Difference between revisions of "2015 AIME II Problems"
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==Problem 2== | ==Problem 2== | ||
− | 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 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | + | In a new school, <math>40</math> percent of the students are freshmen, <math>30</math> percent are sophomores, <math>20</math> percent are juniors, and <math>10</math> percent are seniors. All freshmen are required to take Latin, and <math>80</math> percent of sophomores, <math>50</math> percent of the juniors, and <math>20</math> percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
[[2015 AIME II Problems/Problem 2 | Solution]] | [[2015 AIME II Problems/Problem 2 | Solution]] |
Revision as of 15:57, 26 March 2015
2015 AIME II (Answer Key) | AoPS Contest Collections • PDF | ||
Instructions
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1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Contents
Problem 1
Let be the least positive integer that is both percent less than one integer and percent greater than another integer. Find the remainder when is divided by .
Problem 2
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 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 .
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by 2014 AIME I, 2014 AIME II |
Followed by 2016 AIME | |
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.