Goods pass successively through the hands of three traders and each of them sells his goods at a profit of $25\%$ of his cost price. If the last trader sold the goods for ₹ 250, then how much did the first trader pay for them? (S.S.C., 2005)

### Concept & Successive Profit Calculation When an article is sold sequentially at a profit, the effective selling price is a product of successive multipliers. $$ \text{Final SP} = \text{Initial CP} \times \left(1 + \frac{P_1}{100}\right) \times \left(1 + \frac{P_2}{100}\right) \times \left(1 + \frac{P_3}{100}\right) $$ ### Step-by-Step Solution * Given: * Three traders each make a profit of $25\%$. * Final selling price (SP) = ₹ 250. * Let the original cost price for the first trader be $x$. * The multiplier for a $25\%$ profit is $1 + \frac{25}{100} = \frac{125}{100} = \frac{5}{4}$. * Formulate the equation: * $x \times \left(\frac{5}{4}\right) \times \left(\frac{5}{4}\right) \times \left(\frac{5}{4}\right) = 250$ * $x \times \frac{125}{64} = 250$ * Solve for $x$: * $x = 250 \times \frac{64}{125}$ * $x = 2 \times 64 = 128$ ### Exam Strategy & Shortcut Memorize fraction equivalents of common percentages. $25\% = \frac{1}{4}$. A profit of $\frac{1}{4}$ means the multiplying factor is $\frac{5}{4}$. Since the profit occurs three times, cube the multiplier: $\left(\frac{5}{4}\right)^3 = \frac{125}{64}$. $\frac{125}{64}$ of CP = 250. It's easy to see 125 goes into 250 exactly twice, so CP = $64 \times 2 = 128$. ### Common Pitfall A common error is treating the total profit as $25\% \times 3 = 75\%$ and finding the cost price using $175\% \text{ of CP} = 250$, which completely ignores compounding. ### Final Answer Therefore, the correct answer is **₹ 128**.