A. As in Simple Addition. A. Directly under the separating points above. More Exercises for the Slate. 2. James bought 2,5 cwt. of sugar, 23,265 cwt. of hay, and 4,2657 cwt. of rice; how much did he buy in all? A. 30,030* cwt. 3. James is 14 years old, Rufus 15%, and Thomas 1610 what is the sum of all their ages? A. 46,5 years. 255 4. William expended for a chaise $255, for a wagon $3700, for a bridle $100, and for a saddle $1100; what did these amount to? A. $304,455. 5. A merchant bought 4 hhds. of molasses; the first contained 62 gallons, the second 7255 gallons, the third 50 gallons, and the fourth 5570 gallons; how many gallons did he buy in the whole? A. 240,6157 gallons. 37 6. James travelled to a certain place in 5 days; the first day he went 40 miles, the second 28 miles, the third 421 miles, the fourth 22 miles, and the fifth 291000 miles; how far did he travel in all? A. 162,0792 miles. 42 7. A grocer, in one year, at different times, purchased the following quantity of articles, viz. 427,2623 cwt., 2789,00065 cwt., 42,000009 cwt., 1,3 cwt., 7567,126783 cwt., and 897,62 cwt.; how much did he purchase in the whole year? A. 11724,309742 cwt. 89 62 8. What is the amount of fo, 245100, 61000, 2451000 1100000 1000, 427100000, 40, 10000, and 1925? A. 2854,492472. 9. What is the amount of one, and five tenths; forty-five, and three hundred and forty-nine thousandths; and sixteen hundredths? A. 47,009. SUBTRACTION OF DECIMALS. ¶ LIV. 1. A merchant, owing $270,42, paid $192,625 ; how much did he then owe? OPERATION. $270,42 Ans. $77,795 For the reasons shown in Addition, we proceed to subtract, and point off, as in Subtraction of Federal Money. Hence we derive the following RULEod L Q. How do you write the numbers down? A. As in Simple Subtraction. More Exercises for the Slate. 1. Bought a hogshead of molasses, containing 60,72 gallons; how much can I sell from it, and save 19,999 gallons for my own use? A. 40,721 gallons. 2. James rode from Boston to Charlestown in 4,75 minutes, Rufus rode the same distance in 6,25 minutes; what was the difference in the time? A. 1,5 min. his years, stated 3. A merchant, having resided in Boston 6,2678 to be 72,625 yrs. age How old was he when he emigrated to that place? A. 66,3572 yrs. ̧ answer, the Note.-The pupil must bear in mind, that, in order to obtain the fgures annexed to each question, are first to be pointed off, supplying ciphers, if necessary, then added together as in Addition of Decimals. 4. From ,65 of a barrel take,125 of a barrel-525; take 2 of a barrel-45; take ,45 of a barel-2; take ,6 of a barrel-5; take,12567 of a barrel-^ 2433; take 26 of a barrel-39. A. 2,13933 barrels. 5. From 4, 0,9 pipes take 126,45 pipes-29445; take ,625 of a pipe-420275; take 20,12 pipes-40078; take 1,62 pipes1928; take 419,89 pipes-101; take 419,8999 pipes-10001. A. 1536,7951 pipes. MULTIPLICATION OF DECIMALS. ¶ LV. 1. How many yards of cloth in 3 pieces, each piece containing 20 yards? 75 OPERATION. 20,75 3 Ans. 62,25 yds. the multiplier also, we In this example, since multiplication is a short way of performing addition, it is plain that we must point off as in addition, viz. directly under the separating points in the multiplicand; and, as either factor may be made the multiplicand, had there been two decimals in must have pointed off two more places for decimals, which, counting both, would make 4. Hence, we must always point off in the product as many places for decimals, as there are decimal places in both the factors. 2. Multiply,25 by,5. OPERATION. 25 5 In this example, there being 3 decimal places in both the factors, we point off 3 places in the product, as before directed. The reason of this will appear more evident by considering both the factors common fractions, and multiplying by ¶ XLI., thus;,25%, and ,5=1; now, 100 X 10: , which, written decimally, is ,125, Ans., as before. Ans.,125 3. Multiply,15 by ,05. OPERATION. , Ans. ,0075 ,05; then 10% as before. In this case, there not being so many figures in the product as there are decimal places in both the factors (viz. 4), we place two ciphers on the left of 75, to make as many. This will appear evident by the following; ,15%, and 100=100oo =,0075, Ans., the same == From these illustrations we derive the following RULE. Q. How do you multiply in Decimals? A. As in Simple Multiplication. Q. How many figures do you point off for decimals in the product? A. As many as are in both the multiplicand and multiplier. Q. If there be not figures enough in the product for this purpose, how would you proceed? A. Prefix ciphers enough to make as many. A. To place after. Q. What is the meaning of prefix? A. To place before. More Exercises for the Slate 4. What will 5,66 bushels of rye cost, at $1,08 a bushel? 