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# variables ds

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Director
Joined: 17 Oct 2005
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31 Jan 2006, 20:39
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is yz>0?

1) (y^2)*z>0

2)(y^3)*z >0

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Intern
Joined: 31 Jan 2006
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31 Jan 2006, 21:05
This is my first post, please be gentle if I party foul in this post:-).

1) does not help, because y could be (-) or (+). If y is (-), yz<0. If y is (+), then yz>0. Insufficient.

2) both y& z must both be (-) or both be (+) for this to be true. In both cases, that would make yz>0.

Is the answer B) second statement sufficient, but not 1st?

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VP
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31 Jan 2006, 21:21
BARIDDLA wrote:
This is my first post, please be gentle if I party foul in this post:-).

Welcome to the GMATclub! Everybody here beat up the new guys Just Kidding. We are all here to help each other and learn.

I also think it is B. Same reason.
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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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Manager
Joined: 31 Jan 2005
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01 Feb 2006, 05:48
giddi77 wrote:
BARIDDLA wrote:
This is my first post, please be gentle if I party foul in this post:-).

Welcome to the GMATclub! Everybody here beat up the new guys Just Kidding. We are all here to help each other and learn.

I also think it is B. Same reason.

wow...came across an easy inequality problem ...after long time..
has to be "B"

I jus HATE inequalities...........
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Joined: 28 Dec 2005
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01 Feb 2006, 20:17
wow, i also got B !

The fact that after cubing y and multiplying it by z still gives a positive product must mean that y is positive (if it was negative, cubing y would be negative), and z must also be positive in order for the product to be positive

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GMAT Club Legend
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02 Feb 2006, 18:53
1) insufficient. if y = -1, z = 2, then y^2*z = 2 > 0, and yz < 0. But if y=2, z=2, then y^2*z = 8 > 0 and yz > 0.

2) sufficient. y cannot be negative and z must always be positive for y^2*z to be positive. yz is always positive.

Ans B

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02 Feb 2006, 18:53
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