Three buses make the same journey every day on a route that

ardz24 wrote:Three buses make the same journey every day on a route that has bus stops at exactly equal distances apart. The buses all travel at constant (but different) speeds. At the first stop after the starting point, if you just miss the fast bus (Fbus), then you wait 20 minutes for the medium-speed bus (Mbus), or 30 minutes for the slow bus (Sbus). At the second stop, missing Fbus would result in a wait of 30 minutes for Mbus, or 1 hour 10 minutes for Sbus. 
Which of the following departure times from the start of the route could explain these waiting times?

A Fbus 07:30, Mbus 07:40, Sbus 07:50 

B Fbus 07:40, Mbus 07:50, Sbus 07:30 

C Fbus 07:50, Mbus 07:30, Sbus 07:40 

D Fbus 07:40, Mbus 07:30, Sbus 07:50 


Is there a strategic approach to this question? Can any experts help?

When I refer to time "lost" below I really mean wait time or additional wait time...

Start with the medium bus. At the first stop it had lost 20 minutes to the fast bus, explained by its time over distance (speed) and departure time relative to fast bus. At the second stop, its cumulative loss was 30 minutes against the fast bus, explained by its time over the TOTAL distance and its departure time.

So, the medium bus lost an additional 10 minutes between the first and second stops, meaning that it also lost that same amount of time to the first stop, meaning that 20-10 or 10 minutes must be due to the medium bus departing 10 minutes later than the fast bus. So this eliminates C and D above.

Let's look at the slow bus. At the first stop it had lost 30 minutes against the fast bus. By the second stop it had lost a total of 70 minutes. This means that it lost 70-30 or 40 minutes due to its slower speed compared to the fast bus. But, it had lost only 30 minutes to the first stop ! Meaning that the slow bus left 40 - 30 or 10 minutes earlier than the fast bus. So the order of the buses' departures are

Slow Bus, Fast Bus, Medium Bus, AnswerB