Answer: B
Let the principal \((p) = x\) rupees ∴ Simple interest \((I) = \frac{x \times 2 \times 4}{100} = \frac{2x}{25}\) rupees ∴ \(\frac{2x}{25} = 80\) ⇒ \(x = 80 \times \frac{25}{2} = 1000\) ∴ Principal = ₹1000 Compound amount for 2 years at 4% on ₹1000 = \(1000(1 + \frac{4}{100})^2 = 1000(1 + \frac{1}{25})^2 = 1000\left(\frac{26}{25}\right)^2 = 1000 \times \frac{26}{25} \times \frac{26}{25} = ₹1081.60\) ∴ Compound interest = ₹1081.60 − ₹1000 = ₹81.60
Let the principal \((p) = x\) rupees ∴ Simple interest \((I) = \frac{x \times 2 \times 4}{100} = \frac{2x}{25}\) rupees ∴ \(\frac{2x}{25} = 80\) ⇒ \(x = 80 \times \frac{25}{2} = 1000\) ∴ Principal = ₹1000 Compound amount for 2 years at 4% on ₹1000 = \(1000(1 + \frac{4}{100})^2 = 1000(1 + \frac{1}{25})^2 = 1000\left(\frac{26}{25}\right)^2 = 1000 \times \frac{26}{25} \times \frac{26}{25} = ₹1081.60\) ∴ Compound interest = ₹1081.60 − ₹1000 = ₹81.60