To cover 720 km, a car travelling at its usual speed takes 2 hours more than the expected time. If the speed of the car had been 25% more, then it would have taken 1 hour less than the expected time taken. If the car travelled 6 km/h faster than usual, the car would take how much time more/less than the expected time?

Let the usual speed of the car = ‘4x’ km/h Then, increased speed of the car = 4x × 1.25 = ‘5x’ km/h Let the expected time taken to cover the distance of 720 km = ‘t’ hours Then, according to the question (720/4x) = t + 2 = (180/x) …… (I) Also, (720/5x) = t - 1 = (144/x) …… (II) Equation (I) - Equation (II), we get (t + 2) - (t - 1) = (180/x) - (144/x) or, 3 = (36/x) so, x = 12 so, usual speed of the car = 12 × 4 = 48 km/h expected time taken = (720/48) - 2 = 15 - 2 = 13 hours time taken when speed increases by 6 km/h = 720 ÷ (48 + 6) = 720/54 = 13⅓ hours so, the car takes 20 minutes more than the expected time.