A man buys 1 litre of milk for ₹12. He mixes water equal to 20% of the milk quantity (i.e., adds 0.2 litre water to 1 litre milk), and then sells the entire resulting mixture at ₹15 per litre. What is his percentage gain (profit percentage) on the transaction?

Introduction / Context:
This is a profit-and-adulteration problem. The trick is that water is assumed free, so adding water increases the saleable quantity without increasing cost. If the selling price is per litre of the final mixture, revenue must be calculated on the entire mixture volume (milk + water). Then compare profit against the original cost of milk to get gain percentage.

Given Data / Assumptions:

• Milk bought = 1 litre
• Cost price for milk = ₹12 for 1 litre
• Water added = 20% of milk quantity = 0.2 litre (free)
• Total mixture volume = 1 + 0.2 = 1.2 litres
• Selling price = ₹15 per litre of the mixture

Concept / Approach:
Compute total revenue from selling 1.2 litres at ₹15 per litre. Profit = revenue - cost. Profit% = (profit/cost)*100.

Step-by-Step Solution:

Verification / Alternative check:
You can also compute the effective cost per litre of the mixture: cost per litre = 12/1.2 = ₹10. Selling at ₹15 gives gain = (15-10)/10 *100 = 50%. Same result.

Why Other Options Are Wrong:

Common Pitfalls:
Many students incorrectly compute revenue as ₹15 total (instead of ₹15 per litre), or forget that he sells 1.2 litres, not just 1 litre. Another mistake is calculating profit% on selling price rather than on cost price, which changes the percentage.

Final Answer:
50%