Let's break down this problem about milk mixtures and percentages step by step. We start with a certain quantity of milk that contains a specific percentage of water. We want to add pure milk to decrease the water percentage to a new target value. The key idea here is that the actual amount of water in the mixture does not change when we add pure milk; only the total volume of the mixture increases, thus reducing the water percentage relative to the total volume.

Understanding the Initial Milk Mixture

• Initial quantity of milk mixture = 8 liters
• Initial percentage of water in the mixture = 5%

First, we need to find the actual quantity of water in the initial 8 liters of milk mixture.

Quantity of water = 5% of 8 liters

In mathematical terms, this is:

\( \text{Quantity of water} = \frac{5}{100} \times 8 \) liters

\( \text{Quantity of water} = 0.05 \times 8 \) liters

\( \text{Quantity of water} = 0.4 \) liters

So, the initial mixture contains 0.4 liters of water.

Understanding the Change

We are adding pure milk to this mixture. Pure milk contains 0% water. Let's say we add \(x\) liters of pure milk.

• Quantity of pure milk added = \(x\) liters
• Quantity of water added with pure milk = 0% of \(x\) liters = 0 liters

After adding \(x\) liters of pure milk, the new total volume of the mixture will be the initial volume plus the added volume.

New total volume of mixture = Initial volume + Quantity of pure milk added

New total volume of mixture = \(8 + x\) liters

Understanding the Final Milk Mixture

In the new mixture, the total amount of water remains the same as in the initial mixture (because pure milk has no water). The amount of water is still 0.4 liters.

The target percentage of water in the new mixture is 4%.

This means that the 0.4 liters of water now represent 4% of the new total volume (\(8+x\) liters).

We can write this as an equation:

Quantity of water = 4% of New total volume

\( 0.4 = \frac{4}{100} \times (8 + x) \)

\( 0.4 = 0.04 \times (8 + x) \)

Solving for the Quantity of Pure Milk (\(x\))

Now, we need to solve this equation for \(x\), the quantity of pure milk added.

\( 0.4 = 0.04 \times 8 + 0.04 \times x \)

\( 0.4 = 0.32 + 0.04x \)

Subtract 0.32 from both sides:

\( 0.4 - 0.32 = 0.04x \)

\( 0.08 = 0.04x \)

Divide both sides by 0.04:

\( x = \frac{0.08}{0.04} \)

\( x = 2 \)

So, 2 liters of pure milk should be added.

Verification

Let's check if adding 2 liters of pure milk results in a 4% water concentration.

• Initial mixture: 8 liters with 0.4 liters water.
• Add 2 liters pure milk.
• New total volume = 8 + 2 = 10 liters.
• Total water in new mixture = 0.4 liters (since no water was added).
• New water percentage = \( \frac{\text{Total water}}{\text{New total volume}} \times 100\% \)
• New water percentage = \( \frac{0.4}{10} \times 100\% = 0.04 \times 100\% = 4\% \)

This matches the target percentage, confirming our calculation is correct.

Summary of Calculation

Therefore, 2 liters of pure milk must be added.