Ex. XXXVII. Find the difference between (1) £3. 13s. 93d., and £2. 15s. 51d. (3) £23. 13s. 7d., and £19. 19s. 73d. (4) £416. 10s. 51⁄2d., and £305. 11s. 9ğd. COMPOUND MULTIPLICATION. 121. COMPOUND MULTIPLICATION is the method of finding the amount of any proposed compound number, that is, of any number composed of different denominations, but all of the same kind, when it is repeated a giver number of times. RULE. "Place the multiplier under the lowest denomination of the multiplicand; multiply the number of the lowest denomination by the multiplier, and find the number of units of the next denomination contained in this first product; if there be a remainder, place it down, adding on the number of units just found to the second product; for this second product, multiply the number of the next denomination in the multiplicand by the multiplier, and after carrying on to it the above-mentioned number of units, proceed with the result as with the first product; carry this operation through with all the different denominations of the multiplicand." d. multiplied by 5 is the same as (1+1+1+1+1)d.=5 half-pence =24d.; we therefore put down d., and carry on 2d. to the denomination of pence: 6d. multiplied by 5=30d.; therefore (2+6 × 5)d.=32d. = (2 × 12+8)d. =2s.+8d.; we therefore put down 8d., and carry on 2s. to the denomination of shillings: 48. multiplied by 5=20s.; therefore (2+4×5)s. = 22s. = (20+2)s. = £1 +28. ; we therefore put down 2s., and carry on £1 to the denomination of pounds: Now by Simple Multiplication £56×5= £280; therefore £(1+56 × 5) =£(1+280) = £281. Therefore the total amount is £281. 2s. 81d. 122. When the multiplier exceeds 12 it will be the easiest method to split the multiplier into factors, or into factors and parts: thus 15=3 × 5; 17=3×5+2; 23=4×5+3; 240=4× 6 × 10: and so on. 1112. 15. 5 = value of £222. 11s. 1d. multiplied by 5, or of £55. 128. 91d. multiplied by (4 × 5, or 20). 166. 18. 33= value of £55. 12s. 91d. multiplied by 3. £1279 . 13. 82= value of £55. 12s. 91d. multiplied by (20+3), or 23. Note 1. When the multiplicand contains farthings, if one of the factors of the multiplier be even, it will often be advantageous to use it first, as the farthings may disappear. Note 2. Should the multiplier consist of many factors, it will be found in that case convenient to reduce the multiplicand to the lowest denomination contained in it, then to multiply this result by the multiplier, and then to reduce the result back again. Multiply Ex. XXXVIII. (1) £11. 13s. 6d. separately by 2 and 5. (9) £1875. 13s. 8d. separately by 21 and 64. (13) £2579. Os. O d. separately by 147, 155, 474, and 2331. (16) 45 lbs., 7 oz., 3 drs., 2 sc. separately by 12 and 68. (19) 67 ro., 38 po., 27 yds., 2 ft. separately by 11 and 112. (23) 5 wks., 6 d., 18 h., 14 m. separately by 11 and 339. (24) 84 hhds., 43 gals., 1 pt. of wine separately by 27 and 364. (25) 43 bar., 13 gals., 1 qt., 1 pt. of beer separately by 39 and 764. (26) A person buys 67 lambs at £1. Os. 91d. each; 73 sheep at £2. 2s. 111d. each; 12 cows at the average of £37. Os. 2 d. for every 3 of them; and 17 horses at 37 guineas cach: the expenses of getting them all home amount to 17 guineas. What money must he draw from his bankers to pay for the whole outlay? (27) There are 7 chests of drawers: in each chest there are 18 drawers; and in each drawer 8 divisions; and in each division there is placed £16. 6s. 8d. How much money is deposited in the chests? 123. If the multiplicand contain, instead of farthings, some other fraction of a penny, the process is exactly the same as the above: thus, Ex. 1, if we had to multiply £22. 15s. 45d. by 43; (1) Multiply £6. 12s. 85d. separately by 3, 11, and 57. COMPOUND DIVISION. 124. COMPOUND DIVISION is the method of dividing a compound number, that is, a number composed of several denominations, but all of the same kind, into as many equal parts as the divisor contains units; and also of finding how often one compound number is contained in another of the same kind. When the Divisor is an abstract number. RULE. "Place the numbers as in Simple Division: then find how often the divisor is contained in the highest denomination of the dividend ; put this number down in the quotient; multiply as in Simple Division and subtract; if there be a remainder, reduce that remainder to the next inferior denomination, adding to it the number of that denomination in the dividend, and repeat the division: carry on this process through the whole dividend." Ex. 1. Divide £199. 6s. 8d. by 130. Proceeding by the Rule given above, We first subtract £1 taken 130 times, from £199. 6s. 8d., and there remains £69. 6s. 8d. Now £69. 6s. 8d.=1386s. 8d.; from this amount we subtract 10s. taken 130 times, and there remains 86s. 81. Again, 86s. 8d.=1040d.; from this amount we subtract 8d. taken 130 times, and nothing remains. Therefore £1. 10s. 8d. is contained 130 times in £199. 6s. 8d. Therefore the result is £2. Os. 10d., and there remains 247 farthings to be divided by 527, which division will clearly not give so much as one farthing. |