Difference between revisions of "2014 AMC 8 Problems/Problem 24"

Revision as of 15:04, 17 January 2022

Problem

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

https://www.youtube.com/watch?v=qMb_GCP0mbw

Video Solution

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

Solution

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 $\boxed{\text{(C) }3.5}$

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.

@@ Line 5: / Line 5: @@
 ==Video Solution for Problems 21-25==
-https://www.youtube.com/watch?v=6S0u_fDjSxc
+https://www.youtube.com/watch?v=qMb_GCP0mbw
 ==Video Solution==

Page

Toolbox

Search