Average, Mixture and Alligation Question 6 - CAT 2024
A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had a 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had an 80% acid concentration. The ratio of water and acid in the original solution was:
Step-by-Step Explanation
Work backward from the intermediate state to streamline the calculation.
- Analyze the Second Transition:
Let the volume of the solution after the first step (adding 2 litres of water) be V. At this stage, the acid concentration is 50%, which means the mixture contains 0.5V acid and 0.5V water.
Now, 15 litres of pure acid is added, increasing the final acid concentration to 80% (which equals 4 / 5):
New Acid Volume / New Total Volume = 4 / 5
(0.5V + 15) / (V + 15) = 4 / 5
5(0.5V + 15) = 4(V + 15)
2.5V + 75 = 4V + 60
15 = 1.5V
⟹ V = 10 litres
- Determine Component Volumes in the Intermediate State:
Since the volume V is 10 litres, the intermediate mixture contains:
Acid = 5 litres, Water = 5 litres
- Revert to the Original State:
To find the original mixture, subtract the 2 litres of water that were added in the first step:
Original Acid = 5 litres
Original Water = 5 − 2 = 3 litres
Ratio of Water to Acid = 3 : 5
Correct Option: B
💡 FundaMakers Tip:
When a multi-stage mixture problem defines a clean intermediate percentage state (like a 50% split), treat that point as your base variable location. It ensures your algebra handles a balanced 1:1 ratio, which simplifies the equation compared to starting with the unknown original values. Always re-read the final line to ensure you arrange the required ratio in the right order (Water to Acid vs. Acid to Water).