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]]}} | ||
− | ==Problem | + | ==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> | ||
+ | <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 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>. | ||
− | <math> | + | 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>. |
− | + | Thus, our answer is <math>2020.5-44=\boxed{\textbf{(C)}\ 1976.5}</math>. | |
− | |||
~aryam | ~aryam | ||
+ | |||
+ | ~Additions by BakedPotato66 | ||
== Solution 2 == | == 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. | + | 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. |
− | + | ||
− | 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 == | ||
− | We want to know the <math>2020</math> th term and the <math>2021</math> | + | We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median. |
− | + | ||
− | We know that <math>44^2=1936</math> | + | 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>. |
− | So numbers <math>1^2, 2^2, | + | |
− | 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. | + | 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. |
− | Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. | + | |
− | Then the <math>2020</math> th term will be <math>1936+40 = 1976</math>, and similarly <math>2021</math>th term will be <math>1977</math>. | + | Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. |
− | Solving for the median of the two numbers, we get <math>\boxed{\textbf{(C) } 1976.5}</math> ~toastybaker | + | |
+ | 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 | ||
+ | |||
+ | == 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 <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 | ||
− | ==Video Solution== | + | ==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 1== | ||
+ | https://youtu.be/luMQHhp_Rfk | ||
+ | |||
+ | Education, The Study of Everything | ||
+ | |||
+ | ==Video Solution 2== | ||
https://youtu.be/ZGwAasE32Y4 | https://youtu.be/ZGwAasE32Y4 | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/B0RPkcjdkPU | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 15:22, 21 October 2021
- The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.
Contents
Problem
What is the median of the following list of numbers
Solution 1
We can see that which is less than 2020. Therefore, there are of the numbers greater than . Also, there are numbers that are less than or equal to .
Since there are duplicates/extras, it will shift up our median's placement down . Had the list of numbers been , the median of the whole set would be .
Thus, our answer is .
~aryam
~Additions by BakedPotato66
Solution 2
As we are trying to find the median of a -term set, we must find the average of the th and st terms.
Since is slightly greater than , we know that the perfect squares through are less than , and the rest are greater. Thus, from the number to the number , there are terms. Since is less than and less than , we will only need to consider the perfect square terms going down from the th term, , after going down terms. Since the th and st terms are only and terms away from the th term, we can simply subtract from and from to get the two terms, which are and . Averaging the two, we get
Solution 3
We want to know the th term and the st term to get the median.
We know that . So, numbers are in between and .
So, the sum of and will result in , which means that is the th number.
Also, notice that , which is larger than .
Then the th term will be , and similarly the th term will be .
Solving for the median of the two numbers, we get
~toastybaker
Solution 4
We note that , which is the first square less than , which means that there are additional terms before . This makes the th term. To find the median, we need the th and st term. We note that every term before is one less than the previous term (That is, we subtract to get the previous term.). If is the th term, than is the th term. So, the th term is , and the st term is , and the average of these two terms is the median, or .
~primegn
Solution 5 (Decreasing Order)
To find the median, we sort the numbers in decreasing order, then average the th and the st numbers of the sorted list.
Since and the first numbers of the sorted list are from which the answer is
~MRENTHUSIASM
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.