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NUMBER.

4, 5, 6, &c. are each equal to 1. Then, as the Numerator of a Proper Fraction is never as great as its Denominator, A PROPER FRACTION IS ALWAYS LESS THAN A UNIT. Of course, AN IMPROPER FRACTION IS NEVER LESS THAN A UNIT.

is an Improper Fraction. It may be separated into the two Fractions and t. 4, as we have seen, is equal to 1. Therefore

is equal to 1 and 4, or, as it is commonly written, 1}. A whole number and a Fraction, written thus, are together called a MIXED

is an Improper Fraction, and may be separated into į and

But each of these is equal to 1. Therefore, they are together equal to 2. is more than . Hence, į=25.

Hence it appears, that, as often as the Denominator of an Improp. er Fraction is contained in the Numerator, so many whole ones, or Integers are contained in the Fraction; and, that if the Denominator will not dividë the Numerator exactly, a Proper Fraction will re. inain. Hence, also, any Improper Fraction may be changed, or reduced to a whole or mixed number; and in order thus to reduce it, we must,

DIVIDE THE NUMERATOR BY THE DENOMINATOR ; WRITE THE DAINDER, IF THERE BE ANY, OVER THE DENOMINATOR, AND ANNEX THE FRACTIOX, THUS FORMED, TO THE QUOTIENT.

EXAMPLES FOR PRACTICE. 1. Reduce o to a whole or mixed number. Ans. 93. 2. Reduce. Ans. 93. 8. Ans. 94. 13. Ans. 15).

Ans. 23. 3. Reduce 158. Ans. 523.245. Ans. 565.21925. Ans. 2425. 4. Reduce 81. 6278, 518432, 91548 73.

9 15873. 132 5 9 65

. 5. Reduce

, 7000 70007.

600304002. 6. Reduce 7112 34 5 4 8 0. 4956460217. 3 322211136. 5 9 2 4 3 3 2 1 76 8.

In the following examples, this process is reversed. 1. How many. 4ths. in 1 ? How many in 11? In 12 ? In 19?

2. How many 5ths. in 1 ? In 5 ? In 11? In 13? In 73 ? 3. How many 7ths. in 7? In 8? In 12? In 7* ? In 5% ? 4. How many 12ths. in 9 ts? In 79.? In 3 12 ? In 5

? In 8 ? 5. How many 6ths. in 3? In 4? In 53? In 7 4? In 8 ? In 9 4 ? In 12 ?

6. How many 27ths. in 3? In 2? In 5 - ? Ans. St.

.

9 8 7 6 5 4 3 2 1

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6

12

14_4
27

7. How many 19ths. in 15 ?: In 13 i.? In 17 19 ? Ans. 285 ? 250,

4:

19

Here we multiply the whole number by the number expressing the parts, and adå in the additional parts, if there be any. By this process, we obtain an Improper Fraction. Hence, the process is called, reducing a whole or mixed number to an Improper Fraction, The operation, as performed above, is as follows :

MULTIPLY THE WHOLE NUMBER, BY THE NUMBER EXPRESSING THE PARTS, AND ADD THE NUMERATOR OF THE FRACTION, IF THERE BE ANY,

THE SUM WILL BE THE NUMERATOR AND THE MULTVPLIER, THE DENOMINATOR OF THE RESULTING IMPROPER FRACTION.

TO TUE PRODUCT.

$ XXVIII. From the nature of Fractions, it is evident that Division may be expressed by a Fraction. For, in Division, the dividend is to be separated into a number of parts, denoted by the divisor. Hence,

DivisION MAY BE INDICATED BY WRITING THE DIVISOR UNDER THE DIVIDEND IN THE FORM OF A FRACTION. This mode of indicating Division was given § xxv. Hence, also the general principle,

ALL FRACTIONS ARE INSTANCES OF DIVISION, IN WHICH THE NUMERATOR IS THE DSVIDEND, AND THE DENOMINATOR THE DIVISOR. OS course,

THE VALUE OF A FRACTION IS THE QUOTIENT, WHICH ARISES FROBI DIVIDING THE NUMERATOR BY THE DENOMINATOR.

When the dividend is not less than the divisor, division may be both indicated and performed. When it is less however, Division can only be indicated. We may therefore give a brief rule, for the case when the dividend is less than the divisor. We are the rath. er inclined to give it, though the case is very simple, because we have often seen pupils much perplexed by it.

