Difference between revisions of "2010 AMC 10B Problems/Problem 17"
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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 == | + | == Solution 1 == |
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 21:23, 9 February 2014
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 1
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: 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
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
