# Difference between revisions of "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

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 $6+19=\boxed{025}$

## See also

 2015 AIME II (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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

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