5. Compound Multiplication.

Compound Multiplication is the method of finding the amount of a given number, consisting of different denominations, by repeating it a proposed number of times.

Rule.*

Write the multiplier under the lowest denomination of the multiplicand. Multiply the several denominations successively by the multiplier, setting down the excess and carrying from each denomination to the next higher, as in Compound Addition.

Proof.

The method of proof is the same as in Simple Multiplication.

Examples.

1. What will 5lb. of tea cost at 1 dol. 2 dimes, 7 cts. per pound ?

$ d. cts. or 1 2 7

5

$cts. or cts.

1 27

5

6 35

127

5

635 cts.

By this example it will be seen that the operations by this rule in Federal Money are precisely the same as in Simple Multiplication, 635 one dollar, 2 dimes, 7 cts. being just equal to 1 dol. 27 cts. or to 127 cts. Hence the 127 cents multiplied by 5 the answer is 635 cts.=6 dols. 35 cts.=6 dols. 3 dimes, 5 cts. and the given numbers may in all cases be expressed as a simple number in the lowest denomination mentioned, or as a compound number.

2. What is the cost of 6 lb. of tobacco, at 2s. 6d. 2qrs. per lb?

s. d. qrs.
262

6

15s. 3d. 0

Here 6 times 2 is 12, but 12qrs. are equal to 3d. therefore set down 0 and carry 3. Then 6 times 6 is 36 and 3 to carry is 39d.=3s. 3d. set down 3d. and say 6 times 2 is 12 and 3 to carry is 15s. which set down.

7. What will 5lb. of loaf sugar cost, at 1s. 3d. per pound? Ans. 6s. 3d.

8. What will 8 bushels of corn cost, at 5d. 7cts. or 57cts. per bushel ?

Ans. 84 5d. 6 cts. or $4 56cts. 9. What will 9 yards of cloth cost at 5s. 4d. per yard? Ans. £2 8s. 10. What will 12 gallons of brandy cost, at 9s. 6d. per gal. P Ans. £5 14s.

*The product of a number consisting of different denominations by a simple number, is evidently expressed by the several products of the different parts multiplied by the simple number. Thus, £2 6s. 4d. multiplied by 6, the seve ral products will be £12 36s. 24d. (by taking the shillings from the pence and the pounds from the shillings and placing them in the shillings and pounds respectively) to £13 18s. Od. which is agreeable to rule; and the same will be true when the multiplicand is any compound number whatever.

When the multiplier exceeds 12, and is a composite number, the component parts may be employed successively, as in Simple Multiplication, instead of multiplying by the whole number at once.

2. When the multiplier cannot be produced by the multiplication of two small numbers, take two such numbers as come the nearest to it, and then find the value of the odd parts and add or subtract as the case requires.

3. When the multiplier exceeds 100, find the cost of 100, multiply it by the number of hundreds, and to this product add the cost of the odd parts and their sum will be the answer required.

Duodecimals.

DUODECIMALS are so called because the denominations decrease by 12 from the place of feet towards the right hand, as in the follow

Write the several terms of the multiplier under the corresponding terms of the multiplicand; then multiply the whole multiplicand by the several terms of the multiplier successively, beginning at the right hand, and placing the first term of each of the partial products under its respective multiplier, remembering to carry one for every 12 from a lower to the next higher denomination, and the sum of these partial products will be the answer, the left hand term being feet, and those towards the right primes, seconds, &c.

This is a very useful rule in measuring wood, boards, &c. and for artificers in finding the contents of their work.

*The rule may be expressed in general terms thus. When feet are concerned, the product is of the same denomination as the term multiplying the feet; and when feet are not concerned, the name of the product will be expressed by the sum of the indices of the two factors, or of the strokes over them. Thus 4′ x 2"-8". Here one of the factors is inches, the other seconds, and the indices or strokes over them amount to 3, hence the product, 8, is thirds. And in the same manner 8"x3"-24"" or, divide by 12,-2" The reason of the rule may be shown by the first example. The 4' are 4 twelfths of a foot and the 8' are 8 twelfths of a foot, and ÷×&=44 or 4 of or 32", which reduced gives 2′ 8′′; putting down the 8" we reserve the 2' to be added of 10ft. by 8', or which product is, to which 2 being added, we have or 6ft. 10'. Next multiplying 4' or by 7 we have 2 or 2ft. 4', which added to the product of 10 by 7 gives 72ft. 4', and these results added together give 79ft 0′ 8′′ for the product. The same reasoning may be extended to cases where there is a greater number of denominations.

8

TT

6ft. 7' long, Sft. 5' high, and 3ft. 8 feet, 4 inches long, 3 feet 9 in

QUESTIONS.

6. Compound Division.

COMPOUND DIVISION is the method of finding how often one number is contained in another of different denominations.

Rule.*

Place the numbers as in Simple Division, and divide the several denominations of the dividend successively by the divisor.

If there be a remainder after dividing any denomination, it must be reduced to the next lower, adding the number in the lower denomination. Divide the sum as usual; and so on till the whole is finished.

Proof.

The method of proof is the same as in Simple Division.

4. If 35 yards of cloth cost £57 5s. 5d. what is it per yard? £ s. d. £ s. d. qrs. 35 ( 57 5 5 ( 1 12 8 2 Ans.

35

and then the op

22

20

the same as in

Simple Division.

35) 445 (12s.

35

-

95
70

25

12

35) 305 ( 8d.

280

25

4

35) 100 (2qrs.
70

30

In 57 I find 35 once and 22 over. I then reduce 22 to shillings, adding the 5s. and in the sum 445s. I find 35 12 times and 25 over. I then reduce 25 to pence, adding 5d. and in the sum 305 I find 35 8 times and 25 over. Again, I reduce 25 to farthings, and divide by 35, and the quotient is 2qrs. and 30 remains, which is = of another farthing.

*The division of numbers of different denominations, or compound numbers, depends upon the same principles as Simple Division. This must be sufficiently obvious, when each of the several parts of the dividend can be divided without a remainder. And when there are remainders, the truth of the rule will