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

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If zy < xy < 0 is |x-z| + |x| = |z| [#permalink]

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17 Jan 2010, 11:42
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Question Stats:

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

(1) z < x
(2) y > 0
[Reveal] Spoiler: OA

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Manager
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Re: please help me with this ds problem [#permalink]

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17 Jan 2010, 11:48
This should be asked in the DS forum.

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Re: please help me with this ds problem [#permalink]

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17 Jan 2010, 14:14
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Expert's post
bharat2384 wrote:
If zy<xy<0 is |x-z|+|x|=|z|
1.z<x
2.y>0

This question was discussed before here is my post from there:

This is not a good question, as neither of statement is needed to answer the question, stem is enough to do so.

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

Look at the inequality $$zy<xy<0$$:

We can have two cases:

A. If $$y<0$$ --> when reducing we should flip signs and we'll get: $$z>x>0$$.
In this case: as $$z>x$$ --> $$|x-z|=-x+z$$; as $$x>0$$ and $$z>0$$ --> $$|x|=x$$ and $$|z|=z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$-x+z+x=z$$ --> $$0=0$$, which is true.

And:

B. If $$y>0$$ --> when reducing we'll get: $$z<x<0$$.
In this case: as $$z<x$$ --> $$|x-z|=x-z$$; as $$x<0$$ and $$z<0$$ --> $$|x|=-x$$ and $$|z|=-z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$x-z-x=-z$$ --> $$0=0$$, which is true.

So knowing that $$zy<xy<0$$ is true, we can conclude that $$|x-z|+|x| = |z|$$ will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that $$z<x$$, hence we have case B.

Statement (2) says that $$y>0$$, again we have case B.

Hope it helps.
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Re: If zy<xy<0 is |x-z|+|x|=|z| 1.z<x 2.y>0 A. [#permalink]

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09 Jan 2013, 11:30
If zy<xy<0 is |x-z|+|x|=|z|
1.z<x
2.y>0

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Bunuel is right we don't need stmts (1) or (2) or both to solve this problem. Here how can we come to it.

squre both sides of the equation:

|x-z|+|x|=|z|
|x-z|=|z|-|x|
x^2-2xz+z^2=x^2-2|xz|+z^2
after that we have:

-2xz=-2|xz| divding this equation by -2 we have:
xz=|xz| which means that question asks us: is xz=|xz|? here everything is clear
i.e.If zy<xy<0 y must be positive or negative. If y positive x and z must be both negative or vice versa which means that if y is negative both x and z must be positive where the inequality zy<xy<0 will be correct.

So we don't need any statements to solve this problem

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Re: If zy < xy < 0 is |x-z| + |x| = |z| [#permalink]

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01 Feb 2013, 05:57
Given that zy < xy < 0 . Thus, xy-zy>0 = y(x-z)>0.

(1) z < x . Thus, from above y has to be positive. Hence, sufficient.

(2) y > 0. Similarly from above, (x-z)>0. Also sufficient.

D.
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Re: If zy < xy < 0 is |x-z| + |x| = |z| [#permalink]

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06 Feb 2013, 00:20
zy<xy<0 (for y<0).....z..................x...................|..................x..................z....(for y>0)

so |x-z| + |x| = |z| -----> |delta| + |x| = |z| -------> LHS=RHS (take a look at the no. line i have drawn above)
This is a conditional identity and will always hold true

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Re: If zy < xy < 0 is |x-z| + |x| = |z|   [#permalink] 06 Feb 2013, 00:20
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# If zy < xy < 0 is |x-z| + |x| = |z|

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