Two towns A and B are 400 km apart. A motor cyclist travelling at an average speed of 60km/h left town A for town B at 1:20 pm. A matatu travelling at an average speed of 80km/h also left town A for town B at 2:00 pm. Calculate: a) the distance covered by the motor cyclist by 2:00pm, b) the distance from town A in km, where the matatu caught up with the motor cyclist, c) time when the matatu caught up with the motor cyclist.

The motor cyclist left town A at 1:20 pm.
The time elapsed until 2:00 pm is $2:00 \text{ pm} - 1:20 \text{ pm} = 40 \text{ minutes}$.

Speed of motor cyclist = 60 km/h.
Distance = Speed $\times$ Time

The distance covered by the motor cyclist by 2:00 pm is $\boxed{40 \text{ km}}$.

Let $t$ be the time in hours from 2:00 pm when the matatu catches up with the motor cyclist.
At 2:00 pm, the motor cyclist has already covered 40 km (from part i).
Distance of motor cyclist from A at time $t$ (after 2:00 pm):

The matatu left town A at 2:00 pm.
Distance of matatu from A at time $t$ (after 2:00 pm):

When the matatu catches up with the motor cyclist, their distances from town A are equal.

This means they meet 2 hours after 2:00 pm.

Substitute $t=2$ hours into either distance equation. Using the matatu's distance:

The distance from town A where the matatu caught up with the motor cyclist is $\boxed{160 \text{ km}}$.

The matatu caught up with the motor cyclist 2 hours after 2:00 pm.

The matatu caught up with the motor cyclist at $\boxed{4:00 \text{ pm}}$.