Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10B Problems/Problem 13"
m (→‎Video Solution)
(5 intermediate revisions by 5 users not shown)
Line 11: Line 11:
 
The mean is <math>\frac{4+6+8+17+x}{5}=\frac{35+x}{5}</math>.
 
The mean is <math>\frac{4+6+8+17+x}{5}=\frac{35+x}{5}</math>.
  
−
There are 3 possibilities: either the median is 6, 8, or x.
+
There are three possibilities for the median: it is either <math>6</math>, <math>8</math>, or <math>x</math>.
  
−
Let's start with 6.
+
Let's start with <math>6</math>.
  
−
<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.
+
<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.
  
−
Now let the mean=8
+
Now let the median be <math>8</math>.
  
−
<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.
+
<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.
  
−
Finally we let the mean=x
+
Finally we let the median be <math>x</math>.
  
−
<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.
+
<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.
  
−
So the only option for x is <math>\boxed{-5}.</math>
+
Hence the only possible value of <math>x</math> is <math>\boxed{\textbf{(A) }-5}.</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/mXvetCMMzpU?t=448
 +
 
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
Line 32: Line 37:
 
{{AMC12 box|year=2019|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2019|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
−
SUB2PEWDS
 

Revision as of 22:37, 7 September 2020

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/mXvetCMMzpU?t=448

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_13&oldid=133349"