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

Revision as of 17:23, 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 impossible because the set is $4,5,6,8,17$ .
For case 3, $x=\frac{35}{4}$ . This is impossible 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

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.

@@ Line 10: / Line 10: @@
 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 2, <math>x=5</math>. This is an extraneous case because the set is <math>4,5,6,8,17</math>.<br>
+For case 2, <math>x=5</math>. This is impossible because the set is <math>4,5,6,8,17</math>.<br>
-For case 3, <math>x=\frac{35}{4}</math>. This is an extraneous case because the set is <math>4,6,8,\frac{35}{4},17</math>.<br>
+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>
 Only case 1 yields a solution, <math>x=-5</math>, so the answer is <math>\textbf{(A) } -5</math>.<br>
 <math>Q.E.D \blacksquare</math>

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

Revision as of 17:23, 14 February 2019

Contents

Problem

Solution 1

Solution 2

See Also