26. Find the sum of the above dividends, and multiply it by the sum of the divisors. 27. A man gave 324 dollars to some boys, giving 6 dollars to each; how many boys were there? SOLUTION. If to receive 6 dollars requires one boy, to receive 324 dollars requires as many times one boy as 6 is contained times in 324, which are 54. Therefore, etc. OPERATION. Dol. Boy. 324 324.. = 54. 6 6)324 54 Ans. 63 books. 28. If a book cost 4 dollars, how many books, at the same rate, can you buy for 252 dollars? 29. There are 3 feet in one yard; how many yards in 291 feet? Ans: 97 yards. 30. There are 8 quarts in one peck; how many pecks are there in 1728 quarts? Ans. 216 pecks. 31. There are 7 days in one week; how many weeks in 364 days? Ans. 52 weeks. 32. If one sheep cost 9 dollars, how many sheep can you buy for 1935 dollars? Ans. 215 sheep. 33. If there are 12 pence in one shilling, how many shillings are there in 571836 pence? Ans. 47653. LONG DIVISION. 49. Long Division is the method of dividing when the partial dividends are written. It is generally used when the divisor is greater than twelve. 1. Divide 5848 by 23. OPERATION. 23)5848(254 46 124 115 98 92 SOLUTION.-23 is not contained in 5 thousands any thousands times, hence there are no thousands in the quotient. 5 thousands and 8 hundreds equal 58 hundreds. 23 is contained in 58 hundreds 2 hundreds times; 2 hundreds times 23 equals 46 hundreds, which subtracted from 58 hundreds leaves 12 hundreds. 12 hundreds and 4 tens equal 124 tens. 23 is contained in 124 tens 5 tens times; 5 tens times 23 equals 115 tens, which subtracted from 124 tens leaves 9 tens. 9 tens and 8 units equal 98 units. 23 is contained in 98 units 4 times; 4 times 23 equals 92, subtracting there is a remainder of 6, which will not contain 23; hence the quotient is 2 hundreds, 5 tens, and 4 units, or 254, with & remainder of 6. From the above solution we derive the following -- 6 RULE.-I. Draw curve lines at both sides of the dividend, and place the divisor at the left. II Divide the number expressed by the least number of figures on the left that will contain the divisor, and place the quotient on the right. III. Multiply the divisor by this quotient, write the product under the partial dividend, and subtract, and to the remainder annex the next figure of the dividend. IV. Divide as before, and thus continue until all the figures of the dividend have been used. V. If any partial dividend will not contain the divisor, place a cipher in the quotient, annex the next figure of the dividend, and proceed as before. PROOF.-Multiply the quotient by the divisor, and add the product to the remainder; if the work is right, the result will equal the dividend. NOTES.-I. The pupils will notice that there are five operations: 1st. Write the numbers; 2d. Divide; 3d. Multiply; 4th. Subtract; 5th. Bring down. II. Pupils often have difficulty in finding the correct quotient figure; this difficulty can be greatly diminished by attention to the following suggestions: 1st. Notice how often the left hand figure of the divisor is contained in the figure or figures of the partial dividend, as far from the right hand figure as the left hand figure in the divisor is from the right hand figure. 2d. If, when we multiply, the product is greater than the partial dividend, the quotient figure must be diminished. 3d. If, when we subtract, the remainder is greater than the divisor, the quotient figure must be increased. III. We commence at the left to divide, so that the remainder can be united to the number of units of the next lower order, giving a new partial dividend. The sign, is used to denote a remainder. 24. Find the sum of the above dividends and divisors. 62. Divide 4223745376 by 23456. 63. Divide 4542636786 by 12376. 64. Divide 170627676887 by 413071. 65. Divide 129652565329 by 360073. 66. Find the sum of the dividends and in each of the previous columns. PROBLEMS IN DIVISION. Ans. 180071. Rem. 1234. Rem. 25846. Ans. 360073. also of the divisors 50. CASE I. To find the number of equal parts of a number. 1. At 25 dollars each, how many cows can be bought for 575 dollars? SOLUTION. If 25 dollars will buy one cow, 575 dollars will buy as many cows as 25 dollars are contained times in 575 dollars, which are 23. Therefore, etc. OPERATION. Dol. Cow. 575 .. =23, Ans. 25 2. In one hogshead there are 63 gallons; how many hogsheads in 15435 gallons? Ans. 245. 3. How many horses can you get for 1824 dollars, at the rate of 152 dollars each? Ans. 12 horses. 4. If a boat sails 25 miles an hour, how long will it be in sailing 1800 miles? Ans. 72 hours. 5. How many years must a person labor to earn 13140 dollars, at the rate of 730 dollars a year? Ans. 18 yrs. 6. How many acres of land can you purchase for 11696 dollars, at the rate of 86 dollars an acre? Ans. 136. Ans. 188 oxen. 7. How many cows, at 37 dollars each, can be bought for 74 horses, at 150 dollars each? Ans. 300 cows. 8. How many oxen, at 54 dollars each, can be bought for 108 mules, at 94 dollars each? 9. A labors 72 weeks, at the rate of 14 dollars a week; how much wheat at 42 cents a bushel will pay him? Ans. 2400. 10. The circumference of the earth is 25000 miles ; how long would it take a vessel to sail around it, going at the rate of 125 miles per day? Ans. 200 days. 11. The distance from the earth to the sun is 95000000 miles; |