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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:

CAT · 2023 · Quant Slot 2
Question:
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:

Options

6
10
4
7
Answer: 7

Explanation:
Step 1: Understand the process Initial milk = 40 liters Each time, 4 liters of the mixture is removed and replaced with water After each replacement, the fraction of milk remaining = 1 - 4/40 = 0.9 Let M_n = amount of milk after n steps Then: M_n = 40 * (0.9)^n Step 2: Condition for milk < water Total liquid = 40 liters Water = 40 - M_n We want milk < water, i.e., M_n < 40 - M_n This gives: 2 * M_n < 40, so M_n < 20 Step 3: Solve for n 40 * (0.9)^n < 20 (0.9)^n < 0.5 Taking natural logarithms: n * ln(0.9) < ln(0.5) n > ln(0.5)/ln(0.9) ln(0.5) ≈ -0.6931, ln(0.9) ≈ -0.1054 n > -0.6931 / -0.1054 ≈ 6.57 Step 4: Smallest integer n Since n > 6.57, the smallest integer is 7 Answer: 7 times

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