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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    Let the percentage of mixture withdrawn first time be ‘p%’.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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