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2001 AMC 10 Problem 1

Problem 1 of 25EasierAlgebra

The median of the list n, n+3, n+4, n+5, n+6, n+8, n+10, n+12, n+15 \begin{gathered} n,\ n+3,\ n+4,\ n+5,\ n+6, \\ \ n+8,\ n+10,\ n+12,\ n+15 \end{gathered} is 10.10. What is the mean?

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Solution

The list has 99 numbers in increasing order, so the median is the 55th term, n+6.n+6. Setting n+6=10n+6=10 gives n=4.n=4. The sum of the terms is 9n9n +(3+4+5+6+8+10+12+15)\small {}+(3+4+5+6+8+10+12+15) =9n+63=9n+63 =99,=99, so the mean is 999=11.\frac{99}{9}=11. Thus, the correct answer is E.

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Concepts: median (data) · mean

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.