Simple and Compound Interest Aptitude Questions and Answers:


Overview:


Questions and Answers Type:MCQ (Multiple Choice Questions).
Main Topic:Quantitative Aptitude.
Quantitative Aptitude Sub-topic:Simple and Compound Interest Aptitude Questions and Answers.
Number of Questions:10 Questions with Solutions.

  1. A man deposited \(Rs.6000\) in a bank for two years. Find the amount of interest if rate of interest is \(10 \ \%\) compounded annually?

    1. \(Rs.1200\)
    2. \(Rs.1250\)
    3. \(Rs.1260\)
    4. \(Rs.1280\)


Answer: (c) \(Rs.1260\)

Solution: Given, principal amount (P) = \(Rs.6000\)

rate of interest (R) = \(10 \ \%\)

time period (n) = \(2\) years

then compound interest, $$ CI = P \ \left(1 + \frac{R}{100}\right)^n - P $$ $$ = 6000 \ \left(1 + \frac{10}{100}\right)^2 - 6000 $$ $$ = 6000 \times \frac{11}{10} \times \frac{11}{10} - 6000 $$ $$ CI = Rs.1260 $$

  1. A women lent \(Rs.2000\) at compound interest of \(20 \ \%\) per annum for three years. Find the total amount after three years?

    1. \(Rs.3456\)
    2. \(Rs.3535\)
    3. \(Rs.3628\)
    4. \(Rs.3650\)


Answer: (a) \(Rs.3456\)

Solution: Given, principal amount (P) = \(Rs.2000\)

rate of interest (R) = \(20 \ \%\)

time period (n) = \(3\) years

then the total amount, $$ A = P \ \left(1 + \frac{R}{100}\right)^n $$ $$ = 2000 \ \left(1 + \frac{20}{100}\right)^3 $$ $$ = 2000 \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} $$ $$ A = Rs.3456 $$

  1. If the interest compounded at the rate of \(5 \ \%\) per annum for two years is \(Rs.500\). find the principal amount?

    1. \(Rs.4545\)
    2. \(Rs.4878\)
    3. \(Rs.4656\)
    4. \(Rs.4965\)


Answer: (b) \(Rs.4878\)

Solution: Given, compound interest (CI) = \(Rs.500\)

rate of interest (R) = \(5 \ \%\)

time period (n) = \(2\) years, $$ CI = P \ \left(1 + \frac{R}{100}\right)^n - P $$ $$ 500 = P \ \left(1 + \frac{5}{100}\right)^2 - P $$ $$ 500 = P \ \left[\frac{21}{20} \times \frac{21}{20} - 1\right] $$ $$ 500 = P \ \frac{41}{400} $$ $$ P = Rs.4878 $$

  1. A boy deposited \(Rs.1000\) in a bank for two years. If the rate of interest is \(20 \ \%\) per annum compounded half-yearly. Find the total amount after two years?

    1. \(Rs.1464.1\)
    2. \(Rs.1568.6\)
    3. \(Rs.1666.2\)
    4. \(Rs.1325.5\)


Answer: (a) \(Rs.1464.1\)

Solution: Given, principal amount (P) = \(Rs.1000\)

rate of interest (R) = \(20 \ \%\)

time period (n) = \(2\) years

then total amount after two years, $$ A = P \ \left(1 + \frac{R/2}{100}\right)^{2n} $$ $$ = 1000 \ \left(1 + \frac{10}{100}\right)^4 $$ $$ = 1000 \times \left(\frac{11}{10}\right)^4 $$ $$ A = Rs.1464.1 $$

  1. The amount \(Rs.2000\) is deposited in a bank for one year at the rate of \(40 \ \%\) per annum compounded quarterly. Find the total amount after one year?

