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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