2023 AMC 8 Problems/Problem 20

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

Video Solution by WhyMath

https://youtu.be/lCVPFN1EK_M

~savannahsolver

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.

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