Compound Interest Questions Answers

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

  1. 3%
  2. 4%
  3. 5%
  4. 6%

  Answer And Explanation

  Answer: Option D

  Explanation:

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

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

  1. 4 years
  2. 5 years
  3. 6 years
  4. 7 years

  Answer And Explanation

  Answer: Option 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
  

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• 10. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is

  1. Rs 1650
  2. Rs 1750
  3. Rs 1850
  4. Rs 1950

  Answer And Explanation

  Answer: Option B

  Explanation:

  $$\begin{aligned} C.I. = (4000 \times(1+\frac{10}{100})^2 - 4000) \\ = 4000 * \frac{11}{10} * \frac{11}{10} - 4000 \\ = 840 \\ \text{So S.I. = } \frac{840}{2} = 420\\ \text{So Sum = } \frac{S.I. * 100}{R*T} \\ = \frac{420 * 100}{3*8} \\ = Rs 1750 \end{aligned}$$

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• 11. In what time will Rs.1000 become Rs.1331 at 10% per annum compounded annually

  1. 2 Years
  2. 3 Years
  3. 4 Years
  4. 5 Years

  Answer And Explanation

  Answer: Option 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

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• 12. If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, what will be the compound interest on same values

  1. Rs.51.75
  2. Rs 51.50
  3. Rs 51.25
  4. Rs 51

  Answer And Explanation

  Answer: Option C

  Explanation:

  $$\begin{aligned} S.I. = \frac{P*R*T}{100} \\ P = \frac{50*100}{5*2} = 500\\ Amount = 500(1+\frac{5}{100})^2 \\ 500(\frac{21}{20} * \frac{21}{20}) \\ = 551.25 \\ C.I. = 551.25 - 500 = 51.25 \end{aligned}$$

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• 13. What will be the difference between simple and compound interest @ 10% per annum on the sum of Rs 1000 after 4 years

  1. Rs 62.10
  2. Rs 63.10
  3. Rs 64.10
  4. Rs 65.10

  Answer And Explanation

  Answer: Option C

  Explanation:

  $$\begin{aligned} S.I. = \frac{1000*10*4}{100} = 400 \\ C.I. = [1000(1+\frac{10}{100})^4 - 1000] \\ = 464.10 \end{aligned}$$
  
  So difference between simple interest and compound interest will be 464.10 - 400 = 64.10

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• 14. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum

  1. Rs 600
  2. Rs 625
  3. Rs 650
  4. Rs 675

  Answer And Explanation

  Answer: Option B

  Explanation:

  Let the Sum be P
  $$\begin{aligned} S.I. = \frac{P*4*2}{100} = \frac{2P}{25}\\ C.I. = P(1+\frac{4}{100})^2 - P \\ = \frac{676P}{625} - P \\ = \frac{51P}{625} \\ \text{As, C.I. - S.I = 1}\\ => \frac{51P}{625} - \frac{2P}{25} = 1 \\ => \frac{51P - 50P}{625} = 1 \\ P = 625 \end{aligned}$$
  

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