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Difference between revisions of "2001 AIME II Problems/Problem 2"
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== Problem ==
 
== Problem ==
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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>.  
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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 ==
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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>fish</math> be the number of students who study both.
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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 \cap 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]],
  
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<math>S+F-fish=100</math>
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<cmath>S+F- S \cap F = S \cup F = 2001 </cmath>
  
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For m to be the smallest, S and F must also be the smallest.
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For <math>m = S \cap F</math> to be smallest, <math>S</math> and <math>F</math> must be minimized.
  
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<math>80+30-fish=100</math>
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<cmath>1601 + 601 - m = 2001 \Longrightarrow m = 201</cmath>
  
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<math>fish=0=m</math>
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For <math>M = S \cap F</math> to be largest, <math>S</math> and <math>F</math> must be maximized.
  
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For m to be the largest, S and F must also be the largest.
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<cmath>1700 + 800 - M = 2001 \Longrightarrow M = 499</cmath>
  
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<math>85+40-fish=100</math>
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Therefore, the answer is <math>M - m = 499 - 201 = \boxed{298}</math>.
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<math>fish=25=M</math>
 
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Therefore, the smallest number of students is 0, and the largest number of students is <math>\lfloor \frac{2001}{4}\rfloor =500</math>. <math>500-0=\boxed{500}</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}}
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[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 23:54, 9 January 2024

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 $S \cap F$ 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

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