# SOLUTION: A bicyclist took a ride to the country and it took 5 hours. To get back she went 5mph faster and it took 4 hours. What were the speeds, going and coming?

Algebra ->  Algebra  -> Rate-of-work-word-problems -> SOLUTION: A bicyclist took a ride to the country and it took 5 hours. To get back she went 5mph faster and it took 4 hours. What were the speeds, going and coming?      Log On

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 Click here to see ALL problems on Rate-of-work-word-problems Question 34948: A bicyclist took a ride to the country and it took 5 hours. To get back she went 5mph faster and it took 4 hours. What were the speeds, going and coming?Answer by AnlytcPhil(1276)   (Show Source): You can put this solution on YOUR website!``` A bicyclist took a ride to the country and it took 5 hours. To get back she went 5mph faster and it took 4 hours. What were the speeds, going and coming? Make this chart Distance Rate Time Going | | | | Coming | | | | Let x = the speed going in mph >>...To get back she went 5mph faster...<< So the speed coming back = x+5 mph. So fill in these speeds Distance Rate Time Going | | x | | Coming | | x+5 | | >>...A bicyclist took a ride to the country and it took 5 hours...<< >>...To get back...it took 4 hours...<< So fill in 5 hours and 4 hours for the times: Distance Rate Time Going | | x | 5 | Coming | | x+5 | 4 | Now use Distance = Rate × Time to fill in the distances Distance Rate Time Going | 5x | x | 5 | Coming | 4(x+5) | x+5 | 4 | Now since the distance going equals the distance back, we form an equation by setting the two distances equal 5x = 4(x+5) Solve that and get x = 20 mph going, then since the speed returning is 5mph more, the returning speed is 25 mph. Edwin AnlytcPhil@aol.com```