# In a survey of 100 persons it was found that 28 read magazine A,

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?

Solution:

Let A, B C be the sets of the persons who read magazines A, B and C, respectively. Also, let U denote the universal set.

We have: n(U)  = 100

$n(\mathrm{~A})=28, n(\mathrm{~B})=30, n(\mathrm{C})=42, n(\mathrm{~A} \cap \mathrm{B})=8, n(\mathrm{~A} \cap \mathrm{C})=10, n(\mathrm{~B} \cap \mathrm{C})=5$ and $n(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})=3$

Now,

Number of persons who read none of the three magazines $=n\left(\mathrm{~A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right)$

$=n(\mathrm{~A} \cup \mathrm{B} \cup \mathrm{C})^{\prime}$

$=n(\cup)-n(\mathrm{~A} \cup \mathrm{B} \cup \mathrm{C})$

$=n(\mathrm{U})-\{n(\mathrm{~A})+n(\mathrm{~B})+n(\mathrm{C})-n(\mathrm{~A} \cap \mathrm{B})-n(\mathrm{~A} \cap \mathrm{C})-n(\mathrm{~B} \cap \mathrm{C})+n(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})\}$

$=100-(28+30+42-8-10-5+3)$

$=20$

Number of students who read magazine $C$ only $=n\left(C \cap A^{\prime} \cap B^{\prime}\right)$

$=n\left\{C \cap(A \cup B)^{\prime}\right\}$

$=n(\mathrm{C})-n\{\mathrm{C} \cap(\mathrm{A} \cup \mathrm{B})\}$

$=n(\mathrm{C})-n\{(\mathrm{C} \cap \mathrm{A}) \cup(\mathrm{C} \cap \mathrm{B})\}$

$=n(\mathrm{C})-n\{(\mathrm{C} \cap \mathrm{A})+(\mathrm{C} \cap \mathrm{B})-(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})\}$

$=42-(10+5-3)$

$=30$