Further mathf

In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:
The number of students that play football;
The number of students that play football but not hockey;
The number of students that play hockey but not footbal.

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  1. number who play hockey ---- x
    number who play football -- 2x+5

    x + 2x+5 = 80
    x = 25

    so 25 play hockey, 55 play football, and 15 play both

    now we can use a Venn diagram, filling in as follows:
    15 in the intersection
    55-15 or 40 in the football only part of the circle
    25-15 or 10 in the hockey only part of the circle

    All your questions can now be easily answered.

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  2. solve it for me

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  3. pls i really need the working and answer,can u help me with it

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  4. I don't get what reiny is trying to say

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  5. In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:
    The number of students that play football;
    The number of students that play football but not hockey;
    The number of students that play hockey but not football. Full working sir

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  6. U={80}
    5 more than twice of the number of hockey players plays football.
    Let the hockey players(H) be X
    Football players(F) be 2X+5
    To find X
    2X+5=FnH'+15. Equation1 X=HnF'+15. Equation2
    80=(2X+5-15)+15+(X-15)
    X=30

    From here substitute X into equation1 and 2 to get F only nd H only

    And remember that F will become 2(30)+5=65

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  7. In the department of mathematics, unilag, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If 15 students play both games and every student in school plates at least one game. Find the number of students that play football but not hockey

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  8. H=x
    F=5+2x
    Therefore, x+5+2x+15=80
    3x+20=80
    3x=60
    X=20

    Proof: x+5+2x+15
    20+5+40+15=80

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  9. I got lost on Alexander explanation starting from how to get x

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  10. this maths is confusing especially on how to get the number of those who played football when hockey is 5 and hockey only is 20. i.e H=5+2x+15+x=80. 5+15+2x+x=80 20+3x=80 3x=80-20 3x=60 hockey only=20 hockey=20-15 =5

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  11. D response I'm getin isn't helpin .d ansa from alexandra nd anonymous ren't same.wich should I chose x=20 or x=30

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  12. Equation on number of students that play football only

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  13. To get x = 30 just add the number of two games and the no that play both and equate it to no of the student, and no of the student is 80
    X-15+15+(2x+5)-15=80
    X+2x+5-15=80
    3x-10=80
    3x=80+10
    3x=90
    X=30

    To find no. (ii) and (iii)
    ii. n(F)only= total no of F-- none
    And total no of F is 2x+5
    =2(30)+5
    =65
    Therefore, n(F)only=65-15 =50

    iii. n(H)only = total no of H -- none
    And total no of H is x which is 30
    Therefore, n(H)only=30-15
    =15

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  14. Total number of student =80
    N(F n H)=15
    N(F)=5+2x
    N(H)=x
    (I). The number that play football
    To get that , we need to sum up all the n( F) and n(H) to 80
    I.e 5+2x-15 +15+X-15=80
    3x-10=80
    3x=90
    X=30
    Therefore; n(F)=5+2x
    =5+60=65.
    (ii). The number that play football but not hockey
    N(F n H')
    To get that we need to subtract the intercept from the to ta number that play football .and the intercept is 15
    Therefore; 65-15=50
    (ii). The number that play hockey but not football
    N(F' n H)
    We already know that x =30 and the intercept is 15
    Therefore ; 30-15=15.

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  15. The number of those that played football but not hokey

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  16. This was so helpful

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