# 7.3 Applications of proportions

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses applications of proportions. By the end of the module students should be able to solve proportion problems using the five-step method.

## Section overview

• The Five-Step Method
• Problem Solving

## The five-step method

In [link] we noted that many practical problems can be solved by writing the given information as proportions. Such proportions will be composed of three specified numbers and one unknown number represented by a letter.

The first and most important part of solving a proportion problem is to deter­mine, by careful reading, what the unknown quantity is and to represent it with some letter.

## The five-step method

The five-step method for solving proportion problems:
1. By careful reading, determine what the unknown quantity is and represent it with some letter. There will be only one unknown in a problem.
2. Identify the three specified numbers.
3. Determine which comparisons are to be made and set up the proportion.
4. Solve the proportion (using the methods of [link] ).
5. Interpret and write a conclusion in a sentence with the appropriate units of measure.

Step 1 is extremely important. Many problems go unsolved because time is not taken to establish what quantity is to be found.

When solving an applied problem, always begin by determining the unknown quantity and representing it with a letter.

## Sample set a

On a map, 2 inches represents 25 miles. How many miles are represented by 8 inches?

• The unknown quantity is miles.
Let $x=$ number of miles represented by 8 inches
• The three specified numbers are
2 inches
25 miles
8 inches
• The comparisons are
2 inches to 25 miles → $\frac{\text{2 inches}}{\text{25 miles}}$
8 inches to x miles → $\frac{\text{8 inches}}{\text{x miles}}$
Proportions involving ratios and rates are more readily solved by suspending the units while doing the computations.
$\frac{2}{\text{25}}=\frac{8}{x}$
• $\begin{array}{cccc}\hfill \frac{2}{25}& =& \frac{8}{x}\hfill & \text{Perform the cross multiplication.}\hfill \end{array}$
$\begin{array}{cccc}\hfill 2\cdot x& =& 8\cdot \mathrm{25}\hfill & \\ \hfill 2\cdot x& =& 200\hfill & \text{Divide 200 by 2.}\hfill \\ \hfill x& =& \frac{200}{2}\hfill & \\ \hfill x& =& 100\hfill & \end{array}$
In step 1, we let $x$ represent the number of miles. So, $x$ represents 100 miles.
• If 2 inches represents 25 miles, then 8 inches represents 100 miles.

An acid solution is composed of 7 parts water to 2 parts acid. How many parts of water are there in a solution composed of 20 parts acid?

• The unknown quantity is the number of parts of water.
Let $n=$ number of parts of water.
• The three specified numbers are
7 parts water
2 parts acid
20 parts acid
• The comparisons are
7 parts water to 2 parts acid → $\frac{7}{2}$
$n$ parts water to 20 parts acid → $\frac{n}{\text{20}}$
$\frac{7}{2}=\frac{n}{\text{20}}$
• $\begin{array}{cccc}\hfill \frac{7}{2}& =& \frac{n}{\mathrm{20}}\hfill & \text{Perform the cross multiplication.}\hfill \end{array}$
$\begin{array}{cccc}\hfill 7\cdot \mathrm{20}& =& 2\cdot n\hfill & \\ \hfill 140& =& 2\cdot n\hfill & \text{Divide 140 by 2.}\hfill \\ \hfill \frac{140}{2}& =& n\hfill & \\ \hfill 70& =& n\hfill & \end{array}$
In step 1 we let $n$ represent the number of parts of water. So, $n$ represents 70 parts of water.
• 7 parts water to 2 parts acid indicates 70 parts water to 20 parts acid.

A 5-foot girl casts a $3\frac{1}{3}$ -foot shadow at a particular time of the day. How tall is a person who casts a 3-foot shadow at the same time of the day?

• The unknown quantity is the height of the person.
Let $h=\text{height of the person}$ .
• The three specified numbers are
5 feet ( height of girl)
$3\frac{1}{3}$ feet (length of shadow)
• The comparisons are
5-foot girl is to $3\frac{1}{3}$ foot shadow → $\frac{5}{3\frac{1}{3}}$
h -foot person is to 3-foot shadow → $\frac{h}{3}$
$\frac{5}{3\frac{1}{3}}=\frac{h}{3}$
• $\begin{array}{cccc}\hfill \frac{5}{3\frac{1}{3}}& =& \frac{h}{3}\hfill & \end{array}$
$\begin{array}{cccc}\hfill 5\cdot 3& =& 3\frac{1}{3}\cdot h\hfill & \\ \hfill 15& =& \frac{10}{3}\cdot h\hfill & \text{Divide}\phantom{\rule{2px}{0ex}}15\phantom{\rule{2px}{0ex}}\text{by}\phantom{\rule{2px}{0ex}}\frac{10}{3}\hfill \\ \hfill \frac{15}{\frac{10}{3}}& =& h& \\ \hfill \frac{\stackrel{3}{\overline{)15}}}{1}\cdot \frac{3}{\underset{2}{\overline{)10}}}& =& h\hfill & \\ \hfill \frac{9}{2}& =& h\hfill & \\ \hfill h& =& 4\frac{1}{2}\hfill & \end{array}$
• A person who casts a 3-foot shadow at this particular time of the day is $4\frac{1}{2}$ feet tall.

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