Question

Jaidee made her trip to the train station and back. The trip there took two hours and the trip back took five hours. She averaged 30 mph faster on the trip there than on the return trip. What was Jaidee's average speed on the outbound trip?

Answer 1

The trip to the train station: t 1 = 2 h, v 1 = d / t 1, v 1 = d / 2, where d is the distance from the train station.
The trip back:
t 2 = 5 h, v 2 = d / t 2, v 2 = d / 5
Also: v 2 = v 1 - 30
d / 5 = d / 2 - 30
d / 2 - d / 5 = 30
0.3 d = 30
d = 30 : 0.3
d = 100 miles.
v 1 = 100 / 2 = 50 mph
We can prove it: v 2 = 50 - 30 = 20 mph
100 miles = 50 mph * 2 h = 20 mph * 5 h
Answer: Average speed on the ongoing trip was 50 mph.