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

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Director
Joined: 06 Sep 2006
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If xyz > 0, is x*y^2*z^3 < 0? 1) y < 0 2) x > 0 [#permalink]  10 Jul 2007, 14:53
If xyz > 0, is x*y^2*z^3 < 0?

1) y < 0

2) x > 0
Director
Joined: 08 Feb 2007
Posts: 610
Location: New Haven, CT
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Kudos [?]: 17 [0], given: 0

a. says nothing about x or z
b. nothing about y or z

combine, we know that y is negative so x or z must also be negative in order for xyz > 0. since x > 0 then z must be negative making x*y^2*z^3 < 0.
Director
Joined: 08 Feb 2007
Posts: 610
Location: New Haven, CT
Followers: 5

Kudos [?]: 17 [0], given: 0

You are right, it is A. I jumped to a conclusion on that one.

If y<0, then x or z have to be negative in the first equation. If x is negative then in the second equation, (-x) * y^2 will be negative and since z must be positive in the first, the second equation is less than zero. If z if negative then x has to be positive and the same result.

Last edited by lanter1 on 10 Jul 2007, 15:29, edited 1 time in total.
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Quote:
If xyz > 0, is x*y^2*z^3 < 0?

1) y <0> 0

I loath DS but that's why I decided to give this a shot:

original stem: xyz > 0 means all are positive or 2 out of the three are negative.

(1) y is negative. Tells me that either x or z is negative. if that's the case, then x*y^2*z^3 would have to be negative since y^2 removes one of the negatives from the equation. Sufficient.

(2) x is positive. Doesn't give much as y and z could both be positive or both be negative. if y and z are negative, the answer is "yes" but if y and z are positive, then the answer is "no". not sufficient.

my choice: A
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