A regular train service operates between Jayford and Kayton, a 16 km journey which takes 19 minutes. The trains travel at a constant speed of 60 km per hour in both directions except through a tunnel, where they are limited to 20 km per hour. Trains travelling towards Kayton enter the tunnel 4 km after setting off from Jayford.

This entire question can be solved by a single equation, but to construct it we must understand that:

If the entire train was to theoretically travel at 60km/h, the trip would take 16 minutes.

But we know that for a segment of the trip, which we can call x, the train traveled 3 times slower. From 60km every 60 minutes into 20 km every 60 minutes.

We must remove that distance x from the theoretical 16 minutes, and add it back at the rate at which it actually goes- which we know is 3 times slower.