Examples.

1. How much land is there in a township, consisting of 385 farms, each containing 129 A. 3 R. 6 P. 15 sq. yds. 51 s. f.?

In the multiplier 385, it is easy to recognize the 3 factors 5, 7, and 11. We therefore multiply by these in succession, and find that the township contains 78 square miles, 49 acres, 1 rood, 28 poles, 10 square yards, and of a square foot, or 108 square inches.

2. How much land is there in a township, consisting of 495 farms, each containing 568 A. 3 R. 22 P. 6 s. yds. 61 s. f. ? Answer, 440 s. m.

323. When the multiplier cannot be resolved into convenient factors, take the nearest composite number inferior to it, by which multiply, as in the preceding article: then multiply the given compound number by the difference between the composite number and the given multiplier. This last product, added to the product given by the composite number, is the required product.

Examples.

1. What weight of coffee is there in 236 bags, each containing 1 cwt. 1 qr. 4 lbs. 5 oz. 6 dr. ?

As the factors of 236 are not convenient, seeing that one of them is the prime 59, we take the composite number 231, which differs by only 5 units, and of which it is easy to see that the factors are 3, 7, and 11. By these we multiply as in the preceding article, and to the product 14 t. 17 cwt. 2 qrs. 21 lbs. 9 oz. 10 dr., which is 231 times the multiplicand, we add 6 cwt. 1 qr. 21 lbs. 10 oz. 14 dr., which is 5 times that number. The sum, therefore, 15 t. 4 cwt. 0 qrs. 15 lbs. 4 oz. 8 dr., is 236 times the multiplicand, or whole weight required. 2. How far will a man travel in 534 days, at the average rate of 25 m. 5 fur. 39 p. 4 yds. 2 ft. 6 in. per day?

Answer, 13750 m. 2 fur. 15 p. 1 yd. 1 ft. 6 in. 3. How much grain is there in 785 bags, each containing 2 bush. 3 pks. 1 gal. 3 qts. 1 pt. ; and how high will it cover a floor that is 20 ft. long and 15 ft. wide, allowing 2150 cubic inches to the bushel?

Answer. The quantity of grain is 2342 bush. 2 pks. 1 gal. 3 qts. 1 pt. ; and the depth to which it will cover the floor is 9 ft. 832 in., or 9 ft. 83 in., nearly.

9

55296

324. When a compound number is reduced to its lowest denomination, or to a fraction, proper or improper, of its principal unit, it becomes simple; and, in this form, susceptible of all the fundamental operations, as an abstract number.

For example, to multiply 13 cwt. 3 qrs. 16 lbs. by 191, we may proceed thus: 13 cwt. 3 qrs. 16 lbs. 1556 lbs.: then, 1556 × 191 = 297196 lbs. 132 t. 13 cwt. 2 qrs. 4 lbs.

=

Or thus: 13 cwt. 3 qrs. 16 lbs.

=

74299 560

=

132 t. 13 cwt. 2 qrs. 4 lbs., as before

By the usual method :

What weight of iron is there in 171 loads, each weighing 19 cwt. 3 qrs. 12 lbs. ? Answer, 169 t. 15 cwt. 2qrs. 8lbs.

Multiplication of a Compound Number by a Compound Number:

325. From the above, it is manifest that the multiplication of two compound numbers may be reduced to the multiplica

tion of a fraction by a fraction, thus: Reduce each of the given numbers to a fraction of its principal unit; multiply the two fractions together, and reduce the product to the denominations required by the nature of the question.

Examples.

1. What is the cost of 6 cwt. 3 qrs. 14 lbs. of sugar, at £3 5 s. 6 d. per cwt.? 3 qrs. 14 lbs. 55cwt. 5 s. 6 d.

3 qrs.

=

40

40

=

11

2 qrs. = cwt., and 67 cwt. 51 s. s. = 1}£, and 311 £=-431-£. Now, as -55 is the given quantity in cwts., and 131 £, the cost of 1 cwt., it is evident that 131-X-55-131 1 1111£= £22 10 s. 33 d., is the cost of the whole, as required. The student may, by the preceding methods, perform the following examples, and prove the work by vulgar and by decimal fractions:

14

64

8

8. 11 cwt. 1 qr. 14 lbs. of sugar, at £3 15 s. 6 d. per cwt., amount to £42 18 s. 9 d.

