price of one unit, to find the value of the whole, the price being compound. It is plain that, in this case, we must multiply the price of a unit by the number of units. There are several methods of effecting this : that, however, which seems to be most naturally suggested is, to multiply by the given multiplier, each order of units, from the lowest to the highest; and, when any product exceeds in value a unit of the next higher order, to reduce it to that order, as in addition. For example, we find the value of 17 yds. at 6 s. 114 d. per yd., thus : 6 11% 17 £5 18 7 value of 17 yds. Beginning with the farthings, we say, 17 X 3=51 qrs. = 12 d., we write , and carry 12: then 17 times 11 is 187, and 12 is 199 d.=16 s. 7 d., we write 7, and carry 16 : lastly, 17 times 6 is 102 and 16 is 118 s. = £5 18 s., which we write, and the work is done. The work may be proved thus :-6 s. 114 d. =335 qrs.= £35: then £ €335 X 17 596 = £5 18 s. 73 d. 960 Or thus : 114 d. = 11,75 d. =,97916 s., and 6,97916 s. = £ ,3489583: then £ ,3489583 X 17= £ £5,9322916 £5 18 s. 7 d. By both these methods the student may prove the following Examples 1. Required the cost of 19 yds. of cloth, at 17 s. 6 d. per yd. Answer, £16 12 s. 6 d. 2. What is the cost of 13 yds. at 18 s. 9 d. per yd. ? Answer, £12 4 s. 6 d. 3. What is the weight of 17 hhds. of sugar, each containing 9 cwt. 3 qrs. 14 lbs. 8 oz. ? Answer, 8 t. 7 cwt. 3 qrs. 22 lbs. 8 oz. 4. What is the content of a farm consisting of 11 fields, each containing 39 A. 3 R. 23 P. 29 s. yds. 8 s. f. 111 s. in. ? Answer, 438 A. 3 R. 23 P. 27 s. yds. 1 s. f. 141 s. in. -322. When the multiplier is too large to multiply by at once, and can be resolved into 2 or more convenient factors, we mula tiply, successively, by all the factors, taking them in any order which may seem most convenient.. 8. m. 51 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. 54 s. f. ? A. R. P. s. yds. s.f. 0 129 3 6 15 5 7.11=385 1 8 3 32 17 33 7 7 62 2 28 0 81 11 78 491 28 10 03 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. 64 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. ? T. cwt. qr. lbs. oz. 3.7.11 +5=236 7 11 6 1 21 10 14 5 times dr. 6 15 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. O 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 cubio 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. 83 urt in., or 9 ft. 8 in., nearly. 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 x 191 = 297196 lbs. 132 t. 13 cwt. 2 qrs. 4 lbs. 389 X 191 Or thus : 13 cwt. 3 qrs. 16 lbs. = 388 t. : then, 560 74782 132 t. 13 cwt. 2 qrs. 4 lbs., as before By the usual method : cwt. gr. lbs. 0 13 3 16 191 132 13 2 4 What weight of iron is there in 171 loads, each weighing 19 cwt. 3 qrs. 12 lbs. ? Answer, 169 t. 15 cwt. 2 qrs. 8 lbs. 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. qrs. 523 yds. 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 Ibs. = 31 qrs. cwt., and 67 cwt. 55 cwt. 5 s. 6 d.=518.=-1-8.=i£, and 311 £=481 £. Now, as 5,6 is the given quantity in cwts., and 42301- £, the cost of 1 cwt., it is evident that 432 X 56=131 X 1911£= £22 10 s. 31 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: 2. 187 yds. at 3 s. 6 d. per yd., amount to £32 14 s. 6 d. 3. 4 s. 8 d. 122 0 8 4. 315 yds." 58. 3 d.“ 83 13 51 5. 735 yds.“ 19 s. 11 d.“ 702 16 103 6. 1962 yds. “ 23 s. 5d.“ 2303 6 11 7. 2627 yds.“ 22 s. 11 d. “ 3012 16 92 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. 21 d. 10. 7 cwt. O qr. 19 lbs. of sugar, at £3 14 s. 81 d. per cwt., amount to £26 15 s. 71 d. 11. 13 cwt. 2 qrs. 7 lbs. of sugar, at £2 3 s. 98 d. per cwt., amount to £29 14 s. 21 d. 12. 9 cwt. 1 qr. 12 lbs. of sugar, at £5 11 s. 64 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. 297t., at £13 12 8. 01 d. per t., amount to £4039 16 s. 41 d. 15. 921 t., at £9 19 s. 11 d. per t., amount to £9169 14 s. 17 d. Division of Compound Numbers. 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 divi. sion 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. 41 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 |