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Difference between revisions of "2010 AMC 10B Problems/Problem 17"

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== Solution ==
 
== Solution ==
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Let the <math>n</math> be the number of schools, <math>3n</math> be the number of contestants, and <math>x</math> be Andrea's score. Since the number of participants divided by three is the number of schools, <math>n\geq\frac{64}3=21\frac13</math>. Since Andrea received a higher score than her teammates, <math>x\leq36</math>. Since <math>36</math> is the maximum possible median, then <math>2*36-1=71</math>is the maximum possible number of participants. Therefore, <math>3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23</math>. This yields the compound inequality: <math>21\frac13\leq n\leq
+
Let the <math>n</math> be the number of schools, <math>3n</math> be the number of contestants, and <math>x</math> be Andrea's score. Since the number of participants divided by three is the number of schools, <math>n\geq\frac{64}3=21\frac13</math>. Andrea received a higher score than her teammates, so <math>x\leq36</math>. Since <math>36</math> is the maximum possible median, then <math>2*36-1=71</math>is the maximum possible number of participants. Therefore, <math>3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23</math>. This yields the compound inequality: <math>21\frac13\leq n\leq
 
  23\frac23</math>. 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, <math>n</math> cannot be even. <math>\boxed{\mathrm {(B)} 23}</math> is the only other option.
 
  23\frac23</math>. 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, <math>n</math> cannot be even. <math>\boxed{\mathrm {(B)} 23}</math> is the only other option.

Revision as of 01:15, 25 January 2011

Problem

Every high school in the city of Euclid sent a team of $3$ 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 $37$th and $64$th, respectively. How many schools are in the city?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$

Solution

Let the $n$ be the number of schools, $3n$ be the number of contestants, and $x$ be Andrea's score. Since the number of participants divided by three is the number of schools, $n\geq\frac{64}3=21\frac13$. Andrea received a higher score than her teammates, so $x\leq36$. Since $36$ is the maximum possible median, then $2*36-1=71$is the maximum possible number of participants. Therefore, $3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23$. This yields the compound inequality: $21\frac13\leq n\leq 23\frac23$. 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, $n$ cannot be even. $\boxed{\mathrm {(B)} 23}$ is the only other option.

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