A 60 L mixture has acid : water = 2 : 1 (i.e., 40 L acid and 20 L water). How many liters of water must be added so that the final ratio becomes acid : water = 1 : 2?

Introduction / Context: We are changing the composition of a mixture by adding only water until the new ratio is reached. This is a direct proportion setup.

Given Data / Assumptions:

• Total = 60 L with acid : water = 2 : 1 ⇒ acid = 40 L, water = 20 L.
• Add x liters of water.
• Target ratio: acid : water = 1 : 2.

Concept / Approach: Acid remains 40 L (no acid added). Final water becomes 20 + x. Enforce the target ratio 40 : (20 + x) = 1 : 2 and solve for x.

Step-by-Step Solution: 40 / (20 + x) = 1 / 2 ⇒ 2 * 40 = 20 + x. 80 = 20 + x ⇒ x = 60 L.

Verification / Alternative check: New mixture: 40 L acid, 80 L water ⇒ ratio 1 : 2 as desired.

Why Other Options Are Wrong: 55, 50, 45, 40 do not produce 1 : 2 when added to the 20 L of water.

Common Pitfalls: Changing acid accidentally or using total volumes instead of component-wise ratio.

Final Answer: 60