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Anestetic
22 days ago
15

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play

ed both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball
Mathematics
1 answer:
lawyer [9.2K]22 days ago
3 0

Answer:

Cricket only = 30

Volleyball only = 15

Hockey only = 25

Explanation:

The number of cricket players is represented as n(C).

The count of hockey players as n(H).

The total volleyball participants as n(V).

The available data indicates;

n(C) = 50, n(H) = 50, n(V) = 40

For students playing both cricket and hockey, we denote as n(C∩H).

Participants in both hockey and volleyball as n(H∩V).

Players in both cricket and volleyball as n(C∩V).

Students engaging in all three sports as n(C∩H∩V).

From the data given;

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Thus, the total number of students who play at least one sport:

n(CᴜHᴜV) = n(C) + n(H) + n(V) - n(C∩H) - n(H∩V) - n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 - 15 - 20 - 15 + 10

Consequently, the overall total number of students, n(U) = 100.


Note: n(U) represents the universal set.

Let a = number of individuals who only played cricket and volleyball.

Let b = number of individuals who only played cricket and hockey.

Let c = number of individuals who only participated in hockey and volleyball.

Let d = number who played all three sports.

This allows us to establish:

d = n(CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

From here, we deduce:

a = 15 - 10 = 5

b = 15 - 10 = 5

c = 20 - 10 = 10

Thus, for the cricket-only player count;

n(C) - [a + b + d] = 50 - (5 + 5 + 10) = 30

For hockey-only players:

n(H) - [b + c + d] = 50 - (5 + 10 + 10) = 25

And for volleyball-only participants:

n(V) - [a + c + d] = 40 - (10 + 5 + 10) = 15

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