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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