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Difference between revisions of "2023 AMC 8 Problems/Problem 20"

(→‎Video Solution (CREATIVE THINKING!!!))
(→‎Solution for AMC 8 Problem 20. Easy and understandable. Enjoy Type in the comments if you have any questions!)
 
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<math>\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61</math>
 
<math>\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61</math>
  
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==Solution==
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==Solution 1 ==
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To double the range, we must find the current range, which is <math>28 - 3 = 25</math>, to then double to: <math>2(25) = 50</math>. Since we do not want to change the median, we need to get a value less than <math>8</math> (as <math>8</math> would change the mode) for the smaller, making <math>53</math> fixed for the larger. Remember, anything less than <math>3</math> is not beneficial to the optimization. So, taking our optimal values of <math>7</math> and <math>53</math>, we have an answer of <math>7 + 53 = \boxed{\textbf{(D)}\ 60}</math>.
+
To double the range, we must find the current range, which is <math>28 - 3 = 25</math>, to then double to: <math>2(25) = 50</math>. Since we do not want to change the median, we need to get a value less than <math>8</math> (as <math>8</math> would change the mode) for the smaller, making <math>53</math> fixed for the larger. Anything less than <math>3</math> is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of <math>7</math> and <math>53</math>, we have an answer of <math>7 + 53 = \boxed{\textbf{(D)}\ 60}</math>.
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 +
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021
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 +
==Video Solution by STEM Station(Quick and Easy to Understand)==
 +
https://www.youtube.com/watch?v=qZO9JLezr-U
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2294
 +
~hsnacademy
 +
 
 +
==Video Solution==
 +
https://youtu.be/BdkBQppueWY
 +
 
 +
Please like and subscribe
 +
 
 +
==Solution for AMC 8 Problem 20. Easy and understandable. Enjoy! Type in the comments if you have any questions!==
 +
 
 +
https://youtu.be/YeCldZn5qD8
 +
 
 +
Subscribe for more videos and please like!
 +
 
 +
==Video Solution (CREATIVE THINKING(Very fast paced)!!!)==
 +
https://youtu.be/NpVLhU3AgNg
 +
 
 +
~Education, the Study of Everything
  
−
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower
 
 
==Animated Video Solution==
 
==Animated Video Solution==
 
https://youtu.be/ItntB7vEafM
 
https://youtu.be/ItntB7vEafM
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https://youtu.be/-N46BeEKaCQ?t=3136
 
https://youtu.be/-N46BeEKaCQ?t=3136
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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https://youtu.be/1bA7fD7Lg54?t=1970
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https://youtu.be/DBqko2xATxs&t=2625
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
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~savannahsolver
 
~savannahsolver
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==Video Solution==
 
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https://youtu.be/FsT5WyQJbQ0
 
−
 
−
Please like and subscribe!!
 
  
 
==Video Solution by harungurcan==
 
==Video Solution by harungurcan==
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~harungurcan
 
~harungurcan
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==Video Solution by Dr. David==
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https://youtu.be/mMU-uvgwVzg
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=19|num-a=21}}
 
{{AMC8 box|year=2023|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 15:50, 24 July 2025

Problem

Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?

$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$

Solution 1

To double the range, we must find the current range, which is $28 - 3 = 25$, to then double to: $2(25) = 50$. Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Anything less than $3$ is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of $7$ and $53$, we have an answer of $7 + 53 = \boxed{\textbf{(D)}\ 60}$.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021

Video Solution by STEM Station(Quick and Easy to Understand)

https://www.youtube.com/watch?v=qZO9JLezr-U

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2294 ~hsnacademy

Video Solution

https://youtu.be/BdkBQppueWY

Please like and subscribe

Solution for AMC 8 Problem 20. Easy and understandable. Enjoy! Type in the comments if you have any questions!

https://youtu.be/YeCldZn5qD8

Subscribe for more videos and please like!

Video Solution (CREATIVE THINKING(Very fast paced)!!!)

https://youtu.be/NpVLhU3AgNg

~Education, the Study of Everything

Animated Video Solution

https://youtu.be/ItntB7vEafM

~Star League (https://starleague.us)

Video Solution by OmegaLearn (Using Smart Sequence Analysis)

https://youtu.be/qNsgNa9Qq9M

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=3136

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=2625

Video Solution by WhyMath

https://youtu.be/lCVPFN1EK_M

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s

~harungurcan

Video Solution by Dr. David

https://youtu.be/mMU-uvgwVzg

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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