2010 AMC 10B Problem 17

Say there are $n$ schools, so $3n$ students, all with different scores. Turn each fact into a condition on $n$.

1. Andrea is the median. She is one of the students, so there must be one single middle student. That only happens when the number of students $3n$ is odd, which means $n$ is odd. For example, with $69$ students the middle one is $35$th, with $34$ above and $34$ below. In general her place is $\frac{3n + 1}{2}$.
2. Andrea beat Beth, who was $37$th. So Andrea's place is better than $37$th:
$$\frac{3n + 1}{2} < 37 \quad\Longrightarrow\quad 3n + 1 < 74 \quad\Longrightarrow\quad 3n < 73.$$
So $n$ is at most $24$.
3. Carla was $64$th. So there are at least $64$ students: $3n \ge 64$. Since $3 \cdot 21 = 63$ is too few, $n$ is at least $22$.
4. So $n$ is $22$, $23$ or $24$, and $n$ must be odd. Only $n = 23$ is left.

Check with $n = 23$: there are $69$ students, Andrea is $35$th (ahead of Beth in $37$th), and a $64$th place exists for Carla.

The answer is $\boxed{\textbf{(B)}\ 23}$.

Each sentence of the problem gives one piece of information about $n$: Beth's place gives an upper limit, Carla's place gives a lower limit, and the median being a real person forces $n$ to be odd. Squeezing $n$ from both sides and then using the odd/even condition leaves only one value.

Check the answer choices. $n = 22, 24, 26$ give $66, 72, 78$ students, an even number, so there is no single middle student. $n = 25$ gives $75$ students, and the middle one is $38$th, which is behind Beth in $37$th. Only $n = 23$ works.