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

Author Message
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Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
Followers: 0

Kudos [?]: 90 [0], given: 87126

If x is negative, is x < -3? (1) x2 > 9 (2) x3 < -9 [#permalink]  25 Aug 2004, 16:27
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(N/A)

Question Stats:

67% (02:03) correct 33% (02:12) wrong based on 7 sessions
If x is negative, is x < -3?
(1) x2 > 9
(2) x3 < -9
CIO
Joined: 09 Mar 2003
Posts: 464
Followers: 2

Kudos [?]: 34 [0], given: 0

B. Must be that way. 1 could be greater than 3 or less than -3.
Senior Manager
Joined: 25 Jul 2004
Posts: 274
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Kudos [?]: 7 [0], given: 0

A.

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

in B, consider, -2.5 -4
Senior Manager
Joined: 07 Oct 2003
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Re: DS: Algebra [#permalink]  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
CIO
Joined: 09 Mar 2003
Posts: 464
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Kudos [?]: 34 [0], given: 0

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, 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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