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. |

- A man deposited \(Rs.6000\) in a bank for two years. Find the amount of interest if rate of interest is \(10 \ \%\) compounded annually?
- \(Rs.1200\)
- \(Rs.1250\)
- \(Rs.1260\)
- \(Rs.1280\)

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

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

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

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

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

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

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

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

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

- A women lent \(Rs.2000\) at compound interest of \(20 \ \%\) per annum for three years. Find the total amount after three years?
- \(Rs.3456\)
- \(Rs.3535\)
- \(Rs.3628\)
- \(Rs.3650\)

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

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

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

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

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

- If the interest compounded at the rate of \(5 \ \%\) per annum for two years is \(Rs.500\). find the principal amount?
- \(Rs.4545\)
- \(Rs.4878\)
- \(Rs.4656\)
- \(Rs.4965\)

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

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

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

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

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

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

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

- 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?
- \(Rs.1464.1\)
- \(Rs.1568.6\)
- \(Rs.1666.2\)
- \(Rs.1325.5\)

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

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

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

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

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

- 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?
- \(Rs.2835.4\)
- \(Rs.2768.5\)
- \(Rs.2928.2\)
- \(Rs.2564.8\)

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

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

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

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

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

- Calculate the simple interest on the principal amount \(Rs.15000\) at the rate of interest \(2 \ \%\) for two years?
- \(Rs.2000\)
- \(Rs.2800\)
- \(Rs.3000\)
- \(Rs.3400\)

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

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

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

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

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

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

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

- 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?
- \(Rs.4687.5\)
- \(Rs.4258.6\)
- \(Rs.4863.4\)
- \(Rs.4755.2\)

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

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

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

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

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

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

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

- 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??
- \(Rs.3260\)
- \(Rs.3587\)
- \(Rs.3658\)
- \(Rs.3750\)

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, $$ A = P + SI $$ $$ A = 2500 + 1250 $$ $$ A = Rs.3750 $$

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, $$ A = P + SI $$ $$ A = 2500 + 1250 $$ $$ A = Rs.3750 $$

- 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?
- \(Rs.45,000\)
- \(Rs.50,000\)
- \(Rs.52,000\)
- \(Rs.55,000\)

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, $$ \frac{P}{100} = 500 $$ $$ P = Rs.50,000 $$

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, $$ \frac{P}{100} = 500 $$ $$ P = Rs.50,000 $$

- If a man paid \(Rs.12,000\) after two years including interest compounded \(10 \ \%\) annually, then find the principal amount?
- \(Rs.9865.12\)
- \(Rs.9917.35\)
- \(Rs.8989.62\)
- \(Rs.8957.24\)

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

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

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

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

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

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

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

Lec 1: Introduction to Simple Interest
Exercise-1
Lec 2: Compound Interest
Exercise-2
Lec 3: Compound Interest for Successive years
Exercise-3
Exercise-4
Exercise-5