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periods of the given number, for a divisor. Divide, and reserve the quotient.

To 6 times the reserved quotient, add the index of the root, plus 1, for a second dividend.

To 6 times the reserved quotient, add 4 times the index of the root, subtract 2 from the sum, and multiply by the reserved quotient for a second divisor. Divide, add 1 to the quotient, and multiply by the as. certained root for the true root nearly. If greater accuracy is desired, repeat the process with the root thus found.

By this rule, the number of figures in surd roots, may generally be tripled at each operation.

The following is the application of the rule, in extracting the 5th root of 659901.

Ascertained root 14.
145 = 537824

given no. 659901
index
5

145 537824

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dividend 2689120

1st divisor 122077 2689120 • 122077 = 22.02806, reserved quotient. reserved quotient 22.02806

6

[blocks in formation]

second dividend 138.16836 6 X by reserved quotient 132.16836

4 X 5 - 2 = 18.

150.1 6836 Multiply by 22.02806

2nd divisor 3307.91764 138.16836 • 3307.91764 .041768

1.041768 x 14 = 14.584752, approximate root, correct to the fourth decimal place.

This contraction is of use in extracting the higher roots. Any root below the 10th may be obtained in the usual way, nearly as readily, and with much greater accuracy. 58. Find convenient fractional approximations to V3.

59. What are the approximate fractional values of 5?
60. Extract the 13th root of 1.08.
61. Extract the 17th root of 1.004.
62. Extract the 100th root of 1.07.
63. Extract the 45th root of 1.2.
64. Reduce 13 to a continued fraction.

65. What are the approximate values of the continued fraction which is equivalent to /27?

CHAPTER XXIV.

ANALYSIS.

All the operations of Arithmetic have for their object, the discovery of one or more unknown quantities; and the great difficulty in complicated questions, is to perceive the application of the simple rules which will lead to this discovery.

The examination of any question, in order to determine the relation of the different quantities to each other, is called ANALYSIS. To keep the unknown terms more con. stantly in view, letters are frequently employed to represent them, and the work expressed in the statement of the question, is performed on these letters, as if their value was known, and we were proving the truth of the answer.

EXAMPLE FOR THE BOARD.

There is a fish whose head weighs 9 pounds; his tail weighs as much as his head and half his body; and his body weighs as much as his head and tail both. What is the weight of the fish ?

Let 2 = 9 +

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2

[ocr errors]

2

2

+18.

the weight of the body. In this example, if the the weight of the tail.

weight of the body were 9 = the weight of the head. known, the answer would

be readily obtained. We X = 9+ + 9

therefore represent this + 18

weight, which is the un= 18

known quantity by x. The

tail weighing as much as x = 36, the weight of the body. the head and half the body, 9 + 27, the weight of the tail.

will be represented by 9, the weight of the head. 9 + } of x.

But as } of any number is 72, the weight of the fish. the number divided by 2,

1 of x will be x = 2 or i. The body weighing as much as the head and tail both,

x = 9, (w't of head) + 9+ (wit of tail.) But

9 + 9 = 18, therefore, x = She ha 등 Now, if 18 added to the half of z gives x, 18 must itself be equal to

, and twice 18, or 36, to t. Having found the value of £, we easily obtain the remaining values.

The pupil may analyse the following examples, either with or without the aid of letters.

1. A father's age is 7 times that of his son, and the sum of their ages is 40. What is the age of each?

2. In a certain school there are 45 scholars, and there are twice as many boys as girls. Required the number of each?

3. A man performed a journey of 135 miles, going twice as far the second day as on the first, and three times as far the third day as on the second. How far did he travel each day?

4. A., B., and C., entered into partnership, contributing in the whole, $4833. B. paid twice as much as A., and C. paid twice as much as A. and B. How much did each contribute ?

5. In a certain school of 70 scholars, three times as many

study Arithmetic as study Latin, and twice as many learn to read, as study Arithmetic. How many are there in each study?

6. An estate of $7000 was so divided that the widow re. ceived $500 more than the daughter, and the son $1100 more than the widow. What was the share of each ?

7. Divide the number 97 into four such parts that the second may be twice the first, the third 7 more than the second, and the fourth 18 more than the first.

8. A thief travels at the rate of 6 miles an hour, and after he has been absent 5} hours, a constable starts in pursuit, at the rate of 9 miles an hour. In what time will the thief be overtaken?

9. A man when he was married, was three times as old as his wife, but after they had lived together 15 years, he was only twice as old. How old was each at the time of marriage ?

10. A farmer bought some cows and some calves for $461, giving $23 apiece for the cows, and $9 apiece for the calves, and there were twice as many calves as cows. How many were there of each ?

11. At a certain election, the successful candidate had 163 votes more than his opponent, and the whole number of votes polled was 1125. How

many did each receive ? 12. What sum of money will yield $123,50 in 2 years, at 5 per cent. simple interest ?

13. A gentleman distributed $1.95 among 3 beggars, giving the second 25 cents more than the first, and the third twice as much as the second. How much did each receive ?

14. If from three times a certain number 17 be subtracted, the remainder will be 112. What is the number? 15. A's age is 21 times B's, and the sum of their ages

is 49. Required the age of each.

EXAMPLE FOR THE BOARD.

A man being asked his age, replied, “If ], , , and 3

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[ocr errors]

+

+

of my age and 19 years more be added to my age, the sum will be 3 times my age.” What was his age ? Let x = his age.

3.3 X +

+

+ 19 = 3x,
2 3 4 5
30x 20x 15x 36x
X + + + + + 19 = 3x,
60
60 60

60
161x

180.0
+ 19 =
60

60
192
60

19 =

[ocr errors][merged small][merged small]

60 =

2

60 =

or X.

60'

After stating the question, we reduce all the fractions to a common denominator, and add them as in ordinary Addition of fractions. We then find that

161x

180x
+ 19 =
60

60
Therefore, 19 must be the difference between
1613
180x

19x
and

which is
60
60

60
19x

60x If 19 then 1

and 60

60 16. A merchant owes two of his creditors $1575, and he owes the second but as much as the first. What is the amount of each debt?

17. In a certain school į the boys learn to read, learn to write, is learn Algebra, 1 learn drawing, and the re. maining 4 study Latin. How many are there in the school?

18. One-third of a certain pole is painted green, and of it is painted white, the remainder, which is 8 feet, being in the ground. What is the length of the pole?

19. A man going to market, was met by another, who said : “Good morrow, neighbor, with your hundred geese.”

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