
In 48 pence how many shillings?

159. In what does reduction consist? How is the value of pence and farthings expressed in farthings? How do we find the pence in any number of shillings and pence?

160. How do we obtain the value of any denomination in higher terms? Of what denomination is the remainder ?

If in dividing there be a remainder, it will be of the same kind as the number divided.

3. In 30 pence how many shillings? 30d.+12=2s. and 6d. over, Ans.

4. In 40 pence how many shillings and pence? In 56d.? 5. In 37s. how many pounds? 37s.÷20=£1 17s., Ans. In 43s. how many pounds? In 73s.? In 87s.? In 93s.?

§ 161. From the preceding examples and remarks, it is obvious that there are two processes of reduction, viz.: One, by which the value of a higher denomination is fully expressed in terms of a lower value; and a second, by which a number is found expressing a given value in terms higher than those of the given denomination.

The former of these processes, viz.: reducing higher denominations to lower, is called, Reduction Descending; while the latter, which consists in bringing low denominations to those of a higher value, is called Reduction Ascending.

GENERAL RULES.

1. To reduce high denominations to those that are lower: multiply the highest of the given denominations by that number which expresses the units required of the denomination next lower to make one of the higher, and to the product, add what is given of the lower denomination. Multiply the number thus obtained by the number requisite to bring it to the next lower denomination, and add whatever is given of that lower denomination; and so continue to do till the required denomination is obtained.

2. To reduce low denominations to those that are higher : divide the given denomination by the number required of this denomination to make one of the next higher; divide the number thus obtained, in the same manner; and after each division, carefully observe the remainder.

161. How many processes of reduction? How reduce high denominations to their value in lower denominations? What is done when any thing of a lower denomination is given? How proceed? How reduce low denominations to high? Of what denomination will the quotient be in each instance ?

The quotient will, in each case, be a higher denomination, and the remainder, if any, will be of the same name as the number divided. This process continued till the required denomination is obtained, will determine the

answer.

The former of these rules relates entirely to bringing high denominations into low; and the latter, to bringing low denominations into high. By neither operation is the given value affected.

§ 162. PROOF. The work in each case is proved by reversing the operation.

1. Reduce £16 to shillings, pence, and farthings.

A higher denomination is to be reduced to a lower one. We therefore multiply.

OPERATION.

£16
20

32 0s.; since £1=20s., therefore £16=320s.
12

3 8 4 0d.; since 1s.=12d.; 320s.=3840d.

4

1536 0qr.; since 1d.=4gr.; 3840d.=15360gr.

Hence, £16-320s.=3840d.=15360gr. Each of these denominations are of the same value, and differ only in the number of units required of each to express that value. 2. In 15360 farthings, how many pounds?

OPERATION.

4) 15 3 6 0gr.

12) 3 8 4 Od.

20 ) 3 2 0s.
16£.

Here we bring the value given in the previous sum, but expressed in a lower denomination, back to its original denomination by division.

What is the difference between the two rules given? Is the given value changed by either operation?

162. How prove the work?

3. Reduce £15 12s. 8d. 3gr. to farthings.

1 50 1 1=farthings in 3752d. and 3qr., and is the re

quired answer.

We here reduce the £15 to shillings, and add the 12s. given; we then reduce the shillings to pence and add the given pence; and, finally, we reduce the pence to farthings, and add the given farthings. It will be observed that as we reduce the higher denominations to lower, we, in each instance, add whatever is given of the lower denomination.

4. Reduce 17685 farthings to pence.

Ans. £18 8s. 5d. 1qr.

5. Reduce £27 15s. 8d. to pence and farthings.

Ans. 6668d., 26672qr. 6. In 26672qr. how many pounds, shillings, and pence? Ans. £27 15s. 8d.

Observe first, with what denomination the reduction commences, and whether it is required to bring high denominations to those of lower value, or low denominations to those of higher value.

Observe, also, that the divisor or multiplier required in each instance is determined by the number of units required of the given denomination to make one of the next higher or lower denomination.

7. Reduce 17 cwt. 3 qr. 10 lb. 7 oz. to ounces.

Ans. 28567 oz.

8. Reduce 36 cwt. 2 qr. to pounds and ounces.

Ans. 3650 lb., 58400 oz.

What is to be observed? How is the divisor or multiplier to be determined?