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
- 70
- 96
- 48
- 84
Option: 4
Solution:
Solution:
Required weight of \(C = \left( {\frac{{7 \times 6}}{{3 \times 5 + 4 \times 2 + 7 \times 6}}} \right) \times 130 = 84\) kg
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