# 1.4: Proportions

- Page ID
- 130921

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In the previous section, we learned that a *ratio* is a comparison of two quantities. However, many of the problems we solved involved comparing multiple ratios, and often required finding an equality between two ratios. These sorts of problems are most easily solved using *proportions*, which are the subject of this section.

- Recognize and set up proportion problems
- Apply the Cross Multiplication and Division undoes Multiplication methods to solve proportion problems
- Use the Compare to the Whole method to solve problems involving proportions

## Proportions: Definition and Basic Methods

Let's recall our starting example from the previous section, in which we had a recipe that called for \(2\) cups of flour and \(1\) cup of sugar. We asked the question: if we wanted to make a larger version of the same recipe using \(3\) cups of sugar, how much flour should we use?

We know now that ratios can be expressed as fractions, and whenever ratios are equivalent, the fractional representations of those ratios are equal. So, let's call the number of cups of flour \(x\) and set up the following equality:

\[\frac{2 \text{ cups of flour}}{1 \text{ cup of sugar}} = \frac{x \text{ cups of flour}}{3 \text{ cups of sugar}}\]

We need to figure out the value of \(x\) that makes this equation true. That is, what number can be substituted in place of \(x\) so that these fractions truly are equal? In this case, the easiest way is to guess-and-check, which is a perfectly valid solution method with small numbers. If you just try some small numbers, you can find that \(x = 6\) is the solution, because \(\frac{6}{3}\) reduces to \(\frac{2}{1}\), and thus \(\frac{6}{3}\) and \(\frac{2}{1}\)are equivalent ratios.

In this chapter, we'll learn how to solve problems like this generally, including those that can't be solved using guess-and-check. Our primary tool will be a *proportion.*

A **proportion** is an equality between ratios.

This means that, in a strictly mathematical sense, a proportion is an equation. For example,

\[\frac{5}{2} = \frac{40}{16}\]

is a proportion, because it contains two ratios that are equal to one another. You may have heard the word proportion used in other ways, and that's because the word "proportion" has a different mathematical meaning than its typical English usage. That doesn't mean either usage is wrong; rather, it is dependent upon context. In this book, we will use the word *proportion *to mean any equation that looks like this:

\[\frac{a}{b} = \frac{c}{d}\]

where \(a, b, c\), and \(d\) will usually be numbers or variables.

The reasons we care about proportions is that they give us a way to *find an unknown part of one of the ratios involved***.** Recall the following example, which we saw in the Ratios section:

On the Western Oregon University website, the total enrollment is listed as \(3752\) students, and the student-faculty ratio is listed as \(13 \colon 1\). You want to know how many faculty there are at WOU. How might you find this out, and how do you explain your answer?

In solving this problem before, we set up two ratios

\[3752 \colon x \quad \text{and} \quad 13 \colon 1 \]

Why did we do this? Well, it turns out that all proportion problems can be solved using a method from algebra known as *cross multiplication*. While this text mostly stays away from algebra, this procedure is essential. The good news is that it works the same way every time, and it's not very complicated.

If you have a proportion of the form:

\[\frac{a}{b} = \frac{c}{d}\]

then "cross multiplication" refers to rewriting the equation in the following equivalent way:

\[c \times b= a \times d\]

In other words, we are "crossing" from \(a\) to \(d\) in the \(\searrow\) direction and from \(c\) to \(b\) in the \(\swarrow\) direction.

A quick mathematical note: what we're really doing is multiplying both sides by \(b\) and \(d\), and then canceling common factors -- but calling it cross multiplication seems to make it easier for students to understand and remember.

