Percentages and Percent Changes | Using Percentages with Ratios |

Ratios and Proportions | More Practice |

Unit Multipliers (Dimensional Analysis) |

**Note**: For more problems with percents and ratios, see the **Algebra Word Problems** section.

## Percentages and Percent Changes

Percentages are something you are probably quite familiar with because of shopping; for example, when a store has everything $ 20\%$ off? The easiest example of percentages is $ 50\%$ off, which means that the item is half price.

Percentages really aren’t that difficult if you truly understand what they are. The word “percentage” comes from the word “per cent”, which means “per hundred” in Latin. Remember that “per” usually means “over”. So, “per cent” literally means “over $ 100$” or “divided by $ 100$”. And remember what “of” typically means in math? I’ll write it again, since it’s so important:

**OF = TIMES**

When we say “$ 20\%$ **off** of something”, translate it to “$ 20$ over (or divided by) $ 100$ — then times the original price”, and that will be the amount to **subtract** from the original price.

** Remember that we cannot use a percentage in math; we need to turn it into a decimal. **To turn a percentage into a decimal, move the decimal $ 2$ places to the

**left**(to divide by $ 100$), and to turn a decimal back into a percentage, move the decimal $ 2$ places to the

**right**(to multiply by $ 100$). (

**I like to think of it this way**: when taking away the $ \%$, we are

**afraid**of it, so we move $ 2$ decimal places

**away**from it, or to the

**left**. When we need to turn a number into a $ \%$, we

**like**it, so we move $ 2$ decimals

**towards**it, or to the

**right**.)

Let’s get back to our percentage example. If there’s an item we like for say $ \$50$**,** and it’s $ 20\%$ off (“off” means take-away or minus!), we’ll do the math to figure out the sales price. This is called a **percent change**** **problem.

**Amount of sale**: $ \displaystyle 20\%\,\,\text{of }\,\$50=.2\times \$50=\$10.\,\,\,\,\$50-\$10=\$40$. The item would be $ \$40$. See how we had to turn the $ 20\%$ into a decimal by taking away the $ \%$ sign and moving $ 2$ decimals to the left, or away from it?

We could have also multiplied the original price by $ \displaystyle 80\%\,(100\%-20\%)$, or $ \displaystyle \frac{{80}}{{100}}$, since that’s what we’ll be paying if we get $ 20\%$ off ($ 100\%$ full price minus $ 20\%$ discount equals $ 80\%$ discounted price):

**Price of discounted item**: $ \displaystyle 80\%\,\,\text{of }\,\$50=.8\times \$50=\$40$. This method has fewer steps.

This shopping example is a **percent decrease **problem; the following is the formula for that. Make sure you relate this formula back to the example above.

$ \displaystyle \begin{array}{l}\text{New }\,\text{lower}\,\,\text{price =}\,\,\text{original}\,\,\text{price}\,-\\\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{original}\,\,\text{price}\,\,\times \,\,\left. {\displaystyle \frac{{\text{percentage}\,\,\text{off}}}{{100}}} \right)} \right.\end{array}$

$ \displaystyle \begin{array}{l}\$50-\left( {\$50\,\,\times \,\,\left. {\displaystyle \frac{{20}}{{100}}} \right)} \right.\,\\\,\,\,\,\,\,\,\,\,=\,\,\$50-\$10=\$40\end{array}$

**Notice that we worked the math in the parentheses first **(we will get to this in more detail later)**.**

Now let’s talk about a **percent increase **problem, which is also a percent change problem. A great example of a percent increase is the tax you pay on this item. Tax is typically a percentage that you add on to what you pay so we can continue driving on the streets free and going to public school “free”.

If we need to add on $ 8.25\%$ sales tax to the $ \$40$ that we are going to spend on the item, we’ll have to know the percent increase formula, but let’s first figure it out without the formula. Tax is the amount we have to add that is based on a percentage of the price that we’re paying for the item.

The tax is $ 8.25\%$ or $ .0825$ (remember – we don’t like the %, so we take it away and move away from it?) **times** the price of the item and then **add it back** to the price.

**Total price with tax:**

$ \displaystyle \begin{array}{l}\$50+(8.25\%\times 50)=\$50+(.0825\times 50)\\=\$50+\$4.125\\=\$54.125=\$54.13\end{array}$

Note that we rounded up to two decimal places, since we’re dealing with money. Note also that we did the math inside the parentheses first. The total price of the item is $ \$54.13$.

