A survey shows that 63% of Americans like cheese whereas 76% like apples. If x% of Americans like bo

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1.Set Theory

hard

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

A

$x = 39$

B

$x = 63$

C

$39 \le x \le 63$

D

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, $39 \le n\,(A \cap B) \le 63$ ==> $39 \le x \le 63$.

Standard 11

Mathematics

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