Two cyclists left simultaneously from cities A and B heading towards each other at constant rates and met in 5 hours. The rate of the cyclist from A was 3 mph less than the rate of the other cyclist. If the cyclist from B had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from A. What is the distance between A and B, in miles?

Answer 1

Let d = distance from A to B
If a is A's speed, then a+3 = B's speed
d = 5a + 3(a+3)

since distance = speed * time, then at time t hours from when A started out
at = 31.8
(a+3)(t-1/2) = d-31.8

Solve those three equations to find d