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

$48$

$70$

$84$

$96$

✅ Correct Option: 3

We have an alloy made by mixing three metals A, B, and C. We're given two different ratios: volume ratio (how much space each metal takes up) and weight ratio (how heavy the same volume of each metal is).

We need to find the actual weight of metal C in 130 kg of this alloy.

Let's organize our information clearly:

Metal A: Volume Ratio 3, Weight per Unit Volume 5

Metal B: Volume Ratio 4, Weight per Unit Volume 2

Metal C: Volume Ratio 7, Weight per Unit Volume 6

Key Insight: To find the actual weight contribution of each metal, we need to multiply the volume ratio by the weight per unit volume ratio.

For each metal, we calculate: Volume Ratio × Weight per Unit Volume

Metal A: $3 \times 5 = 15$

Metal B: $4 \times 2 = 8$

Metal C: $7 \times 6 = 42$

Why do we multiply? Think of it this way: if metal A takes up 3 parts by volume, and each part weighs 5 units, then metal A contributes $3 \times 5 = 15$ units to the total weight.

The actual weight ratio of A:B:C in the alloy is:

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

Total weight parts $= 15 + 8 + 42 = 65$ parts

Metal C represents $\frac{42}{65}$ of the total weight.

In 130 kg of alloy:

Weight of C $= \frac{42}{65} \times 130$ kg

$= \frac{42 \times 130}{65} = \frac{5460}{65} = 84$ kg

The weight of metal C in 130 kg of the alloy is 84 kg.

Key Takeaway: When dealing with alloy problems involving both volume and weight ratios, we always multiply the volume ratio by the weight per unit volume ratio to get the actual weight contribution of each component. This is the most efficient approach for such problems.

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CAT Past Year Questions (Topic-Wise)

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