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20 Dec 2017, 06:18
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:38) correct
33%
(01:46)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr\) ?
(1) \(yq = r^2\) and \(zr = q^2\)
(2) \(x = p\)
(1) \(yq = r^2\) and \(zr = q^2\)
(2) \(x = p\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
20 Dec 2017, 06:20
Kudos
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21 Dec 2017, 11:34
Kudos
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Bunuel
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.
(C) is our answer.
