SOLUTION: Sarawang drove to the lake and back. The trip there took two hours and the trip back took three hours. He averaged 20mph faster on the trip there than on the return trip. What was
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Question 1142949: Sarawang drove to the lake and back. The trip there took two hours and the trip back took three hours. He averaged 20mph faster on the trip there than on the return trip. What was Sarawong's average speed on the outbound trip? Found 2 solutions by solver91311, ikleyn:Answer by solver91311(24713) (Show Source):
The distance traveled either way is , the speed going is , the time going is 2, the speed returning is and the time returning is 3.
So the outbound trip is described by
And the return trip is described by
Since the distance is the same both ways:
Solve for and then calculate
Of course, this solution will only be valid for Sarawang. I have no idea about Sarawong's return speed. Lesson: Proofread your post BEFORE you click on Submit.
John
My calculator said it, I believe it, that settles it
Let x be the average speed of the trip TO the lake, in miles per hour.
Then the average speed on the returning trip was (x-20) mph.
The distance to the lake is 2*x miles; the distance from the lake to the home was 3*(x-20).
Both distances are the same, which gives you an equation
2x = 3*(x-20).
Simplify and solve for x
2x = 3x - 60
60 = 3x - 2x
x = 60.
ANSWER. Sarawong's average speed on the outbound trip was 60 mph.
CHECK. The distance to the lake was 2*60 = 120 miles.
The distance back was 3*(60-20) = 3*40 = 120 miles.
Both distances are the same ----> hence, the solution is correct.
