In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% families buy B and C and 4% buy A and C. If 2% families buy all three newspapers, find the number of families which buy (i) A only (ii) B only (iii) none A, B or C.

$n\left(A\right)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} A=40\% \mathrm{of} 10,000=4000$$n\left(B\right)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} B=20\% \mathrm{of} 10,000=2000$$n\left(C\right)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} C=10\% \mathrm{of} 10,000=1000$$n(A\cap B)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} A \mathrm{and} B=5\% \mathrm{of} 10,000=500$$n(B\cap C)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} B \mathrm{and} C=3\% \mathrm{of} 10,000=300$$n(A\cap C)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} A \mathrm{and} C=4\% \mathrm{of} 10,000=400$$n(A\cap B\cap C)= \mathrm{no} \mathrm{of} \mathrm{people} \mathrm{reading} \mathrm{newspaper} A,B \mathrm{and} C=2\% \mathrm{of} 10,000=200$$$$\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{families} \mathrm{reading} \mathrm{newspapers}=n(A\cup B\cup C)$$=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(C\cap B)-n(A\cap C)+n(A\cap B\cap C)$$=4000+2000+1000-500-300-400+200=6000$$\left(i\right) \mathrm{Families} \mathrm{reading} \mathrm{only} \mathrm{paper} A$$n(B\cup C)=n\left(B\right)+n\left(C\right)--n(C\cap B)=2000+1000-300=2700$$\mathrm{Number} \mathrm{of} \mathrm{families} \mathrm{reading} A= \mathrm{total} \mathrm{no} \mathrm{of} \mathrm{families} \mathrm{reading} \mathrm{paper} -\mathrm{no} \mathrm{of} \mathrm{families} \mathrm{reading} B \mathrm{and} C=6000-2700=3300$$$$\left(\mathrm{ii}\right) \mathrm{Families} \mathrm{reading} \mathrm{only} \mathrm{paper} B$$n(A\cup C)=n\left(A\right)+n\left(C\right)--n(C\cap A)=4000+1000-400=4600$$\mathrm{Number} \mathrm{of} \mathrm{families} \mathrm{reading} \mathrm{only} B= \mathrm{total} \mathrm{no} \mathrm{of} \mathrm{families} \mathrm{reading} \mathrm{paper} -\mathrm{no} \mathrm{of} \mathrm{families} \mathrm{reading} A \mathrm{and} C=6000-4600=1400$$$$\left(\mathrm{iii}\right) \mathrm{Families} \mathrm{reading} \mathrm{none} \mathrm{of} A,B \mathrm{and} C$$=10,000-6000=4000$$$$$">