S99-08

Official Solution:


The first thing to do is rephrase the question, setting up cases that are based on the value of \(x\).

If \(x \gt 60\), then \(Max(x, 60) = x\), and \(Min(40, x) = 40\). So then, with \(x\) in this range, the average we are asked for equals \(\frac{x + 40}{2}\).

If \(40 \le x \le 60\), then \(Max(x, 60) = 60\) and \(Min(40, x) = 40\). So then, with \(x\) in this range, the average we are asked for equals \(\frac{60 + 40}{2} = 50\).

If \(x \le 40\), then \(Max(x, 60) = 60\) and \(Min(40, x) = x\). So then, with \(x\) in this range, the average we are asked for equals \(\frac{x + 60}{2}\).

Statement (1): INSUFFICIENT. If \(Min(x, 60) = x\), then \(x \le 60\). However, the average we are asked for does not have a fixed value. If \(x\) is between 40 and 60, then the average is 50, but if \(x\) is below 40, the average is \(\frac{x + 60}{2}\), which does not equal 50.

Statement (2): INSUFFICIENT. If \(Max(40, x) = x\), then \(x \gt 40\). By similar reasoning as we used for Statement (1), we know that the average does not have a fixed value.

Statements (1) and (2) together: SUFFICIENT. We know that \(x \le 60\) AND \(x \ge 40\). Thus, \(x\) is in the range in which the average we are asked for equals 50.


Answer: C