2021 AMC 10B Problem 3

Let $J$ be the number of juniors and $S$ the number of seniors, so $J + S = 28$.

The team has $\frac{1}{4}J$ juniors and $\frac{1}{10}S$ seniors, and these are equal:
$$\frac{J}{4} = \frac{S}{10} \quad\Longrightarrow\quad 10J = 4S \quad\Longrightarrow\quad S = \frac{5}{2}J.$$

So juniors to seniors is $2 : 5$, and $28$ split in the ratio $2 : 5$ gives $J = \frac{2}{7} \cdot 28 = 8$ and $S = 20$.

Check: $25\%$ of $8$ is $2$, and $10\%$ of $20$ is $2$. Equal, as required.

The answer is $\boxed{\textbf{(C)}\ 8}$.

"Equal numbers from each class" is an equation between two percentages of two unknowns. That equation fixes the ratio of the unknowns, and the total then fixes their actual values. Percent-of-unknown problems almost always reduce to a ratio plus a total.

$25\%$ of the juniors must be a whole number, so $J$ is a multiple of $4$: only $8$ and $20$ survive among the choices. With $J = 20$, $S = 8$ and $10\%$ of $8$ is not an integer; with $J = 8$, $S = 20$ and $2 = 2$ works.

• Answering 20 (the number of seniors) or 2 (the number of juniors on the team) instead of the number of juniors in the program.
• Setting $0.25J = 0.10S$ but then solving $J + S = 28$ with the ratio inverted ($5 : 2$), giving $J = 20$.