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
Explanation:
First car starts at 10 am : Let the speed and time taken be a and t respectively.
Second car starts at 11 am : Let the speed be b. Time taken = t − 1.
Required percentage = [(b − a)/a] × 100 = [(b/a) − 1] × 100
Required percentage will be highest when b/a is highest.
Now, since distances covered by both cars are same, so D = at = b(t − 1)
∴ b/a = t/(t − 1) = 1/[1 − (1/t)]
(b/a) is maximum when t is minimum.
tmin = 6 (given)
∴ b/a = 6/5.
Maximum required percentage = [(6/5) − 1] × 100 = 20%.
Hence, option (d).
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