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Dividing with the Same Units, Such As Money

Date: 06/03/2008 at 15:44:52
From: Karen
Subject: Can you divide money by money

What is $30.00 divided by $5.00?

I did not think you could divide money by money.  $30.00 divided by 5 
equals $6.00, but does $30.00 divided by $5.00 equal $6.00 also?



Date: 06/03/2008 at 16:48:53
From: Doctor Achilles
Subject: Re: Can you divide money by money

Hi Karen,

Thanks for writing to Dr. Math.  Great question!

You can think of this at least three ways:

1) Dollars (and other units like inches, gallons, horses, etc.) can be
treated just like a variable.  So:

  30 dollars
 ------------
      6

equals:

  30/6 dollars

or:

  5 dollars

Just like:

  30x
 -----
   6

equals

  (30/6)x

or:

  5x

So in the question you had:

  30 dollars
 ------------
   6 dollars

equals

  (30/6)*(dollars/dollars)

and whenever you divide anything by itself, you get 1.  So you have:

  5*1

or just:

  5

Similarly:

  30x/6x = (30/6)(x/x) = 5

Notice that when you divide dollars by dollars you lose the unit.

So 30 dollars divided by 6 dollars is equal to 5.  Just the number 5.
NOT, NOT, NOT 5 dollars!!!

One way this is commonly said is:

  "The ratio of 30 dollars to 6 dollars is 5 to 1"

or sometimes just:

  "The ratio of 30 dollars to 6 dollars is 5"

2) Put it into a word problem:  "I have 30 dollars.  I give 6 dollars
to each person.  How many people can I give money to?"

To solve this:

  30 dollars
 ------------
   6 dollars

comes out to:

  5

That's 5 people, NOT 5 dollars!  The answer we want is a number (of
people), not a number of dollars.

Compare that to the word problem: "I have 30 dollars.  I want to
divide this money evenly between 5 people.  How much money does each
person get?"

To solve:

  30 dollars
 ------------
      5

equals:

  6 dollars

Here we didn't divide by dollars, so the dollars stick around.  The
answer is in dollars.

3) We can both agree that this is a true equation:

  (30 dollars)/5 = 6 dollars

What happens if you multiply both sides of that equation by 5, you get:

  30 dollars = 5*(6 dollars)

Now, divide both sides by "6 dollars":

  (30 dollars)/(6 dollars) = 5

That must, according to the rules of algebra, also be a true equation.


No matter which way you think of it, this is a very important
distinction and keeping track of units (dollars, cows, feet, square
feet, etc.) is vitally important.

Consider this word problem: "The area of a rectangular field is 600
square feet.  The field is 30 feet wide.  What is the length of the
field?"

We know:

  area = length * width

So:

  length = area / width

A lot of people would solve it like this:

  length = 600 / 30

  length = 20

But I would say this is, strictly speaking, WRONG.  Why?  You have to
keep track of units!

The correct solution is:

  length = (600 ft^2)/(30 ft)

  length = (600/30)*(ft^2/ft)

Just like x^2/x is equal to x, ft^2/ft is equal to ft.  Does that make
sense?

So you have:

  length = 20 ft

A lot of people just throw "ft" in at the end, so they do the problem
like this:

  length = 600 / 30

  length = 20 ft

But this is sloppy and can get you into trouble in two ways.

The first way this gets you into trouble is if you have mixed units. 
What if we change the question to be:  "A field has an area of 600
square feet and a width of 10 yards, what is the length?"

  length = 600 / 10

  length = 60

But what's the unit?

It's better to do the problem this way:

            600 ft^2
  length = ----------
             10 yd

Because:

  1 yd = 3 ft

We can divide both sides of this equation by "3 ft" and get:

  1 yd
 ------ = 1
  3 ft

Which is really, really useful.  We can always multiply anything by 1
without changing the value.  So let's take our equation:

            600 ft^2
  length = ---------- * 1
             10 yd

But we know that 1 is equal to (1 yd)/(3 ft), so we can do some
substitution:

            600 ft^2     1 yd
  length = ---------- * ------
             10 yd       3 ft

Or:

            600*1 yd*ft^2
  length = ---------------
             10*3 yd*ft

or:

            600 yd*ft^2
  length = -------------
             30 yd*ft

And, just like yx^2/yx is equal to x, yd*ft^2/(yd*ft) = ft

  length = 20 ft

Why go through all that trouble?  I admit that it's more than is
typically necessary for most algebra problems.  And certainly it's
fine to first convert everything to feet.  But it is important to keep
track of units at every step.

Which brings me to the second way that being sloppy about units can
get you into trouble.

Let's take this simple physics problem: "A force of 50 kg*m/s^2 is
applied to a mass of 10 kg.  What is the acceleration?"

The appropriate equation is force = mass * acceleration.  But what if
I forget the equation?  What if I mess up and think the equation is
supposed to be force * mass = acceleration?

Well, if I make that mistake, here is how I would solve the problem
being sloppy about my units:

  force * mass = acceleration

  50 * 10 = 500

So, I would say, acceleration is 500.  I might remember that the units
for acceleration are m/s^2, so I would just write:

  500 m/s^2

and think I was done.

BUT, let's say I make that SAME mistake, but I'm careful about my
units.  I still use the WRONG equation:

  force * mass = acceleration

  50 kg*m/s^2 * 10 kg = acceleration

  500 kg^2*m/s^2 = acceleration

Maybe I remember right away that the units for acceleration are m/s^2
and so I see right away that I made a mistake.  Or, maybe I don't, but
I at least notice that there's a kg^2 in my answer.  I can stop and
wonder what kg^2 could possibly have to do with acceleration. 
Acceleration has to do with speed, not mass.  There's no reason for kg
to be there, and certainly not kg^2!  I must have messed up.

So, by keeping track of my units along the way, I have caught an error
that I would have missed.

In fact, if I know before I start that the units for acceleration are
m/s^2, I can just look at the units and make a good guess at the
equation.  Sometimes you'll still have some trouble since the units
alone wouldn't tell you that there is a 1/2 in an equation like:

  kineticEnergy = 1/2 * mass * velocity^2

But in general, being careful about the units will at least help you
get on the right track and catch some common mistakes like with the
example above.

For more information, check out:

  Unit Analysis and Conversion Factors
    http://mathforum.org/library/drmath/view/62176.html 

Hope this helps.  If you have other questions or you'd like to talk
about this some more, please write back.

- Doctor Achilles, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Middle School Terms/Units of Measurement

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