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Math Help - Train Sum

  1. #1
    Bar0n janvdl's Avatar
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    Train Sum

    Could someone please confirm the answer for me?

    A Passenger train is x times faster than a Goods train.
    If the trains move in the same directrion, it takes them x times longer to pass each other than when they're moving in opposite directions. What is x?

    OK, this is what i did:

    Passenger train speed = xv
    Goods train speed: v

    Moving in the same direction:
    xv + v = tx
    Thus:
    xv = tx - v ---- (1)

    Moving in the opposite direction:
    xv - v = t
    Thus:
    xv = t + v ---- (2)

    Set (1) = (2)

    tx - v = t + v
    tx - t = 2v
    x = (2v/t) + 1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by janvdl View Post
    Could someone please confirm the answer for me?

    A Passenger train is x times faster than a Goods train.
    If the trains move in the same directrion, it takes them x times longer to pass each other than when they're moving in opposite directions. What is x?

    OK, this is what i did:

    Passenger train speed = xv
    Goods train speed: v

    Moving in the same direction:
    xv + v = tx
    And what is t, and why is this true?

    RonL
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  3. #3
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    Hello, janvdl!

    You should get a number for the value of x.


    A Passenger train is x times faster than a Goods train.
    If the trains move in the same directrion, it takes them x times longer to pass each other
    than when they're moving in opposite directions. .What is x?

    Let v = speed of the Goods train.
    Let xv = speed of the Passenger train.


    Going in opposite directions, their relative speed is: .xv + v mph.
    It is as if the G-train is stopped,
    . . and the P-train is approaching at xv + v mph.
    Code:
          * - - - - - *       * - - - - - *
          | Passenger |   →   | Passenger |
          * - - - - - *       * - - - - - *
                      * - - - *
                      | Goods |
                      * - - - *
    To pass the G-train, the P-train must move the length of the G-train (LG)
    . . plus the length of the P-train (L
    P).

    Suppose it takes t hours for this to happen.
    . . Then we have: .(xv + v)t .= .L
    G + LP .[1]


    Going in the same direction, their relative speed is: .xv - v mph
    It is as if the G-train is stopped,
    . . and the P-train is approaching at xv - v mph.

    To pass the G-grain, the P-train must move the length of the G-train
    . . plus the length of the P-train.

    We are told that this takes xt hours.
    . . Then we have: .(xv - v)xt .= .L
    G + LP .[2]


    Equate [2] and [1]: .(xv - v)xt .= .(xv + v)t . . . . v(x - 1)xt .= .v(x + 1)t

    . . Divide by vt: .(x - 1)x .= .x + 1

    . . and we have the quadratic: .x - 2x - 1 .= .0

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
    Using the Quadratic Formula, the positive root is: .x .= .1 + √2

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  4. #4
    Bar0n janvdl's Avatar
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    So can we not say that for them to pass each other equals a certain time?

    xv + v = xt
    And xv - v = t
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post

    . . Then we have:.(xv + v)t .= .LG + LP .[1]


    . . Then we have:.(xv - v)xt .= .LG + LP .[2]

    Sorry Soroban, you might a slight error in your calculation.

    Moving in the same direction it takes tx

    So shouldn't it be:
    (xv + v)xt [1]
    (xv - v)t [2]

    And then x = root(v)
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