2025 AMC 10A Problem 4

Let there be $s$ students and $t$ teachers, so $s + t = 15$, and let $A$ be Ash's age.

The students' ages total $12s$. Adding Ash makes $s + 1$ people with average $14$:
$$12s + A = 14(s+1) \implies A = 2s + 14.$$
The teachers' ages total $55t$. Adding Ash makes $t + 1$ people with average $52$:
$$55t + A = 52(t+1) \implies A = 52 - 3t.$$
Substituting $t = 15 - s$ in the second expression gives $A = 52 - 3(15 - s) = 3s + 7$. Equating the two,
$$2s + 14 = 3s + 7 \implies s = 7,$$
so $t = 8$ and $A = 2(7) + 14 = 28$.

Check both teams: the students total $7 \cdot 12 = 84$, and $(84 + 28)/8 = 14$; the teachers total $8 \cdot 55 = 440$, and $(440 + 28)/9 = 52$.

The answer is $\boxed{\textbf{(A)}\ 28}$.

An average statement is useless until it is converted into a total, and once both statements are totals the problem is an ordinary two-equation system with the head count $s + t = 15$ as the third fact. The intermediate forms $A = 14 + 2s$ and $A = 52 - 3t$ also carry a reading worth keeping: a newcomer who lifts the average of $s$ people by $2$ must be worth the new average plus $2$ for every person already there, and one who drops it by $3$ is short by $3$ for every person already there.

Test the choices for integrality. Ash's age forces $s = (A - 14)/2$ and $t = (52 - A)/3$, both positive integers summing to $15$. Only $A = 28$ makes both integers, giving $s = 7$ and $t = 8$, which do sum to $15$; $A = 29$ and $A = 33$ fail on $s$, and $A = 30$ and $A = 32$ fail on $t$.

• Dividing by the old team size instead of the new one, for example setting (12s + A)/s = 14, which leaves Ash out of the head count.
• Guessing a near-even split such as $8$ students and $7$ teachers, which produces $30$, choice (C).
• Counting Ash inside the given total of $15$, when the $15$ describes the teams before he joins.