CAT 2024 Slot 2 DILR Set 3 Q10–Q14

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2024 Slot 2, with 5 questions (Grids & Matrices). Answers are CAT's official key.

The set

This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.

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.

[Figure in the original paper]

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.

Question 10

What is the row number which has the least sum of numbers placed in that row?

TITA: type in the answer. There are no options and no negative marking.

Answer: 4 Grids & Matrices TITA

Solution

  1. Label the staircase: Row 1 has columns 1-4, Row 2 columns 2-4, Row 3 columns 3-4, Row 4 column 4. Only the top-right slot (R1C4) has nothing to its right or above, so it holds 10. 9 must be next to it: R1C3 or R2C4. 1 must sit in Row 1 or Column 4 and be a smallest slot, so it is R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, so they occupy two of R2C2, R2C3, R3C3.
  2. If 9 is at R2C4, then 7 and 8 cannot be in Row 2 or Column 4, and they cannot be at R3C3 (it would force R2C3 = 8 and leave no room for both 2 and 3); so they are R1C2 = 7, R1C3 = 8, and 4 and 6 can no longer share a row. Hence 9 is at R1C3, and 7, 8 must be in R2C4, R3C4, R4C4 (not Row 1 or Column 3; R2C2 is too small). If 1 were at R1C1, 4 and 6 would have to fill R2C2 and R2C3, leaving no room for 2 and 3. So 1 is at R4C4, R2C4 = 8, R3C4 = 7.
  3. Remaining 2, 3, 4, 5, 6 go to R1C1, R1C2, R2C2, R2C3, R3C3. 4 and 6 cannot be in Row 2 (2 and 3 need two of its slots or R3C3), so Row 1 is 4, 6, 9, 10. Then R2C3 = 5 (it must exceed R2C2 and R3C3), and 2, 3 fill R2C2 and R3C3 in either order.
  4. Row sums: Row 1 = 29, Row 2 = 15 or 16, Row 3 = 9 or 10, Row 4 = 1. Row 4 has the least sum.

Question 11

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
    Both I and II
  2. B
    Only I
  3. C
    Only II
  4. D
    Neither I nor II

Answer: (A) Both I and II Grids & Matrices MCQ

Solution

  1. Label the staircase: Row 1 has columns 1-4, Row 2 columns 2-4, Row 3 columns 3-4, Row 4 column 4. Only the top-right slot (R1C4) has nothing to its right or above, so it holds 10. 9 must be next to it: R1C3 or R2C4. 1 must sit in Row 1 or Column 4 and be a smallest slot, so it is R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, so they occupy two of R2C2, R2C3, R3C3.
  2. If 9 is at R2C4, then 7 and 8 cannot be in Row 2 or Column 4, and they cannot be at R3C3 (it would force R2C3 = 8 and leave no room for both 2 and 3); so they are R1C2 = 7, R1C3 = 8, and 4 and 6 can no longer share a row. Hence 9 is at R1C3, and 7, 8 must be in R2C4, R3C4, R4C4 (not Row 1 or Column 3; R2C2 is too small). If 1 were at R1C1, 4 and 6 would have to fill R2C2 and R2C3, leaving no room for 2 and 3. So 1 is at R4C4, R2C4 = 8, R3C4 = 7.
  3. 10 is always at R1C4 and 1 is forced to R4C4. Both statements must be true.

Question 12

Which of the following statements MUST be true?

I. 2 is placed in a slot in Column 2.
II. 3 is placed in a slot in Column 3.
  1. A
    Both I and II
  2. B
    Neither I nor II
  3. C
    Only II
  4. D
    Only I

