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### Biking and Walking

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Date: 06/02/97 at 18:59:12
From: Robert p Kris
Subject: Biking and walking

I need help with an algebra problem. I keep coming up with 8/3 as
an answer, but it does not seem to check out when I put it back into
the problem.

Here is the original problem:

Juan can bike twice as fast as he can run.  One day, he biked 12 miles
to a friend's house, and ran back.  The entire trip took 3 hours.
How fast did Juan run?

I tried to set up this problem as follows

running rate = x            total distance = 24 miles
biking rate  = 2x           time = 3 hours

rate * time = distance --->  3x * 3 = 24
3x = 8
x = 8/3

When I put 8/3 back into the problem to check, it doesn't work out.
```

```
Date: 06/02/97 at 19:28:21
From: Doctor Wilkinson
Subject: Re: Biking and walking

The reason why 8/3 doesn't work is because you can only apply the
formula rate x time = distance when the rate is constant throughout
the problem.

Here you have two different rates, which you quite properly call x and
2x.  Now what should you do?  What's the information you're given?
It's 12 miles out and 12 miles back, and the total time is 3 hours. So
on the way out we have 12 miles at 2x miles per hour, so the time is
12/2x.  On the way back we have 12 miles at x miles per hour so the
time is 12/x.  The total time is 3 hours, so:

12/2x + 12/x = 3
6 + 12 = 3x  Multiplying by x
18 = 3x
6 = x

Presumably Juan leaves his bike at his friend's house and spends no
time visiting, but these word problems never do make too much sense!

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
Middle School Word Problems

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