Difference between revisions of "2004 AMC 8 Problems/Problem 11"

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==See Also==
 
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Latest revision as of 00:55, 5 July 2013

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 $(4+9)/2 = \boxed{\textbf{(C)}\ 6.5}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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