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Difference between revisions of "2020 AMC 10A Problems/Problem 11"
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==Solution==
 
==Solution==
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We can see that <math>44^2</math> is less than 2020. Therefore, there are <math>1976</math> of the <math>4040</math> numbers after <math>2020</math>. Also, there are <math>2064</math> numbers that are under and equal to <math>2020</math>. Since <math>44^2</math> is <math>1936</math> it, with the other squares will shift our median's placement up <math>44</math>. We can find that the median of the whole set is <math>2020.5</math>, and <math>2020.5-44</math> gives us <math>1976.5</math>. Our answer is <math>\boxed{\textbf{(D) } 1976.5}</math>.
+
We can see that <math>44^2</math> is less than 2020. Therefore, there are <math>1976</math> of the <math>4040</math> numbers after <math>2020</math>. Also, there are <math>2064</math> numbers that are under and equal to <math>2020</math>. Since <math>44^2</math> is <math>1936</math> it, with the other squares will shift our median's placement up <math>44</math>. We can find that the median of the whole set is <math>2020.5</math>, and <math>2020.5-44</math> gives us <math>1976.5</math>. Our answer is <math>\boxed{\textbf{(C) } 1976.5}</math>.
  
 
~aryam
 
~aryam

Revision as of 21:59, 31 January 2020

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

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
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

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