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Maths about ratio

  1. Apr 18, 2005 #1
    1)If (x+y-z)/3
    =(y+z-x)/4
    =(z+x-y)/5,
    then x:y:z=
    A.3:4:5
    B.8:7:9
    C.9:16:25
    D.15:12:20

    The answer for the above question is option B. But I don't know how to solve this kind of questions. Can anyone tell me what should I do in the first step? Thanks for answering my question.
     
  2. jcsd
  3. Apr 18, 2005 #2
    step 1: 3*4*5=60
    step 2: 20x+20y-20z=15y+...
     
  4. Apr 18, 2005 #3
    Step2:20x+20y-20z=15y+15z-15x=12z+12x-12z
    Step3:35x+32y=27y+35z=32z+27x

    x=4(z-y)
    y=7(z-x)
    z=9(x-y)

    That's what I solve from the above equation. After that, what should I do? Thanks.
     
  5. Apr 18, 2005 #4

    HallsofIvy

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    Did you see AntonVrba's response? Since you asked to choose among different possible answers, the simplest thing to do is try each.
    x= 3, y= 4, z= 5 ARE in ratio 3:4:5. Do those numbers satisfy the original equations?

    If not try x= 8, y=7, z= 9, etc.
     
  6. Apr 18, 2005 #5
    no need for trial and error of trying given posibilities, you can calculate
    from the three relationships x,y,z that you already have solved by substituting first y and then for second z in the first of your equations.

    x = 4z - 4y
    = 4z - 28z + 28x
    or
    24z = 27x
    8z = 9 x
    8/9 = x/z or 8:9 = x:z

    and

    x = 4z - 4y
    x = 36x - 36y - 4y
    ........
     
  7. Apr 18, 2005 #6

    HallsofIvy

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    Yeah, but it works and thought it would be simpler for fork.
     
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