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ATQ, Let 'A's normal speed be ‘s’ km/h. Time taken to reach the Bus stop at normal speed = (130/s) hour Now, after increasing the normal speed by 5 km/h. New speed will be = (s + 5) km/h Time taken to reach the Bus stop at new speed = 130/(s + 5) hour According to the given information, 130/s – 130/(s + 5) = 1/6 s2 + 5s – 3900 = 0 s2– 60s + 65s – 3900 = 0 s(s – 60) + 65(s – 60) = 0 (s –60)(s + 65) = 0 Therefore, s = 60 km/h Therefore, his increased speed on that day = 60 + 5 = 65 km/h
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