An 80 litre milk-water solution contains 10% milk. Water is added to this solution.\nHow many litres of water must be added so that the new solution becomes a 5% milk solution (i.e., milk percentage reduces to 5%)?

Introduction:
This is a dilution problem where only water is added. Milk amount remains the same, but the total volume increases, reducing milk percentage. To reach a target milk percentage, we keep milk fixed and solve for the new total volume, then subtract the original volume to get water added.

Given Data / Assumptions:

• Initial solution = 80 litres
• Initial milk percentage = 10%
• Milk amount stays constant (only water is added)
• Target milk percentage = 5%

Concept / Approach:
Milk amount initially = 10% of 80. Let x litres water be added.\nThen milk% condition:\n(milk) / (80 + x) = 0.05.

Step-by-Step Solution:
Initial milk = 10% of 80 = 8 litresLet water added = x litresNew total volume = 80 + xMilk remains = 8 litresTarget condition: 8 / (80 + x) = 0.058 = 0.05(80 + x) = 4 + 0.05x4 = 0.05xx = 80 litres

Verification / Alternative Check:
After adding 80 litres water, total becomes 160 litres. Milk is still 8 litres. Milk% = 8/160 = 0.05 = 5%. This matches exactly, so the result is correct.

Why Other Options Are Wrong:
10, 20, 40 litres: not enough dilution, milk% remains greater than 5%.60 litres: gives milk% = 8/140 ≈ 5.71%, still higher than 5%.

Common Pitfalls:
Reducing milk amount when only water is added.Trying to subtract 5% from 10% directly (percentages don't change linearly with volume).Using 5% of 80 instead of 5% of (80 + x).

Final Answer:
80 litres