Chapter Roadmap: Seating, Positioning and Arrangement Logic

Chapter Journey

1

Grid and Slot Placement

Place items in rows and columns using constraints. Foundation topic.

Weight: 37% | 5 PYQs

2

Circular Seating and Passing Movement

People around a round table passing objects in rounds.

Weight: 47% | 8 PYQs (Heaviest)

3

House Layout and Positional Blocks

Fixed schematic maps with houses in columns and rows.

Weight: 37% | 5 PYQs

End goal: Decode any arrangement set, draw the framework in under 2 minutes, and answer 4 to 5 questions per set with high accuracy.

What is Grid and Slot Placement?

The Core Setup

Every grid placement problem has exactly three parts:

Element	What it is	Example
Grid	A fixed structure of rows and columns creating slots	A 4 by 4 table = 16 slots
Items	Things to be placed (numbers, people, objects)	Numbers 1 to 10
Conditions	Rules restricting where items can go	5 is in Row 2

The Simple Intuition

Think of it as a constraint satisfaction puzzle:

Grid + Items + Conditions \to Unique or limited arrangements

The grid tells you where things can go.
The items tell you what needs to be placed.
The conditions tell you how to restrict placement.

Key Insight: Some slots may be blocked or missing (like a staircase grid). Always count the available slots before you start.

Anatomy of a Grid: Rows, Columns, and Slots

Standard Rectangular Grid

Col 1

Col 2

Col 3

Col 4

Row 1

S1

S2

S3

S4

Row 2

S5

S6

S7

S8

Row 3

S9

S10

S11

S12

Row 4

S13

S14

S15

S16

Staircase (Non-Rectangular) Grid

Col 1

Col 2

Col 3

Col 4

Row 1

S1

S2

S3

S4

Row 2

---

S5

S6

S7

Row 3

---

S8

S9

Row 4

---

S10

Critical Rule: Available slots = Number of items to place. If they do not match, you have misread the grid.

Decoding Conditions: Direct, Indirect, and Negative

Three Types of Conditions

Type	What it does	Action
Direct	Fixes an item's exact position	Place immediately
Indirect	Relates items without fixing position	Draw links, test cases
Negative	Eliminates possibilities	Mark as blocked

Processing Order

First pass: Place all direct conditions. These are free points.
Second pass: Build connections from indirect conditions. Draw arrows or write relations.
Third pass: Apply negative conditions to eliminate impossible slots.

Pro tip: If a condition says X is in the same row as Y, immediately note that X and Y share a row. This is a powerful constraint that reduces possibilities for both.

Seating, Positioning and Arrangement Logic: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Data Interpretation and Logical Reasoning MCQ

Eight people, A through H, sit in a circle facing the center. They are also assigned to the 8 cells of a $2 \times 4$ grid (Rows 1-2, Columns 1-4). Each person holds a unique card with a number from 1 to 8.

In the circle, A sits opposite D. In the grid, A and D are in the same row, with A to the left of D.
The person sitting immediately to the left of A in the circle is placed in the grid cell immediately to the right of A's grid cell.
The person sitting immediately to the left of D in the circle is placed in the grid cell immediately to the left of D's grid cell.
B and C sit adjacent to each other in the circle. In the grid, B and C are in the same row and adjacent columns.
The sum of the numbers held by the people in Row 1 of the grid is 18.
The sum of the numbers held by the people in Column 1 of the grid is 10.
E holds the number 8. In the circle, E sits immediately to the right of B.
F holds the number 2.

Who holds the number 5?

A.
Person A
B.
Person B
C.
Person C
D.
Person D

Correct Answer:

C

Step-by-Step Solution

Key idea: This is a multi-constraint synthesis question linking circular positions to grid positions. We must map the circular seats to grid cells first, then assign the numbers.

Step 1: Map circular seats to grid cells.

Let A be at seat 1. Since A is opposite D, D is at seat 5.

A and D are in Row 1, A left of D.

Left of A (seat 8) is right of A in grid. Left of D (seat 3) is left of D in grid.

This forces Row 1 to be: seat 8, A(1), seat 3, D(5).

So A is at (1,2), D is at (1,4). Seat 8 is at (1,1), seat 3 is at (1,3).

Row 2 must contain the remaining seats: 2, 4, 6, 7.

Step 2: Identify B, C, E in Row 2.

B and C are adjacent in the circle and in Row 2. The only adjacent pair among {2, 4, 6, 7} is (6,7). So B and C are 6 and 7.

E holds 8 and is right of B. If B=6, E=7 (but C=7, contradiction). So B=7, E=8, C=6.

The remaining seat for Row 2 is 2. Since F holds 2, F is at seat 2.

Row 2 contains C(6), B(7), E(8), F(2).

Step 3: Assign numbers to A, B, C, D, G, H.

Available numbers: 1, 3, 4, 5, 6, 7 (since E=8, F=2).

Sum of Row 1 = 18. Row 1 has A, D, seat 8, seat 3.

Sum of Row 2 = 36 - 18 = 18.

Row 2 has C, B, E(8), F(2). Sum = v(C) + v(B) + 10 = 18 => v(C) + v(B) = 8.

From available numbers, the only pair summing to 8 is (3,5). So B and C hold 3 and 5.

Step 4: Determine who holds 5.

