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Math Help - Problem Solving with Quadratics

  1. #1
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    Problem Solving with Quadratics

    Hi guys,
    I need some help with a question that i just cant figure out how to solve. I have tried a lot but it does not seem to help. I am looking around and there is some complex math so i am guessing this might be a no brain-er for you guys.
    Anyways the question is,

    Two Trains travel a 105 km track each day. The express travels 10kmph faster and takes 30 minutes less than the normal train does. Find the speed of the express.
    The answer is : 51.1 km/h ( the answer book tells me that)

    I need to know how to solve to get to that answer. I cant find a way.
    Thanks in advance, please get to me as soon as you guys can.
    Thanks.
    - Cluelessgeorge
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Use  d = vt .

    So: Let  v and  t be the speed and time of the original train.

    Then  (v+10)(t-0.5) = 105 and  vt = 105 . So  v = \frac{105}{t} . Substitute this into the equation to solve for  v,t .
    Last edited by tukeywilliams; August 1st 2007 at 08:01 AM.
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  3. #3
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    Hey Turkeywilliams,
    Thanks for the suggestion, I tried it out and it did not seem to generate the answer i needed.
    Look below at my solution and see if anything else comes to your mind. Or anybody else, please consider looking and see if you can give me a solution.

    SOLUTION :

    d= vt
    Let v and t represent speed and time.

    (1) (v + 10)(t - 0.5) = 105
    (2) vt = 105 ...v = 105/t

    Sub Equation 2 into 1

    (105/t + 10)(t - 0.5) = 105
    105 - 55/t + 10t - 5 = 105
    105t - 5 +10t^2 - 5t = 105t
    10t^2 - 5t- 55 = 0
    5 ( 2t^2 - t - 11) = 0

    2t^2 - t - 11= 0

    Using a graphing calculator...
    t = -2.108 OR t = 2.608
    As this is a real life situation, negative roots are eliminated as there is no negative time. Hence the only solution for time is 2.608.

    Substitute t = 2.608 into Equation 1

    (v + 10)(2.608 - 0.5) = 105
    (v + 10)(2.108) = 105
    v + 10 = 49.81
    v = 39.81

    This is my solution and according to my book it is wrong. Because my book says that the answer to this question should be 51.1 km/h.

    So please reconsider your original thought process and see what you can figure out.
    Thanks guys,
    I really appreciate it.

    - cluelessgeorge
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  4. #4
    Senior Member tukeywilliams's Avatar
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    You are trying to find  v+10 or the speed of the express train. You solved for the speed of the original train. Also your  t is a bit off. Adding  10 to your  v we get  49.81 which is close to  51.1 .

    I get  t = 2.554 . Plug this into the equation and solve for  v+10 . You should get  51.1 \ \frac{\text{km}}{\text{h}} .
    Last edited by tukeywilliams; August 1st 2007 at 07:58 AM.
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  5. #5
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    Thanks

    Thank you very much
    I did not consider that i was looking for V +10.
    and I am a little surprised that you got different values for x than me.
    I will trust you but i was wondering, why would that happen?
    Thanks again for the help.

    - cluelessgeorge
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  6. #6
    Senior Member tukeywilliams's Avatar
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    I used Maple to get the values. Did you just graph the equation and 'guess' what the value was?
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  7. #7
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    I dont know what maple is but I didn't guess the values.
    I have this computer graphing program which graphs and tells me the intercepts.
    It could have been wrong, I am guessing i should use a graphing calculator, a proper one for accuracy.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    Use  d = vt .

    So: Let  v and  t be the speed and time of the original train.

    Then  (v+10)(t-0.5) = 105 and  vt = 105 . So  v = \frac{105}{t} . Substitute this into the equation to solve for  v,t .
    Why use a graphing calculator for an estimate?
    \left ( \frac{105}{t} + 10 \right ) \left ( t - \frac{1}{2} \right )  = 105<-- Multiply both sides by 2t

    (105 + 10 t) ( 2t - 1)  = 210t

    210t - 105 + 20t^2 - 10t = 210t

    20t^2 - 10t - 105 = 0

    4t^2 - 2t - 21 = 0 <-- Use the quadratic formula

    t = \frac{1 \pm \sqrt{85}}{4}

    Use the "=" solution, as the "-" solution gives a negative time.
    t = \frac{1 + \sqrt{85}}{4}

    So what's v + 10?
    v = \frac{105}{t}

    v = \frac{105}{ \frac{1 + \sqrt{85}}{4} }

    v = \frac{420}{1 + \sqrt{85}} \approx 41.0977

    So v + 10 \approx 51.0977

    -Dan
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