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Percentage problem.

  1. Jun 14, 2006 #1
    This is pretty much the worst problem I've seen this week. It's not quite a homework, but I think it fits requirements of homework-like problems. I spent 15 minutes on it and couldn't come up with any way to solve it and yet I need to know how to solve similar problems for tomorrow, and I don't want to spend day on working out just one problem.

    Here's a problem:
    Nine hundred students were asked whether they thought their school should have a dress code. A circle graph was constructed to show the results. The central angles for two of the three sectors are shown in the diagram. What is the number of students who felt that school should have no dress code?

    Then there's a cirle that shows:
    25 degrees - Partial dress code
    95 deg. - Full dress code
    X deg. - No dress Code

    Thanks, and please don't provide me with a clear answer but rather hint or way of doing,.
     
  2. jcsd
  3. Jun 14, 2006 #2

    Hootenanny

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    HINT: The angle is proportional to the frequency (number of students) of the group. How many degrees in a full circle? What is the value of X is degrees? How many students does one degree represent?
     
  4. Jun 14, 2006 #3
    The angles are in someway proportional to the number of the students for or against the dress code. A certain fraction of a full circle is equal to a fraction of the total number of students against/for the dress code. I think you can easily find the value of X.

    Does that help?

    EDIT: I've been a bit slow in doing things for the past few days. :(
     
  5. Jun 14, 2006 #4
    Thanks guys for quick responses, my error in solving was that I came to a point where both partial dress code and full dress code would give out fractional number of students, and that's not possible. But anyway, I solved this differently and it came up that 'no dress code' = 600, however method seems not to apply for the other two. 25 degrees is a fractional percent of 360, same as 95, what then leads to a fractional number of these two (out of 900). Any ideas?
     
  6. Jun 14, 2006 #5

    Hootenanny

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    OKay I'll start for you;

    [tex]\frac{360}{900} = \frac{2}{5}[/tex]

    Therefore, on your chart [itex]1^{\circ} = \frac{2}{5}[/itex] students.

    Can you go from here?
     
  7. Jun 14, 2006 #6
    That would be only possible if there were less than 900 students. I think you wanted to say 2 degrees equals 5 students. But this as above, leads to a contradiction when applying for 25% and 95%.
    (If I'm thinking wrong, there must be just something wrong with my mind, so sorry for arguing, and please correct me)
     
  8. Jun 14, 2006 #7

    Hootenanny

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    I apologise, it is a while since I have done pie charts. The angle is equal to the proportion of students in that group, thus;

    [tex]\theta = \frac{\text{number of students}}{\text{Total}}\times 360[/tex]
     
  9. Jun 15, 2006 #8

    arildno

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    As a general rule for representative charts of some form (for example pie charts) is that PROPORTIONS ARE PRESERVED.
    Rewriting Hooty's equation shows this:
    [tex]\frac{\theta}{360}=\frac{No. students}{Total no.}[/tex]
     
  10. Jun 15, 2006 #9

    HallsofIvy

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    The problem is, of course, that the "full school" is 900 students and the "full circle" is 360 degrees. That is there should be [itex]\frac{900}{360}= \frac{5}{2}[/itex] student per degree. While it is possible for a full number of students to be represented by an even number of degrees, "95 degrees" and "25 degrees" must be round-offs. And probably unnecessary round-offs at that.
     
  11. Jun 15, 2006 #10
    Ok, I took an exam today and lucky I was similar question didn't appear but anyway. Though I have a solution for this problem I still would like to know how many other students there were and make it all right, just to make not only partial problem true solvable but entire problem, including the 25 and 95 degrees.
    I think they wanted just people just to present the skill of knowing proportions and solving similar problems, however it doesn't change the fact that they should either tell about rounding or use actual percentages. Yepp, calculations show that they rounded degrees and number of students could be off by ~1-4, or a lot less likely made a mistake.
    Ok, thank you all guys for helping me :)
     
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