# 2015 AIME II Problems/Problem 2

## Problem

In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution 1

We see that $40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%$ of students are learning Latin. In addition, $30\% \cdot 80\% = 24\%$ of students are sophomores learning Latin. Thus, our desired probability is $\dfrac{24}{76}=\dfrac{6}{19}$ and our answer is $6+19=\boxed{025}$

## Solution 2

Assume that there are 100 students in the school. There are 40 freshman taking Latin, 24 sophomores taking Latin, 10 juniors taking Latin, and 2 seniors taking Latin. We get the probability to be the number of sophomores taking Latin over the total number of students taking Latin, or $\dfrac{24}{76}$. Simplifying, we get $\dfrac{6}{19}$. Adding, we get $\boxed{025}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 