Understanding the Milk and Water Mixture Problem

This problem involves tracking the quantities of milk and water in two containers, X and Y, after a series of transfers and mixing. Container X starts with pure milk, and container Y starts with pure water. We need to find the relationship between the proportion of milk in X (m) and the proportion of water in Y (n) at the end of the process.

Initial State of Containers

• Container X: 100 ml milk, 0 ml water. Total volume = 100 ml.
• Container Y: 0 ml milk, 100 ml water. Total volume = 100 ml.

Step-by-Step Transfer and Mixing Process

Step 1: Transfer from X to Y

20 ml of milk is transferred from container X to container Y.

• Container X after transfer:
  • Milk: \(100 \text{ ml} - 20 \text{ ml} = 80 \text{ ml}\)
  • Water: 0 ml
  • Total Volume: 80 ml
• Container Y after transfer (before mixing):
  • Milk: \(0 \text{ ml} + 20 \text{ ml} = 20 \text{ ml}\)
  • Water: 100 ml
  • Total Volume: \(100 \text{ ml} + 20 \text{ ml} = 120 \text{ ml}\)

Step 2: Mixing in Y

The contents of container Y are mixed well. Now, container Y contains a mixture of milk and water.

• Proportion of milk in Y = \(\frac{\text{Amount of Milk}}{\text{Total Volume}} = \frac{20 \text{ ml}}{120 \text{ ml}} = \frac{1}{6}\)
• Proportion of water in Y = \(\frac{\text{Amount of Water}}{\text{Total Volume}} = \frac{100 \text{ ml}}{120 \text{ ml}} = \frac{5}{6}\)

Step 3: Transfer from Y to X

20 ml of the mixture from container Y is transferred back to container X. The 20 ml mixture will contain milk and water in the proportions calculated in Step 2.

• Amount of milk in the 20 ml mixture transferred: \(20 \text{ ml} \times \frac{1}{6} = \frac{20}{6} \text{ ml} = \frac{10}{3} \text{ ml}\)
• Amount of water in the 20 ml mixture transferred: \(20 \text{ ml} \times \frac{5}{6} = \frac{100}{6} \text{ ml} = \frac{50}{3} \text{ ml}\)

Final State of Containers

Container X after the second transfer:

• Milk: \(80 \text{ ml (from Step 1)} + \frac{10}{3} \text{ ml (from Step 3)} = \frac{240+10}{3} \text{ ml} = \frac{250}{3} \text{ ml}\)
• Water: \(0 \text{ ml (from Step 1)} + \frac{50}{3} \text{ ml (from Step 3)} = \frac{50}{3} \text{ ml}\)
• Total Volume: \(\frac{250}{3} \text{ ml} + \frac{50}{3} \text{ ml} = \frac{300}{3} \text{ ml} = 100 \text{ ml}\) (Note: The total volume in X returns to 100 ml)

Container Y after the second transfer:

The amounts removed from Y are the amounts calculated in Step 3.

• Milk: \(20 \text{ ml (from Step 1)} - \frac{10}{3} \text{ ml (from Step 3)} = \frac{60-10}{3} \text{ ml} = \frac{50}{3} \text{ ml}\)
• Water: \(100 \text{ ml (from Step 1)} - \frac{50}{3} \text{ ml (from Step 3)} = \frac{300-50}{3} \text{ ml} = \frac{250}{3} \text{ ml}\)
• Total Volume: \(\frac{50}{3} \text{ ml} + \frac{250}{3} \text{ ml} = \frac{300}{3} \text{ ml} = 100 \text{ ml}\) (Note: The total volume in Y returns to 100 ml)

Calculating Proportions m and n

According to the question:

• m = proportion of milk in X in the final state.
• n = proportion of water in Y in the final state.

Let's calculate m and n:

• \(m = \frac{\text{Amount of Milk in X}}{\text{Total Volume in X}} = \frac{\frac{250}{3} \text{ ml}}{100 \text{ ml}} = \frac{250}{3} \times \frac{1}{100} = \frac{250}{300} = \frac{5}{6}\)
• \(n = \frac{\text{Amount of Water in Y}}{\text{Total Volume in Y}} = \frac{\frac{250}{3} \text{ ml}}{100 \text{ ml}} = \frac{250}{3} \times \frac{1}{100} = \frac{250}{300} = \frac{5}{6}\)

Comparing m and n

We found that \(m = \frac{5}{6}\) and \(n = \frac{5}{6}\).

Therefore, \(m = n\).

Summary of Results

From the final state, \(m = \frac{5}{6}\) and \(n = \frac{5}{6}\). Thus, \(m = n\).

Revision Table: Mixture Problems and Proportions

Additional Information: Symmetric Mixture Transfers

This problem is an example of a common type of mixture problem involving two containers where a specific amount is transferred from container 1 to container 2, mixed, and then the same amount is transferred back from container 2 to container 1. A key insight for such problems, especially when the initial volumes in both containers are equal and the amount transferred in each step is the same, is related to symmetry.

Consider two components, A and B, initially separated in containers X and Y respectively. If an amount 'v' is transferred from X to Y, mixed, and then 'v' is transferred back from Y to X, the amount of component B (initially in Y) that ends up in X will be equal to the amount of component A (initially in X) that ends up in Y. In this problem, component A is milk and component B is water.

Let \(V_X\) and \(V_Y\) be the initial volumes in X and Y (both 100 ml). Let \(v\) be the transfer volume (20 ml). The amount of milk transferred from X to Y is \(v\). The amount of water transferred from Y to X is found to be equal to the amount of milk transferred from X to Y that remains in Y after the second transfer, relative to the volume of the original component in that container.

More simply, in a symmetric transfer of volume 'v' between two containers of initial equal volume 'V', if container X starts with pure A and container Y starts with pure B, the amount of B in X at the end equals the amount of A in Y at the end. Since the total volume in each container returns to V (100 ml in this case), the proportion of B in X will equal the proportion of A in Y.

In our case, the proportion of water (B) in X is \(\frac{50/3}{100} = \frac{1}{6}\), and the proportion of milk (A) in Y is \(\frac{50/3}{100} = \frac{1}{6}\).

However, the question asks for m (proportion of milk in X) and n (proportion of water in Y). Our calculation shows \(m = 5/6\) and \(n = 5/6\), leading to \(m=n\).

This equality m=n arises because the amount of the *other* substance that ends up in each container is equal. Let's verify. Amount of water in X = \(50/3\) ml. Amount of milk in Y = \(50/3\) ml. These amounts are equal. Since the final total volume in both containers is the same (100 ml), the proportion of water in X (\(50/300 = 1/6\)) equals the proportion of milk in Y (\(50/300 = 1/6\)). The proportion of the *original* substance remaining is \(1 - (\text{proportion of the other substance})\). So, proportion of milk in X = \(1 - 1/6 = 5/6\) and proportion of water in Y = \(1 - 1/6 = 5/6\). Hence, \(m=n\).