Question

Which of the following is wrong?

• A. A will be to the immediate right of C
• B. D will be to the immediate left of B
• C. E will be to the immediate right of A
• D. F will be to the immediate left of D
• A. B will be to the immediate left of D
• B. H will be to the immediate right of A
• C. C will be to the immediate right of F
• D. B will be to the immediate left of H
• A. B agrees to change her sitting position
• B. C agrees to change her sitting position
• C. G agrees to change her sitting position
• D. H agrees to change her sitting position
• A. A will be directly facing C
• B. B will be directly facing C
• C. A will be directly facing B
• D. B will be directly facing D
• A. C and B
• B. A and F
• C. D and C
• D. E and G

Answer 1

The Correct Option is C

Step 1: Build the chain using “beside” constraints
A must be adjacent to C and H (a), (b). So A’s two neighbors are exactly \{C, H\}.
From (c), C must also sit beside E. Since C already uses one side for A, the other neighbor of C is E. Thus the local order is either A–C–E or E–C–A.
From (e), E sits beside G. E already sits next to C, therefore E’s other neighbor is G. Extend: A–C–E–G (in some direction).
From (g), G sits beside B. Hence G’s other neighbor is B. Extend: A–C–E–G–B.
From (h), B sits beside D. Therefore B’s other neighbor is D. Extend: A–C–E–G–B–D.
From (f), D sits beside F, so D’s other neighbor is F. Extend: …–B–D–F.
From (d), F sits beside H, thus F’s other neighbor is H. Because H must also be next to A (b), closing the circle gives the unique clockwise order:
\[ A – C – E – G – B – D – F – H – (back to A) \] Step 2: Fix “left” and “right” around a round table
In circular seating, everyone faces the center unless stated otherwise. A person’s immediate right is the seat anti-clockwise from them; the immediate left is the seat clockwise from them. (Stand at a seat and face the center—your right-hand neighbor sits anti-clockwise.)
Step 3: Test each option on the arrangement

(A) A to the immediate right of C: For C, the anti-clockwise neighbor is A (order … A–C–E …). Hence true.
(B) D to the immediate left of B: For B, the clockwise neighbor (left) is D (… G–B–D …). Hence true.
(C) E to the immediate right of A: For A, the anti-clockwise neighbor (right) is H (… F–H–A–C …), not E. Hence false.
(D) F to the immediate left of D: For D, the clockwise neighbor (left) is F (… B–D–F …). Hence true. Conclusion: Only statement (C) is wrong.

Answer 2

Recap of the unique circular order (from Q186 constraints)
Working through the “beside” conditions yields the only consistent clockwise seating:
\[ A – C – E – G – B – D – F – H – (back to A) \] Left/Right convention on round tables
All face the centre. A person’s immediate right is the seat anti-clockwise from them; the immediate left is the seat clockwise from them. Test each option against the arrangement

(A) B to the immediate left of D? For D, the clockwise neighbor (left) is F (order … B–D–F …). Not B. False.
(B) H to the immediate right of A? For A, the anti-clockwise neighbor (right) is H (… F–H–A–C …). True.
(C) C to the immediate right of F? For F, the anti-clockwise neighbor (right) is D (… D–F–H …). Not C. False.
(D) B to the immediate left of H? For H, the clockwise neighbor (left) is A (… F–H–A …). Not B. False. Conclusion: Only statement (B) is correct.

Answer 3

Established clockwise order (from Q186): A – C – E – G – B – D – F – H – (back to A).
Here, A’s neighbours are C and H; F’s neighbours are D and H. The only person sitting between A and F is H.
Therefore, unless H vacates that seat (i.e., H changes position), A and F cannot become adjacent. Changing the seats of B, C, or G does not remove H from between A and F. Hence only option (D) works.

Answer 4

With 8 seats, opposite seats are 4 places apart. Using the fixed order:
Indices (clockwise) $0$ A, $1$ C, $2$ E, $3$ G, $4$ B, $5$ D, $6$ F, $7$ H.
Opposites: A↔B (0↔4), C↔D (1↔5), E↔F (2↔6), G↔H (3↔7).
Thus A faces B, C faces D, etc. Only statement (C) matches this. Statements (A), (B), and (D) contradict the opposite pairs.

Answer 5

From the fixed arrangement A – C – E – G – B – D – F – H – A, the neighbours of H are F (on one side) and A (on the other). Hence H sits between A and F. Other pairs listed do not flank H in the circle.