A shopkeeper buys 26 L of milk at Rs. 16/L and also purchases an inferior milk at Rs. 10/L. How many litres of the cheaper milk must be mixed so that selling at Rs. 14/L results in no loss?

Introduction / Context:
We are to find a mixture whose cost price (weighted average) equals the selling price so that overall profit is zero. This requires solving for the unknown volume of cheaper milk that brings the average cost down to Rs. 14 per litre.

Given Data / Assumptions:

• Cost1 = Rs. 16/L; Quantity1 = 26 L.
• Cost2 = Rs. 10/L; Quantity2 = x L.
• Target mixture price (no loss) = Rs. 14/L.

Concept / Approach:
Weighted average equation: (sum of cost) / (sum of litres) = target price. Solve linear equation for x. Alternatively, alligation could be used, but direct equation is straightforward here.

Step-by-Step Solution:
(26*16 + x*10) / (26 + x) = 14. 416 + 10x = 14(26 + x) = 364 + 14x. 416 − 364 = 14x − 10x ⇒ 52 = 4x ⇒ x = 13 L.

Verification / Alternative check:
Total litres = 39; total cost = 26*16 + 13*10 = 416 + 130 = 546. Average = 546/39 = 14, confirming no profit or loss at Rs. 14/L.

Why Other Options Are Wrong:
12, 14, 16 do not satisfy the weighted average equation; only 13 produces exactly Rs. 14/L.

Common Pitfalls:
Forgetting to divide total cost by total litres or misplacing terms in the equation. Keep track of both totals carefully.

Final Answer:
13 liter