# o level mathematics

in a class of 35, all the pupils play at least 1 game, volleyball, netball and hockey. volleyball only 10, those who play netball only 5, those who play hockey only 3, if 2 play all 3 games how many altogether play volleyball

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The intersection of all 3 circles is 2
The other 3 numbers are placed outside any intersections.

So, that means we have accounted for 20 of the 35 students.

The other 15 can play any combination of two of the games, or just one game.

So, there could be anywhere from 10+2+15=27 to 10+2=12 who play volleyball.

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2. 20 play volley ball altogether

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3. 20

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The intersection of all 3 circles is 2
The other 3 numbers are placed outside any intersections.
no of volleyball only : 10
no of netball only: 5
no of hockey.only: 3

So, that means we have accounted for 20 of the 35 students.

The other 15 can play any combination of two of the games, or just one game.

So, therefore the combination of netball and volleyball is t -2
netball and hockey is t -2
volleyball and hockey is t -2

to find t
35 = 10+5+3+2+(t - 2)+(t - 2)+(t-2)
35 = 20 + 3(t-2)
35 = 20 + 3t - 6
so we therefore collect like terms and make t the subject of the formula
35 - 20 + 6 = 3t
21 = 3t
t = 21 รท 3
t = 7

so therefore
n(V) = 10 +(t-2)+(t - 2)+ 2
if t = 7
n(V) = 10 +(7 -2)+(7 - 2)+2
n(V) = 10 +5 + 5 + 2
n(V) = 22

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5. 15 play equal number of games so the remaining amount after adding the available digits you have will be divided into three for the intersects. 15/3=5. The number of people who play volleyball will be 10 + 2 + 5 + 5 equals 22.

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