GATE CS 2024 Set 2 Q10 — General Aptitude

Let the four cells in the last (fourth) row be denoted as $x_1, x_2, x_3,$ and $x_4$ from left to right. Since the last row contains no crosses ($X$), all these cells contain numbers.

The third row of the grid is given as:

• $C_{3,1} = X$
• $C_{3,2} = 4$ (number)
• $C_{3,3} = X$
• $C_{3,4} = 3$ (number)

We determine the value of each cell in the fourth row by counting its immediate neighboring cells that do not contain a cross ($X$):

1. For $x_1$ (at $C_{4,1}$): Its neighbors are $C_{3,1}$ ($X$), $C_{3,2}$ (number), and $C_{4,2}$ (number). Thus, $x_1 = 2$.
2. For $x_2$ (at $C_{4,2}$): Its neighbors are $C_{3,1}$ ($X$), $C_{3,2}$ (number), $C_{3,3}$ ($X$), $C_{4,1}$ (number), and $C_{4,3}$ (number). Thus, $x_2 = 3$.
3. For $x_3$ (at $C_{4,3}$): Its neighbors are $C_{3,2}$ (number), $C_{3,3}$ ($X$), $C_{3,4}$ (number), $C_{4,2}$ (number), and $C_{4,4}$ (number). Thus, $x_3 = 4$.
4. For $x_4$ (at $C_{4,4}$): Its neighbors are $C_{3,3}$ ($X$), $C_{3,4}$ (number), and $C_{4,3}$ (number). Thus, $x_4 = 2$.

Summing these four values: