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 number of students who study Spanish but not French, let f be the number of students who study French but not Spanish, and let b be the number of students who study both languages. It is given that 1600.8<s+b<1700.85 and 600.3<f+b<800.4; thus 1601≤s+b≤1700 and 601≤f+b≤800. Add the last pair of inequalities to obtain 2202≤s+f+2b≤2500. Because s+f+b=2001, it follows that 201≤b≤499. The triples (s,f,b)=(1400,400,201) and (s,f,b)=(1201,301,499) show that m=201 and M=499, so M−m=298.
OR
By the Inclusion-Exclusion Principla,
M=⌊0.85(2001)⌋+⌊0.40(2001)⌋−2001=1700+800−2001=499
and
m=⌈0.80(2001)⌉+⌈0.30(2001)⌉−2001=1601+601−2001=201
so M−n=499−201=298.
The problems on this page are the property of the MAA's American Mathematics Competitions