Let's practice cross multiplying in our example. We had the proportion:

\[\frac{3752}{x} = \frac{13}{1}\]

Cross multiplying this gives us the following equation:

\[13 \times x = 3752 \times 1\]

What good did that do? Well, note that we can simplify a little bit. It's typical to omit the \(\times\) when a variable is multiplied by a number, so we can rewrite \(13x\) for \(13 \times x\). We also know that \(3752 \times 1 = 3752\). So our equation becomes

\[13x = 3752\]

Now what? We've eliminated the fractions, but we can't yet say what \(x\) is. In order to find \(x\), we need one more algebra procedure, which we will call *Division undoes Multiplication.*

Given an equation of the form \[Ax = B\] where \(A\) and \(B\) are numbers, we can find the value of \(x\) by dividing both sides by \(A\). That is, \[x = \frac{B}{A} = B \div A\]

Once again, we are using properties of fractions here: mathematically, we are dividing both sides by \(A\) and then reducing:

\[\begin{align*} Ax & = B\\ \frac{Ax}{A} & = \frac{B}{A} \\ \frac{\cancel{A}x}{\cancel{A}} & = \frac{B}{A} \\ x & = \frac{B}{A} \\ \end{align*} \]

But this is another procedure used so frequently that it's worth giving it a name.

Back to our example: we had the equation \[13x = 3752\]

We now have a tool to find \(x\) -- the fact that *Division undoes Multiplication*! Using this procedure, we have \[x = \frac{3752}{13} = 3752 \div 13 \approx 288.6\]

This is the same answer we found before, but we used a slightly different method. And keep in mind that, just as the previous section, we would need to round this answer to \(289\) faculty to make sense in context. That said; the main point is now we now have a fool-proof way to solve this type of equation!

While this may seem more complicated at first, you'll find that the following sequence of steps will always work to solve proportions:

- Set up the proportion with exactly one unknown value, called \(x\).
- Apply the Cross Multiplication.
- Apply Division undoes Multiplication.

We will get lots of practice with this procedure in the exercises for this section. Once you practice with the procedures above, you'll find that it's not too bad. The hardest part is often the first step — setting up the proportion correctly. That's the part that depends on reading the question very carefully! In general, the way to set up a proportion involves keeping track of units. Let's see an example to understand.

In an office supply store, \(8\) markers cost a total of \(\$12.00\). Assuming all markers are equally priced, how much would 6 markers cost?

###### Solution

This is a problem that is suitable to be solved using proportions because the markers are all equally priced, meaning that the ratio of *total cost : number of markers purchased* will be the same, no matter how many markers are purchased. That means we can set up the following proportion:

\[\frac{12 \text{ dollars}}{8 \text{ markers}} = \frac{x \text{ dollars}}{6 \text{ markers}}\]

Notice how, in the equation above, we are *labeling the units* of all quantities involved. Moreover, the units on each side match: dollars are on top, markers are on bottom, and the corresponding quantities are grouped on each side of the equation -- \(\$12\) for \(8\) markers, and \(\$x\) for 6 markers. Labeling your units in this way will help you avoid mistakes with units!

Now that we've gotten our proportion set up correctly, we can rewrite it without labels: \[\frac{12}{8} = \frac{x}{6}\]

From here, we'll follow the last two steps: cross multiply, and then use division to find \(x\). Using Cross Multiplication, we have \[x \times 8 = 12 \times 6 \]

On the left, we can rewrite \(x \times 8\) as \(8 x\), since multiplication can always switch orders. Then we can simplify to get \[8x = 72\]

Now we can use Division undoes Multiplication to get

\[x = \frac{72}{8} = 72 \div 8 = 9\]

Therefore, \(x = 9\). Now, we want to make sure our answer actually means something. What are the units on \(x\)? Well, if we look back at our original proportion,

\[\frac{12 \text{ dollars}}{8 \text{ markers}} = \frac{x \text{ dollars}}{6 \text{ markers}}\]

we see that \(x\) is a number of dollars. Thus, we can say that \(x = \$9\), which means that 6 markers will cost \(\$9\).

