Question

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?

Answer 1

Distance = speed x 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)

⇒ 360 = (s + 10)`(360/s − 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