2018 AMC 12A Problem 6

Attempt Problem 6 of the 2018 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

For positive integers $m$ and $n$ such that $m + 10 \lt n + 1,$ both the mean and the median of the set $\{m, m + 4, m + 10,$ $n + 1, n + 2, 2n\}$ are equal to $n.$ What is $m + n?$

$20$

$21$

$22$

$23$

$24$

Answer: B

Concepts:mean median (data)system of equations

Difficulty rating: 1350

Solution:

Because $m + 10 \lt n + 1,$ the six numbers are already increasing, so the median is the average of the middle two: $\frac{(m+10)+(n+1)}{2} = n,$ giving $m = n - 11.$ The mean condition is $\tiny \frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n,$ so $7n - 16 = 6n$ and $n = 16.$ Then $m = 5,$ and $m + n = 21.$

Thus, the correct answer is B.

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