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 the double is <math>2(25) = 50</math> Since we dont want to change the median we need to get a value greater than <math>8</math> (as <math>8</math> would change the mode) for the smaller and <math>53</math> is fixed for the larger as anything less than <math>3</math> is not beneficial to the optimization. So taking our optimal values of <math>53</math> and <math>7</math> we have an answer of <math>53 + 7 = \boxed{\textbf{(D)}\ 60}</math>. | To double the range we must find the current range, which is <math>28 - 3 = 25</math>, to then the double is <math>2(25) = 50</math> Since we dont want to change the median we need to get a value greater than <math>8</math> (as <math>8</math> would change the mode) for the smaller and <math>53</math> is fixed for the larger as anything less than <math>3</math> is not beneficial to the optimization. So taking our optimal values of <math>53</math> and <math>7</math> we have an answer of <math>53 + 7 = \boxed{\textbf{(D)}\ 60}</math>. | ||
| − | ~apex304, SohumUttamchandani, wuwang2002, TaeKim | + | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower |
==Animated Video Solution== | ==Animated Video Solution== | ||
Revision as of 01:08, 30 January 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 the double is 
Since we dont want to change the median we need to get a value greater than 
(as would change the mode) for the smaller and 
is fixed for the larger as 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
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. 
