Q. Repeat liquid measure.

SUBTRACTION OF COMPOUND NUMBERS.

158. SUBTRACTION of Compound Numbers consists in finding how much one compound number exceeds another. Let it be required to subtract £17 19s. 9d. from £24 3s.

4d.

OPERATION.

£ 8. d.

24 3 4

17 19 9

Ans. 6

3 7

Setting like units under each other, we commence as in subtraction of simple numbers on the right. We cannot take 9 pence from 4 pence, but borrowing 1 shilling from the minuend of the next column, or what is the same, 12 pence, and adding 12 to 4, we have 16 pence; from which if we take 9 pence, we have 7 pence left. Set down 7 under the column of pence. Carrying now 1 shilling to the subtrahend of the next column to make up for the 1 we borrowed from the minuend, makes 20 shillings. 20 from 3 we cannot, but borrowing 1 pound or 20 shillings from the pounds of the minuend, and adding them to 3 shillings, makes 23 shillings. 20 from 23 leaves 3. Set down 3 under the column of shillings. Carrying 1 to the subtrahend in the next column, 18 from 24 leaves 6. Set down 6. The remainder is £6 3s. 7d.

From which we deduce the following

RULE.

I. Set like units under each other.

II. Commencing on the right, subtract the number of each kind of unit in the subtrahend from the corresponding number in the minuend, and set the remainders under their respective units.

III. Should the number of any one unit in the subtrahend exceed the corresponding number in the minuend, add as many units to the minuend as make 1 of the next higher unit, and then subtract as before, carrying 1 to the subtrahend of the next higher unit.

Q. What is subtraction of compound numbers? How are the numbers set down? Where do you commence to subtract? Where are the remainders set down? Should the number in the subtrahend exceed the corresponding number over it in the minuend, how do you subtract? When you borrow, what do you carry? Repeat the rule.

MULTIPLICATION OF COMPOUND NUMBERS.

159. MULTIPLICATION of Compound Numbers, like that of simple numbers, consists in repeating the multiplicand as many times as there are units in the multiplier.

Multiply £14 28. 3d. by 2.

Placing the multiplier under the lowest unit of the multiplicand, we say, 2 times 3 pence are 6 pence; set down 6 under the pence: 2 times 2 shillings are 4 shillings; set down the 4 under the shillings: 2 times 4 are 8,

OPERATION. £ S. d. 17 8 9

5

Ans. 87

3

2 times 1 are 2. £28 4s. 6d. will be the answer. Again, multiply £17 8s. 9d. by 5. Setting the numbers down as in the last example, we say 5 times 9 pence are 45 pence; but since 12 pence make 1 shilling, 45 pence contain 15 shillings, or 3 shillings and 9 pence over we set down the remainder 9 under the pence, and retain the 3 shillings to carry; 5 times 8 shillings are 40 shillings, and 3 we carry are 43 shillings, but 20 shillings make 1 pound; there are then 3 pounds in 43 shillings, that is, 2 pounds and 3 shillings over; set down 3 under the shillings, and retain the 2 pounds to carry; 5 times 7 are 35, and 2 we carry are 37; set down 7 and carry 3: 5 times 1 are 5 and 3 are 8. The answer is £87 3s. 9d.

Again, multiply £15 5s. 4d. by 24.

Since the multiplier in this case exceeds 12, and can be divided into the two factors 6 and 4, the operation is shortened by multiplying by these factors separately, as in multiplication of simple numbers (Art. 37).

OPERATION.

£ S.

15 5

d.

946

91 12 0

4

Ans. 366

8 0