There are two containers : the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is removed and is mixed well in the second container. Then three cups of this mixture is removed and is mixed in the first container. Let 'A' denote the proportion of water in the fi^tcontainer and 'B' denote the proportion of alcohol in the second container. Then,

Options

A. `A gt B`
B. `A lt B`
C. `A = B`
D. Cannot be determined

Answer

To solve the problem step by step, we will analyze the situation involving the two containers, one with alcohol and the other with water, and calculate the proportions of water and alcohol after the transfers.

### Step 1: Initial Setup
- **Container 1**: 500 ml of alcohol
- **Container 2**: 500 ml of water

### Step 2: Remove Alcohol from Container 1
- We remove 3 cups of alcohol from Container 1. Assuming 1 cup = 100 ml, then:
\[ \text{Total alcohol removed} = 3 \times 100 \text{ ml} = 300 \text{ ml} \]
- After removing 300 ml of alcohol, the remaining alcohol in Container 1 is:
\[ 500 \text{ ml} - 300 \text{ ml} = 200 \text{ ml} \]

### Step 3: Mix Alcohol with Water in Container 2
- Now, we add the 300 ml of alcohol to Container 2. The new contents of Container 2 are:
\[ \text{Total in Container 2} = 500 \text{ ml (water)} + 300 \text{ ml (alcohol)} = 800 \text{ ml} \]

### Step 4: Calculate Proportions in Container 2
- The proportion of alcohol in Container 2:
\[ B = \frac{\text{Alcohol}}{\text{Total}} = \frac{300 \text{ ml}}{800 \text{ ml}} = \frac{3}{8} \]
- The proportion of water in Container 2:
\[ \text{Water} = 500 \text{ ml} \quad \Rightarrow \quad \text{Proportion of water} = \frac{500 \text{ ml}}{800 \text{ ml}} = \frac{5}{8} \]

### Step 5: Remove Mixture from Container 2
- We now remove 3 cups of the mixture from Container 2. The total volume removed is again 300 ml.
- The mixture consists of both alcohol and water. The amounts of alcohol and water in the 300 ml removed can be calculated as follows:
- Alcohol in 300 ml:
\[ \text{Alcohol in 300 ml} = 300 \text{ ml} \times \frac{3}{8} = 112.5 \text{ ml} \]
- Water in 300 ml:
\[ \text{Water in 300 ml} = 300 \text{ ml} \times \frac{5}{8} = 187.5 \text{ ml} \]

### Step 6: Add Mixture Back to Container 1
- We add the removed amounts back to Container 1:
- New alcohol in Container 1:
\[ \text{New alcohol} = 200 \text{ ml} + 112.5 \text{ ml} = 312.5 \text{ ml} \]
- New water in Container 1:
\[ \text{New water} = 0 \text{ ml} + 187.5 \text{ ml} = 187.5 \text{ ml} \]

### Step 7: Calculate Proportions in Container 1
- Total volume in Container 1:
\[ \text{Total in Container 1} = 312.5 \text{ ml (alcohol)} + 187.5 \text{ ml (water)} = 500 \text{ ml} \]
- Proportion of water in Container 1:
\[ A = \frac{187.5 \text{ ml}}{500 \text{ ml}} = \frac{3}{8} \]

### Step 8: Conclusion
- We have found that:
\[ A = \frac{3}{8} \quad \text{and} \quad B = \frac{3}{8} \]
- Therefore, \( A = B \).