Difference between revisions of "2010 AMC 12B Problems/Problem 8"
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− | == Problem | + | {{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #8]] and [[2010 AMC 10B Problems|2010 AMC 10B #17]]}} |
+ | |||
+ | == 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> | ||
− | == Solution == | + | == Solution 1 == |
− | 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. Thus, the only possible answer is <math>23 \Rightarrow \boxed{B}</math>. | + | 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>. |
+ | |||
+ | == Solution 2 == | ||
+ | Let <math>a</math> be Andrea's place. We know that she was the highest on our 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>. | ||
+ | |||
+ | Also, since <math>2a-1</math> is the rank of the last-place person, and one of Andrea's teammates already got 64th place, <math>2a-1 > 64 \implies a \ge 33</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>. | ||
+ | |||
+ | ~adihaya | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/FQO-0E2zUVI?t=297 | ||
+ | |||
+ | ~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}} | ||
+ | {{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:41, 15 February 2021
- 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 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
There are schools. This means that there are people. Because no one's score was the same as another person's score, that means that there could only have been median score. This implies that is an odd number. cannot be less than , because there wouldn't be a th place if there were. cannot be greater than either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is .
Solution 2
Let be Andrea's place. We know that she was the highest on our team, so .
Since is the median, there are to the left and right of the median, so the total number of people is and the number of schools is . This implies that .
Also, since is the rank of the last-place person, and one of Andrea's teammates already got 64th place, .
Putting it all together: and , so clearly , and the number of schools as we got before is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.