# 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.

#### Compound Interest:

Compound interest is calculated on the principal amount as well as on interest amount of previous time duration.$$CI = P \left[1 + \frac{R}{100}\right]^n - P \qquad (1)$$ $$\bbox[5px,border:2px solid maroon] { CI = P \left[\left(1 + \frac{R}{100}\right)^n - 1\right] } \qquad (2)$$

from the equation $$(1)$$ $$CI + P = P \left[1 + \frac{R}{100}\right]^n \qquad (3)$$

Amount(A) = Principal Amount (P) + Interest $$\bbox[5px,border:2px solid maroon] { A = P \left[1 + \frac{R}{100}\right]^n } \qquad (4)$$

Note: For yearly calculation $$R \rightarrow R$$ and $$n \rightarrow n$$

For halfyearly calculation $$R \rightarrow R/2$$ and $$n \rightarrow 2n$$

For Quarterly calculation $$R \rightarrow R/4$$ and $$n \rightarrow 4n$$

Example (1): A woman lent $$Rs. \ 5000$$ at compound interest of $$10 \ \%$$ per annum. What will be the total amount woman will have to pay after $$2$$ years?

Solution: Given values, $$P = Rs. \ 5000$$, $$R = 10 \ \%$$ per annum, $$n = 2$$ years, then according to equation $$(4)$$, $$A = P \left[1 + \frac{R}{100}\right]^n$$ $$A = 5000 \left[1 + \frac{10}{100}\right]^2$$ $$A = 5000 \times \frac{121}{100} = Rs.6050 \ (Answer)$$

Example (2): Find the simple interest and compound interest on the principal amount of $$10,000 \ Rs$$, at the rate of $$10 \ \%$$ per annum for $$3$$ years, and also find the difference between the amount of simple interest and compound interest?

Solution: Given values, $$P = Rs. \ 10,000$$, $$R = 10 \ \%$$ per annum, $$n = 3$$ years, then simple interest, $$SI = \frac{P \times R \times T}{100}$$ $$SI = \frac{10000 \times 10 \times 3}{100}$$ $$SI = \frac{300000}{100} = 3000 \ Rs.$$ Now compound interest, $$CI = P \left[1 + \frac{R}{100}\right]^n - P$$ $$CI = 10000 \left[1 + \frac{10}{100}\right]^3 - 10000$$ $$CI = 10000 \left[\frac{1331}{1000} - 1\right]$$ $$CI = 10,000 \times 0.331 = 3310 \ Rs.$$ The difference between the simple interest and compound interest $$CI - SI = 3310 - 3000 = 310 \ Rs.$$

Example (3): A man lent $$Rs. \ 10,000$$ at compound interest of $$10 \ \%$$ per annum. What will be the total amount man will have to pay after $$18$$ months, if interest is compounded half-yearly?

Solution: Given values, $$P = Rs. \ 10,000$$, $$R = 10 \ \%$$ per annum, $$n = \frac{3}{2}$$ years, but interest is compounted half-yearly so $$R = \frac{R}{2}$$, and $$n = 2n$$, then $$A = P \left[1 + \frac{R/2}{100}\right]^2n$$ $$A = 10000 \left[1 + \frac{10/2}{100}\right]^{2 \times \frac{3}{2}}$$ $$A = 10000 \left[1 + \frac{5}{100}\right]^3$$ $$A = 10000 \times \left[\frac{21}{20}\right]^3$$ $$A = 10000 \times \frac{9261}{8000}$$ $$A = 11,576.25 \ Rs.$$