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Difference between revisions of "2019 AMC 10B Problems/Problem 13"

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<math>\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}</math>
 
<math>\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}</math>
  
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==Solution 1==
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==Solution==
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There are <math>3</math> cases: <math>6</math> is the median, <math>8</math> is the median, and <math>x</math> is the median. In all cases, the mean is <math>7+\frac{x}{5}</math>.<br>
 
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For case 1, <math>x=-5</math>. This allows 6 to be the median because the set is <math>-5,4,6,8,17</math>.<br>
 
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For case 2, <math>x=5</math>. This is impossible because the set is <math>4,5,6,8,17</math>.<br>
 
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For case 3, <math>x=\frac{35}{4}</math>. This is impossible because the set is <math>4,6,8,\frac{35}{4},17</math>.<br>
 
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Only case 1 yields a solution, <math>x=-5</math>, so the answer is <math>\textbf{(A) } -5</math>.
 
  
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==Solution 2==
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The mean is <math>\frac{4+6+8+17+x}{5}=\frac{35+x}{5}</math>.
  
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The mean is <math>\frac{4+6+8+17+x}{5}=\frac{35+x}{5}</math>.
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There are three possibilities for the median: it is either <math>6</math>, <math>8</math>, or <math>x</math>.
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Let's start with <math>6</math>.
  
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There are 3 possibilities: either the median is 6, 8, or x.
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<math>\frac{35+x}{5}=6</math> has solution <math>x=-5</math>, and the sequence is <math>-5, 4, 6, 8, 17</math>, which does have median <math>6</math>, so this is a valid solution.
  
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Let's start with 6.
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Now let the median be <math>8</math>.
  
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<math>\frac{35+x}{5}=6</math> when <math>x=-5</math> and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.
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<math>\frac{35+x}{5}=8</math> gives <math>x=5</math>, so the sequence is <math>4, 5, 6, 8, 17</math>, which has median <math>6</math>, so this is not valid.
  
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Now let the mean=8
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Finally we let the median be <math>x</math>.
  
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<math>\frac{35+x}{5}=8</math> when <math>x=5</math> and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.
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<math>\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75</math>, and the sequence is <math>4, 6, 8, 8.75, 17</math>, which has median <math>8</math>. This case is therefore again not valid.
  
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Finally we let the mean=x
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Hence the only possible value of <math>x</math> is <math>\boxed{\textbf{(A) }-5}.</math>
  
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<math>\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75.</math> and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go.
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==Video Solution==
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https://youtu.be/eHdp481w9I0
  
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So the only option for x is <math>\boxed{-5}.</math>
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~Education, the Study of Everything
  
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--mguempel
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== Video Solution ==
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https://youtu.be/IziHKOubUI8?t=600
  
 
==See Also==
 
==See Also==

Latest revision as of 10:28, 24 June 2023

The following problem is from both the 2019 AMC 10B #13 and 2019 AMC 12B #7, so both problems redirect to this page.

Problem

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

Solution

The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.

There are three possibilities for the median: it is either $6$, $8$, or $x$.

Let's start with $6$.

$\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.

Now let the median be $8$.

$\frac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$, so this is not valid.

Finally we let the median be $x$.

$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$, and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence the only possible value of $x$ is $\boxed{\textbf{(A) }-5}.$

Video Solution

https://youtu.be/eHdp481w9I0

~Education, the Study of Everything

Video Solution

https://youtu.be/IziHKOubUI8?t=600

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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