Author: Param800
Posted: Sun Dec 30, 2012 6:37 am (GMT -8)
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![Sad]()
I will be waiting for someone to tell me the error in my method.
Posted: Sun Dec 30, 2012 6:37 am (GMT -8)
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.
