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

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== Solution ==
 
== Solution ==
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{{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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<math>S+F-fish=100</math>
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For m to be the smallest, S and F must also be the smallest.
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<math>80+30-fish=100</math>
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<math>fish=0=m</math>
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For m to be the largest, S and F must also be the largest.
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<math>85+40-fish=100</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}}

Revision as of 16:25, 28 January 2008

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 $fish$ be the number of students who study both.

$S+F-fish=100$

For m to be the smallest, S and F must also be the smallest.

$80+30-fish=100$

$fish=0=m$

For m to be the largest, S and F must also be the largest.

$85+40-fish=100$

$fish=25=M$

Therefore, the smallest number of students is 0, and the largest number of students is $\lfloor \frac{2001}{4}\rfloor =500$. $500-0=\boxed{500}$

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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