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Question
An ordinary train takes 3 hours less for a j ourney of 360kms when its speed is increased by 1 Okm/ hr.Fnd the usual speed of the train.
Solution
Let the usual speed of the train be S, Hence speed of the train when speed is increased= S+ l O.
D=360Km, Time = Distance/ Speed. Time difference=3 Hours
Hence, in these two conditions ,
`360/"s" - 360/("s" + 10) = 3`
⇒ 360 x (s+ 10) - 360 x S = S x (S+ 10) x 3
⇒ 3S2 + 30 S - 3600=0
⇒ S2+ 10S - 1200 = 0
⇒ S2 + 40 S -30 S -1200 = 0
⇒ S (S + 40) - 30 (S + 40) = 0
⇒ (S + 40) (S - 30) = 0
As the speed can't be negative, S = 30km/ hr
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