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Let the distance travelled be 'd' km, actual speed of the train be 's' km/h and actual time taken by the train be 't' hours. ATQ; (d/s) = t Or, d = st ...... (I) Also, {d/(s + 30) } = t - 5 Or, d = (t - 5) (s + 30) ...... (II) Using equation (I) and (II) , we have; Or, st = (t - 5) X (s + 30) Or, st = st - 5s + 30t - 150 Or, 30t - 5s = 150 ...... (III) Also, d = (t + 10) X (s - 30) ...... (IV) Using equation (I) and (IV) , we have; Or, st = (t + 10) X (s - 30) Or, st = st + 10s - 30t - 300 Or, 10s - 30t = 300 ....... (V) Adding equation (III) and (V) , we have; 5s = 450 Or, 's' = 90 Therefore, original speed of the train = 's' km/h = 90 km/h
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