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In a certain town 25% families own a phone, 15% own a car, 65% own neither a phone nor a car, 2000 families own both the phone and a car. Find (i) How many families live in the town? (ii) How many families own either a phone or a car.

Mr x , 8 Years ago
Grade 11
anser 2 Answers
Vikas TU
Dear Student,
Let the total population of the town be x.
Let us denote the set of families who own a phone as P, and who own a car as C.
We have n(P) = 25x/100, n(C) = 15x/100, n(P' ∩ C') = 65x/100, and n(P ∩ C) = 2000.
From deMorgan's law, we have (P U C)' = P' ∩ C'.
Therefore n(P' ∩ C') = n(P U C)' = 65x/100.
n(P U C) = x – 65x/100 = 35x/100.
n(P U C) = n(P) + n(C) - n(P ∩ C) => 35x/100 = 25x/100 + 15x/100 – 2000 = 40x/100 – 2000.
=> 2000 = 40x/100 – 35x/100 = 5x/100
=> x = 2000 x 100/5 = 40000.
Therefore the total number of families in the town = 40000.
Number of families which own either a phone or a car = n(P U C) = 35 x 40000/100 = 14000.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Last Activity: 8 Years ago
Rishi Sharma
Dear Student,
Please find below the solution to your problem.

Let the total population of the town be x.
Let us denote the set of families who own a phone as P, and who own a car as C.
We have n(P) = 25x/100, n(C) = 15x/100, n(P' ∩ C') = 65x/100, and n(P ∩ C) = 2000.
From deMorgan's law, we have (P U C)' = P' ∩ C'.
Therefore n(P' ∩ C') = n(P U C)' = 65x/100.
n(P U C) = x – 65x/100 = 35x/100.
n(P U C) = n(P) + n(C) - n(P ∩ C) => 35x/100 = 25x/100 + 15x/100 – 2000 = 40x/100 – 2000.
=> 2000 = 40x/100 – 35x/100 = 5x/100
=> x = 2000 x 100/5 = 40000.
Therefore the total number of families in the town = 40000.
Number of families which own either a phone or a car = n(P U C) = 35 x 40000/100 = 14000.

Thanks and Regards
Last Activity: 5 Years ago
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