Proportions

Elizabeth Pople

16 Proportions

Topics Covered:^[1]

In case you missed something in class, or just want to review a specific topic covered in this Module, here is a list of topics covered:

Use the Definition of Proportion
Using and Applying Proportional Relationships to Solve Problems
Write Percent Equations as Proportions
Direct Variation
Joint Variation
Key Concepts

Use the Definition of Proportion

In the section on Ratios and Rates we saw some ways they are used in our daily lives. When two ratios or rates are equal, the equation relating them is called a proportion.

Proportion

A proportion is an equation of the form $\displaystyle\frac{a}{b}= \frac{c}{d}$ where b ≠ 0, d ≠ 0. The proportion states two ratios or rates are equal. The proportion is read “a is to b, as c is to d.”

The equation $\displaystyle \frac{1}{2} = \frac{4}{8}$ is a proportion because the two fractions are equal. The proportion $\displaystyle \frac{1}{2} = \frac{4}{8}$ is read “1 is to 2 as 4 is to 8”.

If we compare quantities with units, we have to be sure we are comparing them in the right order. For example, in the proportion $\displaystyle \frac{20 \text{ students}}{1 \text{ teacher}} = \frac{60 \text{ students}}{3 \text{ teachers}}$ we compare the number of students to the number of teachers. We put students in the numerators and teachers in the denominators.

Try it!

Write as a proportion:

3 is to 7 as 15 is to 35.
5 hits in 8 at-bats is the same as 30 hits in 48 at-bats.
$1.50 for 6 ounces is equivalent to $2.25 for 9 ounces.

Solution A (click to reveal)

Solution

a.

Steps

Algebraic

Original quantities

3 is to 7 as 15 is to 35.

Write as a proportion.

$\displaystyle \frac{3}{7} = \frac{15}{35}$

Solution B (click to reveal)

b.

Steps

Algebraic

Original quantities

5 hits in 8 at-bats is the same as 30 hits in 48 at-bats.

Write each fraction to compare hits to at-bats.

hits at-bats = hits at-bats

Write as a proportion.

$\displaystyle \frac{5}{8} = \frac{30}{48}$

Solution C (click to reveal)

c.

Steps

Algebraic

Original quantities

$1.50 for 6 ounces is equivalent to $2.25 for 9 ounces.

Write each fraction to compare dollars to ounces.

$\displaystyle \frac{\$}{\text{ounces}} = \frac{\$}{\text{ounces}}$

Write as a proportion.

$\displaystyle \frac{\$1.50}{6} = \frac{\$2.25}{9}$

Using and Applying Proportional Relationships to Solve Problems

Using proportions to solve problems is a very useful method. It is usually used when you know three parts of the proportion, and one part is unknown. Proportions are often solved by setting up like ratios. If $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are two ratios such that $\displaystyle \frac{a}{b} = \frac{c}{d}$ , then the fractions are said to be proportional. Also, two fractions $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are proportional $\displaystyle \left(\frac{a}{b} = \frac{c}{d} \right)$ if and only if $a \times d = b \times c$ .

Look at the proportions $\displaystyle \frac{1}{2} = \frac{4}{8}$ and $\displaystyle \frac{2}{3} = \frac{6}{9}$ . From our work with equivalent fractions, we know these equations are true. But how do we know if an equation is a proportion with equivalent fractions if it contains fractions with larger numbers?

To determine if a proportion is true, we find the cross products of each proportion. To find the cross products, we multiply each denominator with the opposite numerator (diagonally across the equal sign). The results are called a cross product because of the cross formed. If, and only if, the given proportion is true, that is, the two sides are equal, then the cross products of a proportion will be equal.

$Ex. 1 \quad\quad\quad\quad\quad\quad\quad\quad\quad Ex. 2$

$\displaystyle\boldsymbol{\huge{8 \cdot 1 = 8 \;and \; 2 \cdot 4 = 8 \; \quad 9 \cdot 2 = 18\; and \; 3 \cdot 6 = 18} }$

$\displaystyle\boldsymbol{\huge{\frac{1}{\box} \searrow \frac{\box}{8} \;and\; \frac{\box}{2} \nearrow \frac{4}{\box}\qquad\frac{2}{\box} \searrow \frac{\box}{9} \;and\; \frac{\box}{3} \nearrow \frac{6}{\box}}}$

Cross Products of a Proportion

For any proportion of the form $\displaystyle \frac{a}{b} = \frac{c}{d}$ , where $b \neq 0, d \neq 0$ , its cross products are equal.

