Find the difference between the compound interest and simple interest.

Solution:

Given:

$\mathrm{P}=\mathrm{Rs} 50,000$

$\mathrm{R}=10 \%$ p. $\mathrm{a} .$

$\mathrm{n}=2$ years

We know that amount $\mathrm{A}$ at the end of $\mathrm{n}$ years at the rate $\mathrm{R} \%$ per annum when the interest is compounded annually is given by $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)$.

$\therefore \mathrm{A}=\mathrm{Rs} 50,000\left(1+\frac{10}{100}\right)^{2}$

$=$ Rs $50,000(1.1)^{2}$

$=$ Rs 60,500

Also,

$\mathrm{CI}=\mathrm{A}-\mathrm{P}$

$=\operatorname{Rs} 60,500-\operatorname{Rs} 50,000$

$=\operatorname{Rs} 10,500$

We know that:

$\mathrm{SI}=\frac{\mathrm{PRT}}{100}$

$=\frac{50,000 \times 10 \times 2}{100}$

$=\operatorname{Rs} 10,000$

$\therefore$ Difference between CI and SI $=$ Rs $10,500-$ Rs 10,000

$=\mathrm{Rs} 500$