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A survey shows that $63\%$ of the Americans like cheese whereas $76\%$ like apples. If $x\%$ of the Americans like both cheese and apples, then
$x = 39$
$x = 63$
$39 \le x \le 63$
None of these
Solution
(c) Let $A$ denote the set of Americans who like cheese and let $B$ denote the set of Americans who like apples.
Let Population of American be $100$.
Then $n\,(A) = 63,n\,(B) = 76$
Now, $n\,(A \cup B) = n(A) + n(B) – n(A \cap B)$
$ = 63 + 76 – n(A \cap B)$
$\therefore n\,(A \cup B) + n(A \cap B) = 139$
==> $n\,(A \cap B) = 139 – n(A \cup B)$
But $n\,(A \cup B) \le 100$
$\therefore – n\,(A \cup B) \ge – 100$
$\therefore 139 – n\,(A \cup B) \ge 139 – 100 = 39$
$\therefore n(A \cap B) \ge 39$ i.e., $39 \le n(A \cap B)$…..(i)
Again, $A \cap B \subseteq A,A \cap B \subseteq B$
$ \therefore n\,(A \cap B) \le n\,(A) = 63$ and $n\,(A \cap B) \le n\,(B) = 76$
$\therefore n(A \cap B) \le 63$…..(ii)
Then, $39 \le n\,(A \cap B) \le 63$ ==> $39 \le x \le 63$.
