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== | ||
| + | |||
| + | What is the sum of all real numbers <math>x</math> for which the median of the numbers <math>4,6,8,17,</math> and <math>x</math> is equal to the mean of those five numbers? | ||
| + | |||
| + | <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> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | The mean is <math>\frac{4+6+8+17+x}{5}=\frac{35+x}{5}</math>. | ||
| + | |||
| + | There are three possibilities for the median: it is either <math>6</math>, <math>8</math>, or <math>x</math>. | ||
| + | |||
| + | Let's start with <math>6</math>. | ||
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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. | ||
| + | |||
| + | Now let the median be <math>8</math>. | ||
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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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| + | Finally we let the median be <math>x</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 <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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| + | 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== | ||
| + | |||
| + | {{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}} | ||
| + | {{AMC12 box|year=2019|ab=B|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
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.
Contents
Problem
What is the sum of all real numbers 
for which the median of the numbers 
and is equal to the mean of those five numbers?
Solution
The mean is 
.
There are three possibilities for the median: it is either 
, 
, or .
Let's start with .
has solution 
, and the sequence is 
, which does have median , so this is a valid solution.
Now let the median be .
gives 
, so the sequence is 
, which has median , so this is not valid.
Finally we let the median be .
, and the sequence is 
, which has median . This case is therefore again not valid.
Hence the only possible value of is 
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. 
