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. |

- A man deposited \(Rs.1000\) in a bank at the rate of \(5 \ \%\) for two years. Find the simple interest on the principal amount?
- \(Rs.120\)
- \(Rs.112\)
- \(Rs.105\)
- \(Rs.100\)

Answer: (d) \(Rs.100\)

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

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

time period (T) = \(2\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{1000 \times 5 \times 2}{100} $$ $$ SI = Rs.100 $$

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

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

time period (T) = \(2\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{1000 \times 5 \times 2}{100} $$ $$ SI = Rs.100 $$

- If a principal amount become two times in five years, then find the rate of interest on the principal amount?
- \(10 \ \%\)
- \(15 \ \%\)
- \(20 \ \%\)
- \(25 \ \%\)

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

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

after five years the amount = \(Rs.2k\)

interest on principal amount = \(2k - k\) = \(Rs.k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ k = \frac{k \times R \times 5}{100} $$ $$ R = \frac{20 \ k}{k} = 20 \ \% $$

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

after five years the amount = \(Rs.2k\)

interest on principal amount = \(2k - k\) = \(Rs.k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ k = \frac{k \times R \times 5}{100} $$ $$ R = \frac{20 \ k}{k} = 20 \ \% $$

- In how many years a principal amount become double, if rate of interest is \(10 \ \%\)?
- \(5 \ years\)
- \(10 \ years\)
- \(15 \ years\)
- \(20 \ years\)

Answer: (b) \(10 \ years\)

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

simple interest = \(2k - k\) = \(Rs.k\)

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

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

simple interest = \(2k - k\) = \(Rs.k\)

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

- In how many years a principal amount become four times, if rate of interest is \(25 \ \%\)?
- \(20 \ years\)
- \(10 \ years\)
- \(12 \ years\)
- \(15 \ years\)

Answer: (c) \(12 \ years\)

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

simple interest = \(4k - k\) = \(Rs.3k\)

rate of interest (R) = \(25 \ \%\), then $$ SI = \frac{P \ R \ T}{100} $$ $$ 3k = \frac{k \times 25 \times T}{100} $$ $$ T = 12 \ years $$

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

simple interest = \(4k - k\) = \(Rs.3k\)

rate of interest (R) = \(25 \ \%\), then $$ SI = \frac{P \ R \ T}{100} $$ $$ 3k = \frac{k \times 25 \times T}{100} $$ $$ T = 12 \ years $$

- If a principal amount become six times in ten years, then find the rate of interest on the principal amount?
- \(20 \ \%\)
- \(30 \ \%\)
- \(40 \ \%\)
- \(50 \ \%\)

Answer: (d) \(50 \ \%\)

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

after ten years the amount = \(Rs.6k\)

interest on principal amount = \(6k - k\) = \(Rs.5k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ 5k = \frac{k \times R \times 10}{100} $$ $$ R = \frac{50 \ k}{k} = 50 \ \% $$

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

after ten years the amount = \(Rs.6k\)

interest on principal amount = \(6k - k\) = \(Rs.5k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ 5k = \frac{k \times R \times 10}{100} $$ $$ R = \frac{50 \ k}{k} = 50 \ \% $$

- A girl deposited \(Rs.2000\) in a bank at the rate of \(10 \ \%\) for five years. Find the total amount after the maturity period?
- \(Rs.1000\)
- \(Rs.2500\)
- \(Rs.3000\)
- \(Rs.3500\)

Answer: (c) \(Rs.3000\)

Solution: Given, principal amount (P) = \(Rs.2000\)

rate of interest (R) = \(10 \ \%\)

time period (T) = \(5\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{2000 \times 10 \times 5}{100} $$ $$ SI = Rs.1000 $$ total amount after maturity period, $$ A = P + SI $$ $$ = 2000 + 1000 $$ $$ = Rs.3000 $$

Solution: Given, principal amount (P) = \(Rs.2000\)

rate of interest (R) = \(10 \ \%\)

time period (T) = \(5\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{2000 \times 10 \times 5}{100} $$ $$ SI = Rs.1000 $$ total amount after maturity period, $$ A = P + SI $$ $$ = 2000 + 1000 $$ $$ = Rs.3000 $$

- A women deposited a certain amount in a bank at the rate of \(10 \ \%\) for five years. If after five years the women got interest amount \(Rs.1000\), then find the principal amount?
- \(Rs.1000\)
- \(Rs.2000\)
- \(Rs.3000\)
- \(Rs.4000\)

Answer: (b) \(Rs.2000\)

Solution: Given, rate of interest (R) = \(10 \ \%\)

time period (T) = \(5\) years

simple interest (SI) = \(Rs.1000\), then $$ SI = \frac{P \ R \ T}{100} $$ $$ 1000 = \frac{P \times 10 \times 5}{100} $$ $$ P = Rs.2000 $$

Solution: Given, rate of interest (R) = \(10 \ \%\)

time period (T) = \(5\) years

simple interest (SI) = \(Rs.1000\), then $$ SI = \frac{P \ R \ T}{100} $$ $$ 1000 = \frac{P \times 10 \times 5}{100} $$ $$ P = Rs.2000 $$

- If a principal amount become double in three years, then find the rate of interest on the principal amount?
- \(30.5 \ \%\)
- \(32.2 \ \%\)
- \(35.2 \ \%\)
- \(33.3 \ \%\)

Answer: (d) \(33.3 \ \%\)

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

after three years the amount = \(Rs.2k\)

interest on principal amount = \(2k - k\) = \(Rs.k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ k = \frac{k \times R \times 3}{100} $$ $$ R = \frac{100 \ k}{3 \ k} = 33.3 \ \% $$

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

after three years the amount = \(Rs.2k\)

interest on principal amount = \(2k - k\) = \(Rs.k\), $$ SI = \frac{P \ R \ T}{100} $$ $$ k = \frac{k \times R \times 3}{100} $$ $$ R = \frac{100 \ k}{3 \ k} = 33.3 \ \% $$

- The amount \(Rs.5000\) is deposited in a bank at the rate of \(20 \ \%\) for two years. Find the total amount after the maturity period?
- \(Rs.5000\)
- \(Rs.6000\)
- \(Rs.7000\)
- \(Rs.8000\)

Answer: (c) \(Rs.7000\)

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

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

time period (T) = \(2\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{5000 \times 20 \times 2}{100} $$ $$ SI = Rs.2000 $$ total amount after maturity period, $$ A = P + SI $$ $$ = 5000 + 2000 $$ $$ = Rs.7000 $$

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

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

time period (T) = \(2\) years, $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{5000 \times 20 \times 2}{100} $$ $$ SI = Rs.2000 $$ total amount after maturity period, $$ A = P + SI $$ $$ = 5000 + 2000 $$ $$ = Rs.7000 $$

- A man deposited \(Rs.10,000\) in a bank and after maturity period the man got the total amount \(Rs.12,000\), then find the interest amount on the principal?
- \(Rs.1000\)
- \(Rs.2000\)
- \(Rs.3000\)
- \(Rs.4000\)

Answer: (b) \(Rs.2000\)

Solution: Given, principal amount (P) = \(Rs.10,000\)

total amount (A) = \(Rs.12,000\), then $$ SI = A - P $$ $$ SI = 12,000 - 10,000 $$ $$ SI = Rs.2000 $$

Solution: Given, principal amount (P) = \(Rs.10,000\)

total amount (A) = \(Rs.12,000\), then $$ SI = A - P $$ $$ SI = 12,000 - 10,000 $$ $$ SI = Rs.2000 $$

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