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(B) Let $R, T,$ and $V$ represent the sets of families owning Radios,$TVs,$ and $VCRs$ respectively.Given:Total families $= 100$$n(R) = 50, n(T) = 75,

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$35$

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(B) Let $R, T,$ and $V$ represent the sets of families owning Radios,$TVs,$ and $VCRs$ respectively.Given:Total families $= 100$$n(R) = 50, n(T) = 75, n(V) = 25$$n(R \cap T \cap V) = 10$Since every $VCR$ owner also has a $TV$,$V \subset T$,which implies $n(V \cap T) = n(V) = 25$.Families having only $VCR$ and $TV$ (but not Radio) $= n(V \cap T) - n(R \cap T \cap V) = 25 - 10 = 15$.Let $x$ be the number of families having both Radio and $TV$ but not $VCR$.From the Venn diagram,the total number of families is the sum of all disjoint regions:$(40 - x) + x + 10 + 15 + (50 - x) = 100$$115 - x = 100$$x = 15$Now,the number of families having only $TV$ is $n(T) - [n(T \cap R 	ext{ only}) + n(T \cap V 	ext{ only}) + n(R \cap T \cap V)]$$= 75 - [15 + 15 + 10] = 75 - 40 = 35$.

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 $TV$?

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