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

Introduction / Context:
This is a classic mixture-and-ratio question. We know the initial volumes and composition (milk : water = 7 : 2) and we are asked to add only water so that the final mixture has milk : water = 7 : 3. Because only water is added, the milk quantity remains constant throughout the process.

Given Data / Assumptions:

• Total initial mixture = 729 ml.
• Initial ratio milk : water = 7 : 2 (9 parts total).
• Only water is added; milk stays unchanged.
• Target ratio milk : water = 7 : 3 (10 parts total).

Concept / Approach:
Convert the given ratio into actual quantities using parts. Since there are 9 parts initially, each part = 729 / 9 = 81 ml. Milk is 7 parts and does not change. We then enforce the target ratio using the unchanged milk as the reference to compute the new total and hence the water to add.

Step-by-Step Solution:
Initial part size = 729 / 9 = 81 mlInitial milk = 7 * 81 = 567 ml (fixed)For ratio 7 : 3, total parts = 10; milk 7 parts ⇒ total = (567 / 7) * 10 = 81 * 10 = 810 mlWater to add = final total − initial total = 810 − 729 = 81 ml

Verification / Alternative check:
After adding 81 ml, water becomes initial water 162 ml + 81 ml = 243 ml, and milk is 567 ml. Ratio 567 : 243 = 7 : 3 (both divisible by 81). Correct.

Why Other Options Are Wrong:
60, 70, 72, and 90 ml do not yield the exact 7 : 3 ratio when recalculated; the simplified ratio will deviate from 7 : 3.

Common Pitfalls:
Changing milk as well, or averaging ratios instead of working with actual quantities. Always lock the unchanged component and scale the other.

Final Answer:
81 ml