BITSAT 2020 Paper

Let R be the set of families having a radio and T the set families having a T.V.,
then n(R∪T)= The number of families having at least on of the radio and T.V. =1003−63=940
n(R)=794 and n(T)=187
Let x families have both a radio and a T.V.
Then number of families who have only radio = 794−x
And the number of families who have only T.V. =187−x
From Venn diagram, 794−x+x−187−x=940
⇒981−x=940 or x=981−940=41
Hence, the required number of families having both a radio and a T.V. =41