# Compound Interest - Aptitude Questions and Answers - RejinpaulPlacement

8.
If the compound interest on a certain sum at 16   2/3% for 3 years is Rs. 1270, find the simple interest on the same sum at the same rate and for the same period.

Solution:   Let the sum be Rs. x. Then,

C.I. = Rs. [x * [1 +(50/(3 * 100))^3] - x] = [(343x/216) - x] = 127x/216

127x/216 = 1270
=> x = (1270 * 216)/127 = 2160
Thus, the sum is Rs. 2160

S.I. = Rs. [2160 * (50/3) * 3 * (1/100)] = Rs. 1080
9.
The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.

Solution:   Let the sum be Rs. x. Then,

C.I. = x[1 + (10/100)]^2 - x = 21x/100
S.I. = [(x * 10 * 2)/100] = x/5

C.I. - S.I. = [(21x/100) - (x/5)] = x/100
x/100 = 631 <=> x = 63100

Hence, the sum is Rs. 63,100
10.
The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a? [Bank P.O. 2003]

Solution:   Let the rate be R% p.a. Then,

[18000 [1 + (R/100)]^2 - 18000] - [(18000 * R * 2)/100] = 405
=> 18000[[(100 + R)^2 / 10000] - 1 - (2R/100)] = 405
=> 18000[((100 + R)^2 - 10000 - 200R)/10000] = 405
=> (9/5)R^2 = 405
=> R^2 = [(405 * 5)/9] = 225
=> R = 15

Rate = 15%
11.
Did Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.
Answer: c) Rs. 676 and Rs. 625

Solution:   Let the two parts be Rs. x and Rs. (1301 - x).

x[1 + (4/100)]^7 = (1301 -x)[1 + (4/100)]^9
=> x/(1301 - x) =[1 + (4/100)]^2 = [(26/25) * (26/25)]
=> 625x = 676(1301 - x)
=> 1301x = 676 * 1301
=> x = 676.

So, the two parts are Rs. 676 and Rs. (1301 - 676); i.e; Rs. 676 and Rs. 625
12.
A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.

Solution:   S.I. on Rs. 7350 for 1 year = Rs. (8575 - 7350) = Rs. 1225

Rate = [(100 * 1225)/(7350 * 1)] % = 16 2/3%

Let the sum be Rs. x. Then,

x[1 + (50 / (3 * 100))]^2 = 7350
=> x * 7/6 * 7/6 = 7350
=> x = [7350 * (36/49)] = 5400.

Sum = Rs. 5400
13.
A sum of money amounts to Rs. 6690 after 3 years and to Rs. 10,035 after 6 years on compound interest. Find the sum.

Solution:   Let the sum be Rs. P. Then,

P[1 + (R/100)]^3 = 6690 ....... (i)
P[1 + (R/100)]^6 = 10035 ....... (ii)

On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2

Substituting this value in (i), we get:
P * (3/2) = 6690
=> P = [6690 * (2/3)] = 4460

Hence, the sum is Rs. 4460.
14.
A sum of money doubles itself at compound interest in 15 years. In how many years will it become eight times?