2004 AMC 8 Problems/Problem 11

Problem

The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:

        1. The largest isn't first, but it is in one of the first three places. 
        2. The smallest isn't last, but it is in one of the last three places. 
        3. The median isn't first or last.

What is the average of the first and last numbers?

$\textbf{(A)}\ 3.5 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6.5 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$

Solution

From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding their order. Their average is $(4+9)/2 = \boxed{\textbf{(C)}\ 6.5}$ .

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 8 Problems/Problem 11

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