$\displaystyle\boldsymbol{\huge{a \cdot d = b\cdot c} }$

$\displaystyle\boldsymbol{\huge{\frac{a}{\box} \searrow \frac{\box}{d} \;and\; \frac{\box}{b} \nearrow \frac{c}{\box}}}$

Cross products can be used to test whether a proportion is true. To test whether an equation makes a proportion, we find the cross products. If they are both equal, we have a proportion.

To solve a proportion containing a variable, we remember that the proportion is an equation. All of the techniques we have used so far to solve equations still apply. In the next example, we will solve a proportion by multiplying by the Least Common Denominator (LCD) using the Multiplication Property of Equality.

Try it!

Solve: $\displaystyle \frac{x}{63} = \frac{4}{7}$

Solution (click to reveal)

Steps

Algebraic

Equation

$\displaystyle \frac{x}{63} = \frac{4}{7}$

To isolate x, multiply both sides by the LDC, 63.

$\displaystyle \boldsymbol{ \textcolor{myred1}{63}} \left(\frac{x}{63} \right) = \boldsymbol{ \textcolor{myred1}{63} } \left(\frac{4}{7} \right)$

Simplify.

$\displaystyle x = \frac{9 \cdot \cancel{7} \cdot 4}{\cancel{7}}$

Divide the common factors.

$\displaystyle x = 36$

Check: To check our answer, we substitute into the original proportion.

$\displaystyle \frac{x}{63} = \frac{4}{7}$

substitute $x = \boldsymbol{\textcolor{myred1}{36}}$

$\displaystyle \frac{\boldsymbol{\textcolor{myred1}{36}}}{63} \stackrel{?}{=} \frac{4}{7}$

Show common factors.

$\displaystyle \frac{4 \cdot 9}{7 \cdot 9} \stackrel{?}{=} \frac{4}{7}$

Simplify.

$\displaystyle \frac{4}{7} = \frac{4}{7} \checkmark$

When the variable is in a denominator, we’ll use the fact that the cross products of a proportion are equal to solve the proportions.

We can find the cross products of the proportion and then set them equal. Then we solve the resulting equation using our familiar techniques.

Try it!

Solve: $\displaystyle\frac{114}{a} = \frac{9}{4}$

Solution (click to reveal)

Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.

Steps

Algebraic

Proportion

$\displaystyle\frac{114}{a} = \frac{9}{4}$

Find the cross products and set them equal.

$4 \cdot 144 = a \cdot 9$

Simplify.

$576 = 9a$

Divide both sides by 9.

$\displaystyle \frac{576}{9} = \frac{9a}{9}$

Simplify.

$64 = a$

Check your answer.

$\displaystyle \frac{144}{a} = \frac{9}{4}$

substitute a = $\boldsymbol{ \textcolor{myred1}{64}}$

$\displaystyle \frac{144}{\textcolor{myred1}{64}} \stackrel{?}{=} \frac{9}{4}$

Show common factors.

$\displaystyle \frac{9 \cdot 16}{4 \cdot 16} \stackrel{?}{=} \frac{9}{4}$

Simplify.

$\displaystyle \frac{9}{4} = \frac{9}{4} \checkmark$

Another method to solve this would be to multiply both sides by the LCD, 4a. Try it and verify that you get the same solution.

Write Percent Equations as Proportions

Previously, we solved percent equations by applying the properties of equality we have used to solve equations throughout this text. Some people prefer to solve percent equations by using the proportion method. The proportion method for solving percent problems involves a percent proportion. A percent proportion is an equation where a percent is equal to an equivalent ratio.

For example, $\displaystyle 60\% = \frac{60}{100}$ , and we can simplify $\displaystyle \frac{60}{100} = \frac{3}{5}$ . Since the equation $\displaystyle \frac{60}{100} = \frac{3}{5}$ shows a percent equal to an equivalent ratio, we call it a percent proportion. Using the vocabulary we used earlier:

$\displaystyle \frac{\text{amount}}{\text{base}} = \frac{\text{percent}}{{100}}$

$\displaystyle \frac{3}{5} = \frac{60}{100}$

Percent Proportion

The amount is to the base as the percent is to 100.

$\displaystyle \frac{\text{amount}}{\text{base}} = \frac{\text{percent}}{{100}}$

If we restate the problem in the words of a proportion, it may be easier to set up the proportion: The amount is to the base as the percent is to one hundred.

We could also say: The amount out of the base is the same as the percent out of one hundred.

We will practice translating into a percent proportion.

Try it!

Translate to a proportion. What number is 75% of 90?

Solution (click to reveal)

If you look for the word “of,” it may help you identify the base.

Steps

Algebraic

Identify the parts of the percent proportion.

