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      Question

      A mixture contains 95% alcohol and rest water. A man

      withdrew some mixture and replaced it with same quantity of kerosene, such that the quantity of alcohol remaining in the mixture now is 76%. If the man wants to bring down the quantity of alcohol to less than 49%, then how many more times should he repeat the same process?
      A 3 Correct Answer Incorrect Answer
      B 2 Correct Answer Incorrect Answer
      C 1 Correct Answer Incorrect Answer
      D 4 Correct Answer Incorrect Answer
      E Cannot be determined Correct Answer Incorrect Answer

      Solution

      Let the percentage of mixture withdrawn first time be ‘p%’. ATQ; 95 × {1 − (p/100)} = 76 Or,
      1 − (p/100) = 0.8 Or,
      (p/100) = 0.2 So,
      p = 20 So, 20% mixture is drawn every time. Now,
      Quantity of alcohol remaining after withdrawing once more = 76 × 0.8 = 60.8 And,
      Quantity of alcohol remaining after withdrawing once more = 60.8 × 0.8 = 48.64 So, after withdrawing two more times, the quantity of alcohol reaches less than 49%.

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