Question

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 is

• A. 70
• B. 96
• C. 48
• D. 84

Answer 1

Answer: (D)

84

Answer 2

Required weight of C =$\bigg (\frac{7×6}{3×5+4×2+7×6}\bigg)×130$

= $84$ kg