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

I tried solving this question but I was not able to. Also the solutions that I looked up for on the web are not well-explained. Any help will be much appreciated.

Ganit
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1 Answers1

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Say their weights are $5x, 2x, 6x$ for unit volume.

In alloy of volume $14$ units, we have volume of $A$ as $3$ units, of $B$ as $4$ units and of $C$ as $7$ units.

Multiplying by their weights of unit volume, we have weight of $15x, 8x, 42x$ for metals $A, B$ and $C$ respectively. Summing, the total weight of the alloy is $65x$.

Now weight of metal $C$ in $65x$ unit weight of alloy is $42x$, so weight of metal $C$ in $130$ kg of alloy

$ = \frac{42x}{65x} \times 130 = 84 \,$ kg.

Math Lover
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    ohhh...i was doing a silly mistake...i was taking the weights in the ratio of 5 : 2 : 6 for only if the volume is in the ratio of 3 : 4 : 7... i overlooked the statement that Weights of the "same volume" of metals A, B, C are in the ratio 5 : 2 : 6... thanks for helping out... – Ganit Dec 27 '20 at 07:10
  • You are welcome! – Math Lover Dec 27 '20 at 07:12