If a person walks 40% more than of his usual speed, reaches his distance 90 minutes before. If the destination is 441 km away, then the usual speed of a person is (in km/hr)?
Let the usual speed be x km/hr New speed = 140% of x = 7x/5 km/hr Distance = 441 km According to the question 441/x – 441/(7x/5) = 90/60 441x [(7 – 5)/7x] = 3/2 Therefore, x = 441 x (2/7) x (2/3) = 84 km/hr
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