Solving Proportions

Objective: When done with this lesson, you will have demonstrated how to

    • set up and solve proportions
    • determine when proportional reasoning is or is not appropriate to use in a situation

Approximate completion time: 2 hours

 

Suppose you wanted to build a new fence. You know that to build more fence, you'll need more the materials. That's a simple concept. How much more is where we'll focus in this lesson.

If you want twice as much fence, will you need twice as much material?
If you want five times as much fence, will it cost five times as much?
If you want ten times as much fence, will it take ten times as long?

If you answer yes to any question like those, then you are say the situation is "proportionate". (Some people call it "proportional".) To be proportionate two ratios must be equal.

 

Most people like to simplify proportions down to the "naked math" that looks like this:

Proportionate
Disproportionate

 

Here are a few quick ways to test if something is proportionate. That is, here are a few ways to see if two fractions are equal:

"left to right"
Compare numerator to numerator and denominator to denominator—can you see a common factor? If so, then it's proportionate.
"top to bottom"
Compare numerator to denominator and numerator to denominator—can you see a common factor? If so, then it's proportionate.
"cross multiply"
Multiply the numerator of one fraction to the denominator of the other. Repeat with the other "diagonal". If they products are the same, then the ratios are equal and it's proportionate.
"decimal form"
If the decimal forms of two fractions are not equal, then the two fractions are not equal and the situation is disproportionate. If the two fractions are equal then it is proportionate.
"unit rate"
If the unit rates are the same, then it's proportionate.

Look back through the Proportionate/disproportionate picture and table above.
See if you can spot how the ratios are equal with one of these three tests.

 

Now let's try our hand at some problems which can be solved using proportional reasoning.

Problem
Solution

Four light bulbs sell for $1.25, how much would 100 cost?

 

 

Since there would be 25 times as many, then the total for 100 light bulbs would be 25 times as much. That would be 25·($1.25) = $31.25.

Here's the same idea using a proportion:

where x is the cost in $.

Simpler it would be .

Comparing numerators, 4·25 = 100
so x = 1.25·25 = 31.25.

A 32 oz box of cereal sells for $3.50. At the same rate, how much should an 84 oz box of the same cereal cost?

Using proportions, where x is the cost in $.
Simpler, .

"Cross multiplying" we get 32·x = 84·3.50.
Which is 32x = 294.
So x = 294/32.
In decimal form, x = 9.1875.

It should cost about $9.19 for an 84 oz box.

To seed 5000 square feet of lawn, Mr. Greenjeans needs 20 pounds of seed. How much lawn could he seed with 50 pounds of seed?

Using proportions, where x is the number of square feet of seed.

Simpler the proportion is .

Cross-multiplying, we get 5000·50 = 20x.
Or 250,000 = 20x.

So x = 250,000/20 = 12,500.

Mr. Greenjeans could seed 12,500 square feet of lawn with 50 pounds of seed.

(Did you think of comparing denominators? Do you see that 50 is 2 ½ times as big as 20? Taking 5000 times 2 ½ gives 12,500.)

 


Now try the following practice on your own. Assignment 6.03.doc