There are some milk in three vessels.`1/4` part of the first mixture is poured into second vessel. `1/3` part of the second mixture is poured into third vessel. `5/11` part of the third mixture is poured into first vessel. Again `1/4` part of the first mixture is poured into second vessel. Now all three vessels contains 6- 6 ltmilk, then find the initial quantity of milk in all three vessels

Answer

To solve the problem, we will denote the initial quantities of milk in the three vessels as follows:

- Let the initial quantity of milk in vessel A be \( x \) liters.
- Let the initial quantity of milk in vessel B be \( y \) liters.
- Let the initial quantity of milk in vessel C be \( z \) liters.

Now, we will follow the steps of the transfers as described in the problem:

### Step 1: Transfer from Vessel A to Vessel B
1. **Pour \( \frac{1}{4} \) of vessel A into vessel B**:
- Amount transferred = \( \frac{1}{4}x \)
- Remaining in A = \( x - \frac{1}{4}x = \frac{3}{4}x \)
- New quantity in B = \( y + \frac{1}{4}x \)

### Step 2: Transfer from Vessel B to Vessel C
2. **Pour \( \frac{1}{3} \) of vessel B into vessel C**:
- Amount transferred = \( \frac{1}{3}(y + \frac{1}{4}x) \)
- Remaining in B = \( y + \frac{1}{4}x - \frac{1}{3}(y + \frac{1}{4}x) \)
- New quantity in C = \( z + \frac{1}{3}(y + \frac{1}{4}x) \)

### Step 3: Transfer from Vessel C to Vessel A
3. **Pour \( \frac{5}{11} \) of vessel C into vessel A**:
- Amount transferred = \( \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) \)
- Remaining in C = \( z + \frac{1}{3}(y + \frac{1}{4}x) - \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) \)
- New quantity in A = \( \frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) \)

### Step 4: Transfer from Vessel A to Vessel B (again)
4. **Pour \( \frac{1}{4} \) of vessel A into vessel B again**:
- Amount transferred = \( \frac{1}{4}\left(\frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right)\right) \)
- Remaining in A = \( \frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) - \frac{1}{4}\left(\frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right)\right) \)
- New quantity in B = \( y + \frac{1}{4}x + \frac{1}{4}\left(\frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right)\right) \)

### Final Step: Set the quantities equal to 6 liters
After all these transfers, we know that each vessel contains 6 liters of milk. Therefore, we can set up the following equations:

1. For vessel A:
\[ A = \frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) - \frac{1}{4}\left(\frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right)\right) = 6 \]

2. For vessel B:
\[ B = y + \frac{1}{4}x + \frac{1}{4}\left(\frac{3}{4}x + \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right)\right) = 6 \]

3. For vessel C:
\[ C = z + \frac{1}{3}(y + \frac{1}{4}x) - \frac{5}{11}\left(z + \frac{1}{3}(y + \frac{1}{4}x)\right) = 6 \]

Now, we can solve these equations simultaneously to find the values of \( x \), \( y \), and \( z \).