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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