Difference between revisions of "2015 AIME II Problems/Problem 2"

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==Solution==
 
==Solution==
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We see that <math>40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%</math> of students are learning Latin. In addition, <math>30\% \cdot 80\% = 24\%</math> of students are sophomores learning Latin. Thus, our desired probability is <math>\dfrac{24}{76}=\dfrac{6}{19}</math> and our answer is <math>6+19=\boxed{025}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2015|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2015|n=II|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:54, 26 March 2015

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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