2001 AIME II Problems/Problem 2

From AoPSWiki

Revision as of 02:12, 26 July 2008 by Azjps (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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,

$S+F- S \cup F = 2001$

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

$1601 + 601 - m = 2001 \Longrightarrow m = 201$

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

$1700 + 800 - M = 2001 \Longrightarrow M = 499$

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

2001 AIME II (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

2001 AIME II Problems/Problem 2

From AoPSWiki

Problem

Solution

See also