Art of Problem Solving

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

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~ruihl123
~ruihl123
==Solution 3 (Guessing)==
If you don’t have time and you really need points, notice that Ash has a name that’s starts with A. So the answer is <cmath>\boxed{\text{(A)}}</cmath>
Note: This is a horrible solution and I highly recommend you learn how to solve this problem intentionally. This is just a last ditch effort.
~Aeioujyot


==Video Solution by SpreadTheMathLove==
==Video Solution by SpreadTheMathLove==

Revision as of 20:09, 7 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

Solution 3 (Guessing)

If you don’t have time and you really need points, notice that Ash has a name that’s starts with A. So the answer is \[\boxed{\text{(A)}}\]

Note: This is a horrible solution and I highly recommend you learn how to solve this problem intentionally. This is just a last ditch effort.

~Aeioujyot

Video Solution by SpreadTheMathLove

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

Chinese Video Solution

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

~metrixgo

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

See Also

2025 AMC 10A (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America.