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      Question

      A mixture contains 70% acid and rest water. A man

      withdrew some mixture and replaced it with same quantity of kerosene, such that the quantity of acid remaining in the mixture now is 56%. If the man wants to bring down the quantity of acid to less than 36%, 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      Let the percentage of mixture withdrawn first time be ‘p%’.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; 70 × {1 − (p/100)} = 56 Or,
      1 − (p/100) = 0.8 Or,
      (p/100) = 0.2 So,
      p = 20 So, 20% mixture is drawn every time. Now,
      Quantity of acid remaining after withdrawing once more = 56 × 0.8 = 44.8 And,
      Quantity of acid remaining after withdrawing once more = 44.8 × 0.8 = 35.84 So, after withdrawing two more times, the quantity of acid reaches less than 36%.

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