Posted by **chris** on Wednesday, March 24, 2010 at 9:37am.

A cyclist and a jogger start from a town at the same time and headed for a destination 24 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 3hrs ahead of the jogger. Whats the rate of the cyclist?

- math -
**PsyDAG**, Wednesday, March 24, 2010 at 11:29am
C = 2J (rates)

Distance = rate/time

Therefore, time = rate/distance

J/24 - C/24 = 3

Use substitution to solve first for cyclist and then for jogger.

- math -
**Chris**, Wednesday, March 24, 2010 at 11:48am
is the answer 12

- math (oops!) -
**PsyDAG**, Friday, March 26, 2010 at 7:52pm
Sorry, I goofed!

Distance = **rate * time**

Therefore, Time = Distance/rate

24/J - 24/C = 3

Substitute 2j for C.

24/j - 24/2J = 3

Multiply both sides by J.

24 - 12 = 3J

12 = 3j

4 = J

C = 2J = 2 * 4 = 8

Sorry for the goof! This should help.

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