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Math Help - Ratios: Combinining A:B , C:D to yield A:B:C:D.

  1. #1
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    Ratios: Combinining A:B , C:D to yield A:B:C:D.

    Greetings to All!

    This is my very first post on a Math forum. I am studying for GRE and there's one particular problem type on ratios which confuses me totally.. Multiple ratios combining..

    Say I have two ratios with me

    A : B => 21:23
    C : D => 9:7

    How do I determine A : B : C : D?

    Thx a lot for your replies and your time..

    Sincerely,
    Abhijit.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    No, unless you are given more information, you won't be able to find A:B:C: D.

    I could very well have A:B = 1:2 and also C: D = 1:2, but where A = 2, B = 4, C = 3, D = 6. THe ratio stilll holds, but you cannot directly find A:B:C: D unless you have something to compare one of each of the ratios given.
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  3. #3
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    Re: Ratios: Combinining A:B , C:D to yield A:B:C:D.

    Quote Originally Posted by Unknown008 View Post
    No, unless you are given more information, you won't be able to find A:B:C: D.

    I could very well have A:B = 1:2 and also C: D = 1:2, but where A = 2, B = 4, C = 3, D = 6. THe ratio stilll holds, but you cannot directly find A:B:C: D unless you have something to compare one of each of the ratios given.
    Thank you for your reply.
    Correct, and I agree.
    However, please help me solve the below problem

    If A=2B=3C=4D then find
    A : B: C: D

    Thanks,
    Abhijit.
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  4. #4
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    Hello, Abhijit!

    \text{Given: }\:A=2B=3C=4D.\;\;\text{Find }\,A : B: C: D

    We have: . 3C = 4D \quad\Rightarrow\quad C = \tfrac{4}{3}D

    . . 2B = 3C \quad\Rightarrow\quad B = \tfrac{3}{2}C \quad\Rightarrow\quad B = \tfrac{3}{2}\left(\tfrac{4}{3}D\right) . . \Rightarrow\quad B = 2D

    . . A = 2B \quad\Rightarrow\quad A = 2(2D) \quad\Rightarrow\quad A = 4D


    \text{Hence: }\;A:B:C:D \;=\;4D: 2D: \tfrac{4}{3}D: D


    \text{Multiply by }\tfrac{3}{D}\!:\quad A:B:C:D \;=\;12:6:4:3

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  5. #5
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    Aha!!! Cool..

    So you evaluated everything in terms of D and then substituted all the values you derived...

    Thanks a lot...

    Cheers,
    Abhijit.
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