Introduction / Context:
This Data Sufficiency puzzle assigns five distinct persons to floors 1–5 (ground is empty). We must determine exactly who is on floor 3 using the interplay of adjacency and parity constraints.
Given Data / Assumptions:
- Floors available: 1, 2, 3, 4, 5.
- I: C is on an even floor above D; B is immediately above A.
- II: D is on an odd floor. A and B are immediate neighbours; C and E are immediate neighbours; C is not on an odd floor (so C is on 2 or 4).
Concept / Approach:
We test I and II separately for uniqueness and then combine. Use parity (odd/even) and immediate-neighbour constraints to reduce possibilities. ‘‘Above’’ means higher floor number.
Step-by-Step Solution:
1) From I alone: C on an even floor above D implies possibilities such as (D=1,C=2), (D=2,C=4), (D=3,C=4). Also B immediately above A implies a vertical pair (A,B) occupying consecutive floors. Many arrangements satisfy I with different occupants on the 3rd floor, so I alone is insufficient.2) From II alone: D is odd (1,3,5). A–B form a consecutive pair; C–E form a consecutive pair; C is even (2 or 4). Balancing two consecutive pairs across 5 floors yields multiple layouts (e.g., C–E on 2–3 or 4–5; A–B on an adjacent pair elsewhere), so II alone is insufficient.3) Combine I + II: Since C is even (II) and above D (I), valid (D,C) pairs are (1,2) or (3,4). If (D=3, C=4), then floor 4 is C and floor 3 is D. A–B must be adjacent with B above A (I); C–E are adjacent (II). Testing placements quickly shows the only consistent packing across five floors that honours all adjacency and parity constraints leads to a unique occupant on floor 3. One valid resolution is: D=1, C=2 (satisfy ‘‘C even and above D’’), then set A=3, B=4 (‘‘B above A’’), and E=5 while honouring ‘‘C and E are neighbours’’ fails; hence adjust to D=3, C=4 and pack A–B on 1–2 (with B above A ⇒ A=1, B=2) and place E=5. All constraints now hold and uniquely give D on floor 3.
Verification / Alternative check:
Attempting other distributions either violates adjacency (C–E or A–B), parity (C even), or order (C above D). The consistent joint solution fixes the 3rd floor occupant uniquely as D.
Why Other Options Are Wrong:
- A/B/C: Neither I nor II alone yields a unique answer; each allows multiple configurations.
- D: Together they are sufficient, not insufficient.
Common Pitfalls:
Misreading “immediately above” as ‘‘somewhere above’’; ignoring that two separate neighbour-pairs need disjoint consecutive slots across only five floors; overlooking the even-floor constraint for C.
Final Answer:
Both statements together are sufficient, but NEITHER alone is sufficient.
Discussion & Comments