Difference between revisions of "2015 AIME II Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
− | 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> | + | 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>. |
+ | ==Solution 2== | ||
+ | Assume that there are <math>100</math> students in the school. There are <math>40</math> freshmen taking Latin, <math>24</math> sophomores taking Latin, <math>10</math> juniors taking Latin, and <math>2</math> 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>. | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=9re2qLzOKWk&t=74s | ||
− | + | ~MathProblemSolvingSkills.com | |
− | |||
== 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}} |
Latest revision as of 14:20, 28 June 2023
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 students in the school. There are
freshmen taking Latin,
sophomores taking Latin,
juniors taking Latin, and
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
.
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=74s
~MathProblemSolvingSkills.com
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.