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Ratio Problem

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Ratio Problem [#permalink] New post 11 Nov 2012, 06:11
Hi guys,

I don't quite get the following problem:

x + y = 2z
2x + 3y = z

What is the ratio of x : y : z ?

Please give specific solution if possible.

Thanks.
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Re: Ratio Problem [#permalink] New post 11 Nov 2012, 06:58
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Pretty Simple, though I took some time to solve this problem.
Well, I think if solved as linear equations, you will get ratios 5: -3 : 1

I think an expert needs to help to validate. :-)
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Re: Ratio Problem [#permalink] New post 11 Nov 2012, 07:56
Answer is ok.

Could you provide the equation you've used?
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Re: Ratio Problem [#permalink] New post 11 Nov 2012, 08:14
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x+y=2z ---------- (A)
2x+3y=z -------- (B)

[(B) x 2] - (A)
=> 4x+6y-x-y=0
=> 3x+5y=0
=> \frac{x}{y}=\frac{5}{-3} ------------- (X)

(B) - [(A) x 2]
=> 2x+3y-2x-2y=z-4z
=> \frac{y}{z}=\frac{-3}{1} ------------- (Y)

From (X) and (Y),
x:y:z = 5:-3:1
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Re: Ratio Problem [#permalink] New post 11 Nov 2012, 08:55
Thank you th03 for your answer. It's much more appealing to me right know.

But why would you do (B) - [(A) x 2]?

This: [(B) x 2] - (A) I understand well. You assign 0 to have some numeric reference.

But with a second equation? Why exactly this?
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Re: Ratio Problem [#permalink] New post 11 Nov 2012, 10:00
[(B) x 2] - (A) is used to eliminate z and get a ratio for x and y.

Similarly, (B) - [(A) x 2] is done to eliminate x. This gives the ratio of y and z.

Combining the above 2 results, we can arrive at x:y:z.

The manipulation to the equations must be done in such a way that we eliminate 1 variable so as to get a ratio between the remaining two.

Does it make sense?
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Re: Ratio Problem [#permalink] New post 14 Nov 2012, 13:14
Thank you very much th03. Your solution was very helpful.
Re: Ratio Problem   [#permalink] 14 Nov 2012, 13:14
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