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

(Video Solution by WhyMath)
(Solution 1)
 
(18 intermediate revisions by 9 users not shown)
Line 4: Line 4:
 
<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>
  
==Solution==
+
==Solution 1 ==
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. Remember, 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>.
 +
 
 +
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021
 +
 
 +
==Video Solution==
 +
https://youtu.be/BdkBQppueWY
 +
 
 +
Please like and subscribe
 +
 
 +
==Video Solution by Math-X (Let's first Understand the question)==
 +
https://youtu.be/Ku_c1YHnLt0?si=1UUKWUVIPwomTx84&t=4411 ~Math-X
 +
 
 +
==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
Line 19: Line 33:
 
https://youtu.be/-N46BeEKaCQ?t=3136
 
https://youtu.be/-N46BeEKaCQ?t=3136
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
https://youtu.be/1bA7fD7Lg54?t=1970
+
https://youtu.be/DBqko2xATxs&t=2625
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
Line 26: Line 40:
 
~savannahsolver
 
~savannahsolver
  
==Video Solution==
+
==Video Solution by harungurcan==
https://youtu.be/FsT5WyQJbQ0
+
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s
  
Please like and subscribe!!
+
~harungurcan
  
 
==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}}

Latest revision as of 21:07, 19 January 2024

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. Remember, 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

https://youtu.be/BdkBQppueWY

Please like and subscribe

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=1UUKWUVIPwomTx84&t=4411 ~Math-X

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

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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

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