How much water (in litres) must be added to 80 litres solution of milk and water containing `10%` milk, so that it becomes a `5%` milk solution ?

Options

A. `10`
B. `20`
C. `40`
D. `80`

Answer

To solve the problem of how much water must be added to an 80-litre solution of milk and water containing 10% milk to make it a 5% milk solution, we can follow these steps:

### Step 1: Calculate the amount of milk in the original solution.
The original solution is 80 litres, and it contains 10% milk.

\[ \text{Amount of milk} = \text{Total solution} \times \text{Percentage of milk} \]
\[ \text{Amount of milk} = 80 \, \text{litres} \times \frac{10}{100} = 8 \, \text{litres} \]

### Step 2: Determine the amount of water in the original solution.
Since the total solution is 80 litres and it contains 8 litres of milk, the amount of water can be calculated as follows:

\[ \text{Amount of water} = \text{Total solution} - \text{Amount of milk} \]
\[ \text{Amount of water} = 80 \, \text{litres} - 8 \, \text{litres} = 72 \, \text{litres} \]

### Step 3: Set up the equation for the new solution.
We want the final solution to be 5% milk. Let \( x \) be the amount of water to be added. The new total volume of the solution will be \( 80 + x \) litres, and the amount of milk remains 8 litres.

We want the concentration of milk in the new solution to be 5%, which can be expressed as:

\[ \frac{\text{Amount of milk}}{\text{Total solution}} = \frac{8}{80 + x} = \frac{5}{100} \]

### Step 4: Solve the equation.
Cross-multiplying gives us:

\[ 8 \times 100 = 5 \times (80 + x) \]
\[ 800 = 400 + 5x \]
\[ 800 - 400 = 5x \]
\[ 400 = 5x \]
\[ x = \frac{400}{5} = 80 \]

### Conclusion:
Thus, the amount of water that must be added is \( 80 \) litres.

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