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anonymous

  • 4 years ago

2/3 of the men are married to 3/4 of the women in Fractionville. What fraction of the people in Fractionville are married? I've gotten it to 2/3x = 3/4y but I'm not sure what to do from here.

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  1. anonymous
    • 4 years ago
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    you should have two equations, since you have two unknowns. so if you let x+y=1 you should be in good shape. (x and y are the corresponding fractions of the town that are male and female)

  2. phi
    • 4 years ago
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    you need another equation x+y=1 means the men and women add up to the total population

  3. anonymous
    • 4 years ago
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    x = the number of men in fractionville and y = the number of women in fractionville

  4. anonymous
    • 4 years ago
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    how can x + y = 1 though?

  5. anonymous
    • 4 years ago
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    if x is the fraction of the population that is male and y is the fraction of the town that is female, they should add up to 1 which is the fraction of people in the population

  6. anonymous
    • 4 years ago
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    but how is x the fraction of- okay I kinda see, but explain how i get my x and y definitions?

  7. asnaseer
    • 4 years ago
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    @JunkieJim is correct

  8. anonymous
    • 4 years ago
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    i just dont understand it though shouldnt x + y = P (population)?

  9. anonymous
    • 4 years ago
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    Yes, that's exactly what we're doing, only we're dividing everything by P so you've got \[\frac{x}{Population} +\frac{y}{Population} = \frac{Population}{Population}\] so that everything remains as fractions

  10. phi
    • 4 years ago
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    Think of it this way: some fraction of the people are men and some fraction are women so x+y=1 the fraction of the population that are married is 2/3 x (another fraction)

  11. phi
    • 4 years ago
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    which of course is the same fraction as 3/4 y

  12. anonymous
    • 4 years ago
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    I appreciate how you are taking the time to explain it to me. And I'm understanding it better. I can do the systems from here, but that x + y = 1 is bothering me

  13. anonymous
    • 4 years ago
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    what you said makes sense Phi, I'm just wondering how i got my definitions for x and y

  14. asnaseer
    • 4 years ago
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    @mridrik: another way of looking at this is to call the total population p and the number of men m. then, number of women = p - m and you can write:\[\frac{2m}{3}=\frac{3(p-m)}{4}\]

  15. phi
    • 4 years ago
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    your definitions should be modified x = the fraction of the population that are men in fractionville and y = the fraction of the population that are women in fractionville

  16. anonymous
    • 4 years ago
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    so are my definitions wrong, or are they right but do not help me advance with the problem?

  17. phi
    • 4 years ago
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    I would *not* use x = the number of men in fractionville and y = the number of women in fractionville

  18. anonymous
    • 4 years ago
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    with asnaseers way, we get m = 9/17p, so how could I use that to get the answer?

  19. asnaseer
    • 4 years ago
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    2m/3 are married

  20. phi
    • 4 years ago
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    you can only find m/p (a fraction)

  21. anonymous
    • 4 years ago
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    so 2/3 * 9/17 gives us ?

  22. anonymous
    • 4 years ago
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    now I'm confused again

  23. phi
    • 4 years ago
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    the fraction married

  24. anonymous
    • 4 years ago
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    but we already knew the fraction married 2/3

  25. anonymous
    • 4 years ago
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    i have a feeling i sound pretty stupid right now to you guys

  26. asnaseer
    • 4 years ago
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    no - 2/3 is the fraction of men that are married - not the fraction of the population that are married.

  27. phi
    • 4 years ago
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    double it (we have to count the women)

  28. anonymous
    • 4 years ago
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    so 2/3 * 9/17 * 2?

  29. phi
    • 4 years ago
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    Sounds good. we know 2/3 * 9/17 = 6/17 of the population is married men 3/4 * 8/17 is 6/17 of the population = married women total fraction 12/17

  30. anonymous
    • 4 years ago
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    You guys are genii! (I think thats genius plural) Thank you, so 12/17 of the population of fractionville is married?

  31. asnaseer
    • 4 years ago
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    :) we've just had a bit more practice than you have mridrik thats all. and yes phi has the correct answer.

  32. anonymous
    • 4 years ago
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    It makes me happy to know that people care about another's will to learn.

  33. asnaseer
    • 4 years ago
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    thats what makes this site so wonderful :)

  34. anonymous
    • 4 years ago
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    If i could, i would make a program that gave each of you over 9000 medals

  35. asnaseer
    • 4 years ago
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    :O) - careful - we may drown in glory - lol!

  36. anonymous
    • 4 years ago
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    yeah lol, my math teacher couldnt even get this today in class

  37. asnaseer
    • 4 years ago
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    try and be gentle on him - after all - he is also human :)

  38. anonymous
    • 4 years ago
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    yes, maybe I'll get the chance to explain what you showed me to the class, even though i dont that well with public speaking, only about 20 students in my algebra 2 class

  39. asnaseer
    • 4 years ago
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    good luck my friend...

  40. anonymous
    • 4 years ago
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    thanks for your help, have you ever heard of mathcounts?

  41. asnaseer
    • 4 years ago
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    no - what are they?

  42. anonymous
    • 4 years ago
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    It's a program that my school participtes in with schools from around the United States, we go to regionals first, (district is too small) and usually win that, and then go to state where this one girl always wins, its from 6th to 8th grade (my last year) and its a blast when we go there to lexington this year (KY) but they questions get ridiculously hard towards the end of the competition and i bet you would beast them up

  43. asnaseer
    • 4 years ago
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    I'm sure as time goes by you will become a mathematical ninja and win that contest hands down :) I'm probably far too old for that now :)

  44. anonymous
    • 4 years ago
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    oh no way, the way you win is the top ten scorers in the written test get to go to a quick recall type thing, and she has won that in 6th grade and 7th grade, I got 40 something place and my friend got in the 30s, but there are a LOT of people there so thats pretty good i guess, our school got 10th place overall last year

  45. asnaseer
    • 4 years ago
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    that is an extremely good achievement. and never think that you will not be able to achieve even better goals for yourself - nothing in this world is impossible if you work hard enough at it. you certainly sound like someone who is willing to work hard to achieve high goals in life.

  46. anonymous
    • 4 years ago
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    Thanks, I would like to be an engineer when I grow up, not really sure what they do, but everyone says it involves math so that sounds fun.

  47. asnaseer
    • 4 years ago
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    yes - there is a lot of maths involved in engineering. your main focus should be to do something you really enjoy - after all, you will most likely be working for the rest of your life in one particular job - so might as well pick something you really enjoy. anyway - must go now. good talking to you.

  48. anonymous
    • 4 years ago
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    you too

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