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# 5.2: Percents and Proportions

Difficulty Level: At Grade Created by: CK-12

## Introduction

Football Game Attendance

The student council decided to conduct a survey to see how many middle school students attend football games on Friday nights. They asked each student when they entered the stadium what grade they were in and then recorded the results. The student council members did this for three weeks and figured that this was enough of a sample for a good estimate on how many middle school students attend the games. This is the report that they announced during morning announcements after their survey was completed.

“40% of our students attend football games on Friday nights! Let’s try to get this number up to 50% next seasons so that we can really support the high school players!”

“Wow, forty percent is still pretty high,” Cameron commented to Carla during homeroom.

“Yes, but 50 percent would be even better since there are 380 students at our school.”

“How many students attend the game then?” Cameron asked.

That is a good question. If you understand percents and proportions, you can use the information provided to figure out how many students attend the games if 40% attend. Use the information in this lesson to figure out how to solve this problem. Then you will see it again at the end of the lesson.

What You Will Learn

In this lesson, you will learn how to demonstrate the following skills.

• Use the proportion $\frac{a}{b}$ to find the percent $p$.
• Use the proportion to find the part $a$.
• Use the cross product property of proportions to find the base $b$.
• Solve real-world problems involving percents using proportions.

Teaching Time

I. Use the Proportion $\underline{\frac{a}{b}}$ to find the percent $\underline{p}$

In the last section, you began using proportions to figure out a percent when writing fractions as percents. Remember that proportions involve comparing quantities. A proportion is a comparison between two equal ratios. Percents are also written to compare a quantity to 100. Because both of these are comparing, we can use proportions to help us figure out a percent.

That is a great question. First, we write the proportion using $a$ over $b$.

$\frac{a}{b}$

This is equal to the percent which is out of 100.

$\frac{p}{100}$

Here is the proportion:

$\frac{a}{b} = \frac{p}{100}$

Example

15 out of 30 is what percent?

To work on this problem, first, we write a ratio comparing our given values to the missing percent.

$\frac{15}{30} = \frac{p}{100}$

We know that fifteen is half of thirty, and 50 is half of 100.

Let’s look at using this proportion to solve a problem.

Example

A senator wants to start a program to encourage more people to vote in his state. In County A, 32,100 people voted. In neighboring County B, 57,800 people voted. Which county needs the program more?

Well, it depends on how many people are in each of the counties. We can’t compare the polling rates unless you use a percent. If we know that the first county has a population of 39,150 people and the second county has a population of 81,400 people, we can now find what percent of the people voted. We are comparing the number of people who voted with the population of each county. We are actually going to find two percents here. For each county, we will use the proportion $\frac{a}{b} = \frac{p}{100}$ where $a$ is the number of people that voted and $b$ is the total population.

$& \text{County A} \qquad \qquad \qquad \text{County B}\\& \ \frac{32100}{39150} = \frac{p}{100} \qquad \qquad \ \frac{57800}{81400} = \frac{p}{100}\\& 39150p = 32100 \cdot 100 \quad 81400p = 57800 \cdot 100\\& 39150p = 3210000 \qquad \ 81400p = 5780000\\& \qquad \ \ p = 82 \% \qquad \qquad \qquad \ p=71 \%$

In order to find the percent in each case, we used cross products as we would for any proportion. Now we can see that in County A, 82% of the people voted, while in County B only 71% of the people voted.

The senator should push the program more in County B.

II. Use the Proportion to Find the Part $\underline{a}$

Proportions can be used to compare any equal quantity. Because of this, proportions can be used to find any missing piece. In the last section we found a percent using a proportion. If we know a percent, we can also find a part of the whole.

Let’s look at an example.

Example

At Big Town Middle School, 37% of the students participate in athletic programs. If there are 968 students at the school, how many uniforms do they need to purchase?

Although 37% is a useful number, it does not tell us how many students need uniforms. We know that for every 100 students, 37 will need a uniform. But that is still not enough information to make a purchase. If there are 968 students at the school, use the proportion $\frac{a}{b} = \frac{p}{100}$ again to find the missing value. If we are given the percent, $p$, then the missing value is not the percent but the part of the student body that need uniforms.

$\frac{a}{968} = \frac{37}{100}$ Use cross products now to solve for $a$.

$\frac{a}{968} &= \frac{37}{100}\\100a &= 968 \cdot 37\\100a &= 35816\\a &= 358$

358 uniforms need to be purchased.

Example

A forest range discovered that 25% of the trees in his area were infected with a parasite. If there are 3060 trees in his area, how many trees are infected?

$\frac{a}{3060} &= \frac{25}{100}\\100a &= 3060 \cdot 25\\100a &= 76500\\a &= 765$

765 trees were infected with the parasite.

Before continuing, be sure that you have the proportion $\frac{a}{b}=\frac{p}{100}$ written down in your notebooks.

III. Use the Cross Product Property of Proportions to Find the Base $\underline{b}$

Since we can use a proportion to solve for any missing variable, we can also find the whole if we know a percent and its corresponding part.

Example

At Big Town High School, 58% participate in athletic programs. If 1670 students are involved in an athletics program, then how many students are there in the school?

This time we know the part $a$ and we know the percent $p$. We do not know the whole, $b$. We need to adjust the proportion.

