Decimal Fractions
Differ in form from common fractions, in not having a written denominator; and from whole numbers, by having the decimal point (.) prefixed; which also separates the integral part from the decimal. The word decimal is derived from the Latin word decem, which signifies ten. The denominator of a decimal is always 10, or some power of 10, as 100, 1000, etc.
A Complex Decimal is a decimal with a common fraction at the right, as,.121⁄2.
A Mixed Decimal is a whole number with a decimal fraction to its right, as, 34.5.
The denominations of United States money are based on the decimal system—the dollar occupying the unit’s place, the dime the tenth’s place, the cent the hundredth’s place, and the mill the thousandth’s place.
The rules given for addition, subtraction, and so on, also apply to decimals.
Addition in Decimals
Example: 27.295 +.0287 + 591.68 + 9.1846.
| 27 | .295 | |
| .0287 | ||
| 591 | .68 | |
| 9 | .1846 | |
| 628 | .1883 | Ans. |
Write the numbers so that the same powers of 10 come under one another, or, what is the same thing, write the numbers so that the decimal points come under one another. Then, adding the ten-thousandths first, 6, 13, carry 1, etc.
Subtraction in Decimals
Example: Subtract.07295 from 21.651.
| 21 | .651 | |
| .07295 | ||
| 21 | .57805 | Ans. |
Write the first number under the second, so that the point comes under the point. Remember that we may consider there are 0’s above the 9 and 5, since in 21.651 there are no ten-thousandths and no hundred-thousandths.
Say, mentally 5 and 5 make 10, carry 1.
Say, mentally 10 and 0 make 10, carry 1.
Say, mentally 3 and 8 make 11, carry 1, etc.
Multiplication in Decimals
Rule.—Multiply as in whole numbers, and point off from the right of the product as many places as there are decimal places in both multiplier and multiplicand—prefixing ciphers if necessary.
Example 1: Multiply 87.432 by 564.
| 87 | .432 | |
| 564 | ||
| 43716 | .0 | |
| 5245 | .92 | |
| 349 | .728 | |
| 49311 | .648 | Ans. |
Place the multiplier so that its unit’s digit comes under the right-hand digit of the multiplicand. Then place the first figure of each product underneath the multiplying digit. The decimal point of the answer will then be directly under the decimal point of the multiplicand.
Example 2: Multiply 31.56 by 5.49.
| 31.56 | ||
| 5 | .49 | |
| 157.80 | ||
| 12.62 | 4 | |
| 2.84 | 04 | |
| 173.26 | 44 | Ans. |
As before, place the unit’s figure of the multiplier—that is, the 5—under the right-hand digit of 31.56, and proceed as above.
Note.—The number of decimal places in the product will always be equal to the sum of the number of decimal places in the multiplier and the multiplicand. Thus, in Example 2, there are two places of decimals (i.e. two figures to the right of the point) in 31.56, and two places of decimals in 5.49; and we found 2 + 2 = 4 places in the product 173.2644.
To multiply a decimal by 10, 100, etc.
Rule.—Remove the (.) as many places to the right as there are ciphers in the multiplier.
| Work: | 8.75 | × | 10 | = | 87. | 5 |
| 8.75 | × | 100 | = | 875. | ||
| 8.75 | × | 1000 | = | 8750. |
Division of Decimals
Rule.—Divide as in whole numbers, annexing ciphers to the dividend, if necessary; then point off from the right of the quotient as many places as the decimal places in the dividend exceed those in the divisor—prefixing ciphers if necessary.
(a) Division of a decimal by a whole number.
Example 1: Divide 18.2754 by 4.
| 4 | ) | 18.2758 | |
| 4.5688 | 5 |
We divide 4 into 18 (units) and have 4 (units) quotient and 3 units remainder. Since the 4 is the unit’s figure of the quotient, we write the decimal point immediately after it. Then, the 2 units remainder and the 2 tenths of the dividend make 22 tenths to be divided by 4, and so on. Having reached the 4 (ten-thousandths) of the dividend, we find 8 (ten-thousandths) quotient and 2 remainder. This remainder is 20 hundred-thousandths, which when divided by 4 gives 5 (hundred-thousandths) and no further remainder.
Example 2: Divide 18.2758 by 11.
| 11 | ) | 18.2758 | |
| 1.6614 | 3636 |
Here we find the digits 3, 6 repeated indefinitely in the quotient. Decimals of this sort will be fully considered later.
