Difference between revisions of "2020 AMC 10A Problems/Problem 11"

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<math> \textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5 </math>
 
<math> \textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5 </math>
  
==Solution==
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==Solution 1==
 
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>.
 
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
 +
 +
== Solution 2 ==
 +
As we are trying to find the median of a <math>4040</math>-term set, we must find the average of the <math>2020</math>th and <math>2021</math>st terms. <br>
 +
<br>
 +
Since <math>45^2 = 2025</math> is slightly greater than <math>2020</math>, we know that the <math>44</math> perfect squares <math>1^2</math> through <math>44^2</math> are less than <math>2020</math>, and the rest are greater. Thus, from the number <math>1</math> to the number <math>2020</math>, there are <math>2020 + 44 = 2064</math> terms. Since <math>44^2</math> is <math>44 + 45 = 89</math> less than <math>45^2 = 2025</math> and <math>84</math> less than <math>2020</math>, we will only need to consider the perfect square terms going down from the <math>2064</math>th term, <math>2020</math>, after going down <math>84</math> terms. Since the <math>2020</math>th and <math>2021</math>st terms are only <math>44</math> and <math>43</math> terms away from the <math>2064</math>th term, we can simply subtract <math>44</math> from <math>2020</math> and <math>43</math> from <math>2020</math> to get the two terms, which are <math>1976</math> and <math>1977</math>. Averaging the two, we get <math>\boxed{\textbf{(C) } 1976.5}.</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:23, 1 February 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 1

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

Solution 2

As we are trying to find the median of a $4040$-term set, we must find the average of the $2020$th and $2021$st terms.

Since $45^2 = 2025$ is slightly greater than $2020$, we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$, and the rest are greater. Thus, from the number $1$ to the number $2020$, there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$, we will only need to consider the perfect square terms going down from the $2064$th term, $2020$, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two, we get $\boxed{\textbf{(C) } 1976.5}.$

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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