# Compound Interest : Aptitude Test (65 Questions with Explanation)

#### Basic Computer Knowledge Test Questions and Answers

1.

What would be the compound interest on Rs.7700/- at 15 ¼% per annum for 2 years compounded annually

A.) Rs.2725.75/-
B.) Rs.2527.57/-
C.) Rs.2227.57/-
D.) Rs.2520.57/-

Principal = Rs. 7700/-
time = 2 years
rate = 15 ¼%
Amount = P(1+R/100)n
= 7700 × (1 + 61/(4 × 100)2)
= 7700 × [(1 + 61/400)2]
= 7700 × [(461/400)2)]
= 7700 × 461/400 × 461/400
= 7700 × 1.1525 × 1.1525
= 7700 × 1.32825625
= 10227.573125
C.I = 10227.573125 - 7700 = 2527.57/-

#### Basic Computer Knowledge Test Questions and Answers

2.

Certain sum becomes 3 times it self at compound interest in 10 years. In how many years it becomes 9 times?

A.) 25 years
B.) 27 years
C.) 30 years
D.) 20 years

P(1 + R/100)10 = 3P
=> P(1 + R/100)10 = 3
Let P(1 + R/100)n = 9P
=> (1 + R/100)n = 9
=> 32 = [(1 + R/100)10]2
=> (1 + R/100)n => (1 + R/100)20
=> n = 20 Years.

#### Basic Computer Knowledge Test Questions and Answers

3.

An amount at compound interest sums to Rs.17640/- in 2 years and to Rs.18522/- in 3 years at the same rate of interest. Find the rate percentage?

A.) 5%
B.) 6%
C.) 4%
D.) 10%

The difference of two successive amounts must be the simple interest in 1 year on
the lower amount of money.
S.I = 18522/- - 17640/- = Rs. 882/-
Rate of interest = (882/17640) × (100/1) => 8820/1764 = 5%
Principal = Amount/(1 + R/100)n
= 17640/(1 + 5/100)2
= 17640/(21/20 × 21/20)
= 17640/(1.05 × 1.05)
= 17640/1.1025
= 16000

#### Basic Computer Knowledge Test Questions and Answers

4.

Certain loan amount was repaid in two annual installments of Rs.1331/- each. If the rate of interest be 10% per annum Compounded annually the sum borrowed was?

A.) Rs.121/-
B.) Rs.2130/-
C.) Rs.2310/-
D.) Rs.1331/-

Principal = (P.W of Rs. 1331/- due 1 year hence) + (P.W of Rs. 1331/- due 2 years hence)
= [1331/(1 + 10/100) + 1331/(1 + 10/100)2
= [1331/(110/100) + 1331/(110/100 × 110/100)]
= 13310/11 + 133100/121 = 1210 + 1100 = Rs.2310/-

#### Basic Computer Knowledge Test Questions and Answers

5.

The difference between the compound interest and simple interest on a certain sum of money at 5% per annum for 2 years is 45. Then the original sum is?

A.) Rs.16000/-
B.) Rs.15000/-
C.) Rs.18000/-
D.) Rs.20000/-

For 2 years = (1002D)/R2
= (1002 × 45)/(5 × 5) = (10000 × 45)/25 = Rs.18000/-

#### Basic Computer Knowledge Test Questions and Answers

6.

The Compound interest in a particular amount for the first year at 8% is Rs.50/-.The compound interest for 2 years at the same rate on the amount will be?

A.) Rs.52/-
B.) Rs.104/-
C.) Rs.102/-
D.) Rs.54/-

Rs.104/-

#### Basic Computer Knowledge Test Questions and Answers

7.

The present worth of Rs.9826/- due 2 years at 6 ¼% per annum compound interest is:

A.) Rs.7804/-
B.) Rs.8704/-
C.) Rs.8760/-
D.) Rs.8504/-

Rs.8704/-

#### Basic Computer Knowledge Test Questions and Answers

8.

The Compound interest of Rs.10240/- at 6 ¼% per annum for 2 years 292days is:

A.) Rs.1898/-
B.) Rs.1798/-
C.) Rs.1688/-
D.) Rs.1698/-

Rs.1898/-

#### Basic Computer Knowledge Test Questions and Answers

9.

A person borrows a certain amount from his friend at the rate of 15% per annum compound interest, interest being compounded annually and agrees to return it in 2 equal yearly installments of Rs.529/- each. Find the amount borrow.

A.) Rs.820/-
B.) Rs.880/-
C.) Rs.860/-
D.) Rs.840/-

Rs.860/-

#### Basic Computer Knowledge Test Questions and Answers

10.

Hari lended a sum of Rs.8000 for 20% per annum at compound interest then the sum of the amount will be Rs.13824 is obtained. After how many years he will get that amount?

A.) 2 years
B.) 1 year
C.) 4 years
D.) 3 years

Let Principal = P, Rate = R% per annum, Time = n years
When interest is compounded annually, total amount can be calculated by using the formula
Compound Amount = P ( 1 + R / 100)n
Given that, P = Rs.8000, R = 20% per annum
Compound Amount = Rs. 13824
We have to find the time period during which the amount will be Rs.13824
=> Rs.13824 = 8000 x (1 + 20/100)n
=> (13824 /8000) = (120 / 100)n
=> (24 / 20)3 = (12 / 10)n
=> (12 /10)3 = (12 /10 )n
Therefore, n = 3.
Hence the required time period is 3 years.