A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more. If the interest was payable half yearly than if it was payable annually, the sum is

Let the amount payable be P

Compound interest when compounded half-yearly = $$P(1 + \frac{R}{2 \times 100})^{2t} - P$$

$$\Rightarrow P(1 + \frac{10}{100})^{4} - P$$

$$\Rightarrow P[(\frac{11}{10})^{4} - 1] \Rightarrow P[\frac{14641 -10000}{10000}] = \frac{4641}{10000} P$$

Compound interest when compounded annually = $$P(1 + \frac{R}{100})^{t} - P$$

$$\Rightarrow P(1 + \frac{20}{100})^{2} - P$$

$$\Rightarrow P(1 + \frac{1}{5})^{2} - P \Rightarrow P(\frac{6}{5})^{2} - P = P[(\frac{36}{25}) - 1]$$

$$\Rightarrow P(\frac{11}{25})$$

According to the given question,

$$P(\frac{4641}{10000}) - P(\frac{11}{25}) = 482$$

$$P(\frac{4641- 4400}{10000}) = 482$$

$$P = \frac{4820000}{241}$$ (or) $$P = 20,000$$.

Hence, option B is the correct answer.