III. THESE PRODUCTS, ADDED AS THEY STAND, WILL GIVE THE TOTAL PRODUCT, WITH THE REQUIRED NUMBER OF DECIMAL PLACES. NOTE. If the multiplier, when thus written, extends below the multiplicand, fill out the multiplicand, or suppose it to be filled out with cyphers annexed. 3. Multiply 27.14986 by 92.41035, retaining four decimal places in the A. 2,508.9280 product. 4. Multiply 480.14936 by 2:72416, retaining four decimal places. A. 1,308.0037 A. 341.80097 5. Multiply 73.8429753 by 4.628754, retaining five decimal places. 6. Multiply 8,634.875 by 843.7527, retaining only the integers in the product. A. 7,285,699 § XLIV. 1. What is the value of .75 of a pound sterling in shillings, pence, &c.? £0.75 I£ contains 20s. Therefore of a pound contains of a shilling, and of a pound contains 20 times as many hundredths of a shilling=1500 of a shitling=15s. This is performing the operation as in vulgar Fractions. It is obviously the same, whether the denominator 100 be expressed or not. 20 A. 15.00s. 2. Reduce .17525£ to lower denominations. Hence, to reduce decimals of higher, to whole numbers, and decimals of lower denominations, MULTIPLY AS IN REDUCTION DESCENDING, RESERVING THE WHOLE NUMBERS OF EACH DENOMINATION, AND CONTINUING THE REDUCTION OF THE DECIMAL ONLY. A. 9 d. A. 11 s. 5 d. 1.504 qrs. A. 17 s. 0 d. 2.4 qrs. 0 A. 9 d. A. 55 gal. O qt. 1 pl. A. 3 qrs. 2 na. A. 28 rds. 2 yds. 1 ft. 11.04 in. 11. Reduce .089 mile. A. A. A. 51.999725 days. 207 d. 16 h. 26 m. 24 sec. A. 2 qrs. 19 lb. 5 oz. 2 hhd. 27 gal. 2 qts. 1 pt, A. 16 s. 7 d. 2.999936_qrs. In reducing decimals of a £. we may evidently reverse the rule thus, § LX, DOUBLE THE TENTHS' FIGURE FOR SHILLINGS, AND IF THE HUNDREDTHS BE 5, OR MORE, DEDUCT THE 5, AND ADD ANOTHER SHILLING. CALL THE REMAINING FIGURES, IN THE HUNDREDTHS' AND THOUSANDTHS' PLACES, FARTHINgs, decreasing THEM BY 1, IF BETWEEN 12 AND 36, AND BY 2, If above 36. NOTE. The decimal below thousandths is neglected. But if it amount to more than half a thousandth, it is best to increase the 1,000ths. by 1. 21. Reduce .8971 £. .13763 cwt. .19843 ml. .15634 yd. .71 lb. Troy. 71 lb. Avoirdupois. .71 . Apothecaries'. .8934 week. .9193 month of 30 d. .7346 rd. .9874 hhd. of wine. .9874 hhd. ale. .7759 hhd. beer. .8557 oz. Troy. .9365 £. .59347 £. SUBTRACTION. § LXV. 1. I have a debt of $95 763, and I make a payment of $87.665. How much is remaining due? Of course, dimes must be taken from dimes, cts. from cts. &c. Hence, the numbers must be written as in Addition, and the point placed in the same manner. Hence, to subtract decimal numbers, $95.763 8.098 Ans. WRITE THE NUMBERS AS IN ADDITION, subtract AS IN WHOLE NUMBERS, AND PLACE THE SEPARATRIX AS IN ADDITION. 2. From a piece of cloth containing 473 yds. a merchant sold 23 yds. How much was left? A. 24.015 3. On a debt of $383.00 there was paid $47.25. How much remained unpaid? A. $335.75 4. On a debt of $1933, there was paid $875. How much remained unpaid? A. $105.775 75. From 1,153 tons of iron, there were sold 684,5 tons. How much remained unsold? A. 468.8312 tons. 6. From 37 gals. of oil, were sold 28 gals. How much remained? A. 9.1372 gals. 7. A man having 759921 bu. of wheat, sold to one person 47% bu.; to another, 8715; to another, 94, ; to another, 387. How much had he left? A. 143 bu. 8. A man had $16,73% of which he spent as follows: for a load of hay, $6; for a load of grain, $7,7%; for 3 bu. of corn, 817 pr. bu.; and for a load of wood, $2, What had he left? A. $0.18 9. A merchant sold a barrel of flour for $25; 5 gals. of molasses for $1 pr. gal.; and 6 gals. of wine for $1 pr. gal. In payment, he received a load of wood. worth $22, and 2 bu. of wheat worth $13 pr. bu.; and the rest in money. How much money did he receive? A. $2.428, nearly. 