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# 5.5: Solving Proportions Using Cross Products

Difficulty Level: At Grade Created by: CK-12

## Introduction

Remember Manuel who was reading all of the medieval books on knights? Well after he finished reading the series, he loaned it to his friend Rafael. Rafael is enjoying the series as much as Manuel did.

In five weeks, Rafael had already finished 8 of the 12 books. It took Manuel 7.5 weeks to read all 12 books. Will Rafael and Manuel finish the series in the same amount of time? Are they reading at the same rate? How can you figure this out?

To figure this out, you will need to know how to determine if two ratios form a proportion. If the reading rate of the boys is the same, then the ratios will form a proportion.

In this lesson you will learn how to solve proportions using cross products. Then you will be able to determine if they are reading at the same rate.

What You Will Learn

In this lesson, you will learn how to do the following:

• Recognize the cross products property of proportions.
• Solve proportions with a variable in the denominator using cross products.
• Recognize equivalent proportions representing the same situation with the variable in the denominator.
• Model and solve real-world problems using the cross products property of proportions.

Teaching Time

I. Recognize the Cross Products Property of Proportions

You already know that a proportion states that two ratios are equivalent. For example, here are two examples of proportions.

$\frac{a} {b} = \frac{c} {d} \qquad \text{or} \qquad a : b = c : d$

In a proportion, the means are the two terms that are closest together when the proportion is written with colons. So, in $a : b = c : d$, the means are $b$ and $c$.

The extremes are the terms in the proportion that are furthest apart when the proportion is written with colons. So, in $a : b = c : d$, the extremes are $a$ and $d$.

The diagram below shows how to identify the means and the extremes in a proportion.

In the last lesson, you learned how to solve proportions by using proportional reasoning. We can also solve a proportion for a variable in another way. This is where the cross products property of proportions comes in.

What is the Cross Products Property of Proportions?

The Cross Products Property of Proportions states that the product of the means is equal to the product of the extremes in a proportion. You can find these cross products by cross multiplying, as shown below.

$\frac{a}{b} &= \frac{c}{d}\\b \cdot c &= a \cdot d$

Now let’s look at an example where we use the Cross Products Property of Proportions.

Example

$\frac{a}{4} = \frac{6}{8}$

To solve this, we can multiply the means and the extremes.

$a \cdot 8 &= 4 \cdot 6 \\8a & = 24$

Next, we solve the equation for the missing variable. To do this, we use the inverse operation. Multiplication is in the problem, so we use division to solve it. We divide both sides by 8.

$\frac{8a}{8} &= \frac{24}{8}\\ a &= 3$

5J. Lesson Exercises

Solve for each variable in the numerator by using cross products.

1. $\frac{x}{5} = \frac{6}{10}$
2. $\frac{a}{9} = \frac{15}{27}$
3. $\frac{b}{4} = \frac{12}{16}$

Take a few minutes to check your work with a friend.

II. Solve Proportions with a Variable in the Denominator using Cross Products

In our last section, you learned to use cross products to solve when the variable is in the numerator. What about when the variable is in the denominator?

You can use cross products to solve these proportions as well. Let’s look at an example.

Example

Use the cross products property of proportions to solve for r: $\ \frac{2}{8} = \frac{7}{r}$.

The cross product property of proportions states that the product of the means is equal to the product of the extremes.

The means in this proportion are 8 and 7. The extremes are 2 and $r$. Find the cross products and solve for $r$.

$\frac{2}{8} &= \frac{7}{r}\\8 \cdot 7 &= 2 \cdot r\\56 &= 2r$

In $2r$, the 2 and the $r$ are being multiplied. The opposite of multiplication is division. So, to find the value of $r$, divide both sides of the equation by 2.

$56 &= 2r\\\frac{56}{2} &= \frac{2r}{2}\\28 &= r$ The value of $r$ is 28.

