Problem Analysis:- Total chess club members = 22
- Total swim team members = 10
- Members of both clubs = 2
Calculating Exclusive Memberships:- Chess only members = Total chess - Both clubs = \(22 - 2 = 20\)
- Swim only members = Total swim - Both clubs = \(10 - 2 = 8\)
Total Members in Clubs:- Total club members = Chess only + Swim only + Both = \(20 + 8 + 2 = 30\)
Gender-based Team Membership and Ratios:- All females (F) are on the swim team.
- Number of males (M) is twice the number of females (M = 2F).
Total Population Breakdown:- Let
x be the number of students neither in chess nor swim.
- Total students = Club members + Neither = \(30 + x\)
Relationship of Total Males and Females to Club Membership:- Total swimmers (including both clubs) = 10, hence \(F \leq 10\)
- Using the ratio of males to females, \(M + F = 30 + x\) and \(M = 2F\), we substitute to find:
- \(2F + F = 30 + x \Rightarrow 3F = 30 + x\)
Determine Maximum Females on Swim Team:- Since \(3F = 30 + x\) and \(F \leq 10\), this implies \(3F \leq 30\), setting \(x + 30 \leq 30 \Rightarrow x \leq 0\).
Concluding the Value of x:- Given
x must be a non-negative integer and
x \leq 0, the only possibility is \(x = 0\).
Answer:- The number of students who are members of neither the chess club nor the swim team is
0.
Correct Answer: A. 0Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of PrizesOf all the students in a certain school, 22 are members of a chess club, 10 are members of a swim team, and 2 are members of both. If all females are members of the swim team and the number of males is twice the number of females, how many students are members of neither the chess club nor the swim team?
A. 0
B. 3
C. 6
D. 9
E. 12