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

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{{duplicate|[[2022 AMC 10A Problems/Problem 8|2022 AMC 10A #8]] and [[2022 AMC 12A Problems/Problem 6|2022 AMC 12A #6]]}}
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==Problem==
 
==Problem==
  
A data set consists of <math>6</math> not distinct) positive integers: <math>1</math>, <math>7</math>, <math>5</math>, <math>2</math>, <math>5</math>, and <math>X</math>. The
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A data set consists of <math>6</math> (not distinct) positive integers: <math>1</math>, <math>7</math>, <math>5</math>, <math>2</math>, <math>5</math>, and <math>X</math>. The average (arithmetic mean) of the <math>6</math> numbers equals a value in the data set. What is the sum of all possible values of <math>X</math>?
average (arithmetic mean) of the <math>6</math> numbers equals a value in the data set. What is
 
the sum of all positive values of <math>X</math>?
 
  
 
<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)==
  
'''Case 1: the mean is <math>5</math>'''
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First, note that <math>1+7+5+2+5=20</math>. There are <math>3</math> possible cases:
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'''Case 1: the mean is <math>5</math>.'''
  
 
<math>X = 5 \cdot 6 - 20 = 10</math>.
 
<math>X = 5 \cdot 6 - 20 = 10</math>.
  
'''Case 2: the mean is <math>7</math>'''
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'''Case 2: the mean is <math>7</math>.'''
  
 
<math>X = 7 \cdot 6 - 20 = 22</math>.
 
<math>X = 7 \cdot 6 - 20 = 22</math>.
  
'''Case 3: the mean is <math>X</math>'''
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'''Case 3: the mean is <math>X</math>.'''
  
<math>\frac{20+X}{6} = X \Rightarrow X=4</math>.
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<math>X= \frac{20+X}{6} \Rightarrow X=4</math>.
  
Hence, adding up the cases, 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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==Video Solution 1 (Quick and Simple)==
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https://youtu.be/8s6SngtEBY4
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~Education, the Study of Everything
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==Video Solution 2 (Don't fall into the trap)==
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https://youtu.be/7yAh4MtJ8a8?si=r_qxJ_xhfngz4Xu6&t=1143
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~Math-X
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==Video Solution 3==
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https://www.youtube.com/watch?v=GX-jmRUadik
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-paixiao
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==Video Solution 4==
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https://youtu.be/35cuytqS9iw
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== See Also ==
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{{AMC10 box|year=2022|ab=A|num-b=7|num-a=9}}
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{{AMC12 box|year=2022|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 12:14, 25 December 2023

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)

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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