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 $4+7=\boxed{\textbf{(E) }11}$.

Video Solution by Daily Dose of Math

https://youtu.be/GQNnGl3nFok?si=wnPOHufsNE5QbaXY

~Thesmartgreekmathdude

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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