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
(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.