SOLUTION: An Air Force plane made a trip to Moscow and back. The trip there took ten hours and the trip back took nine hours. It averaged 50 mph faster on the return trip than on the outboun

Question 1046364: An Air Force plane made a trip to Moscow and back. The trip there took ten hours and the trip back took nine hours. It averaged 50 mph faster on the return trip than on the outbound trip. What was the Air Force plane's average speed on the outbound trip?
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52888) (Show Source): You can put this solution on YOUR website!
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An Air Force plane made a trip to Moscow and back. The trip there took ten hours and the trip back took nine hours.
It averaged 50 mph faster on the return trip than on the outbound trip.
What was the Air Force plane's average speed on the outbound trip?
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Let D be an unknown distance. Then you have this equation = 50. The first term in the left side is the average speed of the return trip. The second term in the left side is the average speed for outbound trip. The difference is 50 mph, according to the condition. To solve the equation, multiply both sides by 9*10. You will get 10D - 9D = 50*90, D = 4500. So, we found the distance. It is 4500 miles. Hence, the outbound speed was 4500/10 = 450 mph. Answer. The average speed on the outbound trip was 450 mph.

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Let r = speed of the plane on the outbound trip.
===> r + 50 = speed of the plane on the return trip
===> 10r = 9(r+50), assuming linear distance is the same for both outbound and return trips.
10r = 9r +450
===> r = 450 mph, the average speed on the outbound trip.
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