# 3.5: Ratios and Proportions

**At Grade**Created by: CK-12

## Learning Objectives

- Write and understand a ratio.
- Write and solve a proportion.
- Solve proportions using cross products.
- Solve problems using scale drawings.

## Introduction

Nadia is counting out money with her little brother. She gives her brother all the nickels and pennies. She keeps the quarters and dimes for herself. Nadia has four quarters and six dimes. Her brother has fifteen nickels and five pennies and is happy because he has more coins than his big sister. How would you explain to him that he is actually getting a bad deal?

## Write a ratio

A **ratio** is a way to compare two numbers, measurements or quantities. When we write a ratio, we divide one number by another and express the answer as a fraction. There are two distinct ratios in the problem above. For example, the ratio of the **number** of Nadia’s coins to her brother’s is \begin{align*}\frac{4 + 6}{15 + 5}\end{align*}, or \begin{align*}\frac{10}{20} = \frac{1}{2}\end{align*}. (Ratios should always be simplified.) In other words, Nadia has half as many coins as her brother.

Another ratio we could look at is the **value** of the coins. The value of Nadia’s coins is \begin{align*}(4 \times 25) + (6 \times 10) = 160 \ cents\end{align*}. The value of her brother’s coins is \begin{align*}(15 \times 5) + (5 \times 1) = 80 \ cents\end{align*}. The ratio of the **value** of Nadia’s coins to her brother’s is \begin{align*}\frac{160}{80} = \frac{2}{1}\end{align*}. So the value of Nadia’s money is twice the value of her brother’s.

Notice that even though the denominator is one, we still write it out and leave the ratio as a fraction instead of a whole number. A ratio with a denominator of one is called a **unit rate**.

**Example 1**

*The price of a Harry Potter Book on Amazon.com is $10.00. The same book is also available used for $6.50. Find two ways to compare these prices.*

**Solution**

We could compare the numbers by expressing the difference between them: \begin{align*}\$10.00 - \$6.50 = \$3.50\end{align*}.

We can also use a ratio to compare them: \begin{align*}\frac{10.00}{6.50} = \frac{100}{65} = \frac{20}{13}\end{align*} (after multiplying by 10 to remove the decimals, and then simplifying).

So we can say that **the new book is $3.50 more than the used book**, or we can say that **the new book costs** \begin{align*}\frac{20}{13}\end{align*} **times as much as the used book**.

**Example 2**

*A tournament size shuffleboard table measures 30 inches wide by 14 feet long. Compare the length of the table to its width and express the answer as a ratio.*

**Solution**

We could just write the ratio as \begin{align*}\frac{14 \ feet}{30 \ inches}\end{align*}. But since we’re comparing two lengths, it makes more sense to convert all the measurements to the same units. 14 feet is \begin{align*}14 \times 12 = 168 \ inches\end{align*}, so our new ratio is \begin{align*}\frac{168}{30} = \frac{28}{5}\end{align*}.

**Example 3**

*A family car is being tested for fuel efficiency. It drives non-stop for 100 miles and uses 3.2 gallons of gasoline. Write the ratio of distance traveled to fuel used as a* **unit rate**.

**Solution**

The ratio of distance to fuel is \begin{align*}\frac{100 \ miles}{3.2 \ gallons}\end{align*}. But a unit rate has to have a denominator of one, so to make this ratio a unit rate we need to divide both numerator and denominator by 3.2. \begin{align*}\frac{\frac{100}{3.2} \ miles}{\frac{3.2}{3.2} \ gallons} = \frac{31.25 \ miles}{1 \ gallon}\end{align*} or **31.25 miles per gallon.**

## Write and Solve a Proportion

When two ratios are equal to each other, we call it a proportion. For example, the equation \begin{align*}\frac{10}{5} = \frac{6}{9}\end{align*} is a proportion. We know it’s true because we can reduce both fractions to \begin{align*}\frac{2}{3}\end{align*}.

(Check this yourself to make sure!)

We often use proportions in science and business—for example, when scaling up the size of something. We generally use them to solve for an unknown, so we use algebra and label the unknown variable \begin{align*}x\end{align*}.

**Example 4**

*A small fast food chain operates 60 stores and makes $1.2 million profit every year. How much profit would the chain make if it operated 250 stores?*

**Solution**

First, we need to write a **ratio:** the ratio of profit to number of stores. That would be \begin{align*}\frac{\$1,200,000}{60}\end{align*}.

Now we want to know how much profit 250 stores would make. If we label that profit \begin{align*}x\end{align*}, then the ratio of profit to stores in that case is \begin{align*}\frac{x}{250}\end{align*}.

