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

Latest revision as of 13:20, 28 June 2023

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 1

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 $6+19=\boxed{025}$ .

Solution 2

Assume that there are $100$ students in the school. There are $40$ freshmen 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 $\dfrac{24}{76}$ . Simplifying, we get $\dfrac{6}{19}$ . Adding, we get $\boxed{025}$ .

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=74s

~MathProblemSolvingSkills.com

@@ Line 4: / Line 4: @@
 ==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
-==Solution 2==
+~MathProblemSolvingSkills.com
-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 ==
 {{AIME box|year=2015|n=II|num-b=1|num-a=3}}
 {{MAA Notice}}

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

Page

Toolbox

Search