Logical Puzzles questions with answers and concepts

Six people A, B, C, D, E, F live on six different floors of a building (1=ground, $6=top$). D lives on an even-numbered floor. B lives immediately above C. E lives two floors above A. F does not live on floor 2 or 6. C lives below D. Who lives on floor 5?

Step 1: D is on an even floor (2, 4, or 6). F is not on floor 2 or 6, so F must be on 1, 3, 4, or 5.
Step 2: B lives immediately above C, so (C, B) occupy consecutive floors (C on floor k, B on floor k+1).
Step 3: E lives two floors above A, so floor(E) - floor(A) = 2 (positions like 1-3, 2-4, 3-5, 4-6).
Step 4: C lives below D (floor(C) < floor(D)).
Step 5: Test $D=2$. Then C < 2 requires $C=1$, giving $B=2$. But floor 2 is already occupied by D. Contradiction.
Step 6: Test $D=4$. Since C < 4 and B is immediately above C, (C, B) could be (1, 2) or (2, 3).
- If (C, B) = (1, 2), remaining floors are 3, 5, 6 for A, E, F. E is two floors above A requires $A=3$, $E=5$. Then F must take the remaining floor 6, which contradicts clue 1 (F does not live on floor 6).
- If (C, B) = (2, 3), remaining floors are 1, 5, 6 for A, E, F. Two-floor gap between A and E requires $A=1$, $E=3$, but floor 3 is already occupied by B. Contradiction.
Step 7: Therefore D must be on floor 6.
Step 8: With $D=6$, remaining even floor for B/C/F leaves (C, B) possibilities below 6: (1, 2), (2, 3), (3, 4), or (4, 5).
- If (C, B) = (4, 5), remaining floors for A, E, F are 1, 2, 3. E two floors above A requires $A=1$, $E=3$, leaving $F=2$, which contradicts clue 1 (F != 2).
- If (C, B) = (2, 3), remaining floors are 1, 4, 5. No pair has a gap of 2 ($5-1=4$, $4-1=3$, $5-4=1$).
- If (C, B) = (3, 4), remaining floors are 1, 2, 5. No pair has a gap of 2.
- If (C, B) = (1, 2), remaining floors are 3, 4, 5. E two floors above A uniquely gives $A=3$, $E=5$. Then F occupies floor 4 (valid, since F != 2 and F != 6).
Step 9: Final floor assignment: Floor $1=C$, Floor $2=B$, Floor $3=A$, Floor $4=F$, Floor $5=E$, Floor $6=D$.
Answer: E lives on floor 5.

Five people P, Q, R, S, T sit in a row facing north. P sits at one of the ends. S sits second to the right of P. Q sits to the immediate left of T. R sits to the immediate left of S. Who sits in the middle?

Step 1: P is at position 1 or 5 (one of the ends).
Step 2: S is second to the right of P.
Case A: $P=1$, then $S=3$ (second to right means +2 positions).
Case B: $P=5$, then S would be at position 7 - impossible.
Step 3: So $P=1$, $S=3$.
Step 4: R is immediately left of S, so $R=2$.
Step 5: Current arrangement: $P=1$, $R=2$, $S=3$, and positions 4,5 remain for Q,T.
Step 6: Q is immediately left of T, so $Q=4$, $T=5$.
Final arrangement: P-R-S-Q-T at positions 1-2-3-4-5.
Answer: S sits in the middle (position 3).

Four friends W, X, Y, Z have different professions: Doctor, Engineer, Lawyer, Teacher. W is neither the Doctor nor the Lawyer. The Engineer is neither X nor Z. Y is neither the Lawyer nor the Teacher. X is the Teacher. Who is the Doctor?

Step 1: Create a 4x4 grid matching people (W, X, Y, Z) to professions (Doctor, Engineer, Lawyer, Teacher).
Step 2: X is the Teacher (given). We can cross out Teacher for W, Y, Z and cross out other professions for X.
Step 3: The Engineer is neither X nor Z, so the Engineer must be W or Y.
Step 4: W is neither the Doctor nor the Lawyer (given), nor the Teacher (X is Teacher). Therefore, W must be the Engineer.
Step 5: Since W is the Engineer, Y cannot be the Engineer. Y is neither the Lawyer nor the Teacher (given). Therefore, Y must be the Doctor.
Step 6: The remaining profession, Lawyer, must belong to Z.
Final assignment: W = Engineer, X = Teacher, Y = Doctor, Z = Lawyer.
Answer: Y is the Doctor.

Five people A, B, C, D and E sit around a circular table facing the centre. A is between B and C. D is to the immediate right of A. E is not adjacent to B. Who is to the immediate left of A?

A is between “C” and “D”. With D immediately right of A (so, “C” is to the right), C sits to the immediate left of A.

Six boxes numbered 1 to 6 are stacked vertically, one above another. Box 4 is below Box 2 but above Box 6. Box 1 is at the top. Box 3 is immediately above Box 6. Which box is at the bottom?

Step 1: Box 1 is at the top.
Step 2: Box 4 is below Box 2 but above Box 6, so the order is 2 above 4 above 6.
Step 3: Box 3 is immediately above Box 6, so 3 must be directly above 6.
Step 4: Thus 4 cannot be immediately above 6 if 3 is already there, so we interpret the order as 2 above 4 and 4 above the level of 6 overall, but Box 3 is directly on top of 6.
Step 5: The lowest definite box among the options is 6, but because 3 is immediately above 6, the bottom is 6.
Step 6: Among the given choices, the intended bottom box is 6.
Answer: 6

Five friends J, K, L, M and N occupy flats 1 to 5 on a floor, from left to right. K lives immediately left of L. J lives at one end. N does not live next to J. M lives to the right of L. Who lives in flat 3?

Step 1: K lives immediately left of L, so K-L are consecutive.
Step 2: J lives at one end.
Step 3: N does not live next to J.
Step 4: M lives to the right of L, so M is somewhere to the right of the K-L pair.
Step 5: If J is in flat 1, then N cannot be in flat 2. The pair K-L can be in flats 2-3 or 3-4. If K-L are in 2-3, M must be in 4 or 5 and N in the remaining slot, which works with L in flat 3.
Step 6: Hence flat 3 is occupied by L.
Answer: L