Since we’re assuming the profit is proportional to the number of stores, the ratios are equal and our proportion is \begin{align*}\frac{1,200,000}{60} = \frac{x}{250}\end{align*}.

(Note that we can drop the units – not because they are the same in the numerator and denominator, but because they are the same on both sides of the equation.)

To solve this equation, first we simplify the left-hand fraction to get \begin{align*}20,000 = \frac{x}{250}\end{align*}. Then we multiply both sides by 250 to get \begin{align*}5,000,000 = x\end{align*}.

**If the chain operated 250 stores, the annual profit would be 5 million dollars.**

**Example 5**

*A chemical company makes up batches of copper sulfate solution by adding 250 kg of copper sulfate powder to 1000 liters of water. A laboratory chemist wants to make a solution of identical concentration, but only needs 350 mL (0.35 liters) of solution. How much copper sulfate powder should the chemist add to the water?*

**Solution**

The ratio of powder to water in the first case, in kilograms per liter, is \begin{align*}\frac{250}{1000}\end{align*}, which reduces to \begin{align*}\frac{1}{4}\end{align*}. In the second case, the unknown amount is how much powder to add. If we label that amount \begin{align*}x\end{align*}, the ratio is \begin{align*}\frac{x}{0.35}\end{align*}. So our proportion is \begin{align*}\frac{1}{4} = \frac{x}{0.35}\end{align*}.

To solve for \begin{align*}x\end{align*}, first we multiply both sides by 0.35 to get \begin{align*}\frac{0.35}{4}=x\end{align*}, or \begin{align*}x = 0.0875\end{align*}.

**The mass of copper sulfate that the chemist should add is 0.0875 kg, or 87.5 grams.**

## Solve Proportions Using Cross Products

One neat way to simplify proportions is to cross multiply. Consider the following proportion:

\begin{align*}\frac{16}{4} = \frac{20}{5}\end{align*}

If we want to eliminate the fractions, we could multiply both sides by 4 and then multiply both sides by 5. But suppose we just do both at once?

\begin{align*} 4 \times 5 \times \frac{16}{4} &= 4 \times 5 \times \frac{20}{5}\\ 5 \times 16 &= 4 \times 20\end{align*}

Now comparing this to the proportion we started with, we see that the denominator from the left hand side ends up being multiplied by the numerator on the right hand side. You can also see that the denominator from the *right* hand side ends up multiplying the numerator on the *left* hand side.

In effect the two denominators have multiplied across the equal sign:

becomes \begin{align*}5 \times 16 = 4 \times 20\end{align*}.

This movement of denominators is known as **cross multiplying**. It is extremely useful in solving proportions, especially when the unknown variable is in the denominator.

**Example 6**

*Solve this proportion for \begin{align*}x\end{align*}:* \begin{align*}\frac{4}{3} = \frac{9}{x}\end{align*}

**Solution**

Cross multiply to get \begin{align*}4x = 9 \times 3\end{align*}, or \begin{align*}4x = 27\end{align*}. Then divide both sides by 4 to get \begin{align*}x = \frac{27}{4}\end{align*}, or \begin{align*}x = 6.75\end{align*}.

**Example 7**

*Solve the following proportion for \begin{align*}x\end{align*}:* \begin{align*}\frac{0.5}{3} = \frac{56}{x}\end{align*}

**Solution**

Cross multiply to get \begin{align*}0.5x = 56 \times 3\end{align*}, or \begin{align*}0.5x = 168\end{align*}. Then divide both sides by 0.5 to get \begin{align*}x = 336\end{align*}.

## Solve Real-World Problems Using Proportions

**Example 8**

*A cross-country train travels at a steady speed. It covers 15 miles in 20 minutes. How far will it travel in 7 hours assuming it continues at the same speed?*

**Solution**

We’ve done speed problems before; remember that speed is just the ratio \begin{align*}\frac{\text{distance}}{\text{time}}\end{align*}, so that ratio is the one we’ll use for our proportion. We can see that the speed is \begin{align*}\frac{15 \ miles}{20 \ minutes}\end{align*}, and that speed is also equal to \begin{align*}\frac{x \ miles}{7 \ hours}\end{align*}.

To set up a proportion, we first have to get the units the same. 20 minutes is \begin{align*}\frac{1}{3}\end{align*} of an hour, so our proportion will be \begin{align*}\frac{15}{\frac{1}{3}} = \frac{x}{7}\end{align*}. This is a very awkward looking ratio, but since we’ll be cross multiplying, we can leave it as it is.

Cross multiplying gives us \begin{align*}7 \times 15 = \frac{1}{3}x\end{align*}. Multiplying both sides by 3 then gives us \begin{align*}3 \times 7 \times 15 = x\end{align*}, or \begin{align*}x = 315\end{align*}.

