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Today’s post was written by Rich Zwelling.

One of the most common areas of frustration for my GMAT students is rate problems. This seems a general extension of the challenges that word problems overall pose to students, but rate problems are particularly tricky. They require intensive setup and often rely on your realizing an implicit piece of information.

As a basic example, suppose I tell you that two joggers run a single lap around the same track. Aaron runs at a rate of 5 meters per second and takes 80 seconds to complete the lap. I then ask you to calculate the differences in the two runners’ times for a single lap if Ben runs at a rate of 4 meters per second. On any rate problem involving distance, rate, and time, you can always default to standard formula (d = r*t) and use that as your guide. But even knowing that is only part of the battle; you must then recognize the implicit information that helps you set up the problem.

In this case, the implicit information is relatively easy to identify: the two runners each ran a single lap around the same track, which implies that the distances are equal. So if we were setting up the equation, it would look something like this:

d_aaron = d_ben (i.e. the distance Aaron runs is the same as the distance Ben runs)

Since d = r*t, we can do a substitution:

(r_aaron) * (t_aaron) = (r_ben) * (t_ben)

(5 m/sec) * (80 sec) = (4 m/sec) * (t_ben)

When we isolate Ben’s time, we find that it is equal to 100 seconds, leading to a 20-second difference in the two runners’ times.

Now, let’s apply the same principle to a question from the official GMAT materials and see how this common-sense approach can aid us:

Car A is 20 miles behind Car B, which is traveling in the same direction along the same route as Car A. Car A is traveling at a constant speed of 58 miles per hour and Car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

(A) 1.5

(B) 2.0

(C) 2.5

(D) 3.0

(E) 3.5

If you set this up immediately with d=r*t, you could label the times for each car as equal, since we start at the same instant (when Car A is 20 miles behind) and finish at the same instant (Car A is 8 miles ahead). But you’d then have to acknowledge that the distance Car A travels is 28 miles greater than the distance B travels, and that leads you to something like d_a = d_b + 28. Setting this all up immediately in a system seems like a daunting and confusing prospect.

But before we overwhelm ourselves with variables, let’s look at the prompt a little more closely and see if some good old-fashioned common sense can bail us out:

We know Car A and Car B travel at 58 mph and 50 mph, respectively. So the difference in speed is 8 mph. Now, what does that mean in real-world terms? It means that in one hour of travel, Car A will gain 8 miles on Car B. We also already know that in the amount of time in question, Car A travels a distance 28 miles greater than that of Car B. So really, we’ve changed the frame of reference to include differences in distance and rate, rather than individual distances and rates.

So applying the standard formula to these changes in distance and rate:

d = r * t

28 mi = (8 mi/hr) * t

t = 28/8 hr = 3.5 hr (final answer: E)

Added note for all the science enthusiasts: This is a basic illustration of the Relativity of Simultaneity principle explored by Einstein and Lorentz. Those of you who have done higher math might have seen Lorentz Transformations.

Now for some homework! Take a look at this OG question, and before you set up any equations, see if you can find a clever way to short-circuit the problem:

Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed the was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3

(B) 1

(C) 1 1/3

(D) 1 3/5

(E) 3

Did you find a short-cut for this problem? Let us know in the comments!