2025 AMC 12A Problems/Problem 3

The following problem is from both the 2025 AMC 10A #4 and 2025 AMC 12A #3, so both problems redirect to this page.

Problem

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?

$\textbf{(A)}~28\qquad\textbf{(B)}~29\qquad\textbf{(C)}~30\qquad\textbf{(D)}~32\qquad\textbf{(E)}~33$

Solution 1

When Ash joins a team, the team's average is pulled towards his age. Let $A$ be Ash's age and $N$ be the number of people on the student team. This means that there are $15-N$ people in the teacher team. Let us write an expression for the change in the average for each team.

The students originally had an average of $12$ , which became $14$ when Ash joined, so there was an increase of $2$ . The term $A-12$ represents how much older Ash is compared to the average of the students'. If we divide this by $N+1$ , which is the number of people on the student team when Ash joins, we get the average change per team member once Ash is added. Therefore, $\frac{A-12}{N+1} = 2.$

Similarly, for teachers, the average was originally $55$ , which decreased by $3$ to become $52$ when Ash joined. Intuitively, $55-A$ represents how much younger Ash is than the average age of the teachers. Dividing this by the expression $(15-N)+1$ , which is the new total number of people on the teacher team, represents the average change per team member once Ash joins. We can write the equation

$\frac{55-A}{16-N} = 3.$

To solve the system, multiply equation (1) by $N+1$ , and similarly multiply equation (2) by $16-N$ . Then add the equations together, canceling $A$ , leaving equation $43=50-N$ . From this we get $N=7$ and $A= \boxed{28}.$

~lprado

Solution 2

As shown above, we get the equation $\frac{55-A}{16-N} = 3.$ Rearranging we get $\frac{55-A}{3} = 16-N.$ Therefore, since N must be an integer, $55-A$ must be divisible by 3. Plugging in values gets us $A= \boxed{28}.$

Solution 3

Another way is to say that there are S students and T teachers and Ash's age is A. We can create the equation $(1) S+T=15$ . We can also create the equations $(2) \frac{12S+A}{S+1}=14$ and $(3) \frac{55T+A}{T+1}=52$ Equation (2) simplifies to $A=2S+14$ and equation (3) simplifies to $A=-3T+52$ . Multiply (2) by $3$ to get equation $(4) 3A=6S+42$ and (3) by $-2$ to get $(5) -2A=6T-104$ Add (4) and (5) to get $A=6(S+T)-62$ . After substituting equation (1) and simplifying, you get $A=\boxed{28}$ or answer choice $\boxed{A}$ .

~Champions247

Chinese Video Solution

https://www.bilibili.com/video/BV18V2uBtEHt/

~metrixgo

Video Solution (Very Fast and Intuitive)

https://youtu.be/ZqswJsf2Odo?si=c9Nx3kD657-9tWcY ~ Pi Academy

Video Solution

https://youtu.be/QBn439idcPo?si=z3_v6d_ZmpZGbm4N&t=243

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/_LKnEMhTAu4

~ Education, the Study of Everything

2025 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
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2025 AMC 12A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

Chinese Video Solution

Video Solution (Very Fast and Intuitive)

Video Solution

Video Solution by SpreadTheMathLove

Video Solution

Video Solution

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