A passenger train takes three times as long to pass a freight train when they are both going in the same direction as when they are going in opposite directions. Assuming that the trains travel at uniform speeds, how many times faster than the freight train is the passenger train going?
Note: What happens if you change the "three times" to "n times"?

**Hint:** Draw a picture to illustrate your understanding of the problem.

Do the lengths of each train matter? Why or why not?

Take a guess for the number of "times faster" and then see what happens when testing your guess out in the problem.

**Solution Commentary:** When all else fails, one can use the power of algebra. But, in addition to using the algebra, make sure you look at the every end of this commentary where an "aha" solution is discussed.

From the problem, we can let r_{p}=nr_{f} and t_{same}=3t_{opp}.

Then, the problem statement can be represented by the equation (r_{p}-r_{f})t_{same}=(r_{p}+r_{f})t_{opp}.

Using the time relationship, we get (r_{p}-r_{f})3t_{opp}=(r_{p}+r_{f})t_{opp}.

Bringing everything on the left side and simplifying, we get 2r_{f}t_{opp}(n-2)=0. And, since neither r_{f} nor t_{opp} can equal 0, we know (n-2)=0 or that n=2. That is, the passenger train is two times faster than the freight train.

Now, let's leave the algebra behind. Think about the following simple picture, which represents the time t_{opp} needed: === === ===. Visualize the trains going in opposite directions...they will only need one-third the time (or ===). Think about this carefully and try to pull out the necessary relationship of r_{p}=2r_{f}, remembering that in one case you are working with the combined speeds of the train and in the other you are working with the difference of the train's rates.

A final note and some consolation: My thanks to a former student for the visual solution, as it brought me out of the "mire" of my algebraic solution.