Example 3: Divide 354.43 by 184.
| 184 | ) | 3 | 5 | 4. | 4 | 3 | ( | 1.92625 Ans. | ||
| 1 | 7 | 0 | 4 | |||||||
| 4 | 8 | 3 | ||||||||
| 1 | 1 | 5 | 0 | [16] | ||||||
| 4 | 6 | 0 | ||||||||
| 9 | 2 | 0 | ||||||||
Here we find the first figure of the quotient is obtained by dividing 184 into 354 units. Having now reached the decimal point in the dividend we also put the decimal point in the answer, and go on as before.
[16] At this stage there is a remainder 115 hundredths. We bring down 0 from the dividend, and obtain 1150 thousandths, etc.
(b) Division of a decimal.
Example 4: Divide 10.6603 by 7.85.
Thus:
| 785 | ) | 1 | 0 | 6 | 6. | 0 | 3 | ( | 1.358 Ans. | ||
| 2 | 8 | 1 | 0 | ||||||||
| 4 | 5 | 5 | 3 | ||||||||
| 6 | 2 | 8 | 0 | ||||||||
Here 7.85 is 785 hundredths, and 10.6603 is 1066.03 hundredths; so that the required quotient is obtained by dividing 1066.03 by 785.
Therefore, to divide by a decimal, move the point as many places to the right as will make the divisor a whole number; move the point in the dividend the same number of places to the right. Then proceed as in Example 3.
Example 5: Divide 176.4 by.00012.
| 12 | ) | 17640000 |
| Ans. | 1470000 | |
Here, to make the divisor a whole number, we have to move the point 5 places. Therefore we also move the point 5 places to the right in the dividend, first writing enough 0’s after the 176.4 to enable us to do so.
To divide a decimal by 10, 100, etc.
Rule.—Remove the (.) as many places to the left as there are ciphers in the divisor.
| Work: | ||||||
| 62.5 | ÷ | 10 | = | 6 | .25 | |
| 62.5 | ÷ | 100 | = | .625 | ||
| 62.5 | ÷ | 1000 | = | .0625 | ||
Expression of decimal fractions as common fractions.
Example: Express 5.375 as a common fraction.
Example:.375 = 375 thousandths.
Therefore 5.375 = 53751000 = 538 Ans.
Rule.—Take the digits of the decimal for numerator; for the denominator put down 1 followed by as many ciphers as there are digits in the decimal. Reduce this fraction to its lowest terms.
Expression of common fractions as decimals.
We have seen that a common fraction represents the quotient of the numerator divided by the denominator. Therefore, to convert a common fraction to a decimal fraction, we divide the numerator by the denominator.
Example: Express 332 as a decimal.
| 4 | ) | 3 | .0 | |
| 8 | ) | .75 | ||
| .09375 | Ans. | |||
It will be found in many cases that there is always a remainder, so that the quotient can be continued indefinitely.
Circulating Decimals
The learner has already discovered that some common fractions cannot be changed to exact decimal fractions, as—
1⁄3 =.33333 on to infinity.
2⁄3 =.66666 on to infinity.
7⁄33 =.212121, etc.
These decimals are known as Circulates, Recurring or Circulating decimals.
The part which recurs is called the Repetend.
This is marked by putting a dot over the first and last figures of it. For instance, if we write the 21 in the last case above, this way: 2̊1̊, it indicates that, if written out, the result would be 21212121, etc., on to infinity.
Where a circulating decimal occurs in work, it is best to reduce it to a common fraction. If need be, it may be expressed in the result as a circulate to any number of decimal places.
To change a pure circulate to a common fraction.
Rule.—Omit the (.) and write the figures of the repetend for the numerator, and as many 9’s for the denominator as there are places in the repetend.
Examples: Change the pure circulates.3̊,.2̊7̊,.1̊42857̊, to common fractions.
.3̊, (39 = 13) Ans. 1⁄3.
.2̊7̊, (2799 = 311) Ans. 3⁄11.
.1̊42857̊, (142857999999 = 17) Ans. 1⁄7.
To change a mixed circulate to a common fraction.
Rule.—From the whole decimal subtract the finite part, and make the remainder the numerator. For the denominator, write as many 9’s as there are figures in the repetend, and annex as many 0’s as there are finite places.
Example: Change the mixed circulates.16̊ and.416̊ to common fractions.
16 - 1 = 15, 1590 = 16. Ans. 1⁄6.
416 - 41 = 375, 375900 = 512. Ans. 5⁄12.
