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 lent \(Rs.100\) at compound interest of \(10 \ \%\) per annum for first year and \(20 \ \%\) per annum for second year compounded annually. Find the total amount after two years?
- \(Rs.130\)
- \(Rs.132\)
- \(Rs.135\)
- \(Rs.138\)

Answer: (b) \(Rs.132\)

Solution: Given, principal amount (P) = \(Rs.100\)

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(20 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2}{100}\right) $$ $$ A = 100 \ \left(1 + \frac{10}{100}\right) \ \left(1 + \frac{20}{100}\right) $$ $$ = 100 \times \frac{11}{10} \times \frac{6}{5} $$ $$ A = Rs.132 $$

Solution: Given, principal amount (P) = \(Rs.100\)

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(20 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2}{100}\right) $$ $$ A = 100 \ \left(1 + \frac{10}{100}\right) \ \left(1 + \frac{20}{100}\right) $$ $$ = 100 \times \frac{11}{10} \times \frac{6}{5} $$ $$ A = Rs.132 $$

- Rahul deposited \(Rs.500\) in a bank at the compound interest of \(20 \ \%\) per annum for first year compounded annually and \(50 \ \%\) per annum for second year compounded half-yearly. Find the total amount after two years Rahul will get from the bank?
- \(Rs.937.5\)
- \(Rs.942.3\)
- \(Rs.946.8\)
- \(Rs.936.6\)

Answer: (a) \(Rs.937.5\)

Solution: Given, principal amount (P) = \(Rs.500\)

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

rate of interest for second year \((R_2)\) = \(50 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2/2}{100}\right)^2 $$ $$ A = 500 \ \left(1 + \frac{20}{100}\right) \ \left(1 + \frac{25}{100}\right)^2 $$ $$ = 500 \times \frac{6}{5} \times \frac{5}{4} \times \frac{5}{4} $$ $$ A = Rs.937.5 $$

Solution: Given, principal amount (P) = \(Rs.500\)

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

rate of interest for second year \((R_2)\) = \(50 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2/2}{100}\right)^2 $$ $$ A = 500 \ \left(1 + \frac{20}{100}\right) \ \left(1 + \frac{25}{100}\right)^2 $$ $$ = 500 \times \frac{6}{5} \times \frac{5}{4} \times \frac{5}{4} $$ $$ A = Rs.937.5 $$

- A girl deposited \(Rs.1000\) in a bank at the compound interest of \(10 \ \%\) per annum for first year compounded half-yearly and \(40 \ \%\) per annum for second year compounded quarterly. Find the total amount after two years the girl will get from the bank?
- \(Rs.1524.37\)
- \(Rs.1614.17\)
- \(Rs.1725.12\)
- \(Rs.1847.57\)

Answer: (b) \(Rs.1614.17\)

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

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(40 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1/2}{100}\right)^2 \ \left(1 + \frac{R_2/4}{100}\right)^4 $$ $$ A = 1000 \ \left(1 + \frac{5}{100}\right)^2 \ \left(1 + \frac{10}{100}\right)^4 $$ $$ = 1000 \times \left(\frac{21}{20}\right)^2 \times \left(\frac{11}{10}\right)^4 $$ $$ = 1000 \times \frac{441}{400} \times \frac{14641}{10000} $$ $$ A = Rs.1614.17 $$

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

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(40 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1/2}{100}\right)^2 \ \left(1 + \frac{R_2/4}{100}\right)^4 $$ $$ A = 1000 \ \left(1 + \frac{5}{100}\right)^2 \ \left(1 + \frac{10}{100}\right)^4 $$ $$ = 1000 \times \left(\frac{21}{20}\right)^2 \times \left(\frac{11}{10}\right)^4 $$ $$ = 1000 \times \frac{441}{400} \times \frac{14641}{10000} $$ $$ A = Rs.1614.17 $$

- A man lent \(Rs.700\) from the bank at the compound interest of \(40 \ \%\) per annum compounded half-yearly. Find the amount, the man will have to pay after two years?
- \(Rs.1226.15\)
- \(Rs.1258.46\)
- \(Rs.1451.52\)
- \(Rs.1652.89\)

