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2012 AMC 12A Problem 8

Problem 8 of 25EasierProbability & StatisticsProblem-Solving Techniques

An iterative average of the numbers 1,1, 2,2, 3,3, 4,4, and 55 is computed in 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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Solution

For the order a,b,c,d,e,a, b, c, d, e, the iterative average is a+b+2c+4d+8e16.\frac{a + b + 2c + 4d + 8e}{16}. The later positions carry the most weight. The largest value uses (a,b,c,d,e)=(1,2,3,4,5),(a,b,c,d,e) = (1,2,3,4,5), giving 6516,\dfrac{65}{16}, and the smallest uses (5,4,3,2,1),(5,4,3,2,1), giving 3116.\dfrac{31}{16}. The difference is 6516−3116=3416=178.\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}. Thus, the correct answer is C.
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Tagged: mean · optimization

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