ORCCA Percentages

Section A.4 Percentages

Percent-related problems arise in everyday life. This section reviews some basic calculations that can be made with percentages.

Figure A.4.0.1. Alternative Video Lesson

Subsection A.4.1 Converting Percents, Decimals, and English

In many situations when translating from English to math, the word “of” translates as multiplication. Also the word “is” (and many similar words related to “to be”) translates to an equals sign. For example:

\begin{align*} \text{One third }\amp\text{of thirty is ten.}\\ \frac{1}{3}\amp\cdot30=10 \end{align*}

Here is another example, this time involving a percentage. We know that “$2$ is $50\%$ of $4\text{,}$” so we can say:

\begin{align*} 2\amp\text{ is } 50\% \text{ of } 4\\ 2\amp= 0.5\cdot4 \end{align*}

Example A.4.1.2.

Translate each statement involving percents below into an equation. Define any variables used. (Solving these equations is an exercise).

How much is $30\%$ of $\$24.00\text{?}$
$\$7.20$ is what percent of $\$24.00\text{?}$
$\$7.20$ is $30\%$ of how much money?

Explanation

Each question can be translated from English into a math equation by reading it slowly and looking for the right signals.

The word “is” means about the same thing as the equals sign. “How much” is a question phrase, and we can let $x$ be the unknown amount (in dollars). The word “of” translates to multiplication, as discussed earlier. So we have:

\begin{equation*} \pinover{x}{how\ much}\pinover{=}{is}\pinover{0.30}{30\%}\pinover{\cdot}{of}\pinover{24}{\$24} \end{equation*}
Let $P$ be the unknown value. We have:

\begin{equation*} \pinover{7.2}{\$7.20}\pinover{=}{is}\pinover{P}{what\ percent}\pinover{\cdot}{of}\pinover{24}{\$24} \end{equation*}

With this setup, $P$ is going to be a decimal value ($0.30$) that you would translate into a percentage ($30\%$).
Let $x$ be the unknown amount (in dollars). We have:

\begin{equation*} \pinover{7.2}{\$7.20}\pinover{=}{is}\pinover{0.30}{30\%}\pinover{\cdot}{of}\pinover{x}{how\ much} \end{equation*}

Checkpoint A.4.1.3.

Solve each equation from Example A.4.1.2.

Subsection A.4.2 Setting up and Solving Percent Equations

An important skill for solving percent-related problems is to boil down a complicated word problem into a simple form like “$2$ is $50\%$ of $4\text{.}$” Let's look at some further examples.

Example A.4.2.4.

In Fall 2016, Portland Community College had $89{,}900$ enrolled students. According to Figure A.4.2.5, how many black students were enrolled at PCC in Fall 2016?

a pie chart that indicates white students are 68%; Hispanic students are 11%; Asian students are 8%; black students are 6%; and students of other ethnicities make up 7% — Figure A.4.2.5. Racial breakdown of PCC students in Fall 2016

Explanation

After reading this word problem and the chart, we can translate the problem into “what is $6\%$ of $89{,}900\text{?}$” Let $x$ be the number of black students enrolled at PCC in Fall 2016. We can set up and solve the equation:

\begin{align*} \pinover{x}{what}\amp\pinover{=}{is}\pinover{0.06}{6\%}\pinover{\cdot}{of}\pinover{89900}{89,900}\\ x\amp=5394 \end{align*}

There was not much “solving” to do, since the variable we wanted to isolate was already isolated.

As of Fall 2016, Portland Community College had $5394$ black students. Note: this is not likely to be perfectly accurate, because the numbers we started with ($89{,}900$ enrolled students and $6\%$) appear to be rounded.

Example A.4.2.6.

The bar graph in Figure A.4.2.7 displays how many students are in each class at a local high school. According to the bar graph, what percentage of the school's student population is freshman?

a bar graph indicating there are 134 freshmen, 103 sophomores, 96 juniors, and 86 seniors — Figure A.4.2.7. Number of students at a high school by class

Explanation

The school's total number of students is:

\begin{equation*} 134+103+96+86=419 \end{equation*}

With that calculated, we can translate the main question:

“What percentage of the school's student population is freshman?”

into:

“What percent of $419$ is $134\text{?}$”

Using $P$ to represent the unknown quantity, we write and solve the equation:

\begin{align*} \pinover{P}{what\ percent}\pinover{\cdot}{of}\pinover{419}{419}\amp\pinover{=}{is}\pinover{134}{134}\\ \divideunder{P\cdot 419}{419}\amp=\divideunder{134}{419}\\ P\amp\approx0.3198\\ P\amp\approx31.98\% \end{align*}

Approximately $31.98\%$ of the school's student population is freshman.

