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If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of

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If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of  [#permalink]

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New post 13 Aug 2018, 03:43
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Question Stats:

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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. \(u < -3v\)

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If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of  [#permalink]

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New post 13 Aug 2018, 09:00
Bunuel 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. \(u < -3v\)


Given, -3v is greater than zero, implies that v<0.

Given, \(\sqrt{u} < \sqrt{-3v}\)
we can square both sides of inequality since they are positive, Squaring both sides, we have
u < -3v-------------(equation 1)-----------------(Option E)

Adding -3v on both sides of (1), we have
u+3v<0--------------(Option D wrong)

Dividing 3 on both sides of (1), we have
\(\frac{u}{3} < -v\)------------(Option A)

Dividing v on both sides of (1), we have
\(\frac{u}{v} > -3\)------------(Option B) (Flip of sign since v is negative)---(2)

Multiplying -1 on both sides of (2), we have
\(\frac{u}{v} < -3\), take square root.
Or, \(\sqrt{\frac{u}{-v}} < \sqrt{3}\)------------(Option C)

Ans. (D)
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If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of &nbs [#permalink] 13 Aug 2018, 09:00
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If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of

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