$\boldsymbol{ \underbrace{What\; number}_{{\color{mypurple1}{amount}}} \quad is \quad \underbrace {75\%}_{{\color{mypurple1}{percent}}}\quad {\color{myred1}{of}} \quad \underbrace{90}_{{\color{mypurple1}{90}}}?}$

Restate as a proportion.

$\text{What number out \textcolor{myred1}{of} 90 is the same as 75 out of 100?}$

Set up the proportion. Let n = number.

$\displaystyle \frac{n}{90} = \frac{77}{100}$

Direct Variation

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities.

Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation

s = 15h

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

Direct Variation

For any two variables x (input) and y (output), y varies directly with x if

y = k·x, where k ≠ 0

The constant k is called the constant of variation.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y. Then we can use that equation to find values of y for other values of x.

Try it – How to Solve Direct Variation Problems

If y varies directly with x and y = 20 when x = 8, find the equation that relates x and y.

Solution (click to reveal)

Steps

Algebraic

Step 1. Write the formula for direct variation. The direct variation formula is y = kx.

$y = kx$

Step 2. Substitute the given values for the variables. $\boldsymbol{\text{We are given } y = \textcolor{myblue1}{20},\ x = \textcolor{myred1}{8}.}$

$\boldsymbol{\textcolor{myblue1}{20} = k \cdot \textcolor{myred1}{8}}$

Step 3. Solve for the constant of variation. Divide both sides of the equation by 8, then simplify.

$\displaystyle\frac{20}{8} = k$
$k = 2.5$

Step 4. Write the equation that relates x and y. Rewrite the general equation with the value we found for k.

$y = 2.5x$

We’ll list the steps below.

How to Solve direct variation problems.

Step 1. Write the formula for direct variation.
Step 2. Substitute the given values for the variables.
Step 3. Solve for the constant of variation.
Step 4. Write the equation that relates x and y.

Try it! – Applications of Variation

When Raoul runs on the treadmill at the gym, the number of calories, c, he burns varies directly with the number of minutes, m, he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

Write the equation that relates c and m.
How many calories would he burn if he ran on the treadmill for 25 minutes?

Solution A (click to reveal)

a. The number of calories, c, varies directly with the number of minutes, m, on the treadmill, and c = 315 when m = 18.

Steps

Algebraic

Write as a proportion.

$\displaystyle \frac{3}{7} = \frac{15}{35}$

Write the formula for direct variation.

y = kx

We will use c in place of y and m in place of x.

c = km

Substitute the given values for the variables.

$\boldsymbol{\displaystyle {\color{mypurple1} 315}=k\cdot {\color{mypurple1} 18}}$

Solve for the constant of variation.

$\displaystyle \boldsymbol{\frac{{\color{mypurple1}{315}}}{{\color{mypurple1}{18}}} = \frac{k\cdot {\color{mypurple1} {18}}}{{\color{mypurple1}{18}}}}$

17.5 = k

Write the equation that relates c and m.

c = km

Substitute in the constant of variation.

c = 17.5m

Solution B (click to reveal)

b. Find c when m = 25

Steps

Algebraic

Write the equation that relates c and m.

c = 17.5m

Substitute the given value for m.

c = 17.5 $\left({\boldsymbol{\color{mypurple1} 25}\right)$

Simplify

c = 437.5

Solution

Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

The number of calories, c, burned varies directly with the amount of time, t, spent exercising. Arnold burned 312 calories in 65 minutes exercising.

Write the equation that relates c and t.
How many calories would he burn if he exercises for 90 minutes?

Solution (click to reveal)

a. c = 4.8t

b. 432 calories

In the previous example, the variables c and m were named in the problem. Usually, that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.

Try it!

The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.

Write the equation that relates the number of gallons of gas used to the number of miles driven.
How many gallons of gas would Eunice’s car use if she drove 1000 miles?.

Solution A (click to reveal)

a. The number of gallons of gas varies directly with the number of miles driven.

Steps

Algebraic

First, we will name the variables.

Let g = number of gallons of gas.

m = number of miles driven

Write the formula for direct variation.

y = kx

We will use g in place of y and m in place of x.

g = km

Substitute the given values for the variables.

$\displaystyle\boldsymbol{ {\color{mypurple1} {14.5}}=k\cdot {\color{mypurple1} {469.8}}}$

Solve for the constant of variation.

We will round to the nearest thousandth.