$\frac{1670}{b} = \frac{58}{100}$ Use cross products to solve for $b$.

$58b &= 1670 \cdot 100\\58b &= 167000\\b &= 2879$

Sometimes, you won’t have a word problem to solve you will just simply have a proportion that needs to be solved for a missing value. Look at this example.

Example

$\frac{90}{b} = \frac{30}{100}$

We use cross products to solve for $b$.

$30b &= 90(100)\\30b &= 9000\\b &= 300$

We can also use proportions and percents to solve other real – world problems.

IV. Solve Real – World Problems Involving Percents Using Proportions

As you can see, percents apply to countless real-life situations. When we solve them, we can create a proportion that will help us to find the answer.

However, we must be careful to place the appropriate values in the $\frac{a}{b} = \frac{p}{100}$ proportion. The $p$ value will always be the percentage which is always placed over 100. The $a$ value will be the part that the percent represents and the $b$ value will always be the whole.

Example

Jared needs to get 92% on his test in order to maintain his A in history. If there are 60 points on the test, how many does he need to earn to keep his A?

We can use the proportion $\frac{a}{b}=\frac{p}{100}$ to solve this problem. The variable $p$ will be the percent, $b$ is the whole test, and $a$ is our unknown.

$\frac{a}{b} & = \frac{p}{100} \longrightarrow \frac{a}{60}=\frac{92}{100}\\\frac{a}{60} &= \frac{92}{100}\\100a &= 60 \cdot 92\\100a &= 5520\\a &= 55.2$

The proportion shows that he needs 55.2 points. In the context of a test, he could most likely receive only whole number points. If he gets 55 points, it would be too low—it wouldn’t be 92%. We must round up, then, to 56 points.

Jared needs 56 points to maintain his grade.

Now let’s go back and apply what we have learned to the problem from the introduction.

## Real-Life Example Completed

Football Game Attendance

Here is the original problem once again. First, reread the problem. Then write a proportion to show how you would solve this problem. Finally, figure out how many students attend the football game given the 40% statistic. There are two parts to your answer.

The student council decided to conduct a survey to see how many middle school students attend football games on Friday nights. They asked each student when they entered the stadium what grade they were in and then recorded the results. The student council members did this for three weeks and figured that this was enough of a sample for a good estimate on how many middle school students attend the games. This is the report that they announced during morning announcements after their survey was completed.

“40% of our students attend football games on Friday nights! Let’s try to get this number up to 50% next seasons so that we can really support the high school players!”

“Wow, forty percent is still pretty high,” Cameron commented to Carla during homeroom.

“Yes, but 50 percent would be even better since there are 380 students at our school.”

“How many students attend the game then?” Cameron asked.

Solution to Real – Life Example

Now let’s write a proportion using the given information.

40% of students attend. We can change this to a ratio out of 100.

$\frac{40}{100}$

There are 380 students in the middle school. That is our whole. We need to figure out what part of whole is 40%. Here is the second ratio.

$\frac{x}{380}$

Now we can write this as a proportion.

$\frac{40}{100} = \frac{x}{380}$

Next, we solve the proportion for the number of students who attend the football games.

$100x=15,200$

152 students attend Friday night football games.

## Vocabulary

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

Proportion
two equal ratios form a proportion.
Percent
a part of a whole out of 100.

## Time to Practice

Directions: Find $p$ in the given problems using cross products. Round to the nearest tenths place.

1. $\frac{7}{15}=\frac{p}{100}$
2. $\frac{52}{3810}=\frac{p}{100}$
3. $\frac{16}{17}=\frac{p}{100}$
4. $\frac{3}{4}=\frac{p}{100}$
5. $\frac{3}{5}=\frac{p}{100}$
6. $\frac{1}{5}=\frac{p}{100}$
7. A dentist filled cavities in 8 of his 30 patients on Tuesday. What percent of his patients had cavities filled?

Directions: Solve the following proportions for the $a$ value.

1. $\frac{a}{23}=\frac{85}{100}$
2. $\frac{a}{1500}=\frac{7}{100}$
3. $\frac{a}{5}=\frac{61}{100}$
4. $\frac{a}{4}=\frac{75}{100}$
5. $\frac{a}{5}=\frac{40}{100}$
6. A small car company sold 65,000 cars last year. Ninety-five percent of those cars had airbags. How many cars had airbags?

Directions: Solve the following proportions for the $b$ value. Round to the nearest tenths place.

1. $\frac{88}{b}=\frac{22}{100}$
2. $\frac{600}{b}=\frac{74}{100}$
3. $\frac{1}{b}=\frac{85}{100}$
4. $\frac{2}{b}=\frac{66}{100}$
5. A recent government survey shows that in New City, people spend 35% of their income on rent. If average rent is \$780, what is the average income?

Directions: Use the proportion $\frac{a}{b}=\frac{p}{100}$ to solve the following problems.

1. A fluorescent light bulb uses 35% as much energy as an incandescent bulb. If an incandescent uses 75 watts, how much does the fluorescent use?
2. An average human weighs 1.3% as much as an average elephant. If an average human weighs 160lbs, how much does an average elephant weigh?
3. An average elephant eats about 350lbs of food per day. Using your calculation from problems #14, what percent of its own weight does it consume each day?

Jan 14, 2013

Jan 13, 2015

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