Two solutions have milk and water in the ratios 7 : 5 and 6 : 11 respectively. In what proportion should these two solutions be mixed so that the resulting solution has 1 part milk and 2 parts water, or is this impossible with the given solutions?

Introduction / Context:
This is a mixture and ratio question involving two different milk-water solutions. We are given their initial ratios and asked to find the mixing proportion that would produce a final mixture with a target milk-to-water ratio of 1 : 2. An important concept in such problems is that the resulting concentration of milk must lie between the concentrations of the individual solutions if we only mix them without adding anything else.

Given Data / Assumptions:
• Solution 1 has milk : water = 7 : 5.
• Solution 2 has milk : water = 6 : 11.
• The target final ratio is milk : water = 1 : 2.
• We mix only these two solutions; no extra milk or water is added separately.

Concept / Approach:
First, convert each ratio into the fraction of milk in that solution. For 7 : 5, total parts = 12, so the milk fraction is 7/12. For 6 : 11, total parts = 17, so the milk fraction is 6/17. The target ratio 1 : 2 corresponds to a milk fraction of 1/(1 + 2) = 1/3. When you mix two solutions of concentrations C1 and C2, the resulting concentration must lie between C1 and C2 provided you use positive quantities of each. If the target concentration is outside this range, it is impossible to obtain by mixing only those two solutions.

Step-by-Step Solution:
Step 1: Compute the milk fraction in solution 1: C1 = 7/12 ≈ 0.5833.Step 2: Compute the milk fraction in solution 2: C2 = 6/17 ≈ 0.3529.Step 3: Compute the target milk fraction for ratio 1 : 2: Ct = 1 / (1 + 2) = 1/3 ≈ 0.3333.Step 4: Compare Ct with C1 and C2. We have C1 ≈ 0.5833 and C2 ≈ 0.3529, both of which are greater than 0.3333.Step 5: Since both available solutions are richer in milk than the target, any convex combination (mixture) of them will also have a milk fraction between 0.3529 and 0.5833, never as low as 0.3333.Step 6: Therefore, it is impossible to obtain a mixture with milk : water = 1 : 2 using only these two solutions.

Verification / Alternative check:
We can attempt to set up an equation. Suppose we mix x litres of solution 1 and y litres of solution 2. Milk in mixture = (7/12)x + (6/17)y. Total volume = x + y. The required condition is [(7/12)x + (6/17)y] / (x + y) = 1/3. When we simplify this equation, we end up with a condition that cannot be satisfied with positive x and y simultaneously, reinforcing the argument that the target concentration lies outside the achievable range.

Why Other Options Are Wrong:
Options 7 : 12, 8 : 13 and 9 : 4 all correspond to some assumed mixing ratios, but none of them can produce the required final milk fraction of 1/3 because any mixture of the two given solutions remains between 7/12 and 6/17 in milk concentration. Thus, these numeric mixing ratios do not satisfy the concentration requirement.

Common Pitfalls:
Students often try to solve for specific x : y without first checking whether the target concentration is feasible. Another common error is to average the ratios incorrectly or to think that any target between the two water ratios, rather than milk fractions, is achievable. Always compare the actual concentration fractions and ensure that the target lies between them before attempting detailed algebra.

Final Answer:
The desired mixture with milk : water = 1 : 2 cannot be obtained by mixing only the two given solutions, so the correct choice is “Not possible with these two solutions”.