# Compound Interest Questions Answers

1) Find the compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annuall

- 1) Rs. 610
- 2) Rs. 612
- 3) Rs. 614
- 4) Rs. 616

**Ans.**B

**Explanation :**

\begin{aligned} Amount = [7500 \times (1+ \frac{4}{100})^2] \ = (7500 \times \frac{26}{25} \times \frac{26}{25}) \ = 8112 \ \end{aligned} So compound interest = (8112 - 7500) = 612

2) Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposi

- 1) Rs. 8510
- 2) Rs. 8620
- 3) Rs. 8730
- 4) Rs. 8820

**Ans.**D

**Explanation :**

\begin{aligned} => (8000 \times(1+\frac{5}{100})^2) \ => 8000 \times \frac{21}{20}\times \frac{21}{20} \ => 8820 \end{aligned}

3) What will be the compound interest on Rs. 25000 after 3 years at the rate of 12 % per ann

- 1) Rs 10123.20
- 2) Rs 10123.30
- 3) Rs 10123.40
- 4) Rs 10123.50

**Ans.**A

**Explanation :**

\begin{aligned} (25000 \times (1 + \frac{12}{100})^3) \=> 25000\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25} \=> 35123.20 \\end{aligned} So Compound interest will be 35123.20 - 25000 = Rs 10123.20

4) A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 year

- 1) Rs 662
- 2) Rs 662.01
- 3) Rs 662.02
- 4) Rs 662.03

**Ans.**C

**Explanation :**

\begin{aligned} [200(1+\frac{5}{100})^3 + 200(1+\frac{5}{100})^2+ \\ 200(1+\frac{5}{100})] = [200(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20})\\ + 200(\frac{21}{20}\times\frac{21}{20})+200(\frac{21}{20})] \ = 662.02 \end{aligned}

5) Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarter

- 1) Rs 2520
- 2) Rs 2521
- 3) Rs 2522
- 4) Rs 2523

**Ans.**C

**Explanation :**

Please remember, when we have to calculate C.I. quarterly then we apply following formula if n is the number of years \begin{aligned} Amount = P(1+\frac{\frac{R}{4}}{100})^{4n} \end{aligned} Principal = Rs.16,000; Time=9 months = 3 quarters; Rate = 20%, it will be 20/4 = 5% So lets solve this question now, \begin{aligned} Amount = 16000(1+\frac{5}{100})^3 \= 18522\C.I = 18522 - 16000 = 2522 \end{aligned}

6) The present worth of Rs.169 due in 2 years at 4% per annum compound interest

- 1) Rs 155.25
- 2) Rs 156.25
- 3) Rs 157.25
- 4) Rs 158.25

**Ans.**B

**Explanation :**

In this type of question we apply formula \begin{aligned} Amount = \frac{P}{(1+\frac{R}{100})^n} \ Amount = \frac{169}{(1+\frac{4}{100})^2} \ Amount = \frac{169 * 25 * 25}{26*26} \ Amount = 156.25 \end{aligned}

7) Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annual

- 1) Rs 312
- 2) Rs 412
- 3) Rs 512
- 4) Rs 612

**Ans.**D

**Explanation :**

Please apply the formula \begin{aligned} Amount = P(1+\frac{R}{100})^n \\text{C.I. = Amount - P} \end{aligned}

8) At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 yea

- 1) 3%
- 2) 4%
- 3) 5%
- 4) 6%

**Ans.**D

**Explanation :**

Let Rate will be R% \begin{aligned} 1200(1+\frac{R}{100})^2 = \frac{134832}{100} \ (1+\frac{R}{100})^2 = \frac{134832}{120000} \ (1+\frac{R}{100})^2 = \frac{11236}{10000} \ (1+\frac{R}{100}) = \frac{106}{100} \ => R = 6\% \end{aligned}

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

- 1) 4 years
- 2) 5 years
- 3) 6 years
- 4) 7 years

**Ans.**A

**Explanation :**

As per question we need something like following \begin{aligned} P(1+\frac{R}{100})^n > 2P \ (1+\frac{20}{100})^n > 2 \ (\frac{6}{5})^n > 2 \ \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\times\frac{6}{5} > 2 \end{aligned} So answer is 4 years

10) In what time will Rs.1000 become Rs.1331 at 10% per annum compounded annua

- 1) 2 Years
- 2) 3 Years
- 3) 4 Years
- 4) 5 Years

**Ans.**B

**Explanation :**

Principal = Rs.1000; Amount = Rs.1331; Rate = Rs.10%p.a. Let the time be n years then, \begin{aligned} 1000(1+\frac{10}{100})^n = 1331 \ (\frac{11}{10})^n = \frac{1331}{1000} \ (\frac{11}{10})^3 = \frac{1331}{1000} \ \end{aligned} So answer is 3 years