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A trick question

  1. Nov 30, 2009 #1
    R=(3065-2965)/(3064+2964)

    find the range of R?

    can any please help me out..
    what approach i need to attack this problem?
     
  2. jcsd
  3. Nov 30, 2009 #2

    tiny-tim

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    Hi vishal007win! :smile:

    I don't understand :redface:

    (3065-2965)/(3064+2964) is a constant, so how can there be a range? :confused:
     
  4. Nov 30, 2009 #3
    sorry the question was..
    what is the range in which R lies?
    is it
    1. 0< R <0.1
    2. 0.1< R < 0.5
    3. 0.5< R <0.7
    4 0.7< R <1
     
  5. Nov 30, 2009 #4
    I think the question is asking for something else.

    If you find the value of R, it ends up being approximately 24 and doesn't lie in any of those choices.
     
  6. Nov 30, 2009 #5
    @retracell
    how did you approached to this solution??
     
  7. Nov 30, 2009 #6
    yep, approximately 24.

    The easiest way to see that it's on the order of 30 is to divide the numerator and the denominator by 30^65 and to use the approximation [itex](1+x)^y \approx e^{xy}[/itex], which holds with good accuracy if x is small and y is large.
     
  8. Dec 1, 2009 #7
    yup after dividin by 30^65 both num. and den.
    i get something like this
    R=(1-y65)/30*(1+y64)

    where y=29/30

    now writing y=(1-x)
    where x=1/30

    [itex]
    (1-(1-x)^{65})/30*(1+(1-x)^{64})
    [/itex]
    now using this
    [itex]
    (1+x)^y \approx e^{xy}
    [/itex]
    i got

    [itex]
    (e^{65x} -1)/30*e^x*(e^{64x}+1)
    [/itex]

    which finally gives the answer R=0.024

    now please check the solution...n point out my mistakes
     
  9. Dec 1, 2009 #8
    No, it should be

    [tex] 30 * (1-e^{-65x}) / (1+e^{-64x}) \approx 30 * 0.9 / 1.1 \approx 24[/tex]
     
  10. Dec 1, 2009 #9
    :cry: calculation mistake..

    thnx
    now i got it...
     
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