Data Sufficiency — Floors (Who Lives on the Topmost, Floor 6?) Among P, Q, R, S, T, V each on a distinct floor (1 lowest … 6 highest), who lives on floor 6? I. There is exactly one floor between R and Q. P lives on an even-numbered floor. II. T does not live on an even floor. Q lives on an even floor and not on the topmost floor. III. S lives on an even floor. There are two floors between S and P. T lives immediately above R.

Introduction / Context:
We must pinpoint the topmost resident using vertical constraints. The original had three statements but binary DS options; we normalize to a standard three-statement sufficiency choice set.

Given Data / Assumptions:

• I: gap(R,Q)=1 (one floor between); P on even.
• II: T on odd; Q on even and not on 6.
• III: S on even; gap(S,P)=2; T immediately above R.

Concept / Approach:
Pairwise testing shows (II+III) tightly binds Q to even non-top, places T over R, and cross-filters even floors among P and S via the two-gap rule, collapsing to a unique stack and topmost occupant.

Step-by-Step Solution:

Verification / Alternative check:
Backtracking quickly shows only II+III prune to one solution; adding I is not necessary.

Why Other Options Are Wrong:
I+II or I+III allow multiple tops; all three needed is overkill; “even all three not sufficient” is false.

Common Pitfalls:
Forgetting floors are 1..6; misplacing “immediately above”.

Final Answer:
II and III together suffice; others do not.