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A survey shows that $63 \%$ of the people watch a news channel whereas $76 \%$ watch another channel. If $\mathrm{x} \%$ of the people watch both channel, then
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The correct answer is:
$39 \leq x \leq 63$
$39 \leq x \leq 63$
Let A and B be the two sets of news channel such that $\mathrm{n}(\mathrm{A})=63, \mathrm{n}(\mathrm{B})=76, \mathrm{n}(\mathrm{A} \cup \mathrm{B})=100$
Also, $n(A \cap B)=x$
Using, $\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
$\Rightarrow 100=63+76-x \quad \Rightarrow x=139-100=39$
Again, $\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{n}(\mathrm{A}) \quad \Rightarrow \mathrm{x} \leq 63$
$\therefore \quad 39 \leq x \leq 63$.
Also, $n(A \cap B)=x$
Using, $\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
$\Rightarrow 100=63+76-x \quad \Rightarrow x=139-100=39$
Again, $\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{n}(\mathrm{A}) \quad \Rightarrow \mathrm{x} \leq 63$
$\therefore \quad 39 \leq x \leq 63$.
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