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

## Answer this Question

## Related Questions

- Math - A cyclist and a jogger start from a town at the same time and head for a ...
- math - A cyclist and a jogger start from a town at the same time and headed for ...
- maths - a cyclist and jogger are 20 miles apart. the cyclist rides at 17 mph and...
- Math - A cyclist traveled at a rate of 32 mph to visit a nearby town. The ...
- Algebra - Two cyclists leave towns apart at the same time and travel toward each...
- Algebra - Two cyclists leave towns apart at the same time and travel toward each...
- Algebra - Two cyclists leave towns apart at the same time and travel toward each...
- physics - A jogger runs down the street at a rate of 1.1 m/s. She sees the ...
- algebra - A cyclist traveled at a rate of 32 mph to visit a nearby town. The ...
- Physics - A cyclist starts from rest and coasts down a 6.5∘{\rm ^\circ} ...