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

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{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #8]] and [[2020 AMC 10A Problems|2020 AMC 10A #11]]}}
 
{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #8]] and [[2020 AMC 10A Problems|2020 AMC 10A #11]]}}
  
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==Problem 11==
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==Problem==
  
 
What is the median of the following list of <math>4040</math> numbers<math>?</math>
 
What is the median of the following list of <math>4040</math> numbers<math>?</math>
 +
<cmath>1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2</cmath>
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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>
  
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<cmath>1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2</cmath>
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==Solution 1==
 +
We can see that <math>44^2=1936</math> which is less than 2020. Therefore, there are <math>2020-44=1976</math> of the <math>4040</math> numbers greater than <math>2020</math>. Also, there are <math>2020+44=2064</math> numbers that are less than or equal to <math>2020</math>.
  
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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>
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Since there are <math>44</math> duplicates/extras, it will shift up our median's placement down <math>44</math>. Had the list of numbers been <math>1,2,3, \dots, 4040</math>, the median of the whole set would be <math>\dfrac{1+4040}{2}=2020.5</math>.
  
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==Solution 1==
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Thus, our answer is <math>2020.5-44=\boxed{\textbf{(C)}\ 1976.5}</math>.
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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 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>.
 
  
 
~aryam
 
~aryam
 +
 +
~Additions by BakedPotato66
  
 
== Solution 2 ==
 
== Solution 2 ==
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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>
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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.
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<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]]
+
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]]
  
 
== Solution 3 ==
 
== Solution 3 ==
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We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median. <br>
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We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median.
−
<br>
+
 
−
We know that <math>44^2=1936</math> <br>
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We know that <math>44^2=1936</math>. So, numbers <math>1^2, 2^2, \ldots,44^2</math> are in between <math>1</math> and <math>1936</math>.
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So numbers <math>1^2, 2^2, ...,44^2</math> are in between <math>1</math> to <math>1936</math>. <br>
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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>
+
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.
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Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. <br>
+
 
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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>
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Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>.
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Solving for the median of the two numbers, we get <math>\boxed{\textbf{(C) } 1976.5}</math> <br>
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 +
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>.
 +
 
 +
Solving for the median of the two numbers, we get <math>\boxed{\textbf{(C)}\ 1976.5}</math>
 +
 
 
~toastybaker
 
~toastybaker
  
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==Video Solution==
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== Solution 4 ==
 +
We note that <math>44^2 = 1936</math>, which is the largest square less than <math>2020</math>, which means that there are <math>44</math> additional terms before <math>2020</math>. This makes <math>2020</math> the <math>2064</math>th term. To find the median, we need the <math>2020</math>th and <math>2021</math>st term. We note that every term before <math>2020</math> is one less than the previous term (That is, we subtract <math>1</math> to get the previous term.). If <math>2020</math> is the <math>2064</math>th term, than <math>2020 - 44</math> is the <math>(2064 - 44)</math>th term. So, the <math>2020</math>th term is <math>1976</math>, and the <math>2021</math>st term is <math>1977</math>, and the average of these two terms is the median, or <math>\boxed{\textbf{(C)}\ 1976.5}</math>.
 +
 
 +
~primegn
 +
 
 +
==Solution 5 (Decreasing Order)==
 +
To find the median, we sort the <math>4040</math> numbers in decreasing order, then average the <math>2020</math>th and the <math>2021</math>st numbers of the sorted list.
 +
 
 +
Since <math>45^2=2025</math> and <math>44^2=1936,</math> the first <math>2021</math> numbers of the sorted list are <cmath>\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },</cmath> from which the answer is <math>\frac{1977+1976}{2}=\boxed{\textbf{(C)}\ 1976.5}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Video Solution by Education, The Study of Everything==
 +
https://youtu.be/luMQHhp_Rfk
 +
 
 +
==Video Solution by TheBeautyOfMath==
 
https://youtu.be/ZGwAasE32Y4
 
https://youtu.be/ZGwAasE32Y4
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/B0RPkcjdkPU
 +
 +
~savannahsolver
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/xqo0PgH-h8Y?t=363
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 13:36, 22 June 2024

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

Problem

What is the median of the following list of $4040$ numbers$?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution 1

We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$. Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$.

Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$. Had the list of numbers been $1,2,3, \dots, 4040$, the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$.

Thus, our answer is $2020.5-44=\boxed{\textbf{(C)}\ 1976.5}$.

~aryam

~Additions by BakedPotato66

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 $1^2, 2^2, \ldots,44^2$ are in between $1$ and $1936$.

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 largest square less than $2020$, which means that there are $44$ additional terms before $2020$. This makes $2020$ the $2064$th term. To find the median, we need the $2020$th and $2021$st 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 $2064$th term, than $2020 - 44$ is the $(2064 - 44)$th term. So, the $2020$th term is $1976$, and the $2021$st term is $1977$, and the average of these two terms is the median, or $\boxed{\textbf{(C)}\ 1976.5}$.

~primegn

Solution 5 (Decreasing Order)

To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$th and the $2021$st numbers of the sorted list.

Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{\textbf{(C)}\ 1976.5}.$

~MRENTHUSIASM

Video Solution by Education, The Study of Everything

https://youtu.be/luMQHhp_Rfk

Video Solution by TheBeautyOfMath

https://youtu.be/ZGwAasE32Y4

~IceMatrix

Video Solution by WhyMath

https://youtu.be/B0RPkcjdkPU

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/xqo0PgH-h8Y?t=363

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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