SOLUTION: Wilbur drove to the town hall and back. The trip there took 1.8 hours and the trip back took 2.2 hours. He averaged 9 mph faster on the trip there than on the return trip. What was
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Question 440961: Wilbur drove to the town hall and back. The trip there took 1.8 hours and the trip back took 2.2 hours. He averaged 9 mph faster on the trip there than on the return trip. What was Wilbur's average speed on the outbound trip?
Found 2 solutions by jorel1380, josmiceli:
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
1.8(x+9)=2.2x
1.8x+16.9=2.2x
16.9=.4x
16.9/.4=x
x=40.5
x+9=49.5
Wilbur's average speed outgoing was 49.5mph..
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = his average speed for the return trip.
given:
To town hall:
(1)
The trip back:
(2)
Set (1) equal to (2)
Trip back:
To town hall:
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