A survey shows that 63 % of the people in a c | vedclass

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(D) Let $n(A) = 63$ and $n(B) = 76$. Let $x = n(A \cap B)$.We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 63 + 76 - x = 139 - x$.Since the to

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$55$

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(D) Let $n(A) = 63$ and $n(B) = 76$. Let $x = n(A \cap B)$.We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 63 + 76 - x = 139 - x$.Since the total percentage of people cannot exceed $100 \%$,we have $n(A \cup B) \leq 100$,which implies $139 - x \leq 100$,so $x \geq 39$.Also,the number of people reading both newspapers cannot exceed the number of people reading either newspaper,so $x \leq n(A)$ and $x \leq n(B)$. Thus,$x \leq 63$.Therefore,the possible range for $x$ is $39 \leq x \leq 63$.Among the given options,only $55$ lies within this range.

$A$ survey shows that $63 \%$ of the people in a city read newspaper $A$ whereas $76 \%$ read newspaper $B$. If $x \%$ of the people read both the newspapers,then a possible value of $x$ can be

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