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AIME 2015 II · 第 2 题

AIME 2015 II — Problem 2

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

In a new school 4040 percent of the students are freshmen, 3030 percent are sophomores, 2020 percent are juniors, and 1010 percent are seniors. All freshmen are required to take Latin, and 8080 percent of the sophomores, 5050 percent of the juniors, and 2020 percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n.

解析

Solution 1

We see that 40%100%+30%80%+20%50%+10%20%=76%40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\% of students are learning Latin. In addition, 30%80%=24%30\% \cdot 80\% = 24\% of students are sophomores learning Latin. Thus, our desired probability is 2476=619\dfrac{24}{76}=\dfrac{6}{19} and our answer is 6+19=0256+19=\boxed{025}.

Solution 2

Assume that there are 100100 students in the school. There are 4040 freshmen taking Latin, 2424 sophomores taking Latin, 1010 juniors taking Latin, and 22 seniors taking Latin. We get the probability to be the number of sophomores taking Latin over the total number of students taking Latin, or 2476\dfrac{24}{76}. Simplifying, we get 619\dfrac{6}{19}. Adding, we get 025\boxed{025}.

Video Solution

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

~MathProblemSolvingSkills.com