NIMCET 2011 Logical Reasoning Question 70 with Solution

Question 70

Logical Reasoning Puzzles Easy

Questions 69 and 70 are based on the following : \( P, Q, R, S, T, U, V \) and \( W \) are sitting round the circle and are facing the centre. \( P \) is second to the right of \( T, T \) is the neighbour of \( R \) and \( V . S \) is not the neighbour of \( \mathrm{P}, \mathrm{V} \) is the neighbour of \( \mathrm{U}, \mathrm{Q} \) is not between \( S \) and \( W \) , and \( W \) is not between \( U \) and \( S \) . What is the position of \( S \) ?

(A) Between \( U \) and \( V \)

(B) Second to the right of \( P \)

(C) To the immediate right of W

(D) Data inadequate

View Dynamic Solution & Explanation ▼

Correct Solution: Option C

Step-by-step Solution:

Quick Solution

This is a circular seating arrangement puzzle. The best way to solve it is to place the friends one by one, starting with the most concrete clues. We will assume "right" means clockwise, as is common in these puzzles when people face the center.

1. Building the Core Arrangement

Clue 2 ("T is the neighbour of R and V") and Clue 1 ("P is second to the right of T") combine to give us a solid block of four people in a row: V-T-R-P (in clockwise order).
Clue 4 ("V is the neighbour of U") means U must sit next to V, extending our block to five people: U-V-T-R-P.

2. Placing the Final Friends and Finding the Order

The remaining people (Q, S, W) must sit in the three empty seats between P and U. A full analysis of the final clues reveals that the only consistent clockwise sequence for this section is P, Q, W, S, U.

This gives us the complete, final clockwise arrangement:

P → Q → W → S → U → V → T → R → (P)

3. Answering the Question

The question asks for the position of S. Let's look at our final arrangement:

In the sequence ...Q, W, S, U..., we can see that S is sitting immediately after W in the clockwise direction.

Therefore, S is to the immediate right of W.

Final Answer: The position of S is To the immediate right of W.