Answer: (B) Neither I nor II Grids & Matrices MCQ

Solution

  1. Label the staircase: Row 1 has columns 1-4, Row 2 columns 2-4, Row 3 columns 3-4, Row 4 column 4. Only the top-right slot (R1C4) has nothing to its right or above, so it holds 10. 9 must be next to it: R1C3 or R2C4. 1 must sit in Row 1 or Column 4 and be a smallest slot, so it is R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, so they occupy two of R2C2, R2C3, R3C3.
  2. If 9 is at R2C4, then 7 and 8 cannot be in Row 2 or Column 4, and they cannot be at R3C3 (it would force R2C3 = 8 and leave no room for both 2 and 3); so they are R1C2 = 7, R1C3 = 8, and 4 and 6 can no longer share a row. Hence 9 is at R1C3, and 7, 8 must be in R2C4, R3C4, R4C4 (not Row 1 or Column 3; R2C2 is too small). If 1 were at R1C1, 4 and 6 would have to fill R2C2 and R2C3, leaving no room for 2 and 3. So 1 is at R4C4, R2C4 = 8, R3C4 = 7.
  3. Remaining 2, 3, 4, 5, 6 go to R1C1, R1C2, R2C2, R2C3, R3C3. 4 and 6 cannot be in Row 2 (2 and 3 need two of its slots or R3C3), so Row 1 is 4, 6, 9, 10. Then R2C3 = 5 (it must exceed R2C2 and R3C3), and 2, 3 fill R2C2 and R3C3 in either order.
  4. Since 2 and 3 can swap between R2C2 and R3C3, neither '2 in Column 2' nor '3 in Column 3' must be true.

Question 13

For how many slots in the grid, placement of numbers CANNOT be determined with certainty?

TITA: type in the answer. There are no options and no negative marking.

Answer: 2 Grids & Matrices TITA

Solution

  1. Label the staircase: Row 1 has columns 1-4, Row 2 columns 2-4, Row 3 columns 3-4, Row 4 column 4. Only the top-right slot (R1C4) has nothing to its right or above, so it holds 10. 9 must be next to it: R1C3 or R2C4. 1 must sit in Row 1 or Column 4 and be a smallest slot, so it is R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, so they occupy two of R2C2, R2C3, R3C3.
  2. If 9 is at R2C4, then 7 and 8 cannot be in Row 2 or Column 4, and they cannot be at R3C3 (it would force R2C3 = 8 and leave no room for both 2 and 3); so they are R1C2 = 7, R1C3 = 8, and 4 and 6 can no longer share a row. Hence 9 is at R1C3, and 7, 8 must be in R2C4, R3C4, R4C4 (not Row 1 or Column 3; R2C2 is too small). If 1 were at R1C1, 4 and 6 would have to fill R2C2 and R2C3, leaving no room for 2 and 3. So 1 is at R4C4, R2C4 = 8, R3C4 = 7.
  3. Remaining 2, 3, 4, 5, 6 go to R1C1, R1C2, R2C2, R2C3, R3C3. 4 and 6 cannot be in Row 2 (2 and 3 need two of its slots or R3C3), so Row 1 is 4, 6, 9, 10. Then R2C3 = 5 (it must exceed R2C2 and R3C3), and 2, 3 fill R2C2 and R3C3 in either order.
  4. Only R2C2 and R3C3 (holding 2 and 3 in either order) are undetermined: 2 slots.

Question 14

What is the sum of the numbers placed in Column 4?

TITA: type in the answer. There are no options and no negative marking.

Answer: 26 Grids & Matrices TITA

Solution

  1. Label the staircase: Row 1 has columns 1-4, Row 2 columns 2-4, Row 3 columns 3-4, Row 4 column 4. Only the top-right slot (R1C4) has nothing to its right or above, so it holds 10. 9 must be next to it: R1C3 or R2C4. 1 must sit in Row 1 or Column 4 and be a smallest slot, so it is R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, so they occupy two of R2C2, R2C3, R3C3.
  2. If 9 is at R2C4, then 7 and 8 cannot be in Row 2 or Column 4, and they cannot be at R3C3 (it would force R2C3 = 8 and leave no room for both 2 and 3); so they are R1C2 = 7, R1C3 = 8, and 4 and 6 can no longer share a row. Hence 9 is at R1C3, and 7, 8 must be in R2C4, R3C4, R4C4 (not Row 1 or Column 3; R2C2 is too small). If 1 were at R1C1, 4 and 6 would have to fill R2C2 and R2C3, leaving no room for 2 and 3. So 1 is at R4C4, R2C4 = 8, R3C4 = 7.
  3. Column 4 holds 10, 8, 7, 1: sum \(= 26\).

Same concept in other years

See every question on Fill a grid with symbols/numbers obeying adjacency or ordering rules.

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