9. 4 cwt. 3 qrs. 14 lbs. of sugar, at £2 10 s. 6 d. per cwt., amount to £12 6 s. 24 d.

4

10. 7 cwt. 0 qr. 19 lbs. of sugar, at £3 14 s. 81 d. per cwt., amount to £26 15 s. 71⁄2 d.

11. 13 cwt. 2 qrs. 7 lbs. of sugar, at £2 3 s. 93 d. per cwt., amount to £29 14 s. 21 d.

12. 9 cwt. 1 qr. 12 lbs. of sugar, at £5 11 s. 63 d. per cwt., amount to £52 3 s. 10₫ d.

13. 29 cwt. 3 qrs. 17 lbs. of sugar, at £7 1 s. 41 d. per cwt., amount to £211 7 s. 41 d.

14. 297 t., at £13 12 s. 04 d. per t., amount to £4039 16 s. 41 d.

15. 921 t., at £9 19 s. 1 d. per t., amount to £9169 14 s. 11 d.

Division of Compound Numbers.

Division of a Compound Number by a Simple Number: 326. If the dividend and divisor represent quantities of different kinds, we first divide the highest order of units in the dividend by the divisor. We then reduce the remainder of this division to the next lower denomination, adding the units of the dividend which are of this denomination: after which we divide and proceed with the remainder of this second division as with that of the first. We continue thus to the lowest denomination, and, collecting the several quotients, we have the total quotient or compound number sought. If the highest order in the dividend does not contain the divisor, we reduce it to the next lower order, taking care to add the units of this denomination.

Examples.

1. If 30 yds. of cloth cost £10 9 s. 4 d., what is the cost per yd?

As 30 is not contained in 10, we reduce the £10 to shillings, and add the 9 s. of the dividend. The sum 209 s. we divide by 30, which gives 6 s. for the quotient and 29 s. for remainder. This remainder we reduce to pence, and add the 4 d. of the dividend. The sum 352 d. we divide by 30, which gives 11 d. for the quotient and 22 d. for remainder. This last we reduce to farthings, and add the 2 farthings of the dividend. The sum 90, we divide by 30, and have 3 farthings

for the quotient, without remainder. Lastly, we collect the several quotients, and have 6 s. 113 d. for the cost per yd., as was required.

We may also divide by the factors of 30, thus:

2. If 185 yds. cost £604 6 s. 8 d., what is the cost per yd.?

Answer, £3 5 s. 4 d.

3. If 284 yds. cost £700 4 s. 9 d., what is the cost per yd.?

4. If 316 t. cost £4034 5 s. 4 d.,

5. If 297 A. cost £4039 16 s. acre?

6. If 921 cwt. cost £9169 14 s. cwt.?

Answer, £2 9 s. 33 d. what is the cost per ton? Answer, £12 15 s. 4 d. 41 d., what is the cost per Answer, £13 12 s. 01 d. 11 d., what is the cost per Answer, £9 19 s. 1 d.

7. Divide 78 s. m. 49 A. 1 R. 28 P. 10 s. yds. 03 s. f. by 385. Answer, 129 A. 3 R. 6 P. 15 s. yds. 51 s. f.

For further practice, the student may reverse the 6 examples which immediately follow the first example of the preceding article; in each of which he may divide the whole cost by the number of yards, which will give the cost per yard.

327. When the dividend and divisor represent quantities of the same kind, we must observe whether the quotient should or should not also be of the same kind. If it should, we operate as in the preceding article. For example, suppose that, with £15, we have gained £27 10 s. 33 d., and that we would find the gain on each pound. It is plain that this gain should be the fifteenth part of the whole gain, and therefore of the same name with the divisor and dividend: consequently, we divide £27 10 s. 33 d. by 15, and have £1 16 s. 81 d. for the required gain.

328. But, if the nature of the question requires that the quotient should be of a different kind, we reduce the divisor and dividend to the lowest denomination of the dividend; after which, considering the units of the dividend as being of the same kind with those of the quotient that we seek, we divide as in Art. 326.