Example

Use the cross products property of proportions to solve for s:$\ \frac{4}{s} = \frac{10}{15}$.

The means in this proportion are $s$ and 10. The extremes are 4 and 15. Find the cross products and solve for $s$.

$\frac{4}{s} &= \frac{10}{15}\\s \cdot 10 &= 4 \cdot 15\\10s &= 60\\\frac{10s}{10} &= \frac{60}{10}\\s &= 6$

The value of $s$ is 6.

5K. Lesson Exercises

1. $\frac{4}{7} = \frac{16}{x}$
2. $\frac{5}{9} = \frac{35}{y}$
3. $\frac{9}{a} = \frac{18}{33}$

Write down some notes on how to solve a proportion in your notebook. When finished, continue with the next section.

III. Recognize Equivalent Proportions Representing the Same Situation with the Variable in the Denominator

We can use proportions to help us solve for unknowns in real life. We can use our knowledge about proportions in these situations when working with equivalent values. We can set up a proportion to help us understand that two situations are equivalent. Let’s look at an example where it makes sense to use equivalent proportions.

Example

Each pencil sold at the school store costs the same price. Caryn paid $1.12 for 14 pencils at the school store. Arnob also bought pencils, but he paid$0.48 for the pencils he buys. Write a proportion to represent $n$, the number of pencils that Arnob bought.

When setting up a proportion, we must be sure to use consistent terms. For example, one way to set up a proportion would be to write two equivalent ratios, each comparing the total price to the number of pencils.

We know that Caryn paid $1.12 for 14 pencils. So, one ratio could be this one. $\frac{price}{pencils} = \frac{1.12}{14}$ We know that Arnob paid$0.48. The number of pencils he bought is unknown, so we can use $n$ to represent that number.

$\frac{price}{pencils} = \frac{0.48}{n}$

Since these two ratios are equivalent, we can put them together to form a proportion.

$\frac{1.12}{14} = \frac{0.48}{n}$

The proportion above could be used to find $n$, the number of pencils Arnob bought.

If Jake joined the two in shopping for pencils and purchased some of the same pencils, he would also have a ratio that would be equivalent to these two ratios. The situation is the same. Everyone is purchasing pencils. The pencils cost the same amount. Therefore, we have equivalent proportions as we compare the number of pencils purchased with pencil price.

Now we looked at writing a proportion to represent an equivalent situation. Next, let’s look at how to solve a proportion that describes a real-life situation.

IV. Model and Solve Real-World Problems Using the Cross Products Property of Proportions

Now we can apply cross products to solving proportions written about real-life situations.

Example

The ratio of boys to girls in the school chorus is 4 to 5. There are a total of 20 boys in the chorus. How many total students are in the chorus?

Be careful. The ratio given, 4 to 5, compares boys to girls. However, the question asks for the total number of students in the chorus.

One way to set up a proportion for this problem would be to write two equivalent ratios, each comparing boys to total students.

The ratio of boys to girls is 4 to 5. We can use this ratio to find the ratio of boys to total students.

$\frac{boys}{total} = \frac{boys}{boys + girls} = \frac{4}{4 + 5} = \frac{4}{9}$

You know that there are 20 boys in the chorus. The total number of students is unknown, so represent that as $x$.

$\frac{boys}{total} = \frac{20}{x}$

Get those ratios equal to form a proportion. Then cross multiply to solve for $x$.

$\frac{4}{9} &= \frac{20}{x}\\9 \cdot 20 &= 4 \cdot x\\180 &= 4x\\\frac{180}{4} &= \frac{4x}{4}\\45 &= x$

So, there are a total of 45 students in the school chorus.

Sometimes, you will use proportional reasoning to solve a proportion and sometimes you will use cross products. When in doubt, cross products will never lead you astray. Now let’s use what we have learned to solve our real-world example.

## Real-Life Example Completed

Here is the original problem once again. Reread it and underline any important information.

