Question 1

Six people—A, B, C, D, E, and F—live in a building with 6 floors numbered 1 to 6 from bottom to top. Each person lives on a different floor. Based on the following clues, who lives on Floor 3?

Clue 1: A lives on an odd-numbered floor.
Clue 2: B lives immediately above C.
Clue 3: D lives on Floor 5.
Clue 4: E lives on Floor 1.
Clue 5: F lives on an even-numbered floor below D.
Clue 6: C lives on Floor 2.

Accepted Answer

From Clue 6: C is on Floor 2. From Clue 2: B is immediately above C, so B is on Floor 3. From Clue 3: D is on Floor 5. From Clue 5: F is on an even floor below Floor 5, so F is on Floor 4 (since Floor 2 is taken by C). From Clue 1: A is on an odd floor. Remaining odd floors are 1 and 3. From Clue 4: E is on Floor 1. Therefore, A must be on Floor 3. Wait—B is on Floor 3. Let me recalculate: E on Floor 1, C on Floor 2, B on Floor 3, F on Floor 4, D on Floor 5, A on Floor 6. But Clue 1 says A is on an odd floor. Contradiction. Re-reading: A must be odd. Floors 1, 3, 5 are odd. E is on 1, D is on 5. So A is on 3. But B is on 3 (immediately above C on 2). This means the puzzle requires A on an odd floor not yet assigned. Since E=1, D=5, and B=3, A cannot satisfy Clue 1. Let me reinterpret: Perhaps C is not on Floor 2. Testing: If C is on Floor 4, B is on Floor 5—but D is on Floor 5. Contradiction. If C is on Floor 3, B is on Floor 4. E on 1, C on 3, B on 4, D on 5, F on 2 (even, below 5), A on 6. But A must be odd. If C on Floor 1—but E is on 1. Contradiction. Correct solution: E on 1, C on 2, B on 3, F on 4, D on 5, A on 6 violates Clue 1. Recheck clues: A on odd means A ∈ {1,3,5}. E=1, so A ∈ {3,5}. D=5, so A=3. But B=3. **The puzzle as stated has no solution.** Correcting: Assume Clue 1 is 'A lives on an even-numbered floor.' Then: E=1, C=2, B=3, F=4, D=5, A=6. **Answer: B lives on Floor 3.**

Question 2

Five friends—P, Q, R, S, and T—live on five different floors of a building (Floors 1–5, bottom to top). Each person lives on exactly one floor.

Clue 1: P lives on Floor 2.
Clue 2: Q lives on a floor higher than P.
Clue 3: R lives on Floor 4.
Clue 4: S lives immediately below T.
Clue 5: T is not on Floor 5.

Who lives on Floor 5?

Accepted Answer

From Clue 1: P is on Floor 2. From Clue 3: R is on Floor 4. From Clue 4: S is immediately below T, so (S, T) occupy consecutive floors. From Clue 5: T is not on Floor 5, so T ∈ {1, 2, 3, 4, 5} but not 5. Since P=2 and R=4, T cannot be 2 or 4. So T ∈ {1, 3}. If T=1, then S=0 (impossible). If T=3, then S=2—but P=2. Contradiction. Let me reconsider: T ∈ {1, 3, 5} but Clue 5 says T ≠ 5. So T ∈ {1, 3}. If T=3, S=2 (conflict with P). If T=1, S=0 (impossible). Hmm, this puzzle also seems unsolvable as stated. Let me reinterpret Clue 5: Perhaps it means T is not the only person on Floor 5 (unlikely). Or Clue 4 is 'S lives immediately above T' instead. Testing: S immediately above T. If T=1, S=2 (conflict with P). If T=3, S=4 (conflict with R). If T=4, S=5. But R=4. Contradiction. Let me assume Clue 1 is 'P lives on Floor 1' instead. Then: P=1, R=4. S immediately below T, T ≠ 5. If T=3, S=2. If T=2, S=1 (conflict with P). If T=5, violates Clue 5. So T=3, S=2. Remaining: Q on Floor 5. From Clue 2: Q > P (1), so Q=5 works. **Answer: Q lives on Floor 5.**

Question 3

Six employees—M, N, O, P, Q, and R—work on six different floors of an office building (Floors 1–6, bottom to top). Each employee is on exactly one floor.

Clue 1: M is on Floor 1.
Clue 2: O is on Floor 6.
Clue 3: N is on an even-numbered floor.
Clue 4: P is on a floor higher than M but lower than O.
Clue 5: Q is on Floor 3.
Clue 6: R is on a floor higher than N.

Who lives on Floor 4?

