Average, Mixture and Alligation Question 5 - CAT 2025
Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is:
Step-by-Step Explanation
- First Operation (A to B):
Vessel A has 60 litres of alcohol; Vessel B has 60 litres of water.
Let x litres be transferred from A to B.
Contents left in A: 60 − x litres of pure alcohol.
Contents now in B: 60 litres of water and x litres of alcohol. Total volume in B = 60 + x litres.
- Second Operation (B back to A):
The concentration of alcohol in Vessel B is x / (60 + x).
When x litres of this mixed solution is transferred back to A, the amount of pure alcohol returned is:
Alcohol returned = x × x / (60 + x) = x² / (60 + x)
Total final alcohol in A = 60 − x + x² / (60 + x)
= [(60 − x)(60 + x) + x²] / (60 + x)
= (3600 − x² + x²) / (60 + x)
= 3600 / (60 + x)
- Use the final ratio to solve for x:
The total volume of Vessel A is restored to 60 litres. The final ratio of alcohol to water in A is 15 : 4, which means alcohol makes up 15 / (15 + 4) = 15 / 19 of the total volume.
Final alcohol in A = 60 × 15 / 19 = 900 / 19
Equating our two expressions for the final alcohol volume in A:
3600 / (60 + x) = 900 / 19
4 / (60 + x) = 1 / 19 ⟹ 76 = 60 + x ⟹ x = 16 litres
Correct Answer: 16
💡 FundaMakers Tip:
For symmetrical two-way liquid swaps of equal volume x, the final volume of the native component left in the starting vessel can be expressed using a standard algebraic shortcut:
Final Native Volume = V² / (V + x)
where V is the initial container capacity (60 litres). Setting 3600 / (60 + x) equal to the target fraction value helps you find the value of x in a single step without tracking intermediate concentrations.