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# Is x a negative number? (1) x2 is a positive number. (2) x

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Manager
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Is x a negative number? (1) x2 is a positive number. (2) x [#permalink]

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31 May 2009, 06:23
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Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.

Manager
Joined: 08 Feb 2009
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31 May 2009, 07:22
I think the answer is C, assuming statement 1 refers to $$x^2$$

(1) $$\Rightarrow$$ x = $$\pm$$

(2) If x = -ve, then x . |y| = Not Positive $$\Rightarrow$$ x = -ve
If x = 0, then x. |y| = Not Positive $$\Rightarrow$$ x = Not -ve.

Combining both the statements,

x = -ve.
Manager
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31 May 2009, 07:59
Even I marked it as C but the OA is E.

REally not sure why.. Manhattan provided OA as E

goldeneagle94 wrote:
I think the answer is C, assuming statement 1 refers to $$x^2$$

(1) $$\Rightarrow$$ x = $$\pm$$

(2) If x = -ve, then x . |y| = Not Positive $$\Rightarrow$$ x = -ve
If x = 0, then x. |y| = Not Positive $$\Rightarrow$$ x = Not -ve.

Combining both the statements,

x = -ve.
Manager
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31 May 2009, 08:41
prinits wrote:
Even I marked it as C but the OA is E.

REally not sure why.. Manhattan provided OA as E

goldeneagle94 wrote:
I think the answer is C, assuming statement 1 refers to $$x^2$$

(1) $$\Rightarrow$$ x = $$\pm$$

(2) If x = -ve, then x . |y| = Not Positive $$\Rightarrow$$ x = -ve
If x = 0, then x. |y| = Not Positive $$\Rightarrow$$ x = Not -ve.

Combining both the statements,

x = -ve.

Answer can be E only if we consider the possibility that y = 0.
Manager
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31 May 2009, 08:49
Stmt 1 - Tells us that X is either -ve or +ve.
Stmt 2 - Tells us that X is either -ve or 0.
By combining the 2 stmt. we can only negate that X is not positive (due to stmt b) however still it can be -ve or 0.
Hence E.
Manager
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31 May 2009, 08:56
mdfrahim wrote:
Stmt 1 - Tells us that X is either -ve or +ve.
Stmt 2 - Tells us that X is either -ve or 0.
By combining the 2 stmt. we can only negate that X is not positive (due to stmt b) however still it can be -ve or 0.
Hence E.

Why can't we negate that X is not equal to 0.
Manager
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01 Jun 2009, 19:19
Thx. goldeneagle94 for pointing a mistake in my explanation.
I agree with you that option E is possible if we consider that y can be 0.And we had to consider that ?
Manager
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02 Jun 2009, 03:39
mdfrahim wrote:
Thx. goldeneagle94 for pointing a mistake in my explanation.
I agree with you that option E is possible if we consider that y can be 0.And we had to consider that ?

I am not completely sure if we can consider a |y| to be 0.
Senior Manager
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03 Jun 2009, 23:49
Why not .??

|Y| when equated to Zero, the value of X varies..

hence IMO the answer shud be E.
Manager
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06 Jun 2009, 15:50
Quote:
Why not .??

|Y| when equated to Zero, the value of X varies..

hence IMO the answer shud be E.

can 0 be considered positive number?
Quote:
Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.

Stat 1 and 2
As x2 is a positive number and x.|y| is not a positive number - x should be a negative number?

I think the answer should be C. correct me if I am wrong..
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Manager
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08 Jun 2009, 19:12
I have a mistake when ignore y. Yes y can be 0 and from 2) x can be +ve, 0 or -ve
With 1) x can be +ve or -ve
together x can be +ve or -ve, not suf
Hence E
Director
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08 Jun 2009, 19:46
prinits wrote:
Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.

1)x^2 positive

insuff could be - or+

2) x-|y| negative
x<|y|

x could be 4 and |y| could be 5 or x could be -4 and |y|could be 5

insuff

E
Manager
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08 Jun 2009, 20:10
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prinits wrote:
Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.

Remember, $$0$$ is neither positive nor negative on the GMAT (but it is even).

(1) $$x^{2} > 0$$
$$\longrightarrow x<0 OR x>0$$
INSUFFICIENT

(2) $$x|y| \leq 0$$
$$\longrightarrow y=0 or x \leq 0$$
INSUFFICIENT

(1&2) $$either y=0 & x>0 or x<0$$

Final Answer, $$E$$.
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Re: Is x a negative number? (1) x2 is a positive number. (2) x [#permalink]

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21 Nov 2016, 16:14
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Re: Is x a negative number? (1) x2 is a positive number. (2) x   [#permalink] 21 Nov 2016, 16:14
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