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Difference between revisions of "2010 AMC 12B Problems/Problem 8"

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== Problem 8 ==
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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #8]] and [[2010 AMC 10B Problems|2010 AMC 10B #17]]}}
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== Problem ==
 
Every high school in the city of Euclid sent a team of <math>3</math> 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 <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?
 
Every high school in the city of Euclid sent a team of <math>3</math> 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 <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?
  
 
<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>
  
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== Solution ==
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== Solution 1 ==
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There are <math>x</math> schools. This means that there are <math>3x</math> people. Because no one's score was the same as another person's score, that means that there could only have been <math>1</math> median score. This implies that <math>x</math> is an odd number. <math>x</math> cannot be less than <math>23</math>, because there wouldn't be a <math>64</math>th place if there were. <math>x</math> cannot be greater than <math>23</math> either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is <math>23 \Rightarrow \boxed{B}</math>.
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There are <math>x</math> schools. This means that there are <math>3x</math> people. Because no one's score was the same as another person's score, that means that there could only have been <math>1</math> median score. This implies that <math>x</math> is an odd number. <math>x</math> cannot be less than <math>23</math>, because there wouldn't be a <math>64</math>th place if <math>x</math> was. <math>x</math> cannot be greater than <math>23</math> either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is <math>23 \Rightarrow \boxed{B}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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Let <math>a</math> be Andrea's score. We know that she was the highest on our team, so <math>a < 37</math>.  
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Let <math>a</math> be Andrea's place. We know that she was the highest on her team, so <math>a < 37</math>.  
  
 
Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>.  
 
Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>.  
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Putting it all together: <math>33 \le a < 37</math> and <math>a \equiv 2 \pmod{3}</math>, so clearly <math>a = 35</math>, and the number of schools as we got before is <math>(2a-1)/3 = 69/3 = \boxed{23}</math>.
 
Putting it all together: <math>33 \le a < 37</math> and <math>a \equiv 2 \pmod{3}</math>, so clearly <math>a = 35</math>, and the number of schools as we got before is <math>(2a-1)/3 = 69/3 = \boxed{23}</math>.
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~adihaya
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==Video Solution==
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https://youtu.be/FQO-0E2zUVI?t=297
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}}
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}}
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{{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:41, 29 October 2024

The following problem is from both the 2010 AMC 12B #8 and 2010 AMC 10B #17, so both problems redirect to this page.

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 1

There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$, because there wouldn't be a $64$th place if $x$ was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \Rightarrow \boxed{B}$.

Solution 2

Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$.

Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$. This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$.

Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$.

Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$, so clearly $a = 35$, and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$.

~adihaya

Video Solution

https://youtu.be/FQO-0E2zUVI?t=297

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 12 Problems and Solutions
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

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