If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2.

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If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2. [#permalink]

02 Feb 2007, 19:01

If zy<xy<0 is |x-z| + |x| = |z|?

1. z<x>0
2. y>0

answer d.

why not a?

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02 Feb 2007, 19:40

If you could post the 2nd statement, we could see if it is suff.

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02 Feb 2007, 20:08

ah! yes.

2. y>0

Thanks.

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02 Feb 2007, 20:28

since y>0 z and x are -ve and z<x
|x-z| + |x| = |z|?

Assuming z=-2 and x=-1
|-1+2|+|-1|=2=|-2|
so suff

Still having problem posting inequalities. Tried using '=>' and am missing certain lines in the post.

Guess statement 1 has to be edited.

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Re: Gmatprep absolute value [#permalink]

04 Feb 2007, 02:51

successstory wrote:

If zy<xy<0 is |x-z| + |x| = |z|?

1. z<x>0
2. y>0

answer d.

why not a?

1. if x>0 and xy<0, y<0. But since zy<0>0. Also, as x>z, x-z>0.

So, |x-z|+|x|=x-z+x=2x-z
|z|=z

If |x-z| + |x| = |z|, 2x-z=z i.e x=z, which we know from (1) is not true.

Thus |x-z| + |x| is not = |z| SUFF

2. COntradicts (1) ???

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04 Feb 2007, 03:51

Sumithra wrote:

Assuming z=-2 and x=-1
|-1+2|+|-1|=2=|-2|

why assuming? why do you need to make an assumption here?
in general picking numbers is a bad way to prove sufficiency (it is an excellent way to show insufficiency). maybe it works for -1 and -2.... but what about -20,-24? and -100,-102? would you go to check all possible answers? i guess not....
in the moderate-difficult gmat questions, it would happen that the most straightforward "number picks" would work... but there will be "corner cases" which wouldn't, and are more difficult to find.

i wouldn't suggest using picking numbers to show sufficiency. it is logically incorrect - and will inevitably lead you to make mistakes.

here is how i'd show sufficiency of st2:
sumithra starts right by inferring from stem, that if y is positive then
z<x<0 and hence |x-z|=x-z
also: x<0 so |x|=-x, z<0 so |z|=-z
according to this it is always true that:
|x-z|+|x| = x-z-x = -z = |z|

simple algebra that proved the sufficiency of st.2

[#permalink] 04 Feb 2007, 03:51

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If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2.

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If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2.

If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2.

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