You may be thinking: there is a much faster way to do that! And that may be true for you. Once again, the point is not to mimic a particular method for problem solving here — these notes will show some good ways of solving a problem, but they cannot cover every good solution. They are intended to highlight themes and strategies that will work for many types of situations. Other ways you may have solved the problem above include:

- Calculate the cost per marker to be \(\$1.50\), and multiply that number by 6 markers to get \(\$9\).
- Calculate that \(6\) is \(\frac{3}{4}\) of \(8\), so the cost of \(6\) markers would be \(\frac{3}{4}\) the cost of \(8\) markers, and \(\frac{3}{4}\) of \(\$12\) is \(\$9\).
- Set up a different initial proportion, such as \(\frac{12 \text{ dollars}}{x \text{ dollars}} = \frac{8 \text{ markers}}{6 \text{ markers}}\) or \(\frac{8 \text{ markers}}{12 \text{ dollars}} = \frac{6 \text{ markers}}{x \text{ dollars}}\) and then solved that proportion.

What's amazing about the last point above is that both of those proportions — which were different than the method used in the solution above — still give the same answer! This shows that there are *many* different ways of approaching the same problem. All you need to do is find the one that works for you, and be able to explain your work.

**Comparing to the Whole**

Sometimes a problem involving proportions will be less straightforward. For example, consider the following:

In a rainforest in Panama, the ratio of two-toed sloths to three-toed sloths is \(10 \colon 3\). There are \(741\) total sloths in the rainforest. How many of them are two-toed?

In the problem above, we are given one ratio that compares the quantity of two-toed versus three-toed sloths. However, we are not given any information about the actual numbers of either two- or three-toed sloths. We simply know the comparison between them. Instead, we are just given the total number of sloths, but no actual breakdown into how many fall into each category. How are we supposed to find the number of two-toed sloths from just this information? We can't readily write down a proportion like we were able to in the previous example, because the units would be wrong; we need to compare like quantities. This situation calls for one more procedure.

Assume there are two quantities, \(x\) and \(y\), neither of which you know. However, you know two things about them

- The total \(x + y\) (the total number of both quantities)
- The ratio of quantity \(x\) to quantity \(y\) is \(a \colon b\)

Then you can use the Compare to the Whole method. This says that, to find quantity \(x\), you use the proportion \[\frac{a}{a+b} = \frac{x}{x+y}\] and then find \(x\). Note: you already know \(x+y\), since it is the total number of both quantities.

Let's see how this procedure can be applied to the sloth example.

In a rainforest in Panama, the ratio of two-toed sloths to three-toed sloths is \(10 \colon 3\). There are \(741\) total sloths in the rainforest. How many of them are two-toed?

###### Solution

In this question, our two quantities \(x\) and \(y\) are the number of two- and three-toed sloths, respectively. We are asked to find \(x\), the number of two-toed sloths. Our known ratio is \(10 \colon 3\), so using the notation of the Compare to the Whole method, we have \(a = 10\) and \(b = 3\), and \(a +b = 13\). We also know that the total number of sloths is 741, so \(x + y = 741\). So we'll set up the following proportion -- pay close attention to the labels!

\[\frac{10 \text{ two-toed sloths}}{13 \text{ total sloths}} = \frac{x \text{ two-toed sloths}}{741 \text{ total sloths}}\]

On the righthand side of the proportion above, the ratio \(\frac{x}{741}\) represents the actual number of sloths, in which there are \(x\) two-toed sloths out of a total of \(741\) total sloths.

On the lefthand side, the ratio \(\frac{10}{13}\) represents an imaginary "smaller but proportional rainforest," in which there are only \(10\) two-toed and \(3\) three-toed sloths, for a total of \(13\) sloths in our imaginary smaller rainforest.

Proportionality says that these proportions must be equal, but since we don't know the breakdown of the total number of sloths, we must *compare to the whole*, which means we must compare the total number of sloths on each side. We get a total number of \(13\) on the left by computing \(10 + 3\), and on the right, we know the total to be \(741\).

