Difference between revisions of "2014 AMC 8 Problems/Problem 24"
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<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== | ||
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+ | 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 <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 08:55, 23 July 2024
Contents
[hide]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
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
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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