7. In a survey of 100 people it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A B, 10 read magazine AC 5 read magazine B and C and 3 read all the three. Find:i) How many read none of the magazines?ii) How many read magazine C only?

Symbols used in ans:-

Union = U

Intersection = ^

.

Let set of Magazine A readers be A, B be B, and C be C.

.

Therefore,

n(U) = 100 ............ [ UNIVERSAL SET including both readers and non-readers ]

n(A) = 28

n(B) = 30

n(C) = 42

n(A^B) = 8

n(A^C) = 10

n(B^C) = 5

n(A^B^C) = 3

.

By the formula:-

n(AUBUC) = n(A) + n(B) + n(C) - n(A^B) - n(A^C) - n(B^C) + n(A^B^C)

n(AUBUC) = 28 + 30 + 42 - 8 - 10 - 5 + 3

Therefore total no. of readers = 80

.

i) Hence total no. of people who dont read any magazine = 100 - 80 = 20

ii) No. of people who read only C = n(C) - n(A^C) - n(B^C) = 42 - 10 - 5 = 27

.

  • 23

Symbols used in ans:-

Union = U

Intersection = ^

.

Let set of Magazine A readers be A, B be B, and C be C.

.

Therefore,

n(U) = 100 ............ [ UNIVERSAL SET including both readers and non-readers ]

n(A) = 28

n(B) = 30

n(C) = 42

n(A^B) = 8

n(A^C) = 10

n(B^C) = 5

n(A^B^C) = 3

.

By the formula:-

n(AUBUC) = n(A) + n(B) + n(C) - n(A^B) - n(A^C) - n(B^C) + n(A^B^C)

n(AUBUC) = 28 + 30 + 42 - 8 - 10 - 5 + 3

Therefore total no. of readers = 80

.

i) Hence total no. of people who dont read any magazine = 100 - 80 = 20

ii) No. of people who read only C = n(C) - n(A^C) - n(B^C) = 42 - 10 - 5 = 27

.

  • 8
What are you looking for?