SOLUTION: My question is, if a bicyclist rode into the country for 5 h. Once returning, what was the speed if the person was traveling 5 mi/h faster and the trip took 4 h. What was the speed

Algebra ->  Inequalities -> SOLUTION: My question is, if a bicyclist rode into the country for 5 h. Once returning, what was the speed if the person was traveling 5 mi/h faster and the trip took 4 h. What was the speed      Log On


   

Question 54008: My question is, if a bicyclist rode into the country for 5 h. Once returning, what was the speed if the person was traveling 5 mi/h faster and the trip took 4 h. What was the speed going each way?
Found 2 solutions by Nate, checkley71:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
1st rate = r
1st time = 5
rate*time = distance
5r = d
2nd rate = r + 5
2nd time = 4
4r + 20 = d
The distances are the same.
4r + 20 = 5r
20 = r
1st rate = 20
2nd rate = 25

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
D=R*T THEN DISTANCE GOING IS R*5 AND THE SAME DISTANCE RETURNING IS (R+5)4 OR
5R=4(R+5) OR 5R=4R+20 OR R=20 MPH GOING FOR 5 HOURS EQUALS 100 MILES.
THE RETURN TRIP WAS @ 25 MPH FOR 4 HOURS THUS 25*4=100 MILES.