Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution 2 (What?))
 
(One intermediate revision by the same user not shown)
Line 16: Line 16:
  
 
Therefore, the answer is <math>M - m = 499 - 201 = \boxed{298}</math>.
 
Therefore, the answer is <math>M - m = 499 - 201 = \boxed{298}</math>.
 +
 +
=== Solution 2 (What?) ===
 +
I have no clue what is going on here. Let <math>x</math> percentage study only Spanish, <math>y</math> study  only French and <math>z</math> study both. We have: <cmath> 80 \leq x + z \leq 85 </cmath> <cmath> 30 \leq y + z \leq 40 </cmath> <cmath> x + y + z = 100 </cmath> <cmath> \implies 10 \leq z \leq 25 </cmath>
 +
Thus the desired answer should be <math>2001 \cdot (\frac{25}{100} - \frac{10}{100}) = 300.15</math>. 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 ==
 
== See also ==

Latest revision as of 04:30, 1 September 2025

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 only Spanish, $y$ study only 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_2&oldid=259490"