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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>
  
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==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 equal to <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>.
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 +
~aryam
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 +
== 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> ~[[User:emerald_block|emerald_block]]
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 +
== Solution 3 ==
 +
We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median. <br>
 +
<br>
 +
We know that <math>44^2=1936</math> <br>
 +
So numbers <math>1^2, 2^2, ...,44^2</math> are in between <math>1</math> to <math>1936</math>. <br>
 +
So the sum of <math>44</math> and <math>1936</math> will result in <math>1980</math>, which means that <math>1936</math> is the <math>1980</math>th number. <br>
 +
Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. <br>
 +
Then the <math>2020</math>th term will be <math>1936+40 = 1976</math>, and similarly the <math>2021</math>th term will be <math>1977</math>.<br>
 +
Solving for the median of the two numbers, we get <math>\boxed{\textbf{(C) } 1976.5}</math> <br>
 +
~toastybaker
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 +
== Solution 4 ==
 +
We note that <math>44^2 = 1936</math>, which is the first square less than <math>2020</math>, which means that there are 44 addition terms before <math>2020</math>. This makes <math>2020</math> the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before <math>2020</math> is one less than the previous term (that is, we subtract 1 to get the previous term). If <math>2020</math> is the 2064th term, than <math>2020 - 44</math> is the (2064 - 44)th term. So, the 2020th term is <math>1976</math>. The next term (term 2021) is <math>1977</math>, and the average of these two terms is the median, or <math>\boxed{\textbf{(C) } 1976.5}</math>.
 +
~ primegn
 +
 
 +
==Video Solution==
 +
 
 +
Education, The Study of Everything
 +
 
 +
https://youtu.be/luMQHhp_Rfk
 +
 
 +
 
 +
https://youtu.be/ZGwAasE32Y4
 +
 
 +
~IceMatrix
 +
 
 +
https://youtu.be/B0RPkcjdkPU
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2020|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2020|ab=A|num-b=10|num-a=12}}
 +
{{AMC12 box|year=2020|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:48, 7 November 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 equal to $1936$, 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 $\boxed{\textbf{(C) } 1976.5}.$ ~emerald_block

Solution 3

We want to know the $2020$th term and the $2021$st term to get the median.

We know that $44^2=1936$
So numbers $1^2, 2^2, ...,44^2$ are in between $1$ to $1936$.
So the sum of $44$ and $1936$ will result in $1980$, which means that $1936$ is the $1980$th number.
Also, notice that $45^2=2025$, which is larger than $2021$.
Then the $2020$th term will be $1936+40 = 1976$, and similarly the $2021$th term will be $1977$.
Solving for the median of the two numbers, we get $\boxed{\textbf{(C) } 1976.5}$
~toastybaker

Solution 4

We note that $44^2 = 1936$, which is the first square less than $2020$, which means that there are 44 addition terms before $2020$. This makes $2020$ the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before $2020$ is one less than the previous term (that is, we subtract 1 to get the previous term). If $2020$ is the 2064th term, than $2020 - 44$ is the (2064 - 44)th term. So, the 2020th term is $1976$. The next term (term 2021) is $1977$, and the average of these two terms is the median, or $\boxed{\textbf{(C) } 1976.5}$. ~ primegn

Video Solution

Education, The Study of Everything

https://youtu.be/luMQHhp_Rfk


https://youtu.be/ZGwAasE32Y4

~IceMatrix

https://youtu.be/B0RPkcjdkPU

~savannahsolver

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
2020 AMC 12A (Problems • Answer Key • Resources)
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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