Mixing water with milk but selling at cost price: In what ratio should water be mixed with milk so that selling the resulting mixture at the original cost price of milk yields a profit of 50/3 %?

Introduction / Context:
This is a classic adulteration/mixture scenario. If a milk seller adds free water and still sells at the milk’s cost price, all the revenue from the water portion becomes profit. The profit percentage directly relates to the water-to-milk ratio.

Given Data / Assumptions:

• Let milk quantity be m and added water be w.
• Selling price per unit equals milk’s cost price per unit.
• Profit % = (50/3)% = 16.666…%.

Concept / Approach:
Because water is free, profit comes from selling w units at milk’s unit cost. Hence Profit% = (w / m) * 100. Set (w / m) * 100 = 50/3 and solve for w : m, then invert if needed to express milk : water.

Step-by-Step Solution:
(w / m) * 100 = 50 / 3 ⇒ w / m = 1 / 6.Therefore, milk : water = m : w = 6 : 1.

Verification / Alternative check:
If milk = 6 L and water = 1 L, total = 7 L. Revenue at milk’s unit cost equals 7 units of cost while cost is only 6 units → profit = 1/6 of cost → 16.666…%, as required.

Why Other Options Are Wrong:

• 1:2, 2:3, and 2:5 do not produce a 16.666…% profit when selling at cost price.

Common Pitfalls:

• Treating profit% as w / (m + w) instead of w / m (since selling at cost price of milk).

Final Answer:
6:1