8. Multiply 6.711 by 6543. Ans. 43910.073. 9. Multiply .0009 by .0009. Ans. .00000081. 10. What is the product of one thousand and twenty-five, multiplied by three hundred and twenty-seven ten-thousandths ? Ans. 33.5175. 11. What is the product of seventy-eight million two hundred five thousand and two, multiplied by fifty-three hundredths ? Ans. 41448651.06. 12. Multiply one hundred and fifty-three thousandths by one hundred twenty-nine millionths. Ans. .000019737. 13. What will 26.7 yards of cloth cost, at $5.75 a yard ? Ans. $ 153,525. 14. What will 14.75 bushels of wheat cost, at $ 1.25 a bushel ? Ans. $ 18.4375. 15. What will 375.6 pounds of sugar cost, at $ 0.125 per pound? 16. What will 26.58 cords of wood cost, at $ 5.625 a cord ? Ans. $ 149.512. 17. What will 28.75 tons of potash cost, at $ 125.78 per ton ? Ans. $ 3616.175. CONTRACTIONS IN MULTIPLICATION OF DECIMALS. 272. To multiply a decimal by 10, 100, 1000, &c. Re. move the decimal point as many places to the right as there are ciphers in the multiplier, annexing ciphers if required. Thus, 1.25 X 10 12.5 ; and 1.6 X 100 160. EXAMPLES. 1. Multiply 131.634 by 1000. Ans. 131634. 2. Multiply 3478.9 by 100. Ans. 347890. 3. Multiply one thousandth by one thousand. 4. What is the profit on one million yards of cotton cloth, at $0.007 per yard. Ans. $ 7000. 273.° When it is not necessary that all the decimal places of the product should be retained, tedious multiplications may often be obviated, by contracting the work as follows: Write the units' place of the multiplier under that figure of the multiplicand whose place it is proposed to retain in the product, and dispose of all the other figures of the multiplier in an order contrary to the usual one. Then, in multiplying, begin, for each partial product, with that figure of the multiplicand which stands above the multiplying figure, observing to add to the product the number nearest to that which would have been carried if the places at the right had not been rejected. Write down the several partial products, so that the right-hand figure of each shall be in the same column, and their sum will be the product required. EXAMPLES. 1. Multiply 3.141592 by 52.7438, retaining only four places for decimals in the product. Ans. 165.6995 FIRST OPERATION. SECOND OPERATION. 3.1 41 5 9 2 Multiplicand 3.1 4 1 5 9 2 8 34 7.2 5 Multiplier reversed. 5 2.7 438 1 5 7 0 7 9 6 Product by 5, +1 2 5 1 3 2 7 3 6 6 2 8 3 2 Product by 2, + 2 9 4 2 4 7 7 6 2 1 9 91 Product by 7, + 4 1 2 5 6 6 3 6 8 1 2 57 Product by 4, +1 2 1 9 9 11 44 94 Product by 3, 1 6 2 8 3 1 84 2 5 Product by 8, 1 15 7 0 7 9 60 1 6 5.6 9 95 Product sought. 1 6 5.6 9 9 5/0 0 1 2 9 6 By comparison of the two methods of solution, it will be seen that the common one, as shown in the second operation, gives ten places of decimals, or six more than are required by the question, thus rendering unnecessary the several figures on the right of the vertical line. By the contrasted way, the multiplier, for convenience, has its figures reversed, or placed contrary to the usual order, so that the product of each figure by the one of the multiplicand above it, must be of the order of ten-thousandths. The first figure, at the right, of each partial product, being of the order of ten-thousandths, is written in the same column. To the product by 5 we add 1, since, if the 2 in the multiplicand had not been rejected, there would have been 1 to carry to the product of the 9 by the 5; to the product by 2 we add 2, since the product of the rejected figures, 92, by 2, approximates to 2 hundred, which would require 2 to be carried to the product by 7 we add 4, since the product of the two rejected figures, 59, by 7, would require 4 to be carried ; to the pro duct by 4 we add 1, since the product of the two rejected figures, 15, by 4, approximates to 1 hundred, which would require 1 to be carried; and so on, it being sufficient to increase the partial product only by such a number as approximates most nearly to that which would have been carried, provided the two rejected figures next to the figure of the multiplicand had been retained. 2. Multiply 325.701428 by .7218393, retaining only three places of decimals in the product. Ans. 235.104. 3. Multiply 56.7534916 by 5.376928, retaining only five places of decimals in the product. 