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

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Joined: 31 Dec 1969
Location: India
Concentration: General Management, Entrepreneurship
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
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If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9 [#permalink] New post 25 Aug 2004, 16:27
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A
B
C
D
E

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(N/A)

Question Stats:

67% (02:03) correct 33% (02:12) wrong based on 8 sessions
If x is negative, is x < -3?
(1) x2 > 9
(2) x3 < -9
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 [#permalink] New post 25 Aug 2004, 16:43
B. Must be that way. 1 could be greater than 3 or less than -3.
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 [#permalink] New post 25 Aug 2004, 16:52
A.

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

in B, consider, -2.5 -4
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Re: DS: Algebra [#permalink] New post 25 Aug 2004, 16:53
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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 [#permalink] New post 25 Aug 2004, 17:20
:oops: 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] New post 30 Nov 2011, 13:14
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

Ans:
[Reveal] Spoiler:
A
Re: If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9   [#permalink] 30 Nov 2011, 13:14
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