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Proportions in Percent Equations

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Use Proportions to Solve Percent Problems
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Have you ever conducted a survey? Take a look at this dilemma.

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 Concept to figure out how to solve this problem.

Guidance

Proportions can be used to compare any equal quantity. Remember that proportions are created when two ratios are equal. Because of this, proportions can be used to find any missing piece. If we know a percent, we can also find a part of the whole.

Take a look at this situation.

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.

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

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.

Take a look at this situation.

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

There are about 2879 students.

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.

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

We use cross products to solve for b .

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

Solve each proportion for the missing part.

Example A

\frac{9}{b}=\frac{20}{100}

Solution:  b = 45

Example B

\frac{a}{10}=\frac{40}{100}

Solution:  a = 4

Example C

\frac{a}{3}=\frac{18}{100}

Solution:  a = .54

Now let's go back to the dilemma from the beginning of the Concept.

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

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

Guided Practice

Here is one for you to try on your own.

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.

Video Review

Using Proportions to Solve Percent Problems

Practice

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. \frac{3}{b}=\frac{75}{100}
  6. A recent government survey shows that in New City, people spend 35% of their monthly 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?

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