A 729 ml mixture contains milk and water in the ratio 7 : 2. How much additional water must be added so that the new mixture has milk and water in the ratio 7 : 3?

Introduction / Context:
The mixture is changed by adding only water. Since no milk is added or removed, the quantity of milk remains constant while water increases until the desired ratio is reached.

Given Data / Assumptions:

• Total initial mixture = 729 ml.
• Initial milk : water = 7 : 2.
• Final milk : water = 7 : 3 (milk unchanged).
• No evaporation or losses.

Concept / Approach:
Compute the initial milk and water. Because milk stays constant, compute the water required to make water = (3/7) of milk (as per the target ratio).

Step-by-Step Solution:
Initial milk = (7/9)*729 = 567 mlInitial water = (2/9)*729 = 162 mlTarget water for 7 : 3 = (3/7)*milk = (3/7)*567 = 243 mlAdditional water to add = 243 − 162 = 81 ml

Verification / Alternative check:
New totals: milk = 567 ml, water = 243 ml; ratio 567 : 243 = 7 : 3 (divide by 81).

Why Other Options Are Wrong:
60, 72, and 96 ml do not produce water = 243 ml. Only 81 ml gets the exact 7 : 3 ratio.

Common Pitfalls:
Altering both milk and water or using 7 : 3 as milk : total instead of milk : water. Keep milk fixed and adjust water only.

Final Answer:
81 ml