2012 AMC 10A Problem 13

Question

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?

\dfrac{31}{16}

2

\dfrac{17}{8}

3

\dfrac{65}{16}

Answer: C

Concepts:mean algebraic manipulation optimization

Difficulty rating: 1600

Small hint:

Write the final average as a weighted sum

Big hint:

Later entries receive larger weights

Solution:

Let the order of the numbers be

a, b, c, d, e.

Then the iterative average is

\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}

= \dfrac{a + b + 2c + 4d + 8e}{16}.

To minimize this, we make the order

5, 4, 3, 2, 1,

which gives us a sum of

\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.

To maximize it, we have to reverse this order to get an average of

\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.

The difference between these is

\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.

Thus, C is the correct answer.