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An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal C 
    • a)
      72
    • b)
      36
    • c)
      84
    • d)
      92
    Correct answer is option 'C'. Can you explain this answer?
    Verified Answer
    An alloy is prepared by mixing three metals A, B and C in the proporti...
    Required weight of C=(7×63×5+4×2+7×6)×130=84C=(7×63×5+4×2+7×6)×130=84 kg
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    Most Upvoted Answer
    An alloy is prepared by mixing three metals A, B and C in the proporti...
    To solve this problem, we need to use the concept of proportions and ratios.

    Let's break down the information given in the question:

    1. The alloy is prepared by mixing three metals A, B, and C in the proportion 3:4:7 by volume.
    2. The weights of the same volume of the metals A, B, and C are in the ratio 5:2:6.

    To find the weight of metal C in the alloy, we need to determine the weight of one unit of volume of the alloy and then calculate the weight of metal C based on its proportion in the alloy.

    Let's assume that the volume of metal A in the alloy is 3x, the volume of metal B is 4x, and the volume of metal C is 7x.

    Weight of metal A in one unit of volume = 5
    Weight of metal B in one unit of volume = 2
    Weight of metal C in one unit of volume = 6

    Now, we can calculate the weight of the alloy per unit volume:

    Weight of the alloy per unit volume = (Weight of metal A + Weight of metal B + Weight of metal C) = (5 * 3x + 2 * 4x + 6 * 7x) = (15x + 8x + 42x) = 65x

    Given that the weight of the alloy is 130 kg, we can set up a proportion to find the value of x:

    65x = 130
    x = 2

    Now, we can calculate the weight of metal C in the alloy:

    Weight of metal C = 7x = 7 * 2 = 14 kg

    Therefore, the weight of metal C in 130 kg of the alloy is 14 kg, and the correct answer is option C.
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    An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal Ca)72b)36c)84d)92Correct answer is option 'C'. Can you explain this answer?
    Question Description
    An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal Ca)72b)36c)84d)92Correct answer is option 'C'. Can you explain this answer? for Class 8 2024 is part of Class 8 preparation. The Question and answers have been prepared according to the Class 8 exam syllabus. Information about An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal Ca)72b)36c)84d)92Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 8 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal Ca)72b)36c)84d)92Correct answer is option 'C'. Can you explain this answer?.
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