CAT 2024 Slot 2 · DILR · Question 15

CAT 2024 Slot 2 previous year question on Quant Based Puzzles and Optimisation, with the answer and a step-by-step solution.

Quant Based Puzzles and OptimisationLOGICAL REASONING - High LevelMultiple choiceMedium

Set 4: Numbers 1 to 10 in a Staircase Grid — this question belongs to a set. Read the full set with all 5 questions →

The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are placed in ten slots of the following grid based on the conditions below.

1. Numbers in any row appear in an increasing order from left to right.

2. Numbers in any column appear in a decreasing order from top to bottom.

3. 1 is placed either in the same row or in the same column as 10.

4. Neither 2 nor 3 is placed in the same row or in the same column as 10.

5. Neither 7 nor 8 is placed in the same row or in the same column as 9.

6. 4 and 6 are placed in the same row.

Which of the following statements MUST be true?

I. 10 is placed in a slot in Row 1.

II. 1 is placed in a slot in Row 4.

  1. A

    Only II

  2. B

    Both I and II

  3. C

    Only I

  4. D

    Neither I nor II

Pick an option, then check.
Answer & solution

Answer: Option B

Solving the set

The grid has Row 1: Columns 1–4; Row 2: Columns 2–4; Row 3: Columns 3–4; Row 4: Column 4 only. Write (r, c) for the slot in Row r, Column c.

Step 1 – Place 10. Every slot (r, c) is smaller than (r, 4) (rows increase to the right), and (r, 4) is smaller than (1, 4) (columns decrease downwards). So (1, 4) is the largest slot: 10 is at (1, 4).

Step 2 – Place 2 and 3. They are not in Row 1 or Column 4 (fact 4), so they lie in (2, 2), (2, 3) or (3, 3). Now (2, 3) is larger than both (2, 2) and (3, 3). If 3 were at (2, 3), the other two slots would need numbers below 3, i.e., 1 and 2, but 1 must be in Row 1 or Column 4 (fact 3). So 2 and 3 are at (2, 2) and (3, 3) in some order.

Step 3 – Place 9. Only (1, 3) and (2, 4) have no larger slot other than (1, 4), so 9 is at one of them.

• If 9 is at (2, 4): 7 and 8 cannot be in Row 2 or Column 4 (fact 5), so they must be in Row 1, Columns 1–3. The numbers left (1, 4, 5, 6) go into one free Row 1 slot, (2, 3), (3, 4) and (4, 4): no row has two free slots, so 4 and 6 cannot share a row (fact 6). Not possible.

• So 9 is at (1, 3). Then 7 and 8 are not in Row 1 or Column 3, so they are in Column 4, rows 2–4.

Step 4 – Place 1. The smallest number must be at the bottom of a column, i.e., (1, 1), (2, 2), (3, 3) or (4, 4); with fact 3 it is (1, 1) or (4, 4).

• If 1 is at (1, 1): 4 and 6 need a row with two free slots. Row 1 has only (1, 2) free, so they go to Row 2: (2, 3) = 4 and (2, 4) = 6. But 7 and 8 must then be below 6 in Column 4, which breaks the decreasing order. Not possible.

• So 1 is at (4, 4). Then 7 and 8 are at (3, 4) and (2, 4); decreasing order gives (2, 4) = 8, (3, 4) = 7.

Step 5 – The numbers 4, 5, 6 remain for (1, 1), (1, 2) and (2, 3). 4 and 6 are in the same row, so they are (1, 1) = 4 and (1, 2) = 6, and (2, 3) = 5. All conditions are satisfied.

 

Column 1

Column 2

Column 3

Column 4

Row 1

4

6

9

10

Row 2

—

2 or 3

5

8

Row 3

—

—

3 or 2

7

Row 4

—

—

—

1

This question

From Step 1, 10 is at (1, 4), so I is true. From Step 4, 1 is at (4, 4), so II is true.

Hence, the answer is option 2.

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