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Thread: 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:

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

    $\displaystyle \frac{x}{30} = t + \frac{1}{10}$ .... (2)
    Last edited by mr fantastic; Dec 21st 2009 at 03:37 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 $\displaystyle x$ km then the time of the journey at $\displaystyle 40$ km/hr is $\displaystyle x/40$ hr, and at $\displaystyle 30$ km/hr is $\displaystyle x/30$ hr.

    So:

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

    (that is the difference in journey times at the two speeds is $\displaystyle 6$ minutes or $\displaystyle 1/10$ 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 $\displaystyle 6$ minutes or $\displaystyle 1/10$ of an hour).
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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 $\displaystyle D$ = distance (in kilometers).
    Let $\displaystyle t$ = time for a normal run (in hours).

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

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

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

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

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


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

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


    But Captain Black has an elegant solution!

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