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# Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?

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Math Expert
Joined: 02 Sep 2009
Posts: 43361
Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ? [#permalink]

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20 Dec 2017, 05:18
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Difficulty:

55% (hard)

Question Stats:

61% (00:55) correct 39% (01:06) wrong based on 32 sessions

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

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ? [#permalink]

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20 Dec 2017, 05:20
Bunuel wrote:
Is $$\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr$$ ?

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$

Similar question from OG: https://gmatclub.com/forum/is-x-m-m-2-n ... 99423.html
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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ? [#permalink]

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21 Dec 2017, 10:34
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Bunuel wrote:
Is $$\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr$$ ?

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$

We'll go for a Precise approach of simplification since this looks like a very technical question.

Multiplying by p (p cannot be zero as it is in denominator) gives:
$$xp^2 + xq^2 + xr^2 = xp^2 + yqp + zrp$$. Cancelling common factors and changing sides gives:
$$xq^2 - yqp + xr^2 - rzp = 0$$
Substituting the equations in (1) gives $$xzr - yqp + xyq - zrp = zr(x - p) + yq(x - p) = (zr + yq)(x-p)$$
Without knowing if $$zr + yq = 0$$ or if $$p = x$$ we cannot know if this is 0.
Insufficient!

Substituting the equations in (2) gives
$$xq^2 - yqx + xr^2 - rzx = x(q^2 - yq + r^2 - rz)$$
Since we don't know if either side is 0, this is also insufficient.

But, if we substitute $$x = p$$ into our result from (1) we get a result of 0!
This means that the original equation is true.
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Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?   [#permalink] 21 Dec 2017, 10:34
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