$\displaystyle \boldsymbol{\frac{{\color{mypurple1} {14.5}}}{{\color{mypurple1} {469.8}}} = \frac{k\cdot{\color{mypurple1} {469.8}}}{{\color{mypurple1} {469.8}}}}$

0.031 = k

Write the equation that relates c and m.

g = km

Substitute the constant of variation.

g = 0.031m

Solution B (click to reveal)

b.

Steps

Algebraic

Write the equation that relates g and m.

g = 0.031m

Substitute the given value for m.

g = 0.031 $\left({\boldsymbol{\color{mypurple1} 1000}\right)$

Simplify

g = 31

Solution

Eunice’s car would use 31 gallons of gas if she drove it 1000 miles.

Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.

The distance that Brad travels varies directly with the time spent traveling. Brad traveled 660 miles in 12 hours,

Write the equation that relates the number of miles traveled to the time.
How many miles could Brad travel in 4 hours?

Solution (click to reveal)

a. m = 55t

b. 220 miles

In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is y = kx². We solve these applications just as we did the previous ones, by substituting the given values into the equation to solve for k.

Try it!

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.

Write the equation that relates the maximum load to the cross-section.
What is the maximum load that can be supported by a beam with a diagonal of 8”?

Solution A (click to reveal)

a. The maximum load varies directly with the square of the diagonal of the cross-section.

Steps

Algebraic

Name the variables.

Let L = maximum load.
c = the diagonal of the cross-section

Write the formula for direct variation, where y varies directly with the square of x.

y = kx²

We will use L in place of y and c in place of x.

L = kc²

Substitute the given values for the variables.

$\boldsymbol{\displaystyle {\color{mypurple1} 75}=k\cdot {\color{mypurple1} 4^2}}$

Solve for the constant of variation.

We will round to the nearest thousandth.

$\displaystyle\boldsymbol{ \frac{75}{16} = \frac{k\cdot 16}{16}}$

4.6875 = k

Write the equation that relates L and c.

L = kc²

Substitute in the constant of variation.

L = 4.6785c²

Solution B (click to reveal)

b.

Steps

Algebraic

Write the equation that relates L and c.

L = 4.6785c²

Substitute the given value for c.

L = 4.6785 $\left({\boldsymbol{\color{mypurple1} 8}\right)^2$

Simplify

L = 300

Solution

A beam with a diagonal of 8″ could support a maximum load of 300 pounds.

Joint Variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d.

Joint Variation

Joint variation occurs when a variable varies with multiple variables.

For instance, if x varies directly with both y and z, we have x = kyz. Notice that we only use one constant in a joint variation equation.

Try it!

Use the information to find the unknown variable: y varies jointly as x, z, and w. When x = 2, z = 1, and w = 12, then y = 72. Find y when x = 1, z = 2, and w = 3.

Solution (click to reveal)

Steps

Algebraic

Write an equation to show the relationship between the variables: y varies jointly with x, z, and w.

y = kxzw

Substitute the given values for the variables to find the constant k.

$\boldsymbol{\displaystyle {\color{mypurple1} 72}=k\cdot {\color{mypurple1} 2\cdot1\cdot12}}$

Solve for the constant of variation.

$\displaystyle\boldsymbol{ \frac{72}{24} = \frac{k\cdot 24}{24}}$

3 = k

Substitute in the constant of variation.

y = 3 ·x ·z ·w

To find y, substitute the values into the equation.

$\displaystyle y = 3 \cdot \boldsymbol{{\color{mypurple1} 1\cdot2\cdot3}}$

Solution

y = 18

The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute (rpm) and the cube of its diameter. If the shaft of a certain material 3 inches in diameter can transmit 45 hp at 100 rpm, what must the diameter be to transmit 60 hp at 150 rpm?

Solution (click to reveal)

2.88 inches

Key Concepts

A proportion is an equation of the form $\displaystyle \frac{a}{b} = \frac{c}{d}$ , where $b\neq 0,d \neq 0$ .
Percent Proportion
- $\displaystyle \frac{\text{amount}}{\text{base}} = \frac{\text{percent}}{{100}}$
Direct variation:
- For any two variables x (input) and y (output), y varies directly with x if y = k·x, where k ≠ 0

Derived from Openstax Pre-algebra, Access for free at https://openstax.org/books/prealgebra-2e/pages/1-introduction. Contemporary Math, Access for free at https://openstax.org/books/contemporary-mathematics/pages/1-introduction. Elementary Algebra, Access for free at https://openstax.org/books/elementary-algebra-2e/pages/1-introduction. College Algebra 2e with Corequisite Supports, Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites ↵

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Use the Definition of Proportion

Using and Applying Proportional Relationships to Solve Problems

Write Percent Equations as Proportions

Direct Variation

Joint Variation

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