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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]  12 Aug 2018, 10:09
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Question Stats:

47% (02:04) correct 52% (00:59) wrong based on 19 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]  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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# If u and –3v are greater than 0, and

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