2025 AMC 10A Problem 4

1. An average is hard to work with directly, but a total is easy: total = average $\times$ number of people. So change every average into a total.

1. Let $s$ be the number of students, $t$ the number of teachers, and $A$ Ash's age. We know $s + t = 15$.

3. Ash joins the students. Before: $s$ people averaging $12$, so their ages total $12s$. After: $s + 1$ people averaging $14$, total $14(s+1)$. The only new age is Ash's, so
$$12s + A = 14s + 14 \quad\Longrightarrow\quad A = 2s + 14.$$

4. Ash joins the teachers. Before: total $55t$. After: $t + 1$ people averaging $52$, total $52(t+1)$. So
$$55t + A = 52t + 52 \quad\Longrightarrow\quad A = 52 - 3t.$$

1. Now use $s + t = 15$. Try a small table of splits and see where the two formulas agree:

They agree when $s = 7$ and $t = 8$, giving $A = 28$.

1. (Algebra version: put $t = 15 - s$ into $52 - 3t$ to get $3s + 7$. Then $2s + 14 = 3s + 7$, so $s = 7$.)

1. Check: $7$ students total $84$ years; with Ash, $84 + 28 = 112$, and $112 \div 8 = 14$. $8$ teachers total $440$; with Ash, $468 \div 9 = 52$. Both work.

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

Whenever someone joins or leaves a group and the average changes, write "total before + new person = total after". That turns average statements into simple equations. There is also a nice way to read $A = 2s + 14$: to raise the average of $s$ people by $2$, the newcomer must be the new average, $14$, plus $2$ extra years for each person already there.

Test the answer choices. From step 3, the number of students is $s = (A - 14) \div 2$; from step 4, the number of teachers is $t = (52 - A) \div 3$. Both must be whole numbers adding to $15$. For $A = 28$: $s = 7$, $t = 8$, and $7 + 8 = 15$. Works. For $29$ or $33$, $s$ is not a whole number; for $30$ or $32$, $t$ is not a whole number.

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 $\boxed{\textbf{(A)}\ 28}$.