Sophia is selecting a planning committee for her schools annual talent show. The committee must consist of 12 students, with members from both the freshman and sophomore classes. At least two students must be from each class. Which of the following is a viable solution for this situation? (1 point)

Answer 1

To form a planning committee of 12 students with the requirement that at least two members must come from each class (freshmen and sophomores), we can denote the number of freshmen as \( F \) and the number of sophomores as \( S \).

We have the following constraints:

1. \( F + S = 12 \)
2. \( F \geq 2 \)
3. \( S \geq 2 \)

With these constraints, let's analyze the possible combinations of \( F \) and \( S \):

Starting with the minimum number from each class:

• If \( F = 2 \), then \( S = 10 \) (2 freshmen, 10 sophomores)
• If \( F = 3 \), then \( S = 9 \) (3 freshmen, 9 sophomores)
• If \( F = 4 \), then \( S = 8 \) (4 freshmen, 8 sophomores)
• If \( F = 5 \), then \( S = 7 \) (5 freshmen, 7 sophomores)
• If \( F = 6 \), then \( S = 6 \) (6 freshmen, 6 sophomores)
• If \( F = 7 \), then \( S = 5 \) (7 freshmen, 5 sophomores)
• If \( F = 8 \), then \( S = 4 \) (8 freshmen, 4 sophomores)
• If \( F = 9 \), then \( S = 3 \) (9 freshmen, 3 sophomores)
• If \( F = 10 \), then \( S = 2 \) (10 freshmen, 2 sophomores)

All of the above combinations satisfy the requirement of at least 2 students from each class while totaling 12 students.

A viable solution could be:

• 4 freshmen and 8 sophomores. This satisfies all the constraints:
• Total students = 4 + 8 = 12
• Freshmen (4) ≥ 2
• Sophomores (8) ≥ 2

Therefore, one viable solution for the committee composition is 4 freshmen and 8 sophomores.

Answer 2

• 6 freshmen and 4 sophomores
O 5 freshmen and 7 sophomores
O o freshmen and 12 sophomores
• 1 freshman and 11 sophomores

Answer 3

To determine which of the options provided meets the requirements that Sophia's committee must consist of 12 students with at least 2 students from each class (freshmen and sophomores), let’s evaluate each option:

1. 6 freshmen and 4 sophomores

   • Total = 6 + 4 = 10
   • This does not satisfy the requirement of 12 total students.

2. 5 freshmen and 7 sophomores

   • Total = 5 + 7 = 12
   • Freshmen (5) ≥ 2
   • Sophomores (7) ≥ 2
   • This option satisfies all the requirements.

3. 0 freshmen and 12 sophomores

   • Total = 0 + 12 = 12
   • Freshmen (0) < 2
   • This does not meet the requirement.

4. 1 freshman and 11 sophomores

   • Total = 1 + 11 = 12
   • Freshmen (1) < 2
   • This does not meet the requirement.

Based on the analysis:

• The only viable solution from the options given is 5 freshmen and 7 sophomores.