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Math Help - Problem based on ratio

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    Problem based on ratio

    Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..


    The answer is 4:11. But I am not getting how to get it.
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    Quote Originally Posted by siddscool19 View Post
    Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..


    The answer is 4:11. But I am not getting how to get it.

    If we assume the first ship has a volume of VL then:

    In ship one there will be V\frac{2}{2+3} of coke and so V(1-\frac{2}{2+3}) of soda

    In ship two there will be 2V\frac{3}{10} of coke and 2V(1-\frac{3}{10}) of soda

    In ship three there will be 3V\frac{1}{5} of coke and 3V(1-\frac{1}{5})

    Add the sum of the coke from the three ships and factor out V.
    <br />
V(\frac{2}{5} + 2\frac{3}{10} + \frac{1}{5}) = \frac{8V}{5}

    Add the sum of the soda from the three ships and factor out V

    V(1-\frac{2}{5} + 2(1-\frac{3}{10}) + 3(1-\frac{1}{5})) = \frac{22V}{5}

    To find the ratio we divide the cola by the soda:

    \frac{8V}{5} \div \frac{22V}{5} = \frac{8V}{5} \times \frac{5}{22V} = \frac{8}{22} = \frac{4}{11}

    Now we have a ratio of \frac{4}{11} : 1 so we multiply by 11 to get an integer ratio of 4:11

    (if you're having trouble visualising it try putting V=5)
    Last edited by e^(i*pi); March 30th 2009 at 07:51 PM. Reason: clarifying my maths and using a general volume V
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  3. #3
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    Quote Originally Posted by siddscool19 View Post
    Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..
    The first vessel is coke:soda = 2:3, or five parts.

    The second vessel is coke:soda = 3:7, or ten parts.

    The third vessel is coke:soda = 1:4, or five parts.

    Overall, their volumes are 1st:2nd:3rd = 1:2:3, but "5:10:5" doesn't match this. The first bit, the "5:10", matches the "1:2", but the second "5" doesn't match the "3". So make it match:

    . . . . .third vessel: 3:12, or fifteen parts

    Now you have:

    The first vessel is coke:soda = 2:3, or five parts.

    The second vessel is coke:soda = 3:7, or ten parts.

    The third vessel is coke:soda = 3:12, or fifteen parts.

    This gives you the right ratio, assuming all the "parts" are of the same volume, so the mixture will then be:

    . . . . .fourth vessel: (2+3+3) : (3+7+12)

    ...or 8:22, which simplifies as 4:11.

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