Two vessels contain milk-water mixtures.\n• Vessel 1 contains 64% milk (and the rest is water).\n• Vessel 2 contains 26% water (so the remaining percentage is milk).\nIn what ratio should the mixtures from vessel 1 and vessel 2 be mixed so that the new mixture contains exactly 68% milk?

Introduction:
This is a classic alligation problem based on percentage concentration. One mixture has milk percentage below the target and the other has milk percentage above the target. The required ratio is found by comparing how far each mixture's concentration is from the desired concentration.

Given Data / Assumptions:

• Vessel 1 milk = 64%
• Vessel 2 water = 26%, so milk = 100% - 26% = 74%
• Target milk percentage = 68%

Concept / Approach:
Using alligation for concentrations:\nRatio (lower concentration mixture : higher concentration mixture) = (higher - target) : (target - lower). Here, milk% are compared (64% and 74%) to reach 68%.

Step-by-Step Solution:
Lower milk% = 64Higher milk% = 74Target milk% = 68Difference (higher - target) = 74 - 68 = 6Difference (target - lower) = 68 - 64 = 4Required ratio (Vessel 1 : Vessel 2) = 6 : 4 = 3 : 2

Verification / Alternative Check:
Take 3 parts from vessel 1 and 2 parts from vessel 2.\nMilk in 3 parts of vessel 1 = 3 * 64 = 192 (in percent-parts)\nMilk in 2 parts of vessel 2 = 2 * 74 = 148\nTotal milk = 192 + 148 = 340 out of total parts 5.\nAverage milk% = 340 / 5 = 68%, exactly as required.

Why Other Options Are Wrong:
2:1 or 5:4: gives milk% closer to 64% than needed.1:2 or 2:3: uses too much of the stronger (74%) mixture, pushing milk% above 68%.

Common Pitfalls:
Forgetting to convert 26% water into 74% milk.Mixing up which difference goes with which vessel in alligation.Trying to average percentages directly without using ratio logic.

Final Answer:
3 : 2