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# variables - ds

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Manager
Joined: 11 Jan 2008
Posts: 54
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variables - ds [#permalink]  04 Jan 2009, 22:16
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Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Attachment:

varibles-ds.GIF [ 2.08 KiB | Viewed 607 times ]

can the q be re-written as : |x - y| = |z| - |x| ??

doesnt stmt 1 is exactly what is given to us...

zy < xy < 0 which is z < x < 0

i know there is a rule that u cannot divide or multiply an inequality by a varible unless u know the sign..

so confused....
SVP
Joined: 17 Jun 2008
Posts: 1570
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Re: variables - ds [#permalink]  04 Jan 2009, 22:31
"zy < xy < 0 which is z < x < 0" will be true only if y > 0.

However, xy > zy or, (x-z)y > 0
That means either (x-z) > 0 and y > 0
or, (x-z) < 0 and y < 0.

Both stmt1 and 2 tell that (x-z) > 0 and y > 0
Also, if y > 0 then x < 0 and z < 0.

Hence, |x-z| + |x| = x - z - x = -z = |z|

Hence, D.
Manager
Joined: 11 Jan 2008
Posts: 54
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Kudos [?]: 20 [0], given: 0

Re: variables - ds [#permalink]  06 Jan 2009, 17:33
Just want to make sure i got this one correct,,,

Q : |x - z| + |x| = |z| ??
Given : zy < xy < 0

which can be written as xy-zy > 0 --> y(x-z) > 0
From above either y > 0 then (x-z) > 0 or y <0 or (x-z) <0

So we dont know which to take..?

From 1 we get , z < x --> x - z > 0 so y >0 and from that x < 0, z < 0..
|x - z| + |x| = |z|
x - z - x = -z hence true (A suff..)

Similarly, we can prove from B...

Does both stmts suff....(D).

Is this considered to be a 500 - 600 level q on gmat?
Re: variables - ds   [#permalink] 06 Jan 2009, 17:33
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