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

(Created page with "==Problem 8== 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...")
 
(→‎Problem 8)
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==Problem 8==
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==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
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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>
  
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[[2022 AMC 10A Problems/Problem 8|Solution]]
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==Solution==
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'''Case 1: the mean is <math>5</math>'''
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<math>X = 5 \cdot 6 - 20 = 10</math>.
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'''Case 2: the mean is <math>7</math>'''
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<math>X = 7 \cdot 6 - 20 = 22</math>.
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'''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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Hence, adding up the cases, the answer is <math>10+22+4=\boxed{\textbf{(D) }36}</math>.
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 +
~MrThinker

Revision as of 21:20, 11 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

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

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