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==Video Solution by harungurcan==
 
==Video Solution by harungurcan==
 
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s
 
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s
 +
 +
~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}}
 
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{{MAA Notice}}

Revision as of 06:20, 1 May 2023

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

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. 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

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/1bA7fD7Lg54?t=1970

Video Solution by WhyMath

https://youtu.be/lCVPFN1EK_M

~savannahsolver

Video Solution

https://youtu.be/FsT5WyQJbQ0

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Video Solution (CREATIVE THINKING!!!)

https://youtu.be/NpVLhU3AgNg

~Education, the Study of Everything

Video Solution by harungurcan

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

~harungurcan

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

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

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