$2n(A \setminus B) = n(B \setminus A)$ and $5n(A \cap B) = n(A) + 3n(B)$,where $P \setminus Q = P \cap Q^C$. If $n(A \cup B) \leq 10$,then the value of $\frac{n(A) \cdot n(B) \cdot n(A \cap B)}{8}$ is:

Let $A$ denote the set of all $2$-digit numbers in base $10$ that are equal to four times the sum of the factorials of their digits. The sum of the numbers in $A$ 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:

Out of all the patients in a hospital $89\%$ are found to be suffering from heart ailment and $98\%$ are suffering from lungs infection. If $K\%$ of them are suffering from both ailments,then $K$ cannot belong to the set :

The sum of all integers from $1$ to $100$ that are divisible by $2$ or $5$ is...........

There are $200$ individuals with a skin disorder. $120$ had been exposed to the chemical $C_{1}$,$50$ to chemical $C_{2}$,and $30$ to both the chemicals $C_{1}$ and $C_{2}$. Find the number of individuals exposed to chemical $C_{1}$ or chemical $C_{2}$.