2025 AMC 10A Problem 4

P1-10~3 minPrint

Ratios, Percents & Averages Systems of Equations Set up the equation/formula and compute; no special trick needed Substitute to simplify (u = x+1/x, shifting, scaling)

A team of students is going to compete against a team of teachers in a trivia contest. The total number of students and teachers is $15$ . Ash, a cousin of one of the students, wants to join the contest. If Ash plays with the students, the average age on that team will increase from $12$ to $14.$ If Ash plays with the teachers, the average age on that team will decrease from $55$ to $52$ . How old is Ash?

Show solution without answering

Key insight

Turn each average into a total. Then Ash's age is 14 plus 2 per student, and also 52 minus 3 per teacher. Both agree at 7 students.

Solution

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.

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
$55t + A = 52t + 52 \quad\Longrightarrow\quad A = 52 - 3t.$

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

$s$	$t$	$2s + 14$	$52 - 3t$
$6$	$9$	$26$	$25$
$7$	$8$	$28$	$28$
$8$	$7$	$30$	$31$

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

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

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)