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variables ds

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variables ds [#permalink] New post 31 Jan 2006, 19:39
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
is yz>0?

1) (y^2)*z>0

2)(y^3)*z >0
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 [#permalink] New post 31 Jan 2006, 20:05
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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 [#permalink] New post 31 Jan 2006, 20:21
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 :wink: Just Kidding. We are all here to help each other and learn.

I also think it is B. Same reason.
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 [#permalink] New post 01 Feb 2006, 04:48
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 :wink: 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........... :evil:
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 [#permalink] New post 01 Feb 2006, 19:17
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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 [#permalink] New post 02 Feb 2006, 17:53
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
  [#permalink] 02 Feb 2006, 17:53
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