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

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Current Student
Joined: 31 Aug 2007
Posts: 371
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if zy<xy<0, is |x-z| + |x| = |z|? strange, for some [#permalink]  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.
Manager
Joined: 03 Sep 2006
Posts: 233
Followers: 1

Kudos [?]: 5 [0], given: 0

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
Manager
Joined: 25 Jul 2007
Posts: 108
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: DS abs values [#permalink]  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.
Current Student
Joined: 31 Aug 2007
Posts: 371
Followers: 1

Kudos [?]: 53 [0], given: 1

Re: DS abs values [#permalink]  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?
Manager
Joined: 25 Jul 2007
Posts: 108
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: DS abs values [#permalink]  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.
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 98 [0], given: 0

(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.
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