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Final percentage of water after replacement

CAT 2018 Slot 2 · Quantitative Ability · Medium · Mixtures & Alligations

This is a medium Quantitative Ability question from the CAT 2018 Slot 2 paper. It tests Mixtures & Alligations. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

  1. A.

    25.4

  2. B.

    20.5

  3. C.

    30.3

  4. D.

    35.2

Answer

D

Explanation

Total volume =175+700=875 ml= 175 + 700 = 875\text{ ml}. Initial alcohol quantity =700 ml= 700\text{ ml}.

Each operation removes 10%10\% of the mixture and replaces it with pure water. So the fraction of alcohol remaining after two operations is: Final Alcohol=700×(1−0.10)2=700×(0.9)2=700×0.81=567 ml\text{Final Alcohol} = 700 \times \left(1 - 0.10\right)^2 = 700 \times (0.9)^2 = 700 \times 0.81 = 567\text{ ml}

Percentage of alcohol now =567875×100=64.8%= \frac{567}{875} \times 100 = 64.8\%. Percentage of water now =100%−64.8%=35.2%= 100\% - 64.8\% = 35.2\%.

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