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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 cords in each range? ? Ans. 35 cords.

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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.

86. 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.

FIRST OPERATION.

3 12 6 X 14

12504

4 37 64 Ans.

=

SECOND OPERATION.

3126
14

437 64 Ans.

Ans. 43764.

In the first operation, we multiply the multiplicand by the 4 units of the multiplier, and 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 obtain 43764, 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 8,2 (carried) +6 (from the multiplicand) 16, of which we write down the 6 and carry the 1. 1X4=4,+1+ 2 = 7, which we write down. 3 X 4 12, +1= 13 of which we write down the 3 and carry the 1. 3+1 4, which we write down; and have as the entire result 43764 as before.

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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.

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88. When the multiplier is 101, 102, etc., or 1 with one or more ciphers and a significant figure annexed.

Ex. 1. Multiply 8107 by 103.

OPERATION.

8107 X 103
24321

835021 Ans.

Ans. 835021.

We multiply by 3, the units' figure of the multiplicand, and write the product under the multiplicand, two places to the right, so that the multiplicand, as it stands over this 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.

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EXAMPLES.

2. Multiply 6651 by 108.

3. Multiply 111223 by 104.

4. Multiply 2042 by 1009.

Ans. 718308.

Ans. 11567192.

89. When the multiplier is 21, 31, etc., or 1 with a signifi

cant figure prefixed.

Ex. 1. Multiply 3113 by 41.

OPERATION.

3 1 1 3 × 41 12452

Ans. 127633.

We multiply by 4, the tens' figure of the multiplier, and write the product under the multiplicand, one place to the left, so that the multiplicand, as it stands over this partial product, will represent the product of the multiplicand by the 1 unit of the multiplier; and, adding these, obtain the result required.

127 6 3 3 Ans.

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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.

3. Multiply 71389 by 21.

4. Multiply 12062 by 91.

90.

Ans. 679167.

Ans. 1097642.

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.

OPERATION.

1 4 1 1 8 X 6 0 1. 84708

8 48 49 18 Ans.

multiplicand by the 1 unit have the answer required.

RULE.

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Ans. 8484918.

We multiply by 6, the hundreds' figure of the multiplier, and write the product under the multiplicand, two places to the left, so that the multiplicand, as it stands over this partial product, will represent the product of the of the multiplier; and adding these we

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.

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91. When the multiplier or multiplicand has a fraction annexed.

Ex. 1. Multiply 426 by 7. By 83. Ans. 3124; 3692.

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In multiplying 426 by 73, we first obtain the product of 426 by 7, and then the product of 426 by, and, adding these two partial products together, have 3124, the product by 7. In multiplying by, we take one third of the multiplicand, by dividing it by 3; thus, 426 X 426 ÷ 3 = 142. In multiplying 426 by 83, we proceed as in the other case, 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 × (426 ÷ 3) × 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%.

3. Multiply 12244 by 18.

Ans. 20496.

Ans. 220344.

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 = 4 of 10; 3 = § of 10; 12 = of 100; 163 = 1 of 100; 25 of 100; 33 of 100; 125 of 1000; 1663 = of 1000; 250 =

of 1000; 333

-

of 1000.

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Ex. 1. Multiply 785643 by 25.

OPERATION.

4) 785 64300

19641075 Product.

Ans. 19641075.

We multiply by 100, by annexing two ciphers to the multiplicand (Art. 65), and obtain a product 4 times as large as it should 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 31, 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, of which the multiplier is a part, and, of the product thus found, take the same part.

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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 1663 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 3334 times 28044?

Ans. 9348000.

93. When a part of the multiplier is a factor of another

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Ans. 1853384. We regard the multiplier as separated into two parts, 56 tens and 8 units, or 560 +8; of which the smaller part is evidently a factor of the larger, since the 56 tens, or 560, is equal to 7 tens × 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 × 7 tens 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 so on 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 have the units of the different orders stand in their proper places for adding.

EXAMPLES.

2. Multiply 112345678 by 288144486.

Ans. 32371787641631508.

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