Question

In a survey relating to reading habits of people in a town
it was observed that 60% read
magazine A, 50% magazine B, 50% magazine C, 30% A and B, 20% B and C, 30% C and
A and 10% all the threes. In the fitness of the thing, state (i) what percentage read exactly
two magazines ? (ii) what percentage do not read any of the threes ?

Answer 1

In In a survey relating to reading habits of people in a town

it was observed that 60% read

magazine A, 50% magazine B, 50% magazine C, 30% A and B, 20% B and C, 30% C and

A and 10% all the threes. In the fitness of the thing, state (i) what percentage read exactly

two magazines ? (ii) what percentage do not read any of the threes ?

Answer 2

Total number of persons =100%

Number of persons who read magazineA=n(A)=80%

Number of persons who read magazine B=n(B)=50%

Number of persons who read magazine C=n(C)=50%

Number of persons who read both magazine A and B =n(A∩B)=30%

Number of persons who read both magazine B and C =n(B∩C)=20%

Number of persons who read both magazine C and A =n(A∩C)=30%

Number of persons who read all three magazine =n(A∩B∩C)=10%

Number of people who do not read magazine are,

 ⇒ n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B) −n(B∩C)−n(C∩A)+n(A∩B∩C)

  ⇒n(A∪B∪C)=80 + 50 + 50 - 30 - 20 - 30 + 10

  ⇒ 110 persons

Number of readers who read magazines are = 110%

So, number of readers who do not read any magazine are =100−110= -10

Hence, readers who read none of the magazines are 0 because negative numbers aren't considered.

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