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Thread: Fractions & ratios when items are removed

  1. #1
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    Fractions & ratios when items are removed

    I have a word equation i cannot figure out! anyone have any ideas?

    There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

    Im stumped!
    I've written that:

    Red - n = 3/4 Red
    blue - n = 2/5 blue
    n is the number of counters removed as it the same number removed from both.

    therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!




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  2. #2
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    Re: Fractions & ratios when items are removed

    Hint:
    r = 24, b = 10, n = 6

    r: 24 - 6 = 18 : 3/4 left
    b: 10 - 6 = 04 : 2/5 left
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  3. #3
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    Re: Fractions & ratios when items are removed

    Quote Originally Posted by Jessic17 View Post
    I have a word equation i cannot figure out! anyone have any ideas?

    There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

    Im stumped!
    I've written that:

    Red - n = 3/4 Red
    blue - n = 2/5 blue
    n is the number of counters removed as it the same number removed from both.

    therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!

    we have

    $\displaystyle r=4n$ and $\displaystyle b=\frac{5}{3}n$

    the fraction we are looking for is

    $\displaystyle \frac{r+b-2n}{r+b}$
    Thanks from topsquark
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  4. #4
    Member Cervesa's Avatar
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    Re: Fractions & ratios when items are removed

    $n = \dfrac{r}{4} = \dfrac{3b}{5}$

    $\dfrac{3b}{5} \div \dfrac{r}{4} = \dfrac{n}{n} \implies \dfrac{12}{5} \cdot \dfrac{b}{r} = 1 \implies \dfrac{b}{r} = \dfrac{5}{12}$

    $b = 5k$ and $r=12k$ for some positive integer $k$

    ratio of counters remaining

    $\dfrac{\frac{3}{4} \cdot 12k + \frac{2}{5} \cdot 5k}{12k+5k}= \dfrac{11}{17}$
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