Let $A = {1, 3, 4, 6, 9}$ and $B = {2, 4, 5, 8, 10}$. Let $R$ be a relation defined on $A \times B$ such that $R = {((a_1, b_1), (a_2, b_2)): a_1 \le b_2 \text{ and } b_1 \le a_2}$. Then the number of elements in the set R is :

Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If $A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$ $\epsilon S\}$, then the sum of all the elements in the set $T-A$ is equal to __________.

Two newspapers A and B are published in a city. It is known that $25$% of the city populations 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 and $40$% of those who read B but not A also look into advertisements, while $50$% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is :-

For $\alpha \in N$, consider a relation $R$ on $N$ given by $R = \{(x, y) : 3x + \alpha y$ is a multiple of $7\}$. The relation $R$ is an equivalence relation if and only if :

Let $A = { (\alpha, \beta ) \in R \times R : |\alpha - 1| \leq 4$ and $|\beta - 5| \leq 6 }$ and $B = { (\alpha, \beta ) \in R \times R : 16(\alpha - 2)^{2}+ 9(\beta - 6)^{2} \leq 144 }$. Then