2010 AMC 10B Problems/Problem 17
Contents
Problem
Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schools are in the city?
Solution
Let the be the number of schools, be the number of contestants, and be Andrea's place. Since the number of participants divided by three is the number of schools, . Andrea received a higher score than her teammates, so . Since is the maximum possible median, then is the maximum possible number of participants. Therefore, . This yields the compound inequality: . Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, cannot be even. is the only other option.
Solution 2
Let = the number of schools that participated in the contest.
Then students participated in the contest.
Since Andrea's score was the median score of all the students, the number of students in the contest must be odd - we know that no two students can have the same score (from the problem), and we know that if the number of students was even, the middle two scores would have to be the same to get a whole number for the median.
So, we can now write an expression for the median score (Andrea's score) in terms of the number of students who participated in the contest (i.e. ):
Also, since we know that this expression must be smaller than (Andrea, whose score was the median, got a better score than Beth and Carla), we can write an inequality and solve it for :
Solving this inequality for , we get that:
$n<\frac{37}{3} \mathrm{(which is a little over 24)}$ (Error compiling LaTeX. ! Package inputenc Error: Unicode char \u8: not set up for use with LaTeX.)
So this eliminates answer choices and .
But we already know that must be odd, implying that must also be odd! So at this point, the only odd answer choice is , and we are done.
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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