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