4. Multiply 843.7527 by 8634.175, retaining only the integers in the product. Ans. 7285109. DIVISION OF DECIMALS. OPERATION. 1_728 1000 274. Ex. 1. Divide 1.728 by 1.2. Ans. 1.44. We divide as in whole numbers, 1.2) 1.7 2 8 ( 1.4 4 Ans. and, since the divisor and quotient 12 are the two factors, which, being mul tiplied together, produce the dividend, 5 2 we point off two decimal figures in the 48 quotient, to make the number in the 48 two factors equal to the number in the product or dividend. 48 The reason for pointing off will also be seen by performing the example with the decimals in the form of common fractions. Thus, 1.728 1778; and 1.2 : 1% = 16. Then 1776 : ii 1768 x 18 = 12888 = 1dt = 1.4=1.44, Ans. as before. 2. Divide 36.6947 by 589. Ans. .0623. We divide as in whole 58 9 ) 3 6.6 9 47 (.0 6 2 3 Ans. numbers, and since we have 3 5 34 but three figures in the quo tient, we place a cipher be. 1 3 54 fore them, and thus make the 117 8 decimal places in the divisor 17 6 7 and quotient equal to those of the dividend. 17 67 The reason for prefixing the cipher will appear more obvious by solving the example in the form of common fractions. Thus, 36.6947 36 6,947 ; and 589 514. Then 36.694,7 ; 582 1888 = .0623, Ans. as before. Hence the following OPERATION. 36,69 47 10000 3 6 6 947 10000 3 6 6 9 4 7 5890000 RULE. Divide as in whole numbers, and point off as many figures in the quotient as the number of decimal places in the dividend exceeds the number in the divisor ; but if there are not as many, supply the deficiency by prefixing ciphers. NOTE 1. - When the decimal places in the divisor exceed those in the dividend, make them equal by annexing ciphers to the dividend, and the quotient will be a whole number. NOTE 2. — When there is a remainder after dividing the dividend, ciphers may be annexed, and the division continued; the ciphers thus annexed being segarded as decimals of the dividend; and to indicate in any case that the division does not terminate, the sign plus (+) can be used. Proof. - The proof is the same as in division of whole numbers. EXAMPLES. 3. Divide 780.516 by 2.43. Ans. 321.2. 4. Divide 7.25406 by 9.57. Ans. .758. 5. Divide .21318 by .38. 6. Divide 7.2091365 by .5201. Ans. 13.861+. 7. Divide 56.8554756 by .0759. Ans. 749.084. 8. Divide 119109094.835 by 38123.45. Ans. 3124.3. 9. Divide 1191090.94835 by 3812345. 10. Divide 11910909483.5 by 38.12345. 11. Divide 11.9109094835 by 381234.5. 12. Divide 1191.09094835 by 3.812345. 13. Divide 11910909483.5 by .3812345. 14. Divide 1.19109094835 by 3.812345. 15. Divide .119109094835 by .3812345. 16. Divide 30614.4 by .9567. Ans. 32000. 17. Divide .306144 by 9567. Ans. .000032. 18. Divide four thousand three hundred twenty-two,and four thousand five hundred seventy-three ten-thousandths by eight thousand, and nine thousandths. Ans. .5403+. 19. How many yards of calico at $ 0.0775 per yard can be purchased for $ 10.85 ? 20. What costs 1 acre of woodland when 19.65 acres are sold for $ 982.50 ? Ans. $ 50. 21. Divide three hundred twenty-three thousand seven hun. dred sixty-five by five millionths. Ans. 64753000000. CONTRACTIONS IN DIVISION OF DECIMALS. 275. To divide a decimal by 10, 100, 1000, &c. Remove the decimal point as many places to the left as there are ciphers in the divisor, and if there be not figures enough in the number, prefix ciphers. Thus, 2.15 • 10 = .215; and 1.9 ; 100 .019. EXAMPLES. 1. Divide 31.675 by 10. Ans. .0001965. 10. If $ 3500 are paid for 1000 yards of broadcloth, what is it a yard ? Ans. $ 3.50. 11. When $ 1025 are paid for 40 boxes of sugar, each containing 250 pounds, what is the cost of 1 pound ? Ans. $ 0.101 276. When the divisor contains many decimal places, and only a certain number of decimals are required to be retained in the quotient, the work may be contracted as follows : First consider how many figures, in all, it is necessary for ihe quotient to contain. Then, by using the same number of figures from the left of the divisor, find the first figure of the quotient, and, instead of bringing down a new figure from the dividend, or annexing a cipher to the remainder, reject a figure on the right of the divisor at each successive division, and make the other figures a divisor. In multiplying such a divisor by the quotient figure, observe to add to the product the number nearest to that which would have been carried if no figures had been rejected. |