110 members of a sports club play atleast one of the games, football, basketball and volleyball. If 20 play football and basketball only, 15 play football and volleyball only, 26 play basketball and volleyball only, x play all the three games, 2x each play only one game, how many play basketball altogether?
Let the Sports Club be denoted by U,
.
F = Football
.
B = Basketball
.
V = Volleyball
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n(U) = 110
.
n(FnBnC') = 20
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n(FnVnB') = 15
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n(BnVnF') = 26
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n(FnBnV) = x
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n(FnB'nV') = n(F'nBnV') = (F'nB'nV) = 2x
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Note: B', F' and V' denote the complements of the various sets(clubs).
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n(U) = n(FuBuV) + n(FuBuV)'
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In this case, n(FuBuV)' is empty.
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110 = 2x + 20 + x + 15 + 2x + 26 + 2x
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110 = 61 + 7x
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49 = 7x
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x = 7
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n(B) = 20 + 26 + x + 2x
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n(B) = 20 + 26 + 3x
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n(B) = 20 + 26 + 21
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n(B) = 67
.
67 members play Basketball altogether.
~Darematics(07034495462)