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Push yourself again and again. Don't give an inch until the final buzzer sounds. -Larry Bird Success isn't something that just happens - success is learned, success is practiced and then it is shared. -Sparky Anderson -S

Last edited by Bunuel on 26 Dec 2013, 03:53, edited 1 time in total.

statement 1) says xyz> 0 from this we conclude that either all three are positive or two are negative. statement 2 says that yz> 0 it means either both are negative or positive....by combining two statements we come to know that x is positive...x cant be negative in order to make the statement one right....xyz>0 and yz> 0....so x is positive...y and z are either positive or negative....x(y/z) >0 if y and z are negative then negative signs cancel each other and answer could be greater than 0...i am so much confused regarding this question... _________________

Push yourself again and again. Don't give an inch until the final buzzer sounds. -Larry Bird Success isn't something that just happens - success is learned, success is practiced and then it is shared. -Sparky Anderson -S

guys, dont think E is the correct answer... As per me , its A

here is the explanation:

1) it says xyz>0 which means either two of these variables are negative or all three are positive...you can take out any combinations from these and you will find out that every combination gives you x (y/z) >0.. so, this seems sufficient

2) yz>0 doesn't give any information about x so insufficient

We need to know whether expression E= {x * y * (1/y)} positive or negative. Let E = A*B*C where { A=x, B=y, C=1/y } Expression will be -ve if, 1 OR 3 terms among A,B & C are –ve. That is.. 1. Any 1 term out of of A,B and C is –ve and 2 are +ve. OR 2. All 3 A, B & C are –ve.

Option 1: xyz>0.. Again we can write this as ABC>0.. So, this option is sufficient. Multiplication and Division provide similar signs.

Option 2: yz>0. This means, Both y & z are –ve Or Both are +ve. We can’t determine whether x(y/z)>0.

Multiplication or division doesn't change the sign of a term. So x(y/z) or xyz or any formation of xyz will be same as long as one of that is established.

In option 1. xyz>0 is sufficient to determine thus that x(y/z) > 0. Suff

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