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Math Help - Distance, speed and time problem

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    Distance, speed and time problem

    A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?
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    Quote Originally Posted by saberteeth View Post
    A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?
    Solve the following two equations simultaneously for x:

    \frac{x}{40} = t .... (1)

    \frac{x}{30} = t + \frac{1}{10} .... (2)
    Last edited by mr fantastic; December 21st 2009 at 04:37 AM. Reason: Fixed a typo
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    Quote Originally Posted by saberteeth View Post
    A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?
    If the distance between the stations is x km then the time of the journey at 40 km/hr is x/40 hr, and at 30 km/hr is x/30 hr.

    So:

    \frac{x}{30} - \frac{x}{40} = \frac{1}{10}

    (that is the difference in journey times at the two speeds is 6 minutes or 1/10 of an hour).

    CB
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    Quote Originally Posted by CaptainBlack View Post

    (that is the difference in journey times at the two speeds is 6 minutes or 1/10 of an hour).
    Thanks!
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    Hello, saberteeth!

    A train running between two stations arrives at its destination
    10 minutes late when it travels at 40 km/hr
    and 16 minutes late when it travels at 30 km/hr.
    Find the distance between the two stations.

    Let D = distance (in kilometers).
    Let t = time for a normal run (in hours).

    We will use: . \text{Distance} \:=\: \text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}

    At 40 kph, it is 10 minutes \left(\tfrac{1}{6}\text{ hr}\right) late.
    . . \frac{D}{40} \:=\:t + \frac{1}{6} \quad\Rightarrow\quad D \:=\:40t + \frac{20}{3}\quad [1]

    At 30 kph, it is 16 minutes \left(\tfrac{4}{15}\text{ hr}\right) late.
    . . \frac{D}{30} \:=\:t + \frac{4}{15} \quad\Rightarrow\quad D \:=\:30t + 8\quad[2]

    Equate [1] and [2]: . 40t + \frac{20}{3} \:=\:30t + 8

    Hence: . 10t \:=\:\frac{4}{3} \quad\Rightarrow\quad t \:=\:\frac{2}{15}


    Substitute into [1]: . D \:=\:40\left(\frac{2}{15}\right) + \frac{20}{3}

    Therefore: . D \:=\:12\text{ km}


    But Captain Black has an elegant solution!

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