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Note.-Bonds are made in great variety both as to form and content; but in all, indebtedness is acknowledged, and the amount, rate of interest, and time of payment for both principal and interest, named. The above is a very short and concise form of Bond (much reduced in size) and is an exact copy of one prepared for actual use.

1. Examine the above Bond. If the time it is to run is five years, how many coupons should be attached ?

2. If the Bond is dated Jan. 1, 1898, what date should be written in each coupon ?

3. If the face of the Bond is $100 and the rate 5 %, what sum should be written in each coupon?

4. If the rate is 5% per annum, what is the half-yearly interest on a $10,000 bond?

Algebra.

272. ALGEBRA APPLIED TO SOME PROBLEMS IN INTEREST.

EXAMPLE
At what rate per cent will $500 gain $55 in 2 yrs ? *
Let

the rate,

1000 x then of 500, or

or 10 x = the interest, 100

100 and

10 x = Dividing

53

2 x

55

X =

PROBLEMS.

1. At what rate per cent will $450 gain $72 in 2 years? † 2. At what rate per cent will $320 gain $48 in 3 years?

3. At what rate per cent will $560 gain $84 in 2 years 6 months?

4. At what rate per cent will $600 gain $75 in 2 years 6 months?

5. At what rate per cent will $600 gain $114 in 2 years 4 months 15 days? 2 yr. 4 mo. 15 da.

23 years.

23 x
Let x = the rate. Then of 600 = 114.

100

Note. Problem 5 may be solved arithmetically by finding the interest of $600 for 2 yr. 4 mo. 15 da. at 6 %. Divide this interest by 6 (to find the interest at 1 %) and find how many times the quotient is contained in $114.

* The arithmetical solution of this problem is as follows: The interest of $500 for 2 years at 1% is išo of $500. išo of $500 $10. To gain $55 in 2 years, $500 must be loaned at as many per cent as $10 is contained times in $55. It is contained 54 times; so $500 must be loaned at 54% to gain $55 in 2 years. Observe that by this method we divide the given interest by the interest of the principal for the given time at one per cent.

+To The PUPIL.-Prove each answer by finding the interest on the given prin. cipal for the given time at the rate obtained.

Algebra.

273. ALGEBRA APPLIED TO SOME PROBLEMS IN INTEREST.

EXAMPLE.

In how long a time will $650 gain $97.50 at 6 %?

Let

the number of years.

6 x

then

of 650 97.50

100 Simplifying

39 x = 97.50 Dividing

2.5*

PROBLEMS.

1. In how long a time will $400 gain $30 at 5%?
2. In how long a time will $600 gain $96 at 6 %?
3. In how long a time will $800 gain $68 at 6 % ?
4. In how long a time will $500 gain $56 at 6 %?
5. In how long a time will $400 gain $29 at 6 %?

REVIEW PROBLEMS.

6. What principal at 8 % will gain $124.80 in 3 years? (See page 167.)

7. What principal at 7 % will amount to $410.40 in 2 years? (See page 168.)

8. At what rate per cent will $900 gain $72 in 2 years? (See page 177.)

9. In how long a time will $1000 gain $160 at 6 per cent? (See above.)

TO THE PUPIL.—Prove each answer by finding the interest on the given principal at the given rate for the time obtained.

*The arithmetical solution of this problem is as follows: The interest of 8850 for one year at 6 % is $39. As many years will be required to gain $97.50 as $39.00 is contained times in $97.50. It is contained 242 times; so in 24 years $850 will gain $97.50. Observe that by either method we divide the given interest by the interest of the principal for 1 year at the given rate.

[blocks in formation]

In the figure here given cf is the base and ab the altitude.

2. Convince yourself by measurements and by papercutting that from every rhomboid there may be cut a triangle, (abc), which when placed upon the opposite side, (def), converts the rhomboid into a rectangle (adeb).

Observe that the base of the rectangle is equal to the base of the rhomboid, and the altitude of the rectangle equal to the altitude of the rhomboid.

3. A rhomboid is equivalent to a rectangle having the same base and altitude. Hence, to find the area of a rhomboid, find the area of a rectangle whose base and altitude are the same as the base and altitude of the rhomboid. Or, as the rule is given in the older books, -"Multiply the base by the altitude."

PROBLEM.—If the above figure represents a piece of land, and is drawn on a scale of inch to the rod, how many acres of land ?

* The statements upon this page apply to the rhombus as well as to the rhomboid.

[blocks in formation]

1. Mr. Watson purchased 15 shares of C., B. & Q. R. R. stock at 12 % discount. (a) How much did he pay for the stock? (b) When a 3% dividend is declared and paid, how much does he receive ? *

2. James Cooper bought 12 shares of stock in the Sugar Grove Creamery at 8 % below par, and a few days after sold the stock at 5 % above par. How much more did he receive for the stock than he gave for it ?

3. A certain city borrowed a large sum of money and issued therefor 10-year 5 % bonds with the interest payable semi-annually. (a) How many coupons were attached to each bond? (b) On a $1000 bond, each coupon should call for how much money?

4. Sometimes such bonds as those described in problem 3, are offered for sale to the highest bidder, in “ blocks” of $10000, $20000, or $50000. If a $20000 “block” is “bid off” at 21% premium, how much should the city receive for the “block"?

5. What must be the nominal value of 5% bonds that will yield to their owner an annual income of $750 ?


Let x = the nominal value; then - $750.

100 6. What must be the nominal value of 4 % bonds that will yield to their owner an annual income of $720 ?

7. A owns $6000 of 5% bonds ; B owns $8000 of 41 % bonds. How much greater is the annual income from B's bonds than from A's?

8. Find the area of a piece of land in the form of a rhomboid, whose base is 32 rods and whose altitude is 15 rods.

9. Find the area of a piece of land in the form of a rectangle, whose base is 32 rods and whose altitude is 15 rods.

* The par value of each share of stock menticned on this page is $100.

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