If the car catches up with the motorcycle after time t, the motorcycle will have travelled 80t km. In the hour between 10pm and 11pm the motorcycle has travelled 80km before the car has left.
This can be represented by s=80t, where s is the distance from the start.
The car is delayed by an hour so s=120(t-1). The car has to cover the same distance in less time.
Therefore 80t=120(t-1), 80t=120t-120, (120-80)t=120, 40t=120, t=120/40=3 hr.
s=80t=80×3=240km.
The car catches up with the motorcycle after 3hr, (2hr of driving time for the car), so the time would be 10pm+3=1:00am, 240km from the start.
CHECK
Between 11pm and 1am the car has been travelling for 2hrs in which time it has gone 2×120=240km.
Between 10pm and 1am the motorcycle has been travelling for 3hrs in which time it has gone 3×80=240km. Therefore the car catches up with the motorcycle 240km from the start at 1am.
[Apologies for my initial error--I misread 80kph as 88kph.]