Let $A = {1, 2, 3, ..., 100}$ and $R$ be a relation on $A$ such that $R = {(a, b) : a = 2b + 1}$. Let $(a_1, a_2), (a_2, a_3), (a_3, a_4), ..., (a_k, a_{k+1})$ be a sequence of $k$ elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k , for which such a sequence exists, is equal to :

Consider the relations $R_1$ and $R_2$ defined as $aR_1b \Leftrightarrow a^2 + b^2 = 1$ for all $a, b \in R$ and $(a, b)R_2(c, d) \Leftrightarrow a+ d = b + c$ for all $(a, b), (c, d) \in N \times N$. Then:

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 :-

Let $A=\{1,2,3,4,5,6,7\}$. Define $B=\{T \subseteq A$ : either $1 \notin T$ or $2 \in T\}$ and $C=\{T \subseteq A: T$ the sum of all the elements of $T$ is a prime number $\}$. Then the number of elements in the set $B \cup C$ is ________________.

A survey shows that $73$% of the persons working in an office like coffee, whereas $65$% like tea. If $x$ denotes the percentage of them, who like both coffee and tea, then $x$ cannot be :