https://file.aiquickdraw.com/mathbot/file/1726197915496kfxrfv6e.jpg

Let's analyze the problem step by step.

### Problem Breakdown:
1. **Containers X and Y:**
   - \( X \): Initially contains **100 ml of milk**.
   - \( Y \): Initially contains **100 ml of water**.

2. **First transfer:**
   - **20 ml of milk** is transferred from \( X \) to \( Y \).
   - Now, \( Y \) contains a mixture of 100 ml of water and 20 ml of milk. Thus, the total volume of liquid in \( Y \) is 120 ml.

3. **Mixing in \( Y \):**
   - After the transfer, the liquids in \( Y \) are mixed well, so the mixture in \( Y \) consists of water and milk in the ratio:
     \[
     \frac{\text{Milk}}{\text{Total Volume in Y}} = \frac{20 \, \text{ml}}{120 \, \text{ml}} = \frac{1}{6}.
     \]
     Hence, the milk forms \( \frac{1}{6} \) of the mixture in \( Y \), and the water forms \( \frac{5}{6} \).

4. **Second transfer:**
   - 20 ml of the mixture from \( Y \) (which contains \( \frac{1}{6} \) milk and \( \frac{5}{6} \) water) is transferred back to \( X \).
     - The amount of milk in this 20 ml mixture is \( \frac{1}{6} \times 20 \, \text{ml} = \frac{20}{6} \, \text{ml} \approx 3.33 \, \text{ml} \).
     - The amount of water in this 20 ml mixture is \( \frac{5}{6} \times 20 \, \text{ml} = \frac{100}{6} \, \text{ml} \approx 16.67 \, \text{ml} \).

5. **Final state of \( X \):**
   - Initially, \( X \) had 100 ml of milk. After transferring 20 ml of pure milk out, it has \( 100 \, \text{ml} - 20 \, \text{ml} = 80 \, \text{ml} \) of milk left.
   - When 20 ml of the mixture from \( Y \) is transferred back to \( X \), \( X \) receives \( \approx 3.33 \, \text{ml} \) of milk and \( \approx 16.67 \, \text{ml} \) of water.

   Hence, the final amount of milk in \( X \) is:
   \[
   80 \, \text{ml} + 3.33 \, \text{ml} = 83.33 \, \text{ml}.
   \]
   The final total volume in \( X \) is still 100 ml (since 20 ml of mixture was added back).

   Therefore, the proportion of milk in \( X \), denoted by \( m \), is:
   \[
   m = \frac{83.33 \, \text{ml}}{100 \, \text{ml}} = 0.8333.
   \]

6. **Final state of \( Y \):**
   - After the second transfer, \( Y \) loses 20 ml of its mixture. So, the remaining liquid in \( Y \) is:
     \[
     120 \, \text{ml} - 20 \, \text{ml} = 100 \, \text{ml}.
     \]
   The amount of water left in \( Y \) is:
   \[
   100 \, \text{ml of water} - 16.67 \, \text{ml} \approx 83.33 \, \text{ml}.
   \]
   The proportion of water in \( Y \), denoted by \( n \), is:
   \[
   n = \frac{83.33 \, \text{ml}}{100 \, \text{ml}} = 0.8333.
   \]

### Conclusion:
We have \( m = 0.8333 \) and \( n = 0.8333 \), so \( m = n \).

Thus, the correct answer is option **(a) \( m = n \)**.

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Would you like more details or have any other questions?

### Related Questions:
1. What would happen if 30 ml of milk were transferred instead of 20 ml?
2. How would the proportions change if the initial amounts of water and milk were unequal?
3. What is the effect of transferring different amounts back from \( Y \) to \( X \)?
4. How would the result change if the mixture in \( Y \) wasn't perfectly mixed?
5. How could we generalize this process for any two liquids with different volumes?

### Tip:
Mixing and transferring liquids involves careful consideration of proportions. Always track the liquid content in each container separately to avoid confusion.

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? (a) m = n, (b) m > n, (c) m < n, (d) Cannot be determined due to insufficient data.

Learn how to solve a fluid mixture transfer problem involving two containers, one with milk and one with water. Understand the proportions of milk and water after transfers and mixing. Suitable for Grade 9-11 students.

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