If xyz > 0, is (x)(y^2)(z^3) < 0 ? (1) y < 0 (2) x : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 11:33

### 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 xyz > 0, is (x)(y^2)(z^3) < 0 ? (1) y < 0 (2) x

Author Message
Senior Manager
Joined: 28 Jun 2007
Posts: 319
Followers: 1

Kudos [?]: 37 [0], given: 0

If xyz > 0, is (x)(y^2)(z^3) < 0 ? (1) y < 0 (2) x [#permalink]

### Show Tags

11 Nov 2007, 08:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If xyz > 0, is (x)(y^2)(z^3) < 0 ?

(1) y < 0

(2) x > 0
Manager
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 36 [0], given: 0

### Show Tags

11 Nov 2007, 09:18
Two cases:
A) Two of these variables are negative
B) All variables are positive

1) Only gives us y, which will be positive regardless
INSUFF
2) Only gives us x, we need to know z to determine the answer
INSUFF

Both:
SUFFICIENT C.
If we know what y is -ve and x is +ve and xyz > 0, then z must be -ve.

Last edited by yuefei on 11 Nov 2007, 10:20, edited 1 time in total.
Manager
Joined: 01 Nov 2007
Posts: 69
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

11 Nov 2007, 09:24
Here xyz > 0 is given

Now for the above to hold true either all x,y & z must be positive or 2 out of x,y & z must be negative...

Since in statement 1 it is given that y < 0 then it is clear that either of x or z must be negative. In any case (x)(y^2)(z^3) <0> 0.
In this case y and z can both be greater than 0 or less than 0.

So (x)(y^2)(z^3) can be both positive or negative depending on Z..

Hence it should be A
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 283 [0], given: 2

### Show Tags

11 Nov 2007, 10:18
yup A makes sense to me as well..
Manager
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 36 [0], given: 0

### Show Tags

11 Nov 2007, 10:19
I got it. Answer should be A.
Thanks sporty!
11 Nov 2007, 10:19
Display posts from previous: Sort by