10. A. Holmes sold 14 boxes of soap, each containing 24 pounds, at 9 cents a pound, and received for pay 63 barrels of ashes, each containing 3 bushels. What was allowed a bushel for the ashes? Ans. 16 cents. 11. M. Gardner sold 5 piles of brick, each containing 12 thousand, at 7 dollars a thousand, and was paid in wood, 3 ranges, at 4 dollars a cord. How many cords in each range? Ans. 35 cords. 12. A merchant exchanged 8 cases of shoes, each containing 60 pairs, at 75 cents a pair, for a certain number of casks of molasses, each containing 90 gallons, at 40 cents a gallon. How many casks did he get? Ans. 10 casks. CONTRACTIONS IN MULTIPLICATION. 86i A Contraction is the process of shortening any operation. 87. When the multiplier is 13, 14, etc., or 1 with a significant figure annexed. Ex. 1. Multiply 3126 by 14. Ans. 43764. riRST Operation. Second OPERATION. Ll the first Operation , 3126X14 3126 we multiply the multi 1 2 5 0 4 1 4 plicand by the 4 units 4 3 7 6 4 Ans. 4 3 7 6 4 Ans. write the prodUct under the multiplicand one place to the right. To this partial product we add the multiplicand, since, as it stands, it represents the product of the multiplicand by the 1 ten of the multiplier ; and obtai n43764,the answer required. In the second operation, we add in the multiplicand taken as the product by the 1 ten of the multiplier, as we multiply by the 4 units; thus, 6 X 4 = 24, of which we write down the 4 and carry the 2. 2 X 4 = 8, -j- 2 (carried) -\- 6 (from the multiplicand) = 16, of which we write down the 6 and carry the 1. 1X4 = 4, -4-1-42 = 7, which we write down. 3X4 = 12, -[-1 = 13 of which we write down the 3 and cany the 1. 3 -\- 1 = 4, which we write down; and have as the entire result 43764 as before. Rule. — Write the product by the units' figure of the multiplier under the multiplicand, one place to the right, and add them together. Or, Multiply each figure of the multiplicand by the units' figure of the multiplier, and, after the units' place, add in the preceding figure of the multiplicand. Examples. 2. Multiply 63013 by 17. Ans. 1071221. 3. Multiply 79245 by 19. 4. Multiply 32067812 by 16. Ans. 513084992. 88. When the multiplier is 101, 102, etc., or 1 with one ov more ciphers and a significant figure annexed. Ex. 1. Multiply 8107 by 103. Ans. 835021. Operation. We multiply by 3, the units' figure of the 8107 X103 multiplicand, and write the product under 2 4 3 2 1 the multiplicand, two places to the right, so that the multiplicand, as it stands over 8 6 O 0 I 1 Ans. thjs partial product, will represent the pro duct of the multiplicand by the 1 hundred of the multiplier; and, adding these, we obtain 835021, the result required. For the reason given, if the multiplier had been such as to have contained one more intervening cipher, we should have written the product by the units' figure three places to the right, and so on, one place farther to the right for every additional intervening cipher. Rule. — Write the product by the units' figure of the multiplier under the multiplicand, as many places to the right as there are in the multiplier intervening ciphers plus 1; and add them together. Examples. 2. Multiply 6651 by 108. Ans. 718308. 3. Multiply 111223 by 104. Ans. 11567192. 4. Multiply 2042 by 1009. 89i When the multiplier is 21, 31, etc., or 1 with a significant figure prefixed. Ex. 1. Multiply 3113 by 41. Ans. 127633. opeRation. We multiply by 4, the tens' figure of the 3 113X41 multiplier, and write the product under the 1 2 4 5 2 multiplicand, one place to the left, so that — the multiplicand, as it stands over this par1 2 7 6 3 3 Ans. t|al product, will represent the product of the multiplicand by the 1 unit of the multiplier; and, adding these, obtain the result required. Rule. — Write the product by the tens' figure of the multiplier under the multiplicand, one place to the left, and add them together Examples. 2. Multiply 13317 by 51. Ans. 679167. 3. Multiply 71389 by 21. 4. Multiply 12062 by 91. Ans. 1097642. 90. When the multiplier is 201, 301, etc., or 1 with one or more ciphers and a significant figure prefixed. Ex. 1. Multiply 14118 by 601. Ans. 8484918. Operation . We multiply by 6, the hundreds' 14118X60 1. figure of the multiplier, and write the 8 4 7 0 8 product under the multiplicand, two -~ places to the left, so that the multiplies 4 o 4 9 1 o Ans. cand, as it stands over this partial product, will represent the product of the multiplicand by the 1 unit of the multiplier; and adding these we have the answer required. Rule. — Write the product by the hundreds' figure of the multiplier under the multiplicand, as many places to the left as there are in the multiplier intervening ciphers plus 1; and add them together. m Examples. 