| 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
$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \times \left(1 + \frac{R_3/4}{100}\right)^4 $$
$$ A = P \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2/2}{100}\right)^2 \\ \times \left(1 + \frac{R_3/4}{100}\right)^4 $$
$$ A = 5000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10/2}{100}\right)^2 \times \left(1 + \frac{20/4}{100}\right)^4 $$
$$ A = 5000 \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10/2}{100}\right)^2 \\ \times \left(1 + \frac{20/4}{100}\right)^4 $$
$$= 5000 \left(\frac{21}{20}\right) \times \left(\frac{21}{20}\right)^2 \times \left(\frac{21}{20}\right)^4 $$ $$ A = Rs.7035.50 \ (Answer) $$