|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 28 Jun 2007
Posts: 322
Followers: 1
Kudos [?]:
6
[0], given: 0
|
If xyz > 0, is (x)(y^2)(z^3) < 0 ? (1) y < 0 (2) x [#permalink]
 11 Nov 2007, 09:44
If xyz > 0, is (x)(y^2)(z^3) < 0 ?
(1) y < 0
(2) x > 0
|
|
|
|
|
|
|
Manager
Joined: 02 Aug 2007
Posts: 151
Followers: 1
Kudos [?]:
4
[0], given: 0
|
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:
INCORRECT - See answer below----
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, 11:20, edited 1 time in total.
|
|
|
|
|
|
Manager
Joined: 01 Nov 2007
Posts: 69
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
135
[0], given: 2
|
yup A makes sense to me as well..
|
|
|
|
|
|
Manager
Joined: 02 Aug 2007
Posts: 151
Followers: 1
Kudos [?]:
4
[0], given: 0
|
I got it. Answer should be A.
Thanks sporty!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
