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]]}} | ||
+ | |||
==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> | ||
− | ==Solution== | + | ==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== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Let's start with 6. | ||
+ | |||
+ | <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. | ||
+ | |||
+ | Now let the mean=8 | ||
+ | |||
+ | <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. | ||
+ | |||
+ | Finally we let the mean=x | ||
+ | |||
+ | <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. | ||
+ | |||
+ | So the only option for x is <math>\boxed{-5}.</math> | ||
+ | |||
+ | --mguempel | ||
+ | |||
==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}} | ||
+ | {{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.
Contents
[hide]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 1
There are cases: is the median, is the median, and is the median. In all cases, the mean is .
For case 1, . This allows 6 to be the median because the set is .
For case 2, . This is an extraneous case because the set is .
For case 3, . This is an extraneous case because the set is .
Only case 1 yields a solution, , so the answer is .
Solution by a1b2
Solution 2
The mean is .
There are 3 possibilities: either the median is 6, 8, or x.
Let's start with 6.
when and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.
Now let the mean=8
when and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.
Finally we let the mean=x
and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go.
So the only option for x is
--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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.