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Mathematics Question on Sets

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% buy B and C and 4% buy A and C. If 2% families buy all the newspapers, then

A

3,300 families buy A only

B

1,400 families buy B only.

C

4000 families buy none of A, B and C

D

All are correct

Answer

All are correct

Explanation

Solution

Let P = set of families buying A, Q = set of families buying B and R = set of families buying C. ∴n(P)40%\therefore n\left(P\right) 40\% of 10,000=4,00010,000 = 4,000, similarly n(Q)=2,000,n(R)=1,000n\left(Q\right) = 2,000 , n\left(R\right) = 1,000 n(P∩Q)=500,n(Q∩R)=300n\left(P \cap Q\right) = 500, n \left(Q \cap R\right) = 300 n(P∩R)=400n \left(P \cap R\right) = 400 and n (P∩Q∩R)=200\left(P \cap Q\cap R\right) = 200 (i)\left(i\right) Number of families buying only A=n(P∩Q′∩R′)A = n\left(P \cap Q' \cap R'\right) =n(P∩(Q∪R)′)=n(P)−n(P∩(Q∪R))= n \left(P \cap \left(Q \cup R\right)'\right) = n\left(P\right) -n\left(P \cap \left(Q \cup R\right)\right) =n(P)−[n(P∩Q)+n(P∩R)−n((P∩Q)I(P∩R))]= n\left(P\right) -\left[n\left(P\cap Q\right) + n\left(P\cap R\right) - n\left(\left(P\cap Q\right)I\left(P\cap R\right)\right)\right] =n(P)−n(P∩Q)−n(P∩R)+n(P∩Q∩R)= n\left(P\right) - n\left(P\cap Q\right) - n\left(P\cap R\right) + n\left(P\cap Q\cap R\right). =4,000−500−400+200=3,300= 4,000 - 500 - 400 + 200 = 3,300. (ii)\left(ii\right) Number of families buying only B =n(Q)−n(P∩Q)−n(Q∩R)+n(P∩Q∩R)= n\left(Q\right) - n\left(P\cap Q\right) - n\left(Q\cap R\right) + n\left(P\cap Q\cap R\right) [see(i)]\left[see \left(i\right)\right] =2,000−500−300+200=1,400= 2,000 - 500 - 300 + 200 = 1,400. (iii)\left(iii\right) Number of families buying none of A, B and C=n(P′∩Q′∩R′)=n(P′∩(Q∪R)′)C = n\left(P'\cap Q'\cap R'\right) = n\left(P'\cap \left(Q\cup R\right) '\right) =n(P∪(Q∪R))′=10000−n(P∪Q∪R)= n\left(P\cup \left(Q\cup R\right)\right) ' =10000 - n\left(P\cup Q\cup R\right) =10,000−[n(P)+n(Q)+n(R)−n(P∩Q)−n(Q∩R)−n(P∩R)+(P∩Q∩R)]= 10,000- \left[n\left(P\right) +n\left(Q\right) +n\left(R\right)- n\left(P \cap Q\right)-n\left(Q \cap R\right) - n\left(P \cap R\right) + \left(P\cap Q \cap R\right)\right] =10,000−[4,000+2,000+1,000−500−300−400+200]= 10,000 - \left[4,000 + 2,000 + 1,000 - 500 - 300 - 400 + 200\right] =10,000−6,000=4,000.= 10, 000 - 6,000 = 4,000. Note : For sets A, B, we have (A∩B)∪(A∩B′)=A∩(B∪B′)=A∩U=A\left(A\cap B\right)\cup \left(A\cap B'\right) = A\cap \left(B\cup B'\right) = A\cap U = A and (A∩B)∩(A∩B′)=A∩(B∩B′)=A∩f=f\left(A\cap B\right)\cap \left(A\cap B'\right) = A\cap \left(B\cap B'\right) = A\cap f = f ∴n(A)=n(A∩B)+n(A∩B′)orn(A∩B′)\therefore n\left(A\right) = n\left(A\cap B\right) + n\left(A\cap B'\right) or n\left(A\cap B'\right) =n(A)−n(A∩B)= n\left(A\right) - n\left(A\cap B\right) Replacing A by P and B by Q∪RQ \cup R, we have n(P∩(Q∪R)′)=n(P)−n(P∩(Q∪R))n \left(P\cap\left(Q\cup R\right) '\right) = n\left(P\right) - n\left(P\cap \left(Q\cup R\right)\right) etc. Hence all options are correct.