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Math Help - Train journey problem

  1. #1
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    Train journey problem

    I can't do this at all, please explain clearly. Many many thanks!


    A train does the first part of a journey at 30mph, then the second part at 60 mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?
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  2. #2
    Member u2_wa's Avatar
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    Quote Originally Posted by Natasha1 View Post
    I can't do this at all, please explain clearly. Many many thanks!


    A train does the first part of a journey at 30mph, then the second part at 60 mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?
    I got 5/12 and this is how I arrived at my answer:

    T=time for journey with speed 30Kmph
    t=time for journey with speed 60Kmph


    Think of two trains: train 1 is moving at avg. speed and Train 2 is the train given in the question.

    During first part the difference in the distance covered by these trains would be:
    (42 6/17 - 30)T

    During second part the difference in the distance covered by these trains would be:
    (60 - 42 6/17)t

    As we know both trains cover the same distance finally so net effect of the differences would be zero.

    (42 6/17 - 30)T-(60 - 42 6/17)t =0


    T/t=10/7
    D/d=(10*30)/(7*60) (I multiplied time with speed to calculate distance)

    D/(D+d)=5/12
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  3. #3
    Member pflo's Avatar
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    Quote Originally Posted by u2_wa View Post
    I got 5/12
    I concur, using a different method.

    d_{1} is the distance of the first part
    t_{1} is the time of the first part

    d_{2} is the distance of the second part
    t_{2} is the time of the second part

    D is the total distance and D=d_1+d_2
    T is the total time and T=t_1+t_2

    The question is asking for \frac{d_1}{D} and we know \frac{d_1}{D}+\frac{d_2}_D=1.

    Simply using d=rt, you can come up with the following formula: 30t_1+60t_2=42\frac{6}{7}T. You can solve this equation for [tex]\frac{t_2}{T}[tex]:

    30t_1+30t_2+30t_2=42\frac{6}{7}T
    30(t_1+t_2)+30t_2=42\frac{6}{7}T
    30T+30t_2=42\frac{6}{7}T
    30t_2=12\frac{6}{7}T
    \frac{t_2}{T}=\frac{7}{17}

    So, \frac{t_1}{T}=\frac{10}{17}

    Again using d=rt, you can perform the following calculation:
    \frac{d_1}{D}=\frac{30*10}{42\frac{6}{17}*17}=\fra  c{5}{12}.
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