CAT 2024 Slot 2 DILR Set 1: Arrangements (5 Questions) with Solutions

The shape from the image: Row 1 has columns 1-4, Row 2 has columns 2-4, Row 3 has columns 3-4, Row 4 has only column 4. Call the cells a1-a4 (Row 1), b2-b4 (Row 2), c3-c4 (Row 3) and d4 (Row 4).

Rows increase to the right and columns decrease downward. So a4 is larger than every other cell, and $a_4 = 10$.

9 must sit directly below 10 in the order, so it is in a3 or b4.

Condition 4 keeps 2 and 3 out of Row 1 and Column 4. Only b2, b3 and c3 remain. Condition 3 puts 1 in a1 or d4, so 2 and 3 take b2 and c3 in some order.

If 9 is in b4, condition 5 forces 7 and 8 into a2 and a3. Then 4 and 6 have no common row, which breaks condition 6. So 9 is in a3.

With 9 in a3, condition 5 pushes 7 and 8 into c4 and b4, so $c_4 = 7$ and $b_4 = 8$.

Condition 6 now needs 4 and 6 in Row 1, so $a_1 = 4$ and $a_2 = 6$. That leaves $d_4 = 1$ and $b_3 = 5$.

Final grid: Row 1: 4, 6, 9, 10. Row 2: 2 or 3, 5, 8. Row 3: 3 or 2, 7. Row 4:

9. Row sums: Row 1 = 29, Row 2 = 15 or 16, Row 3 = 10 or 9, Row 4 = 1.

Name the slots by position, for example R1C4 for Row 1, Column 4. Rows rise left to right and columns fall top to bottom. So R1C4 is larger than every slot in its row and column, and every other slot sits below it in a chain (R1C3 > R2C3 > R3C3, R1C4 > R2C4 > R3C4 > R4C4, R1C2 > R2C2, and so on). Hence 10 is in R1C4.

Condition 3 puts 1 in Row 1 or Column 4. Only R1C1 (Row 1) and R4C4 (Column 4) are possible, because R2C2 and R3C3 are not in either line.

Condition 4 keeps 2 and 3 out of Row 1 and Column 4. They must therefore be in R2C2, R2C3 or R3C3. R2C3 is larger than both R2C2 and R3C3, so it cannot hold 2 or 3 without forcing a 1 into R2C2 or R3C3. So 2 and 3 are in R2C2 and R3C3.

The only slots that can hold 9 are those just below 10, namely R1C3 and R2C4. Try 9 in R2C4. Then condition 5 keeps 7 and 8 out of Row 2 and Column 4, so they must be in R1C1, R1C2, R1C3, and they would be R1C2 and R1C3. That leaves 4, 5, 6 and 1 for R1C1, R2C3, R3C4, R4C4. No row has room for two of them (R1C1, R2C3 and R3C4 are each alone in their row among these slots), so condition 6 fails. Hence 9 is in R1C3.

With 9 in R1C3, 7 and 8 must avoid Row 1 and Column 3, so they fall in R2C4 and R3C4, giving R2C4 = 8 and R3C4 = 7. The leftover numbers 1, 4, 5, 6 go to R1C1, R1C2, R2C3, R4C4. Condition 6 needs 4 and 6 in the same row, which only Row 1 allows: R1C1 = 4, R1C2 = 6. Then R2C3 = 5 and R4C4 = 1.

Final grid: Row 1: 4, 6, 9, 10. Row 2: (2 or 3), 5, 8. Row 3: (3 or 2), 7. Row 4:

7.

10 is in R1C4, so it is in Row 1. True.

1 is in R4C4, so it is in Row 4. True.

R1C4 is larger than every slot in its row and column, and every other slot sits below it in a chain, so 10 is in R1C4.

Condition 3 puts 1 in Row 1 or Column 4, so 1 is in R1C1 or R4C4. Condition 4 keeps 2 and 3 out of Row 1 and Column 4, so they are in R2C2, R2C3 or R3C3. R2C3 exceeds both R2C2 and R3C3, so it cannot be 2 or 3. Hence 2 and 3 occupy R2C2 and R3C3 in some order.

9 can only sit just below 10, in R1C3 or R2C4. With 9 in R2C4, 7 and 8 would have to go in Row 1 (R1C2, R1C3), and no row could then hold both 4 and 6, which breaks condition 6. So 9 is in R1C3, 7 and 8 are in R3C4 and R2C4, and condition 6 forces 4 and 6 into R1C1 and R1C2.

The final grid is: Row 1: 4, 6, 9, 10. Row 2: (2 or 3), 5, 8. Row 3: (3 or 2), 7. Row 4:

5. Nothing in the conditions decides which of R2C2 and R3C3 holds 2. Both 2 in R2C2 with 3 in R3C3, and 3 in R2C2 with 2 in R3C3, satisfy the row order, the column order and conditions 3 to 6.

Statement I (2 is in Column 2) is false in the second arrangement, where 2 is in Column 3. Statement II (3 is in Column 3) is false in the first arrangement, where 3 is in Column 2. Neither statement must be true.

The shape from the image: Row 1 has 4 cells, Row 2 has 3, Row 3 has 2, Row 4 has 1, giving 10 cells.

Rows increase rightward and columns decrease downward, so 10 is in the top-right cell, and 9 is in a3 or b4.

If 9 were in b4, condition 5 would force 7 and 8 into a2 and a3. Then 4 and 6 would need a common row, and none is available. So 9 is in a3.

Condition 5 then forces $b_4 = 8$ and $c_4 = 7$.

Condition 6 forces 4 and 6 into Row 1, so $a_1 = 4$ and $a_2 = 6$. That makes $d_4 = 1$, which is in Column 4 with 10 as condition 3 requires. The remaining cell gives $b_3 = 5$.

Condition 4 keeps 2 and 3 out of Row 1 and Column 4. They fill b2 and c3, and nothing decides which is which.

The cells with an undetermined number are b2 and c3, which makes 2 slots. The other 8 are fixed.

The shape from the image: Row 1 has columns 1-4, Row 2 has columns 2-4, Row 3 has columns 3-4, Row 4 has column 4 only. Column 4 therefore has four cells.

Rows increase rightward and columns decrease downward, so 10 is in the top-right cell, which is the top of Column 4.

9 is in a3 or b4. If 9 were in b4, condition 5 would force 7 and 8 into a2 and a3, and 4 and 6 could not share a row. So 9 is in a3.

Condition 5 then forces 8 and 7 into b4 and c4, so $b_4 = 8$ and $c_4 = 7$ (Column 4 decreases downward).

Condition 6 puts 4 and 6 in Row 1, so $a_1 = 4$ and $a_2 = 6$. That leaves 1 for the last cell of Column 4, $d_4 = 1$. This also satisfies condition 3, since 1 shares Column 4 with 10.

Column 4 from top to bottom is 10, 8, 7, 1, and its sum is $10 + 8 + 7 + 1 = 26$.