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

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VP
Joined: 06 Jun 2004
Posts: 1059
Location: CA
Followers: 2

Kudos [?]: 46 [0], given: 0

If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]  17 Dec 2005, 22:49
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If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0
Manager
Joined: 12 Nov 2005
Posts: 78
Followers: 1

Kudos [?]: 5 [0], given: 0

XYZ > 0

Statement I
XY >0
Implies X & Y are both Positive or Both negative.......Not sufficient

Statement II
XZ >0
Implies X & Z are both Positive or Both negative.......Not sufficient

Statement I & II
XY > 0
XZ > 0 From the main statement (XYZ > 0), This is possible only if all three are positive

Director
Joined: 17 Dec 2005
Posts: 552
Location: Germany
Followers: 1

Kudos [?]: 11 [0], given: 0

[i]Disagree..

xyz>0

(1) If xy>0, than both x and y have to be either negative or positive

(2) If xz>0, than both x and z have to be either positive or negative

=> Thus we do not know if all integers are positive or negative, therefore we have to choose E, because the product of three negative numbers is negative, that of three positive is positive.[/i]

Last edited by allabout on 19 Dec 2005, 06:42, edited 2 times in total.
Intern
Joined: 14 Dec 2005
Posts: 14
Followers: 0

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

If xy > 0 then z > 0
If xz > 0 then y > 0

If z > 0 and y > 0 then x > 0.

Hence C.
Director
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

great question.

At first i chose E but now i see that C makes sense
Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: GMAT Prep - DS XYZ [#permalink]  19 Dec 2005, 10:40
TeHCM wrote:
If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0

C..
possible arrangements

x y z
1) + + +
- - +

2) + + +
- + -

Only + + + is common to both, thus x>0
Senior Manager
Joined: 14 Apr 2005
Posts: 419
Location: India, Chennai
Followers: 1

Kudos [?]: 6 [0], given: 0

Re: GMAT Prep - DS XYZ [#permalink]  20 Dec 2005, 04:27
TeHCM wrote:
If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0

From 1 x could be +ve or -ve hence insuff.
From 2 x could still be + ve or -ve hence insuff.

Combining both. we get x^2yz > 0
or x (xyz) > 0
We are given that xyz > 0 which means x must be +ve to make
x(xyz) > 0. Hence C.
Senior Manager
Joined: 11 Nov 2005
Posts: 332
Location: London
Followers: 1

Kudos [?]: 8 [0], given: 0

IMO C.

if xyz>0

1. xy>0, therefore either x y are positive or negative, not sufficient

however if xy are positive or negative, in any case, z is positive which satisfy the original statament.

2. xz >0 alone is insufficient to say x>0, but combined with the above observation (z>0) x must be positive. and the answer C.
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