Here’s the formula:

Another way we can figure percent increase is to multiply the original amount by $ 1$ plus the tax rate, so we include the original amount plus the tax:

If we need to figure out the actual **percent decrease or increase** (**percent change**), we can use the following formulas:

$ \displaystyle \text{Percent Increase}=\frac{{\text{New Price}-\text{Old Price}}}{{\text{Old Price}}}\,\times 100$

$ \displaystyle \text{Percent Decrease}\,=\frac{{\text{Old Price}-\text{New Price}}}{{\text{Old Price}}}\,\,\times \,100$

For example, say we want to work backwards to get the percentage of sales tax that we pay (percent increase). If we know that the original (old) price is $ \$50$, and the price we pay (new price) is $ \$54.13$, we could get the $ \%$ we pay in tax this way (note that since we rounded to get the $ 54.13$, our answer is off a little):

$ \displaystyle \begin{align}\text{Percent Increase (Tax)}\,\,&=\frac{{54.13-50}}{{50}}\,\,\times \,100\,\\&=\,8.26\%\end{align}$

Sometimes we have to work a little backwards in the problem to get the right answer. For example, we may have a problem that says something like this:

Your favorite pair of shoes are on sale for $ 30\%$ off. The sale price is $ \$62.30$. What was the original price?

Notice that if the shoes are on sale for $ 30\%$, we need to pay $ 70\%$ for them. Also remember that “**of = times**”. We can set it up this way:

$ \displaystyle \,\,.7\,\,\times \,\,?=\$62.30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,? = \frac{{62.30}}{{.7}} = \$89$

The original price of the shoes would have been $ \$89$ before tax.

In the **Algebra** sections, we will address solving the following types of percentage problems, but I’ll briefly address them here if you need to do them now. If you don’t totally follow how to get the answers, don’t worry about it, since we’ll cover “**Algebra Word Problems**” later!

One other way to address percentages is the “$ \displaystyle \frac{{\text{is}}}{{\text{of}}}$” trick, which we’ll address below.

## Ratios and Proportions

**Ratios** are just a comparison of two numbers. They look a little scary since they involve fractions, but they really aren’t bad at all. Again, they are typically used when you are **comparing** two things — like cost of one pair of shoes to another pair, or maybe even the number of shirts you have compared to the number of jeans you have.

Let’s use that as an example. Let’s say you have about $ 5$ shirts for every $ 1$ pair of jeans you have, and you figure this same ratio is pretty typical among your friends. You can write your ratio in a fraction like $ \displaystyle \frac{5}{1}$, or you can use a colon in between the two numbers, like $ 5:1$ (spoken as “$ 5$ to $ 1$”).

The fractions over $ 1$ is actually a **rate** (this word is related to the word ratio!), for example, just like when you think of miles per hour. Our rate is shirts per one pair of jeans – $ 5$ shirts for every pair of jeans. Also note that this particular ratio is a **unit rate**, since the second number (denominator in the fraction) is $ 1$.

Here’s an example. Let’s say you know your friend Alicia has $ 7$ pairs of jeans and you’re wondering how many shirts she has, based on the ratio or rate of $ 5$ shirts to one pair of jeans. We can do this with math quite easily by setting up the following **proportion**, which is an **equation** (setting two things equal to one another) with a ratio on each side:

$ \displaystyle \frac{{\text{shirts}}}{{\text{jeans}}}=\frac{5}{1}=\frac{?}{7}$

How do we figure out how many shirts Alicia has? One way is just to think about reducing or expanding fractions. Let’s expand the fraction $ \displaystyle \frac{5}{1}$ to another fraction that has $ 7$ on the bottom:

$ \displaystyle \frac{{\text{shirts}}}{{\text{jeans}}}\,=\,\frac{5}{1}\,=\,\frac{5}{1}\,\times \,1\,=\frac{5}{1}\,\,\times \,\,\frac{7}{7}\,=\frac{{35}}{7}$ Alicia would have $ 35$ shirts.

I’m going to also show you a concept called **cross-multiplying**, which is very, very useful, even when we get into Algebra, Geometry and up through Calculus! This is a much easier way to do these types of problems.

Remember the “butterfly” concept when we’re comparing fractions, and remember how the fractions are equal when the “butterfly” products are equal?

We’re going to use this concept to set the fractions or ratios equal so we know how many shirts Alicia has:

We know that $ 5\times 7=35$, so we need to know what multiplied by $ 1$ will give us $ 35$. $ 35$!! Alicia has $ 35$ shirts!!! See how easy that was? If we didn’t have the $ 1$ as a factor to get to $ 35$, we’d have to divide $ 35$ by the number under the $ 5$ to get the answer. This is because dividing “undoes” multiplying.

There’s also something called a “**WON**” method for proportions. To do this, set up a table with **WON** at the top. “**W**” stands for **Words**, “**O**” stands for **Original** or **Old**, and “**N**” stands for **New** (in this example, for Alicia). Put the words and numbers in the table, and then cross multiply like we did earlier. Again, we get that Alicia has $ 35$ **shirts**, based on the proportion of $ 5$ shirts to every pair of jeans, and the fact that she has $ 7$ pairs of jeans.

