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Difference between revisions of "2012 AMC 10A Problems/Problem 13"
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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?
 
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?
  
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<math> \textbf{(A)}\ \frac{31}{16}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{17}{8}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \frac{65}{16} </math>
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<math> \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} </math>
  
 
== Solution ==
 
== Solution ==

Revision as of 15:30, 9 February 2012

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 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AMC 10 Problems and Solutions
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