Question

Difficulty: MediumAnalytical Puzzles and Grouping

Six executives—P, Q, R, S, T, and U—are to be assigned to two committees: Committee 1 and Committee 2, each consisting of exactly three members. The assignments must satisfy the following conditions:
• P and Q must be in the same committee.
• R and S cannot be in the same committee.
• T is assigned to Committee 1.
• If P is assigned to Committee 1, then U must be assigned to Committee 2.
• U and S must be in the same committee.

Which of the following executives must be assigned to Committee 2 along with P and Q?

  1. RAnswer
  2. B
    S
  3. C
    U
  4. D
    T

Answer

R
Since P and Q must be together in Committee 2, and U and S must be together in Committee 1 to respect the 3-member capacity per committee, R is the only remaining executive who can join P and Q in Committee 2 while satisfying the rule that R and S are in different committees.

Step-by-Step Solution

1
Analyze committee capacities and fixed placements
Committee 1 has T. Both Committee 1 and Committee 2 need exactly 3 members.
Total executives are 6, split equally into two groups of 3.
2
Evaluate placement of P and Q
If P and Q are in Committee 1, Committee 1 has {T, P, Q} (full). Then Committee 2 must have {R, S, U}. But R and S cannot be together, so P and Q cannot be in Committee 1. Thus, P and Q are in Committee 2.
Placing P and Q in Committee 1 forces R and S into Committee 2 together, violating the rule that R and S must be in different committees.
3
Determine placements for remaining members U, S, and R
Committee 2 currently has {P, Q} (2 members). Committee 1 currently has {T} (1 member). Since U and S must be together, they must go to Committee 1 to avoid overflowing Committee 2. Thus, Committee 1 = {T, U, S}. The remaining person R must go to Committee 2.
Putting U and S in Committee 2 would give Committee 2 four members ({P, Q, U, S}), which exceeds the limit of 3.

Key Concept

Multi-constraint grouping and deduction based on capacity limits
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