If the dividend be less than the divisor, WRITE THE DIVISOR UNDER THE DIVIDEND, AND THE FRACTION FORMED WILL DE THE ANSWER.

EXAMPLES FOR PRACTICE. Divide 37 by 45. 81 by 83. 97 by 120. 16 by 31. 17 by 19. 27 by 245. 383 by 384. 2 by 231. 3 by 756. 16 by 165.4 by 71. 326 by 525. 482 by 491. 374 by 1,693.

The pupil will observe, that where a less number is divided by a greater, the fractional quotient found, shows what part the dividend is ofthe divisor. Thus,l is the seventh part of 7, and 1:7=. 2 is two seventlis of 7 and 2:7=4, and so on.

It is common, likewise to extend the term part, to numbers, greater than those, of which they are said to be parts ; thus. When the question is asked what part of 3 is 1, the answer is, one third,=}, When it is asked what part of 3 is 2, we say two thirds,=s. In like manner, when it is asked, what part of 3 is 4, the answer is four thirds, f. When it is asked what part of 3 is 5, the answer is five thirds, and so on. Hence,

THE

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Every Fraction shows what part the Numerator is of the Denomin. ator. Of course, INDICATING DIVISION IS FINDING WHAT PART OF THE DIVISOR IS EQUAL TO THE DIVIDEND. Therefore, to find what part one number is of another, MAKR THE NUMBER CALLED THE PART

NUMERATOR OF FRACTION, AND THE OTHER NUMBER THE DENOMINATOR.

Note. This rule must be strictly followed, without regarding which number is the greater. This is sometimes called, finding the ratio of one number to another. ($xc.) The Fraction obtained expresses the RATIO of the denominator to the numerator.

EXAMPLES OF FINDING PARTS, OR RATIOS. 1. What part of 5 is 1 ? A. }. What part is 12 of 5 ? A. ! 2. What part of 9 is 6 ? A... Is 3 ? is 15 ? is 19? 3. What part is 16 of 25? What part of 25 is 16 ? 4. What part of 3,684 is 27,942 ? What part is 3,684 of 27,942 ?

XXIX. 1. A Gentleman paid 6,372 dollars, for 36 acres of land. How much did he give an acre ?

Here our divisor consists of two figures. We 4)6372 cannot therefore, divide by the last rule. But we know that 4 times 9 are 36 ; and that 4 times 9 acres 9)1593 are 36 acres. If we divide, then, by 4, we obtain the price of 9 acres. We can then divide this price by A. 177 9 and obtain the price of 1 acre.

2. If 56 hogsheads of molasses cost 2,016 dollars, 2)2016 what cost 1 hogshead ? 2X4X7=56. Therefore, if we divide by 2, we shall obtain the price of half 4)1008 as many hogsheads, that is, of 28 hogsheads. If we divide this result by 4, we shall obtain the price 7)252 of 6 hogsheads. If we divide this result by 7, we shall obtain the price of 1 hogshead, which is required. A. 36

Hence, to divide by a composite number, Resolve THE DIVISOR INTO FACTORS, AND DIVIDE BY THOSE FACTORS SUCCESSIVELY.

EXAMPLES FOR PRACTICE. 3. At 15 dollars a hhd., how many hhds. of sugar can be bought for $4,500 ? A. 300.

4. At $18 a ton, how many tons of iron can be bought for $2,250 ? A. 125.

5. If an acre of ground cost $25, what number of acres will $9,375 buy? A. 375.

6. If a hhd. of molasses cost $27, what number of hhds, will $5,940 buy? A. 220.

7. Divide 6,894 by 18 | 12. Divide 273,045 by 15
8. Divide. 57,960 by 36 | 13. Divide 714,357 by 21
9.

39,942 · 18 | 14. 2,295,495 - 45
10. 265,824 " 24 | 15. 3,575,635 " 35
11, 333,225

66 25 16. " 27,966,232 ( 56

66

It is to be observed, that, in many instances, we have remainders when the division is completed. The pupil may not readily discov. er, in dividing by composite numbers, how to find the true remainder. We will endeavour to explain the manner, by an example.

