If 80 liters of milk solution contains 60\% milk, how many liters of pure milk should be added to make the milk content 80\% in the new solution?

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**Solution:**

1. Initial milk in the solution = 60\% of 80 liters = $80 	imes 0.60 = 48$ liters.

2. Let the amount of pure milk to be added be $x$ liters.

3. After adding $x$ liters of pure milk, total volume = $80 + x$ liters.

4. The new solution should have 80\% milk, so:
   $$\frac{48 + x}{80 + x} = 0.80$$

5. Multiply both sides by $(80 + x)$:
   $$48 + x = 0.80(80 + x)$$

6. Expand the right side:
   $$48 + x = 64 + 0.80x$$

7. Rearranging terms:
   $$x - 0.80x = 64 - 48$$
   $$0.20x = 16$$

8. Solve for $x$:
   $$x = \frac{16}{0.20} = 80$$

**Answer:** $80$ liters of pure milk should be added.

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