**The train will travel 315 miles in 7 hours.**

**Example 9**

*In the United Kingdom, Alzheimer’s disease is said to affect one in fifty people over 65 years of age. If approximately 250000 people over 65 are affected in the UK, how many people over 65 are there in total?*

**Solution**

The fixed ratio in this case is the 1 person in 50. The unknown quantity \begin{align*}x\end{align*} is the total number of people over 65. Note that in this case we don’t need to include the units, as they will cancel between the numerator and denominator.

Our proportion is \begin{align*}\frac{1}{50} = \frac{250000}{x}\end{align*}. Each ratio represents \begin{align*}\frac{\text{people with Alzheimer's}}{\text{total people}}\end{align*}.

Cross multiplying, we get \begin{align*}1 \cdot x = 250000 \cdot 50\end{align*}, or \begin{align*}x = 12,500,000\end{align*}.

**There are approximately 12.5 million people over the age of 65 in the UK.**

For some more advanced ratio problems and applications, watch the Khan Academy video at http://www.youtube.com/watch?v=PASSD2OcU0c.

## Scale and Indirect Measurement

One place where ratios are often used is in making maps. The **scale** of a map describes the relationship between distances on a map and the corresponding distances on the earth's surface. These measurements are expressed as a fraction or a ratio.

So far we have only written ratios as fractions, but outside of mathematics books, ratios are often written as two numbers separated by a colon (:). For example, instead of \begin{align*}\frac{2}{3}\end{align*}, we would write 2:3.

Ratios written this way are used to express the relationship between a map and the area it represents. For example, a map with a scale of 1:1000 would be a map where one unit of measurement (such as a centimeter) on the map would represent 1000 of the same unit (1000 centimeters, or 10 meters) in real life.

**Example 10**

*Anne is visiting a friend in London, and is using the map below to navigate from Fleet Street to Borough Road. She is using a 1:100,000 scale map, where 1 cm on the map represents 1 km in real life. Using a ruler, she measures the distance on the map as 8.8 cm. How far is the real distance from the start of her journey to the end?*

**Solution**

The scale is the ratio of distance on the map to the corresponding distance in real life. Written as a fraction, it is \begin{align*}\frac{1}{100000}\end{align*}. We can also write an equivalent ratio for the distance Anne measures on the map and the distance in real life that she is trying to find: \begin{align*}\frac{8.8}{x}\end{align*}. Setting these two ratios equal gives us our proportion: \begin{align*}\frac{1}{100000} = \frac{8.8}{x}\end{align*}. Then we can cross multiply to get \begin{align*}x = 880000\end{align*}.

That’s how many *centimeters* it is from Fleet Street to Borough Road; now we need to convert to kilometers. There are 100000 cm in a km, so we have to divide our answer by 100000.

\begin{align*}\frac{880000}{100000} = 8.8.\end{align*}

**The distance from Fleet Street to Borough Road is 8.8 km.**

In this problem, we could have just used our intuition: the \begin{align*}1 \ cm = 1 \ km\end{align*} scale tells us that any number of cm on the map is equal to the same number of km in real life. But not all maps have a scale this simple. You’ll usually need to refer to the map scale to convert between measurements on the map and distances in real life!

**Example 11**

*Antonio is drawing a map of his school for a project in math. He has drawn out the following map of the school buildings and the surrounding area*

*He is trying to determine the scale of his figure. He knows that the distance from the point marked A on the baseball diamond to the point marked B on the athletics track is 183 meters. Use the dimensions marked on the drawing to determine the scale of his map.*

**Solution**

We know that the real-life distance is 183 m, and the scale is the ratio \begin{align*}\frac{\text{distance on map}}{\text{distance in real life}}\end{align*}.

To find the distance on the map, we use Pythagoras’ Theorem: \begin{align*}a^2 + b^2 = c^2\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the horizontal and vertical lengths and \begin{align*}c\end{align*} is the diagonal between points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.

\begin{align*} 8^2 + 14^2 &= c^2\\ 64 + 196 &= c^2\\ 260 &= c^2\\ \sqrt{260} &= c\\ 16.12 & \approx c\end{align*}

So the distance on the map is about 16.12 cm. The distance in real life is 183 m, which is 18300 cm. Now we can divide:

\begin{align*}\text{Scale} = \frac{16.12}{18300} \approx \frac{1}{1135.23}\end{align*}

**The scale of Antonio’s map is approximately 1:1100.**

Another visual use of ratio and proportion is in **scale drawings**. Scale drawings (often called **plans**) are used extensively by architects. The equations governing scale are the same as for maps; the scale of a drawing is the ratio \begin{align*}\frac{\text{distance on diagram}}{\text{distance in real life}}\end{align*}.

