Biking and WalkingDate: 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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