let, speed of Ali=X mph and Faizer=Y mph

Now, relative speed when Ali is walking int the same direction with Faizer= (X-Y)mph
hence, according to question, 27/(X-Y)=9 ; solving X-Y=3; --- (1)

Now relative speed when walking in the opposite direction= (X+Y) mph
hence, according to question, 27/(X+y)=3; solving X+Y=9; --- (2)

solving 1 and 2 we get X=6 and Y=3, where X is the speed of Ali.

In 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 - 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min.

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800 x 18 60 x 5= 48 sec.

Speed on return trip = 150% of 50 = 75 km/hr.

Average speed = (2 x 50 x 75)/(50 + 75) = 60 km/hr.

Average speed = Total distance / Total time.

Total distance traveled by Joel = Distance covered in the first 3 hours + Distance covered in the next 5 hours.

Distance covered in the first 3 hours = 3 x 60 = 180 miles
Distance covered in the next 5 hours = 5 x 24 = 120 miles

Therefore, total distance traveled = 180 + 120 = 300 miles.

Total time taken = 3 + 5 = 8 hours.

Average speed = 300/8 = 37.5 mph.

we know that 1 mile = 1.6 kms

=> 37.5 miles = 37.5 x 1.6 = 60 kms

Average speed in Kmph = 60 kmph.

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts.