Topic Included: | Formulas, Definitions & Exmaples. |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Simple Interest and Compound Interest Aptitude Notes & Questions. |

Questions for practice: | 10 Questions & Answers with Solutions. |

**Case (1):** If the rate of interest changes successively after a period of time, then,

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \times ..... $$

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \\ \times \left(1 + \frac{R_3}{100}\right) \times ..... $$

**Example (1):** Mr. Kamal lent \(Rs. \ 2000\) at compound interest of \(10 \ \%\) per annum for 1st year and \(20 \ \%\) per annum for 2nd year. What will be the total amount Mr. Kamal will have to pay after \(2\) years?

**Solution:** Given values, \(P = Rs. \ 2000\), \(R_1 = 10 \ \%\), \(R_2 = 20 \ \%\), \(n = 2\) years, then $$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) $$ $$ A = 2000 \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{20}{100}\right) $$ $$ A = 2000 \left(\frac{11}{10}\right) \times \left(\frac{6}{5}\right) $$ $$ A = 2000 \times \frac{66}{50} = Rs.2640 \ (Answer) $$

**Example (2):** A man lent \(Rs. \ 4000\) at compound interest of \(5 \ \%\) per annum for 1st year, \(10 \ \%\) per annum for 2nd year and \(20 \ \%\) per annum for 3rd year. What will be the total amount the man will have to pay after \(5\) years?

**Solution:** Given values, \(P = Rs. \ 4000\), \(R_1 = 5 \ \%\), \(R_2 = 10 \ \%\), \(R_3 = 20 \ \%\), \(n = 5\) years, then

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

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

$$ A = 4000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{20}{100}\right) $$

$$ A = 4000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10}{100}\right) \\ \times \left(1 + \frac{20}{100}\right) $$

$$ A = 4000 \left(\frac{21}{20}\right) \times \left(\frac{11}{10}\right) \times \left(\frac{6}{5}\right) $$ $$ A = Rs.5544 \ (Answer) $$

**Case (2):** If the rate of interest changes successively after a period of time, but for first year interest compounded yearly, for second year interest compounded half-yearly and for third year interest compounded quarterly then,

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \times \left(1 + \frac{R_3/4}{100}\right)^4 \times ..... $$

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \\ \times \left(1 + \frac{R_3/4}{100}\right)^4 \times ..... $$

**Example (1):** A man lent \(Rs. \ 10000\) at compound interest of \(20 \ \%\) per annum for 1st year, compounded yearly and \(40 \ \%\) per annum for 2nd year, compounded half-yearly. What will be the total amount the man will have to pay after \(2\) years?

**Solution:** Given values, \(P = Rs. \ 10000\), \(R_1 = 20 \ \%\), \(R_2 = 40 \ \%\), \(n = 2\) years, then $$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 $$ $$= 10000 \left(1 + \frac{20}{100}\right) \times \left(1 + \frac{40/2}{100}\right)^2 $$ $$ A = 10000 \left(\frac{6}{5}\right) \times \left(\frac{6}{5}\right)^2 $$ $$ A = Rs.17280 \ (Answer) $$

**Example (2):** A women lent \(Rs. \ 5000\) at compound interest of \(5 \ \%\) per annum for 1st year, compounded yearly and \(10 \ \%\) per annum for 2nd year, compounded half-yearly and \(20 \ \%\) per annum for third year, compounded quarterly. What will be the total amount the women will have to pay after \(3\) years?

**Solution:** Given values, \(P = Rs. \ 5000\), \(R_1 = 5 \ \%\), \(R_2 = 10 \ \%\), \(R_3 = 20 \ \%\), \(n = 3\) years, then

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \times \left(1 + \frac{R_3/4}{100}\right)^4 $$

$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \\ \times \left(1 + \frac{R_3/4}{100}\right)^4 $$

$$ A = 5000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10/2}{100}\right)^2 \times \left(1 + \frac{20/4}{100}\right)^4 $$

$$ A = 5000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10/2}{100}\right)^2 \\ \times \left(1 + \frac{20/4}{100}\right)^4 $$

$$= 5000 \left(\frac{21}{20}\right) \times \left(\frac{21}{20}\right)^2 \times \left(\frac{21}{20}\right)^4 $$ $$ A = Rs.7035.50 \ (Answer) $$

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