SOLUTION: A student drove a distance of 135 miles at an average speed of 50 mph. How much faster would he have to drive on the return trip to save 30 minutes of driving time?

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Question 348380: A student drove a distance of 135 miles at an average speed of 50 mph. How much faster would he have to drive on the return trip to save 30 minutes of driving time?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
135=50T
T=135/50
T=2.7 HOURS.
135=R(2.7-.5)
135=2.2R
R=135/2.2
R=61.364 MPH.