Difference between revisions of "2015 AIME II Problems/Problem 2"
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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 percent of the students are freshmen, percent are sophomores, percent are juniors, and percent are seniors. All freshmen are required to take Latin, and percent of the sophomores, percent of the juniors, and percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is , where and are relatively prime positive integers. Find .
Solution
We see that of students are learning Latin. In addition, of students are sophomores learning Latin. Thus, our desired probability is and our answer is
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
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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