Question bank/JEE Main/Mathematics

There are 100 families in a society, 40 families buy newspaper A,
30 families buy newspaper B, 30 families buy newspaper C, 10 families
buy newspaper A and B, 8 families buy newspaper B and C, 5 families buy
newspaper A and C, 3 families buy newspaper A, B and C, then the number
of families who do not buy any newspaper is

A20B80C0DNone of these

Show answer and solution▾

Given that n(A) = 40, n(B) = 30, n(C) = 30
$n(A \cap B) = 10,n(B \cap C) = 8,n(C \cap A) = 5,n(A \cap B \cap C) = 3$
$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A)$
$+ n(A \cap B \cap C)$
= 40 + 30 + 30 - 10 - 8 - 5 + 3
= 80 (number of families reading at least one newspapers A,B and C)
Total number of families = 100
So, 20 families do not read any newspaper.
II Method
(Using venn diagram)
[IMAGE] = 100 - (28 + 7 + 15 + 2 + 3 + 5 + 20) = 20

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx