# 2001 AMC 10 Problems/Problem 1

## Problem

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$. What is the mean? $\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

## Solution

The median of the list is $10$, and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$.

We substitute the median for $10$ and the equation becomes $n+6=10$.

Subtract both sides by 6 and we get $n=4$. $n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$.

The mean of those numbers is $\frac{9n+63}{9}$ which is $n+7$.

Substitute $n$ for $4$ and $4+7=\boxed{\textbf{(E) }11}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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