Question 67

# 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

Solution

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

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