2020 AMC 10A Problems/Problem 11

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The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.

Problem 11

What is the median of the following list of $4040$ numbers$?$

\[1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2\]

$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution

We can see that $44^2$ is less than 2020. Therefore, there are $1976$ of the $4040$ numbers after $2020$. Also, there are $2064$ numbers that are under and equal to $2020$. Since $44^2$ is $1936$ it, with the other squares will shift our median's placement up $44$. We can find that the median of the whole set is $2020.5$, and $2020.5-44$ gives us $1976.5$. Our answer is $\boxed{\textbf{(C) } 1976.5}$.

~aryam

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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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