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# GRE Math — Do Rates Freak You Out?

Two trains are headed along parallel tracks in the same direction at varying rates of 72 mph and 47 mph. In how many hours will the faster train be 100 miles in front of the slower train?

(A) 2 hrs

(B) 2.5

(C) 3.5

(D) 4

(E) Cannot be determined by information provided.

Last time, in rates, I mentioned two trains headed towards each other at varying rates. In that case, we had to add the train’s respective rates to find a combined rate.

Now, the trains are not headed towards each other but are going in the same direction. We could solve the problem by finding how long it would take each train to reach the other side. Then we could subtract the times by each other. That would take a very long time, especially with the numbers 72 and 47.

Instead, let’s imagine that one train is stationary, and the other moving. How? Let’s say two cyclists (just to mix it up) are headed in the same direction. One is going 10 kilometer per hour, and the other is going 15 kilometers per hour. After one hour, the faster cyclist is going to be 5 kilometers ahead of the slower one. That’s the same distance as if the slower cyclist decided not to bike at all, and the faster cyclist moved at 5 kph.

The takeaway: when two entities (trains, cyclists, cars) are headed in the same direction, the difference between the two rates is the amount the faster entity is outpacing the slower one per hour (in the case that the rate is expressed in terms of per hour).

Returning to the original problem, which asks for trains headed in the same direction, we want to take the difference of the two rates. This gives us 72 – 47 = 25. Therefore, for every hour the faster one is 25 miles ahead of the slower train.

How long, then, will it take until the faster train is 100 miles ahead? 100/25 = 4 hours.

Notice that the numbers in this question—72 and 47—were very daunting. But, if you imagine the slower train not moving at all, then the faster train is moving 25 mph. This number easily divides into 100. And, like that, we have the answer.

Of course rate questions on the GRE–and the New GRE–can get even more complicated than this one. Indeed, I’ll be posting a challenging rate question very soon, involving a peculiar animal.

### 3 Responses to GRE Math — Do Rates Freak You Out?

1. KSM January 7, 2018 at 10:14 pm #

SIR IF TWO BODIES ARE GOING APART FROM EACH OTHER AT SPEED V1 AND V2. AT WHAT TIME DSISTANCE BETWEEN THEM IS D. HOW TO SOLVE IT??

• Magoosh Test Prep Expert January 8, 2018 at 12:57 pm #

Hi there,

Happy to help 🙂 I recommend that you read this GMAT Club posting which has some great information on how to set up and solve all types of rate problems (The GMAT Math section is very similar to the GRE math section, so these sorts of resources are super helpful for both tests.)

Let’s think about this question logically. I always like to do a quick drawing to illustrate what is happening in the question and label the rates and distances so that I can see what I am doing. Try doing that with this The bodies start at the same place and move apart from each other. That means that we will have a straight line of distance D. It also means that they are in motion for the same period of time (since they started at the same time). As the bodies move away from each other, they are both adding distance based on their own velocity. This means that we have to add their rates together in order to get the total distance. So, we know that D=V1*T+V2*T. Remember: they are in motion for the same amount of time T, so the variable T is the same for both. We can re-arrange this equation in the following way. First, we factor out the T:
D=T(V1+V2)
Then we divide both sides by V1+V2:
D/(V1+V2)=T