Difference between revisions of "2022 AMC 10A Problems/Problem 8"

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<math>\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40</math>
 
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40</math>
  
==Solution==
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==Solution (Casework)==
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First, note that <math>1+7+5+2+5=20</math>. Then, there are 3 cases.
  
 
'''Case 1: the mean is <math>5</math>'''
 
'''Case 1: the mean is <math>5</math>'''

Revision as of 08:06, 14 November 2022

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 $1+7+5+2+5=20$. Then, there are 3 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 $X$

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

Hence, adding up the cases, 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

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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