data sufficiency question

[sana.noor]

Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

E[/spoiler]

[Param800]

What's the source of this question ?

I am not able to get the correct answer. I am getting C.

[sana.noor]

its from original gmat paper and i believe it should be A

[Param800]

hmm... I see... ok this is how I was approaching it and because of which I believe C is the correct answer.

We have to show xy/z >0
which means either xy and z are both positive
or xy and z are both negative.

St (1) - (xy)z >0 ...which means xy and z are both positive OR xy and z are both negative.

but we can also say this ... x(yz)>0 ...which will mean either x and yz are both positive or x and yz both negative.

CORRECTION
Since there would always be two negative signs and one positive signs...when we will perform xy/z we will always get a positive number ... thus answer is should be A

because of this St(1) is insufficient.

St (2) -- yz > 0 ....doesn't tell about the sign of x...so insufficient.

Combination-

xyz >0 and yz >0 ...if we substitute second in first ..we should get... x >0

now... yz > 0 means either y >0 and Z >0
OR y<0 and z <0

Plugging in the main equation --- xy/z ... first case - y>0 , z >0 and x >0

xy/z will be >0

second case -- y<0 , z< 0 and X >0

xy/z will be >0

So.. sufficient and thus answer is C

But, doesn't match the OA

I will be waiting for someone to tell me the error in my method.

[sana.noor]

How can you multiply "x(yz) >0?? the original question asks X into Y divided by Z...but i couldnt understand that why you multiply y with z?? this question seems tricky

[Param800]

I agree that the question asks xy/z but the first statement says xyz > 0 which can be interpreted as follows-

(i) (xy)z>0
OR
(ii) x (yz) >0
OR
(iii) (xz)y>0

So...that's why I am multiplying x(yz)

And..yes this question is tricky...I am thinking why the OA is E..but I am not getting it...I hate when I am stuck like this.. lol

[sana.noor]

let take x as 2 or -2
y as 8 or -8
Z as 4 or -4

i) (xy)z taking xy negative we will get (-2)(-8) and 4 = 64
ii) x(yz) taking yz negative we will again get a positive answer
iii) and if we take (xz) negative and y positive we again get positive answer....

In X (y/z) whatever two integers if taken (-ve) the answer will be greater than 0 and if we take some non-integer number still the answer will be greater than 0...

[Param800]

YES.... i think you are right...it should be A... it makes sense... but lets see if someone else comes with another answer.

[Brent@GMATPrepNow]

Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

Target question: Is x(y/z) > 0?
Notice that x(y/z) = (xy)/z = (x)(y)(1/z) [you'll see why I did this later]

Rephrased target question: Is (x)(y)(1/z) positive?

IMPORTANT CONCEPTS:
If z is positive, then 1/z is positive.
If z is negative, then 1/z is negative.
In other words, z and 1/z must have the same sign.

Statement 1: xyz > 0
In other words, the product (x)(y)(z) is positive.
Since z and 1/z have the same sign (positive or negative) we can conclude with certainty that (x)(y)(1/z) is positive
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: yz > 0
There are several sets of numbers that meet this condition. Here are two:
Case a: x=1, y=1, z=1, in which case (x)(y)(1/z) is positive
Case b: x=-1, y=1, z=1, in which case (x)(y)(1/z) is negative
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent