# 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. If a certain principal amount becomes $$Rs.12,000$$ after two years and $$Rs.20,000$$ after four years on compound interest, then find the principal amount?

1. $$Rs.7000$$
2. $$Rs.7200$$
3. $$Rs.7500$$
4. $$Rs.7800$$

Answer: (b) $$Rs.7200$$

Solution: Given, Let the principal amount = $$Rs.P$$, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$12000 = P \ \left(1 + \frac{R}{100}\right)^2......(1)$$ and $$20000 = P \ \left(1 + \frac{R}{100}\right)^4......(2)$$ by dividing equation (2) from (1), $$\frac{20000}{12000} = \left(1 + \frac{R}{100}\right)^2$$ $$\frac{5}{3} = \left(1 + \frac{R}{100}\right)^2.....(3)$$ by putting the value from equation (3) to equation (1), we get $$12000 = P \times \frac{5}{3}$$ $$P = Rs.7200$$

1. If a principal amount becomes $$9$$ times in two years on compound interest. Find the rate of interest?

1. $$110 \ \%$$
2. $$100 \ \%$$
3. $$150 \ \%$$
4. $$200 \ \%$$

Answer: (d) $$200 \ \%$$

Solution: Given, total amount (A) = $$Rs.9P$$

time period (n) = $$2$$ years, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$9P = P \ \left(1 + \frac{R}{100}\right)^2$$ $$(3)^2 = \left(1 + \frac{R}{100}\right)^2$$ $$3 = 1 + \frac{R}{100}$$ $$R = 200 \ \%$$

1. If a principal amount becomes $$Rs.2000$$ after four years and $$Rs.5000$$ after $$8$$ years on compound interest, then find the principal amount?

1. $$Rs.700$$
2. $$Rs.750$$
3. $$Rs.800$$
4. $$Rs.850$$

Answer: (c) $$Rs.800$$

Solution: Given, Let the principal amount = $$Rs.P$$, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$2000 = P \ \left(1 + \frac{R}{100}\right)^4......(1)$$ and $$5000 = P \ \left(1 + \frac{R}{100}\right)^8......(2)$$ by dividing equation (2) from (1), $$\frac{5000}{2000} = \left(1 + \frac{R}{100}\right)^4$$ $$\frac{5}{2} = \left(1 + \frac{R}{100}\right)^4.....(3)$$ by putting the value from equation (3) to equation (1), we get $$2000 = P \times \frac{5}{2}$$ $$P = Rs.800$$

1. In what time the amount $$Rs.150$$ will become $$Rs.300$$. If simple interest is being calculated at the rate of $$25 \ \%$$ per annum?

1. $$2 \ years$$
2. $$3 \ years$$
3. $$4 \ years$$
4. $$5 \ years$$

Answer: (c) $$4 \ years$$

Solution: Given, principal amount (P) = $$Rs.150$$

total amount (A) = $$Rs.300$$

simple interest (SI) = $$Rs.150$$

rate of interest (R) = $$25 \ \%$$, then $$SI = \frac{P \ R \ T}{100}$$ $$150 = \frac{150 \times 25 \times T}{100}$$ $$T = 4 \ years$$

1. Vishal lent $$Rs.500$$ to Shikhar for one year and $$Rs.600$$ to Rohit for two years and from both, he received $$Rs.150$$ as interest. Find the rate of interest, if simple interest is being calculated?

1. $$8.82 \ \%$$
2. $$8.15 \ \%$$
3. $$9.12 \ \%$$
4. $$9.95 \ \%$$

Answer: (a) $$8.82 \ \%$$

Solution: Given, $$P_1 = Rs.500$$

$$P_2 = Rs.600$$

$$T_1 = 1 \ year$$

$$T_2 = 2 \ years$$

amount of interest = $$Rs.150$$, then $$\frac{P_1 \ R_1 \ T_1}{100} + \frac{P_2 \ R_2 \ T_2}{100} = SI$$ $$\frac{500 \times R \times 1}{100} + \frac{600 \times R \times 2}{100} = 150$$ $$5 \ R + 12 \ R = 150$$ $$R = 8.82 \ \%$$

1. If a principal amount becomes two times in two years, then find in how many years the principal amount becomes $$8$$ times?

