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    The vessel contains 60 liters of a mixture that contains 80% milk and the rest is water. Firstly, 20 liters of mixture are replaced by water, and then
    Question

    The vessel contains 60 liters of a mixture that contains 80% milk and the rest is water. Firstly, 20 liters of mixture are replaced by water, and then, 15 liters of mixture are removed from the resultant mixture. If x liters of water are added to the remaining mixture, then the ratio of milk to water becomes equal in the final mixture. Find the value of x.

    A.

    3

    B.

    4

    C.

    5

    D.

    6

    E.

    2

    Correct option is A

    Given

    Total mixture = 60 liters

    percentage of milk = 80%

    The ratio of milk to water in resultant mixture is same

    Explanation:

    Quantity of milk in the initial mixture=60100×80=48 litresQuantity of water in the initial mixture=6048=12 litresAfter first replacement,Quantity of milk=4820×45=32 litresQuantity of water=1220×15+20=28 litresRatio of milk to water=32:28=8:7Now,3215×8152815×715+x=1124=21+xx=3\text{Quantity of milk in the initial mixture} = \frac{60}{100} \times 80 = 48 \, \text{litres} \\\text{Quantity of water in the initial mixture} = 60 - 48 = 12 \, \text{litres} \\\text{After first replacement,} \\\text{Quantity of milk} = 48 - 20 \times \frac{4}{5} = 32 \, \text{litres} \\\text{Quantity of water} = 12 - 20 \times \frac{1}{5} + 20 = 28 \, \text{litres} \\\text{Ratio of milk to water} = 32:28 = 8:7 \\\text{Now,} \\\frac{32 - 15 \times \frac{8}{15}}{28 - 15 \times \frac{7}{15} + x} = \frac{1}{1} \\24 = 21 + x \\x = 3

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