If zy < xy < 0, is |x-z| + |x| = |z|? : GMAT Data Sufficiency (DS)
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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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28 Jan 2010, 01:50
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If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z < x
(2) y > 0

[Reveal] Spoiler:
For zy < xy < 0 to be true, I am counting two possible scenarios

x y z
-ve +ve -ve -------------------1
+ve -ve +ve--------------------2

Statement 1
rules out scenario 2 but scenario 1 is possible.
Now when i substitute the signs of x and z and take them out from the modulus, i get -

(-x + z) + (-x) = (-z)
-2x = -2z
x=z - - -
Therefore in the original equation, |x-z| = 0
and |x| = |z|

-hence sufficient

Statement 2

Again eliminates the possibility of the scenario 2
and hence is sufficient

Am i correct here?
[Reveal] Spoiler: OA
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28 Jan 2010, 02:51
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An easy way to approach this ineq will be to analyze that:
|x-z| + |x| = |z| means
|z-x| = |z| - |x|.
|z-x| is the distance between z and x on number line. It can only be equal to |z| - |x| if both z and x have the same signs.

a) z < x - implies that y > 0 because zy < xy.
If y > 0 then z < x < 0. Therefore both have same signs.
SUFF

b) y>0 then z < x < 0. Therefore both have same signs.
SUFF

D is the answer i think
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28 Jan 2010, 05:14
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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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31 Jan 2010, 11:40
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wow, I solved this question by myself:)
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Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]

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27 Nov 2013, 07:32
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zaarathelab wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z < x
(2) y > 0

[Reveal] Spoiler:
For zy < xy < 0 to be true, I am counting two possible scenarios

x y z
-ve +ve -ve -------------------1
+ve -ve +ve--------------------2

Statement 1
rules out scenario 2 but scenario 1 is possible.
Now when i substitute the signs of x and z and take them out from the modulus, i get -

(-x + z) + (-x) = (-z)
-2x = -2z
x=z - - -
Therefore in the original equation, |x-z| = 0
and |x| = |z|

-hence sufficient

Statement 2

Again eliminates the possibility of the scenario 2
and hence is sufficient

Am i correct here?

Rearrange question |-(-x+z)|=|z-x| = |z|-|x|?
By property they will be equal when both x and z have the same sign

Statement 1
If x>z, then with zy < xy < 0, x and z are both positive. Same sign. Suff

Statement 2.
If y>0 then same with zy < xy < 0, x and z both positive. Same sign Suff

Hope it helps
Cheers!
J
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Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]

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17 Dec 2013, 03:17
I solved this question on number line, we can first analyze this eq |x-z| + |x| = |z|?
This equality is only possible, if x and z are on the same side of the number line.

(-) ----z----x----0--- or ---0-----x----z---- (+)

|distance between xandz| + |distance of x from origin| = |distance of z from origin|

given zy<xy<0 which can happen in two case.

z x y
- - + zy negative xy negative < both less than 0
+ + - zy negative xy negative < both less than 0

Therefore we don't even need option 1 and 2 to validate this |x-z| + |x| = |z|.

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

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16 Apr 2014, 21:28
(1) z < x => y > 0
(2) Same as (1)

(1) or (2)
x < 0
z < 0

0 < x-z
abs(x-z) + abs(x) = abs(z)?
x-z + -x = -z
S

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

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10 May 2014, 11:04
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using Statement 1 :
from the question , zy < xy
y(z-x)<0 ------- (A)
Now statement 1 tells me that (z-x)< 0 . This implies Y>0
So, if zy < xy < 0 and Y>0
This implies Z & X < 0
mod (x-z) + mod (x) = X-Z (since Z-X<0) + (-X) = -Z = |Z|

Using Statement 2 :
From (A), y(z-x)<0
Since from Statement 2 we know that y > 0
That implies (z-x)< 0 .

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

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16 May 2015, 06:49
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Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]

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19 Jun 2016, 00:46
Hello from the GMAT Club BumpBot!

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Re: If zy < xy < 0, is |x-z| + |x| = |z|?   [#permalink] 19 Jun 2016, 00:46
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