Question
If a person walks 20% more than of his usual speed,
reaches his distance 90 minutes before. If the destination is 531 km away, then the usual speed of a person is (in km/hr)?Solution
Let the usual speed be x km/hr New speed = 120% of x = 6x/5 km/hr Distance = 531 km According to the question 531/x – 531/(6x/5) = 90/60 531 x [(6 – 5)/5x] = 3/2 Therefore, x = 531 x (1/5) x (2/3) = 59 km/hr
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