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Question No. 9
In a society, 70 families have television sets and 40 have radio sets. What is the total number of families in the society if no family has both television and radio sets?
A. 70
B. 110
C. 150
D. 160

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{A} \mathrm{stands} \mathrm{for} \mathrm{Television} \\ \mathrm{B} \mathrm{stand} \mathrm{for} \mathrm{radio} \\ \mathrm{As} \mathrm{per} \mathrm{question} \\ \mathrm{n}\left(\mathrm{A}\right)=70 \\ \mathrm{n}\left(\mathrm{B}\right)=40 \\ \mathrm{As} \mathrm{no} \mathrm{family} \mathrm{has} \mathrm{both} \mathrm{television} \mathrm{and} \mathrm{radio} \mathrm{sets}, \mathrm{hence} \\ \mathrm{n}\left(\mathrm{A}\cap \mathrm{B}\right)=0 \\ \mathrm{Now} \mathrm{number} \mathrm{of} \mathrm{famiies} \mathrm{in} \mathrm{society}=\mathrm{n}\left(\mathrm{A}\cup \mathrm{B}\right). \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{n}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}\left(\mathrm{A}\cap \mathrm{B}\right) \\ =70+40-0 \\ =110 \left[\mathrm{Answer}\right] \end{gathered}$$

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