Difference between revisions of "2001 AIME II Problems/Problem 2"
m (→See also) |
|||
| (3 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == 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 <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>. | + | 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 == | ||
{{AIME box|year=2001|n=II|num-b=1|num-a=3}} | {{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 
students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between 
percent and 
percent of the school population, and the number who study French is between 
percent and 
percent. Let 
be the smallest number of students who could study both languages, and let 
be the largest number of students who could study both languages. Find 
.
Solution
Let 
be the percent of people who study Spanish, 
be the number of people who study French, and let 
be the number of students who study both. Then 
, and 
. By the Principle of Inclusion-Exclusion,
![\[S+F- S \cap F = S \cup F = 2001\]](http://latex.artofproblemsolving.com/d/f/f/dfffe8b65ca37116896e9856176dce5699be5e96.png)
For 
to be smallest, and must be minimized.
![\[1601 + 601 - m = 2001 \Longrightarrow m = 201\]](http://latex.artofproblemsolving.com/5/2/1/521cb1d0f3bae37b6e1849cdd217e8d8c9b3ef21.png)
For 
to be largest, and must be maximized.
![\[1700 + 800 - M = 2001 \Longrightarrow M = 499\]](http://latex.artofproblemsolving.com/6/9/3/6939ec2d19f43a3df574f2f09431fd3a2df1acc8.png)
Therefore, the answer is 
.
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. 
