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

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/-

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.

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

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/-

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/-

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/-

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/-

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/-

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/-

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.

11.

What will be the amount if sum of Rs.10,00,000 is invested at compound interest for 3 years with rate of interest 11%, 12% and 13% respectively?

A.) Rs.14,04,816
B.) Rs.16,00,816
C.) Rs 12,14,816
D.) Rs. 11, 13,816

Given
Here, P = Rs.10,00,000, R1 = 11 , R2 = 12, R3 = 13.
Each rate of interest is calculated for one year.
Hence, N = 1 year.
Amount after 3 years,
= P(1 + R1/100) (1 + R2/100) (1 + R3/100)
= 10,00,000 × (1 + 11/100) × (1 + 12/100) × (1 + 13/100)
= 10,00,000 × (111/100) × (112/100) × (113/100)
= 111 x 112 x 113
= 14,04,816
Hence the total amount after 3 years is Rs.14,04,816

12.

Akarsh left a will of Rs. 16,400 for his two sons whose age are 17 and 18 years.They must get equal amounts when they are 20 years at 5% compound interest. Find the present share of the younger son.

A.) Rs. 7,000
B.) Rs. 8,000
C.) Rs. 5,000
D.) Rs. 11,000

Given, total amount (to be shared by two sons at the age of 20 on Compound interest) = Rs. 16,400
Let the Present share (Principal amount) for 17 year old son = "X"
Then the Present share (Principal amount) for 18 year old son = (16,400 - X)
To attain 20 years of age,
=> 17 year old son takes 3 years (N = 3 years on Compound interest)
=> 18 year old son takes 2 years (N = 2 years on Compound interest)
Given, Rate of interest (R) = 5%
Given that, at the age of 20, two sons get equal amount
=> Compound Amount of 17 year old son = Compound Amount of 18 year old son
W.K.T, Formula for Compound Amount = P [1 + (R/100)]^N
=> X (1 + 5/100)^3 = (16,400 - X) (1 + 5/100)^2
=> X (1 + 5/100) = (16,400 - X)
=> (105/100) X = (16,400 - X)
=> [(105/100) X] + X = 16,400
=> 205 X = 16,400 * 100
=> X = 16,40,000 / 205
=> X = 8,000
Therefore, Present share for 17 year old son = Rs. 8,000

13.

A sum of money is invested at 10% per annum compounding annually for 2 years. If the interest received is Rs. 210, find the principal.

A.) 2500
B.) 1000
C.) 1900
D.) 2100

Given, r = 10%
n = 2 years
Compound interest, C.I = Rs. 210
Compound Interest = Amount – Principal
=> C.I = P {[1 + (r/100)]n - 1}
=> 210 = P {[1 + (10/100)]2 - 1}
=> 210 = P {[1 + (1/10)]2 - 1}
=> 210 = P {[(10 + 1)/10]2 - 1}
=> 210 = P {[11/10]^2 - 1}
=> 210 = P {[121 / 100] – 1}
=> 210 = P {(121 – 100) / 100}
=> 210 = P {21 / 100}
=> P = (210 × 100) / 21
=> P = 1000 Rs.
Thus, Principal = Rs. 1000

14.

Rs. 10000 is borrowed at compound interest at the rate of 4 % annum. What will be the amount to be paid after 2 years?

A.) 10816
B.) 10800
C.) 10808
D.) 10826

Principal : P = 10000 Rs.
Rate of Interest : r = 4 %
Number of years : n = 2
Amount = P x (1 + r/100)n
Amount = 10000 x (1+4/100)2
=10000 x (1+1/25)2
=10000 x (26/25) x (26/25) =10816

15.

What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 9 % per annum?

A.) 940.5
B.) 5940.5
C.) 5990.5
D.) 980.5

Given principal = 5000
No. of years = 2
Rate of interest = 9
Amount = P x (1+r/100)n,
We get Amount = 5000 x (1+9/100)2 = 5000 x (109/100) x (109/100) = 5940.5
Compound Interest, C. I = Amount - Principal = 5940.5 - 5000 = 940.5

16.

The difference between the simple interest on a certain sum at the rate of 10% p.a. for 2 years and compound interest which is compounded every 6 months is Rs.124.05 .What is the principal sum?

A.) Rs.12,000
B.) Rs.8000
C.) Rs.10,000
D.) Rs.6000

Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
=P(1+ (R / 2)100) 2T−P= P(1+(10 / 2)100) 2× 2− P
=P(1+120)4−P=P(21 / 20) 4− P
Simple Interest on P at 10% for 2 years
=P × R × T /100
=P×10×2100
= P / 5
Then P[(1+5 / 100)4-1] - P x 10 x 2/100 = 124.05
⇒ P[(21/20)4 - 1 - 1/4] = 124.05
⇒ P[(194481/160000) - (6/5)] = 12405 /100
⇒ P[194481-192000 / 160000] = 12405 /100
⇒ P = [(12405/100) x (160000/2481)]
= 124.05 x 64.490
= 7999.9845
= 8000.

17.

The difference between compound interest and simple interest compounded annually on a certain sum of money for 2 years at 4% p.a. is Re.1 The sum (in Rs) is:

A.) 625
B.) 525
C.) 635
D.) 685

625

18.

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs.12,000 after 3 years at the same rate?

A.) Rs.3972
B.) Rs.2160
C.) Rs.3120
D.) Rs.6240

Let the Principal be, P = Rs. 100
Given, S.I = 60% of 100 = Rs. 60, n = 6 years
Then, Rate of Interest, r = (S.I × 100 )/ (p × n)
=> r = (60 × 100) / (100 × 6)
=> r = 10 % p.a
Now, P = Rs. 12,000, n = 3 years, r = 10% p.a
C.I = P {[1 + (r/100)]n - 1}
= 12,000 × {[1 + (10/100)]3-1}
= 12,000 × [(11/10)3-1]
= 12,000 × [(1331/1000) - 1]
= 12,000 × (331/1000)
= 12 × 331
= 3972
Thus, Compound Interest = Rs. 3972

19.

The compound interest on a certain sum for 2 years at 10% p.a. is Rs.525.The simple interest on the same sum for double the time at the half the rate percent per annum is:

A.) Rs.600
B.) Rs.500
C.) Rs.400
D.) Rs.800

Given, n = 2 years
r = 10 %
Compound interest (C.I) = Rs. 525
Compound interest (C.I) = P {[1 + (r/100)]n – 1}
=> P {[1 + (10/100)]2– 1} = 525
=> P {[1 + (1/10)]2 – 1} = 525
=> P {[11/10]2 – 1} = 525
=> P {[121/100] – 1} = 525
=> 21 × P / 100 = 525
=> P = 2500
Now for Simple Interest, S.I = p × n × r / 100
P = Rs. 2500
n = 4 years
r = 5%
S.I = (2500 × 4 × 5) / 100
=> S.I = 500 Rs.

20.

What will be compounded interest on a sum of Rs.25,000 after 3 years at the rate of 12 p.c.p.a.?

A.) Rs.9000.30
B.) Rs.10123.20
C.) Rs.9720
D.) Rs.9820