To add, subtract, multiply and divide circulates, reduce them to common fractions, then apply the respective rules.
Short Methods in Merchandising
When one of the numbers is an aliquot part of 100, the process of multiplication and division can often be very much shortened, as shown below.
Find cost of 27 yards of goods at 162⁄3c ($1⁄6) per yard. At $1 per yard, 27 yards cost $27; at $1⁄6, (27 ÷ 6), $41⁄2. Ans. $41⁄2.
Find cost of a bale of cotton, 528 pounds at 81⁄3c ($1⁄12) per pound. At $1 per pound, 528 pounds cost $528; at $1⁄12 (528 ÷ 12) $44. Ans. $44.
Find cost of 1845 pounds of iron, at 31⁄3c ($1⁄30) per pound. Take 1⁄30 of 1845, since 31⁄3c is 1⁄30 of $1. (1845 ÷ 30 = 611⁄2). Ans. $611⁄2.
Find cost of 16 pounds of butter at 371⁄2c ($3⁄8) per pound. Here we take 3⁄8 of 16. Say 1⁄8 of 16 is 2, and 3⁄8 is (2 × 3) 6. Or say 3 times 16 is 48, and 1⁄8 of 48 is 6. Ans. $6.
Find cost of 171⁄2 bushels of apples at 75c ($3⁄4) per bushel. The shortest way to find 3⁄4 of $17.50 is to diminish it by 1⁄4 of itself.
| 4 | ) | 17 | .50 | at $1 | |
| 3 | .371⁄2 | at $1⁄4 | |||
| 13 | .121⁄2 | at $3⁄4 | |||
Ans. $13.121⁄2.
At 61⁄4c per pound how much sugar will $5 buy? As 61⁄4c is 1⁄16 of $1, evidently each dollar will buy 16 pounds. Ans. 80 pounds.
In multiplying by a fraction, write the quantity in a line with the numerator and cancel common factors.
Find cost of 72 yards of carpet, at 871⁄2c ($7⁄8) a yard. Cancel 8, also 72 and write 9 instead. Ans. $63.
78 × 72 9 = 63
Of 28 pounds of coffee, at 183⁄4c ($3⁄16) per pound. Cancel 28 and 16, write 7 and 4. Ans. $51⁄4.
316 ×4 28 7 = 214 or 51⁄4
At 662⁄3c ($2⁄3) per bushel, how many bushel of wheat will $34 buy? Ans. 51 bushel.
32 × 34 17 = 51
In division, invert terms of fraction.
How much syrup, at 412⁄3c ($5⁄12) per gallon can be bought for $15? Ans. 36 gallons.
125 × 15 3 = 36
Table of Aliquot Parts of 100
| 3 | 1⁄3 | is | 1⁄30 |
| 6 | 1⁄4 | is | 1⁄16 |
| 8 | 1⁄3 | is | 1⁄12 |
| 12 | 1⁄2 | is | 1⁄8 |
| 16 | 2⁄3 | is | 1⁄6 |
| 18 | 3⁄4 | is | 3⁄16 |
| 20 | is | 1⁄5 | |
| 25 | is | 1⁄4 | |
| 31 | 1⁄4 | is | 5⁄16 |
| 33 | 1⁄3 | is | 1⁄3 |
| 37 | 1⁄2 | is | 3⁄8 |
| 40 | is | 2⁄5 | |
| 41 | 2⁄3 | is | 5⁄12 |
| 43 | 3⁄4 | is | 7⁄16 |
| 50 | is | 1⁄2 | |
| 56 | 1⁄4 | is | 9⁄16 |
| 58 | 1⁄3 | is | 7⁄12 |
| 60 | is | 3⁄5 | |
| 62 | 1⁄2 | is | 5⁄8 |
| 66 | 2⁄3 | is | 2⁄3 |
| 68 | 3⁄4 | is | 11⁄16 |
| 75 | is | 3⁄4 | |
| 80 | is | 4⁄5 | |
| 81 | 1⁄4 | is | 13⁄16 |
| 83 | 1⁄3 | is | 5⁄6 |
| 87 | 1⁄2 | is | 7⁄8 |
| 91 | 2⁄3 | is | 11⁄12 |
| 93 | 3⁄4 | is | 15⁄16 |
This table embodies all the aliquot parts of 100 and their equivalent fractions which are generally used in practical calculations.
Problems in Grain, Stock, Cotton, Coal, Hay, Lumber, etc.
