Three vessels whose capacities are as `3 : 2 : 1 `are completely filled with milk mixed with water. The ratio of milk to water in the mixture of vessels are as` 5 : 2, 4 : 1 and 4` : 1 respectively. Find the percentage of water in the new mixture obtained when 1/3 of first, 1/2 of second and 1/7 of the third vessel is taken out and mixed together.

Answer

To solve the problem step by step, we will follow the process of calculating the quantities of milk and water in each vessel, the amounts taken out, and then find the percentage of water in the new mixture.

### Step 1: Define the capacities of the vessels
Let the capacities of the three vessels be represented as:
- Vessel 1: \(3x\)
- Vessel 2: \(2x\)
- Vessel 3: \(x\)

### Step 2: Determine the amounts of milk and water in each vessel
1. **Vessel 1** (Ratio of milk to water = 5:2)
- Total parts = 5 + 2 = 7
- Amount of milk = \(\frac{5}{7} \times 3x = \frac{15x}{7}\)
- Amount of water = \(\frac{2}{7} \times 3x = \frac{6x}{7}\)

2. **Vessel 2** (Ratio of milk to water = 4:1)
- Total parts = 4 + 1 = 5
- Amount of milk = \(\frac{4}{5} \times 2x = \frac{8x}{5}\)
- Amount of water = \(\frac{1}{5} \times 2x = \frac{2x}{5}\)

3. **Vessel 3** (Ratio of milk to water = 4:1)
- Total parts = 4 + 1 = 5
- Amount of milk = \(\frac{4}{5} \times x = \frac{4x}{5}\)
- Amount of water = \(\frac{1}{5} \times x = \frac{x}{5}\)

### Step 3: Calculate the amounts taken out from each vessel
1. **From Vessel 1**: Taking out \(\frac{1}{3}\) of \(3x\)
- Amount taken out = \(\frac{1}{3} \times 3x = x\)
- Milk taken out = \(\frac{5}{7} \times x = \frac{5x}{7}\)
- Water taken out = \(\frac{2}{7} \times x = \frac{2x}{7}\)

2. **From Vessel 2**: Taking out \(\frac{1}{2}\) of \(2x\)
- Amount taken out = \(\frac{1}{2} \times 2x = x\)
- Milk taken out = \(\frac{4}{5} \times x = \frac{4x}{5}\)
- Water taken out = \(\frac{1}{5} \times x = \frac{x}{5}\)

3. **From Vessel 3**: Taking out \(\frac{1}{7}\) of \(x\)
- Amount taken out = \(\frac{1}{7} \times x = \frac{x}{7}\)
- Milk taken out = \(\frac{4}{5} \times \frac{x}{7} = \frac{4x}{35}\)
- Water taken out = \(\frac{1}{5} \times \frac{x}{7} = \frac{x}{35}\)

### Step 4: Total amounts of milk and water taken out
- **Total Milk**:
\[ \text{Total Milk} = \frac{5x}{7} + \frac{4x}{5} + \frac{4x}{35} \]
To add these fractions, we find a common denominator, which is 35:
\[ = \frac{25x}{35} + \frac{28x}{35} + \frac{4x}{35} = \frac{57x}{35} \]

- **Total Water**:
\[ \text{Total Water} = \frac{2x}{7} + \frac{x}{5} + \frac{x}{35} \]
Again, using a common denominator of 35:
\[ = \frac{10x}{35} + \frac{7x}{35} + \frac{x}{35} = \frac{18x}{35} \]

### Step 5: Total volume of the new mixture
\[ \text{Total Volume} = x + x + \frac{x}{7} = 2x + \frac{x}{7} = \frac{14x}{7} + \frac{x}{7} = \frac{15x}{7} \]

### Step 6: Calculate the percentage of water in the new mixture
- Total mixture = Total milk + Total water = \(\frac{57x}{35} + \frac{18x}{35} = \frac{75x}{35}\)

- Percentage of water:
\[ \text{Percentage of Water} = \left(\frac{\text{Total Water}}{\text{Total Mixture}}\right) \times 100 = \left(\frac{\frac{18x}{35}}{\frac{75x}{35}}\right) \times 100 = \left(\frac{18}{75}\right) \times 100 = 24\% \]

### Final Answer
The percentage of water in the new mixture is **24%**.

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