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
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
Updated On: Aug 15, 2024
- 70
- 96
- 48
- 84
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The Correct Option is D
Approach Solution - 1
Required weight of C =\(\bigg (\frac{7×6}{3×5+4×2+7×6}\bigg)×130\)
= \(84\) kg
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Approach Solution -2
Let the volumes of Metals \(A, B\) and \(C\) be \(3x, 4x\) and \(7x\) respectively.
The weight ratios for the given volumes are \(5y, 2y\) and \(6y\).
\(15xy+8xy+42xy=130\)
\(⇒ 65xy=130\)
\(⇒ xy=2\)
The weight of metal \(C\)
\(= 42xy\)
\(= 42 \times 2\)
\(=84\) \(kg\)
So, the correct option is (D): \(84\)
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