Question

out of 400 boys in a school: 112 played cricket,120 played Hockey, 168 played football, 32 played both football and hockey, 40 played both football and cricket, 20 played hockey and cricket and 12 played all the games. How many boys did not paly any game? And how many played only one game? Draw a venn diagram corresponding to this data.

Answer 1

Who play no any game=400-44-8-12-80-20-28-20-108
=80.


Who play only one game= 44+80+108
=232

Answer 2

A Ven diagram clarifies grouped worked problems on sets. The detailed set are encompassed in an universal set and the intersections of the shapes in the diagrams represent the terms common in each of the sets.

Let U = 400, A = Boys who played cricket

B = Boys who played Hockey and

C = Boy who played football

Given that,

n(A) = 112; n(B) = 120; n(C) = 168.

n(A∩C) = 40

n(A∩B) = 20

n(B∩C) = 32

n(A∩B∩C) = 12.

To find: U - n(A∪B∪C).

We know,

n(A∪B∪C) = n(A) + n(B) + n(C) + n(A∩B∩C) - n(A∩C) - n(A∩B) - n(B∩C)

∴ n(A∪B∪C) = 112 + 120 + 168 + 12 - 40 - 20 - 32

∴ n(A∪B∪C) = 320

∴ Boys who did not play any game = U - n(A∪B∪C)

∴ Boys who did not play any game = 400 -320

∴ Boys who did not play any game = 80

Now,

n'(A) = n(A) - (A∩B∩C) - n(A∩C) - n(A∩B)

∴ n'(A) = 112 - 12 - 40 - 20

∴ n'(A) = 40

n'(B) = n(B) - (A∩B∩C) - n(A∩B) - n(B∩C)

∴ n'(B) = 120 - 12 - 32 - 20

∴ n'(B) = 56

n'(C) = n(C) - (A∩B∩C) - n(B∩C) - n(A∩C)

∴ n'(C) = 168 - 12 - 32 - 40

∴ n'(C) = 84

∴ Boys who played only one game = n'(A) + n'(B) + n'(C)

∴ Boys who played only one game = 40 + 56 + 84

∴ Boys who played only one game = 180