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    Question

    In a 448 ml mixture, the ratio of copper to gold is 9:5. When 'a + 12' ml of copper and 'a - 10' ml of gold are added to this mixture, the ratio of copper to gold in the final mixture becomes 12:7.find the value of 'a + 32' under these conditions.

    A 100 Correct Answer Incorrect Answer
    B 95 Correct Answer Incorrect Answer
    C 65 Correct Answer Incorrect Answer
    D 85 Correct Answer Incorrect Answer
    E 92 Correct Answer Incorrect Answer

    Solution

    we can say that  Quantity of Copper in 448 ml of mixture = (9/14) × 448 = 288 ml Quantity of Gold in 448 ml of mixture = 448 – 288 = 160 ml According to question; (288 + a + 12)/(160 + a – 10) = 12/7 Or, (300 + a)/(a + 150) = 12/7 Or, 2100 + 7a = 12a + 1800 Or, 5a = 300 Or, a = 60 So, a + 32 = 60 + 32 = 92

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