Two equal sums of money were lent at simple interest at 11% p.a. for $3\frac{1}{2}$ years and $4\frac{1}{2}$ years respectively. If the difference in interests for two periods was ₹ 412.50, then each sum is:

### Concept & Difference in Time When identical sums are invested at the same rate but for different time periods, the difference in the simple interest earned is solely due to the interest earned during the difference in time. $$\Delta SI = \frac{P \times R \times \Delta T}{100}$$ ### Step-by-Step Solution * **Given:** $P_1 = P_2 = P$, Rate ($R$) = $11\%$, $T_1 = 3.5$ years, $T_2 = 4.5$ years, Difference in Interest = ₹ $412.50$. * **Calculation:** Find the difference in the time periods ($\Delta T$): * $\Delta T = 4.5 - 3.5 = 1$ year. * This means the extra interest of ₹ 412.50 was generated in exactly 1 year on the principal $P$ at $11\%$. * Apply the simple interest formula for this 1 year difference: * $412.50 = \frac{P \times 11 \times 1}{100}$ * Multiply by 100: * $41250 = 11P$ * Solve for $P$: * $P = \frac{41250}{11} = 3750$. ### Exam Strategy & Shortcut Observe that the time difference is exactly 1 year. This implies that $11\%$ of the principal is exactly equal to the difference in interest. $11\%$ of $P = 412.50$ $P = 412.50 \times \frac{100}{11} = 3750$. This skips writing out the full equation twice and subtracting. ### Common Pitfall Calculating the total interest for $3.5$ years and $4.5$ years as two separate long equations creates unnecessary room for arithmetic errors. Use the difference in time instead. ### Final Answer Therefore, the correct answer is **₹ 3750**.