Difference between revisions of "2014 AMC 8 Problems/Problem 24"
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==Solution== | ==Solution== | ||
In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are <math>100</math> people, the median will be the average of the <math>50\text{th}</math> and <math>51\text{st}</math> largest amount of cans per person. To minimize the first <math>49</math>, they would each have one can. Subtracting these <math>49</math> cans from the <math>252</math> cans gives us <math>203</math> cans left to divide among <math>51</math> people. Taking <math>\frac{203}{51}</math> gives us <math>3</math> and a remainder of <math>50</math>. Seeing this, the largest number of cans the <math>50</math>th person could have is <math>3</math>, which leaves <math>4</math> to the rest of the people. The average of <math>3</math> and <math>4</math> is <math>3.5</math>. Thus our answer is <math>\boxed{\text{(C) }3.5}</math> | In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are <math>100</math> people, the median will be the average of the <math>50\text{th}</math> and <math>51\text{st}</math> largest amount of cans per person. To minimize the first <math>49</math>, they would each have one can. Subtracting these <math>49</math> cans from the <math>252</math> cans gives us <math>203</math> cans left to divide among <math>51</math> people. Taking <math>\frac{203}{51}</math> gives us <math>3</math> and a remainder of <math>50</math>. Seeing this, the largest number of cans the <math>50</math>th person could have is <math>3</math>, which leaves <math>4</math> to the rest of the people. The average of <math>3</math> and <math>4</math> is <math>3.5</math>. Thus our answer is <math>\boxed{\text{(C) }3.5}</math> | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/TkZvMa30Juo?t=2342 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=23|num-a=25}} | {{AMC8 box|year=2014|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:27, 25 December 2022
Contents
Problem
One day the Beverage Barn sold 
cans of soda to 
customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
Video Solution for Problems 21-25
https://www.youtube.com/watch?v=qMb_GCP0mbw
Video Solution
https://www.youtube.com/watch?v=FE0u3Y1FCGk
https://youtu.be/VlJzQ-ZNmmk ~savannahsolver
Solution
In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are people, the median will be the average of the 
and 
largest amount of cans per person. To minimize the first 
, they would each have one can. Subtracting these cans from the cans gives us 
cans left to divide among 
people. Taking 
gives us 
and a remainder of 
. Seeing this, the largest number of cans the th person could have is , which leaves 
to the rest of the people. The average of and is 
. Thus our answer is 
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=2342
~ pi_is_3.14
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
