There are two containers X and Y. X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of milk from X is transferred to Y. After mixing well, 20 ml of the mixture in Y is transferred back to X. If m denotes the proportion of milk in X and n denotes the proportion of water in Y, then which one of the following is correct?

The correct option is m = n.

Let us understand this result step-by-step by tracking the volumes of milk and water in both containers throughout the process.

Step 1: Initial State
Container X contains 100 ml of milk and 0 ml of water (Total volume = 100 ml).
Container Y contains 0 ml of milk and 100 ml of water (Total volume = 100 ml).

Step 2: First Transfer
20 ml of milk is transferred from X to Y.
After this transfer:
In Container X: Milk remaining = 100 - 20 = 80 ml. Water = 0 ml. (Total volume = 80 ml).
In Container Y: Water = 100 ml, Milk added = 20 ml. (Total volume = 120 ml).
The concentration of milk in Y is:
$$\frac{20}{120}=\frac{1}{6}$$
The concentration of water in Y is:
$$\frac{100}{120}=\frac{5}{6}$$

Step 3: Second Transfer (Back to X)
After mixing Y thoroughly, 20 ml of this mixture is transferred back to X.
Since the mixture is homogeneous, the 20 ml transferred contains milk and water in the ratio of their concentrations in Y:
Milk in the transferred 20 ml = $20\times \frac{1}{6}=\frac{20}{6}=\frac{10}{3}$ ml.
Water in the transferred 20 ml = $20\times \frac{5}{6}=\frac{100}{6}=\frac{50}{3}$ ml.

Step 4: Final State Calculation
Let us calculate the quantities in each container after the second transfer:
In Container X:
Final volume of X is 80 ml + 20 ml = 100 ml.
Final Milk in X = 80 + $\frac{10}{3}=\frac{250}{3}$ ml.
The proportion of milk in X (denoted by m) is:
$$m=\frac{\text{Milk in X}}{\text{Total Volume of X}}=\frac{\frac{250}{3}}{100}=\frac{250}{300}=\frac{5}{6}$$
In Container Y:
Final volume of Y is 120 ml - 20 ml = 100 ml.
Final Water in Y = 100 - $\frac{50}{3}=\frac{250}{3}$ ml.
The proportion of water in Y (denoted by n) is:

Comparing the values, we find that:
$$m=n=\frac{5}{6}$$
Therefore, the relationship m = n always holds true, regardless of the intermediate concentrations, because the total volume in both containers remains identical to the starting volumes (100 ml each), ensuring that any milk displaced from X must be replaced by an equal volume of water from Y.