Repeated dilution of milk

CAT 2023 Slot 2 · QA · Easy · Arithmetic

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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

Answer

7

Explanation

Fraction of milk remaining after each replacement =1−440=0.9= 1 - \frac{4}{40} = 0.9.

After nn operations, volume of milk =40×(0.9)n= 40 \times (0.9)^n.

We want milk < water, which means milk < 20 liters (half of total volume 40): 40×(0.9)n<20  ⟹  (0.9)n<0.540 \times (0.9)^n < 20 \implies (0.9)^n < 0.5

Calculating powers of 0.90.9:

  • 0.91=0.90.9^1 = 0.9
  • 0.92=0.810.9^2 = 0.81
  • 0.93=0.7290.9^3 = 0.729
  • 0.94=0.65610.9^4 = 0.6561
  • 0.95=0.590490.9^5 = 0.59049
  • 0.96=0.5314410.9^6 = 0.531441
  • 0.97=0.4782969<0.50.9^7 = 0.4782969 < 0.5

So the smallest number of times is 77.

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