DS 6 : DS Archive
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 09 Dec 2016, 05:00

### 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

# DS 6

Author Message
SVP
Joined: 16 Oct 2003
Posts: 1810
Followers: 4

Kudos [?]: 132 [0], given: 0

### Show Tags

26 May 2004, 19:20
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Attachments

d3.JPG [ 6.19 KiB | Viewed 865 times ]

SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 97 [0], given: 0

### Show Tags

26 May 2004, 19:59
A) xyz > 0 if y = -1 z = 1 and x = -1 then xyz > 0 but x(y+z) = 0 (Insufficient)
B) yz > 0 y = -1, z = -1 yz = 1 > 0 if x = 1 then xyz = 1 > 0 but if x = -1 then xyz < 0 (Insufficient)

Combine these make y = -1 z = -2 then x has to be +ve because xyz > 0
Let x = 1 1(-1+-2) = -3
if y = 1 x = 2 and x = 1 then 1(1+2) = 3
(Insufficient)

Director
Joined: 05 May 2004
Posts: 577
Location: San Jose, CA
Followers: 2

Kudos [?]: 59 [0], given: 0

### Show Tags

28 May 2004, 17:12
Bhai wrote:

Ans: E

yz>0
xyz>0
which means x is positive
but y,z can be either positive or negative!
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 15 [0], given: 0

### Show Tags

29 May 2004, 02:30
Bhai wrote:

1 and 2 together are not sufficient, because

x = 1, y = -2, z = -3 satisfy 1 and 2 but do not satisfy x(y+z)>0.
Intern
Joined: 26 Jan 2003
Posts: 28
Location: Reality
Followers: 1

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

### Show Tags

29 May 2004, 11:36
Bhai wrote:

(2) y>0, z>0 or y<0, z<0 (insufficient)
(1) x>0, yz>0 or x<0, yz<0 (insufficient)
(1)and(2) x>0, yz>0, y?0, z?0 => (y+z)?0

Re: DS 6   [#permalink] 29 May 2004, 11:36
Display posts from previous: Sort by