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?
14 answers
i need answer
52
52 is the answer
Let the Sports Club be denoted by U,
.
F = Football
.
B = Basketball
.
V = Volleyball
.
n(U) = 110
.
n(FnBnC') = 20
.
n(FnVnB') = 15
.
n(BnVnF') = 26
.
n(FnBnV) = x
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n(FnB'nV') = n(F'nBnV') = (F'nB'nV) = 2x
.
Note: B', F' and V' denote the complements of the various sets(clubs).
.
n(U) = n(FuBuV) + n(FuBuV)'
.
In this case, n(FuBuV)' is empty.
.
110 = 2x + 20 + x + 15 + 2x + 26 + 2x
.
110 = 61 + 7x
.
49 = 7x
.
x = 7
.
n(B) = 20 + 26 + x + 2x
.
n(B) = 20 + 26 + 3x
.
n(B) = 20 + 26 + 21
.
n(B) = 67
.
67 members play Basketball altogether.
~Darematics(07034495462)
.
F = Football
.
B = Basketball
.
V = Volleyball
.
n(U) = 110
.
n(FnBnC') = 20
.
n(FnVnB') = 15
.
n(BnVnF') = 26
.
n(FnBnV) = x
.
n(FnB'nV') = n(F'nBnV') = (F'nB'nV) = 2x
.
Note: B', F' and V' denote the complements of the various sets(clubs).
.
n(U) = n(FuBuV) + n(FuBuV)'
.
In this case, n(FuBuV)' is empty.
.
110 = 2x + 20 + x + 15 + 2x + 26 + 2x
.
110 = 61 + 7x
.
49 = 7x
.
x = 7
.
n(B) = 20 + 26 + x + 2x
.
n(B) = 20 + 26 + 3x
.
n(B) = 20 + 26 + 21
.
n(B) = 67
.
67 members play Basketball altogether.
~Darematics(07034495462)
67 members play basketball altogether
Pls how was the answer 52 gotten
Pls Ayodeji And Edidiong Umanah How Did U Get 52 To Be The Answer
How many played football and how many played volleyball. Please I need a straight answer
Pls how was 52 gotten
pls how did you get 52
Thank you so much
Have been trying it all day but still didn't get
Have been trying it all day but still didn't get
Pls can u represent in on a venn diagram
How did you arrive at 52
Pls can u represent it in a Venn diagram
4 answers