# 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. Rahul lent $$Rs.2000$$ to Sachin for two years and $$Rs.2000$$ to Vikash for three years and from both he received $$Rs.100$$ as interest. Find the rate of interest, if simple interest is being calculated?

1. $$5.05 \ \%$$
2. $$3.25 \ \%$$
3. $$1.25 \ \%$$
4. $$1.50 \ \%$$

Answer: (c) $$1.25 \ \%$$

Solution: Given, $$P_1 = Rs.1000$$

$$P_2 = Rs.2000$$

$$T_1 = 2 \ years$$

$$T_2 = 3 \ years$$

amount of interest = $$Rs.100$$, then $$\frac{P_1 \ R_1 \ T_1}{100} + \frac{P_2 \ R_2 \ T_2}{100} = SI$$ $$\frac{1000 \times R \times 2}{100} + \frac{2000 \times R \times 3}{100} = 100$$ $$80 \ R = 100$$ $$R = 1.25 \ \%$$

1. In what time the amount $$Rs.5000$$ will convert into $$Rs.20,000$$. If simple interest is being calculated at the rate of $$15 \ \%$$ per annum?

1. $$12 \ years$$
2. $$15 \ years$$
3. $$20 \ years$$
4. $$22 \ years$$

Answer: (c) $$20 \ years$$

Solution: Given, principal amount (P) = $$Rs.5000$$

total amount (A) = $$Rs.20,000$$

simple interest (SI) = $$Rs.15,000$$

rate of interest (R) = $$15 \ \%$$, then $$SI = \frac{P \ R \ T}{100}$$ $$15000 = \frac{5000 \times 15 \times T}{100}$$ $$T = 20 \ years$$

1. A man borrowed $$Rs.1200$$ for two years at the rate of $$5 \ \%$$ per annum and $$Rs.1500$$ for the same duration at the rate of $$8 \ \%$$ per annum. Find the interest, if simple interest is being calculated?

1. $$Rs.300$$
2. $$Rs.325$$
3. $$Rs.350$$
4. $$Rs.360$$

Answer: (d) $$Rs.360$$

Solution: Given, $$P_1 = Rs.1200$$

$$P_2 = Rs.1500$$

$$T_1 = 2 \ years$$

$$T_2 = 2 \ years$$

$$R_1 = 5 \ \%$$

$$R_2 = 8 \ \%$$, then $$SI = \frac{P_1 \ R_1 \ T_1}{100} + \frac{P_2 \ R_2 \ T_2}{100}$$ $$= \frac{1200 \times 5 \times 2}{100} + \frac{1500 \times 8 \times 2}{100}$$ $$SI = Rs.360$$

1. If the simple interest on a certain amount at the rate of $$10 \ \%$$ per annum for ten years is $$Rs.200$$, then at what rate of interest the same amount of interest can be received on the same principal amount after five years?

1. $$10 \ \%$$
2. $$15 \ \%$$
3. $$20 \ \%$$
4. $$25 \ \%$$

Answer: (c) $$20 \ \%$$

Solution: from the first condition, $$SI = \frac{P \ R \ T}{100}$$ $$200 = \frac{P \times 10 \times 10}{100}$$ $$P = Rs.200$$ now the interest rate after five years, $$SI = \frac{P \ R \ T}{100}$$ $$200 = \frac{200 \times R \times 5}{100}$$ $$R = 20 \ \%$$

1. If a sum of money becomes four times in two years at compound interest. Find the rate of interest?

1. $$100 \ \%$$
2. $$75 \ \%$$
3. $$50 \ \%$$
4. $$25 \ \%$$

Answer: (a) $$100 \ \%$$

Solution: Given, total amount (A) = $$Rs.4P$$

time period (n) = $$2$$ years, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$4P = P \ \left(1 + \frac{R}{100}\right)^2$$ $$(2)^2 = \left(1 + \frac{R}{100}\right)^2$$ $$2 = 1 + \frac{R}{100}$$ $$R = 100 \ \%$$

1. If a sum of money becomes $$16$$ times in four years at compound interest. Find the rate of interest?

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

Answer: (b) $$100 \ \%$$

Solution: Given, total amount (A) = $$Rs.16 \ P$$

time period (n) = $$4$$ years, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$16P = P \ \left(1 + \frac{R}{100}\right)^4$$ $$(2)^4 = \left(1 + \frac{R}{100}\right)^4$$ $$2 = 1 + \frac{R}{100}$$ $$R = 100 \ \%$$

1. If a sum of money becomes two times in two years, then find in how many years the same sum of money becomes four times?

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

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

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 four times then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$4P = P \ \left(1 + \frac{R}{100}\right)^n$$ $$(2)^2 = \left(1 + \frac{R}{100}\right)^n$$ from the equation (1), $$\left[\left(1 + \frac{R}{100}\right)^2\right]^2 = \left(1 + \frac{R}{100}\right)^n$$ $$\left(1 + \frac{R}{100}\right)^4 = \left(1 + \frac{R}{100}\right)^n$$ Hence, $$n = 4 \ years$$

1. In what time the amount $$Rs.1000$$ will convert into $$Rs.5000$$. If simple interest is being calculated at the rate of $$10 \ \%$$ per annum?

1. $$10 \ year$$
2. $$20 \ years$$
3. $$30 \ years$$
4. $$40 \ years$$

Answer: (d) $$40 \ years$$

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

total amount (A) = $$Rs.5000$$

simple interest (SI) = $$Rs.4000$$

rate of interest (R) = $$10 \ \%$$, then $$SI = \frac{P \ R \ T}{100}$$ $$4000 = \frac{1000 \times 10 \times T}{100}$$ $$T = 40 \ years$$

1. If a sum of money becomes $$9$$ times in two years at compound interest, then find the rate of interest?

1. $$100 \ \%$$
2. $$150 \ \%$$
3. $$200 \ \%$$
4. $$250 \ \%$$

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

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

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 sum of money becomes three times in two years, then find in how many years the same sum of money becomes nine times?

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

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

Solution: Given, total amount (A) = $$3P$$

time period (n) = $$2$$ years, then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$3P = P \ \left(1 + \frac{R}{100}\right)^2$$ $$3 = \left(1 + \frac{R}{100}\right)^2....(1)$$ sum becomes nine times then $$A = P \ \left(1 + \frac{R}{100}\right)^n$$ $$9P = P \ \left(1 + \frac{R}{100}\right)^n$$ $$(3)^2 = \left(1 + \frac{R}{100}\right)^n$$ from the equation (1), $$\left[\left(1 + \frac{R}{100}\right)^2\right]^2 = \left(1 + \frac{R}{100}\right)^n$$ $$\left(1 + \frac{R}{100}\right)^4 = \left(1 + \frac{R}{100}\right)^n$$ Hence, $$n = 4 \ years$$