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  • Help me answer: In a town of 10000 families.It was found that 40% buys newspaper A,20%buys B and 10% buys C, 5% buy both A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers,then find the number of families which bu

In a town of 10000 families.It was found that 40% buys newspaper A,20%buys B and 10% buys C, 5% buy both A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers,then find the number of families which buy only A

  • Option 1)

    3100

  • Option 2)

    3300

  • Option 3)

    2900

  • Option 4)

    1400

 

Answers (1)

best_answer

As we learnt

 

Number of Elements in Union A , B & C -

 

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C)

- wherein

Given A, B and C be any finite sets. then Number of Elements in union A , B & C is given by this formula.

 

 n(A)=40%of10000=4000

n(B)=2000, n(C)=1000,n(A\capB)=500,n(B\capC)=300,n(A\capC)=400,n(A\capB\capC)=200

We need 

n\left ( A\cap B{}'\cap C{}' \right )= n\left ( A\cap \left ( B\cup C \right ){}' \right )

= n(A)-n(A\cap \left ( B\cup C \right ))

= n(A)-n\left [ \left ( A\cap B \right )\cup \left ( A\cap C \right ) \right ]

= n(A)-n\left [ \left ( A\cap B \right )+ \left ( A\cap C \right ) -n(A\cap B\cap C) \right ]

= 4000-[500+400-200]=3300

 


Option 1)

3100

Option 2)

3300

Option 3)

2900

Option 4)

1400

Posted by

Himanshu

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