Let speed in the return journey = V
? Speed in onward journey = (125/100) x V = 5V/4 km/h
Average speed = (2x 5V/4 x V) / (5V/4 + V) = 10V/9 km/h
? 1600 x (9/10V) = 30 [? halts for 2 h ]
? V = 1600 x 9/30 x 10 = 48 km/h
? Speed in onward journey = 5V/4 = 5/4 x 48 = 60 km/h
Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h
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