divisor, and place a point below it, divide 341 by 273, the quotient is 1, multiply 273 by 1, and subtract the product from 341, the remainder is 68. Omit the 3 from 273, and divide 68 by 27, the quotient is 2 and the remainder 13. Reject 7 from 27 and divide 13 by 2, the quotient is 4, since 4×2+3 to carry from 28, make 11; the last remainder, 2, is neglected. The quotient is found to be 45124, from which we point out the three decimal places required, and for which we annexed three figures in the beginning; the answer is then 45.124. In comparing both operations, it is easily perceived that the figures rejected exercise no influence on the figures of the quotient. Instead of preparing the dividend as we did, we may notice that the number of figures of the divisor used at first, is the . same as the number of figures of the quotient; therefore, we may begin by rejecting as many figures to the right of the divisor as its own number of figures exceeds that of the quotient, then find how many times the divisor, thus prepared, will be contained in the first figures of the dividend, and proceed as explained above. By a single observation, it is generally easy to see how many places of whole numbers the quotient will contain. In this example it is seen, by dividing 1234 by 27, to be two places, and there are three decimal places to be found, so together we have to determine five places in the quotient. Determine the quotient of 229.4703568 and 7.3596 correct to four places. The quotient will contain two places of integers and four decimal places, together six places; but as the divisor contains only five figures, a cipher is annexed to it, and we proceed as explained before. 7.35940)229.4703568(31.1805 2207820 86883 73594 13289 7359 5930 5887 43 37 Divide .625 by .428571 correct to six places of decimals. .428571).6250000(1.458335 4285712 1964288 1714285 250003 214286 35717 34280 1437 1286 151 128 23 21 2 Divide .3474 by .8, retaining six decimal places. .666666).3474474(.521170 3333330 141144 133333 7811 6667 1144 667 467 467 209. EXERCISES ON THE DIVISION, &C., OF DECIMALS. 1. Required, the quotient of 67.4903 and 8.04; of .3495 and .735; of .026705 and .0834; of .0572 and .0014; of .0594 and 26356; of 1 and 10.1673. 2. Divide 2.295 by .297; 2.14535 by .07; 234.6 by .7; 41.5432 by 34.48; 27.2179487 by .115346; .0563215 by 1.34562, 3. Find the sum, difference, product, and quotient of .5736 and .23. 4. 3.562+16.71-3.05+.394 X.62÷3.45. REDUCTION OF DECIMALS. 210. The knowledge which the learner must now have acquired about vulgar and decimal fractions, will very much facilitate the study of this section. First, we propose to find the value of any decimal of a given quantity. = Example: What is the value of .86875 of £1, or £.86875? £.86875.86875 x 20s. 17.37500 shillings. .375s.=.375 × 12d.=4.500 pence. .5d.=.5 x 4f.=2.0 farthings. Or briefly thus: £.86875 20 17.37500 12 4.500 4 2.0 Therefore, £.86875=17s. 44d. Example Convert .00213 of a day to positive terms: = = 3.06720 min. ; 4.032 sec. .05112 hrs. 60 3.06720 min. 60 4.03200 = Therefore, .00213 day 3 min, 4.032 sec. Example: What is the value of 3.765 of a cwt? Here 3.765 cwt.=3,85 cwt.=3 cwt. 3 qrs. 1 lb. 12.25 oz. From these examples, it follows that, to express by means of known units the value of decimals, multiply the given decimal by the numbers which would reduce it to the lower denominations were it an integer, and the integral parts of the products, pointed off as they occur, will be the required value. If the decimal be expressed as part of more than one denomination, reduce it to one, and then apply the law. In case of recurring decimals, it is generally better to reduce it to a common fraction and to find the value. 211. EXERCISES. 1. Find the value of .64 of £1; .6325 of £1; 1.626 of 1 guinea. 2. Express 32.08 of £5; 2.15506 of £50; .205 of 6s. 81d. 3. What is the value of 1.265 of 1 ton; of .275 of 1 mile; .025 of 1 yard? 4. Required, the value of 36.8 of 24 tons 2 cwt.; of .375 of 6s. 8d. 5. Determine the value of 6 of 3s. 9d.+.037 of 27s.-.081 of £3. 5s. 212. Secondly, we have to consider the manner of reducing a given quantity to the decimal of another quantity. Example: Express 16s. 94d. as the decimal of £1. Since d. is the of £1; therefore, 16s. 94d, or 493d. are 403 × of £1, or £483; or £.839583; or thus, d..5d.; 94d. 9.5; 9.5d. reduced to the decimal of 1 shilling is .7916; and 16.7916s., expressed = as a decimal of £1, is 167916.839583. 30 The operation is here exhibited : 12)9.5d. £.839583 Example: Reduce 13s. 6d. to the decimal of 15s. 6d. and 1d. is the T of 15s. 6d. ; therefore, 13s. 63d., or 162.75d. are 162.75 × 1, or 1675, or 875 of 15s. 6d. ; thus, 3f..75d. 12)6.75 15.5)13.5625 .875 of 15s. 6d. Hence, to reduce a given quantity to the decimal of another quantity, bring both quantities to the same denomination, and divide the first by the second; or, when convenient, divide the lowest denomination by the number which connects it with the next, and affix to the left of the quotient the number of this denomination; reduce the result to a decimal of the next higher denomination, and so on, till the required denomination is obtained. 213. EXERCISES. 1. Reduce 7s. 6d. to the decimal of £1. 2. Reduce 17 cwt. 2 qrs. 15 lb. to the decimal of 1 ton. 3. Reduce 6 oz. 5 dwt. to the decimal of 1 lb. 4. Reduce 5 days 15 hrs. 16 min. 5 sec. to the decimal of 1 week. 5. Express 2 r. 16 p. as the decimal of 1 acre. 214. MISCELLANEOUS EXERCISES IN DECIMALS. We have given here a great variety of examples, in order to familiarize the pupil with the several parts explained before. He has already experienced that some problems are susceptible of several solutions, but there is generally one which is preferable to others; let it, therefore, be his aim to choose the one best adapted to the conditions of the question. 1. A merchant owes a certain sum of money, he pays severally £248.40, £150.60, £545.75, and lastly, a bill of £500, and receives back £147.75. What was the debt? |