W (Words) | O (Old) | N (New) |

Shirts | $ 5$ | ? $ 7\times 5\div 1=35$ |

Jeans | $ 1$ | $ 7$ |

Here’s a cooking example with proportions. Sometimes a recipe might give you the amounts in tablespoons, for example, and you only have a measuring spoon with teaspoons. We know from the **Fractions** section that $ 1$ tablespoon = $ 3$ teaspoons, and let’s say the recipe calls for $ 2$ tablespoons; how many teaspoons would you need?

This seems pretty easy to do without the proportion, but let’s set it up anyway, so you can see how easy it is to use proportions:

$ \require{cancel} \displaystyle \frac{{\text{teaspoons}}}{{\text{tablespoons}}}\,\,\,\,\,={}^{6}{{\xcancel{{\frac{3}{1}\,\,\,=\,\,\,\frac{?}{2}}}}^{6}}$

We know that $ 3\times 2=6$, so we need to know what multiplied by $ 1$ will give us $ 6$. We would need $ 6$ teaspoons for our $ 2$ tablespoons.

Now let’s go on to a more complicated example that relates back to converting numbers back and forth between the **Metric System** and our customary system here.

Let’s say we have $ 13$ meters of something and we want to know how many feet this is. We can either look up how many feet are in $ 1$ meter, or how many meters are in $ 1$ foot – it really doesn’t matter – but we need a conversion number.

We find that $ 1$ meter equals approximately $ 3.28$ feet; it’s easier to deal with numbers over $ 1$, so this is the best conversion. Let’s set all this up in a proportion. Remember to keep the same unit either on the tops of the proportion, or on the sides; it works both ways:

$ \displaystyle \begin{array}{l}\displaystyle \frac{{\text{meters}}}{{\text{meters}}}\,\,=\,\,\displaystyle \frac{{\text{feet}}}{{\text{feet}}}\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\displaystyle \frac{{\text{meters}}}{{\text{feet}}}\,\,=\,\,\displaystyle \frac{{\text{meters}}}{{\text{feet}}}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\displaystyle \frac{{\text{feet}}}{{\text{meters}}}\,\,=\,\,\displaystyle \frac{{\text{feet}}}{{\text{meters}}}\end{array}$

Let’s solve both two different ways to get the number of feet in $ 13$ meters. Notice that we can turn proportions sideways, with the “$ =$” sideways too; this is how we got from the first equation to the second above.

Here’s an example where we have to do some dividing with our cross multiplying. Try to really understand why we have to divide by $ 2$ to get the answer (it “undoes” the multiplying):

$ \displaystyle \begin{array}{l}\displaystyle \frac{5}{2}\,=\,\displaystyle \frac{?}{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\times \,\,9\,=2\,\times \,?\\\,\,\,\,?\,=\,\displaystyle \frac{{5\,\times \,9}}{2}\,=\,\displaystyle \frac{{45}}{2}\,=\,\,22\displaystyle \frac{1}{2}\end{array}$

## Unit Multipliers

We can also use what we call **unit multipliers** to change numbers from one unit to another. The idea is to multiply fractions that equal $ 1$ to get rid of the units we don’t want. This concept is also called **Dimensional Analysis**, and the multipliers are called **Conversion Ratios** or **Conversion Factors**.

Let’s say we want to use **unit multipliers** to convert $ 58$ inches to yards.

Since we have inches and we want to end up with yards, we can multiply by ratios (fractions) that relate the units to each other. We are really multiplying by “$ 1$”, since the top and bottom amounts will be the same (just the units will be different). Start with the units we have, and end with the units we want, to see what we need to do in between. Make all the terms look like fractions:

$ \displaystyle \frac{{58\text{ inches}}}{1}\,\,\times \,\,\frac{?}{?}\,\,\times \,\,\frac{?}{?}\,\,\times …=\,\,\frac{{?\text{ yards}}}{1}$

We need to get rid of the inches unit on the top and somehow get the yards unit on the top; it’s easiest to include feet to do this and set up the units so the ones not needed can be crossed out:

$ \require{cancel} \displaystyle \frac{{58\text{ }\cancel{{\text{inches}}}}}{1}\,\times \,\frac{{?\text{ }\cancel{{\text{feet}}}}}{{?\text{ }\cancel{{\text{inches}}}}}\,\times \,\frac{{?\text{ }\,\text{yards}}}{{?\text{ }\cancel{{\text{feet}}}}}\,=\,\frac{{\text{? }\,\text{yards}}}{\text{1}}$

