A vessel contains a mixture with water:syrup = 3:5. How much of the mixture must be drawn off and replaced with water so the final mixture is half water and half syrup?

Problem restatement
From an initial ratio water:syrup = 3:5 (total 8 parts), remove some mixture and replace it with water such that final water:syrup becomes 1:1.

Given data

• Initial water fraction = 3/8
• Initial syrup fraction = 5/8
• Final syrup fraction required = 1/2

Concept/Approach
Let the vessel capacity be V and the amount drawn off be x. Syrup remaining after removal is (5/8)(V − x). This must equal V/2.

Step-by-step calculation
(5/8)(V − x) = V/25(V − x) = 4V ⇒ 5V − 5x = 4Vx = V/5 ⇒ Draw off 1/5 of the vessel and replace with water

Verification/Alternative
After replacement: syrup = (5/8)(4V/5) = V/2; water = V − V/2 = V/2 (as required).

Common pitfalls
Equating water amounts instead of syrup amounts; either works if set up consistently, but syrup is simpler here.

Final Answer
1/5 of the vessel