A passenger train takes 3 hours less for a journey of 360 km, if its speed is increased by 10 km / hr from its usual speed. What is the usual speed?

Distance = speed × time


Given, passenger train takes 3 hours less for a journey of 360 km, if its speed is increased by 10 km / hr from its usual speed.


Let the speed be ‘s’ and time be ‘t’.


st = 360


t = 360/s


Also, 360 = (s + 10)(t – 3)



360s = 360s + 3600 -3s2 – 30s


s2 + 10s – 1200 = 0


s2 + 40s – 30s – 1200 = 0


s(s + 40) – 30(s + 40) = 0


(s – 30)(s + 40) = 0


s = 30 km/hr


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