Answer: (c) \(Rs.1451.52\)

Solution: Given, principal amount (P) = \(Rs.700\)

rate of interest (R) = \(40 \ \%\)

time period (n) = \(2\) years

then total amount after two years, $$ A = P \ \left(1 + \frac{R/2}{100}\right)^{2n} $$ $$ A = 700 \ \left(1 + \frac{20}{100}\right)^4 $$ $$ A = 700 \ \left(\frac{6}{5}\right)^4 $$ $$ A = Rs.1451.52 $$

Solution: Given, principal amount (P) = \(Rs.700\)

rate of interest (R) = \(40 \ \%\)

time period (n) = \(2\) years

then total amount after two years, $$ A = P \ \left(1 + \frac{R/2}{100}\right)^{2n} $$ $$ A = 700 \ \left(1 + \frac{20}{100}\right)^4 $$ $$ A = 700 \ \left(\frac{6}{5}\right)^4 $$ $$ A = Rs.1451.52 $$

- If the interest compounded at the rate of \(10 \ \%\) per annum for three years is \(Rs.1000\). Find the principal amount?
- \(Rs.3021.14\)
- \(Rs.3124.61\)
- \(Rs.3358.33\)
- \(Rs.3568.14\)

Answer: (a) \(Rs.3021.14\)

Solution: Given, compound interest (CI) = \(Rs.1000\)

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

time period (n) = \(3\) years,$$ CI = P \ \left(1 + \frac{R}{100}\right)^n - P $$ $$ 1000 = P \ \left(1 + \frac{10}{100}\right)^3 - P $$ $$ 1000 = P \ \left[\left(\frac{11}{10}\right)^3 - 1\right] $$ $$ 1000 = P \ \left[\left(\frac{1331}{1000}\right) - 1\right] $$ $$ P = Rs.3021.14 $$

Solution: Given, compound interest (CI) = \(Rs.1000\)

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

time period (n) = \(3\) years,$$ CI = P \ \left(1 + \frac{R}{100}\right)^n - P $$ $$ 1000 = P \ \left(1 + \frac{10}{100}\right)^3 - P $$ $$ 1000 = P \ \left[\left(\frac{11}{10}\right)^3 - 1\right] $$ $$ 1000 = P \ \left[\left(\frac{1331}{1000}\right) - 1\right] $$ $$ P = Rs.3021.14 $$

- Sachin deposited \(Rs.100\) at the compound interest of \(5 \ \%\) per annum compounded yearly for first year, \(10 \ \%\) per annum compounded yearly for second year and \(20 \ \%\) per annum compounded half-yearly for third year. Find the total amount after three years?
- \(Rs.142.3\)
- \(Rs.141.5\)
- \(Rs.140.6\)
- \(Rs.139.7\)

Answer: (d) \(Rs.139.7\)

Solution: Given, principal amount (P) = \(Rs.100\)

rate of interest for first year \((R_1)\) = \(5 \ \%\)

rate of interest for second year \((R_2)\) = \(10 \ \%\)

rate of interest for third year \((R_3)\) = \(20 \ \%\)

then total amount after three years,

Solution: Given, principal amount (P) = \(Rs.100\)

rate of interest for first year \((R_1)\) = \(5 \ \%\)

rate of interest for second year \((R_2)\) = \(10 \ \%\)

rate of interest for third year \((R_3)\) = \(20 \ \%\)

then total amount after three years,

$$ A = P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right) \left(1 + \frac{R_3/2}{100}\right)^2 $$

$$ A = P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right) \\ \left(1 + \frac{R_3/2}{100}\right)^2 $$

$$ A = 100 \left(1 + \frac{5}{100}\right) \left(1 + \frac{10}{100}\right) \left(1 + \frac{10}{100}\right)^2 $$

$$ A = 100 \left(1 + \frac{5}{100}\right) \left(1 + \frac{10}{100}\right) \\ \left(1 + \frac{10}{100}\right)^2 $$

$$ = 100 \times \left(\frac{21}{20}\right) \times \left(\frac{11}{10}\right)^3 $$ $$ A = Rs.139.7 $$- Calculate the simple interest on the principal amount \(Rs.8000\) at the rate of interest \(6 \ \%\) per annum for three years?
- \(Rs.1360\)
- \(Rs.1400\)
- \(Rs.1418\)
- \(Rs.1440\)

