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CAT 2018 Slot 2 — QA Question 8

Removal & ReplacementEasy

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

Answer & solution

  • A

    25.4

  • B

    20.5

  • 35.2

  • D

    30.3

Solution

Easy

Each operation removes 10%10\% of the mixture and tops up with water, so the alcohol (the substance not added) is multiplied by (1−110)\left(1-\tfrac{1}{10}\right) each time. Track alcohol over two operations, then water =100%−=100\%- alcohol%\%.

1

Initial composition. Water =175=175 ml, alcohol =700=700 ml, total =875=875 ml.

alcohol%=700875×100=80%\begin{aligned} &\text{alcohol}\% = \frac{700}{875}\times 100 = 80\% \end{aligned}
2

Replacement law for alcohol. Removing 10%10\% of the mixture removes 10%10\% of the alcohol; adding water adds none. So after each step alcohol ×910\times\tfrac{9}{10}. After two steps:

alcohol%=80%×(910)2⇒ =80%×81100⇒ =64.8%\begin{aligned} &\text{alcohol}\% = 80\% \times \left(\frac{9}{10}\right)^{2}\\ &\Rightarrow\ = 80\% \times \frac{81}{100}\\ &\Rightarrow\ = 64.8\% \end{aligned}
3

Water percentage.

water%=100%−64.8%=35.2%\begin{aligned} &\text{water}\% = 100\% - 64.8\% = 35.2\% \end{aligned}
water=35.2%(option c)\text{water} = 35.2\%\quad\text{(option c)}
CAT 2018 Slot 2 QA Q8: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitute — Solution | TheCATExam