If $A = \{x \in R : |x| < 2\}$ and $B = \{x \in R : |x - 2| \geq 3\}$,then:

Two finite sets have $m$ and $n$ elements. The total number of subsets of the first set is $56$ more than the total number of subsets of the second set. The values of $m$ and $n$ are

Each set $X_r$ contains $5$ elements and each set $Y_r$ contains $4$ elements and $\bigcup_{r = 1}^{24} X_r = S = \bigcup_{r = 1}^n Y_r$. If each element of set $S$ belongs to exactly $10$ of the $X_r$'s and to exactly $6$ of the $Y_r$'s,then $n$ is equal to:

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. Then the percentage of the population who look into advertisements is:

The domain of the function $f(x) = \frac{1}{\log_{10}(1 - x)} + \sqrt{x + 2}$ is

Set $A$ has $m$ elements and Set $B$ has $n$ elements. If the total number of subsets of $A$ is $112$ more than the total number of subsets of $B$,then the value of $m \times n$ is