1. Aug 27, 2006

kuahji

I'm having trouble just figuring out how to set these problems up. Here is one of them.

A jogger and a walker both cover a distance of 5 miles. The runner is traveling 1.5 times faster than the walker and finishes in 25 minutes less time. How fast is each going?

I know it probably should include the r*t=d

So I "tried" to make a table

jogger = rate = 1.5x time = x-25 = 5
walker I had tried x*x+25= 5 but
I then tried to combine the two & got 2.5x^2-12.5x-10=0

I know the answer is 4mph for the jogger & 6 for the walker. Any help on how to get the equation setup or where I'm going wrong would be appreciated.

2. Aug 27, 2006

HallsofIvy

Staff Emeritus
"jogger = rate = 1.5x time = x-25 = 5 "
What is x? in "rate= 1.5x" it appears to be the rate (speed) of the walker but in "time= x- 25" it appears to be the time the walker takes to walk the 5 miles. And, if that last "5" is the distance walked, "time= 5" makes no sense. Finally, whatever x is, in x*x+ 25= 5, you can't add 25 minutes to any quantity and get 5 miles as the result.

Let r be the speed of the walker. since "r*t= d", t= d/r so the time required for the walker to go 5 miles is t= 5/r. The speed of the jogger is 1.5r so the time required for the jogger to go 5 miles is t= 5/(1.5r)= 10/(3r). The walker took 25 minutes longer and 25 minutes= 25/60= 5/12 hour so 5/r= 10/(3r)+ 5/12. Multiply the equation by 12r to get rid of the fractions. That does not give a quadratic equation.

3. Aug 28, 2006

kuahji

Thanks, that makes sense. What was throwing me off was the fact that all the other problems in the chapter were quadratic equations. So naturally I wanted to "try" & set the problem up as one... but as you can see I just got confused & "tried" to throw something together that would make it quadratic. This here helped a lot. Thanks again.