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    Question

    A mixture contains 80% juice and rest water. A man

    withdrew some mixture and replaced it with same quantity of kerosene, such that the quantity of juice remaining in the mixture now is 64%. If the man wants to bring down the quantity of juice to less than 41%, 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    Let the percentage of mixture withdrawn first time be ‘p%’.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; 80 × {1 − (p/100)} = 64 Or,
    1 − (p/100) = 0.8 Or,
    (p/100) = 0.2 So,
    p = 20 So, 20% mixture is drawn every time. Now,
    Quantity of juice remaining after withdrawing once more = 64 × 0.8 = 51.2 And,
    Quantity of juice remaining after withdrawing once more = 51.2 × 0.8 = 40.96 So, after withdrawing two more times, the quantity of juice reaches less than 41%.

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