SOLUTION: John drove to his friends house and back. It took 0.3 hours longer to go there than it did to come back. The average speed on the trip there was 44.8 mph. The average speed on the

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Question 1014881: John drove to his friends house and back. It took 0.3 hours longer to go there than it did to come back. The average speed on the trip there was 44.8 mph. The average speed on the way back was 48 mph. How many hours did the trip there take?
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
It took 0.3 hours longer to go there than it did to come back.
The average speed on the trip there was 44.8 mph.
The average speed on the way back was 48 mph.
How many hours did the trip there take?
:
let t = time required to get there
Return trip took .3 hrs less, therefore
(t-.3) = return time
:
Write a distance equation; dist = speed * time
44.8t = 48(t-.3)
44.8t = 48t - 14.4
14.4 = 48t - 44.8t
14.4 = 3.2t
t = 14.4/3.2
t = 4.5 hrs to get there
:
:
:
Check this
Find the actual distance each way, obviously, they should be the same.
4.5 * 44.8 = 201.6 mi
4.2 * 48 = 201.6, (return trip took .3 hrs less)

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