**Example 12**

*Oscar is trying to make a scale drawing of the Titanic, which he knows was 883 ft long. He would like his drawing to be at a 1:500 scale. How many inches long does his sheet of paper need to be?*

**Solution**

We can reason intuitively that since the scale is 1:500, the paper must be \begin{align*}\frac{883}{500} = 1.766 \ feet\end{align*} long. Converting to inches means the length is \begin{align*}12(1.766) = 21.192 \ inches\end{align*}.

**Oscar’s paper should be at least 22 inches long.**

**Example 13**

*The Rose Bowl stadium in Pasadena, California measures 880 feet from north to south and 695 feet from east to west. A scale diagram of the stadium is to be made. If 1 inch represents 100 feet, what would be the dimensions of the stadium drawn on a sheet of paper? Will it fit on a standard \begin{align*}8.5 \times 11\end{align*} inch sheet of paper?*

**Solution**

Instead of using a proportion, we can simply use the following equation: (distance on diagram) = (distance in real life) \begin{align*}\times\end{align*} (scale). (We can derive this from the fact that \begin{align*}\text{scale} = \frac{\text{distance on diagram}}{\text{distance in real life}}\end{align*}.)

Plugging in, we get

\begin{align*}\text{height on paper} = 880 \ feet \times \frac{1 \ inch}{100 \ feet} = 8.8 \ inches\end{align*}

\begin{align*}\text{width on paper} = 695 \ feet \times \frac{1 \ inch}{100 \ feet} = 6.95 \ inches\end{align*}

**The scale diagram will be \begin{align*}8.8 \ in \times 6.95 \ in\end{align*}. It will fit on a standard sheet of paper.**

## Lesson Summary

- A
**ratio**is a way to compare two numbers, measurements or quantities by dividing one number by the other and expressing the answer as a fraction. -
**A proportion**is formed when two ratios are set equal to each other. -
**Cross multiplication**is useful for solving equations in the form of proportions. To cross multiply, multiply the bottom of each ratio by the top of the other ratio and set them equal. For instance, cross multiplying results in \begin{align*}11 \times 3 = 5x\end{align*}. -
**Scale**is a proportion that relates map distance to real life distance.

## Review Questions

- Write the following comparisons as ratios. Simplify fractions where possible.
- $150 to $3
- 150 boys to 175 girls
- 200 minutes to 1 hour
- 10 days to 2 weeks

- Write the following ratios as a unit rate.
- 54 hotdogs to 12 minutes
- 5000 lbs to 250 square inches
- 20 computers to 80 students
- 180 students to 6 teachers
- 12 meters to 4 floors
- 18 minutes to 15 appointments

- Solve the following proportions.
- \begin{align*}\frac{13}{6} = \frac{5}{x}\end{align*}
- \begin{align*}\frac{1.25}{7} = \frac{3.6}{x}\end{align*}
- \begin{align*}\frac{6}{19} = \frac{x}{11}\end{align*}
- \begin{align*}\frac{1}{x} = \frac{0.01}{5}\end{align*}
- \begin{align*}\frac{300}{4} = \frac{x}{99}\end{align*}
- \begin{align*}\frac{2.75}{9} = \frac{x}{ \left( \frac{2}{9} \right )}\end{align*}
- \begin{align*}\frac{1.3}{4} = \frac{x}{1.3}\end{align*}
- \begin{align*}\frac{0.1}{1.01} = \frac{1.9}{x}\end{align*}

- A restaurant serves 100 people per day and takes in $908. If the restaurant were to serve 250 people per day, how much money would it take in?
- The highest mountain in Canada is Mount Yukon. It is \begin{align*}\frac{298}{67}\end{align*} the size of Ben Nevis, the highest peak in Scotland. Mount Elbert in Colorado is the highest peak in the Rocky Mountains. Mount Elbert is \begin{align*}\frac{220}{67}\end{align*} the height of Ben Nevis and \begin{align*}\frac{11}{12}\end{align*} the size of Mont Blanc in France. Mont Blanc is 4800 meters high. How high is Mount Yukon?
- At a large high school it is estimated that two out of every three students have a cell phone, and one in five of all students have a cell phone that is one year old or less. Out of the students who own a cell phone, what proportion owns a phone that is more than one year old?
- Use the map in Example 10. Using the scale printed on the map, determine the distances (rounded to the nearest half km) between:
- Points 1 and 4
- Points 22 and 25
- Points 18 and 13
- Tower Bridge and London Bridge

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