Compound Interest Problems Solutions

• 1. 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 :

  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

• 2. Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarterly

  1. Rs 2520
  2. Rs 2521
  3. Rs 2522
  4. Rs 2523

  Answer :

  Option 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}

• 3. Find the compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

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

  Answer :

  Option 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

• 4. 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 :

  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}
  

• 5. The present worth of Rs.169 due in 2 years at 4% per annum compound interest is

  1. Rs 155.25
  2. Rs 156.25
  3. Rs 157.25
  4. Rs 158.25

  Answer :

  Option 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}

• 6. 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 :

  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}

• 7. 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 :

  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