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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}

  • 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

  • 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}

  • 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

  • 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}

  • 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

  • 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}