# Tick (✓) the correct answer

Question:

The compound interest on Rs 10000 at $12 \%$ per annum for $1 \frac{1}{2}$ years, compounded annually, is

(a) Rs 1872

(b) Rs 1720

(c) Rs 1910.16

(d) Rs 1782

Solution:

(a) Rs 1872

Here, $A=P \times\left(1+\frac{R}{100}\right)^{1} \times\left(1+\frac{\frac{1}{2} R}{100}\right)$

$=\operatorname{Rs} 10000 \times\left(1+\frac{12}{100}\right) \times\left(1+\frac{\frac{1}{2} \times 12}{100}\right)$

$=\operatorname{Rs} 10000 \times\left(\frac{100+12}{100}\right) \times\left(\frac{100+6}{100}\right)$

$=\operatorname{Rs} 10000 \times\left(\frac{112}{100}\right) \times\left(\frac{106}{100}\right)$

$=\operatorname{Rs} 10000 \times\left(\frac{28}{25}\right) \times\left(\frac{53}{50}\right)$

$=\operatorname{Rs}(8 \times 28 \times 53)$

$=\operatorname{Rs} 11872$

$\therefore$ Compound interest $=$ amount $-$ principal $=\operatorname{Rs}(11872-10000)=\operatorname{Rs} 1872$
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