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