Arup and Swarup leave point A at 8 AM to point B. To reach B, they have to walk the first 2 km, then travel 4 km by boat and complete the final 20 km by car. Arup and Swarup walk at a constant speed of 4 km/hr and 5 km/hr respectively. Each rows his boat for 30 minutes. Arup drives his car at a constant speed of 50 km/hr while Swarup drives at 40 km/hr. If no time is wasted in transit, when will they meet again?

Both of them row the boat for the same time. Therefore, we can ignore the time taken to row the boat and add it to the final answer. 
Arup will walk 2 km in 2/4 = 30 minutes.
Swarup will walk 2 km in 2/5 = 24 minutes.

Therefore, Swarup will start driving 30-24 = 6 minutes earlier than Arup. 
In 6 minutes, Swarup will gain a lead of 6*40/60 = 4 km. 
Arup drives at 50 kmph (i.e, 10 kmph faster than Swarup). Therefore, Arup will cover the 4 km advantage that Swarup has in 4/10*60 = 24 minutes. 
Therefore, the time after which both of them will meet is 30 (to walk) + 30 (to row the boat) + 24 = 1 hour 24 minutes after Arup starts. 
Since both of them start at 8 AM, they will meet again at 9:24 AM. 
Therefore, option D is the right answer.