Let $A = \{n \in [100, 700] \cap \mathbb{N} : n \text{ is neither a multiple of } 3 \text{ nor a multiple of } 4\}$. Then the number of elements in $A$ is

The set $\{x \in R : [x - |x|] = 5\}$ is equal to

Let $X$ be a set of $5$ elements. The number $d$ of ordered pairs $(A, B)$ of subsets of $X$ such that $A \neq \phi, B \neq \phi, A \cap B \neq \phi$ satisfies:

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:

If sets $A$ and $B$ have $3$ and $6$ elements respectively,what is the minimum number of elements in $A \cup B$?

In a city,$20\%$ of persons read an English newspaper,$40\%$ read a Hindi newspaper,and $5\%$ read both newspapers. The percentage of people who do not read either paper is: (in $\%$)