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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 is
1. 48
2. 70
3. 84
4. 96

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Correct Answer - Option 3 : 84

Calculation:

Let the volumes of A, B and C be 3x, 4x, and 7x

We assume that 1 litre of each of the metals is taken;

When the ratio of A, B, and C is 5 : 2 : 6, in the same volume of the mixture

So, the ratio of the weights of the three metals A, B, and C in the overall alloy of 130 kg, will become:

A ∶ B ∶ C = (3x × 5) ∶ (4x × 2) ∶ (7x × 6) = 15 ∶ 8 ∶ 42

∴ The required weight of the metal C = [42/(15 + 8 + 42)] × 130 = 84 kg

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