Remember Manuel who was reading all of the medieval books on knights? Well after he finished reading the series, he loaned it to his friend Rafael. Rafael is enjoying the series as much as Manuel did.

In five weeks, Rafael had already finished 8 of the 12 books. It took Manuel 7.5 weeks to read all 12 books. Will Rafael and Manuel finish the series in the same amount of time? Are they reading at the same rate? How can you figure this out?

Let’s write a proportion to solve this problem.

$\frac{8 \ books}{5 \ weeks} = \frac{12 \ books}{7.5 \ weeks}$

Next, we can use cross products to see if the two ratios form a proportion. If they do, then the two boys will finish the series in the same amount of time.

$8 \times 7.5 &= 60\\5 \times 12 & = 60$

The two cross products are equal so the two ratios form a proportion. The two boys will finish the series of books in the same amount of time.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Proportion
two equal ratios form a proportion.
Means
the values of a proportion that are close to each other when written in ratio form using a colon.
Extremes
the values of a proportion that are farthest apart from each other when written in ratio form using a colon.
Cross Product Property of Proportions
states that the product of the means is equal to the product of the extremes.

## Technology Integration

Other Videos:

1. http://www.mathplayground.com/mv_proportions.html – This is a Brightstorm video on the basics of proportions.

## Time to Practice

Directions: Use cross products to find the value of the variable in each proportion.

1. $\frac{6}{10} = \frac{x} {5}$

2. $\frac{2}{3} = \frac{x} {9}$

3. $\frac{4}{9} = \frac{a} {45}$

4. $\frac{7}{8} = \frac{a} {4}$

5. $\frac{b}{8} = \frac{5} {16}$

6. $\frac{6}{3} = \frac{x} {9}$

7. $\frac{4}{x} = \frac{8} {10}$

8. $\frac{1.5}{y} = \frac{3} {9}$

9. $\frac{4}{11} = \frac{c} {33}$

10. $\frac{2}{6} = \frac{5} {y}$

11. $\frac{2}{10} = \frac{5} {x}$

12. $\frac{4}{12} = \frac{6} {n}$

13. $\frac{5}{r} = \frac{70} {126}$

14. $\frac{4}{14} = \frac{14} {k}$

15. $\frac{8}{w} = \frac{6} {3}$

16. $\frac{2}{5} = \frac{17} {a}$

17. If Ethan bakes 2 batches of cookies, he will have a total of 28 cookies. He wants to bake 5 batches of cookies for a bake sale. He writes two proportions that he thinks could be used to find $c$, the total number of cookies he will bake.

a. $\frac{2}{28} = \frac{5}{c}$ and b. $\frac{2}{5} = \frac{28}{c}$

Are the two ratios above equivalent? If not, which proportion––a or b––could be used to find the actual number of cookies he will bake?

18. Each notebook sold at the school store costs the same price. Marco pays $2.40 for 2 notebooks at the school store. Keecia also buys notebooks, but she pays$8.40 for the notebooks she buys. Marco wants to determine how many notebooks Keecia bought. He writes two proportions that he thinks could be used to find $n$, the number of notebooks Keecia bought.

a. $\frac{2.40}{2} = \frac{8.40}{n}$ and b. $\frac{8.40}{2.40} = \frac{2}{n}$

Are the two ratios above equivalent? If not, which proportion––a or b––could be used to find the actual number of notebooks Keecia bought?

19. Sheri can make 2 bracelets in 16 minutes. She needs to make 7 bracelets.

a. Set up a proportion that could be used to find $m$, the number of minutes it would take Sheri to make 7 bracelets.

b. How many minutes will it take Sheri to make 7 bracelets?

20. A bag contains only two colors of marbles––silver and gold. The ratio of silver marbles to gold marbles in the bag is 5 : 3. There are a total of 30 silver marbles in the bag.

a. Set up a proportion that could be used to find $t$, the total number of marbles in the bag

Feb 22, 2012

Jan 06, 2016