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An alloy is prepared by mixing three metals $\text{A, B}$ and $\text{C}$  in the proportion $3:4:7$ by volume. Weights of the same volume of the metals $\text{A, B}$ and $\text{C}$ are in the ratio $5:2:6$. In $130$ kg of the alloy, the weight, in kg, of the metal $\text{C}$ is

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

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

Weights of the same volume of the metals $A, B,$ and $C$ are in the ratio $5:2:6.$

So, the ratio of the weights in the overall alloy.

$A:B:C = (3x \times5) : (4x \times 2) : (7x \times 6)$

$\Rightarrow A:B:C = 15 : 8 : 42$

$\therefore$ The weight of metal $C = 130 \times \frac{42}{65} = 84 \; \text{kg}.$

$\textbf{Short Method}:$

$\begin{array}{} \text{} & A & B & C \\ \text{Volume:} & 3: &4: & 6 \\ \text{Weight:} & 5: & 2: & 6(1 \;\text{litre}) \\ \text{Weight:} & 15: & 8: & 42 \end{array}$

Now,

• $65 \longrightarrow 130$
• $1 \longrightarrow 2$
• $42 \longrightarrow 84 \; \text{kg}$

Correct Answer$: \text{D}$

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