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

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If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink] New post 10 Jan 2010, 18:31
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Question Stats:

75% (01:35) correct 25% (00:39) wrong based on 89 sessions
If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0
[Reveal] Spoiler: OA
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Re: If XYZ>0 is X>0 [#permalink] New post 10 Jan 2010, 20:12
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DestinyChild wrote:
If XYZ>0 is X>0

XY > 0
XZ > 0

Obviously the answer is Yes or No;
Get your kudos for reasoning out why the answer is C and keep an eye on time too :)

Triple kudos for someone suggesting quicker technique for DS stems with triple variables...!


xy > 0 both are positive or both negative
xz > 0 both are positive or both negative

if x < 0 then z < 0 and y < 0 but xyz < 0 so this doesn't work
if x > 0 then y > 0 and z > 0 and xyz > 0 so this works and x > 0
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Re: If XYZ>0 is X>0 [#permalink] New post 10 Jan 2010, 20:29
cheers lagomez

will give you one...!
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Re: If XYZ>0 is X>0 [#permalink] New post 11 Apr 2011, 08:51
1) x and y both are positive or negative, so x could be -ve, or +ve insuff.
2) x and z both are positive or negative, so x could be -ve, or +ve insuff.

For C
as xyz > 0, so if y and z both -ve, then x has to be +ve, and
if y and z become +ve then x has to be +ve to become xyz > 0.
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Re: If XYZ>0 is X>0 [#permalink] New post 11 Apr 2011, 10:54
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Following chart lists down the possible scenarios in which XYZ > 0

X Y Z
1) - - +
2) - + -
3) + - -
4) + + +

As per statement I --> 1 & 4 have to be true for XY > 0. So I alone is insufficient as X can be +ve or -ve
As per statement II --> 2& 4 have to be true for XZ > 0 So II alone is insufficient as X can be +ve or -ve
If we combine both statements then only 4 is true and X has to be +ve

Hence C.
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Re: If XYZ>0 is X>0 [#permalink] New post 11 Apr 2011, 12:43
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DestinyChild wrote:
If XYZ>0 is X>0

XY > 0
XZ > 0

Obviously the answer is Yes or No;
Get your kudos for reasoning out why the answer is C and keep an eye on time too :)

Triple kudos for someone suggesting quicker technique for DS stems with triple variables...!


Certainly each statement is not sufficient alone (the logic is identical for each: for S1, x and y can both be negative or both positive, and for S2, x and z can both be negative or both positive). You might notice that S1 tells us z is positive, and S2 tells us y is positive, but that's not what we're looking for.

Now, together, the inequality xy > 0 means 'x and y have the same sign'. So from the two statements we know x and y have the same sign, and x and z have the same sign, so x, y and z all have the same sign. If xyz > 0, they must all be positive, and the answer is C.
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Re: If XYZ>0 is X>0 [#permalink] New post 11 Apr 2011, 17:41
C. For XYZ > 0 either all are positive or two of them negative. From St1 and St2 it is clear that both xy and xz has to be positive or both xz and xy has to be negative. in the latter case XYZ > 0 will be false.
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Re: If XYZ>0 is X>0 [#permalink] New post 11 Apr 2011, 21:53
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XYZ > 0 => X > 0 and YZ > 0 OR x < 0 and YZ < 0


(1)

XY > 0 means x > 0 and Y > 0 OR X < 0 and Y < 0

So (1) is not sufficient

(2)

XZ > 0 means X > 0 and Z > 0 OR X < 0 and Z < 0

So (2) is not sufficient

So if x < 0 then y < 0 and z < 0

=> xyz < 0 which is contrary to what is given in question

=> x > 0

Answer C.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink] New post 18 Nov 2013, 07:29
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0   [#permalink] 18 Nov 2013, 07:29
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