*Two trains starting from cities 300 miles apart head in opposite directions at rates of 70 mph and 50 mph, respectively. How long does it take the trains to cross paths?*

This is a classic problem that sends chills up students’ collective spines. I’m now going to add another bone rattling element: The Empty Box.

That’s right—the GRE will have a fill-in-the blank/empty box math problem. There won’t be too many, judging from the ETS Revised GRE book, but even a few should be enough to discomfit most.

**Solution**

Let’s go back, and attack the above problem the following way. When you have any two entities (trains, bicyclists, cars, etc.) headed towards each other you must add their rates to find the total rates. The logic behind this is the two trains (as is the case here) are coming from opposite directions straight into each other.

This yields 120 mph, a very fast rate (which accounts for the severity of head-on collisions…don’t worry the trains in the problem won’t collide).

To find the final answer, we want to employ our nifty old formula: D = RT, where D stands for distance, R stands for rate, and T stands for time.

We’ve already found R, which is their combined rate of 120 mph. They are 300 miles apart so that is D. Plugging those values in, we get 300 = 120T. Dividing 120 by both sides, we get T = 2.5 hrs.

Now we can confidently fill that box in, and let the trains continue on their ways.

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Hi Chris,

For answers which come in a fraction form, are we supposed to get it to its reduced form ?

Hi Urna,

If you are answering a fraction entry question (it will have two boxes, one for numerator and one for denominator) then you do not need the most reduced form. Any equivalent form will be accepted. 🙂

Hi Chris,

But as we have learned from the video, we can not add the rates?

Thanks

Kavita

Hi Kavita,

We can add rates or subtract rates if the units for the two rates are the same. In this question, for example, we are given both speeds in miles/hour (mph). Since the units of the two speeds are the same, we’re allowed to add or subtract them. We add the two speeds in this case, since the two trains are moving in opposite directions. The sum of the two trains’ speeds is the speed at which the distance between the trains is decreasing. We add the two speeds since the two trains are moving towards each other, contributing positively to the rate at which they approach each other. If, for example, one train was not moving, it would take longer for the trains to cross paths, since only one would be moving towards the other.

Does that make sense, Kavita?

For more review, I’d recommend checking out this video lesson from our YouTube channel on shrinking and expanding rates 🙂

I hope this helps!

Hi,

I get

Does that mean we can add or subtract rates every-time we see the rates with same units, even if we have to find the average rate for two trips?

For instance in below question can we subtract the rates?

On a trip of 408 miles, Don travels first 160 miles in 4 hours. At what rate must he travel the remainder of the trip in order to average 34 miles per hour for the entire trip?

A.) 25 mph, B.) 27 mph, C.) 29 mph, d.) 31 mph

Thanks.

Hi Kavita 🙂

While it’s always possible to add/subtract rates with the same units, that doesn’t mean that we’ll always want to do so. The question you’ve asked about, for example, is a weighted average problem, and we need to keep that in mind when solving it. That said, important part of solving any problem involving rates or any type of units is to keep those units straight! We can add/subtract values with the same units, while we can multiply/divide values regardless of their units. However, the units of the product/quotient will be different from those of the original values. Let’s see what I mean in the context of the question about Don’s trip 🙂

We’re told that Don travels for a total of 408 miles. We’re also told that Don travels the first 160 miles in 4 hours. From that information, we can determine how many miles he travels at the other speed:

408 miles – 160 miles = 248 miles

So, we’re being asked how fast Don must travel during the last 248 miles so that his total average speed is equal to 34mph:

(total distance / total time) = d/t = 34 mph

We can use the information in the question to plug in the total distance into the equation:

408 miles / t = 34 miles/hour

We can multiply both sides of the equation by t and divide both sides by 34 miles/hour to get t alone on its own side of the equation:

408 miles / 34 miles/hour = t

Since we have miles in the numerator and denominator on the left side, those units cancel out. We also move the units “hours” to the numerator since we have 1/hours in the denominator:

408/34 hours = t

12 hours = t = total time

Now that we know the total time traveled, we can determine how long the second part of the trip lasted by subtracting the amount of time Don travels during the first 160 miles, 4 hours:

12 hours – 4 hours = 8 hours = time traveled at 34 mph

We’ve figured out a lot of values so far and we have the values we need to answer the question:

* time traveled during 2nd part = 8 hours

* distance traveled during 2nd part = 248 miles

Rate during 2nd part = [distance traveled during 2nd part] / [time traveled during 2nd part] = 248 miles / 8 hours = 31 mph

So, the answer is D. As you can see, we did not add or subtract rates to determine the answer to this question. Rather, we kept the units in mind for the given values in order to solve this weighted average question.

I hope this helps to clear up some doubts 🙂

Thanks Chris!

This helps a lot.

Regards,

Kavita

Hey if the train goes in opposite directions they will never meet

Then what is the question saying..am i getting something wrong here?

Hi Manan!

Two trains going in opposite directions could be going away from each other, but they could also be going towards each other (which is what we’re assuming here). If you picture a train on the east coast heading west, and another train on the west coast heading east, then you can see that they are traveling opposite directions but will still meet somewhere in the middle 🙂

sorry silly question what is aka?????

aka, also stands for “also know as”. Don’t worry–not a silly question 🙂