Bunuel wrote:
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?
(A) 45
(B) 50
(C) 53
(D) 55
(E) 60
Distance travelled by first car in 10 mins = 40/6 = 20/3 miles
Relative Distance between two cars at the time second car starts = 20/3 miles
Relative speed of two cars = (s-14) where s is the speed of second car
Time taken by second car to catch first car = (9:10 to 10:30) = 1 hour 20 mins = 4/3 hours
Relative speed = Relative Distance / time
(s-40) = (20/3)/(4/3) = 5
i.e. s = 5+40 = 45 miles per hour
Answer: Option A
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