A certain company has the following policy regarding membership in committees A, B, C: each of the committees have exactly...
GMAT Two Part Analysis : (TPA) Questions
A certain company has the following policy regarding membership in committees \(\mathrm{A, B, C}\): each of the committees have exactly 40 members, and each employee must be a member of at least one of the committees. The company currently has 110 employees. Let \(\mathrm{X}\) be the current number of employees who are members of more than one of the committees.
Based on the current total number of employees, select the least value of X that is compatible with the policy and select the greatest value of X that is compatible with the policy. Make only 2 selections one in each column.
Let me solve this step by step.
Given information:
- 3 committees with 40 members each
- Total positions: \(3 \times 40 = 120\)
- Total employees: 110
- Some employees must be on multiple committees
Let's define variables:
- \(\mathrm{n_1}\) = employees on exactly 1 committee
- \(\mathrm{n_2}\) = employees on exactly 2 committees
- \(\mathrm{n_3}\) = employees on exactly 3 committees
Constraints:
- \(\mathrm{n_1} + \mathrm{n_2} + \mathrm{n_3} = 110\) (total employees)
- \(\mathrm{n_1} + 2\mathrm{n_2} + 3\mathrm{n_3} = 120\) (total positions)
Subtracting the first equation from the second:
\(\mathrm{n_2} + 2\mathrm{n_3} = 10\)
We want to find the range of \(\mathrm{X} = \mathrm{n_2} + \mathrm{n_3}\) (employees on more than one committee).
For minimum X:
To minimize the number of people on multiple committees, maximize those on 3 committees.
If \(\mathrm{n_3} = 5\), then \(\mathrm{n_2} = 0\)
So \(\mathrm{X_{min}} = 0 + 5 = 5\)
For maximum X:
To maximize the number of people on multiple committees, maximize those on exactly 2 committees.
If \(\mathrm{n_2} = 10\), then \(\mathrm{n_3} = 0\)
So \(\mathrm{X_{max}} = 10 + 0 = 10\)
Therefore: Least value of \(\mathrm{X} = 5\), Greatest value of \(\mathrm{X} = 10\)