2010 AMC 12B Problems/Problem 8

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

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.

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2010 AMC 12B Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

Video Solution

See also