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