Question

A passenger train traveled to the fueling station and back. The trip there took 10 hours, the trip back took 12 hours, and it averaged 5 mph faster on the trip there than on the return trip. What was the passenger train's average speed on the outbound trip?

Answer 1

The average speed of the passenger train on the outbound trip to the fueling station was 30 mph, calculated using the times for each direction of the journey and the fact that the outbound speed was 5 mph faster than the return speed.

Step-by-step explanation:

To find the average speed of the passenger train on the outbound trip to the fueling station, we use the given information that the train traveled the same distance both ways and it took 10 hours for the outbound trip and 12 hours for the return trip. We also know that the average speed for the outbound trip was 5 mph faster than on the return trip. Let's denote the average speed on the return trip as v mph.

The distance covered in each direction is the same. So, the average speed for the return trip can be expressed as:

distance = speed × time
Let the train's average speed on the return trip be v mph. Then its average speed on the outbound trip was v + 5 mph.

Since the distance is the same in both directions, we can write:
v × 12 = (v + 5) × 10

Now we solve for v:

12v = 10v + 50

12v - 10v = 50

2v = 50

v = 25 mph

So the average speed of the train on the return trip was 25 mph, which means that the average speed on the outbound trip was v + 5 = 30 mph.