Question

In a survey of 100 persons it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read magazines B and C and 3 read all the three magazines. Find :

(i) How many read none of three magazines ?

(ii) How many read magazine C only ?

Answer 1

(i) Let n (P) denote total number of persons

n(A) denote number of people who read magazine A

n(B) denote number of people who read magazine B

and n(C) denote number of people who read magazine C

Then, n(P) = 100, n(A)= 28, n(B) = 30, n(C)= 42, $n(A\cap B)$ = 8,

$n(A\cap C)=10,n(B\cap C)=5,n(A\cap B\cap C)=3$

Now,

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

$n(A\cap B)-n(B\cap C)-n(A\cap C)+n(a\cap B\cap C)$

= 28+30+42-8-10-5+3

= 100-23+3

= 100-20

= 80

$\therefore$ Number of people who read none of the three magazines.

= $n(A\cup B\cup C)$

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

= 100-80

= 20

Hence, 20 people read none of the three magazines.

(ii) n (C only) = 42 - (7+3+2)

= 42-12

= 30