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:

Two newspapers $A$ and $B$ are published in a city. It is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$,while $8\%$ reads both $A$ and $B$. Further,$30\%$ of those who read $A$ but not $B$ look into advertisements,$40\%$ of those who read $B$ but not $A$ look into advertisements,and $50\%$ of those who read both $A$ and $B$ look into advertisements. The percentage of the population who look into advertisements is:

Let $A = \{x : x \in R, |x| < 1\};$ $B = \{x : x \in R, |x - 1| \ge 1\}$ and $A \cup B = R - D,$ then the set $D$ is

Let $A_1, A_2, \ldots, A_m$ be non-empty subsets of $\{1, 2, 3, \ldots, 100\}$ satisfying the following conditions:
$1.$ The numbers $|A_1|, |A_2|, \ldots, |A_m|$ are distinct.
$2.$ $A_1, A_2, \ldots, A_m$ are pairwise disjoint.
(Here $|A|$ denotes the number of elements in the set $A$).
Then,the maximum possible value of $m$ is:

In a town of $10,000$ families,it was found that $40\%$ of families buy newspaper $A$,$20\%$ buy newspaper $B$,and $10\%$ buy newspaper $C$. Also,$5\%$ of families buy $A$ and $B$,$3\%$ buy $B$ and $C$,and $4\%$ buy $A$ and $C$. If $2\%$ of families buy all three newspapers,then the number of families that buy newspaper $A$ only is:

$A$ number $n$ is chosen at random from $S=\{1, 2, 3, \ldots, 50\}$. Let $A=\{n \in S: n+\frac{50}{n} > 27\}$,$B=\{n \in S: n \text{ is a prime}\}$ and $C=\{n \in S: n \text{ is a square}\}$. Then,the correct order of their probabilities is: