A jogger and a cyclist set out at 9 A.M. from the same point headed in the same direction. The average speed of the cyclist is four times the average speed of the jogger. In 2 h, the cyclist is 33 mi ahead of the jogger. How far did the cyclist ride?

Question

anonymous · Answer

I am stuck ! :(

anonymous · Answer

Let x = the average speed of the jogger.
Let y = the average speed of the cyclist.

y = 4x
2y = 2x + 33

Substitute 4x for y and solve for x

2(4x) = 2x + 33
8x = 2x + 33
6x = 33
x = 33/6 = 5.5 <-- this is the average speed of the jogger

Plug 5.5 in for x in the equation y = 4x

y = 4(5.5) = 22 <-- this is the average speed of the cyclist

Now plug both values into the equation 2y = 2x + 33
2(22) = 2(5.5) +33
44 = 11 + 33
44 = 44 <-- this is the total distance traveled by the cyclist