A 100-litre solution contains acid and water in the ratio 3:2. Some quantity of this solution is removed and replaced with pure acid. If the final ratio of acid to water becomes 7:3, how many litres of solution were replaced?

Correct — B. Track the water: it only leaves with the removed solution, and none is added back.

Start: 3 : 2 of 100 L = 60 L acid, 40 L water
End: still 100 L at 7 : 3, so water = 3⁄10 × 100 = 30 L

Removing x litres takes out 2⁄5 of it as water = 0.4x
Water left: 40 − 0.4x = 30
Solve: 0.4x = 10 → x = 25 litres → option (b)

Why the others are wrong

• (a)
  35 litres — Replacing 35 L leaves 40 − 14 = 26 L water and 74 L acid, a 37 : 13 mix, stronger than 7 : 3.
• (c)
  50 litres — Replacing 50 L halves the water to 20 L, with 80 L acid: 4 : 1, not 7 : 3.
• (d)
  20 litres — Replacing 20 L leaves 40 − 8 = 32 L water and 68 L acid: 17 : 8, not yet 7 : 3.

In a remove-and-replace question the removed portion carries every component in the original ratio. Taking out x litres of a 3 : 2 mix removes 0.6x acid and 0.4x water.

The component that is never added back, water here, is the easy one to follow: it falls from 40 L to 40 × (1 − x⁄100) L.

Setting that equal to the final 30 L gives 1 − x⁄100 = 3⁄4, so x = 25.

Key facts

• Removed solution leaves in the original ratio: 25 L of a 3 : 2 mix takes out 15 L acid and 10 L water.
• The total stays 100 L because every litre removed is replaced.
• Final check: acid 60 − 15 + 25 = 70 L and water 30 L, so 70 : 30 = 7 : 3.

Common traps

• Treating the removed x litres as pure water, which gives x = 40 − 30 = 10 L.
• Adding x litres of acid without first taking out the 0.6x litres of acid that left with the solution.

No directly related past PYQ was found.