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Using Units and Variables to Make Sense of Word Problems

Date: 05/24/2004 at 17:05:29
From: Leadra
Subject: word problems

I am having a hard time changing a word problem into a formula in 
order to get the results.  For example:

Apples cost $1.50 per pound and pears cost $2.30 per pound.  A total
of 8 pounds of apples and pears were bought, and the price was $14.00.
How many pounds of pears were bought?

I don't know how to start it.

apples = x (1.5)
pears = y (2.3)

1.5x + 2.3y = 8
x + y = $14.00

Can you help me?

Date: 05/25/2004 at 17:00:44
From: Doctor Ian
Subject: Re: word problems

Hi Leadra,

You're on the right track, but you are getting confused about the 
units and how they relate to your equations.  Let's look at a few
ideas that might make the problem easier for you.

The first thing I'd do is express everything in pennies, rather than
dollars, because that would let me work with integers instead of
decimals, and they tend to be better behaved.  So we have

  apples at 150 cents per pound
  pears  at 230 cents per pound

  8 pounds of apples and pears combined for 1400 cents
  How many pounds of pears? 

Well, the amount you spend on apples is the price per pound, times the
number of pounds:

  (pounds of apples)(price per pound of apples)

And the amount you spend on pears is the price per pound, times the
numbers of pounds:

  (pounds of pears)(price per pound of pears)

And this all has to add up to 1400 cents:

  1400 cents  =  (pounds of apples)(price per pound of apples)
               + (pounds of pears)(price per pound of pears)

We know the prices per pound:  

  1400 cents  =  (pounds of apples)(150 cents/pound)
               + (pounds of pears)(230 cents/pound)

And we know something else:  If we have P pounds of pears, then we
have (8 - P) pounds of apples!  Do you see why?  So we have

  1400 cents  =  (8 - P pounds)(150 cents/pound)
               + (P pounds)(230 cents/pound)

After checking to see that the units work out (if we multiply pounds
by cents per pound, we get cents; and we'll add cents to cents on the
right, which matches cents on the left), we can get rid of them:

  1400 = (8 - P)(150) + (P)(230)

And now we just have to solve for P, to find what we were looking for. 

A couple of things to note here:

First, I used words instead of variables for as long as possible,
because it's usually easier to look at words and see if what you have
makes sense.  It's faster to jump right in with variables, but as they
say, you can't make mistakes fast enough to get the right answer.  :^D

Second, it's usually a good idea to leave units in for as long as
possible, since mismatches in units are one of the best ways to catch
equations that haven't been set up properly.  For example, looking at
your equations, 

  1.5x + 2.3y = 8

     x +    y = $14.00

for the second equation to make sense, the units of x and y would have
to be dollars.  But then the first equation becomes

  (1.5 ___)*(x dollars) + (2.3___)*(y dollars) = 8 pounds
which means that the units on 1.5 and 2.3 must be pounds per dollar,
instead of dollars per pound.  If you'd left the units in, you'd have had

    (1.5 dollars/pound)*(x dollars) 
  + (2.3 dollars/pound)*(y dollars) = 8 pounds

and that would have told you right away that there was something amok
with the equation. 

Does this make sense? 

- Doctor Ian, The Math Forum 
Associated Topics:
Middle School Word Problems

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