If a person walks 25% more than of his usual speed, reaches his distance 90 minutes before. If the destination is 217.5 km away, then the usual speed of a person is (in km/hr)?
Let the usual speed be x km/hr New speed = 125% of x = 5x/4 km/hr Distance = 217.5 km According to the question 217.5/x – 217.5/(5x/4) = 90/60 217.5 x [(5 – 4)/5x] = 3/2 Therefore, x = 217.5 x (1/5) x (2/3) = 29 km/hr
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