A flight leaves A at 4:15 pm and reaches B at 6:45 pm. Another flight leaves B at 2:30pm and reaches A at 8:00pm. If the average speed of the flights are 600kmph then what is the distance between A and B in km?(All the time given are in local time)

I think it is 2400km.
Logic :-
total time = 2.5+5.5 = 8hrs
avg speed = 600kmph
So, total dist = 600*8=4800 km
So, dist b/w A & B =4800/2 =2400km

let time of A ahead of B by t hours then time taken from a to b =2.5 hours + t and from b to a = 5.5-t but dist is equal hence (2.5 +t) *600 = (5.5-t)*600 hence t=1.5 and dist =(2.5+t)*600 =(2.5+1.5)*600 =2400

if you go by the local time , it will create confusion .
let say ,
time at A when it left was 4:15 ( given )
Difference in time on 1st way from A to B is 2.5 hrs
difference in 2nd way - 5.5 hrs
5.5 +2.5= 8
8/2= 4 hrs .
it will take 4 hrs to go one round so by the time it reaches B , A's time would be 8:15 pm
but the local time of B is 6:45 pm
So basically when A's time is 8:15 , B's time is 6.45 pm
A is 1.5 hr ahead of B.
if you go the other way round from B to A keeping again A's time constant you would see the logic fits.
therefore, A to B time taken would be 4 hrs only.
distance = 600* 4= 2400 km . Ans

2400KM ?

2400 Km as if we consider the flight leaves at 12.00 am then it reaches at 02.30 pm and leaves right back and reaches at 8.00 pm so, the total time to make a round trip is 8 hours and hence 4 hours for a oneway trip so therefore 600 x 4 = 2400 km

but it had written that the time it shows are local time show it cannot be 8 hrs i believe