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

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Isolating <imath>s</imath> and <imath>t,</imath> we have <imath>2s=a-14</imath> and <imath>3t=52-a.</imath> We also know that <imath>6s+6t=90,</imath> so substituting, we have <imath>3a-42+104-2a=90.</imath> Therefore, <imath>a=\boxed{\text{(A) 28}}.</imath>
Isolating <imath>s</imath> and <imath>t,</imath> we have <imath>2s=a-14</imath> and <imath>3t=52-a.</imath> We also know that <imath>6s+6t=90,</imath> so substituting, we have <imath>3a-42+104-2a=90.</imath> Therefore, <imath>a=\boxed{\text{(A) 28}}.</imath>
- harshu13
- harshu13
==Video Solution by Daily Dose of Math==
https://youtu.be/LN5ofIcs1kY
~Thesmartgreekmathdude


==See Also==
==See Also==

Revision as of 17:01, 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?

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

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

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

See Also

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

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