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 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 $$ A = P \left[1 + \frac{R/2}{100}\right]^2n $$ $$ A = 10000 \left[1 + \frac{10/2}{100}\right]^{2 \times \frac{3}{2}} $$ $$ A = 10000 \left[1 + \frac{5}{100}\right]^3 $$ $$ A = 10000 \times \left[\frac{21}{20}\right]^3 $$ $$ A = 10000 \times \frac{9261}{8000} $$ $$ A = 11,576.25 \ Rs. $$

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