Difference between revisions of "2023 AMC 8 Problems/Problem 20"
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− | To double the range we must find the current range, which is <math>28 - 3 = 25</math>, to then | + | 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>. |
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower |
Revision as of 09:49, 23 February 2023
Contents
Problem
Two integers are inserted into the list to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
Solution
To double the range, we must find the current range, which is , to then double to:
. Since we do not want to change the median, we need to get a value less than
(as
would change the mode) for the smaller, making
fixed for the larger. Remember, anything less than
is not beneficial to the optimization. So, taking our optimal values of
and
, we have an answer of
.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Smart Sequence Analysis)
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
~savannahsolver
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.