Question
If a person walks 25% more than of his usual speed,
reaches his distance 90 minutes before. If the destination is 495 km away, then the usual speed of a person is (in km/hr)?Solution
Let the usual speed be x km/hr New speed = 125% of x = 5x/4 km/hr Distance = 495 km According to the question 495/x – 495/(5x/4) = 90/60 495 x [(5 – 4)/5x] = 3/2 Therefore, x = 495 x (1/5) x (2/3) = 66 km/hr
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1.      Bhoodan Movement...
