2025 AMC 12A Problems/Problem 3: Difference between revisions

Revision as of 11:58, 8 November 2025

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 Ashley'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)&(5) to get $A=6(S+T)-62$ . After substituting equation (1) and simplifying, you get $A=28$ .

Video Solution (Intuitive, Quick Explanation!)

https://youtu.be/_LKnEMhTAu4

~ Education, the Study of Everything

@@ Line 24: / Line 24: @@
 As shown above, we get the equation <cmath> \frac{55-A}{16-N} = 3.</cmath> Rearranging we get <cmath> \frac{55-A}{3} = 16-N.</cmath> Therefore, since N must be an integer, <imath>55-A</imath> must be divisible by 3. Plugging in values gets us <imath>A= \boxed{28}.</imath>
+==Solution 3==
+Another way is to say that there are S students and T teachers and Ashley's age is A. We can create the equation (1) <imath>S+T=15</imath>. We can also create the equations (2) <imath>\frac{12S+A}{S+1}=14</imath> and (3) <imath>\frac{55T+A}{T+1}=52</imath>. Equation (2) simplifies to <imath>A=2S+14</imath> and equation (3) simplifies to <imath>A=-3T+52</imath>. Multiply (2) by <imath>3</imath> to get equation (4) <imath>3A=6S+42</imath> and (3) by <imath>-2</imath> to get (5) <imath>-2A=6T-104</imath>. Add (4)&(5) to get <imath>A=6(S+T)-62</imath>. After substituting equation (1) and simplifying, you get <imath>A=28</imath>.
 == Video Solution (Intuitive, Quick Explanation!) ==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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