2014 AMC 8 Problems/Problem 24: Difference between revisions
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==Problem== | ==Problem== | ||
One day the Beverage | One day the Beverage Barn sold <math>252</math> cans of soda to <math>100</math> 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? | ||
<math>\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5</math> | <math>\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5</math> | ||
==Video Solutions== | |||
https://youtu.be/VlJzQ-ZNmmk ~savannahsolver | |||
https://www.youtube.com/watch?v=TkZvMa30Juo&t=2342s | |||
~ pi_is_3.141592653589793238462643383279502884197 | |||
==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 49, 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> | 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>. | ||
==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}} | ||
Latest revision as of 06:19, 13 January 2025
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 Solutions
https://youtu.be/VlJzQ-ZNmmk ~savannahsolver
https://www.youtube.com/watch?v=TkZvMa30Juo&t=2342s ~ pi_is_3.141592653589793238462643383279502884197
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 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