4. $6,1128, or $6, 11 c., 21 m. 5. How many gallons of rum in,65 of a barrel, each barrel containing 31 gallons?-20475. In,8 of a barrel ?-252. In ,42 of a barrel?-1323. In 6 of a barrel?-189. In 1126,5 barrels ?-3548475. In 1,75 barrels ?-55125. In 125,626789 barrels ?-39572438535. A. 39574,9238535 gallons. 6. What will 8,6 pounds of flour come to, at $,04 a pound?344. At $,03 a pound?-258. At $,035 a pound?-301. At $,0455 a pound?-3913. At $,0275 a pound?-23650. A. $1,5308. 7. At $,9 a bushel, what will 6,5 bushels of rye cost?-585. What will 7,25 bushels ?-6525. Will 262,555 bushels ?2362995. Will ,62 of a bushel?-558. Will 76,75 bushels ?69075. Will 1000,0005 bushels ?-90000045. Will 10,00005 bushels ?-9000045. A. 1227,307995. DIVISION OF DECIMALS. ¶ LVI. In multiplication, we point off as many decimals in the product as there are decimal places in the multiplicand and multiplier counted together; and, as division proves multiplication, by making the multiplier and multiplicand the divisor and quotient, hence there must be as many decimal places in the divisor and quotient, counted together, as there are decimal places in the dividend. 1. A man bought 5 yards of cloth for $8,75; how much was it a yard? $8,75=875 cents, or 100ths; now, 875÷5=175 cents, or 100ths, $1,75, Ans. = OR By retaining the separatrix, and dividing as in whole numbers, thus: OPERATION. 5) 8,75 Ans. $1,75 As the number of decimal places in the divisor and quotient, when counted together, must always be equal to the decimal places in the dividend, therefore, in this example, as there are no decimals in the divisor, and two in the dividend, by pointing off two decimals in the quotient, the number of decimals in the divisor and quotient will be equal to the dividend, which produces the same result as before. 2. At $2,50 a barrel, how many barrels of cider can I have for $11? $111100 cents, or 100ths, and $2,50=250 cents, or 100ths; then, dividing 100ths by 100ths, the quotient will evidently be a whole number, thus: In this example, we have for an answer 4 barrels, and 88 of another barrel. But, instead of stopping here in the process, we may bring the remain der, 100, into tenths, by annexing a cipher (that is, multiplying by 10), placing a decimal point at the right of 4, a whole num ber, to keep it separate from the 10ths, which are to follow. The separatrix may now be retained in the divisor and dividend, thus: OPERATION. 2,50) 11,00 (4,4 Ans. 1000 1000 1000 We have now for an answer, 4 barrels and 4 tenths. of another barrel. Now, if we count the decimals in the divisor and quotient (being 3), also the decimals in the dividend, reckoning the cipher annexed as one decimal (making 3), we shall find again the decimal places in the divisor and quotient equal to the decimal places in the dividend. We learn, also, from this operation, that, when there are more decimals in the divisor than dividend, there must be ciphers annexed to the dividend, to make the decimal places equal, and then the quotient will be a whole number. Let us next take the 3d example in Multiplication, (TLV.) and see if multiplication of decimals, as well as whole numbers, can be proved by Division. 3. In the 3d example, we were required to multiply,15 by,05; now we will divide the product ,0075 by,15. We have, in this example (before the cipher was placed at the left of 5), four decimal places in the dividend, and two in the divisor; hence, in order to make the decimal places in the divisor and quotient equal to the dividend, we must point off two places for decimals in the quotient. But as we have only one decimal place in the quotient, the deficiency must be supplied by prefixing a cipher. The above operation will appear more evident by common fractions, thus:,0075—100%, and,15=7%; now Toooo is divided by T by inverting Too ( XLVII.), thus, 15 X1000 100-75 =758880=180=,05, Ans., as before. 15 TOO |