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If xyz > 0, is x>0? 1) xy> 0 2) xz>0

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Manager
Manager
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Joined: 08 Feb 2006
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If xyz > 0, is x>0? 1) xy> 0 2) xz>0 [#permalink] New post 24 Feb 2006, 19:04
If xyz > 0, is x>0?

1) xy> 0

2) xz>0
Manager
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Joined: 20 Feb 2006
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 [#permalink] New post 24 Feb 2006, 19:30
Answer is C.
(1) => xy > 0 => x,y both > 0 or x,y both < 0 => 1 is insuff
(2) => xz > 0 => x,z both > 0 or x,z both < 0 => 2 is insuff

If we take two statements together, x should be greater than x should be greater than 0.
SVP
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 [#permalink] New post 24 Feb 2006, 20:46
It should be "C"

Stat1 & Stat2 are individually insufficient.

Together we can say that, (x^2)yz > 0 & we have xyz > 0

It means, yz > 0, i.e. both y & z have same sign.

=> x>0
VP
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Re: XYZ [#permalink] New post 24 Feb 2006, 21:32
jlui4477 wrote:
If xyz > 0, is x>0?

1) xy> 0

2) xz>0


using both statements, they all are +ves.
Re: XYZ   [#permalink] 24 Feb 2006, 21:32
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If xyz > 0, is x>0? 1) xy> 0 2) xz>0

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