Difference between revisions of "2015 AIME II Problems/Problem 2"
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==Solution 2== | ==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 <math>\dfrac{24}{76}</math>. Simplifying, we get <math>\dfrac{6}{19}</math>. Adding, we get <math>\boxed{025}</math> | + | 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 <math>\dfrac{24}{76}</math>. Simplifying, we get <math>\dfrac{6}{19}</math>. Adding, we get <math>\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 20:36, 15 September 2021
Contents
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 1
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 
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 
. Simplifying, we get 
. Adding, we get 
.
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. 
