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

(Solution)
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<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math>
 
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math>
  
== Solution 1 ==
+
== Solution ==
 
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 place. 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
 
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 place. 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{\textbf{(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{\textbf{(B)}\ 23}</math> is the only other option.
== Solution 2: Using the Answer Choices ==
 
First, we know that if Andrea's score is the median, then there must be an odd number of people and therefore, an odd number of schools. We can immediately eliminate answer choices A, C, and E. We can automatically conclude that B is the right answer because the smallest of the remaining two options would put Andrea ahead of her teammates.
 
 
(Solution by Flamedragon)
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2010|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:15, 30 December 2015

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 place. 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. $\boxed{\textbf{(B)}\ 23}$ is the only other option.

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AMC 10 Problems and Solutions

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