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If kmn <> 0 ....

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by jeffedwards » Tue Dec 27, 2011 10:12 pm
See screenshot. I'm not sure how to solve this in less than 2 minutes.
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Source: — Data Sufficiency |
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by jeffedwards » Tue Dec 27, 2011 10:21 pm
I can unravel this in about four and a half minutes, but that wont work on the real test.
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by Anurag@Gurome » Tue Dec 27, 2011 10:28 pm
jeffedwards wrote:See screenshot. I'm not sure how to solve this in less than 2 minutes.
The question is: Is x/m(m² + n² + k²) = xm + yn + zk? OR
xm² + xn² + xk² = xmm + ynm + zkm?
xn² + xk² = ynm + zkm?

(1) z/k = x/m
mz = xk
mzk = xk² but we do not know if xn² = ynm; NOT sufficient.

(2) x/m = y/n
nx = my
xn² = mny but we do not know if mzk = xk²; NOT sufficient.

Combining (1) and (2), we get xn² + xk² = ynm + zkm; SUFFICIENT.

The correct answer is C.
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by jeffedwards » Tue Dec 27, 2011 10:54 pm
Thanks that's pretty good. Any tips on ways to spot the best way to move the variables around...or does that just come with practice?
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