Quick-Solve Checklist for Grid Placement

The Grid Placement Checklist

Before Solving:

Draw the grid exactly as given

Count available slots (mark blocked ones)

Verify: Available slots = Number of items

Reading Conditions:

Classify: Direct, Indirect, Negative

Read all conditions once before solving

Solving Order:

Place all direct conditions

Pick most restrictive indirect condition

Apply negative conditions to eliminate

Look for forced placements (only one option left)

If stuck, backtrack and try next option

For Staircase Grids:

Start from the row with fewest slots

Use column constraints to cross-verify

Final Check:

Verify every condition is satisfied

Ensure no item is placed in a blocked slot

"Draw, count, classify, place direct, build indirect, eliminate negative, backtrack if stuck."

Quick-Solve Checklist for Circular & Movement

The Circular & Movement Checklist

1. Setup:

Draw circle, mark $N$ seats.
Note facing direction (Inside/Outside) $\to$ Lock in Left/Right rules.

2. Placement:

Find the Anchor (most definite clue) and place them.
Place adjacent pairs next.
Treat empty seats as "ghost" entities (write 'E' or '-').

3. Movement / Passing:

Keep the seat framework STATIC.
Move the labels (names or object marker) step-by-step.
Track long passing sequences in a vertical list.

4. Verification:

Check every condition against the final diagram.
For "between" questions, verify which path (shortest vs specified direction) is asked.

One-Line Mantra

"Draw the loop, lock the directions, anchor the knowns, count the ghosts, and move the labels."

Quick Method Checklist

Before solving

Read grid dimensions (rows x columns)

Identify blocks and boundaries

Note road positions

While solving

Start with fixed positions

Apply block & adjacency constraints

Critical checks

Adjacent = sharing a side (not diagonal)

Houses separated by road are NOT adjacent

If You See X, Do Y

If you see...	Do this...
Exact position constraint	Place that house immediately
"Same block"	Eliminate positions in other blocks
"Adjacent"	Check all 4 directions (up, down, left, right)
"Different block"	Ensure houses are in separate blocks
"Faces the road"	House must be in the row facing that road
"Not adjacent"	Eliminate all 4 adjacent positions

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 Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Quick-Solve Checklist for Grid Placement

The Grid Placement Checklist

Quick-Solve Checklist for Circular & Movement

The Circular & Movement Checklist

Quick Method Checklist

Quick Method Checklist

If You See X, Do Y

If You See X, Do Y

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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