2025 AMC 10A Problems/Problem 4: Difference between revisions

Revision as of 23:34, 6 November 2025

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?

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

Video Solution

https://youtu.be/l1RY_C20Q2M

Solution 1

Let $s$ be the number of students, $t$ be the number of teachers, and $a$ be Ash's age. We know that $s+t=15.$ Also, the sum of the ages of the students is $12s$ and the sum of the ages of the teachers are $55t.$ Thus, we have the following equations: $\frac{12s+a}{s+1}=14$ $\frac{55t+a}{t+1}=52.$ Isolating $s$ and $t,$ we have $2s=a-14$ and $3t=52-a.$ We also know that $6s+6t=90,$ so substituting, we have $3a-42+104-2a=90.$ Therefore, $a=\boxed{\text{(A) 28}}.$ - harshu13

Solution 2

Let $s$ and $t$ represent the number of students and teachers respectively. If the average age of each of the students increases by 2 when Ash joins them, then Ash’s age in terms of $s$ is $2s+14$ . Similarly, Ash’s age is $52-3t$ in terms of $t$ . Equating these two equations gives $2s+3t=38$ , and $s+t=15$ is also clearly true. Solving the system yields $s=7$ and plugging this into $2s+14$ gives Ash's age, $\boxed{\textbf{(A) }28}$ .

~ruihl123

Video Solution (Fast and Easy)

https://youtu.be/ZqswJsf2Odo?si=jeXT21MqPOIEj_Rz ~ Pi Academy

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

@@ Line 4: / Line 4: @@
 <imath>\textbf{(A)}~28\qquad\textbf{(B)}~29\qquad\textbf{(C)}~30\qquad\textbf{(D)}~32\qquad\textbf{(E)}~33</imath>
+==Video Solution==
+https://youtu.be/l1RY_C20Q2M
 ==Solution 1==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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