Average, Mixture and Alligation Question 5 - CAT 2023
A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is:
Step-by-Step Explanation
- Formulate the reduction factor:
Each operation removes 4 litres out of the 40-litre total pool, meaning 4 / 40 = 1 / 10 of the volume is replaced. The remaining fraction of milk after each iteration is:
Remaining Multiplier = 1 − 1 / 10 = 9 / 10 = 0.90
- Determine the target boundary:
For the volume of milk to become less than the volume of water, the milk must drop below 50% of the total capacity.
Final Milk Volume < 20 liters
40 × 0.90ⁿ < 20 ⟹ 0.90ⁿ < 0.50
Let us evaluate the powers of 0.90:
0.9² = 0.81
0.9³ = 0.729
0.9⁴ = 0.6561
0.9⁵ = 0.59049
0.9⁶ = 0.531441
0.9⁷ = 0.4782969
Since 0.9⁶ > 0.50 and 0.9⁷ < 0.50, the milk volume drops below the 50% threshold on the 7th operation.
Correct Answer: 7
💡 FundaMakers Tip:
For exponential decay benchmarks like 0.9ⁿ < 0.5, keep standard approximations handy to save time. You can think of this as finding when 1.11ⁿ > 2. Since 1.11 is roughly a 11% compound growth rate, applying the standard Rule of 72 (72 / 11 ≈ 6.5) tells you the crossover point must happen right around 7 iteration.