Milk adulteration and pricing: a seller buys pure milk at Rs x per litre, mixes 2 L of water with every 6 L of milk (total 8 L), and sells the mixture at Rs 2x per litre. Compute the profit percentage on cost.

Introduction:
Adulteration problems require separating cost-bearing and cost-free components. Only milk costs money here; water is assumed free. Revenue is earned on the enlarged volume at the quoted selling rate.

Given Data / Assumptions:

• Milk cost = Rs x per litre; water cost = Rs 0.
• Mixture: 6 L milk + 2 L water = 8 L total.
• Selling price of mixture = Rs 2x per litre.

Concept / Approach:
Total cost is from milk only. Total revenue is selling price times mixture volume. Profit% = (revenue − cost)/cost * 100.

Step-by-Step Solution:
Cost = 6 * x = 6xRevenue = 8 * (2x) = 16xProfit = 16x − 6x = 10xProfit% = 10x / 6x * 100 = 166.66...% (i.e., 166 2/3%)

Verification / Alternative check:
For x = 10, cost = 60, revenue = 160, profit = 100, which is 166.66...% of 60.

Why Other Options Are Wrong:

• 116% / 60% / 100%: all miscompute cost components or the enlarged volume.

Common Pitfalls:

• Including water cost or computing profit% on revenue instead of cost.

Final Answer:
166.66%