SOLUTION: Navigate questions by number Question #10/10 A cyclist travels 70 in 5 hours going against the wind. He travels 110 mi with the wind in the same a

Algebra ->  Linear-equations -> SOLUTION: Navigate questions by number Question #10/10 A cyclist travels 70 in 5 hours going against the wind. He travels 110 mi with the wind in the same a      Log On


   

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Question #10/10

A cyclist travels 70 in 5 hours going against the wind. He travels 110 mi with the wind in the same amount of time. What is the rate of the cyclist in still air and what is the rate of the wind?
Rate of the cyclist in the sill air:
Rate of the wind:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
70=R*5 AGAINST THE WIND.
R=70/5
R=14 MPH.
110=R*5 WITH THE WIND
R=110/5
R=22 MPH.
(22-14)/2=8/2=4 MPH IS THE SPEED OF THE WIND.
70=(R-4)*5
70=5R-20
5R=70+20
5R=90
R=90/5
R=18 MPH FOR THE CYCLIST IN STILL AIR.
110=(R+4)5
110=5R+20
5R=110-20
5R=90
R=18 DITTO.