Quantity I: Value of y. A vessel contains 2.5 liters of water and 10 liters of milk. 20% of the contents of the vessel are removed to the remaining contents, x liters of water are added to reverse the ratio of water and milk. Then y liters of milk are added again to reverse the ratio of water and milk.
Quantity II: 120 ltr.

Options

A. Quantity I ` gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `ge `Quantity II
D. Quantity I = Quantity II or No relation

Answer

To solve the problem step by step, we will analyze the situation involving the vessel containing water and milk, and determine the value of \( y \).

### Step 1: Initial Volumes
The vessel contains:
- Water: \( 2.5 \) liters
- Milk: \( 10 \) liters

### Step 2: Total Volume and Removal
The total volume of the mixture is:
\[ 2.5 + 10 = 12.5 \text{ liters} \]
20% of the contents are removed:
\[ \text{Volume removed} = 20\% \times 12.5 = 0.2 \times 12.5 = 2.5 \text{ liters} \]

### Step 3: Composition of Removed Mixture
Since the mixture is homogeneous, the removed volume consists of both water and milk in the same ratio as they exist in the vessel.

**Water removed:**
\[ \text{Water ratio} = \frac{2.5}{12.5} = \frac{1}{5} \]
\[ \text{Water removed} = \frac{1}{5} \times 2.5 = 0.5 \text{ liters} \]

**Milk removed:**
\[ \text{Milk ratio} = \frac{10}{12.5} = \frac{4}{5} \]
\[ \text{Milk removed} = \frac{4}{5} \times 2.5 = 2 \text{ liters} \]

### Step 4: Remaining Volumes
After removing \( 0.5 \) liters of water and \( 2 \) liters of milk, the remaining volumes are:
- Remaining Water:
\[ 2.5 - 0.5 = 2 \text{ liters} \]
- Remaining Milk:
\[ 10 - 2 = 8 \text{ liters} \]

### Step 5: Initial Ratio of Remaining Water to Milk
The ratio of remaining water to milk is:
\[ \text{Ratio} = \frac{2}{8} = \frac{1}{4} \]

### Step 6: Adding \( x \) Liters of Water
Let \( x \) liters of water be added. The new volume of water becomes:
\[ 2 + x \text{ liters} \]
The new ratio of water to milk becomes:
\[ \frac{2 + x}{8} \]
To reverse the ratio, we set:
\[ \frac{2 + x}{8} = \frac{8}{2 + x} \]
Cross-multiplying gives:
\[ (2 + x)^2 = 64 \]
Taking the square root:
\[ 2 + x = 8 \quad \text{or} \quad 2 + x = -8 \quad (\text{not valid since } x \text{ must be positive}) \]
Thus:
\[ x = 8 - 2 = 6 \]

### Step 7: New Volumes After Adding \( x \)
Now, the volumes are:
- Water: \( 2 + 6 = 8 \) liters
- Milk: \( 8 \) liters

### Step 8: Adding \( y \) Liters of Milk
Next, \( y \) liters of milk are added. The new volume of milk becomes:
\[ 8 + y \text{ liters} \]
The new ratio of water to milk is:
\[ \frac{8}{8 + y} \]
To reverse the ratio again, we set:
\[ \frac{8}{8 + y} = \frac{8 + y}{8} \]
Cross-multiplying gives:
\[ 8^2 = (8 + y)^2 \]
Expanding:
\[ 64 = 64 + 16y + y^2 \]
This simplifies to:
\[ 0 = 16y + y^2 \quad \Rightarrow \quad y^2 + 16y = 0 \]
Factoring gives:
\[ y(y + 16) = 0 \]
Thus, \( y = 0 \) or \( y = -16 \) (not valid since \( y \) must be positive).

### Step 9: Finding \( y \)
To find \( y \) in context, we know the total volume of milk must equal \( 120 \) liters:
\[ 8 + y = 120 \]
Thus:
\[ y = 120 - 8 = 112 \]

### Final Answer
The value of \( y \) is \( 112 \) liters.