SOLUTION: two cyclist start at the same time and travel toward the same direction. at the end of two hours, they are 50 meters APART . if the speed of the cyclist is twice of that the slower

Question 439763: two cyclist start at the same time and travel toward the same direction. at the end of two hours, they are 50 meters APART . if the speed of the cyclist is twice of that the slower one , how fast did each travel?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
two cyclist start at the same time and travel toward the same direction. at the end of two hours, they are 50 meters APART . if the speed of the cyclist is twice of that the slower one , how fast did each travel?
.
Applying the "distance formula" of d=rt
.
Let x = speed of slower cyclist
then
2x = speed of faster cyclist
.
2x+50 = 2(2x)
2x+50 = 4x
50 = 2x
25 meters/hour = x (slower cyclist)
.
faster cyclist:
2x = 2(25) = 50 meters/hour

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