Answer: (d) \(Rs.1440\)

Solution: Given, principal amount (P) = \(Rs.8000\)

rate of interest (R) = \(6 \ \%\)

time period (T) = \(3\) years, then $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{8000 \times 6 \times 3}{100} $$ $$ SI = Rs.1440 $$

Solution: Given, principal amount (P) = \(Rs.8000\)

rate of interest (R) = \(6 \ \%\)

time period (T) = \(3\) years, then $$ SI = \frac{P \ R \ T}{100} $$ $$ = \frac{8000 \times 6 \times 3}{100} $$ $$ SI = Rs.1440 $$

- In how many years a principal amount become three times if the rate of interest is \(4 \ \%\) per annum?
- \(40 \ years\)
- \(45 \ years\)
- \(50 \ years\)
- \(55 \ years\)

Answer: (c) \(50 \ years\)

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

simple interest (SI) = \(3k - k\) = \(Rs.2k\)

rate of interest (R) = \(4 \ \%\)

Hence, $$ SI = \frac{P \ R \ T}{100} $$ $$ 2k = \frac{k \times 4 \times T}{100} $$ $$ T = 50 \ years $$

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

simple interest (SI) = \(3k - k\) = \(Rs.2k\)

rate of interest (R) = \(4 \ \%\)

Hence, $$ SI = \frac{P \ R \ T}{100} $$ $$ 2k = \frac{k \times 4 \times T}{100} $$ $$ T = 50 \ years $$

- Vikash deposited \(Rs.1200\) in a bank at the rate of \(20 \ \%\) per annum compounded quarterly. Find the total amount after one year?
- \(Rs.1458.6\)
- \(Rs.1486.1\)
- \(Rs.1488.6\)
- \(Rs.1492.8\)

Answer: (a) \(Rs.1458.6\)

Solution: Given, principal amount (P) = \(Rs.1200\)

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

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

then total amount after one year, $$ A = P \ \left(1 + \frac{R/4}{100}\right)^{4n} $$ $$ A = 1200 \ \left(1 + \frac{5}{100}\right)^4 $$ $$ A = 1200 \ \left(\frac{21}{20}\right)^4 $$ $$ A = Rs.1458.6 $$

Solution: Given, principal amount (P) = \(Rs.1200\)

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

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

then total amount after one year, $$ A = P \ \left(1 + \frac{R/4}{100}\right)^{4n} $$ $$ A = 1200 \ \left(1 + \frac{5}{100}\right)^4 $$ $$ A = 1200 \ \left(\frac{21}{20}\right)^4 $$ $$ A = Rs.1458.6 $$

- A women lent \(Rs.2200\) from the bank at the compound interest of \(10 \ \%\) per annum for first year compounded annually and \(20 \ \%\) per annum for second year compounded half-yearly. Find the total amount after two years?
- \(Rs.2834.6\)
- \(Rs.2874.4\)
- \(Rs.2928.2\)
- \(Rs.2936.3\)

Answer: (c) \(Rs.2928.2\)

Solution: Given, principal amount (P) = \(Rs.2200\)

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(20 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2/2}{100}\right)^2 $$ $$ A = 2200 \ \left(1 + \frac{10}{100}\right) \ \left(1 + \frac{10}{100}\right)^2 $$ $$ = 2200 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} $$ $$ A = Rs.2928.2 $$

Solution: Given, principal amount (P) = \(Rs.2200\)

rate of interest for first year \((R_1)\) = \(10 \ \%\)

rate of interest for second year \((R_2)\) = \(20 \ \%\)

then total amount after two years, $$ A = P \ \left(1 + \frac{R_1}{100}\right) \ \left(1 + \frac{R_2/2}{100}\right)^2 $$ $$ A = 2200 \ \left(1 + \frac{10}{100}\right) \ \left(1 + \frac{10}{100}\right)^2 $$ $$ = 2200 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} $$ $$ A = Rs.2928.2 $$

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