A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched ₹482 more by way of interest. Calculate the sum of money lent out.

Let the sum lent out be ₹x.

When C.I. is reckoned yearly,

Rate = 20%

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{20}{100}\Big)^2 \\[1em] = x\Big(1 + \dfrac{1}{5}\Big)^2 \\[1em] = x\Big(\dfrac{6}{5}\Big)^2 \\[1em] = \dfrac{36x}{25}$

C.I. = A - P = $\dfrac{36x}{25} - x = \dfrac{11x}{25}.$

When C.I. is reckoned half-yearly,

Rate = $\dfrac{20\%}{2}$ = 10%.

n (no. of conversion periods) = 4 half-years.

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = x\Big(1 + \dfrac{1}{10}\Big)^4 \\[1em] = x\Big(\dfrac{11}{10}\Big)^4 \\[1em] = \dfrac{14641x}{10000}$

C.I. = A - P = $\dfrac{14641x}{10000} - x = \dfrac{4641x}{10000}.$

Given, difference between C.I. in both cases = ₹482.

$\therefore \dfrac{4641x}{10000} - \dfrac{11x}{25} = 482 \\[1em] \Rightarrow \dfrac{4641x - 4400x}{10000} = 482 \\[1em] \Rightarrow \dfrac{241x}{10000} = 482 \\[1em] \Rightarrow x = \dfrac{482 \times 10000}{241} \\[1em] \Rightarrow x = ₹20000.$

Hence, sum lent out = ₹20000.