Difficulty: Medium
Correct Answer: 25%
Explanation:
Introduction / Context:This question tests profit calculation in adulteration problems. The milkman’s cost is only for milk, while water is free. By adding water, he increases the sellable volume without increasing cost. If he sells the entire mixture at the original milk price per litre, he earns extra revenue, which becomes profit.
Given Data / Assumptions:
Concept / Approach:Total cost = cost of milk only. Total revenue = selling price per litre * total litres sold. Profit% = (profit / cost) * 100.
Step-by-Step Solution:
Step 1: Cost of 20 litres milk = 20 * 18 = ₹360 Step 2: Water cost = ₹0, so total cost remains ₹360 Step 3: Total mixture volume sold = 20 + 5 = 25 litres Step 4: Revenue = 25 * 18 = ₹450 Step 5: Profit = 450 - 360 = ₹90 Step 6: Profit% = (90/360) * 100 = 25%Verification / Alternative check:Another way: the extra 5 litres are “free inventory.” Selling them at ₹18 per litre adds ₹90 extra revenue, which is pure profit. Profit on original cost ₹360 is 90/360 = 1/4 = 25%.
Why Other Options Are Wrong:
22.5%: would correspond to a smaller free addition than 5 litres. 20%: would imply profit ₹72, not ₹90. 33.33%: would require profit ₹120 on cost ₹360, not possible here. 18%: does not match the computed ratio of extra revenue to cost.Common Pitfalls:Students sometimes compute profit% on selling price instead of cost price. Others mistakenly treat the mixture selling price as “cost price of mixture,” which would be ₹360/25 = ₹14.4 per litre, not ₹18. Always separate cost of inputs from selling rate.
Final Answer:25%
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