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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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Joined: 02 Sep 2009
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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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13 Aug 2018, 04:43
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Difficulty:

65% (hard)

Question Stats:

56% (02:10) correct 44% (01:50) wrong based on 59 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. $$u < -3v$$

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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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13 Aug 2018, 10:00
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   [#permalink] 13 Aug 2018, 10:00
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