    1. \(Rs.2835.4\)
    2. \(Rs.2768.5\)
    3. \(Rs.2928.2\)
    4. \(Rs.2564.8\)


Answer: (c) \(Rs.2928.2\)

Solution: Given, principal amount (P) = \(Rs.2000\)

rate of interest (R) = \(40 \ \%\)

time period (n) = \(1\) years

then total amount after two years, $$ A = P \ \left(1 + \frac{R/4}{100}\right)^{4n} $$ $$ = 2000 \ \left(1 + \frac{10}{100}\right)^4 $$ $$ = 2000 \times \left(\frac{11}{10}\right)^4 $$ $$ A = Rs.2928.2 $$

  1. Calculate the simple interest on the principal amount \(Rs.15000\) at the rate of interest \(2 \ \%\) for two years?

    1. \(Rs.2000\)
    2. \(Rs.2800\)
    3. \(Rs.3000\)
    4. \(Rs.3400\)


Answer: (c) \(Rs.3000\)

Solution: Given, principal amount (P) = \(Rs.15000\)

rate of interest (R) = \(2 \ \%\)

time period (T) = \(10\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{15000 \times 2 \times 10}{100} $$ $$ SI = Rs.3000 $$

  1. A man deposited \(Rs.3000\) in a bank for two years. If interest compounded annually at the rate of \(25 \ \%\) per annum, then Find the total amount after two years?

    1. \(Rs.4687.5\)
    2. \(Rs.4258.6\)
    3. \(Rs.4863.4\)
    4. \(Rs.4755.2\)


Answer: (a) \(Rs.4687.5\)

Solution: Given, principal amount (P) = \(Rs.3000\)

rate of interest (R) = \(25 \ \%\)

time period (n) = \(2\) years, $$ A = P \ \left(1 + \frac{R}{100}\right)^n $$ $$ = 3000 \ \left(1 + \frac{25}{100}\right)^2 $$ $$ = 3000 \times \frac{5}{4} \times \frac{5}{4} $$ $$ = Rs.4687.5 $$

  1. A man lent \(Rs.2500\) from the bank at the rate of \(10 \ \%\) per annum for five years. Find the total amount after five years the man will have to pay. If simple interest is being calculated??

    1. \(Rs.3260\)
    2. \(Rs.3587\)
    3. \(Rs.3658\)
    4. \(Rs.3750\)


Answer: (d) \(Rs.3750\)

Solution: Given, principal amount (P) = \(Rs.2500\)

rate of interest (R) = \(10 \ \%\)

time period (T) = \(5\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{2500 \times 10 \times 5}{100} $$ $$ SI = Rs.1250 $$ then total amount the man will have to pay after five years, $$ A = P + SI $$ $$ A = 2500 + 1250 $$ $$ A = Rs.3750 $$

  1. The difference between the compound interest and the simple interest on a sum of money for two years at \(10 \ \%\) per annum is \(Rs.500\). Find the principal amount?

    1. \(Rs.45,000\)
    2. \(Rs.50,000\)
    3. \(Rs.52,000\)
    4. \(Rs.55,000\)


Answer: (b) \(Rs.50,000\)

Solution: Given, Let the principal amount = \(Rs.P\)

rate of interest (R) = \(10 \ \%\)

time period (n) = \(2\) years

then compound interest, $$ CI = P \ \left(1 + \frac{R}{100}\right)^n - P $$ $$ CI = P \ \left(1 + \frac{10}{100}\right)^2 - P $$ $$ = P \times \frac{11}{10} \times \frac{11}{10} - P $$ $$ CI = \frac{21 \ P}{100} $$ now the simple interest, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{P \times 10 \times 2}{100} $$ $$ SI = \frac{P}{5} $$ now the difference between CI and SI, $$ \frac{21 \ P}{100} - \frac{P}{5} $$ $$ = \frac{P}{100} $$ Hence, $$ \frac{P}{100} = 500 $$ $$ P = Rs.50,000 $$

  1. If a man paid \(Rs.12,000\) after two years including interest compounded \(10 \ \%\) annually, then find the principal amount?

    1. \(Rs.9865.12\)
    2. \(Rs.9917.35\)
    3. \(Rs.8989.62\)
    4. \(Rs.8957.24\)


Answer: (b) \(Rs.9917.35\)

Solution: Given, total amount (A) = \(Rs.12,000\)

rate of interest (R) = \(10 \ \%\)

time period (n) = \(2\) years, $$ A = P \ \left(1 + \frac{R}{100}\right)^n $$ $$ 12000 = P \ \left(1 + \frac{10}{100}\right)^2 $$ $$ 12000 = P \times \frac{11}{10} \times \frac{11}{10} $$ $$ P = Rs.9917.35 $$