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 $a>0, a \neq 1$. 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 $\mathrm{A}=\{\mathrm{n} \in[100,700] \cap \mathrm{N}: \mathrm{n}$ is neither a multiple of $3$ nor a multiple of 4$\}$. Then the number of elements in $\mathrm{A}$ is

Suppose ${A_1},\,{A_2},\,{A_3},........,{A_{30}}$ are thirty sets each having $5$ elements and ${B_1},\,{B_2}, ......., B_n$ are $n$ sets each with $3$ elements. Let $\bigcup\limits_{i = 1}^{30} {{A_i}} = \bigcup\limits_{j = 1}^n {{B_j}} = S$ and each elements of $S$ belongs to exactly $10$ of the $A_i's$ and exactly $9$ of the $B_j's$. Then $n$ is equal to

Let $A =\{ x \in R :| x +1|<2\}$ and $B=\{x \in R:|x-1| \geq 2\}$. Then which one of the following statements is NOT true ?

Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If

$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in N, a_{1}, a_{2}, a_{3}, \ldots, a_{k} \in S\right\}$ then the sum of all the elements in the set $T - A$ is equal to $......$