A milkman claims to sell at cost price but mixes water with milk and thus gains 25%. What percentage of water is present in the mixture?

Introduction / Context:
When a trader adds a free component (water) to a product sold at the original cost price of the valuable component (milk), the effective gain equals the fraction of free component relative to the paid component. This is a standard result used in mixture-profit problems.

Given Data / Assumptions:

• Selling price equals cost price of pure milk (CP).
• Water is free; profit arises solely from water content.
• Stated gain = 25%.

Concept / Approach:
If water is added to milk and the mixture is sold at the CP of milk, then Profit% = (quantity of water / quantity of milk) * 100. Set this equal to 25% and solve for the water-to-milk ratio, then convert to mixture percentage.

Step-by-Step Solution:
Let milk = 4 units; to get 25% gain, water must be 1 unit (since 1/4 = 0.25). Total mixture = 4 + 1 = 5 units. Water% in mixture = 1 / 5 * 100 = 20%.

Verification / Alternative check:
Selling 5 units (4 milk + 1 water) at CP of 4 units yields revenue equivalent to CP of milk; effectively the extra 1 unit water is pure profit, i.e., 25% gain on milk cost.

Why Other Options Are Wrong:
25% refers to profit, not water percentage. 4% is too small. “None of these” is not needed since 20% is valid.

Common Pitfalls:
Confusing profit% with water%. Always relate profit to the paid component (milk), not to the total mixture directly.

Final Answer:
20%