2. Multiply 8360 by 7001. Ans. 58528360. 3. Multiply 10613 by 801. Ans. 8501013. 4. Multiply 91603 by 2001 91. When the multiple or multiplicand has a fraction annexed. Ex. 1. Multiply 426 by ty. By 8§. Ans. 3124; 3692. OPERATION. OPERATION. 42 6 42 6 U 8f 2 9 8 2 = Product by 1. 3408 = Product by 8. 14 2 = Product by J. 2 8 4 = Product by §. 3124= Product by 7£. 3 6 9 2 = Product by 8§. In multiplying 420 by 7|, we first obtain the product of 426 by 7, and then the product of 426 by J, and, adding these two partial products together, have 3124, the product by 7 J. In multiplying by J, we take one third of the multiplicand, by dividing it by 3; thus, 426 x i = 426-i-3 = 142. In multiplying 426 by 8§, we proceed as in the other cape, except in obtaining the product by the fraction; we take two thirds of the multiplicand, by taking one third of it 2 times; thus, 426 x i = (426 -f- 3) x 2 = 284. If the fraction had been annexed to the multiplicand instead of the multiplier, we then would have found the product of the fraction and multiplier, for a partial product, in like manner as above. Rule. — Multiply the fractional part and the whole number separately, and add the products. Examples. 2. Multiply 915 by 22§. Ans. 20496. 3. Multiply 1224} by 18. Ans. 22034f 4. If 69£ miles make 1 degree, how many miles are 180 degrees? Ans. 12450 miles. 92. When the multiplier is a convenient part of a number of tens, hundreds, or thousands. Note. — The following are some of the convenient parts often occurring as multipliers; 2£ =\ of 10; 3J = J of 10; 124 = { of 100; 16| = ' of 100; 25 = i of 100; 33j = £ of 100; 125 = j of 1000; 166£ = i of 1000; 250 = { of 1000; 333i = J of 1000. Ex. 1. Multiply 785643 by 25. Ans. 19641075. Operation. We multiply by 100, by annex 4)78564300 ing two ciphers to the multipli i n c A i n n e T> J . C£"id (Art. 65), and obtain a pro1 9 6 41 0 7 5 Product. ^ times » , a? it ?h£uM be, since 25, the multiplier, is only one fourth part of 100; we therefore take one fourth part of that product, by dividing by 4, for the true product. Upon the same principle, if the multiplier had been 3J, we should have annexed one cipher and divided by 3, or, if the multiplier had been 125, we should have annexed three ciphers and divided by 8. Hence the following Rule. — Multiply by the number of tens, hundreds, or thousands, oj which the multiplier is a part, and, of the product thus found, take the tame part. Examples. 2. Multiply 68056 by 12£. Ans. 850700. 3. Multiply 17924 by 2£. 4. Multiply 192378 by 16§. Ans. 3206300. 5. Multiply 12345678 by 125. Ans. 1543209750. 6. How much can be earned in one year of 313 working days, at 3£ dollars a day? 7. What will a farm containing 534 acres cost, at 33^ dollars an acre? 8. In a certain large field there are 771 rows of corn, each containing 250 hills; how many hills in all? Ans. 192750 hills. 9. From a port in Louisiana there were exported, in a given time, 9168 boxes of sugar, averaging 166§ pounds to a box; required the whole number of pounds. Ans. 1528000 pounds. 10. An agent has bought for the army, at different times, in the aggregate, 1993 horses, at an average price of 125 dollars each. How much did he pay for them in all? Ans. 249125 dollars. 11. How many are 333$ times 28044? Ans. 9348000. 93° When a part of the multiplier is a factor of another part. Ex. 1. Multiply 3263 by 568. Ans. 1853384. Operation. We regard the multiplier 3 2 6 3 Multiplicand. as separated into two parts, 5 6 8 Multiplier. 5G tens and 8 units, or 560 ———, , . -4- 8; of which the smaller 2 6 1 0 4 = Product by 8 units. part is ev;dentiy a factor 0f 1 8 2 7 2 8 = Product by 56 tens, the larger, since the 56 tens, 1 85 338 4 - Product by 568. or 560, is equal to ItensX 8. * We next multiply by the 8 units, obtaining the product for that part of the multiplier. Now, as this product is the same as that by the factor 8 of the other part of the multiplier, we multiply it by 7 tens, obtaining the product of the multiplicand by 8 X 1 (ens or 56 tens. These products of the parts, 560 and 8, added together, give the true product by 568. Rule. —, Multiply first by the smaller part of the multiplier; and then that partial product by a factor, or factors, of a larger part; and soon with all the parts. The sum of the several partial products will be the product required. Note. — Care must be taken in writing down the partial products to hav» the units of the different orders stand in their proper places for adding. Examples. 2. Multiply 112345678 by 288144486. Ans. 32371787641631508. |