17. How many cubic yards in 369 cubic feet ?

There are 27 cubic feet in a cubic yard. There. 9)369 fore, we must divide by 27=9X3. Divide first by 9 thus,

3)41 There is no remainder-Then by 3; thus,

Here we have 2 remainder-But this does not 13+2 Rem. signify that there are only 2 cubic feet left. For 41, the first quo. tient shows that there are 41 9's in 369. Then each unit of the quotient 41, is 9 units of the dividend, 369. Now, in dividing this 41, there is left a remainder of 2 units. But if 1 of these units is 9 units of the dividend, 2 of them will be twice as much ; that is, 18 of the dividend. Of course there are 18 cubic feet left. Then, when there is a remainder on the last Division, we must multiply that remainder, by the first divisor, to find the true remainder.

18. At 36 dollars a ton, how many tons of iron can I buy with $762 ? 35=7X5 Divide by 7, thus, 7)762

Then by 5, thus, 5)108 +6

21+3 Multiply the last remainder 3 by the first divisor 7, as directed by rule. 7X3=21. Then if there had been no remainder on the first Division, 21 would be the true remainder. But the first divis. ion left 6, which must, therefore, be added to the 21, making 27 the true remainder. Then when the divisor is resolved into two fac. tors, in order to find the true remainder, we must multiply the last remainder by the first divisor and add in the first remainder.

For similar reasons, when the divisor is resolved into several factors, multiply each remainder, arising after the first division, by all the preceding divisors, and add the products to the first remainder.

19. At $25 a barrel, how many barrels of brandy can I buy for $263 ? A. 10. 20. At $27 an acre, how many acres of land can I buy for $988 ?

A. 3614 21. Divide 853.by 54. A. 154 1. 22. Divide 971 by 63. A. 15 kg.

23. Divide 4,761 by 45 | 25. Divide 333,222 by 49
24.

9,893
72 | 26.

641,641 • 36

♡ XXX. 1. At 10 cents a pound, how many lbs. of raisins can I buy for 80 cents ?

We saw in Multiplication, that, if a figure were removed one place to the left, it was increased in a ten-fold proportion ; that is, it was multiplied by 10.

So in Division, which is the reverse of Multiplication, if a figure be removed one place to the right, it is diminished in a ten-fold proportion ; that is, it is divided by 10.

To divide 80 by 10, then, take away the cypher on the right, and the 8 will be one place lower ; that is, it will be in the units' place instead of the tens': of course its value will be diminished ten-fold. The quotient, therefore, is 8.

Now if the dividend had been 89, instead of 80, taking away the 9 would have changed the 8 tens, to 8 units, as before ; that is, would have diminished them ten-fold, or divided them by 10, and as 9 is not great enough to make another ten, it would have been the remainder,

Then,

DIVISOR CONTAINS CYPHERS,

REMOVING THE RIGHT HAND FIGURE OF ANY NUMBER DIVIDES IT BY 10, AND THE FIGURE SO REMOVED IS THE REMAINDER.

The same reasoning wil} show that removing two figures from the right of a number divides it by 100; and that the two figures so removed are the Remainder. Also that removing three, four, five figures, and so on, from the right of a number, divides it by 1,000, 10,000, and so on, and that the figures, so removed, are the Remainders. Hence, to divide by a number, consisting of 1, with cyphers annexed, REMOVE AS MANY FIGURES FROM THE RIGHT OF THE DIVIDEND, AS TILE

THE FIGURES SO REMOVED, WILL BE THE REMAINDER ; THOSE NOT REMOVED, THE QUOTIENT.

Note. It will not be necessary, actually to take the figures away; but merely to place a point, to separate the quotient and remainder.

EXAMPLES FOR PRACTICE. 2. How many dollars in 7,963 cents ? A. $79.63. 3. How many dollars in 87,555 mills ? A. $87.555.

4. 100 men were to share equally a prize of $97,543. How muci: was each man's share ? A. $975.43.

5. In an army of 100,000 men, an amount of pay of $2,775,000 was distributed, each man sharing an equal sum with the rest What was each man's share ? A. $27.75, 6. Divide 33 by 10 | 10. Di ide 81,960 by

100 7.

10 | 11.

230,893 » -100 8.

10

679,821 " 1,000 9.

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45 » 360 » 945

12.

100 | 13. 350,325 ” 10,009

71

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