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Is X negative?

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Source: — Data Sufficiency |
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ans

by crackthetest » Mon Oct 19, 2009 7:23 pm
1) x^3 (1-x^2) < 0
x^3 < 0
1^2 - x^2 < 0
(1+x) (1-x) < 0
1<x or x<-1

x is -ve

A or D

2) x^2 - 1 < 0

x^2 - 1^2 < 0
(x+1) (x-1) < 0
x<1 or x<-1

(corrected)
indeterminant

Ans: A
Last edited by crackthetest on Tue Oct 20, 2009 5:45 am, edited 1 time in total.
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B is no good

by kiennguyen » Mon Oct 19, 2009 8:12 pm
IMO A
B: x^2 -1 < 0
equal: x^2 < 1 => -1<x<1. since x is not an integer, x may be - or +
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by Ozlemg » Fri Jul 22, 2011 5:38 am
No A is not sufficient as long as we do not know whether x is integer or fraction.

OA is C. Borh statements must be used together.

statement 1 is not sufficient
take example as 3 , -3
both satisfy the equation

statement 2 is not possible
anything ^ 2 will always be more than 0
whether it is -4 or 4
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by HeintzC2 » Thu Aug 04, 2011 10:49 am
Statement 2 actually states x^2 < 1 when decomposed.

Decomposing statement 1 we get x^3(1-x^2) < 0 = x^3 - x^5 < 0 = x^3 < x^5
for x^3 < x^5, x could be a positive value > 1 or a negative value > -1
insufficient.

Decomposing statement 2 we get x^2-1 < 0 = x^2 < 1.
for x^2 < 1 x can be -1 < x < 1.
insufficient.

Combining statements 1 & 2 we get;
From 1: x>1 OR 0>x>-1
AND
From 2: -1<x<1
Therefore, x must a negative value greater than -1 but less than 0.

Answer is C
Ozlemg wrote:No A is not sufficient as long as we do not know whether x is integer or fraction.

OA is C. Borh statements must be used together.

statement 1 is not sufficient
take example as 3 , -3
both satisfy the equation

statement 2 is not possible
anything ^ 2 will always be more than 0
whether it is -4 or 4
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by GMATGuruNY » Thu Aug 04, 2011 1:25 pm
Pugalenthi wrote:OG 11th edition DS question:

Is X negative?

(1) X^3(1-x^2)<0
(2) X^2 -1 < 0

Please explain your response.
Statement 1: x³(1-x²)<0.
x=2:
2³(1-2²) < 0.
-24 < 0.

x= -1/2:
(-1/2)³(1- (-1/2)² )<0
(-1/8)(3/4) < 0
-3/32 < 0.

Since it's possible that x=2 or that x=-1/2, we cannot determine whether x<0.
Insufficient.

Statement 2: x² < 1.
x = 1/2:
(1/2)² < 1.
1/4 < 1.

x = -1/2:
(-1/2)² < 1.
1/4 < 1.

Since it's possible that x=1/2 or that x=-1/2, we cannot determine whether x<0.
Insufficient.

Statements 1 and 2:
Since x²-1 < 0, we know that 1-x² > 0.
Thus, statement 1 becomes x³ * positive < 0, implying that x³ < 0 and that x<0.
Sufficient.

The correct answer is C.
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