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# Teaching ratios and proportions

Many times kids do learn how to solve proportion problems in school (they manage to memorize the steps), but that seems to get forgotten in a flash after school is over. Maybe they only remember faintly something about cross multiplying, but that's as far as it goes. How can we educators help them learn and retain?

## Proportions and ratios are NOT some odd way-out mathematical stuff

Truly they aren't. You use them every day, constantly, whether you realize it or not. Do you ever talk about going 55 miles per hour? Or figure how long it takes to travel somewhere with such and such a speed? Have you ever seen prices such as \$1.22 per pound, \$4 per foot, \$2.50 per gallon or similar ones? Have you ever figured how much something costs given the price per pound or per gallon etc.? Ever figured your daily or monthly pay if given the hourly rate? You've used ratios (or rates) and proportions.

## What are proportions all about

Consider the problem: if 2 gallons (of something) costs this much, how much would 5 gallons cost? What is the general idea to solve this problem?

Or, if car travels this much in 3 hours, how long could it travel in 4 hours? 6 hours? 7 hours?

In proportion problems you have two things that both change at the same rate. For example, you have dollars and gallons as your two things. You know the dollars & gallons in one situation (e.g. 2 gallons costs \$5.40), and you know either the dollars or the gallons of another situation, and are asked the missing one. For example, you are asked how much would 5 gallons cost. You know it is "5 gallons" and are asked the amount of dollars.

You can make tables to organize your information:

Example 1: Example 2:
 2 gallons - 5.40 dollars 5 gallons - x dollars
 110 miles - 3 hours x miles - 4 hours

In both examples, there are two things that both change at the same rate. In both examples, you have four numbers (two for one situation, two for the other situation), you are given three of them, and asked the fourth. So, how would we solve these types of problems?

## The many ways to solve a proportion

1. If 2 gallons is \$5.40 and I'm asked how much is 5 gallons, since gallons increased 2.5-fold, I just multiply the dollars by 2.5 too.

2. If 2 gallons is \$5.40, I figure first how much 1 gallon would be, and then how much 5 gllons. Okay, 1 gallons would be half of \$5.40 or \$2.70, and I'll go five times that.

3. I build a proportion like in the math book and solve by cross multiplying:
 5.40 2 gallons = x 5 gallons

Cross-multiplying from that, I get:

5.40 × 5 = 2x

 x = 5.40 × 5 2 =

4. I build a proportion like above but instead of cross-multiplying, I simply multiply both sides of the equation by 5.

5. I build a proportion but this way: (and it still works - you see, you can build the two fractions for your proportion in several different ways)

 5.40 x = 2 gallons 5 gallons

The point is, to solve problems like above, you don't need to remember how to write out the proportion equation and solve it -- you can ALWAYS solve them just by thinking and using common sense (and possibly a calculator). And this is the best approach to teach them to kids, too: make the students understand the basic idea in the problems so well that they can just figure them out without equations. You can cover the equation method too - for completeness' sake.

One good basic idea for solving proportion word problems is to think what would it be for 1 or for some other easy unit, and then multiply to get what is asked. For example: if car goes 110 miles in 3 hours, how far will it go in four hours? Figure out how far it gets in 1 hour, then multiply by 4.

Or, if 1 meter costs \$5.44, how about 0.40 meters? Well, divide the price by 10 to get price for 0.10 meters, then multiply by four.

## How to teach proportions

I hope by this point you yourself understand the basic ideas in these kinds of problems. To introduce them to your child or students, write out a few tables like below to be filled by student. This will help students learn proportional reasoning.

 Miles 45 Hours 1 2 3 4 5

 Dollars 3.3 Pounds 1 2 3 4 5

.. and other similar ones. Just make enough of these tables (first using easy numbers) so that the student gets used to filling in these proportional tables. Tie each table into some real-life situation - you can find many for example by looking at the proportion word problems in your math book.

Then, let them fill in a few tables this where the "givens" are in the middle:

 Dollars 45 Hours 1 2 3 4 5

 Dollars 42 Hours 1 2 3 4 5

 Dollars 15.5 Meters 0.10 0.20 0.30 0.40 0.5

Of course we hope the student notices it is easy if you first figure out how many dollars for 1 hour, and then find the other amounts.

Set up enough of these proportional reasoning tables until the student is familiar with the idea.

### The next step: proportion problems and thinking

After filling out such tables, the student is ready to tackle some problems. Just choose some simple proportion word problems, such as ones above, and let 'em think! Students might very well come up with an answer on their own. They might make a table. Or, they might figure out how to calculate the thing for 1 unit, and then go from there. Or they might discover the need to know how many-fold the one thing increased (or decreased), and then just multiply with that number to get the unknown thing.

If the child or student succeeds in figuring out how to do these type problems, then that method is likely to stick with them much better than some mechanical rote-memorized procedure they don't understand why it works.

That is all there is to these proportion word problems. You don't actually need to build an equation to solve them.

Now, I don't want to put down equations or cross-multiplying; it's just that understanding should come first, and is more important. It is still needful to learn proportions in the equation form for algebra's sake, but maybe that could be pushed off a little, instead of tackling it in 6th grade.

## Definitions

Did you notice I didn't give definitions of the terms ratio and proportion? Well, I didn't want to confuse. Sometimes you don't have to to learn the exact definitions or terms up front.

RATIO is two "things" (numbers or quantities) compared to each other. For example, "3 dollars per gallon" is a ratio. Or, "40 miles per 1 hour". Or, 15 girls versus 14 boys. Or, 569 words in 2 minutes. Or, 23 green balls to 41 blue balls, etc. Your math book might say it is a comparison of two numbers or quantities.

A related term, RATE, is defined as a ratio where the two quantities have different units. Some people differentiate and say that the two things in a ratio have to have a same unit; some people don't differentiate and allow "3 dollars per gallon" to be called a ratio as well.

PROPORTION is when you have two ratios set to be equal to each other. For example, "3 dollars per gallon" equals "6 dollars per two gallons". Or, 2 teachers per 20 kids equals 3 teachers per 30 kids. Or,

 3 liters 48 square meters = 10 liters 160 square meters

Of course, for it to be a problem, one of those four numbers is not given and ends up being the famous x.

Ratio and Proportion in-depth lessons at Purplemath.com

Supposedly Difficult Arithmetic Word Problems - a report that presents simple, conceptual-understanding based methods that will allow students to solve simple rate problems, ratio and proportion problems, work problems and others.

Ratio and Proportion at BBC Skillswise - factsheets, game, quiz, activity, and worksheets.

Math: Ratio and Proportion from MathLeague - summary explanations.

Rate vs. ratio - a discussion on the definition of these two terms.

Ratio and Proportion is applicable to other areas in math. For example: finding the lowest common multiple of two numbers. Try LCM[62,217]. The lowest common proper fraction is relatively prime. So 62/217 = 2/7 (i.e. the ratio is 2:7. Thus one simply cross multiplies the original proper fraction by the relatively prime one. 62 x 7 = 217 x 2. Or 434 = 434. This is the LCM of 62 & 217.
Or try LCM[20, 50]. 20/50 = 2/5. 20 x 5 = 50 x 2 = 100.
LCM[15, 20]. 15/20 = 3/4. 15 x 4 = 20 x 3 = 60. Etc.
John Shotzbarger, BSEE 6/3/06