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

1.

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/-

2.

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

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

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

3.

What would be the compound interest accrued on an amount of 6500 Rs. at the end of 2 years at the rate of 15 % per annum ?

A.) 8596.25
B.) 2096.25
C.) 8589.75
D.) 8589.25

Given principal = 6500
No. of years = 2
Rate of interest = 15
Amount = P [ 1 + ( r / 100 )n]
= 6500 x [ 1 + ( 15 /100 )2]
= 6500 x [ 1 + ( 3 / 20 )2]
= 6500 x [ 23 / 20 ]2
= 6500 x [ 529 / 400 ]
Amount =8596.25
Compound Interest = Amount - Principal

= 8596.25 - 6500
= 2096.25

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.

John invested an amount of Rs. 20000 for 2 years at compound interest at the rate of 6 % per annum. Find the amount he receives at the end of 2 years.

A.) 22120
B.) 22472
C.) 22000
D.) 22372

Principal : P = 20000 Rs.
Rate of Interest : r = 6 %
Number of years : n = 2
Amount = P x (1 + r/100)n
= 20000 x (1+6/100)2
= 20000 x (1+3/50)2
= 20000 x (53/50) x (53/50)
= 22472
Thus, Amount that Johnreceives at the end of 2 years = Rs.22472