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If u and –3v are greater than 0, and

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If u and –3v are greater than 0, and [#permalink] New post 12 Aug 2018, 10:09
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Question Stats:

35% (02:10) correct 64% (00:49) wrong based on 28 sessions
If u and –3v are greater than 0, and \(\sqrt{u} < \sqrt{-3v}\) , which of the following cannot be true ?

A. \(\frac{u}{3} < -v\)

B. \(\frac{u}{v} > -3\)

C. \(\sqrt{\frac{u}{-v}} < \sqrt{3}\)

D. \(u + 3v > 0\)

E. \(u < -3v\)
[Reveal] Spoiler: OA

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Re: If u and –3v are greater than 0, and [#permalink] New post 18 Aug 2018, 20:24
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Carcass wrote:
If u and –3v are greater than 0, and \(\sqrt{u} < \sqrt{-3v}\) , which of the following cannot be true ?

A. \(\frac{u}{3} < -v\)

B. \(\frac{u}{v} > -3\)

C. \(\sqrt{\frac{u}{-v}} < \sqrt{3}\)

D. \(u + 3v > 0\)

E. \(u < -3v\)


A quick glance through all the option, we will see that

option D which is \(u + 3v > 0\) is not possible, so might be the answer, however lets look at the remaing option

option A :: \(\frac{u}{3} < -v\) - This is possible if we divide by 3 on both sides

Option B:: \(\frac{u}{v} > -3\) - This is possible, since -3v > 0 so v < 0

Option C:: \(\sqrt{\frac{u}{-v}} < \sqrt{3}\) - It is possible by dividing \(\sqrt{-v}\) on both sides

Option E:: \(u < -3v\) - It is also possible by squaring on both sides.

** When the GRE writes a root sign, they are indicating a nonnegative root only**
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Re: If u and –3v are greater than 0, and   [#permalink] 18 Aug 2018, 20:24
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