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# if zy<xy<0, is |x-z| + |x| = |z|? strange, for some

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if zy<xy<0, is |x-z| + |x| = |z|? strange, for some [#permalink]

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20 Nov 2007, 17:20
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if zy<xy<0, is |x-z| + |x| = |z|?

strange, for some reason it is not coming out as i type it...

statement 1--z is less than x
statement 2--y is greater than 0

Last edited by young_gun on 20 Nov 2007, 18:01, edited 5 times in total.

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Manager
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20 Nov 2007, 19:48
My ans is C:

1) Z < X
it means than Z and X can either be both +ve or both -ve
***** if +ve:
Z < X
3 < 5
|5 - 3| + |5| = |5| NO

***** if -ve:
Z < X
-5 < -3
|-3 - (-5)| + |-3| = |-5| YES

Hence 1) is insuff

2) Y > 0, this means that Z and Y are -ve
Z < X
-5 < -3
|-3 - (-5)| + |-3| = |-5| YES

Z > X
-1 > -100
|-100 - (-1)| + |-100| = |-1| NO

Hence 2) is insuff

Together: suff

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Manager
Joined: 25 Jul 2007
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Re: DS abs values [#permalink]

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20 Nov 2007, 20:06
young_gun wrote:
if zy<xy<0, is |x-z| + |x| = |z|?

strange, for some reason it is not coming out as i type it...

statement 1--z is less than x
statement 2--y is greater than 0

First of all, we note that x,y or z cannot be equal to 0.

Statement 1: Sufficient.

If z < x, then both 'x' and 'z' have to be negative and 'y' has to be positive. ( since zy < xy < 0 )

In such a scenario, |x-z| + |x| will always be equal to |z|.

Statement 2: Sufficient.

If y is greater than 0, it again implies that both 'x' and 'z' are negative. ( since zy < xy < 0 ).

Therefore answer is D. Both are sufficient.

Kudos [?]: 23 [0], given: 0

Current Student
Joined: 31 Aug 2007
Posts: 366

Kudos [?]: 168 [0], given: 1

Re: DS abs values [#permalink]

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21 Nov 2007, 06:28
jbs wrote:
young_gun wrote:
if zy<xy<0, is |x-z| + |x| = |z|?

strange, for some reason it is not coming out as i type it...

statement 1--z is less than x
statement 2--y is greater than 0

First of all, we note that x,y or z cannot be equal to 0.

Statement 1: Sufficient.

If z < x, then both 'x' and 'z' have to be negative and 'y' has to be positive. ( since zy < xy < 0 )

In such a scenario, |x-z| + |x| will always be equal to |z|.

Statement 2: Sufficient.

If y is greater than 0, it again implies that both 'x' and 'z' are negative. ( since zy < xy < 0 ).

Therefore answer is D. Both are sufficient.

can you/someone pls elaborate the bold section?

Kudos [?]: 168 [0], given: 1

Manager
Joined: 25 Jul 2007
Posts: 107

Kudos [?]: 23 [0], given: 0

Re: DS abs values [#permalink]

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21 Nov 2007, 07:21
young_gun wrote:
jbs wrote:
young_gun wrote:
if zy<xy<0, is |x-z| + |x| = |z|?

strange, for some reason it is not coming out as i type it...

statement 1--z is less than x
statement 2--y is greater than 0

First of all, we note that x,y or z cannot be equal to 0.

Statement 1: Sufficient.

If z < x, then both 'x' and 'z' have to be negative and 'y' has to be positive. ( since zy < xy < 0 )

In such a scenario, |x-z| + |x| will always be equal to |z|.

Statement 2: Sufficient.

If y is greater than 0, it again implies that both 'x' and 'z' are negative. ( since zy < xy < 0 ).

Therefore answer is D. Both are sufficient.

can you/someone pls elaborate the bold section?

Practically, just put in couple of negative values for 'x' and 'z' and you will find out for yourself.

Alternatively, here's the conceptual explanation.

We know that 'x' and 'z' are negative.

Therefore |x-z| is basically the same as |z| - |x|. (e.g: |-2 - (-3)| = |-3| - |-2|
i.e. |x-z| = |z| - |x| + |x| = |z|

Hope this helps.

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SVP
Joined: 01 May 2006
Posts: 1794

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21 Nov 2007, 09:21
(D) for me too

|x-z| + |x| = |z|?

zy<xy<0

Implies that:
o zy - xy < 0
<=> y*(z-x) < 0
<=> y*(x-z) > 0

That means : sign(y) = sign(x-z)

Stat 1
We have :
o z < x
<=> x-z > 0

That means : y > 0.

As y > 0, with zy<xy<0, we now know that z < 0 and x < 0.

Finally,
o |x-z| + |x|
= (x-z) + (-x) as x-z > 0 and -x > 0
= -z
= |z| as z < 0

SUFF.

Stat 2
We have y > 0

That means x-z > 0.

Again, as y > 0, with zy<xy<0, we now know that z < 0 and x < 0.

Finally,
o |x-z| + |x|
= (x-z) + (-x) as x-z > 0 and -x > 0
= -z
= |z| as z < 0

SUFF.

Kudos [?]: 171 [0], given: 0

21 Nov 2007, 09:21
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# if zy<xy<0, is |x-z| + |x| = |z|? strange, for some

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