Question

What is the compound interest on Rs 9,000 invested for 2 years at the rate of 10 p. c. p. a ? *
Rs 10,890
Rs 10,980
Rs 1890
Rs 1980​

Answer 1

We have,

$a \: = p (1 + \frac{r}{100} ) {}^{n}$

Where

• Principal (P) = ₹9000
• Rate (R) = 10
• Number of Years (n) = 2

Finding The Amount :-

$= 9000(1 + \frac{10}{100} ) {}^{2}$

$= 9000(1 + \frac{1}{10}) {}^{2}$

$= 9000( \frac{10 + 1}{10} ) {}^{2}$

$= 9000( \frac{11}{10} ) {}^{2}$

$= 9000 \times \frac{11}{10} \times \frac{11}{10}$

$= 90 \times \frac{11}{1} \times \frac{11}{1}$

$= 90 \times 11 \times 11$

$= 10890$

Therefore,Amount (a) = ₹10890

Compound Interest:-

CI = A - P

CI = ₹(10890 - 9000)

CI = ₹1890