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

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'''Case 3: the mean is <math>X</math>.'''
 
'''Case 3: the mean is <math>X</math>.'''
  
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<math>\frac{20+X}{6} = X \Rightarrow X=4</math>.
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<math>X= \frac{20+X}{6} \Rightarrow X=4</math>.
  
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Hence, the answer is <math>10+22+4=\boxed{\textbf{(D) }36}</math>.
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Therefore, the answer is <math>10+22+4=\boxed{\textbf{(D) }36}</math>.
  
 
~MrThinker
 
~MrThinker
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~Education, the Study of Everything
 
~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==
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https://youtu.be/35cuytqS9iw
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:55, 27 September 2024

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 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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

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