Algebra problem: two speeds

Photo courtesy of Ben Cooper

The speed of a freight train is 14km/h slower than the speed of a passenger train. The freight train travels 330 km, in the same time that it takes a passenger train to travel 360 km. Find the speed of each train.
Again, an algebra problem about speeds. Again, we will make a simple table about the two trains. The table will have columns for speed, distance, and time.

| distance | speed | time
Freight train | 330 | v |
Passenger train| 360 | v + 14 |

Notice the problem says "in the same time". Let's call that t.

| distance | speed | time
Freight train | 330 | v | t
Passenger train| 360 | v + 14 | t

Of course, the goal is to have an equation in a single variable, not in two variables (t and v).

Since the time is the same, we can build an equation of the time t = ... for both trains, and then set those expressions to be equal.

For the freight train, t = distance/speed = 330/v
For the passenger train, t = 360/(v + 14)

Now let's make those equal:

330/v = 360/(v + 14)

Cross multiply:

360v = 330(v + 14)

360v = 330v + 4620

30v = 4620

v = 154

So, the speed of the freight train is 154 km/h and the speed of the passenger train is then 168 km/h.

Check: How long will it take for the freight train to travel 330 km? Well, 330/154 hours, or 2.142857143 hours.

How long will it take for the passenger train to travel 360 km? Well, 360/168 hours, or 2.142857143 hours. Seems to match.


Cassidy Cash said…
Hey! This is a great website. I'm glad you're putting up the examples and sample problems. Some things you might consider adding, though, are some clarification points. For instance, you might state that Distance = Rate(Time) and that Rate is the same as speed for this level of math since speed is being expressed in rate form. A lot of students will be looking at these concepts for the first time and may not realize those things.
shooterboss said…
Hurray! I got this one correct on my first try without looking at any parts of the solution!

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