A fruit seller professes to sell his fruits at cost price, but still gains 25% on his outlay. What weight does he substitute for a kilogram? (P.C.S., 2009)

### Concept & Reverse Engineering False Weight When the overall profit percentage is known and the goods are sold at cost price, the profit is driven solely by the false weight. We can set up the gain equation and solve for the unknown false weight. $$Gain\ \% = \left( \frac{True\ Weight - False\ Weight}{False\ Weight} \right) \times 100$$ ### Step-by-Step Solution * Let the substituted false weight be $x$ grams. * True weight for a kilogram = 1000 grams. * Given Gain \% = 25%. * Substituting into the formula: $25 = \left( \frac{1000 - x}{x} \right) \times 100$ * Divide both sides by 25: $1 = \left( \frac{1000 - x}{x} \right) \times 4$ * $x = 4000 - 4x$ * $5x = 4000$ * $x = \frac{4000}{5} = 800$ grams. ### Exam Strategy & Shortcut A 25% gain is a fraction of $\frac{1}{4}$. In a false weight scenario, the ratio of Gain to Cost is $1:4$. This means for every 4 parts given (cost), 1 part is held back (profit), making the total claimed weight 5 parts. If 5 parts = 1000g, then 1 part = 200g. The weight given (cost) is 4 parts = $4 \times 200 = 800$g. ### Common Pitfall Students often subtract 25% directly from 1000g, calculating $1000 - 250 = 750$g. This is incorrect because the 25% profit is earned on the *false weight*, not the true weight. ### Final Answer Therefore, the correct answer is **800 gm**.