10. From 8744 take 3631. From 927,33 take 17954. From 1693 take 4738. From 82937 take 332. From 334 take 23433. From 973 take 864 1008 999 830 11. From 72.345 take 63.1345. From 39.38463 take .27953. From 125.125125 take 25.025025. From 380. 613401 take 1.7834. From 830.595003 take 000004. From .00001 take .00000001. DIVISION. LXVI. 1. If four gallons of wine cost $8.24, what cost I gallon? We must divide by 4. The fourth part of 8 dollars is 4)8.24 2 dollars, and the fourth part of 24 cts., that is, 24 hundredths of a dollar is 6 cts.=6 hundredths=.06 2.06 2. At $2.06 pr. gal., how many gals. of wine may be bought with $8.24? Here we have decimals both in the divisor and dividend. But the 8.24 is 824 hundredths, and the 2.06 is 206 hundredths. -2.06)8.24(4 8.24 Dividing hundredths by hundredths will evidently give whole numbers, just as dividing pints by pints, quarts by quarts, or units by units, gives whole numbers. 824÷206-4, as seen above. From observation of these two examples, we see that when the divisor is a whole number, the quotient has just as many decimals as the dividend. And when the divisor has as many decimals as the dividend, the quotient is a whole number. 824 3. At $0.206 a yard, how many yds. of calico will $8.24 buy? Here the divisor has three decimal places. If .206)8.240(40 the dividend had as many, we should have a whole number quotient. But we may put a cypher on the right of the dividend, without altering its value. (§ LVIII.) The decimal places will then be equal, and the quotient. 40 will be whole numbers. 00 4. At $30.5 pr. hhd., how many hhds. of molasses can be bought for $76.25? 30.5)76.2,5(2.5 1525 1525 As the divisor has, here, one decimal place, I know that, until one decimal place in the dividend has been emploved, the quotient will be whole numbers, and afterwards, decimals. A comma is placed here, after the 2, to show the limit of whole numbers.. The pupil will find it convenient to employ it. 5. Divide .15 by 7.5. 7.5).1,50(0.02 I find here, that, taking one decimal place in the dividend will give no quotient figure. Hence, the quotient contains no whole numbers. In like manner, by taking two places, I find that the quotient contains no tenths. Í then annex a cypher, and obtain 2 in the quotient, which, of course, is 2 hundredths. Hence, the general principle, THE DECIMAL PLACES IN THE DIVISOR AND QUOTIENT TOGETHER, MUST BE JUST EQUAL TO THE DECIMAL PLACES IN THE DIVIDEND. 1 This results, likewise from the principles of multiplication; for the divisor X the quotient the dividend (§ xxv.), and hence their decimal places together ought to equal those of the dividend (§ LXII.) Hence, the rule, DIVIDE AS IN WHOLE NUMBERS, AND POINT OFF DECIMALS ENOUGH TO MAKE THE PLACES IN THE DIVISOR AND QUOTIENT TOGETHER, EQUAL TO THOSE IN THE DIVIDEND. NOTE. If there be not enough figures in the quotient, prefix cyphers to make cut the number. If the division is performed as in the above illustrations, the cyphers will be placed in the quotient during the division, which is best. A. 18.57 6. At $1 pr. lb., how many lbs. of indigo will $2717 buy? 7. If 1 gal. of wine cost $13, how many gals. will $143 buy? A. 18. 8. If I pay $31615 for 921 cords of wood, what is that a cord? A. $3. 9. If I pay $3531 for 12 lb. 15 oz. of tea, what is that a pound? A. $23. 10. At $31 a yd., how many yds. of broadcloth can I buy for $64, A. 199. 