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
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Distance covered by the car in 1.5 hours ( 10:30 A.M. - 9:00 A.M.) is 1.5 * 40 = 60 miles
Now car A has covered 60 miles by 10:30 A.M , so car B must also cover the same distance in ( 10:30 A.M - 9:10 A.M.) 80/60 hr

So, speed of car B must be = 60 * 6/8 i.e. 45 miles per hour

So, Answer must be (A)
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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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We are given that a car left at 9 a.m. traveling at a rate of 40 mph and another car left 10 minutes later. Since the second car caught up to the first car at 10:30 a.m., the time of the first car is 1.5 hours and the distance is 1.5 x 40 = 60 miles.

If we let r = the rate of the second car, the distance of the second car is:

r x (1 hour 20 minutes)

r x (1 1/3 hours)

r x (4/3) = 4r/3

Since the second car catches up to the first car we can set the distances equal and determine r:

4r/3 = 60

4r = 180

r = 45

Answer: A
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