If x>y>z, is xz>0? 1) xy<0 2) yz>0 : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 01:49

### 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 x>y>z, is xz>0? 1) xy<0 2) yz>0

Author Message
SVP
Joined: 14 Dec 2004
Posts: 1702
Followers: 3

Kudos [?]: 137 [0], given: 0

If x>y>z, is xz>0? 1) xy<0 2) yz>0 [#permalink]

### Show Tags

19 Mar 2006, 06:02
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.

If x>y>z, is xz>0?

1) xy<0
2) yz>0
Manager
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1

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

### Show Tags

19 Mar 2006, 06:13
1. xy < 0 => ((x < 0) && (y > 0)) || ((x > 0) && (y < 0)) with given x>y>z => ((x > 0) && (y < 0)) => z < 0 => xz < 0

2. ((y > 0) && (z > 0)) || ((y < 0) && (z < 0)) with ((y < 0) && (z < 0)) we can't say anything about x.

A
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

19 Mar 2006, 07:06
Tyr wrote:
1. xy < 0 => ((x < 0) && (y > 0)) || ((x > 0) && (y < 0)) with given x>y>z => ((x > 0) && (y < 0)) => z < 0 => xz < 0

2. ((y > 0) && (z > 0)) || ((y < 0) && (z < 0)) with ((y < 0) && (z < 0)) we can't say anything about x.

A

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30

Kudos [?]: 357 [0], given: 0

### Show Tags

19 Mar 2006, 07:07
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

2) yz > 0.

Could be y positive, z positive, then xz would be positive (E.g. y = 10, z = 5 and x > 10)
Or y negative, z negative, then xz would be negative/positive. (E.g. y = -3, z = -5 and x > -3)

Insufficient.

Using 1) and 2)

Could be y negative, z negative, then x must be positive and xz = negative
or
Could be y positive, z positive, then x must be negative (Does not hold since x > y)

Ans C
Manager
Joined: 23 Jan 2006
Posts: 192
Followers: 1

Kudos [?]: 26 [0], given: 0

### Show Tags

19 Mar 2006, 07:13
I say A.

ywilfred wrote:
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

2) yz > 0.

Could be y positive, z positive, then xz would be positive (E.g. y = 10, z = 5 and x > 10)
Or y negative, z negative, then xz would be negative/positive. (E.g. y = -3, z = -5 and x > -3)

Insufficient.

Using 1) and 2)

Could be y negative, z negative, then x must be positive and xz = negative
or
Could be y positive, z positive, then x must be negative (Does not hold since x > y)

Ans C

This can't be if xy < 0. either one could be negative, but not both. (but from given information, we know that x>y, so x is positive and y is negative)
Manager
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1

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

### Show Tags

20 Mar 2006, 02:51
ywilfred wrote:
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

"Or x negative, y negative" negative * negative = positive // flaw.
20 Mar 2006, 02:51
Display posts from previous: Sort by