2014 AMC 8 Problems/Problem 24


One day the Beverage Barn sold $252$ cans of soda to $100$ 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?

$\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$

Video Solution for Problems 21-25


Video Solution

https://www.youtube.com/watch?v=FE0u3Y1FCGk ~David

https://youtu.be/VlJzQ-ZNmmk ~savannahsolver


In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$. Seeing this, the largest number of cans the $50$th person could have is $3$, which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$. Thus our answer is $\boxed{\text{(C) }3.5}$

Video Solution by OmegaLearn


~ pi_is_3.14

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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