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. |

- 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?
- \(5.05 \ \%\)
- \(3.25 \ \%\)
- \(1.25 \ \%\)
- \(1.50 \ \%\)

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 $$ \frac{P_1 \ R_1 \ T_1}{100} + \frac{P_2 \ R_2 \ T_2}{100} = SI $$ $$ \frac{1000 \times R \times 2}{100} + \frac{2000 \times R \times 3}{100} = 100 $$ $$ 80 \ R = 100 $$ $$ R = 1.25 \ \% $$

Solution: Given, \(P_1 = Rs.1000\)

\(P_2 = Rs.2000\)

\(T_1 = 2 \ years\)

\(T_2 = 3 \ years\)

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

- 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?
- \(12 \ years\)
- \(15 \ years\)
- \(20 \ years\)
- \(22 \ years\)

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 $$ SI = \frac{P \ R \ T}{100} $$ $$ 15000 = \frac{5000 \times 15 \times T}{100} $$ $$ T = 20 \ years $$

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 $$ SI = \frac{P \ R \ T}{100} $$ $$ 15000 = \frac{5000 \times 15 \times T}{100} $$ $$ T = 20 \ years $$

- 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?
- \(Rs.300\)
- \(Rs.325\)
- \(Rs.350\)
- \(Rs.360\)

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

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

- 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?
- \(10 \ \%\)
- \(15 \ \%\)
- \(20 \ \%\)
- \(25 \ \%\)

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, $$ SI = \frac{P \ R \ T}{100} $$ $$ 200 = \frac{200 \times R \times 5}{100} $$ $$ R = 20 \ \% $$

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, $$ SI = \frac{P \ R \ T}{100} $$ $$ 200 = \frac{200 \times R \times 5}{100} $$ $$ R = 20 \ \% $$

- If a sum of money becomes four times in two years at compound interest. Find the rate of interest?
- \(100 \ \%\)
- \(75 \ \%\)
- \(50 \ \%\)
- \(25 \ \%\)

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

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

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

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

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

- If a sum of money becomes \(16\) times in four years at compound interest. Find the rate of interest?
- \(110 \ \%\)
- \(100 \ \%\)
- \(150 \ \%\)
- \(200 \ \%\)

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

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

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

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

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

- 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?
- \(1 \ year\)
- \(2 \ years\)
- \(3 \ years\)
- \(4 \ years\)

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, $$ n = 4 \ years $$

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, $$ n = 4 \ years $$

- 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?
- \(10 \ year\)
- \(20 \ years\)
- \(30 \ years\)
- \(40 \ years\)

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 $$ SI = \frac{P \ R \ T}{100} $$ $$ 4000 = \frac{1000 \times 10 \times T}{100} $$ $$ T = 40 \ years $$

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

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

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

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

- If a sum of money becomes \(9\) times in two years at compound interest, then find the rate of interest?
- \(100 \ \%\)
- \(150 \ \%\)
- \(200 \ \%\)
- \(250 \ \%\)

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

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

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

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

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

- 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?
- \(1 \ years\)
- \(2 \ years\)
- \(3 \ years\)
- \(4 \ years\)

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, $$ n = 4 \ years $$

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, $$ n = 4 \ years $$

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