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### Finding and Working with Ratios

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Date: 12/26/2001 at 13:16:54
From: Hima Mehta
Subject: Ratios and proportions

1. What is the ratio of the circumference of a circle to its radius?

2. A snail can move i inches in m minutes. At this rate, how many
feet can it move in h hours?
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Date: 12/26/2001 at 16:50:47
From: Doctor Achilles
Subject: Re: Ratios and proportions

Hi Hima,

Thanks for writing to Dr. Math.

A ratio is just a fraction, where the numerator and denominator
can have different units.  For example, if you travel 60 miles
in 2 hours, we can write that as a ratio:

60 miles
--------
2 hours

and we can simplify that ratio, using the same techniques you've
learned for simplifying fractions:

60 miles   30 miles
-------- = --------
2 hours    1 hour

So to find the ratio of a circumference of a circle to the radius
of the same circle, we can start by writing expressions for those
quantities, and using them to form a fraction.  Suppose we have
a circle with radius r.  The circumference of the circle is pi
times the diameter of the circle, and the diameter is twice the

circumference = 2 * pi * radius

So now let's form the fraction:

circumference   2 * pi * radius
------------- = ---------------

which simplifies to just

2 * pi

So the ratio of the circumference of a circle to its radius is 2*pi.

Your second question asks you "A snail can move i inches in m minutes.
At this rate, how many feet can it move in h hours?"

We're told something about inches and minutes and we're supposed to
find something about feet and hours. How can we do this?  We'll need
to use a special kind of ratio, called a 'conversion ratio'.

A conversion ratio is a fraction where the numerator and denominator
express the same quantity using different units.  The ones we'll need
for this problem are

1 hour               1 foot
----------    and    ---------
60 minutes           12 inches

Note that since each of these has the same quantity in both the
numerator and the denominator, multiplying by either ratio has the
same effect as multiplying by 1 - which is to say, it may change
the appearance of a quantity, but it won't change the value.

(This is sort of the same idea as when we multiply a fraction like
1/2 by a fraction like 3/3 to get 3/6.  Since 3/3 is just another
way to write 1, 1/2 and 3/6 have the same value, even though they
look different.)

Let's go back to the problem. We are told that a snail moves i inches
in m minutes. Then we are asked to figure out something about his
rate.

Here is one of the most important equations in math and physics:

distance = rate * time

Or, dividing both sides by time:

distance / time = rate

The snail's distance is i inches, and the time is m minutes, so let's
put those into the equation:

i inches
----------  = rate
m mins

Now we have a couple of conversion ratios that we can use.
We can multiply the left side of this equation by one of
our ratios:

i inches        1 foot
----------  *  -----------  = rate
m mins        12 inches

Remember that in multiplying fractions, you take the numerator (top)
of the first and multiply it by the numerator of the second, and you
take the denominator (bottom) of the first and multiply it by the
denominator of the second.

i inches * 1 foot
---------------------  = rate
m mins * 12 inches

Now, you can treat units (such as inches, feet, hours, whatever) like
constants. That means they can cancel each other out. So since we have
inches on top and on the bottom of the fraction, it cancels out:

i * 1 foot
-------------  = rate
m mins * 12

Or, more simply:

i feet
-----------  = rate
12*m mins

Let's use our second conversion ratio:

i feet         60 mins
-----------  *  ---------  = rate
12*m mins       1 hour

And so we multiply the numerator by the numerator and the denominator
by the denominator:

i feet * 60 mins
--------------------  = rate
12*m mins * 1 hour

And since "mins" is in both the numerator and the denominator, we can
cancel it out:

i feet * 60
---------------  = rate
12*m * 1 hour

And simplify a bit:

60*i feet
-----------  = rate
12*m hours

Now, let's just look at the numbers for a minute. We have 60 on top
and 12 on the bottom. They both have 6 as a factor, so let's factor 6
out:

6 * 10*i feet
---------------  = rate
6 * 2*m hours

And the 6's cancel:

10*i feet
-----------  = rate
2*m hours

Now, both the top and bottom have 2 as a factor, so let's get that out
also:

2 * 5*i feet
--------------  = rate
2 * m hours

And cancel:

5*i feet
----------  = rate
m hours

So the question we were trying to answer is: how many feet can the
snail move in h hours? Well, we know from this that it can move
5*i feet in m hours. How do we figure out h hours? We need to get an h
on the bottom. The only way to do that is to multiply the top and
bottom of the fraction by h. (Notice that this is just another ratio,
h/h is a fraction that equals one.):

5*i feet       h
----------  *  ---  = rate
m hours       h

Becomes:

5*i feet * h
--------------  = rate
m hours * h

Becomes:

5*i*h feet
------------  = rate
h*m hours

We're almost done. Now we just have to get rid of that pesky little m
on the bottom. There's one more ratio we can multiply by:

5*i*h feet       1/m
------------  *  -----  = rate
h*m hours       1/m

Becomes:

5*i*h feet * 1/m
------------------  = rate
h*m hours * 1/m

Becomes:

5*i*h/m feet
--------------  = rate
h*m/m hours

And finally, we have:

5*i*h/m feet
--------------  = rate
h hours

So our little snail (who is probably all tired out from doing so many
ratios), can move 5*i*h/m feet every h hours.

Be sure you understand all the steps for this. If you understand how
to use conversion ratios, you will find that they are very useful for
any type of math or science.

- Doctor Achilles, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Ratio and Proportion

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