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Sorry Anurag but I think the answer is B

See we can re group (x^7)(y^2)(z^3) as ((xyz)^2)(xz)(x^4)

Applying the common logic as taking out the even powers

checking this just the statement 2 is sufficient to answer

Hence B

Hi ankita1709:

Anurag is right. I got the same answer.
Your logic of even powers is fine and we need to check for xz. However, you are missing a critical piece of information.

we are asked if (x^7)(y^2)(z^3) > 0 (Strictly greater than 0).
This means that we need to ascertain whether even one of x, y, z can be zero. (if one of x, y, z is zero then we can't find the answer from B, which says xz<0).

yz<0 tells us that neither y nor z is 0.
xz<0 tells us that (xz) one of xz is less than 0 and that none of x or z is equal to 0. That's why C is the correct answer.

Now, if the question asked us (x^7)(y^2)(z^3)>=0, then the answer would have been B because we would not have been concerned with any or all of x,y,z being 0s.

Let me know if this helps