2021 AMC 10B Problem 3

1. Name the unknowns: $J$ juniors and $S$ seniors, with $J + S = 28$.
2. The team has $25\%$ of the juniors, which is $\frac{J}{4}$, and $10\%$ of the seniors, which is $\frac{S}{10}$. The problem says these two numbers are equal:
$$\frac{J}{4} = \frac{S}{10}.$$
3. Multiply both sides by $20$ to clear the fractions: $5J = 2S$.
4. This says the juniors and seniors come in the ratio $2 : 5$. For example, $2$ juniors and $5$ seniors fit, since $5 \cdot 2 = 2 \cdot 5$.
5. So the $28$ students split into $2 + 5 = 7$ equal parts, each part being $28 \div 7 = 4$ students. Juniors get $2$ parts: $J = 8$. Seniors get $5$ parts: $S = 20$.
6. Check: $25\%$ of $8$ is $2$, and $10\%$ of $20$ is $2$. Equal, as needed.

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

"The same number from each group" is one equation, and it tells you how the two groups compare (the ratio). The total then tells you the actual sizes. Many percent problems work this way: first a ratio, then use the total.

Use the answer choices. $25\%$ of the juniors has to be a whole number of people, so the number of juniors is a multiple of $4$. Only $8$ and $20$ are.

• If there are $20$ juniors, there are $8$ seniors, and $10\%$ of $8$ is $0.8$ of a person. Impossible.
• If there are $8$ juniors, there are $20$ seniors: $2$ juniors and $2$ seniors on the team. This works.

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 $\boxed{\textbf{(C)}\ 8}$.