A mizture of milk and water contains `80%` milk. X litre of mixture is taken out and replaced with water. Find the value of x if final concentration of milk in mixture is `50%` and initial quantity of mixture is 80 l.

Options

A. A)`35 l`
B. B)`40 l`
C. C)`30 l`
D. D)`18 l`

Answer

To solve the problem step by step, we need to find the value of \( x \) (the amount of mixture taken out) such that after replacing it with water, the final concentration of milk in the mixture becomes \( 50\% \).

### Step 1: Understand the initial conditions
The initial mixture contains \( 80 \) liters of which \( 80\% \) is milk.

- Amount of milk in the initial mixture:
\[ \text{Milk} = 80 \times \frac{80}{100} = 64 \text{ liters} \]

### Step 2: Set up the equation for the final concentration
After taking out \( x \) liters of the mixture and replacing it with water, the total volume remains \( 80 \) liters. The final concentration of milk is given as \( 50\% \).

- Amount of milk in the final mixture:
\[ \text{Final milk concentration} = 50\% \text{ of } 80 = 40 \text{ liters} \]

### Step 3: Calculate the amount of milk after taking out \( x \) liters
When \( x \) liters of the mixture is taken out, the amount of milk removed is:
\[ \text{Milk removed} = x \times \frac{80}{100} = 0.8x \text{ liters} \]

Thus, the amount of milk remaining after removing \( x \) liters is:
\[ \text{Remaining milk} = 64 - 0.8x \]

### Step 4: Set up the equation based on the final condition
We know that after replacing the \( x \) liters with water, the amount of milk remaining is \( 40 \) liters:
\[ 64 - 0.8x = 40 \]

### Step 5: Solve for \( x \)
Now, we can solve the equation:
\[ 64 - 0.8x = 40 \]
Subtract \( 64 \) from both sides:
\[ -0.8x = 40 - 64 \]
\[ -0.8x = -24 \]
Now, divide both sides by \( -0.8 \):
\[ x = \frac{-24}{-0.8} = 30 \]

### Conclusion
The value of \( x \) is \( 30 \) liters.