SOLUTION: A passenger train made a trip to las vegas and back. The trip there took 17 hours and the trip back took ten hours. It averaged 21 mph faster on the return trip than on the outboun
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Question 1046371: A passenger train made a trip to las vegas and back. The trip there took 17 hours and the trip back took ten hours. It averaged 21 mph faster on the return trip than on the outbound trip. Find the passenger train's average speed on the outbound trip. Found 2 solutions by solver91311, josmiceli:Answer by solver91311(24713) (Show Source):
Let be the average speed on the return trip. Then, since distance equals rate times time, the distance from Las Vegas to home is . The average speed on the trip from home to Las Vegas is 21 mph slower, so this speed can be represented by , and the distance from home to Las Vegas must be . Presuming that home and Las Vegas did not move relative to one another during the time that this trip occurred, we can assert that the distance from home to Las Vegas is the same as the distance from Las Vegas to home. Therefore, the two expressions for distance that were derived above must be equal.
Solve for
John
My calculator said it, I believe it, that settles it
You can put this solution on YOUR website! Let = the average speed on the outbound trip in mi/hr = the average speed on the return trip in mi/hr
Let = the one way distance to Las Vegas in miles
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Equation for outbound trip:
(1)
Equation for return trip:
(2)
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Substitute (1) into (2)
(2)
(2)
(2)
(2)
The speed on the outbound trip is 30 mi/hr
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check:
(1)
(1)
(1) mi
and
(2)
(2)
(2)
(2) mi
