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

1.

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

2.

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/-

3.

Find the simple interest on Rs. 1300 at 10 % per annum for 5 years

A.) Rs. 650
B.) Rs. 550
C.) Rs. 525
D.) Rs. 615

Given
Principal : 1300
Rate of interest : 10
Number of years : 5
Simple Interest = pnr / 100
Simple Interest = (1300 x 5 x 10) / 100
= 13 x 5 x 10
= 650
So,Simple Interest = 650

4.

What would be the compound interest accrued on an amount of 4500 Rs. at the end of 2 years at the rate of 10 % per annum ?

A.) 5435
B.) 5445
C.) 5665
D.) 5345

Given principal = 4500
No. of years = 2
Rate of interest = 10
Amount = P x (1+r/100)n,
We get Amount = 4500 x (1+10/100)2
Ans : 5445

5.

What would be the compound interest accrued on an amount of 11500 Rs. at the end of 2 years at the rate of 10 % per annum ?

A.) 13905
B.) 13915
C.) 13925
D.) 13965

Given
Principal = 11500 No. of years = 2 Rate of interest = 10
Amount = P x (1+r/100)n
we get Amount = 11500 x (1+10/100)2
= 11500 x ( 110 / 100 )2
= 115 x 11 x 11
= 115 x 121
Ans : 13915