Remark A.4.2.8.

When solving equations that do not have context we state the solution set. However, when solving an equation or inequality that arises in an application problem (such as the context of the high school in Example A.4.2.6), it makes more sense to summarize our result with a sentence, using the context of the application. This allows us to communicate the full result, including appropriate units.

Example A.4.2.9.

Carlos just received his monthly paycheck. His gross pay (the amount before taxes and related things are deducted) was $\$2{,}346.19\text{,}$ and his total tax and other deductions was $\$350.21\text{.}$ The rest was deposited directly into his checking account. What percent of his gross pay went into his checking account?

Explanation

Train yourself to read the word problem and not try to pick out numbers to substitute into formulas. You may find it helps to read the problem over to yourself three or more times before you attempt to solve it. There are three dollar amounts to discuss in this problem, and many students fall into a trap of using the wrong values in the wrong places. There is the gross pay, the amount that was deducted, and the amount that was deposited. Only two of these have been explicitly written down. We need to use subtraction to find the dollar amount that was deposited:

\begin{equation*} 2346.19-350.21=1995.98 \end{equation*}

Now, we can translate the main question:

“What percent of his gross pay went into his checking account?”

into:

“What percent of $\$2346.19$ is $\$1995.98\text{?}$”

Using $P$ to represent the unknown quantity, we write and solve the equation:

\begin{align*} \pinover{P}{what\ percent}\pinover{\cdot}{of}\pinover{2346.19}{\$2346.19}\amp\pinover{=}{is}\pinover{1995.98}{\$1995.98}\\ \divideunder{P\cdot 2346.19}{2346.19}\amp=\divideunder{1995.98}{2346.19}\\ P\amp\approx0.8507\\ P\amp\approx85.07\% \end{align*}

Approximately $85.07\%$ of his gross pay went into his checking account.

Checkpoint A.4.2.10.

Example A.4.2.11.

According to e-Literate, the average cost of a new college textbook has been increasing. Find the percentage of increase from 2009 to 2013.

a plot over time indicating average textbook price was $62.00 in 2009; $65.11 in 2010; $68.87 in 2011; $72.11 in 2012; and $79.00 in 2013 — Figure A.4.2.12. Average New Textbook Price from 2009 to 2013

Explanation

The actual amount of increase from 2009 to 2013 was $79-62=17\text{,}$ dollars. We need to answer the question “$\$17$ is what percent of $\$62\text{?}$” Note that we are comparing the $17$ to $62\text{,}$ not to $79\text{.}$ In these situations where one amount is the earlier amount, the earlier original amount is the one that represents $100\%\text{.}$ Let $P$ represent the percent of increase. We can set up and solve the equation:

\begin{align*} \pinover{17}{\$17}\amp\pinover{=}{is}\pinover{P}{what\ percent}\pinover{\cdot}{of}\pinover{62}{\$62}\\ 17\amp=62P\\ \divideunder{17}{62}\amp=\divideunder{62P}{62}\\ 0.2742\amp\approx P \end{align*}

From 2009 to 2013, the average cost of a new textbook increased by approximately $27.42\%\text{.}$

Checkpoint A.4.2.13.

Example A.4.2.14.

Enrollment at your neighborhood's elementary school two years ago was $417$ children. After a $15\%$ increase last year and a $15\%$ decrease this year, what's the new enrollment?

Explanation

It is tempting to think that increasing by $15\%$ and then decreasing by $15\%$ would bring the enrollment right back to where it started. But the $15\%$ decrease applies to the enrollment after it had already increased. So that $15\%$ decrease is going to translate to more students lost than were gained.

Using $100\%$ as corresponding to the enrollment from two years ago, the enrollment last year was $100\%+15\%=115\%$ of that. But then using $100\%$ as corresponding to the enrollment from last year, the enrollment this year was $100\%-15\%=85\%$ of that. So we can set up and solve the equation

\begin{align*} \pinover{x}{this\ \text{year's}\ enrollment}\amp\pinover{=}{is}\pinover{0.85}{85\%}\pinover{\cdot}{of}\pinover{1.15}{115\%}\pinover{\cdot}{of}\pinover{417}{enrollment\ two\ years\ ago}\\ x\amp=0.85\cdot1.15\cdot417\\ x\amp=407.6175 \end{align*}

We would round and report that enrollment is now $408$ students. (The percentage rise and fall of $15\%$ were probably rounded in the first place, which is why we did not end up with a whole number.)