M13-19

Official Solution:


Is \(x^2 * y^5 * z \gt 0\) ?

In order for \(x^2 * y^5 * z \gt 0\) to be true, the following conditions must be satisfied:

AND

Thus, only if BOTH conditions are met, we will have \((x^2) * (y^5 * z) = (positive) * (positive) = positive\).

(1) \(\frac{xz}{y} \gt 0\).

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

(2) \(\frac{y}{z} \lt 0 \).

From this statement, we can deduce that \(y\) and \(z\) have opposite signs, which violates the second condition. Thus, the answer to the question is NO. Sufficient.

Side note for (2): \(\frac{y}{z} \lt 0\) does not imply that \(x^2 * y^5 * z \lt 0\). Instead, it implies that \(x^2 * y^5 * z \le 0\) because it's possible for \(x\) to be equal to zero, and in this case, \(x^2 * y^5 * z = 0\). However, in any case, \(x^2 * y^5 * z\) is not greater than zero, so we can answer NO to the question.


Answer: B