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# In a class of 42 students, each play at least one of the three games-cricket,hockey or football.It is found that 14 play cricket, 20 play hockey and 24 play football. 3 play both cricket and football, 2 play both hockey and football and none play all the three games. Find the number of students who play cricket but not hockey.

Asked by Koushtav Chakrabarty 2nd August 2012, 10:03 PM
Let the set of students who play cricket, hockey and football be C, H, F respectively.
Then,
n(C) = 14, n(H) = 20, n(F) = 24, n(C intersection F) = 3, n (H intersection F) = 2, n (C union H union F) = 42  and n (C intersection H intersection F) = 0.

We know:
n (C union H union F) = n(C) + n(H) + n(F) - n(C intersection H) - n(H intersection F) - n(C intersection F) + n (C intersection H intersection F)

Substituting the values, we get,
n(C intersection H) = 11

But, C = (C intersection H) intersection (C union H')
So, n (C) = n (C intersection H) + n (C intersection H')

[Since, (C intersection H) and (C intersection H') are disjoint sets]

Substituting the values, we get
n (C intersection H') = 3

Thus, the number of students who play cricket but not hockey is 3.
Answered by Expert 2nd August 2012, 10:35 PM
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