Difference between revisions of "2019 AMC 12B Problems/Problem 7"

(Problem)
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==Solution==
 
==Solution==
<math>\boxed{-5}</math> if my memory isn't trash
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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}=6</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==
 
{{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}}

Revision as of 13:49, 14 February 2019

Problem

4, 6, 8, 17, x

What is the sum of all values of x such that the mean is equal to the median?

Solution

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

$\frac{35+x}{5}=6$ 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 12B (ProblemsAnswer KeyResources)
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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