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Is x negative ? (1) n3 (1-X2) < 0 (2) X2 - 1 < 0

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Director
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Is x negative ? (1) n3 (1-X2) < 0 (2) X2 - 1 < 0 [#permalink] New post 24 Jul 2005, 08:17
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A
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C
D
E

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Is x negative ?
(1) n3 (1-X2) < 0
(2) X2 - 1 < 0
Director
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 [#permalink] New post 24 Jul 2005, 08:26
Hi Arsene, did you mean to write X^2 and n^3 or 2x and 3n? . thnaks
Senior Manager
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 [#permalink] New post 24 Jul 2005, 10:37
If X2 = X^2, then it doesn't make any sense. The answer is E. X could be anything from -1 to 1.
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 [#permalink] New post 24 Jul 2005, 11:11
Answer is E and here is why

From stmt 1, n^3 (1 – X^2) < 0
n^3 – n^3 * x^2 < 0
n^3 < n^3*x^2
1 < x^2
Insuff cos x could be -2, or 2

From stmt 2, x^2 -1 < 0
X^2 < 1
Means x could be < 1 and -1

Combining both, they are insuff
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Re: DS: X [#permalink] New post 25 Jul 2005, 10:24
Arsene_Wenger wrote:
Is x negative ?
(1) n3 (1-X2) < 0
(2) X2 - 1 < 0


E , both the inequalities would be valid for both "+ and - "x alike.

HMTG.
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 [#permalink] New post 26 Jul 2005, 23:14
sorry guys, that's not what the OA says.
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 [#permalink] New post 27 Jul 2005, 08:57
E

x=-0.01 and x=0.01 both satisfy

n^3(1-x^2) < 0
and
x^2 -1 < 0

of course, n needs to be < 0

Not sure how the OA is not E
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 [#permalink] New post 28 Jul 2005, 07:09
Could you post OA and OE?
And please, check the question again. It seems that you missed an important detail.
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 [#permalink] New post 29 Jul 2005, 02:11
Yeah guys, OA is (E). I am still trying to convince myself on (E) b/c i do not have an OE.

i'll look at qpoo's post now.
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 [#permalink] New post 29 Jul 2005, 03:34
Arsene_Wenger wrote:
Yeah guys, OA is (E). I am still trying to convince myself on (E) b/c i do not have an OE.

i'll look at qpoo's post now.


I am pretty sure you accept why COndition 2 is insufficient

As far as condition 1 is concerned i went like this

n^3 - n^3x^2 <0

subtract n^3 on both sides

- n^3x^2 < -n^3

multiply -1 to both sides (sign in ineq changes)

n^3x^2 > n^3

Divide both sides ny n^3 which could be + or -

If it is positive then we get x^2 > 1 (or x > +/-1) so we cannot say
if it is negative then we get x^2 < 1 (or x < +/- 1) so again we cannot say

COmbining both conditions is useless because condition 1 is a subset of condition 2. Hence E
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  [#permalink] 29 Jul 2005, 03:34
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