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Difference between revisions of "2019 AMC 10B Problems/Problem 13"
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #13]] and [[2019 AMC 12B Problems|2019 AMC 12B #7]]}}
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==Problem==
 
==Problem==
  
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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==
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==Solution 1==
 
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>
 
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>
 
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>
 
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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<math>Q.E.D \blacksquare</math>
 
<math>Q.E.D \blacksquare</math>
 
Solution by [[User:a1b2|a1b2]]
 
Solution by [[User:a1b2|a1b2]]
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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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There are 3 possibilities: either the median is 6, 8, or x.
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Let's start with 6.
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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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Now let the mean=8
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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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Finally we let the mean=x
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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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So the only option for x is <math>\boxed{-5}.</math>
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--mguempel
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}}
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{{AMC12 box|year=2019|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:46, 14 February 2019

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 1

There are $3$ cases: $6$ is the median, $8$ is the median, and $x$ is the median. In all cases, the mean is $7+\frac{x}{5}$.
For case 1, $x=-5$. This allows 6 to be the median because the set is $-5,4,6,8,17$.
For case 2, $x=5$. This is an extraneous case because the set is $4,5,6,8,17$.
For case 3, $x=\frac{35}{4}$. This is an extraneous case because the set is $4,6,8,\frac{35}{4},17$.
Only case 1 yields a solution, $x=-5$, so the answer is $\textbf{(A) } -5$.
$Q.E.D \blacksquare$ Solution by a1b2

Solution 2

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

There are 3 possibilities: either the median is 6, 8, or x.

Let's start with 6.

$\frac{35+x}{5}=6$ when $x=-5$ and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.

Now let the mean=8

$\frac{35+x}{5}=8$ when $x=5$ and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.

Finally we let the mean=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 so no go.

So the only option for x is $\boxed{-5}.$

--mguempel

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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