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If u and –3v are greater than 0, and
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12 Aug 2018, 10:09
12 Aug 2018, 10:09
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45% (02:01) correct
54% (01:40) wrong based on 148 sessions
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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. \(\frac{u}{3} < -v\)
B. \(\frac{u}{v} > -3\)
C. \(\sqrt{\frac{u}{-v}} < \sqrt{3}\)
D. \(u + 3v > 0\)
E. \(u < -3v\)
Re: If u and –3v are greater than 0, and
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18 Aug 2018, 20:24
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. \(\frac{u}{3} < -v\)
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**
Re: If u and –3v are greater than 0, and
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25 May 2019, 07:33
25 May 2019, 07:33
here, u>0, and -3v>0
So, u+(-3v)>0
U-3v >0
But, option D is not as above . So, D.
So, u+(-3v)>0
U-3v >0
But, option D is not as above . So, D.
