Question

In a class of 50 students, 23 play hockey, 15 play basketball and 20 play cricket. 7 play
hockey and basketball, 9 play cricket and basketball, 8 play hockey and cricket, and 13
students do not play any of the games. Find (i) how many students play hockey,
basketball and cricket? (ii) how many students play hockey but not cricket?
(iii) how
how many students play hockey and cricket but not basketball?

Answer 1

Answer:

Let B, C and H be the sets of students that plays Basketball, Cricket and Hockey

n(H) = 23, n(B) = 15, n(C) = 20

n(H intersection B) = 7, n(C intersection B) = 5, n(H intersection C) = 4

n(H union B union C) = 60 – 15 = 45

1. Hence, students that plays all games,

n(H intersection B intersection C) = n(H union B union C) – n(H) – n(B) – n(C) + n(H intersection B) + n(B intersection C) + n(C intersection H)

= 45 – 23 – 15 – 20 + 7 + 5 + 4 = 3

2. Who plays hockey not cricket

n(H) – n(H intersection C) = 23 – 4 = 19

3. Plays cricket and hockey not basketball

n(H intersection C) – n(H intersection C intersection B)

4 -3 = 1

Step-by-step explanation:

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Answer 2

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