If the C.I. on a sum for 2 years at 12 1/2 % per annum is Rs. 510, the S.I. on the same sum at the same rate for the same period of time is?
Answer: D
\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]
Recover the principal from the compound interest. With \( 12\tfrac{1}{2}\% = \tfrac{1}{8} \), the yearly factor is \( \tfrac{9}{8} \).
\[ 510 = P\left[\left(\frac{9}{8}\right)^{2} - 1\right] = P\left(\frac{81}{64} - 1\right) = \frac{17P}{64} \]
\[ P = \frac{510 \times 64}{17} = \text{Rs. } 1920 \]
\[ \text{SI} = 1920 \times 0.125 \times 2 = \text{Rs. } 480 \]
The Rs. 30 by which the compound interest exceeds it is \( P\left(\tfrac{R}{100}\right)^{2} = 1920 \times \tfrac{1}{64} \) - a useful check on the principal.
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