# 2005 AMC 12B Problems/Problem 9

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The following problem is from both the 2005 AMC 12B #9 and 2005 AMC 10B #19, so both problems redirect to this page.

## Problem

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam? $\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$

## Solution

To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$, since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$. The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$. The difference between the mean and median, therefore, is $\boxed{\textbf{(B)}\ 1}$.

## Solution 2

The remaining $30\%$ of the students got $95$ points. The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$. The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$, the median is $85$. The difference between the mean and the median is $\boxed{\textbf{(B)}\ 1}$.

~mobius247

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