CAT 2024 Slot 2 · DILR · Question 17
CAT 2024 Slot 2 previous year question on Quant Based Puzzles and Optimisation, with the answer and a step-by-step solution.
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.
For how many slots in the grid, placement of numbers CANNOT be determined with certainty?
Answer & solution
Answer: 2
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
Only the two slots holding 2 and 3 ((2, 2) and (3, 3)) are undetermined. The other 8 slots are fixed.
Hence, the answer is 2.
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