Let the amount payable be P
Compound interest when compounded half-yearly \(=P(1+\frac{R}{2×100})^{2t}-P\)
⇒ \(P(1+\frac{10}{100})^4-P\)
⇒ \(P[(\frac{11}{10})^4-1]\) ⇒ \(P[\frac{14641-10000}{10000}]=\frac{4641}{10000}P\)
Compound interest when compounded annually \(=P(1+\frac{R}{100})^t-P\)
⇒ \(P(1+\frac{1}{5})^2-P\)
⇒ \(P(1+\frac{1}{5})^2-P\) ⇒ \(P(\frac{6}{5})^2-P=P[(\frac{36}{5})-1]\)
⇒ \(P(\frac{11}{25})\)
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.The correct option is (B): Rs.20,000