146. One of the equal parts of a number is an aliquot part of that number; e.g. 8 oz. is an aliquot part of 16 oz. because 8 oz. is of 16 oz.; 163 cents is an aliquot part of 100 cents because 163 cents = of 100 cents.

Find the number of cents in $1; 8}; 81; $}; $}; $}; $}; $16; 81; $26.

1

The answers you have given are all what kind of parts of a dollar?

Prove the correctness of the following table :

This table should be committed to memory like the multiplication table, because its use will shorten many problems; e.g. 33 books, at $.163, each, will cost 33 × 8} = $5}.

When handkerchiefs are 121⁄2¢ apiece, $3 will buy as many handkerchiefs as $3 $1, or $3 x = 24 handkerchiefs. Ans.

3. 20 cents are what part of a dollar? 40 cents?

80 cents? Which of these is an aliquot part of $1?

60 cents?

4. Mention three aliquot parts of 12; two aliquot parts of 10; five aliquot parts of 64.

5. Give four numbers of which 81 is an aliquot part.

6. What is the cost of 28 pineapples when they are bought at the rate of $.142 apiece?

7. At $.331 a pound how many pounds of butter will $5 buy? 8. A man bought five dozen cans of corn at the rate of 81 cents apiece. What did they cost?

148. Written

1. Find the cost of the following:

a. 166 pounds of pork at 12 cents a pound.

b. 248 lb. of veal at 163 cents a pound.

c. 148 boxes of strawberries at 25 cents a box.
d. 250 lb. of butter at 371 cents a pound.
e. 150 lb. of honey at 25 cents a pound.

f. 640 bars of soap at 61 cents a bar.

g. 960 dozen of eggs at $.163 a dozen.

h. 32 yd. of dress goods at $.331 a yard.

i. 328 grammar school arithmetics at $.621 apiece.

j. 656 steel shovels at $.871 each.

2. At $.331 a yard, how many yards of linen can be bought for $150?

3. How many bushels of barley can be bought for $624, at $.75 a bushel?

4. At $.66% each, how many pocket knives can be purchased for $64?

5. When butter is 25 cents a pound, how many pounds can be bought for $650?

6. How many articles, at 144 cents each, can be purchased for $154?

7. At 87 cents each, how many books can be bought for $1456?

8. How many boxes of berries can be bought for $250, at 163 cents a box?

9. At $.621 each, how many pairs of gloves can be bought for $120?

10. A man bought potatoes at $.621 a bushel and sold them at $.87 a bushel. His profit was $160. How many bushels were sold?

11. A dealer spent $120 for chickens, and the same amount for ducks. The chickens cost him 163 cents, and the ducks 121 cents a pound. How many more pounds of ducks than chickens did he purchase?

SPECIAL CASES IN MULTIPLICATION

149. I. To multiply a number by 10 or a power of 10. Each removal of a figure one place to the left multiplies its value by 10.

Therefore, if the multiplicand is an integer, annex as many ciphers as there are ciphers in the multiplier; if the multiplicand is a decimal, move the decimal point as many places to the right as there are ciphers in the multiplier.

This is the same as moving all the figures to the left.
II. To multiply a number by 25.

25 = 100 ÷ 4

Therefore, multiply the given number by 100 and divide the product by 4. (Apply Case I in multiplying by 100.)

III. To multiply a number by 125.

125 1000 ÷ 8

Therefore, multiply the given number by 1000 and divide the product by 8. (Apply Case I in multiplying by 1000.) IV. To multiply a number by :

a. .333, multiply the given number by . b. .25, multiply the given number by 1. c. .16, multiply the given number by . d. .142, multiply the given number by 4. e. .125, multiply the given number by . V. To multiply a number by 99.

Therefore, multiply the given number by 100 and subtract the multiplicand from the product thus obtained.

How can we multiply a number by 999?

VI. To multiply by a number having one or more ciphers at the right.

Multiply by the significant figures of the multiplier, and annex to the product thus obtained, as many ciphers as there are in the multiplier. Explain.

SPECIAL CASES IN DIVISION

150. I. To divide a number by 10 or a power of 10.

Each removal of a figure one place to the right divides its value by 10.

Therefore, if the dividend is an integer, point off as many decimal places as there are ciphers in the divisor; if the dividend is a decimal, move the decimal point as many places to the left as there are ciphers in the divisor.

This is the same as moving all the figures to the right.

II. To divide a number by 25.

A number divided by 100 equals the number multiplied by , or the number multiplied by 4 and divided by 100. Therefore, multiply the given number by 4, and divide the product by 100. (Apply Case I in dividing by 100.)

III. To divide a number by 125.

Multiply the given number by 8, and divide the product by 1000. Explain.

IV. To divide a number:

a. By 331, point off two decimal places and multiply by 3. b. By 163, point off two decimal places and multiply by 6. c. By 142, point off two decimal places and multiply by 7. d. By .333, divide by 1. By .163? By .125? By .142? By 2.5? By .111?

V. To divide by a number with one or more ciphers at the right.

Point off in the dividend as many decimal places as there are ciphers at the right of the divisor, then divide by the remaining figures.

Explain.

In the following examples, find results by the methods given in sections 149 and 150.