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. |

- 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?
- \(Rs.7000\)
- \(Rs.7200\)
- \(Rs.7500\)
- \(Rs.7800\)

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 $$ 12000 = P \times \frac{5}{3} $$ $$ P = Rs.7200 $$

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 $$ 12000 = P \times \frac{5}{3} $$ $$ P = Rs.7200 $$

- If a principal amount becomes \(9\) times in two years on compound interest. Find the rate of interest?
- \(110 \ \%\)
- \(100 \ \%\)
- \(150 \ \%\)
- \(200 \ \%\)

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

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

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

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

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

- If a principal amount becomes \(Rs.2000\) after four years and \(Rs.5000\) after \(8\) years on compound interest, then find the principal amount?
- \(Rs.700\)
- \(Rs.750\)
- \(Rs.800\)
- \(Rs.850\)

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 $$ 2000 = P \times \frac{5}{2} $$ $$ P = Rs.800 $$

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 $$ 2000 = P \times \frac{5}{2} $$ $$ P = Rs.800 $$

- In what time the amount \(Rs.150\) will become \(Rs.300\). If simple interest is being calculated at the rate of \(25 \ \%\) per annum?
- \(2 \ years\)
- \(3 \ years\)
- \(4 \ years\)
- \(5 \ years\)

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 $$ SI = \frac{P \ R \ T}{100} $$ $$ 150 = \frac{150 \times 25 \times T}{100} $$ $$ T = 4 \ years $$

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

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

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

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

- 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?
- \(8.82 \ \%\)
- \(8.15 \ \%\)
- \(9.12 \ \%\)
- \(9.95 \ \%\)

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 $$ \frac{P_1 \ R_1 \ T_1}{100} + \frac{P_2 \ R_2 \ T_2}{100} = SI $$ $$ \frac{500 \times R \times 1}{100} + \frac{600 \times R \times 2}{100} = 150 $$ $$ 5 \ R + 12 \ R = 150 $$ $$ R = 8.82 \ \% $$

Solution: Given, \(P_1 = Rs.500\)

\(P_2 = Rs.600\)

\(T_1 = 1 \ year\)

\(T_2 = 2 \ years\)

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

- If a principal amount becomes two times in two years, then find in how many years the principal amount becomes \(8\) times?
- \(3 \ years\)
- \(4 \ years\)
- \(5 \ years\)
- \(6 \ years\)

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, $$ n = 6 \ years $$

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, $$ n = 6 \ years $$

- 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?
- \(Rs.2000\)
- \(Rs.2120\)
- \(Rs.2180\)
- \(Rs.2250\)

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

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

- 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?
- \(Rs.2830\)
- \(Rs.3025\)
- \(Rs.3145\)
- \(Rs.3225\)

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

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

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

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

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

- A man deposited \(Rs.1500\) in a bank at the rate of \(20 \ \%\) per annum compounded quarterly. Find the total amount after six months?
- \(Rs.1458.6\)
- \(Rs.1586.2\)
- \(Rs.1653.7\)
- \(Rs.1792.8\)

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

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

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

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

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

- 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?
- \(Rs.3450\)
- \(Rs.3587\)
- \(Rs.3658\)
- \(Rs.3750\)

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, $$ A = P + SI $$ $$ A = 3000 + 450 $$ $$ A = Rs.3450 $$

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, $$ A = P + SI $$ $$ A = 3000 + 450 $$ $$ A = Rs.3450 $$

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