Posted by **Jamie** on Monday, November 23, 2009 at 3:29pm.

A cyclist and a jogger start from a town at the same time and head for a destination 6 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 3 hr ahead of the jogger. find the rate of the cyclist.

This is what i got so far but its not adding up.

r t d

cyclist 2x*3=6

jogger x*-3=6

- Math -
**MathMate**, Monday, November 23, 2009 at 5:49pm
It always helps to define the variables used, with the units, and cite the equations you use.

Distance = speed * time, or

time = distance/speed

Let

x=speed of cyclist, m.p.h.

x/2=speed of jogger, m.p.h.

D=distance, 6 mi.

"The cyclist arrives 3 hr ahead of the jogger. find the rate of the cyclist. "

means that the difference of the time taken by each is 3 hours.

So we write the equation:

(D/(x/2) - D/x = 3 hours

On simplifying,

D/x = 3 hours

x=D/3=2 mph (cyclist)

x/2=2/1=1 mph (jogger)

Check:

Time for cyclist: 6/2= 3 hours

Time for jogger: 6/1=6 hour.

OK

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