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

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(Video Solution by Magic Square)
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==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=3136
 
https://youtu.be/-N46BeEKaCQ?t=3136
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==Video Solution by Interstigation==
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https://youtu.be/1bA7fD7Lg54?t=1970
  
 
==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}}

Revision as of 21:47, 16 February 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 the double is $2(25) = 50$ Since we dont want to change the median we need to get a value greater than $8$ (as $8$ would change the mode) for the smaller and $53$ is fixed for the larger as anything less than $3$ is not beneficial to the optimization. So taking our optimal values of $53$ and $7$ we have an answer of $53 + 7 = \boxed{\textbf{(D)}\ 60}$.

~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

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

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