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 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 $7 + 53 = \boxed{\textbf{(D)}\ 60}$.

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

https://youtu.be/Ku_c1YHnLt0?si=1UUKWUVIPwomTx84&t=4411

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Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=925

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Video Solution

https://youtu.be/BdkBQppueWY

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Animated Video Solution

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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

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Video Solution by harungurcan

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

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Video Solution by Dr. David

https://youtu.be/mMU-uvgwVzg

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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