# Successive Compound Interest Aptitude Formulas, Definitions, & Examples:

#### Overview:

 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.

#### Successive Compound Interest:

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

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

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

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

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

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