Difference between revisions of "2023 AMC 8 Problems/Problem 20"
(→Video Solution (This video got deleted)) |
(→Solution 1) |
||
| Line 7: | Line 7: | ||
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 because you want to get the largest range without changing the mode. 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 because you want to get the largest range without changing the mode. 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, CHECKMATE2021 |
==Video Solution== | ==Video Solution== | ||
Revision as of 21:07, 19 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Video Solution
- 4 Video Solution by Math-X (Let's first Understand the question)
- 5 Video Solution (CREATIVE THINKING(Very fast paced)!!!)
- 6 Animated Video Solution
- 7 Video Solution by OmegaLearn (Using Smart Sequence Analysis)
- 8 Video Solution by Magic Square
- 9 Video Solution by Interstigation
- 10 Video Solution by WhyMath
- 11 Video Solution by harungurcan
- 12 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 because you want to get the largest range without changing the mode. So, taking our optimal values of 
and , we have an answer of 
.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021
Video Solution
Please like and subscribe
Video Solution by Math-X (Let's first Understand the question)
https://youtu.be/Ku_c1YHnLt0?si=1UUKWUVIPwomTx84&t=4411 ~Math-X
Video Solution (CREATIVE THINKING(Very fast paced)!!!)
~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/DBqko2xATxs&t=2625
Video Solution by WhyMath
~savannahsolver
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. 
