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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?
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Solution
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
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