Accepted Answer

From Clue 1: M on Floor 1. From Clue 2: O on Floor 6. From Clue 5: Q on Floor 3. From Clue 3: N on even floor ∈ {2, 4, 6}. Since O=6, N ∈ {2, 4}. From Clue 4: P ∈ {2, 3, 4, 5} (between M=1 and O=6). Since Q=3, P ∈ {2, 4, 5}. From Clue 6: R > N. If N=2, R ∈ {3, 4, 5, 6}. If N=4, R ∈ {5, 6}. Case 1: N=2. Then R ∈ {3, 4, 5, 6}. Since Q=3, O=6, R ∈ {4, 5}. And P ∈ {2, 4, 5}. Since N=2, P ∈ {4, 5}. If P=4, R=5. If P=5, R=4. Both work. Case 2: N=4. Then R ∈ {5, 6}. Since O=6, R=5. And P ∈ {2, 4, 5}. Since N=4, R=5, P=2. All three cases are valid without additional constraints. However, re-reading: the puzzle should have a unique solution. Let me assume Clue 6 is 'R is on a floor higher than N and lower than O.' Then R ∈ {N+1, ..., 5}. Case 1: N=2, R ∈ {3, 4, 5}. Since Q=3, R ∈ {4, 5}. Case 2: N=4, R=5. Testing Case 1 with R=4: P ∈ {2, 4, 5}. Since N=2, R=4, P=5. Floors: M=1, N=2, Q=3, R=4, P=5, O=6. **Answer: R lives on Floor 4.**

Question 4

Four people—A, B, C, and D—live on four different floors of a building (Floors 1–4, bottom to top).

Clue 1: A does not live on Floor 1.
Clue 2: B lives on Floor 2.
Clue 3: C lives on a floor higher than B.
Clue 4: D lives on the floor immediately below C.

Who lives on Floor 1?

Accepted Answer

From Clue 2: B on Floor 2. From Clue 3: C > 2, so C ∈ {3, 4}. From Clue 4: D is immediately below C. If C=3, D=2. But B=2. Contradiction. If C=4, D=3. Then A is on the remaining floor: Floor 1. From Clue 1: A ≠ 1. Contradiction. This puzzle is unsolvable as stated. Let me reinterpret Clue 1 as 'A does not live on Floor 4.' Then: B=2, C ∈ {3, 4}, D immediately below C. If C=3, D=2 (conflict with B). If C=4, D=3. Remaining: A on Floor 1. Clue 1 (A ≠ 4) is satisfied. **Answer: A lives on Floor 1.**

Question 5

Five people—V, W, X, Y, and Z—live on five different floors (Floors 1–5, bottom to top).

Clue 1: V lives on Floor 3.
Clue 2: W lives on a floor lower than V.
Clue 3: X lives on a floor higher than V.
Clue 4: Y lives on Floor 5.
Clue 5: Z lives on a floor lower than X but higher than V.

Who lives on Floor 4?

Accepted Answer

From Clue 1: V on Floor 3. From Clue 4: Y on Floor 5. From Clue 2: W < 3, so W ∈ {1, 2}. From Clue 3: X > 3, so X ∈ {4, 5}. Since Y=5, X=4. From Clue 5: V < Z < X, so 3 < Z < 4. But there is no integer between 3 and 4. Contradiction. Let me reinterpret Clue 5 as 'Z lives on a floor lower than X and higher than or equal to V.' Then 3 ≤ Z < 4, so Z=3. But V=3. Contradiction. Alternatively, Clue 5 as 'Z lives on a floor lower than X and higher than W.' Then W < Z < X. W ∈ {1, 2}, X=4. If W=1, Z ∈ {2, 3}. Since V=3, Z=2. If W=2, Z=3. But V=3. So W=1, Z=2. Remaining: Y=5, V=3, X=4, Z=2, W=1. **Answer: X lives on Floor 4.**

Question 6

Seven people — A, B, C, D, E, F, and G — live on different floors of a 7-floor building (floors numbered 1 to 7 from bottom to top). Based on the following clues, determine who lives on Floor 4.

Clue 1: A lives on an odd-numbered floor.
Clue 2: B lives immediately above C.
Clue 3: D lives on Floor 2.
Clue 4: E lives on Floor 6.
Clue 5: F lives on a floor lower than G.
Clue 6: G lives on Floor 5.
Clue 7: C lives on Floor 3.

Who lives on Floor 4?

Accepted Answer

From Clue 7: C is on Floor 3. From Clue 2: B is immediately above C, so B is on Floor 4. From Clue 3: D is on Floor 2. From Clue 6: G is on Floor 5. From Clue 4: E is on Floor 6. From Clue 5: F lives lower than G (Floor 5), and the only remaining floor is Floor 1 or 7. Since Floors 2, 3, 4, 5, 6 are occupied by D, C, B, G, E respectively, F and A occupy Floors 1 and 7. From Clue 1: A lives on an odd floor. Both 1 and 7 are odd, but since F must be lower than G (Floor 5), F must be on Floor 1, leaving A on Floor 7. Therefore, B lives on Floor 4.

IBPS RRB PO Floor-Based Puzzle

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Key Points to Remember

Exam-Specific Tips

Floor-Based Puzzle — Practice Questions

60-Second Revision — Floor-Based Puzzle