karthikpandian19 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
10 minutes = 1/6 of an hour.
The distance traveled by the first car in 1/6 of an hour = r*t = 40(1/6) = 20/3 miles.
10:30am - 9:10am = 80 minutes = 4/3 of an hour.
Required rate for the second car to catch up by 20/3 miles in 4/3 of an hour = d/t = (20/3)/(4/3) = 5 miles per hour.
Thus, the second car must travel 5 miles per hour FASTER than the first car.
Rate of the second car = rate of the first car + 5 = 40+5 = 45 miles per hour.
The correct answer is
A.
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