provides the basis for problem solving by the ratio and proportion method

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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2023 Jan-.

StatPearls [Internet].

Dose calculation ratio and proportion method.

Tammy J. Toney-Butler ; Samar Nicolas ; Lance Wilcox .

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Last Update: June 20, 2023 .

• Introduction

Three primary methods for calculation of medication dosages exist; Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method. We are going to explore the Ratio-Proportion Method, one of these three methods, in more detail.

Ratio-Proportion Method allows us the ability to compare numbers, units of measurement, or values. [1] [2] [3]

Clinicians must define a ratio and proportion. Ratios, often expressed in fraction format, are mathematical works of art designed in relationship patterns that explore comparisons between units, words, numbers. As in any relationship, key players forge a bond to make the association stronger or manageable. Proportions are those key players formed by the equality of ratios. A complicated relationship simplified by utilization and strategic placement of key players of like units or volumes. Ratios and proportions expressed as fractions, canceled out by cross multiplication or division, provide for ease in problem-solving using this method of drug calculation. 

Numerators (top) numbers or denominators (bottom) numbers multiplied and divided after the same units are canceled out. Some equations or formulas get expressed with a colon (:) or backslash (/) to indicate division and its subsequent deployment in this problem-solving technique.

For ease of calculation, a person should place the numerator of the fraction to the left of the colon or slash. In completion of this relationship, the denominator gets put on the right of the slash or colon. Unknown amounts, unknown quantities, or unknown desired amounts are depicted as an (x) in the equation and solved. The symbol (x) placement is to the left of the equation, making cross multiplication and division for (x) a simple undertaking. Keeping in mind the fundamental principle regarding the same units of measurement, numbers or units on the top and bottom of a fraction possess the ability to cancel each other out. [4] [5]

It is common for healthcare workers to use a calculator to calculate drug dosages. Calculators may be useful to decrease medication errors related to a calculation issue but are not helpful in the recognition of a conceptual error (Savage, 2015).

One study by Boyle and Eastwood found there were (40%) conceptual errors, (60%) arithmetical errors, and (25%) computational errors amongst study participants in a paper-based drug calculation questionnaire of twenty paramedics where no calculator was allowed (Boyle & Eastwood, 2018).

This study highlighted the need for ongoing education modules to improve basic arithmetic skills, for example, proper use of formulas, ability to construct a mathematical equation, and ability to perform long division without the use of a calculator. Calculators are not always available in a prehospital environment, and cell phone coverage with the use of apps may be spotty at best. The study concluded that a basic knowledge of performing manual drug calculations be a part of training modules by educators (Boyle & Eastwood, 2018).

Using 1 of the 3 methods of drug calculation as discussed above will ease the performance of manual drug calculations; Ratio and Proportion, Desired Over Have or Formula, and Dimensional Analysis. Whichever method is employed by the healthcare provider, a second method may be of use as a check for the first method. A second check further decreases the chance of a medication error related to an incorrect dosage calculation.

• Technique or Treatment

Ratio and Proportion Method

The Ratio and Proportion Method has been around for years and is one of the oldest methods utilized in drug calculations. Addition principles are a problem-solving technique that has no bearing on this relationship; only multiplication and division are used to navigate our way through a ratio and proportion problem, not adding. An example listed below will help us provide a better explanation using a fraction or a colon format:

A provider orders lorazepam 4 mg intravenous (IV) push for a CIWA score of 25. On hand, the clinicians has 2 mg/mL vials. How many milliliters are required to carry out the ordered dose? 

• Have on hand/Quantity you have = Desired Amount/x
• 2 mg/1 mL = 4 mg/x

In colon format, you would use H:V::D:X and multiply means DV and Extremes HX.

• Hx = DV, x = DV/H, 2:1::4:x, 2x = (4)(1), x = 4/2, x = 2 mL

Desired Over Have or Formula Method

Desired over Have or Formula Method uses a formula or equation to solve for an unknown quantity (x), much like ratio proportion. Drug calculations require the use of conversion factors, such as when converting from pounds to kilograms or liters to milliliters. Simplistic in design, this method allows us to work with various units of measurement, converting factors to find our answer (as cited in Boyer, 2002) [Lindow, 2004]. Useful in checking the accuracy of the other methods of calculation as above mentioned, thus acting as a double or triple check.

