Practice MCQ Questions and Answer on Compound Interest

21.

An amount of Rs. 10000 becomes Rs. 14641 in 2 years if the interest is compounded half yearly. What is the rate of compound interest p.c.p.a. ?

(A) 10%
(B) 12%
(C) 16%
(D) 20%

Solution: $\begin{aligned} Let the rate be R% p .a . \\ Then, \\ 10000 {(1 + \frac{R}{2 \times 100})}^{4} = 14641 \\ \Rightarrow {(1 + \frac{R}{200})}^{4} = \frac{14641}{10000} = {(\frac{11}{10})}^{4} \\ \Rightarrow 1 + \frac{R}{200} = \frac{11}{10} \\ \Rightarrow \frac{R}{200} = \frac{1}{10} \\ \Rightarrow R = 20% \end{aligned}$

22.

If the compound interest on a sum of money for 3 years at the rate of 5% per annum is Rs. 252.20, the simple interest on the same sum at the same rate and for the same time is ?

(A) Rs. 220
(B) Rs. 240
(C) Rs. 245
(D) Rs. 250

Solution: $\begin{aligned} Rate = 5% \\ Time = 3 years \\ Compound Interest Rs. 252 .20 \\ Effective rate% of CI for 3 years \\ = 15 .7625% \\ Effective rate% of SI for 3 years \\ = 5 \times 3 = 15% \\ Required SI \\ = \frac{252.20}{15.7625} \times 15 \\ = 240 \end{aligned}$

23.

Find the rate percent per annum if Rs. 2000 amounts to Rs. 2315.25 in one and half years interest being compounded half yearly.

(A) 10%
(B) 11.5%
(C) 5%
(D) 20%

Solution: $\begin{aligned} According to the question, \\ compounded half yearly \\ Rate = \frac{R}{2} \\ Time = \frac{2T}{3} \\ Amount = P {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow 2315.25 = 2000 {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow \frac{2315.25}{2000} = {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow \frac{231525}{200000} = {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow \frac{9261}{8000} = {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow {(\frac{21}{20})}^{3} = {(1 + \frac{R}{2 \times 100})}^{3} \\ \Rightarrow 1 + \frac{R}{200} = \frac{21}{20} \\ \Rightarrow R = 10 % \end{aligned}$

24.

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

(A) 3
(B) 4
(C) 5
(D) 6

Solution: $\begin{aligned} P {(1 + \frac{20}{100})}^{n} > 2 P \Rightarrow {(\frac{6}{5})}^{n} > 2 \\ Now, (\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}) > 2 \\ S o, n = 4 years \end{aligned}$

25.

A loan is to be returned in two equal yearly instalments. If the rate of interest is 10% p.a., compounded annually, and each instalment if Rs. 5,808, then the total interest charged in this scheme is:

(A) Rs. 1,563
(B) Rs. 1,536
(C) Rs. 1,632
(D) Rs. 1,602

Solution: $\begin{aligned} 5808 \times \frac{10}{11} \times \frac{21}{11} \\ Principal = 10080 \\ Amount = 5808 \times 2 = 11616 \\ Interest = 1536 \end{aligned}$

26.

What will be the simple to compound interest (compounded annually) ratio on two identical funds put in a bank at an interest rate of 8% (per annum) for three years?

(A) 3 : 5
(B) 1875 : 2029
(C) 3 : 4
(D) 243 : 342

Solution: $8 % = \frac{2}{25}$ $Simple Interest = \frac{15625 \times 8 \times 3}{100} = 3750$ $\begin{matrix} \Rightarrow & Simple Interest & : & Compound Interest \\ 3750 & : & 4058 \\ 1875 & : & 2029 \end{matrix}$

27.

The difference between compound and simple interest on a certain sum for 3 years at 5% per annum is Rs. 122. The sum is = ?

(A) Rs. 16000
(B) Rs. 15000
(C) Rs. 12000
(D) Rs. 10000

Solution: $\begin{aligned} \Rightarrow P [{(\frac{21}{20})}^{3} - 1 - \frac{3}{20}] = 122 \\ \Rightarrow P [\frac{21^{3} - 20^{3} - 3 \times 20^{2}}{20^{3}}] = 122 \\ \Rightarrow P [\frac{9261 - 8000 - 1200}{8000}] = 122 \\ \Rightarrow P \times \frac{61}{8000} = 122 \\ \Rightarrow P = \frac{8000 \times 122}{61} \\ \Rightarrow P = Rs . 16000 \end{aligned}$

28.

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years ?

(A) 6.5%
(B) 4.5%
(C) 6%
(D) 7.5%

Solution: $\begin{aligned} A = P {(1 + \frac{R}{100})}^{n} \\ \Rightarrow 1348.32 = 1200 {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{134832}{120000} = {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{231525}{200000} = {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{2809}{2500} = {(1 + \frac{R}{100})}^{2} \\ \Rightarrow {(\frac{53}{50})}^{2} = {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{53}{50} = 1 + \frac{R}{100} \\ \Rightarrow R = 6% \end{aligned}$

29.

Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?

(A) Rs. 8600
(B) Rs. 8620
(C) Rs. 8820
(D) None of these

Solution: $\begin{aligned} Amount = R s . [8000 \times {(1 + \frac{5}{100})}^{2}] \\ = R s . (8000 \times \frac{21}{20} \times \frac{21}{20}) \\ = R s . 8820 \end{aligned}$

30.

The difference between the amount of compound interest and simple interest accrued on an amount of Rs. 26000 at the end of 3 years is Rs. 2994.134. What is the rate of interest p.c.p.a ?

(A) 17%
(B) 19%
(C) 22%
(D) Cannot be determined

Solution: Let the R% p.a. Then, $[26000 \times {(1 + \frac{R}{100})}^{3} - 26000] -$ $(\frac{26000 \times R \times 3}{100}) =$ $2994.134$ $\Rightarrow 26000 [{(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100}] =$ $2994.134$ $\Rightarrow 26000$ $[\frac{{(100 - R)}^{3} - 1000000 - 30000 R}{1000000}] =$ $2994.134$ $\Rightarrow 26 [{1000000 + R^{3} + 300 R (100 + R) - 1000000 - 30000 R}] = 2994134$ $\Rightarrow R^{3} + 300 R^{2} = \frac{2994134}{26} =$ $115159$ $\Rightarrow R^{2} (R + 300) = 115159$ $\Rightarrow R = 19 %$

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Practice MCQ Questions and Answer on Compound Interest

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