SOLUTION: john drove to a distant city in 6 hour. When he returned there was less trafic and the trip only took 4 hours. If john averaged 38 mph faster on the return trip, how fast did he dr
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-> SOLUTION: john drove to a distant city in 6 hour. When he returned there was less trafic and the trip only took 4 hours. If john averaged 38 mph faster on the return trip, how fast did he dr
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Question 359430: john drove to a distant city in 6 hour. When he returned there was less trafic and the trip only took 4 hours. If john averaged 38 mph faster on the return trip, how fast did he drive each way? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! forward let speed = xmph
time = 6 hours
distance = 6x miles
...
return
speed = x+38 mph
time =4 hours
distance = 4(x+38)
distance = same
6x=4(x+38)
6x=4x+152
6x-4x=152
2x=152
x=76 mph forward journey speed
76+38=114 mph return
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m.ananth@hotmail.ca