# 2022 AMC 10A Problems/Problem 8

The following problem is from both the 2022 AMC 10A #8 and 2022 AMC 12A #6, so both problems redirect to this page.

## Problem

A data set consists of $6$ (not distinct) positive integers: $1$, $7$, $5$, $2$, $5$, and $X$. The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$? $\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$

## Solution (Casework)

First, note that $1+7+5+2+5=20$. There are $3$ possible cases:

Case 1: the mean is $5$. $X = 5 \cdot 6 - 20 = 10$.

Case 2: the mean is $7$. $X = 7 \cdot 6 - 20 = 22$.

Case 3: the mean is $X$. $X= \frac{20+X}{6} \Rightarrow X=4$.

Therefore, the answer is $10+22+4=\boxed{\textbf{(D) }36}$.

~MrThinker

## Video Solution 1 (Quick and Simple)

~Education, the Study of Everything

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