Q.7 P. Q, R, S, T, U, V, and W are seated around a circular table.
1. Sis seated opposite to W.
II. U is seated at the second place to the right of R.
III. T is seated at the third place to the left of R.
IV. V is a neighbour of S.
Which of the following must be true?
(A) Pis a neighbour of R.
(B) Q is a neighbour of R.
(C) P is not seated opposite to Q.
(D) R is the left neighbour of s.

Solving the Circular Seating Arrangement Puzzle: Key Insights and Must-Be-True Statements

Circular seating arrangement puzzles challenge logical reasoning skills, often appearing in competitive exams like bank PO, SSC, and UPSC. This guide breaks down a classic 7-person puzzle involving P, Q, R, S, T, U, V, and W seated around a circular table. We’ll decode the clues, determine the arrangement, evaluate all options, and identify the correct answer with step-by-step reasoning.

Puzzle Clues and Arrangement Breakdown

Seven people—P, Q, R, S, T, U, V, and W—sit around a circular table. In such puzzles:

• Positions are relative (fix one person to avoid rotational symmetry).

• “Right” means clockwise; “left” means anticlockwise (standard convention).

• Opposite means 4 seats apart (since 7 is odd: positions 1 opposite 5, 2 opposite 6, etc.).

Clues:

1. S is seated opposite W.

2. U is seated at the second place to the right of R.

3. T is seated at the third place to the left of R.

4. V is a neighbour of S.

Let’s fix R at position 1 for clarity. Number positions 1 to 7 clockwise.

• Clue 2: U is second right of R → U at position 3 (right of 1: pos 2, then 3).

• Clue 3: T is third left of R → From 1 left (anticlockwise): pos 7, 6, 5 → T at 5.

Positions so far:

• 1: R

• 3: U

• 5: T

Remaining seats: 2, 4, 6, 7. Remaining people: P, Q, S, V, W.

• Clue 1: S opposite W. Possible opposite pairs (with current fixed):

  Position	Opposite
  1 (R)	5 (T)
  2	6
  3 (U)	7
  4	1 (R)

  Valid pairs for S-W: (2,6) or (6,2); (3U,7) invalid (U fixed); (4,1R) invalid; (7,3U) invalid.

• Clue 4: V neighbors S.

Case 1: S at 2, W at 6

• Pos 2: S

• Pos 6: W

• V neighbors S (pos 2) → V at 1 (R—no) or 3 (U—no). Impossible.

Case 2: S at 6, W at 2

• Pos 6: S

• Pos 2: W

• V neighbors S (pos 6) → Neighbors: 5 (T—no), 7 (open).

• So V must be at 7.

Now positions:

• 1: R

• 2: W

• 3: U

• 5: T

• 6: S

• 7: V

• 4: Only P or Q left? Wait, Q and P remain—one seat.

Remaining: P and Q for pos 4. But only one seat left (4). 7 people, all positions should fill. Count: R,W,U,T,S,V— that’s 6. Q and P remain, but seats 1-7: 1R,2W,3U,4?,5T,6S,7V. Yes, pos 4 gets one, but two left? Error in count.

Recount people: P,Q,R,S,T,U,V,W—8 people? No, query says P, Q, R, S, T, U, V, and W—yes 8? Wait: