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272. ALGEBRA APPLIED TO SOME PROBLEMS IN INTEREST.
or 10 x = the interest, 100
PROBLEMS. 1. At what rate per cent will $450 gain $72 in 2 years ? + 2. At what rate per cent will $320 gain $18 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.
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. lio 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 53 times; so $500 must be loaned at 53 % to gain $53 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.
273. ALGEBRA APPLIED TO SOME PROBLEMS IN INTEREST.
In how long a time will $650 gain $97.50 at 6 %?
the number of years.
of 650 = 97.50
39 x =
1. In how long a time will $400 gain $30 at 5%?
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 8650 for one year at 6 % is $39. As many years will be required to gain $97.50 as $39.00 is contained times in 897.50. It is contained 242 times; so in 24 years 8650 will gain 897.50. Observe that by either method we divide the given interest by the interest of the principal for 1 year at the given rate.
274. THE AREA OF A RHOMBOID.*
1. One side of a rhomboid may be regarded as its base. The perpendicular distance from the base to the opposite side, is its altitude.
In the figure here given of 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.
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 ?
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.
276. Ratio is relation by quotient. The two numbers (magnitudes) of which the ratio is to be found are called the terms of the ratio. The first term is called the antecedent and the second term the consequent. The ratio is the quotient of the antecedent divided by the consequent.
The usual sign of ratio is the colon. It indicates that the ratio of the two numbers between which it stands is to be found, the number preceding the col being antecedent and number following it, the consequent. The expression, 12:4= 3, is read, the ratio of 12 to 4 is 3.
Read and complete the following: 1. 12:4
4:12 2. 18:9
9:18 3. 15 : 5
5:15 4. 40:10 =
NOTE.-It will be observed that the sign of ratio is the sign of division ( : ) with the line omitted.
277. Every integral number is a ratio. The number 4 is the ratio of a magnitude 4 inches, ounces, bushels,) to the measuring unit 1 (inch, ounce, bushel). The number 7 is the ratio of 7 yards to 1 yard; of 7 dollars to 1 dollar, or of 7 seconds to 1 second, etc.
Note.-The ratio aspect of numbers is not the aspect most frequently uppermost in consciousness: neither ought it to be. But the pupil should now see that number is ratio; that while it implies aggregation and often stands in consciousness for magnitude, its essence is relation-ratio.