To find the value of articles sold by the unit, hundred or thousand.
Rule.—Multiply the quantity by the price, or vice versa, and point off the proper number of decimal places in the result.
Find the cost of a bale (518 pounds) of cotton at 73⁄8c per pound.
| 518 | × | .07 | = | 36 | .26 |
| „ | × | .003⁄8 | = | 1 | .941⁄4 |
| Ans. | $38 | .201⁄4 | |||
At 7c (.07) per pound, 518 pounds cost $36.26; at 3⁄8c, $1.941⁄4. For 3⁄8 of 518, multiply by 3, and divide product by 8.
Find cost of a lot of hogs, weighing 8740 pounds, at $4.35 per hundredweight.
87.40
4.35
380.1900
The price being $4.35 per 100 pounds and as in 8740 pounds there are 87.40 hundredweight, four decimal places are pointed off. Ans. $380.19.
Find the cost of 2864 feet of lumber, at $171⁄4 per 1000 feet.
Price being dollars per 1000, point off three places. (2.864 × 171⁄4 = 49.404.) Ans. $49.40.
To find the value of articles sold by the ton (2000 pounds).
Rule.—Multiply the weight by the price and take half of the product.
Find the cost of 2680 pounds of hay, at $111⁄2 per ton.
Point off three places, when price is dollars; five if dollars and cents. (2680 × 111⁄2 = 30820; 30820 ÷ 2 = 15.410.) Ans. $15.41.
When the long ton of 2240 pounds is used.
Rule.—Multiply the weight by the price and divide the product by 2.240.
Find the cost of 4800 pounds coal, at $63⁄4 per long ton. (4800 × 63⁄4) ÷ 2.24 = $14.46, Ans.
To find the cost of grain, when the price per bushel and weight is given.
Rule.—Reduce the weight to bushels, and multiply by the price.
Find the cost of 3570 pounds of shelled corn, at 36c per bushel.
| 56 | ) | 3570 | ( | 63 | .75 | bu. |
| .36 | ||||||
| Ans. $22.9500 | ||||||
To reduce pounds of shelled corn to bushels, divide by 56. At 36c per bushel, 63.75 bushels come to $22.95.
Find cost of 2900 pounds of wheat, at 57c per bushel.
To reduce pounds of wheat to bushels divide by 60. 2900 ÷ 60 = 481⁄3 bushels; 481⁄3 ×.57 = $27.55, Ans.
In computing the value of grain, the operation can often be abbreviated by cancellation.
Rule.—Write the weight and price per bushel, on the right of a vertical line, and the number of pounds to the bushel on the left. Then cancel common factors, as explained above.
Find the cost of 3230 bushels of wheat, at 72c per bushel.
| 60 | 3230 | ||
| 72 | 12 | ||
| 323 × 12 | = | 38.76 | |
Here we cancel the 0’s on both sides; then, 6 and 72, which leaves 323 and 12. Their product being the answer.
At 28c per bushel, what will 4080 pounds of oats cost?
| 32 | 4080 | 510 |
| 4 | 28 | 7 |
| Ans. | $35.70 | |
Oats, 32 pounds to the bushel. See table, page 861. Cancel 32 and 4080, then, 4 and 28, leaving the factors 510 and 7.
Other short cuts for computing cost of merchandise, produce, etc.
Find cost of 261⁄2 dozen eggs, at 181⁄2c a dozen.
| 26 × 18 | = | 4.68 | |
| 1⁄2 of 44 | = | .22 | |
| 1⁄2 × 1⁄2 = 1⁄4 | 4.90 | 1⁄4 | |
| Ans. $4.90. | |||
When both fractions are 1⁄2. To product of the whole numbers, add 1⁄2 of their sum, and annex 1⁄4 to answer.
Of 533⁄4 pounds of butter, at 283⁄4c per pound.
| 53 ×.28 | = | 14.84 | |
| 3⁄4 of 81 | = | .60 | 3⁄4 |
| 3⁄4 × 3⁄4 = 9⁄16 | 15.45 | ||
| Ans. $15.459⁄16. | |||
To the product of the whole numbers, add 3⁄4 of their sum, plus the square of 3⁄4.
Of 131⁄4 yards of flannel, at 311⁄4c per yard.
13 ×.31 = 4.03 +.11 = 4.14, Ans.
To 4.03 add.11, 1⁄4 of 44 (13 + 31). The 1⁄16 (1⁄4 × 1⁄4) is disregarded.