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Math Help - fractions problem

  1. #1
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    fractions problem

    Can the following story problem about voters be solved by adding 1/2 + 1/3? If so, explain why. If not explain why not, and solve the problems if they can be solved. Write a differnet story problem about the voters that can be solved by adding 1/2 + 1/3

    Story problem 1 about voters: In Kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. Altogether, what fraction of the voters voted for the referendum in Kneebend County?

    Story Problem 2 about voters: In kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. There are the same number of women voters as men voters in knneben County. Altogether, what fraction of the voters voted for the referendum in Kneebend County.
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    Quote Originally Posted by recca View Post
    Can the following story problem about voters be solved by adding 1/2 + 1/3? If so, explain why. If not explain why not, and solve the problems if they can be solved. Write a differnet story problem about the voters that can be solved by adding 1/2 + 1/3

    Story problem 1 about voters: In Kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. Altogether, what fraction of the voters voted for the referendum in Kneebend County?

    Story Problem 2 about voters: In kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. There are the same number of women voters as men voters in knneben County. Altogether, what fraction of the voters voted for the referendum in Kneebend County.
    Okay, so for this one, i'm not too sure what the difference is.... even though the second one states that "there are the same number of women voters as men voters," it doesn't exactly change anything since it just doesn't matter to the problem... So, yeah. I guess you'd add them together.... I'm not sure. So it would be, \frac{1}{2}+\frac{1}{3}\Rightarrow\frac{3}{6}+\fra  c{2}{6}=\frac{5}{6}
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by blair_alane View Post
    Okay, so for this one, i'm not too sure what the difference is.... even though the second one states that "there are the same number of women voters as men voters," it doesn't exactly change anything since it just doesn't matter to the problem... So, yeah. I guess you'd add them together.... I'm not sure. So it would be, \frac{1}{2}+\frac{1}{3}\Rightarrow\frac{3}{6}+\fra  c{2}{6}=\frac{5}{6}
    I'm sure.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by recca View Post
    Can the following story problem about voters be solved by adding 1/2 + 1/3? If so, explain why. If not explain why not, and solve the problems if they can be solved. Write a differnet story problem about the voters that can be solved by adding 1/2 + 1/3

    Story problem 1 about voters: In Kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. Altogether, what fraction of the voters voted for the referendum in Kneebend County?
    Story problem 1 cannot be answered by adding the fractions. it matters how many women in respect to men there are.

    why? let me give a counter example and you can try to explain why it doesn't work.

    let's say there are 100 female voters and 30 male voters. then there are 130 voters in all.

    so we have 50 women and 10 men that voted for the referendum. Hence, the fraction of voters that voted for the referendum is (50 + 30)/130 = 80/130 = 8/13 \ne 5/6. so adding 1/2 and 1/3 did not work!


    Story Problem 2 about voters: In kneebend County, 1/2 of the women voters and 1/3 of the male voters voted for a certain referendum. There are the same number of women voters as men voters in knneben County. Altogether, what fraction of the voters voted for the referendum in Kneebend County.
    Now let's see if this one works.

    Let (1/2)A represent the number of female voters. So there are (1/2)A male voters as well.

    Thus, there are A voters in all.

    we have (1/4)A votes from the women (this is 1/2 or (1/2)A) and (1/6)A votes from the men (this is 1/3 of (1/2)A).

    in all, there are (1/4)A + (1/6)A = (1/4 + 1/6)A = (10/24)A = (5/12)A

    so again, it is not 5/6 of the votes were made.

    thus, neither scenario can be solved by adding 1/2 and 1/3
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