Let $A = \{n \in N : n \text{ is a } 3 \text{-digit number}\}$. Let $B = \{9k + 2 : k \in N\}$ and $C = \{9k + l : k \in N\}$ for some $l$ $(0 < l < 9)$. If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$,then $l$ is equal to:

Let $A = \{x : |x^{2} - 10| \le 6\}$ and $B = \{x : |x - 2| > 1\}$. Then:

There are $100$ students in a class. In an examination, $50$ of them failed in Mathematics, $45$ failed in Physics, $40$ failed in Biology and $32$ failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing in all the three subjects is:

$20$ teachers of a school either teach mathematics or physics. $12$ of them teach mathematics while $4$ teach both the subjects. Then the number of teachers teaching physics is

Let $S = \{(x, y, z) : x, y, z \in \mathbb{Z}, x + 2y + 3z = 42, x, y, z \geq 0\}$. Find the number of elements in $S$.

Of the members of three athletic teams in a school,$21$ are in the cricket team,$26$ are in the hockey team,and $29$ are in the football team. Among them,$14$ play hockey and cricket,$15$ play hockey and football,and $12$ play football and cricket. Eight play all the three games. The total number of members in the three athletic teams is: