Calculating Pure Milk Addition for Water Reduction
This problem involves calculating the amount of pure milk to add to a mixture to achieve a specific lower water concentration. We need to find the quantity of pure milk to add to 8 L of milk, initially containing 4% water, so that the final mixture has only 2% water.
Initial Mixture Analysis
- Total initial volume of milk = 8 L
- Initial water percentage = 4%
- Calculate the initial amount of water:
Amount of Water = $4\% \times 8 \text{ L} = 0.04 \times 8 \text{ L} = 0.32 \text{ L}$ - The amount of pure milk initially is $8 \text{ L} - 0.32 \text{ L} = 7.68 \text{ L}$.
Calculating Added Pure Milk
Let the quantity of pure milk added be '$x$' litres.
- When pure milk is added, the amount of water remains constant ($0.32$ L), but the total volume increases.
- New total volume = $(8 + x)$ L
- The desired final water percentage is 2%.
- We can set up an equation using the final state:
$\frac{\text{Amount of Water}}{\text{New Total Volume}} = \text{Desired Water Percentage}$
$\frac{0.32 \text{ L}}{(8 + x) \text{ L}} = 2\% = 0.02$
Solving for Added Milk Quantity
- Cross-multiply to solve the equation:
$0.32 = 0.02 \times (8 + x)$ - Distribute the 0.02:
$0.32 = (0.02 \times 8) + (0.02 \times x)$
$0.32 = 0.16 + 0.02x$ - Isolate the term with '$x$':
$0.02x = 0.32 - 0.16$
$0.02x = 0.16$ - Solve for '$x$':
$x = \frac{0.16}{0.02} = \frac{16}{2} = 8$
Therefore, 8 L of pure milk should be added.
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