A survey shows that 63% of the people watch a news channel whereas, 76% watch an entertainment channel at a particular time. lf x% of the people watch both types of
channels, then

Let A be the set of those people who watch news channel and
B be the set of those people who watch entertainment channel.
Then

$\mathrm{n}\left(\mathrm{A}\right)=63,\mathrm{n}\left(\mathrm{B}\right)=76,\text{ and }\mathrm{n}(\mathrm{A}\cap \mathrm{B})=\mathrm{x}$

$\begin{array}{l}\because \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}(\mathrm{A}\cup \mathrm{B})\le 100\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})\le 100\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}63+76-\mathrm{x}\le 100\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\ge 39\end{array}$

Also, $\mathrm{n}(\mathrm{A}\cap \mathrm{B})\le min\cdot \left\{\mathrm{n}\right(\mathrm{A}),\mathrm{n}(\mathrm{B}\left)\right\}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\le 63\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}39\le \mathrm{x}\le 63\end{array}$