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2001 AIME II Problems/Problem 2

Revision as of 04:29, 1 September 2025 by Rolt (talk | contribs) (→‎Solution)

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 $M - m = 499 - 201 = \boxed{298}$.

Solution 2 (What?)

I have no clue what is going on here. Let $x$ percentage study Spanish, $y$ study French and $z$ study both. We have: \[80 \leq x + z \leq 85\] \[30 \leq y + z \leq 40\] \[x + y + z = 100\] \[\implies 10 \leq z \leq 25\] Thus the desired answer should be $2001 \cdot (\frac{25}{100} - \frac{10}{100}) = 300.15$. That's a decimal, and this is an integer type contest. In search of an answer, the individual above has rounded upwards. I do not know if that was specified in the contest, because it was not specified in the problem.

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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