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Is x/p(p 2 + q 2 + r 2 ) = x

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Is x/p(p 2 + q 2 + r 2 ) = x [#permalink] New post 03 Feb 2013, 17:45
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79% (02:43) correct 21% (02:03) wrong based on 43 sessions
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Re: Is x/p(p 2 + q 2 + r 2 ) = x [#permalink] New post 03 Feb 2013, 21:46
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carcass wrote:
Is \frac{x}{p} (p 2 + q 2 + r 2 ) = xp + yq + zr?

(1) q^2 = r^2

(2) yq = zr


There are too many variables here so let's focus on logic.

Is \frac{x}{p} (p 2 + q 2 + r 2 ) = xp + yq + zr?
Is xp + \frac{x}{p}(q 2 + r 2 ) = xp + yq + zr?
Is \frac{x}{p}(q 2 + r 2 ) = yq + zr?

(1) q^2 = r^2
Even if q^2 = r^2, there are lots of other variables e.g. x, y and z which can take different values to make the two sides unequal. But if q = r = 0, the two sides become equal so not sufficient alone.

(2) yq = zr
Again, there are no constraints on x and p and they can take different values to make the two sides of the equation unequal. But if q = r = 0, the two sides become equal so not sufficient alone.

Using both together, there are still no constraints on the values x and p can take so the two sides needn't be equal. But if q = r = 0, the two sides become equal so not sufficient.

Answer (E)
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Expert Post
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Re: Is x/p(p 2 + q 2 + r 2 ) = x [#permalink] New post 04 Feb 2013, 03:59
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Re: Is x/p(p 2 + q 2 + r 2 ) = x [#permalink] New post 04 Feb 2013, 13:16
What additional condition must be present for the equation to be true ?

[Reveal] Spoiler:
xq = yp

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Re: Is x/p(p 2 + q 2 + r 2 ) = x   [#permalink] 04 Feb 2013, 13:16
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