2000 AMC 10 Problem 8

1. Try real numbers first. Suppose $40$ students from each class took the test.
2. Freshmen: $\tfrac25$ of the freshmen is $40$, so $\tfrac15$ is $20$ and there are $100$ freshmen.
3. Sophomores: $\tfrac45$ of the sophomores is $40$, so $\tfrac15$ is $10$ and there are $50$ sophomores.
4. $100$ is twice $50$, so there are twice as many freshmen as sophomores.
5. To see this is always true, call the numbers of freshmen and sophomores $f$ and $s$. The same number took the test from each class:
$$\tfrac25 f = \tfrac45 s.$$
Multiply both sides by $5$: $2f = 4s$. Divide by $2$: $f = 2s$.

The answer is $\boxed{\textbf{(D)}\ \text{There are twice as many freshmen as sophomores.}}$

If a smaller share of one group gives the same number of people as a bigger share of another group, the first group has to be bigger. Writing the equation settles the direction, and then you read $f = 2s$ out loud as "freshmen are double the sophomores."

Let $f$ be the number of freshmen and $s$ the number of sophomores. The contestant counts are $\tfrac25 f$ and $\tfrac45 s$, and these are equal:
$$\tfrac25 f = \tfrac45 s \quad\Longrightarrow\quad 2f = 4s \quad\Longrightarrow\quad f = 2s.$$

So there are twice as many freshmen as sophomores. Sanity check with numbers: $100$ freshmen send $40$ contestants, and $50$ sophomores also send $40$.

The answer is $\boxed{\textbf{(D)}\ \text{There are twice as many freshmen as sophomores.}}$