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# If zy < xy < 0, is abs(x - z) + abs(x) = abs(z)? (1) z

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Manager
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If zy < xy < 0, is abs(x - z) + abs(x) = abs(z)? (1) z [#permalink]

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01 Jun 2008, 05:47
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If zy < xy < 0, is abs(x - z) + abs(x) = abs(z)?

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

OA is D, but if I have z = -9/12 and x = -1/12, the above formula doesn't hold...

I think using integers allows for the equation to work, but the question doesn't state that x and z are integers...

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Re: If zy < xy < 0... DS (spoiler) [#permalink]

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01 Jun 2008, 07:24
$$|z-x|+|x|=|z|$$

for $$z = -\frac{9}{12}$$ and $$x = -\frac{1}{12}$$,

$$|-\frac{9}{12}+\frac{1}{12}|+|-\frac{1}{12}|=|-\frac{9}{12}|$$

$$|-8|+|-1|=|-9|$$

$$9=9$$ So, your example works well

Actually, two conditions say the same:

(1) z < x ---> y has to be positive in order to hold zy < xy < 0 true
(2) y > 0 ---> z<x (zy < xy < 0 --> z<x<0)
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Re: If zy < xy < 0... DS (spoiler) [#permalink]

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01 Jun 2008, 08:38
Givenn If zy < xy < 0,
Asked is abs(x - z) + abs(x) = abs(z)?

Statement 1;
Tells us x is greater than z.

Statement 2:
Tells us y is +ve, which implies x is greater than z (Given in question xy > zy => x > z). This also implies both x, and z are -ve.

Essentially both statements are telling us same thing.

If x is greater than z then Abs (x-z) will be equal to (x-z).

So question is reduced to x-z+Abs(x) is equal to Abs(z) or not?

If x is -ve then x-z+Abs(x) = -z

In this case this value is Abs(Z), so question is answered, so answer is indeed D.

x = -1/12, z=-9/12
x - z = -1/12 + 9/12 = 8/12
Abs(x-z) + Abs(x) = 8/12 + 1/12 = 9/12

Abs(-9/12) = 9/12

So how come your values are not satisfying the given case?
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Re: If zy < xy < 0... DS (spoiler) [#permalink]

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08 Nov 2008, 00:45
zy < xy < 0, => xz > 0

The question: |x - z| + |x| = |z| => with xz > 0, this is only true if (x-z)x < 0
(1) z < x (insuff, don't know x +ve or -ve)
(2) y > 0 => x,z < 0 (insuff, don't know x - z is +-ve)

(1)&(2) suff
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Re: If zy < xy < 0... DS (spoiler) [#permalink]

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29 Mar 2011, 08:48
lylya4 wrote:
zy < xy < 0, => xz > 0

The question: |x - z| + |x| = |z| => with xz > 0, this is only true if (x-z)x < 0
(1) z < x (insuff, don't know x +ve or -ve)
(2) y > 0 => x,z < 0 (insuff, don't know x - z is +-ve)

(1)&(2) suff

=====
Wrong!
|x - z| + |x| = |z| can be written as |x - z| | = |z! - !x!
So the question asks us whether X and Z are of the same sign?

1) z < x
It means that EITHER Y is smth negative and Z and X are positive
OR
Y is smth negative and Z and X are negative. Suff - that is exactly what we are trying to understand - whether X and Z are of the same sign?

(2) y > 0
Read above. That tells us that X and Z are of the same sign. Sufficient.

OA is D
Re: If zy < xy < 0... DS (spoiler)   [#permalink] 29 Mar 2011, 08:48
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