Now just fill in how many inches are in a foot, and how many feet are in a yard, and multiply across:

$ \displaystyle \begin{array}{l}\displaystyle \frac{{58\text{ }\cancel{{\text{inches}}}}}{1}\,\times \,\displaystyle \frac{{1\text{ }\cancel{{\text{foot}}}}}{{12\text{ }\cancel{{\text{inches}}}}}\,\times \,\displaystyle \frac{{1\text{ yard}}}{{3\text{ }\cancel{{\text{feet}}}}}\,\\\,\,\,\,\,\,=\,\displaystyle \frac{{58\times 1\times 1\text{ yards}}}{{1\times 12\times 3}}\,=\,\displaystyle \frac{{58}}{{36}}\text{ }\,\text{yards}\,=\,\displaystyle \frac{{29}}{{18}}\text{ }\,\text{yards}\end{array}$

Here’s another example where we use unit multipliers; **notice how we can take care of the square kilometers by using the unit multiplier twice.**

Use unit multipliers to convert $ 100$ square kilometers to square meters.

## Using Percentages with Ratios

Now let’s **revisit percentages** and show how proportions can help with them too! One trick to use is the $ \displaystyle \frac{{\text{is}}}{{\text{of}}}$ and $ \displaystyle \frac{{\text{part}}}{{\text{whole}}}$ tricks. You can remember these since the word that comes first in the alphabet (“is” and “part”) are on the top of the fractions. You can typically solve percentage problems by using the following formula:

$ \displaystyle \frac{{\text{is}}}{{\text{of}}}=\frac{\text{ }\!\!\%\!\!\text{ }}{{100}}$

What this means is that the number **around **the “is” in an equation is on **top** of the proportion, and the number that comes **after **the “of” in an equation is on **bottom** of the proportion, and the percentage is **over** the $ 100$.

You can also think of this as the following, but you have to remember that sometimes the part may be actually be bigger than the whole (if the percentage is greater than $ 100$):

$ \displaystyle \frac{{\text{part}}}{{\text{whole}}}=\frac{\text{ }\!\!\%\!\!\text{ }}{{100}}$

Here are some examples, using the same problems that we did above. (Later, in the **Algebra Word Problems** section, we’ll learn how to translate math word problems like these word-for-word from English to math.)

- What is $ 20\%$ of $ 100$? Since the $ 20$ is the $ \%$ part, put that over the $ 100$. The $ 100$ comes after the “of”, so put that on the bottom. We’re looking for the “part” of the “whole”.

$ \displaystyle \begin{array}{c}\displaystyle \frac{{\text{is}}}{{\text{of}}}=\,\displaystyle \frac{\%}{{100}}\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\displaystyle \frac{{\text{part}}}{{\text{whole}}}=\displaystyle \frac{\%}{{100}}\\\displaystyle \frac{?}{{100}}=\displaystyle \frac{{20}}{{100}}\,\,\,\,\,\,\,\,\,?=20\end{array}$

- $ 100$ is what percentage of $ 200$? The $ 100$ is close to the “is”, so put that on the top. The $ 200$ comes after the “of”, so put that on the bottom. The $ 100$ is the “part” of the $ 200$ “whole”.

$ \displaystyle \begin{array}{c}\displaystyle \frac{{\text{is}}}{{\text{of}}}=\displaystyle \frac{\%}{{100}}\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\displaystyle \frac{{\text{part}}}{{\text{whole}}}=\displaystyle \frac{\%}{{100}}\\\displaystyle \frac{{100}}{{200}}=\displaystyle \frac{{?\,\,\,\%}}{{100}}\,\,\,\,\,\,\,\,\,?=50\end{array}$

- $ 200$ is $ 50\%$ of what number? The $ 200$ is close to the “is” and we don’t know what the “of” is. The $ 50$ is the percentage. The $ 200$ is the “part”, so we need to find the “whole”.

$ \displaystyle \begin{array}{c}\displaystyle \frac{{\text{is}}}{{\text{of}}}=\displaystyle \frac{\%}{{100}}\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\displaystyle \frac{{\text{part}}}{{\text{whole}}}=\displaystyle \frac{\%}{{100}}\,\\\,\displaystyle \frac{{200}}{?}=\displaystyle \frac{{50}}{{100}}\,\,\,\,\,\,\,\,\,?=400\end{array}$

Here are a few more **problems on rates and percentages**; remember to cross-multiply to find the answers.

**Remember also – if you’re not quite sure what you’re doing, think of the problems with easier numbers and see how you’re doing it! This can help a lot of the time.**

**Learn these rules and practice, practice, practice!**

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On to **Negative Numbers and Absolute Value **– you are ready!!