A sum of money was lent for 2 years at 20% compounded annually.

Solution:

$\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$

Also,

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

Let the sum of money be Rs $\mathrm{x}$.

If the interest is compounded annually, then:

$\mathrm{A}_{1}=\mathrm{x}\left(1+\frac{20}{100}\right)^{2}$

$=1.44 \mathrm{x}$

$\therefore \mathrm{CI}=1.44 \mathrm{x}-\mathrm{x}$

$=0.44 \mathrm{x} \quad \ldots(1)$

If the interest is compounded half - yearly, then :

$\mathrm{A}_{2}=\mathrm{x}\left(1+\frac{10}{100}\right)^{4}$

$=1.4641 \mathrm{x}$

$\therefore \mathrm{CI}=1.4641 \mathrm{x}-\mathrm{x}$

$=0.4641 \mathrm{x} \quad \ldots(2)$

It is given that if interest is compounded half-yearly, then it will be Rs 482 more.

$\therefore 0.4641 \mathrm{x}=0.44 \mathrm{x}+482 \quad[$ From $(1)$ and $(2)]$

$0.4641 \mathrm{x}-0.44 \mathrm{x}=482$

$0.0241 \mathrm{x}=482$

$\mathrm{X}=\frac{482}{0.0241}$

$=20,000$

Thus, the required sum is Rs 20,000 .