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

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

Let $A_1, A_2, \ldots \ldots, A_m$ be non-empty subsets of $\{1,2,3, \ldots, 100\}$ satisfying the following conditions:

$1.$ The numbers $\left|A_1\right|,\left|A_2\right|, \ldots,\left|A_m\right|$ are distinct.

$2.$ $A_1, A_2, \ldots, A_m$ are pairwise disjoint.(Here $|A|$ donotes the number of elements in the set $A$ )Then, the maximum possible value of $m$ is

Let $a>0, a \neq 1$. Then, the set $S$ of all positive real numbers $b$ satisfying $\left(1+a^2\right)\left(1+b^2\right)=4 a b$ is

Let $A=\left\{n \in N \mid n^{2} \leq n+10,000\right\}, B=\{3 k+1 \mid k \in N\}$ and $C=\{2 k \mid k \in N\}$, then the sum of all the elements of the set $A \cap(B-C)$ is equal to $.....$

Let $A=\{n \in N: H . C . F .(n, 45)=1\}$ and Let $B=\{2 k: k \in\{1,2, \ldots, 100\}\}$. Then the sum of all the elements of $A \cap B$ is