Direction
1.

A sum of money at simple interest amounts to 815 in 3 years and to 854 in 4 years. The sum is:


A. 650 B. 690
C. 698 D. 700
Answer and Explanation
Answer: Option C
Explanation: S.I. for 1 year = (854 - 815) = 39.

S.I. for 3 years = (39 x 3) = 117.

Principal = (815 - 117) = 698.

2.

Mr. Thomas invested an amount of 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be 3508, what was the amount invested in Scheme B?

A. 6400 B. 6500
C. 7200 D. 7500
Answer and Explanation
Answer: Option A
Explanation: Let the sum invested in Scheme A be x and that in Scheme B be (13900 - x).

Then, (x x 14 x 2)/100 + ((13900 - x) x 11 x 2)/100 = 3508

28x - 22x = 350800 - (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = (13900 - 7500) = 6400.

3.

A sum fetched a total simple interest of 4016.25 at the rate of 9 %.p.a. in 5 years. What is the sum?

A. 4462.50 B. 8032.50
C. 8925 D. None of these
Answer and Explanation
Answer: Option C
Explanation: Principal = (100 x 4016.25)/(9 x 5)

= 401625/45

= 8925.

4.

How much time will it take for an amount of 450 to yield 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years B. 4 years
C. 4.5 years D. 5 years
Answer and Explanation
Answer: Option B
Explanation: Time = (100 x 81)/(450 x 4.5) years = 4 years.

5.

Reena took a loan of 1200 with simple interest for as many years as the rate of interest. If she paid 432 as interest at the end of the loan period, what was the rate of interest?

A. 3.6 B. 6
C. 18 D. Cannot be determined
Answer and Explanation
Answer: Option B
Explanation: Let rate = R% and time = R years.

Then, (1200 x R x R)/100 = 432

12R² = 432

R² = 36

R = 6.