Setting Up Algebraic Word Problems

Date: 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/