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 positive 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)

https://youtu.be/8s6SngtEBY4

~Education, the Study of Everything

Video Solution 2 (Don't fall into the trap)

https://youtu.be/7yAh4MtJ8a8?si=r_qxJ_xhfngz4Xu6&t=1143

~Math-X

Video Solution 3

https://www.youtube.com/watch?v=GX-jmRUadik

-paixiao

Video Solution 4

https://youtu.be/35cuytqS9iw

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2022 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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