M30-20

Official Solution:


If \(xyz < 0\), is \(x < 0\)?

\(xyz < 0\) implies that either all three unknowns are negative or only one of them, the smallest one, is negative and the other two are positive.

(1) \(x - y < 0\).

This means that \(x < y\). Could \(x\) be negative? Yes, if all of \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(y\) are positive and \(z\) is negative. Not sufficient.

(2) \(x - z < 0\).

Basically the same here. We have \(x < z\). Could \(x\) be negative? Yes, if all of \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(z\) are positive and \(y\) is negative. Not sufficient.

(1)+(2) We have \(x < y\) and \(x < z\). Hence, \(x\) is the smallest of the three. Could \(x\), the smallest of the three, be positive? No, because if \(x\) is positive, then both \(y\) and \(z\) must also be positive. But in this case, \(xyz\) would be positive, not negative as given in the stem. Therefore, \(x\) must be negative. Sufficient.


Answer: C