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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.

Vikas TU
14149 Points
3 years ago
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)
Rishi Sharma
9 months ago
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.

Thanks and Regards