A train travels from station ‘P’ to station ‘Q’ 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 ‘P’ and station ‘Q’ is:

According to the question, Let the original speed of the train be ‘x’ km/h and the original time taken to reach the destination be ‘t’ hours. So, the distance between station ‘P’ and ‘Q’ = x × t = xt km So, xt/(x + 20) = t – 1..........................(1) And, xt/(x + 40) = t – 1 – 0.5 Or, xt/(x + 40) = t – 1.5........................(2) Dividing equation (1) by equation (2), we get (x + 40)/(x + 20) = (t – 1)/(t – 1.5) So, xt – 30 – 5x + 20t = xt + 20t – x – 20 Or, 20t – 5x = 10 Or, x = 4/5 × (20t – 10) Putting the value of ‘x’ in equation (1), we get 40t² – 80t = 40t² – 110t + 70 Or, t = 3 So, x = 4/5 × (60 – 10) = 40 Desired distance = 3 × 40 = 120 km