1. Introduction
  2. Circular arrangement
  3. Step 1: Draw the diagrams
  4. Step 2: Run the Checklist
  5. Step 3: Proceed with questions

Introduction

There are 4 types of ‘Logical-Arrangement’ questions

  1. Circular : Sitting In a Round Table
  2. Linear : Sitting in rectangular table or row
  3. Traits: List of people given, A lives in Hyderabad and prefers T  and owns Maruti etc.
  4. Family trees: A is the father of B who is mother Office…

These problems are quite easy to solve but can be very time-consuming if you donot proceed in a proper manner.

Circular arrangement

It is the easiest of all arrangement questions.
However one point must be kept in mind: “A is sitting adjacent to B”= Either A is on left side of B OR A is on right side of B. In short, you’ve to consider multiple possible cases during the calculation.
Let’s try a question

6 people A to F are sitting in a round table. A is not adjacent to B or C; D is not adjacent to C or E; B and C are adjacent; F is in the middle of D and C.

Step 1: Draw the diagrams

It is compulsory to draw diagram. You can’t solve the ‘circular-arrangement’ with drawing diagrams.
Ok, Lot of people are not sitting adjacent to eachother but we can’t use those “negatives” as starting-point to draw a diagram.
Better take the positive statement “B and C are adjacent” as our starting point and proceed.
There are two possible cases

  • Either C is on the right side of B Or
  • C is on the left side of B.

aptitude CSAT CMAT CAT IBPSaptitude CSAT CMAT CAT IBPS
Now use another positive statement: “F is in the middle of D and C.” Fill this in given cases
New diagrams look like this
aptitude CSAT CMAT CAT IBPSaptitude CSAT CMAT CAT IBPS

Now deal with the negative statements:”A is not adjacent to B or C”
Consider case#1: There are only two seats left: skyblue and orange colored.
But A cannot be seated in skyblue chair, because question statement says “A is not adjacent to B or C”
Therefore A can be seated only in orange chair for case #1 and
Similarly for case #2, A can be seated only in Red chair.
After A is seated, there is only one seat left and only one man (E) is left so we place him in the remaining seat.
aptitude CSAT CMAT CAT IBPSaptitude CSAT CMAT CAT IBPS
Alright all seats are now occupied, which means we’ve to run the checklist. IT serves two purposes

  • Incase you made any mistake in arrangement, you’ll find out at this stage OR
  • In One of the two cases, all checklist will not ‘pass’ thus we can eliminate one possible case and will be left with final perfect correct arrangement.

Step 2: Run the Checklist

Checklist Case 1 Case 2
A is not adjacent to B or C Pass Pass
D is not adjacent to C or E Pass Pass
B and C are adjacent Pass Pass
F is in the middle of D and C. Pass Pass

Damn it, both cases are passing the checklist. Time for some questions

Step 3: Proceed with questions

Q1. If one neighbor of A is D, who is the other one?

aptitude CSAT CMAT CAT IBPSaptitude CSAT CMAT CAT IBPS
Both cases are valid and in both cases A is sitting between D and E.
Therefore if one neighbor of A is D, then other is E. We get same neighbor “E” in both cases.

Ans. E is the other neighbor.

Q2. Who is sitting on the left hand side of A?

Both case are valid and in first case, D is sitting on the left hand side of A.
But second case, E is sitting on the left hand side of A.
So we are not getting ‘one’ definite answer. It can be either D or E. So final answer is
Ans. No conclusion. Cannot be determined. Data insufficient. D or E.