Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Feb 2016, 00:02

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 21 Mar 2007
Posts: 73
Followers: 1

Kudos [?]: 51 [0], given: 0

If zy < xy < 0, is abs(x - z) + abs(x) = abs(z)? (1) z [#permalink]  01 Jun 2008, 04:47
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

67% (02:31) correct 33% (02:53) wrong based on 3 sessions
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...

CEO
Joined: 17 Nov 2007
Posts: 3573
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 457

Kudos [?]: 2598 [0], given: 359

Re: If zy < xy < 0... DS (spoiler) [#permalink]  01 Jun 2008, 06:24
Expert's post
$$|z-x|+|x|=|z|$$

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)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 10 Sep 2007
Posts: 947
Followers: 8

Kudos [?]: 234 [0], given: 0

Re: If zy < xy < 0... DS (spoiler) [#permalink]  01 Jun 2008, 07:38
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?
Manager
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 13 [0], given: 0

Re: If zy < xy < 0... DS (spoiler) [#permalink]  07 Nov 2008, 23:45
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
Senior Manager
Status: Happy to join ROSS!
Joined: 29 Sep 2010
Posts: 279
Concentration: General Management, Strategy
Schools: Ross '14 (M)
Followers: 19

Kudos [?]: 110 [0], given: 48

Re: If zy < xy < 0... DS (spoiler) [#permalink]  29 Mar 2011, 07:48
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, 07:48
Similar topics Replies Last post
Similar
Topics:
6 If zy < xy < 0 is |x - z| + |x| = |z| 16 26 Nov 2011, 09:10
54 If zy < xy < 0, is | x - z | + |x| = |z|? 16 17 Sep 2010, 11:10
22 If zy < xy < 0, is |x-z| + |x| = |z|? 8 28 Jan 2010, 01:50
1 If zy < xy < 0 is |x-z| + |x| = |z| 5 17 Jan 2010, 10:42
4 ZY<XY<0 is |X-Z| + |x| = |Z| 14 25 Apr 2007, 03:18
Display posts from previous: Sort by