• A basic formula, solving for x, guides us in the setting up of an equation: D/H x Q = x, or desired dose (amount) = ordered dose amount/amount on hand x quantity. 

For example, a provider requests lorazepam 4 mg IV Push for a patient in severe alcohol withdrawal. On hand, the clinician has 2 mg/mL vials. How many milliliters should they draw up in a syringe to deliver the desired dose? 

• Dose ordered (4 mg) x quantity (1 mL)/have (2 mg) = amount you want to give (2 mL)

Remember, units of measurement must match such as milliliters and milliliters, or you will need to convert to like units of measurement. In the example above, the ordered dose was in mg, and the have dose was in mg; both would cancel out, leaving milliliters (answer called for milliliters), so no further conversion is required.

Dimensional Analysis Method

An order placed by a provider for lorazepam 4 mg IV PUSH for CIWA score of 25 or higher, follow CAGE Protocol for subsequent dosages based on CIWA scoring.

• On hand, the supply is with 2 mg/mL vials in the automated dispensing unit.
• How many milliliters are needed to arrive at the ordered dose?
• The desired dose gets placed over 1 remember, (x mL) = 4 mg/1 x 1 mL/2 mg x (4)(1)/2 x 4/2 x 2/1 = 2 mL, the clinician kept multiplying/dividing until they got the desired amount, 2 mL in this problem example
• Notice the fraction was set up with mg and mg strategically placed so like units could cancel each other out, making the equation easier to solve for the unit you desired or milliliters.

Zeros can be canceled out in the same way as like units. Look at the example listed below for clarification:

• 1000/500 x 10/5 = 2, the 2 zeros in 1000 and 2 zeros in 500 can be crossed out since like units in numerator and denominator, leaving 10/5, a much easier fraction to solve, and the answer makes sense. 

Having addressed zeros, the following is a look at 1.

• If you multiply a number by a 1, then the number is unchanged.
• In contrast, if you multiply a number by zero, the number becomes zero.
• Examples listed below are as follows: 18 x 0 = 0 or 20 x 1 = 20.
• Clinical Significance

Medication errors can be detrimental and costly to patients. Drug calculation and basic mathematical skills play a role in the safe administration of medications. Pediatric populations are especially vulnerable to medication errors due to the need to calculate dosages incorporating many factors; height, weight, body surface area, and growth and development level. The higher the complexity of the math, the increased the risk potential for dose calculation errors. 

According to a study of intensive care nurses (ICU) in 2016, 80% of nurses considered knowledge on drug dosage calculation essential to decrease medication errors during the preparation of intravenous drugs.

One study by a group of oncology nurses in 3 Swiss hospitals published in 2018 discusses the process of double-checking and its limitations in the current healthcare environment, specifically, increased nurse workload and time constraints, distracting environments, and lack of resources. The study concluded that oncology nurses strongly believed in the effectiveness of double-checking medication despite reporting limitations of the procedure in clinical practice. [6]

• Enhancing Healthcare Team Outcomes

High-risk medications such as heparin and insulin often require a second check on dosage amounts by more than one provider before administration of the drug. Follow institutional policies and recommendations on the double-checking of dose calculations by another licensed provider.

• Review Questions
• Access free multiple choice questions on this topic.
• Comment on this article.

Disclosure: Tammy Toney-Butler declares no relevant financial relationships with ineligible companies.

Disclosure: Samar Nicolas declares no relevant financial relationships with ineligible companies.

Disclosure: Lance Wilcox declares no relevant financial relationships with ineligible companies.

This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( http://creativecommons.org/licenses/by-nc-nd/4.0/ ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.

• Cite this Page Toney-Butler TJ, Nicolas S, Wilcox L. Dose Calculation Ratio and Proportion Method. [Updated 2023 Jun 20]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2023 Jan-.

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Ratio problem solving

Here you will learn about ratio problem solving, including how to set up and solve problems. You will also look at real life ratio word problems.

Students will first learn about ratio problem solving as part of ratio and proportion in 6 th grade and 7 th grade.

What is ratio problem solving?

Ratio problem solving is a collection of ratio and proportion word problems that link together aspects of ratio and proportion into more real life questions. This requires you to be able to take key information from a question and use your knowledge of ratios (and other areas of the curriculum) to solve the problem.

