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

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

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New post 31 May 2009, 07:23
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A
B
C
D
E

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Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.


Please discuss.
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Re: Tricky DS [#permalink]

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New post 31 May 2009, 08:22
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.
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Re: Tricky DS [#permalink]

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New post 31 May 2009, 08:59
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.
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Re: Tricky DS [#permalink]

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New post 31 May 2009, 09:41
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.
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Re: Tricky DS [#permalink]

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New post 31 May 2009, 09:49
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.
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Re: Tricky DS [#permalink]

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New post 31 May 2009, 09:56
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.
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Re: Tricky DS [#permalink]

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New post 01 Jun 2009, 20:19
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 ?
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Re: Tricky DS [#permalink]

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New post 02 Jun 2009, 04:39
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.
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Re: Tricky DS [#permalink]

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New post 04 Jun 2009, 00:49
Why not .??

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

hence IMO the answer shud be E.
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Re: Tricky DS [#permalink]

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New post 06 Jun 2009, 16:50
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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Re: Tricky DS [#permalink]

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New post 08 Jun 2009, 20:12
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
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Re: Tricky DS [#permalink]

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New post 08 Jun 2009, 20:46
prinits wrote:
Is x a negative number?

(1) x2 is a positive number.

(2) x · |y| is not a positive number.


Please discuss.


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

together no added info

E
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Re: Tricky DS [#permalink]

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New post 08 Jun 2009, 21:10
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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.


Please discuss.


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: Tricky DS   [#permalink] 08 Jun 2009, 21:10
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