2012 AMC 10A Problems/Problem 13

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Problem

An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

$\textbf{(A)}\ \frac{31}{16}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{17}{4}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \frac{65}{16}$

Solution

The minimum and maximum can be achieved with the orders $5, 4, 3, 2, 1$ and $1, 2, 3, 4, 5$.

$5,4,3,2,1 \Rightarrow \frac92,3,2,1 \Rightarrow \frac{15}{4},2,1 \Rightarrow \frac{23}{8},1 \Rightarrow \frac{31}{8}$

$1,2,3,4,5 \Rightarrow \frac32,3,4,5 \Rightarrow \frac94,4,5 \Rightarrow \frac{25}{8},5 \Rightarrow \frac{65}{8}$

The difference between the two is $\frac{65}{8}-\frac{31}{8}=\frac{34}{8}=\boxed{\textbf{(C)}\ \frac{17}{8}}$.

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions