Setting Up Algebraic Word ProblemsDate: 08/04/2003 at 15:42:59 From: Nathan Subject: Algebraic word problems: setting them up If each photocopy of a manuscript costs 4 cents per page, what is the cost, in cents, to reproduce x copies of an x-page manuscript? A 4x B 16x C x^2 D 4x^2 E 16x^2 I set the question up like this: 4/p = 4x/xp. I cross multiplied and got 4xp/p = c. I simplified it and got 4x = c. Unfortunately, answer "A," 4x, was not given as the correct answer. The correct answer given was "D," 4x^2. These types of questions seem to be very simple, yet I always have problems setting them up. Any suggestions? Thanks! Date: 08/04/2003 at 21:42:15 From: Doctor Rick Subject: Re: Algebraic word problems: setting them up Hi, Nathan. You haven't explained your reasoning in the set-up phase. Where did 4/p = 4x/xp come from? Here is what I would do. The cost of copies is 4 cents times the number of pages. To make use of this, I need to know how many pages there are. Each copy has x pages, and I want x copies; the total number of pages is x pages/copy * x copies = x^2 pages Now I can finish the work: cost = 4 cents/page * x^2 pages = 4x^2 cents Is there any part of this that you don't think you could do on your own? I'll be happy to discuss it further. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/06/2003 at 15:07:50 From: Nathan Subject: Algebraic word problems: setting them up Wow. Thanks. As I figured, the solution was very simple. So you're basically just multiplying them all together. That makes sense to me now. For some reason I always seem to get stuck on questions like this. The reason I set up the problem in the way that I did may seem a little queer to you, but for some reason whenever I see the word "PER" in a question, I tend to want to divide, without really thinking it through very carefully, for example, as in miles PER hour. Especially if I'm hurrying through it. Once you look at what they're actually asking for though, your approach makes a lot more sense to me now. With problems like this it's almost as if it's so simple that I overlook the obvious. Maybe my problem is just that my mind goes blank whenever I see a word problem. Thanks so much for your help. I guess practice makes perfect. Date: 08/06/2003 at 20:54:19 From: Doctor Rick Subject: Re: Algebraic word problems: setting them up Hi, Nathan. It's okay to make mistakes. Just keep observing the mistakes you make, so you can learn from them. The word "per" generally signals the presence of a RATE; when you see it, you will likely be able to apply the RATE EQUATION to the problem. For instance, when you see "miles per hour," the number with these units is a rate (in this case, a speed), and you can use the formula distance = rate * time However, whether you multiply or divide depends on whether you also know (or can find) the distance or the time. If you know that a car goes 50 miles per hour for 3 hours, then you know a rate and a time, so the formula says to MULTIPLY them to get the distance traveled (150 miles). On the other hand, if you know that the car went 150 miles at a speed of 50 miles per hour, you know a rate and a distance. To find the time it took for the trip, you must solve the rate equation for the time: time = distance / rate Now you see that you must DIVIDE the distance by the speed to find the time (150/50 = 3 hours). It can be done! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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