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Discrete Maths

  1. Mar 2, 2008 #1
    100 students are surveyed about their sports. Each play at least 1 of the following sports. football, cricket and rugby. 56 play football, 51 play cricket, 55 play rugby, 24 play football and cricket, 21 play football and rugby, 28 play cricket and rugby. How many play all 3 games?

    I tried to do this and I got 0 as my answer. A total of 51 people play cricket, 24 play football & cricket, and 28 play cricket & rugby. 24 + 28 = 52 people play cricket, 52 is greater than 51 which can't be write or the question would be wrong.

    Am I misunderstanding the question? I fail miserably when it comes to this area of maths so any help will be appreciated!
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2
    Since 24 + 28 = 52, that should tell you that there is at least one student who is in both sets ((students who play cricket and rugby) and (students who play cricket and football)). Is it possible for there to be more than one student in this special set (the intersection of these two sets)? Draw a Venn diagram if it helps.
     
    Last edited: Mar 2, 2008
  4. Mar 2, 2008 #3
    Well for another set:

    Football & cricket and Football & Rugby -> 24 + 21 = 45 play football, this is less than 56 so this tell us no students are in both sets?

    Since no student are in therse sets, then surely none can be in all three?
     
  5. Mar 2, 2008 #4
    Actually, this tells us that 56 - 45 students play only football. It does not give us any further information about the triple intersection by itself. Draw a Venn diagram of the triple intersection and you should be able to see what you can do.
     
    Last edited: Mar 2, 2008
  6. Mar 2, 2008 #5
    It's at least 0 for the other remaining sets aswell, so is there 1 person playing all 3? :)
     
  7. Mar 2, 2008 #6
    Not necessarily. All we gathered is that there is at least 1. If you draw a Venn diagram and label what you know, you should be able to get an exact figure from one equation.
     
  8. Mar 3, 2008 #7
    I think there's 7. What did you get?
     
  9. Mar 3, 2008 #8
    I couldn't do it :P I left the answer as at least one, and hoping for some marks!
     
  10. Mar 3, 2008 #9
    when you draw out the diagram, You will see that there are 56+51+55=162 total students playing at least one sport. You subtract off the ones playing two sports, so that's 24+28+21= 79 subtracted from the 162. But in subtracting off the kids with 2 sports, you also subtracted off the kids with three sports, so you have to add them back in...3 times.
    If I let F=football, R=rugby and C=cricket (dull, I know), then I get:
    F+C+R-FC+FCR-FR+FCR-CR+FCR=100
    162-79+3FRC=100
    Solve it from there.
    Look at the picture to see why it's true. 3 circles, 3 intersections of 2 things and one triple intersection.
    CC
     
  11. Mar 3, 2008 #10
    Ah ty!
     
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