2010 AMC 12B Problems/Problem 8

Revision as of 21:07, 19 June 2018 by Adihaya (talk | contribs) (Solution 2)

Problem 8

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?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$


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 there were. $x$ 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 score. We know that she was the highest on our 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 $(2a-1)/3 = 69/3 = \boxed{23}$.

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
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

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