1. $$3 \ years$$
2. $$4 \ years$$
3. $$5 \ years$$
4. $$6 \ years$$

Answer: (d) $$6 \ years$$

Solution: Given, total amount (A) = $$2P$$

time period (n) = $$2$$ years, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$2P = P \ \left(1 + \frac{R}{100}\right)^2$$ $$2 = \left(1 + \frac{R}{100}\right)^2....(1)$$ sum becomes $$8$$ times then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$8P = P \ \left(1 + \frac{R}{100}\right)^n$$ $$(2)^3 = \left(1 + \frac{R}{100}\right)^n$$ now from the equation (1), $$\left[\left(1 + \frac{R}{100}\right)^2\right]^3 = \left(1 + \frac{R}{100}\right)^n$$ $$\left(1 + \frac{R}{100}\right)^6 = \left(1 + \frac{R}{100}\right)^n$$ Hence, $$n = 6 \ years$$

1. A man lent $$Rs.1500$$ at compound interest of $$20 \ \%$$ per annum for first year and $$25 \ \%$$ per annum for second year compounded annually. Find the total amount after two years?

1. $$Rs.2000$$
2. $$Rs.2120$$
3. $$Rs.2180$$
4. $$Rs.2250$$

Answer: (d) $$Rs.2250$$

Solution: Given, principal amount (P) = $$Rs.1500$$

rate of interest for first year $$(R_1)$$ = $$20 \ \%$$

rate of interest for second year $$(R_2)$$ = $$25 \ \%$$

then total amount after two years, $$A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2}{100}\right)$$ $$A = 1500 \ \left(1 + \frac{20}{100}\right) \ \left(1 + \frac{25}{100}\right)$$ $$= 1500 \times \frac{6}{5} \times \frac{5}{4}$$ $$A = Rs.2250$$

1. A girl lent $$Rs.2500$$ from the bank at the compound interest of $$20 \ \%$$ per annum compounded half-yearly. Find the amount, the girl will have to pay after one year?

1. $$Rs.2830$$
2. $$Rs.3025$$
3. $$Rs.3145$$
4. $$Rs.3225$$

Answer: (b) $$Rs.3025$$

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

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

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

then total amount after one year, $$A = P \ \left(1 + \frac{R/2}{100}\right)^{2n}$$ $$A = 2500 \ \left(1 + \frac{10}{100}\right)^2$$ $$A = 2500 \ \left(\frac{11}{10}\right)^2$$ $$A = Rs.3025$$

1. A man deposited $$Rs.1500$$ in a bank at the rate of $$20 \ \%$$ per annum compounded quarterly. Find the total amount after six months?

1. $$Rs.1458.6$$
2. $$Rs.1586.2$$
3. $$Rs.1653.7$$
4. $$Rs.1792.8$$

Answer: (c) $$Rs.1653.7$$

Solution: Given, principal amount (P) = $$Rs.1500$$

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

time period (n) = $$1/2$$ year

then total amount after six months, $$A = P \ \left(1 + \frac{R/4}{100}\right)^{4n}$$ $$A = 1500 \ \left(1 + \frac{5}{100}\right)^2$$ $$A = 1500 \ \left(\frac{21}{20}\right)^2$$ $$A = Rs.1653.7$$

1. Rahul lent $$Rs.3000$$ from the bank at the rate of $$5 \ \%$$ per annum for three years. Find the total amount after three years, the man will have to pay after three years. If simple interest is being calculated?

1. $$Rs.3450$$
2. $$Rs.3587$$
3. $$Rs.3658$$
4. $$Rs.3750$$

Answer: (a) $$Rs.3450$$

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

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

time period (T) = $$3$$ years, $$SI = \frac{P \ R \ T}{100}$$ $$= \frac{3000 \times 5 \times 3}{100}$$ $$SI = Rs.450$$ then total amount the man will have to pay after three years, $$A = P + SI$$ $$A = 3000 + 450$$ $$A = Rs.3450$$