M05-37

Official Solution:


Is the average of a set of 5 distinct positive integers {\(a\), \(b\), 4, 6, 2} greater than the median?

First of all, we must remember that the set contains distinct positive integers.

(1) The largest number in the set is 6.

Since 6 is the largest number in the set, both \(a\) and \(b\) must be less than 6. Therefore, \(a\) and \(b\) can be any two from 1, 3, and 5. If \(a = 1\) and \(b = 3\), the median is 3 and the average is 3.2. However, if \(a = 3\) and \(b = 5\), the average and the median are equal, at 4. Not sufficient.

(2) The smallest number in the set is 2.

Since 2 is the smallest number in the set, both \(a\) and \(b\) must be greater than 2. If \(a\) and \(b\) are very large numbers, say 10 and 20, the average will be greater than the median, which will be 6. However, if \(a = 3\) and \(b = 5\), the average and the median are equal, at 4. Not sufficient.

(1)+(2) From the information given, we know that both \(a\) and \(b\) must be greater than 2, and less than 6. Thus, \(a=3\) and \(b=5\), or vice-versa. In either case, the average and the median are equal, at 4. Sufficient.


Answer: C