Division-Simple Numbers. 130. Find the quotient of 576 divided by 4. "Short Division." 4) 576 144 Explanation No. 1. One fourth of 5 hundred is 1 hundred with a remainder of 1 hundred; 1 hundred equals 10 tens; 10 tens plus 7 tens are 17 tens. One fourth of 17 tens is 4 tens with a remainder of 1 ten; 1 ten equals 10 units; 10 units plus 6 units are 16 units. One fourth of 16 units is 4 units. Hence one fourth of 576 is 144. Explanation No. 2. Four is contained in 5 hundred, 1 hundred times, with a remainder of 1 hundred; hundred equals 10 tens; 10 tens and 7 tens are 17 tens. Four is contained in 17 tens, 4 tens (40) times with a remainder of 1 ten; 1 ten equals 10 units; 10 units and 6 units are 16 units. Four is contained in 16 units 4 times. Hence 4 is contained in 576, 144 times. 131. Find the quotient of 8675 divided by 25. 'Long Division.” 25)8675(347 75 117 100 175 175 Explanation No. 1. One twenty-fifth of 86 hundred is 3 hundred, with a remainder of 11 hundred; 11 hundred equal 110 tens. 110 tens plus 7 tens equal 117 tens. One twenty-fifth of 117 tens is 4 tens, with a remainder of 17 tens; 17 tens equal 170 units; 170 units plus 5 units equal 175 units. One twenty-fifth of 175 units is 7 units. Hence one twenty-fifth of 8675 is 347. TO THE PUPIL.—Make another explanation of this process similar to Explanation No. 2, under Art. 130. Division-Decimals. 133. Find the quotient of 785.6 divided by .5. Operation .5)785.6'5 1571.3 Explanation. First place a separatrix (v) after that figure in the dividend that is of the same denomination as the right hand figure of the divisor—in this case, after the figure 6. Then divide, writing the decimal point in the quotient when, in the process of division, the separatrix is reached— in this case, after the figure 1. It was required to find how many times 5 tenths are contained in 7856 tenths. 5 tenths are contained in 7856 tenths 1571 times. There are yet 15 hundredths to be divided. 5 tenths are contained in 15 tenths, 3 times; in 15 hundredths, 3 tenths of a time. NOTE.-By holding the thought for a moment upon that part of the dividend which corresponds in denomination to the divisor, the place of the decimal point becomes apparent at once. 5 apples are contained in 7856 apples, 1571 times. 134. Solve and explain the following problems with special reference to the placing of the decimal point : Division-United States Money. 135. Divide $754.65 by $.27. Operation. 8.27)$754.65 (2795 54 214 189 256 243 135 135 Explanation. This means, find how many times 27 cents are contained in 75465 cents. 27 cents are contained in 75465 cents, 2795 times. PROBLEM. At 27 a bushel, how many bushels of oats can be bought for $754.65? As many bushels can be bought, as $.27 is contained times in $754.65. It is contained 2795 times. Therefore, 2795 bushels can be bought. DENOMINATE NUMBERS. 138. Divide 46 rd. 12 ft. 8 in. by 4. Operation. 4)46 rd. 12ft. 8 in. Explanation. This means, find 1 fourth of 46 rd. 12 ft. 11 rd. 11 ft. 5 in. 8 in. One fourth of 46 rd. is 11 rd. with a re mainder* of 2 rd.; 2rd. equal 33 ft. 33 ft. plus 12ft. equal 45 ft. One fourth of 45 ft. equals 11 ft. with a remainder of 1 ft.; 1 ft. equals 12in.; 12 in. plus 8 in. equals 20 in. One fourth of 20 in. equals 5 in. One fourth of 46 rd. 12 ft. 8 in. equals 11 rd. 11 ft. 5 in. PROBLEM. The perimeter of a square garden is 46 rd. 12 ft. 8 in. How far across one side of it? 139. MISCELLANEOUS. Tell the meaning of each of the following, solve, explain, and state in the form of a problem the conditions that would give rise to each number process. 1. Multiply 64 rd. 14 ft. 6 in. by 8. 2. Divide 37 rd. 15 ft. 4 in. by 5. 3. Divide $675.36 by $48. 4. Divide $675.36 by 48. 5. Divide $675.36 by .48 6. Divide $675.36 by $4.8 7. Divide $675.36 by 4.8 8. Divide $675.36 by $.48 9. Multiply $356.54 by .36 10. Multiply $356.54 by 3.6 11. Multiply $356.54 by 36. 12. Multiply $275.56 by 2.25. 13. Multiply $275.56 by 21. 14. Can you multiply by a number of dollars? 15. Can you divide by a number of dollars? * The word remainder in this connection suggests incomplete division. When the division is complete there can be no remainder. 1. Observe that in the above examples we divide each term of the dividend by the divisor. 2. Prove Nos. 1 and 3, by (1) reducing each dividend to its simplest form, (2) dividing it so reduced, by the divisor, and (3) comparing the result with the quotient reduced to its simplest form. 3. Verify No. 2, by letting a = 3, and b = 4. Verify No. 4, by letting a = 3, b = 5, and c = 7. Observe that to divide one algebraic term by another we must find the quotient of the co-efficients and the difference of the exponents. 6. Verify Problem 5, by letting a = 3, and 6='5. |