A vessel initially contains liquids A and B in the ratio 3 : 2 (by volume). Twenty liters of the mixture are removed and replaced with 20 liters of B, after which the ratio becomes 1 : 4 (A : B). How many liters of A were present initially?

Introduction / Context:
This is a remove-and-replace problem. Removing a fixed volume of mixture takes out each component in proportion to its current fraction. Then pure B is added, altering the ratio further.

Given Data / Assumptions:

• Initial A : B = 3 : 2; let total = T L → A = 3T/5, B = 2T/5.
• Remove 20 L mixture; then add 20 L of B.
• Final A : B = 1 : 4.

Concept / Approach:
Removal takes out components proportionally: A removed = (3/5)*20 = 12; B removed = (2/5)*20 = 8. Update quantities and then add 20 L B. Use the final ratio to solve for T.

Step-by-Step Solution:
A after removal = 3T/5 − 12B after removal and addition = 2T/5 − 8 + 20 = 2T/5 + 12(3T/5 − 12) : (2T/5 + 12) = 1 : 44*(3T/5 − 12) = 2T/5 + 12 ⇒ (12T/5 − 48) = 2T/5 + 1210T/5 = 60 ⇒ 2T = 60 ⇒ T = 30 LInitial A = 3T/5 = 18 L

Verification / Alternative check:
Check final: A = 18 − 12 = 6; B = 12 + (2*30/5) = 12 + 12 = 24; ratio 6 : 24 = 1 : 4.

Why Other Options Are Wrong:
12, 24, 22, 20 do not satisfy the final ratio after the remove-and-add operations.

Common Pitfalls:
Subtracting 20 directly from A or B instead of removing proportionally to the mixture composition.

Final Answer:
18 litres