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

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==Problem==
 
==Problem==
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One day the Beverage Barn sold <math>252</math> cans of soda to <math>100</math> customers, and every customer bough at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
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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>
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 +
==Video Solution==
 +
 +
 +
https://youtu.be/VlJzQ-ZNmmk ~savannahsolver
 +
 
==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 100 people, the median will be the average of the 50th and 51st 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 <math>\frac{203}{51}</math> Gives us 3 and a remainder of 50. Seeing this, the largest number of cans the 50th 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 C.
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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 08:55, 23 July 2024

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

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 $100$ 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 $\boxed{\text{(C) }3.5}$

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

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