Is x^2*y^5*z > 0 ? (1) xz/y > 0 (2) y/z < 0

For \(x^2*y^5*z>0\) to hold true:
1. \(x\) must not be zero;
and
2. \(y\) and \(z\) must be either both positive or both negative.

(1) \(\frac{xz}{y}>0\) --> first condition is satisfied: \(x\neq{0}\), but we don't know aout the second one: \(\frac{xz}{y}>0\) means that either all of them are positive (answer YES) or ANY two are negative and the third one is positive, so it's possible \(y\) and \(z\) to have opposite signs (answer NO). Not sufficient.

(2) \(\frac{y}{z}<0\) --> \(y\) and \(z\) have opposite signs --> second condition is already violated, so the answer to the question is NO. Sufficient.

Answer: B.

Side note for (2): \(\frac{y}{z}<0\) does not mean that \(x^2*y^5*z<0\), it means that \(x^2*y^5*z\leq{0}\) because it's possible \(x\) to be equal to zero and in this case \(x^2*y^5*z=0\). But in any case \(x^2*y^5*z\) is not MORE than zero, so we can answer NO to the question.

Hoe it's clear.