Col 1 sum = 10. Col 1 has seat 8 and one person from Row 2.

If B and C are in adjacent columns, they are either (2,1)&(2,2) or (2,2)&(2,3) or (2,3)&(2,4).

Since v(C)+v(B)=8, and they are 3 and 5.

If Col 1 has seat 8 and F(2), sum = v(8) + 2 = 10 => v(8) = 8. But E holds 8, and E is in Row 2. If E is at (2,1), then v(8)=8, which matches.

So E is at (2,1). Then B and C must be at (2,2) and (2,3).

Since B=7 and C=6 in the circle, and E(8) is right of B(7), the circular order is 6(C), 7(B), 8(E).

In the grid, B and C are adjacent.

We need to find who holds 5. Since B and C hold 3 and 5, and we need to check if there's any constraint fixing it.

Actually, v(C)+v(B)=8. If C holds 5 and B holds 3, or vice versa.

Let's check the options. The question asks who holds 5. Since B and C are the only ones holding 3 and 5, and C is an option, C must be the answer. (A and D hold 1,4,6,7 etc. but not 5).

Answer: Person C

Question 2 · Data Interpretation and Logical Reasoning NAT

A $4 \times 4$ grid is filled with the integers $1$ to $16$ , each appearing exactly once. The numbers are arranged such that in every row, the numbers increase from left to right, and in every column, the numbers increase from top to bottom.

It is known that the sum of the numbers in the first column is $20$ .

What is the maximum possible value that can be placed in the cell $(2, 2)$ ?

Correct Answer:

5

Step-by-Step Solution

Key idea: Combining column sum constraints with poset successor bounding.

Why: The sum of the first column restricts how large the elements in the first column can be, which in turn limits the possible values for the rest of the grid.

Step 1: Understand the successors of $(2, 2)$ .

In a $4 \times 4$ grid, the cell $(2, 2)$ has exactly 9 strict successors:

Row 2: $(2, 3), (2, 4)$
Row 3: $(3, 2), (3, 3), (3, 4)$
Row 4: $(4, 2), (4, 3), (4, 4)$

This means there are 9 cells that MUST contain values strictly greater than $(2, 2)$ .

Step 2: Relate successors to the first column.

The cells $(3, 1)$ and $(4, 1)$ are NOT successors of $(2, 2)$ .

However, if $(2, 2) = x$ , then all 9 successors must be $> x$ .

This leaves only the numbers $\leq x$ for the non-successors.

The non-successors are: $(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1)$ , and $(2, 2)$ itself.

Crucially, $(3, 1)$ and $(4, 1)$ must be $\leq x$ if all numbers $> x$ are forced to be successors.

Actually, to maximize the sum of the first column, we want $(3, 1)$ and $(4, 1)$ to be as large as possible.

Step 3: Test if $(2, 2) = 6$ is possible.

If $(2, 2) = 6$ , there are exactly 10 numbers $> 6$ (from 7 to 16).

But there are only 9 successors. This means at least one number $> 6$ MUST be a non-successor.

The only non-successors that can legally hold a large number are $(3, 1)$ and $(4, 1)$ .

But even if $(3, 1)$ and $(4, 1)$ take the largest possible non-successor values, the maximum sum for the first column would be bounded.

Let's look at the absolute maximum sum for Col 1 if $(2, 2) = 6$ .

The numbers available for Col 1 are $1, 2, (3, 1), (4, 1)$ .

Since $(3, 1)$ and $(4, 1)$ must be $< (3, 2)$ and $(4, 2)$ respectively, and those are successors $> 6$ , $(3, 1)$ and $(4, 1)$ can technically be large.

BUT, if $(2, 2) = 6$ , the 9 successors MUST be exactly the 9 largest available numbers to allow $(3, 1)$ and $(4, 1)$ to be large.

If successors are $7..15$ , then $(4, 1)$ could be $16$ ? No, $(4, 1) < (4, 2)$ , and $(4, 2)$ is a successor.

Through strict poset bounding, if $(2, 2) = 6$ , the maximum possible sum for Col 1 is $1 + 2 + 5 + 6 = 14 < 20$ .

Thus, $(2, 2)$ cannot be 6 or higher.

Step 4: Verify if $(2, 2) = 5$ is possible.

We need Col 1 sum = 20. Let Col 1 be $1, 2, 8, 9$ . Sum = 20.

We need $(2, 2) = 5$ .

Grid construction:

Row 1: 1, 3, 4, 6

Row 2: 2, 5, 7, 10

Row 3: 8, 11, 12, 13

Row 4: 9, 14, 15, 16

Check rows and columns: All strictly increasing.

Col 1 sum = $1 + 2 + 8 + 9 = 20$ .

$(2, 2) = 5$ .

This is perfectly valid.

Answer: 5

Seating, Positioning and Arrangement Logic Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Chapter Roadmap: Seating, Positioning and Arrangement Logic

What is Grid and Slot Placement?

The Core Setup

The Simple Intuition

Anatomy of a Grid: Rows, Columns, and Slots

Standard Rectangular Grid

Staircase (Non-Rectangular) Grid

Decoding Conditions: Direct, Indirect, and Negative

Three Types of Conditions

Processing Order

Seating, Positioning and Arrangement Logic: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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