In a vessel milk is 60% less than water. When x liter milk is added to mixture then ratio of milk and water becomes `3:5`. Now y liter water is extracted from mixture then this ratio becomes `7:10`. Now 30liter milk is added again then new ratio of milk and water becomes `4:5`. Find the value of `(x+y)`?

Options

A. 60
B. 120
C. 180
D. 144

Answer

To solve the problem step by step, let's break it down:

### Step 1: Establish Initial Quantities
Let the amount of water in the vessel be \(100A\). Since the milk is 60% less than the water, the amount of milk will be:
\[ \text{Milk} = 100A - 60\% \text{ of } 100A = 100A - 60A = 40A \]

### Step 2: Adding Milk
When \(x\) liters of milk is added, the new amount of milk becomes:
\[ \text{New Milk} = 40A + x \]
The amount of water remains:
\[ \text{Water} = 100A \]

### Step 3: Set Up the First Ratio
According to the problem, the ratio of milk to water becomes \(3:5\):
\[ \frac{40A + x}{100A} = \frac{3}{5} \]
Cross-multiplying gives:
\[ 5(40A + x) = 3(100A) \]
Expanding and simplifying:
\[ 200A + 5x = 300A \]
\[ 5x = 300A - 200A \]
\[ 5x = 100A \quad \Rightarrow \quad x = 20A \]

### Step 4: Extracting Water
Next, \(y\) liters of water is extracted, making the amount of water:
\[ \text{New Water} = 100A - y \]
The ratio of milk to water now becomes \(7:10\):
\[ \frac{40A + x}{100A - y} = \frac{7}{10} \]
Substituting \(x = 20A\):
\[ \frac{40A + 20A}{100A - y} = \frac{7}{10} \]
This simplifies to:
\[ \frac{60A}{100A - y} = \frac{7}{10} \]
Cross-multiplying gives:
\[ 10 \cdot 60A = 7(100A - y) \]
Expanding and simplifying:
\[ 600A = 700A - 7y \]
\[ 7y = 700A - 600A \]
\[ 7y = 100A \quad \Rightarrow \quad y = \frac{100A}{7} \]

### Step 5: Adding More Milk
Now, 30 liters of milk is added:
\[ \text{New Milk} = 40A + x + 30 = 40A + 20A + 30 = 60A + 30 \]
The amount of water remains:
\[ \text{New Water} = 100A - y = 100A - \frac{100A}{7} = \frac{700A - 100A}{7} = \frac{600A}{7} \]

### Step 6: Set Up the Final Ratio
The new ratio of milk to water becomes \(4:5\):
\[ \frac{60A + 30}{\frac{600A}{7}} = \frac{4}{5} \]
Cross-multiplying gives:
\[ 5(60A + 30) = 4 \cdot \frac{600A}{7} \]
Expanding:
\[ 300A + 150 = \frac{2400A}{7} \]
Multiplying through by 7 to eliminate the fraction:
\[ 7(300A + 150) = 2400A \]
Expanding:
\[ 2100A + 1050 = 2400A \]
Rearranging gives:
\[ 2400A - 2100A = 1050 \]
\[ 300A = 1050 \quad \Rightarrow \quad A = \frac{1050}{300} = \frac{7}{2} \]

### Step 7: Calculate \(x\) and \(y\)
Now substituting \(A\) back to find \(x\) and \(y\):
\[ x = 20A = 20 \cdot \frac{7}{2} = 70 \]
\[ y = \frac{100A}{7} = \frac{100 \cdot \frac{7}{2}}{7} = 50 \]

### Step 8: Find \(x + y\)
Finally, we find:
\[ x + y = 70 + 50 = 120 \]

### Final Answer
\[ \boxed{120} \]