2022 AMC 10A Problems/Problem 8

Revision as of 21:42, 11 November 2022 by MRENTHUSIASM (talk | contribs)

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

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 $X$

$\frac{20+X}{6} = X \Rightarrow X=4$.

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

~MrThinker

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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

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