Once we have that proportion, we can simply solve it using our processes from the previous section. From the proportion \[\frac{10}{13} = \frac{x}{741}\]

we use *Cross Multiplication *to obtain \[13x = 7410\]

and then use *Division undoes Multiplication* to get \[x = \frac{7410}{13} = 570\]

Looking back, we see that \(x\) represents the number of two-toed sloths. Therefore, there are \(570\) two-toed sloths in the rainforest.

That's the best way to think about the Compare to the Whole method -- the ratio you are given represents a "smaller version" of the situation described, and to find the total quantity in the smaller version, you simply add the two parts together. Then compare that to the actual total quantity using a proportion. *If the problem statement contains words like "total," "whole," or "all together," it's likely that you'll need to use the Compare to the Whole method*. However, as always, the most important thing is to read the problem and think critically about what it's asking!

P.S. for this section: You may notice that some of the algebra is becoming less explicit as we see more and more examples. If you are confused about why an algebra or arithmetic step is true, try looking for a similar problem earlier in this book — there is likely an explanation there. If you can't find one, or are still confused, you should ask your instructor or email the author of this book at merrill@wou.edu.

## Exercises

When you are completing these exercises, make sure to show supporting work.

- You can walk 2 miles in 36 minutes. How long will it take you to walk 5 miles? Give you answer as a number of hours plus a number of minutes (that is, you would express 70 minutes as "1 hour and 10 minutes"). Remember that there are 60 minutes in an hour!
- You can mow 1/3 of an acre of lawn in 90 minutes. How long would it take you to mow 2 acres of lawn? Give your answer as a number of hours.
- When brewing an amber ale (a type of beer), recipes typically call for an 8:2 ratio of pale malts to crystal malts (these are types of grain in the beer). If you are brewing a 10 gallon batch of amber ale, you need a total of 22 pounds of malt. How many pounds of each type of malt (pale and crystal) should you buy? Make sure to indicate both answers clearly, and do not round them -- decimals are fine. [Hint: they should add up to 22 pounds!]
- The ratio of the length of the shadow cast by an object to the height of that object is 2-to-3 at a certain time of day.
- How long of a shadow does a 6 foot tall person cast?
- If a shadow of a tree is 20 feet long, how tall is the tree?

- The ratio of registered Democrats to registered Republicans is 47 : 52 in Polk County. There is a total of 8920 registered Democrat and Republican voters. How many of them are Democrats?
- The following is an excerpt* of an article about sea turtle sex ratios (e.g., what proportion of the population is female or male.):
"More than 200,000 sea turtles nest on or near Raine, a tiny 80-acre curl of sand along the northern edge of the Great Barrier Reef, the portion hardest hit by warming waters. The other portion of that sea turtle population nests further from the equator, near Brisbane, where temperature increases have not been as dramatic.

What Allen and Jensen discovered was significant. Older turtles that had emerged from their eggs 30 or 40 years earlier were also mostly female, but only by a 6 to 1 ratio. But younger turtles for at least the last 20 years had been more than 99 percent female. And as evidence that rising temperatures were responsible, female turtles from the cooler sands near Brisbane currently still only outnumber males 2 to 1.

Six weeks after Allen and Jensen published their results, another study from Florida looking at loggerheads revealed that temperature is just one factor. If sands are moist and cool, they produce more males. If sands are hot and dry, hatchlings are more female.

But new research in the last year also offered rays of hope."

- What ratios can you find above? Write them down, stating explicitly what they are comparing.
- What two factors does this article assert affect the sex of sea turtles? List them.
- Given the information in the article, if a randomly selected group of 120 turtles from Brisbane have their sex examined, how many do you expect to be female? Show your work.
- Write a 2-4 sentence reaction to the article excerpt above, and make sure to answer the following question: do you feel that that ratios in the article are presented in a way that makes sense? If not, how else could you present this same information?

[*Note: you can access the article for free if you enter your email when prompted; however, you do NOT need to access the article answer this question.]