Math, asked by shatayusk06, 9 months ago

4 ро
Q17.In a town of 10.000
families it was found that 40%
families buy newspaper A,20%
families buy newspaper B, 10%
families buy newspaper C, 5%
families buy A and B. 3% buy B
and C and 4% buy A and C. If
2% families buy all the thre
newspapers. Then the number
of families which buy none of
A, B and C is equal to​

Answers

Answered by seemaranisana
0

Answer:

Number of families=10000

n(A)=  

100

40×10000

​  

=4000

n(B)=  

100

20×10000

​  

=2000

n(C)=  

100

10×10000

​  

=1000

n(A∩B)=  

100

5×10000

​  

=500

n(A∩C)=  

100

4×10000

​  

=400

n(B∩C)=  

100

3×10000

​  

=300

n(A∩B∩C)=  

100

2×10000

​  

=200

no. of people buy only A, only B & only C=1400+200+3300=4900

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