If $A, B$ and $C$ are three sets such that $A \cap B = A \cap C$ and $A \cup B = A \cup C$,then:

Out of $100$ families in the neighbourhood,$50$ have radios,$75$ have $TVs$ and $25$ have $VCRs$. Only $10$ families have all three and each $VCR$ owner also has a $TV$. If some families have radio only,how many have only $TV$?

If the number of elements in a set $A$ is $n(A) = 40$,$n(B) = 26$,and $n(A \cap B) = 16$,then what is the value of $n(A \cup B)$?

Suppose $A_1, A_2, A_3, \dots, A_{30}$ are $30$ sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ elements. Let $\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s. Then $n$ is equal to:

If $A = \{a, d\}, B = \{b, c, e\}$ and $C = \{b, c, f\},$ then $A \times (B - C) =$

$A$ survey shows that $73 \%$ of the persons working in an office like coffee,whereas $65 \%$ like tea. If $x$ denotes the percentage of them who like both coffee and tea,then $x$ cannot be