2022 AMC 10A Problems/Problem 8: Difference between revisions

Latest revision as of 08:11, 21 June 2025

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 4

https://youtu.be/35cuytqS9iw

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2022 AMC 10A Problems/Problem 8|2022 AMC 10A #8]] and [[2022 AMC 12A Problems/Problem 6|2022 AMC 12A #6]]}}
 ==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
+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 positive 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>
-==Solution==
+==Solution (Casework)==
+First, note that <math>1+7+5+2+5=20</math>. There are <math>3</math> possible cases:
-'''Case 1: the mean is <math>5</math>'''
+'''Case 1: the mean is <math>5</math>.'''
 <math>X = 5 \cdot 6 - 20 = 10</math>.
-'''Case 2: the mean is <math>7</math>'''
+'''Case 2: the mean is <math>7</math>.'''
 <math>X = 7 \cdot 6 - 20 = 22</math>.
-'''Case 3: the mean is <math>X</math>'''
+'''Case 3: the mean is <math>X</math>.'''
-<math>\frac{20+X}{6} = X \Rightarrow X=4</math>.
+<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>.
+Therefore, the answer is <math>10+22+4=\boxed{\textbf{(D) }36}</math>.
 ~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 4==
+https://youtu.be/35cuytqS9iw
 == See Also ==
-{{AMC10 box|year=2022|ab=A|num-b=3|num-a=5}}
+{{AMC10 box|year=2022|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2022|ab=A|num-b=5|num-a=7}}
 {{MAA Notice}}

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