Difference between revisions of "2023 AMC 8 Problems/Problem 20"
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We have the set {3, 3, 8, 16, 28}. 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 8 (as 8 would change the mode) for the smaller and 53 is fixed for the larger as anything less than 3 is not beneficial to the optimization. So taking our optimal values of 53 and 7 we have an answer of <math>53 + 7 = \boxed{\text{(D)}60}</math> | We have the set {3, 3, 8, 16, 28}. 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 8 (as 8 would change the mode) for the smaller and 53 is fixed for the larger as anything less than 3 is not beneficial to the optimization. So taking our optimal values of 53 and 7 we have an answer of <math>53 + 7 = \boxed{\text{(D)}60}</math> | ||
-apex304, SohumUttamchandani, wuwang2002, TaeKim | -apex304, SohumUttamchandani, wuwang2002, TaeKim | ||
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| + | ==Animated Video Solution== | ||
| + | https://youtu.be/ItntB7vEafM | ||
| + | |||
| + | ~Star League (https://starleague.us) | ||
Revision as of 19:25, 24 January 2023
We have the set {3, 3, 8, 16, 28}. 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 8 (as 8 would change the mode) for the smaller and 53 is fixed for the larger as anything less than 3 is not beneficial to the optimization. So taking our optimal values of 53 and 7 we have an answer of 
-apex304, SohumUttamchandani, wuwang2002, TaeKim
Animated Video Solution
~Star League (https://starleague.us)
