times convenient to resort to a process similar to those explained below. How many are 12 times 743? Explanation. Since 12 equals 6 times 2, 12 times a number must be equal to 6 times 2 times the number. Applying this to the above example, the work may be written thus: Again, since 123 times 4, 12 times a number must be equal to 3 times 4 times the number. Applying this to the above example, the work may be written thus: 89163 × (4 × 743) = 12 X 743. Again, since 122 times 2 times 3, 12 times a number must be equal to 2 times 2 times 3 times the number. The work of the above, thus performed, would be written as follows: 8916 = 2 × (6 × 743) 12 × 743. Apply these principles to the following examples : — 1. How many are 879 X 18? 6427 × 42? 7. How many are 2377.17 cwt. 2 qr. 19 lb. 6 oz. 11 dr.X 63? 8. How many are 27 H, 83, 6 3, 2 Ə, 17 gr. × 45? 9. How many are 24 m. 6 fur. 27 rd. 4 yd. 1 ft. 7 in. × 42? 10. How many are 19 w. 5 d. 17 h. 38 m. 29 sec. X 36? 11. What will 12 acres of land cost at 349 dollars per acre? Solution. Since 12: = 2 times 6, 12 acres will cost 2 times The process would be; if 1 acre cost as much as 6 acres. $349, 6 acres will cost 6 times $349, which is found to be $2094, 2 times 6 acres, or 12 acres, will $2094. If 6 acres cost 2 times $2094, which is found to be $4188. The following is a convenient form for writing the work on the slate or blackboard: $349 cost of 1 acre. $2094= 66 $4188 66 6 acres. "C 12. How many bushels of corn are there in 18 bins, each bin containing 198 bushels? How many in 42 bins? in 16 bins? 13. How many pounds of cotton are there in 63 bales, each bale containing 437 pounds? How many in 15 bales? in 30 bales? 14. A man bought 36 bales of cloth, each bale containing 428 yards. How many yards did they all contain? He paid $753.28 for each bale. What did he pay for all? He sold it at $797.36 per bale. How much did he receive for it all? How much did he gain on the cloth? 15. A farmer had 24 stacks of hay, each stack containing 2 T. 14 cwt. 3 qr. 19 lb. How much did they all contain? If he should sell each stack for $36.274 how much would he sell them all for? 16. A jeweller bought of a California gold, each bar weighing 6 oz. 17 dwt. 22 gr. all weigh? He paid $136.75 per bar for cost him? merchant 42 bars of How much did they it. What did it all G. We can find the product of two numbers by multiplying one of them by the parts into which we choose to separate the other, and then adding the products thus obtained together. For instance; 8 times 7 6 times 7+ 2 times 7 3 times 7+ 5 times 77 times 7+ once 4 times 7+ 4 times 7 56. This principle, and the one illustrated in letter F., are ordinarily employed when the multiplier contains several denominations. We usually get 83 times a number, by adding together 80 times the number and 3 times the number. We get 647 times a number by adding together 600 times the number, 40 times the number, and 7 times the number. We get 8009 times a number by adding together 8000 times the number and 9 times the number. It can make no difference which part of a number we multiply by first, provided we multiply by all its parts, yet for the sake of uniformity it may be well generally to begin with the lowest denomination. Suppose that we are required to find the product of 869 by 75. We may find the product of 869 by 5 in the usual method. We then wish to find the product of 869 by 70. Now as 70 =7X 10, the product obtained by multiplying by 70 must be 7 times the product obtained by multiplying by 10. But 869 multiplied by 10=869 tens, 8 thousands, 6 hundreds, and 9 tens, which, multiplied by 7, will give for a product 70 times 869. This will be found to be 60830, and adding it to the product of the multiplication by 5, we have 65175, as the product of 869 by 75. = A good method of writing the work is given below: Since the product of the multiplication by the units is sufficient to fix the place of the figures in the subsequent products, the zero at the right of the second product need not have been written. The product would then stand: Examples in which the multiplier consists of more than two figures are performed in a similar way. 2. What is the product of 78.069 × 285? 78.069 Multiplicand. 285. = Multiplier. 3. What is the product of 80.276 x 39? of 30678 X 427? 4. What is the product of 298.794 X 148? of 5.796 X 238? 5. What is the product of 273.986 .27? of 45.718 X 4.32? 6. What is the product of 494.3 X .0078? of .00876 .074? 7. What is the product of 2.3879 x 27.41? of .478 X 39.65? 8. A piano-forte dealer sold in one month 57 piano-fortes for $337.50 apiece. How much did he receive for them? 9. Bought 337 bales of silk goods at $447.86 per bale. What was the cost of the whole? 10. What will be the cost of building 43.7 miles of rail-road at $29756.84 per mile? 11. If a man works on an average 2745 hours per year, how many hours will he work in 9.76 years? 12. A country trader bought 47 lb. of tea at 38 cents per lb., 28 lb. at 59 cents per lb., 7 cwt. of sugar, at $6.833 per cwt., and 9 kegs of raisins at $4.42 per keg. What was the amount of his purchase? H. When the multiplier consists of more than one denomination, much labor in writing figures may be saved by applying the principles illustrated in the following examples: What is the product of 8356 multiplied by 79? - We It is evident that, in performing the required multiplication, we shall obtain units by multiplying 6 units by 9 units. shall obtain tens by multiplying 5 tens by 9, and 6 units by 7 tens, and we may have some from the product of the units. We shall obtain hundreds by multiplying 3 hundreds by 9, and 5 tens by 7 tens, and we may have some from the former products. We shall obtain the other denominations in a similar manner. We may then proceed thus, writing the figures of each denomination as usual: WRITTEN WORK. 8356. Multiplicand. 660124. Product. Explanation. 9 times 6 units = 54 units. We write 4 units. 9 times 5 tens 45 tens, +5 tens (from the product of the units) 50 tens, +7 tens times 6, or 42 tens, = 92 tens = 9 hundreds and 2 tens. We write 2 tens. 9 times 3 hundreds 27 hundreds, + 9 hundreds (from the previous product) = 36 hundreds, +7 tens times 5 tens, or 35 hundreds, =71 hundreds = 7 thousand and 1 hundred. We write 1 hundred. 9 times 8 thousands 72 thousands, +7 thousands (from the previous product)=79 thousands, +7 tens times 3 hundreds, or 21 thousands, 100 thousands = 10 ten-thousands. We write O in the thousands' place of the product. 7 tens times 8 thousands = 56 ten-thousands, +10 ten-thousands (from the previous product) = 66 ten-thousands, which we write. The multiplication being now completed shows the answer to be 660124. The following will exhibit the method to the eye : — When the pupil clearly understands how the different denominations of the product are obtained, it will be well to shorten the work somewhat by omitting to name the denomination of the factors used: Thus, 9 x 6 = 54 units 5 tens and 4 units. 9X5= 45,550, +7 × 6, or 42, — = 92. 92 tens 9 hundreds and 2 tens. |