Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM. 

Hence, car 2 will take 5 hours. 

Hence a= $$\ \frac{\ d}{6}$$ and b = $$\ \frac{\ d}{5}$$

Hence the speed of car 2 will exceed the speed of car 1 by $$\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$$ = $$\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$$ = 20