If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9

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If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9 [#permalink]

25 Aug 2004, 17:27

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If x is negative, is x < -3?
(1) x2 > 9
(2) x3 < -9

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25 Aug 2004, 17:43

B. Must be that way. 1 could be greater than 3 or less than -3.

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25 Aug 2004, 17:52

A.

X is negative to start. X must be <-3

in B, consider, -2.5 -4

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Re: DS: Algebra [#permalink]

25 Aug 2004, 17:53

Anonymous wrote:

If x is negative, is x < -3?
(1) x2 > 9
(2) x3 < -9

1. if x is -2, or -3, both answers do not result in x^2>9, so the answer needs to be either -4, -5, etc.. sufficient
2. x could be -3 or -4, so 2 is insufficient

hence A

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25 Aug 2004, 18:20

wow, i totally botched that one. been kind of a long day...

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Re: If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9 [#permalink]

30 Nov 2011, 14:14

This is a tricky one.

(1) Gives x \leq -3 as x is negative. Sufficient

(2)
=> x < \sqrt[3]{-9}
=> x < - \sqrt[3]{9}

We know -8 is the nearest perfect cube and -27 is next nearest
so
-\sqrt[3]{27} < -\sqrt[3]{9} < -\sqrt[3]{8}

-3 < -\sqrt[3]{9} < -2

-\sqrt[3]{9} is a point between -3 and -2 on the number line and we got [highlight]x < - \sqrt[3]{9}[/highlight]
So this doesn't answer our question if x \leq -3 as there can be some values of x that can be x \geq -3 using statement 2 rephrase that we got above.
In sufficient

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Re: If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9 [#permalink] 30 Nov 2011, 14:14

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If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9

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If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9

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