Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AIME II Problems/Problem 2"

 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Each of the <math>2001</math> students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between <math>80</math> percent and <math>85</math> percent of the school population, and the number who study French is between <math>30</math> percent and <math>40</math> percent. Let <math>m</math> be the smallest number of students who could study both languages, and let <math>M</math> be the largest number of students who could study both languages. Find <math>M-m</math>.
  
 
== Solution ==
 
== Solution ==
 +
Let <math>S</math> be the percent of people who study Spanish, <math>F</math> be the number of people who study French, and let <math>S \cup F</math> be the number of students who study both. Then <math>\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700</math>, and <math>\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800</math>. By the [[Principle of Inclusion-Exclusion]],
 +
 +
<cmath>S+F- S \cap F = S \cup F = 2001 </cmath>
 +
 +
For <math>m = S \cap F</math> to be smallest, <math>S</math> and <math>F</math> must be minimized.
 +
 +
<cmath>1601 + 601 - m = 2001 \Longrightarrow m = 201</cmath>
 +
 +
For <math>M = S \cap F</math> to be largest, <math>S</math> and <math>F</math> must be maximized.
 +
 +
<cmath>1700 + 800 - M = 2001 \Longrightarrow M = 499</cmath>
 +
 +
Therefore, the answer is <math>M - m = 499 - 201 = \boxed{298}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2001 AIME II Problems]]
+
{{AIME box|year=2001|n=II|num-b=1|num-a=3}}
 +
 
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Revision as of 22:52, 6 March 2014

Problem

Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$.

Solution

Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cup F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$, and $\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800$. By the Principle of Inclusion-Exclusion,

\[S+F- S \cap F = S \cup F = 2001\]

For $m = S \cap F$ to be smallest, $S$ and $F$ must be minimized.

\[1601 + 601 - m = 2001 \Longrightarrow m = 201\]

For $M = S \cap F$ to be largest, $S$ and $F$ must be maximized.

\[1700 + 800 - M = 2001 \Longrightarrow M = 499\]

Therefore, the answer is $M - m = 499 - 201 = \boxed{298}$.

See also

2001 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_2&oldid=60702"