| 1. \[10\,%\] families own both a car and a phone |
| 2. \[35\,%\] families own either a car or a phone |
| 3. 40,000 families live in the town |
| The correct statements are: |
A) 1 and 2
B) 1 and 3
C) 2 and 3
D) 1, 2 and 3
Correct Answer: C
Solution :
\[\operatorname{n}(P)=25\,%,\,\,n(C)=15\,%\] \[\operatorname{n}({{P}^{C}}\cap {{C}^{C}})=\,\,65\,%,\,\,n(P\cap C)=2000\] Since, \[n({{P}^{c}}\cap {{C}^{c}})=\,\,65\,%\] \[\therefore \,\,\,n{{\left( P \cup C \right)}^{c}}= 65\,% \,and n\left( P \cup C \right) = 35\,%\] Now, \[\operatorname{n}\left( P \cup C \right) = n(P) + n(C) - n\left( P \cap C \right)\] \[35=\,\,25+15-n\left( P\cap C \right)\] \[\therefore \,\,n\left( P\cap C \right)=40-35=5.\,\,Thus\,\,\,n\left( P\cap C \right)=\,\,5\,%\] But \[\operatorname{n}\left( P\cap C \right)= 2000\] \[\therefore \,\,Total number of families=\frac{2000\times 100}{5}= 40,000\]Since, \[\operatorname{n}\left( P\cup C \right)= 35%\] and \[\operatorname{total} number of families = 40,000\] and\[n(P\cap C)=5\,%\,\,\therefore \,\,(2)\,\,and\,\,(3)\,\,are\,\,correct\].
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