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# Is |x−z−y|>x−z+y?

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Intern
Joined: 05 Mar 2013
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12 Jun 2013, 06:01
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Difficulty:

65% (hard)

Question Stats:

60% (01:37) correct 40% (01:43) wrong based on 105 sessions

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Is |x−z−y|>x−z+y?

(1) 0<x<z<y
(2) (x–z–y) is negative

Hi,

I have been doing various GMAT problems, and one major confusion was pertaining to absolute sign problems in data sufficiency (Specifically the one mentioned below)

Using the standard approach for absolute problems, I usually interpreted the question as if x-z-y >0, Then y<0 OR if x-z-y<0, Then x<z.
However my point of confusion starts, when i start looking at the two given statements.

As per the above problem, statement 1 states that x<z - Is this only information sufficient or do we also need the condition that x-z-y<0. I am really confused at this part, in some problems, both the statements are used, and in some statements only one statement is sufficient. I am really not sure how should i go about dealing with the DS statements for these type of absolute problems.

Please clarify my confusion or let me know if i am missing out on something. Thanks you.
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jun 2013, 06:07, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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12 Jun 2013, 06:17
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Is |x−z−y|>x−z+y?

(1) 0<x<z<y. This implies that $$x-z-y$$ is negative (positive number x minus greater number z minus greater number y must be less than 0). Thus $$|x-z-y|=-(x-z-y)=y+z-x$$. So, the question becomes: is $$y+z-x>x-z+y$$? --> is $$2z>2x$$? --> is $$z>x$$? The statement gives an YES answer to the question. Sufficient.

(2) (x–z–y) is negative. Almost the same here: $$|x-z-y|=-(x-z-y)=y+z-x$$. So, the question becomes: is $$y+z-x>x-z+y$$? --> is $$2z>2x$$? --> is $$z>x$$? We don't know that. Not sufficient.

Theory on absolute value is HERE.
PS absolute value questions are HERE.
DS absolute value questions are HERE.
700+ absolute value questions HERE.
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27 Jun 2013, 17:06
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Is |x−z−y|>x−z+y?

(1) 0<x<z<y
|x−z−y|>x−z+y
Because x is less than z is less than y, |x-z-y| is negative, so:
-(x-z-y)>x-z+y
-x+z+y>x-z+y
z+y>2x-z+y
2z>2x
z>x
z>x which is given to us in the statement.
SUFFICIENT

(2) (x–z–y) is negative
|x−z−y|>x−z+y
-(x-z-y)>x-z+y
-x+z+y>x-z+y
2z>2x
z>x
INSUFFICIENT

(A)

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03 Feb 2015, 05:33
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Re: Is |x−z−y|>x−z+y?   [#permalink] 03 Feb 2015, 05:33
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