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247. TO FIND THE TIME BETWEEN TWO DATES.

EXAMPLE How many years, months, and days from Sept. 25, 1892, to June 10, 1896 ?

THE USUAL METHOD.

1896 - 6-10
1892 - 9 - 25

From the 1896th yr., the 6th mo., and the 10th day, subtract the 1892nd yr., the 9th mo., and the 25th day. Regard a month as 30 days.

3-8-15

A BETTER METHOD.

From Sept. 25, 1892 to Sept. 25, 1895, is 3 years.
From Sept. 25, 1895 to May 25, 1896, is 8 months.
From May 25, 1896 to June 10, 1896, is 16 days.

NOTB.- The two methods will not always produce the same results. The greatest possible variation is two days. Find the time from Jan. 22, 1895, to March, 10, 1897, by each method, and compare results. The difference arises from the fact that the month as a measure of time is a variable unit-sometimes 28 days, sometimes 21. The usual methodregards each month as 30 days ; the better methodcounts first the whole years, then the whole months, then the clays remaining. By the “usual method,” the time from Feb. 28, 1897 to March 1, 1897, is 3 days; by the “better method,” it is 1 day.

PROBLEMS.

Find the time by both methods and compare the results.

1. From March 15, 1894 to Sept. 10, 1897.
2. From March 15, 1894 to Sept. 20, 1897.
3. From May 25, 1895 to Sept. 4, 1898.
4. From May 25, 1895 to Oct. 4, 1898.
5. From June 28, 1894 to Mch. 1, 1898.
6. From June 28, 1894 to May 1, 1898.
7. From Jan. 10, 1892 to Jan. 25, 1898.
8. From Jan. 10, 1892 to Dec. 25, 1898.
9. From April 15, 1893 to Aug. 15, 1898.

248. ALGEBRA APPLIED TO PROBLEMS IN PERCENTAGE.

EXAMPLE
75 is 15 per cent of what number?
Let

the number sought.
15
Then
of x, or

75. 100

100 Multiplying by 100,

7500, Dividing by 15,

500. Ans.

X =

15 x

15 x

x =

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7. James has $54.20, and James's money equals 40 per cent of Henry's money. How much money has Henry ?

8. Mr. Williams's annual expenses are $791.20 ; this is 92 per cent of his annual income. How much is his annual income?

9. Mr. Randall has 450 sheep; these equal 125 per cent of Mr. Evans's sheep. How many sheep has Mr. Evans ?

* To The'TEACHER.-Encourage pupils to solve the 7th, 8th, and 9th problems in three ways: 1st, by the method given above; 2nd, arithmetically (Prob. 7– Henry's money equals 100 40ths of James's money); 3rd, by the application of the formula,

100 n

= r.

249. ALGEBRA APPLIED TO PROBLEMS IN PERCENTAGE

EXAMPLE.

60 is what per cent of 75 ? Let

the per cent (number

of hundredths).
X
Then of 75, or

60.
100

100 Multiplying by 100, 75 x = 6000 Dividing by 75,

Ans.

75 x

x = 80.

PROBLEMS. 1. 180 is what per cent of 200? 2. 17 is what per cent of 340? 3. $87.50 is what per cent of $250 ? 4. 81 is what per cent of 540 ? 5. $75.60 is what per cent of $630? 6. n is what per cent of b? * Let

the per cent.

bx
Then
of b or,

= n.
100

100 Multiplying by 100,

bx

100 n

100 n Dividing by b,

X =

b

X =

X

To THE TEACHER.-Require the pupils to solve the first five prob. lems in four ways:

(1) Let x = the per cent and solve as the "example" is solved.

(2) Using one hundredth of each base as a divisor and the other number mentioned in the problein as a dividend, find the quotient.

(3) Find what part the first number mentioned in each problem 15 of the base, and change the fraction thus obtained to hundredths.

100 n (4) Apply the formula,

6

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* The b in problem 6, and in the formula may be thought of as standing for the base.

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Geometry. 250. SOME INTERESTING FACTS ABOUT SQUARES, TRIANGLES, AND HEXAGONS. 1. Four equal squares may be so joined

cover all the space about a point. Each angle whose vertex is at the central point of the figure is an angle of 90°. Four tinies 90°

degrees. 2. Six equal equilateral triangles may be so joined as to cover all the space about a point. Each angle whose vertex is at the central point of the figure is an angle of 60°. Six times 60o = degrees.

NOTE.—Cut from paper 6 equal equilateral triangles and join them as shown in the figure.

3. Three equal hexagons may be so joined as to cover all the

space

about a point. Each angle whose vertex is at the central point of the figure is an angle of 120°. Three times 1200 degrees.

=

NOTE.—Cut from paper 3 equal hexagons and join them as shown in the figure.

*

4. Since every regular hexagon may be divided into six equal equilateral triangles, * as shown in the figure, it follows that the side of a regular hexagon is exactly equal to the radius of the circle that circumscribes the hexagon.

* See Observation, page 149.

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251. MISCELLANEOUS REVIEWS. 1. Find the interest of $ 250 from Sept. 5, 1896, to Jan. 17, 1893, at 6%.

2. Find the amount of $310 from April 19, 1895, to Oct. 1, 1896, at 6%.

3. Find the amount of $500 from May 20, 1897, to Feb. 28, 1898, at 5%.

4. Find the amount of $630 from July 1, 1896, to Nov. 1, 1896, at 7%.

5. Find the amount of $800 from Jan. 1, 1897, to Jan. 25, 1897, at 6%.

6. If a 60-day bill of $100 is discounted 2% for cash, how much ready money will be required to pay the bill?

7. Find the amount of $392 for 60 days (two months) at 6%.

NOTE --Observe that $392 is the answer to problem 6. The result of problem 7, then, is the amount that the goods mentioned in problem 6, would cost at the end of 60 days if the purchaser borrowed the money at 6% to pay for them. Compare this result with $400. How much does the purchaser of the bill save by borrowing the money at 6% to pay the bill instead of letting the bill run 60 days and then paying $400?

8. Find the cost of goods, the list price being $46, and the discounts “50 and 10 off and 2 off for cash.”

9. My remuneration for selling goods on a commission of 40% amounted to $56.20. How much should the man for whom the goods were sold receive?

NOTE. --$56.20 is 40% of the selling price of the goods. The man for whom the goods were sold should receive 60% of the selling price, When goods are sold on a commission of 40%, what the agent receives equals what part of what the employer receives? What the employer receives is how many times what the agent receives?

K. The sum of all the angles of 6 triangles equals how many right angles ?

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