Question

Among 1000 families of a city, 40% read newspaper a, 20% read newspaper b, 10% read newspaper c, 5% read both a and b, 3% read both b and c, 4% read a and c and 2% read all three newspapers. The number of families which read only newspaper a is -

Answer 1

Answer:

total families = 1000

% of pepople reading newspaper a = 40%

thus no. of people reading newspaper a = 40/100×1000

= 400peoples

people reading both a and b= 5%

this no. of people reading newspaper a and b= 5/100×1000

=50

no of reading a and c=2/100×1000=20

thus no. if people reading a only a=400-50-20

=330 (answer)

Answer 2

Answer:

Total no. of Families =1000

A be the no of families buy newspaper A=40/100 ×1000=400

B be the no. of families buy newspaper B=20/100 ×1000=200

C be the no. of farilies buy newspaper C=10/100 ×1000=100

n(A)=400,n(B)=200,n(C)=100

n(A∩B)=50,n(B∩C)=30,n(A∩C)=40

n(A∩B∩C)=20

Number of families which buy only A is

n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C)

=400−50−40+20=330

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