Difference between revisions of "2023 AMC 8 Problems/Problem 20"
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<math>\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61</math> | <math>\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61</math> | ||
| − | ==Solution== | + | ==Solution 1 == |
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>. | 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 | ||
| + | |||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== | ||
https://youtu.be/NpVLhU3AgNg | https://youtu.be/NpVLhU3AgNg | ||
Revision as of 00:26, 12 October 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Video Solution (CREATIVE THINKING!!!)
- 4 Animated Video Solution
- 5 Video Solution by OmegaLearn (Using Smart Sequence Analysis)
- 6 Video Solution by Magic Square
- 7 Video Solution by Interstigation
- 8 Video Solution by WhyMath
- 9 Video Solution
- 10 Video Solution by harungurcan
- 11 See Also
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 1
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
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
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
Video Solution
Please like and subscribe!!
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s
~harungurcan
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. 
