If a person walks 20% more than of his usual speed, reaches his distance 90 minutes before. If the destination is 459 km away, then the usual speed of a person is (in km/hr)?
Let the usual speed be x km/hr New speed = 120% of x = 6x/5 km/hr Distance = 459 km According to the question 459/x – 459/(6x/5) = 90/60 459 x [(6 – 5)/6x] = 3/2 Therefore, x = 459 x (1/6) x (2/3) = 51 km/hr
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