DS question...with a twist

Simplify the question stem first
x^7.y^2.z^3 > 0 is the same as (x^6.y^2.z^2).(x.z) > 0
we know that even powers of any number is always positive i.e. > 0
hence the stem can be rewritten as is xz>0
and without any further work B is the obvious answer

Its not the answer..
See the topic..with a twist :roll:

IMO C

From information statment 2 , Qtn can be rephrased as Is xz*y^2>0

(x and z both same sign hence both cannot be zero, dividing both sides by x^6*z^2)

From stem 1; y and z are of different sign....however term y^2 will yield positive and also y cant be 0

From stem 2, xz>0

Collectively,we know that xz*y^2>0

Yes..the OA is C

I soon realized that I assumed the positivity of y that was left out. Another reminder not to read what is not stated!!

This is a nice question rahulg83. Thanks for posting.

To check whether (x^7)(y^2)(z^3) > 0, we need to consider the possibility that neither x or y or z equals zero.

Stmt1: This statement tells us that the signs of y & z are different (i.e. when y-ve, z+ve and when y+ve,z-ve). We do not know anything about x. x can be +ve,-ve or even 0. So, this statement is not sufficient.

Stmt2: This statement tells us that the signs of x & z are same (i.e. when y-ve, z-ve and when y+ve,z+ve). We do not know anything about y. y can be +ve,-ve or even 0. When y is +ve or -ve, the answer to the question is YES, but when y=0, the answer to the question is NO. So, this statement is not sufficient.

Together both the statements are sufficient to answer the question. Hence C