Question
A train travels from station βPβ to station βQβ
at a constant speed. If its speed was increased by 20 km/h, it would have taken 1 hour less to cover the distance. It would have taken further 30 minutes less if the speed was further increased by 20 km/h. The distance between station βPβ and station βQβ is:Solution
According to the question, Let the original speed of the train be βxβ km/h and the original time taken to reach the destination be βtβ hours. So, the distance between station βPβ and βQβ = x Γ t = xt km So, xt/(x + 20) = t β 1..........................(1) And, xt/(x + 40) = t β 1 β 0.5 Or, xt/(x + 40) = t β 1.5........................(2) Dividing equation (1) by equation (2), we get (x + 40)/(x + 20) = (t β 1)/(t β 1.5) So, xt β 30 β 5x + 20t = xt + 20t β x β 20 Or, 20t β 5x = 10 Or, x = 4/5 Γ (20t β 10) Putting the value of βxβ in equation (1), we get 40tΒ² β 80t = 40tΒ² β 110t + 70 Or, t = 3 So, x = 4/5 Γ (60 β 10) = 40 Desired distance = 3 Γ 40 = 120 km
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