11. If 71% yds. of cloth cost $16871, what is that a yard? A. $225. 12. 1f 16 lb. 9 oz. of sugar cost $2, what is that pr. pound? 13. Divide 226.827 by 8.13. 14. Divide 1 by 1.25. 15. Divide 1 by 562.5. 16. Divide 3.464 by 2,706.25. 17. Divide 313.5 by 861.3. A. $1. A. 27.9. A. .8. A. .0017. A. .00128. 913.41 by 3,864. 21. Divide 298.7461 by 10. A. 29.87461. 101,001 by .00001. 21 by 269,843.21. By 100. A. 2.987461. NOTE. The above example shows, that dividing by 10 does not alter the significant figures, but only removes the decimal point one place towards the left. In like manner, dividing by 100, removes the separatrix two places to the left; by 1,000, three places, and so on. Hence, when the div sor is a unit, with cyphers annexed, REMOVE THE SEPARATRIX AS MANY PLACES TO THE LEFT AS THE DIVISOR CONTAINS CYPHERS, PREFIXING CYPHERS, IF NECESSARY. NOTE. If the divisor be any number with cyphers annexed, it may be treat ed as a composite number, one of whose factors is 10, 100, 1,000, or a similar number; and we may first divide by this factor, according to the above rule, and afterwards by the other. (§ XXIX.) § LXVII. As we know, (LXVI.) that the quotient will be whole numbers. until as many decimals are employed in the dividend as there are in the divisor, and that all figures, afterwards obtained in the quotient, are, of course, decimals, we can always tell what will be the place of the first quotient figure; that is, whether it will be hundreds, tens, units, tenths, hundredths, thousandths, &c. Now, for ordinary purposes, it has been remarked, we frequently do not need many decimal places. A shorter mode of division, in such cases, may be employed. 1. Divide 82.931453 by 78.124. COMMON METHOD. 78.124)82.931,453(1.061 78 12 4 480 745 120013 4 1889 CONTRACTION. 78.12 481 12 7 5 My divisor, I observe, has 78.12,4)82.93,1453(1.061 three decimal places. Taking three, likewise, in the dividend, I find the divisor will go, and therefore, conclude that there will be one place of whole numbers in the quotient. I wish to retain three decimal places in the quotient. The quotient, then, will consist of four figures. Now, though not perfectly accurate, it is suffi ciently so, to take four figures on the left of the divisor, and divide by them, instead of using the whole. Thus, I obtain one quotient figure. By the common method, I should now bring down another figure of the dividend. But it will be sufficiently accurate to shorten the divisor one figure; that is, cut off a figure from the right of it. For this makes the divisor ten times less, (§ Xxx.) as annexing a figure to the number divided, would make it ten times greater, nearly. (SXV.) In like manner, for the third figure of the quotient, I take off another figure from the divisor, instead of bringing down another from the dividend, and so on. In multiplying the divisor by the quotient figure, we must carry, as in multiplication, (SLXIII.) to the first figure of the product, for the nearest number of tens, from the product of the last neglected figure of the divisor. Thus, in the "above example, in multiplying by C, we should have 6X8=48, but to this we carry 1, because 6X1 (the last figure cut off,)=6, which is nearer 10, than 0. Hence, the first figure of the product is 9, instead of 8. Hence, the rule, I. TAKE FROM THE LEFT OF THE DIVISOR AS MANY FIGURES AS YOU WISH IN THE QUOTIENT, AND, FOR THE FIRST QUOTIENT FIGURE, DIVIDE BY THESE AS USUAL. II. FOR EACH FOLLOWING QUOTIENT FIGURE, DIVIDE THE LAST REMAINDER BY THE PRECEDING DIVISOR, DIMINISHED ONE FIGURE ON THE RIGHT |