1. ## Train Sum

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

2. Originally Posted by janvdl

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

3. 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: .x² - 2x - 1 .= .0

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

4. So can we not say that for them to pass each other equals a certain time?

xv + v = xt
And xv - v = t

5. Originally Posted by Soroban

. . 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)