A ratio is a relationship between two or more quantities. They are usually written in the form a : b where a and b are two quantities. When problem solving with a ratio, the key facts that you need to know are:

• What is the ratio involved?
• What order are the quantities in the ratio?
• What is the total amount / what is the part of the total amount known?
• What are you trying to calculate ?

As with all problem solving, there is not one unique method to solve a problem. However, this does not mean that there aren’t similarities between different problems that you can use to help you find an answer.

The key to any problem solving is being able to draw from prior knowledge and use the correct piece of information to allow you to get to the next step and then the solution.

Let’s look at a couple of methods you can use when given certain pieces of information.

When solving ratio word problems, it is very important that you are able to use ratios. This includes being able to use ratio notation.

For example, Charlie and David share some sweets in the ratio of 3 : 5. This means that for every 3 sweets Charlie gets, David receives 5 sweets.

Charlie and David share 40 sweets, how many sweets do they each get?

You use the ratio to divide 40 sweets into 8 equal parts.

40 \div 8=5

Then you multiply each part of the ratio by 5.

3\times 5:5\times 5=15 : 25

This means that Charlie will get 15 sweets and David will get 25 sweets.

There can be ratio word problems involving different operations and types of numbers.

Here are some examples of different types of ratio word problems:

Common Core State Standards

How does this relate to 6 th grade math?

• Grade 6 – Ratios and Proportional Relationships (6.RP.A.3) Use ratio and rate reasoning to solve real-world and mathematical problems, for example, by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
• Grade 7 – Ratio and Proportional Relationships (7.RP.A.2) Recognize and represent proportional relationships between quantities.

How to do ratio problem solving

In order to solve problems including ratios:

Identify key information within the question.

Know what you are trying to calculate.

Use prior knowledge to structure a solution.

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Ratio problem solving examples

Example 1: part:part ratio.

Within a school, the total number of students who have school lunches to packed lunches is 5 : 7. If 465 students have a school lunch, how many students have a packed lunch?

Within a school, the number of students who have school lunches to packed lunches is \textbf{5 : 7} . If \textbf{465} students have a school lunch, how many students have a packed lunch?

Here you can see that the ratio is 5 : 7, where the first part of the ratio represents school lunches (S) and the second part of the ratio represents packed lunches (P).

You could write this as:

Where the letter above each part of the ratio links to the question.

You know that 465 students have school lunch.

2 Know what you are trying to calculate.

From the question, you need to calculate the number of students that have a packed lunch, so you can now write a ratio below the ratio 5 : 7 that shows that you have 465 students who have school lunches, and p students who have a packed lunch.

You need to find the value of p.

3 Use prior knowledge to structure a solution.

You are looking for an equivalent ratio to 5 : 7. So you need to calculate the multiplier.

You do this by dividing the known values on the same side of the ratio by each other.

465\div 5 = 93

This means to create an equivalent ratio, you can multiply both sides by 93.

So the value of p is equal to 7 \times 93=651.

There are 651 students that have a packed lunch.

Example 2: unit conversions

The table below shows the currency conversions on one day.

Use the table above to convert £520 \; (GBP) to Euros € \; (EUR).

Use the table above to convert \bf{￡520} \textbf{ (GBP)} to Euros \textbf{€ } \textbf{(EUR)}.

The two values in the table that are important are \text{GBP} and EUR. Writing this as a ratio, you can state,

You know that you have £520.

You need to convert GBP to EUR and so you are looking for an equivalent ratio with GBP=£520 and EUR=E.

To get from 1 to 520, you multiply by 520 and so to calculate the number of Euros for £520, you need to multiply 1.17 by 520.

1.17 \times 520=608.4

So £520=€608.40.

Example 3: writing a ratio 1 : n

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the 500 \, ml of concentrated plant food must be diluted into 2 \, l of water. Express the ratio of plant food to water, respectively, in the ratio 1 : n.

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the \bf{500 \, ml} of concentrated plant food must be diluted into \bf{2 \, l} of water. Express the ratio of plant food to water respectively as a ratio in the form 1 : n.

Using the information in the question, you can now state the ratio of plant food to water as 500 \, ml : 2 \, l. As you can convert liters into milliliters, you could convert 2 \, l into milliliters by multiplying it by 1000.

2 \, l=2000 \, ml

So you can also express the ratio as 500 : 2000 which will help you in later steps.

You want to simplify the ratio 500 : 2000 into the form 1:n.

You need to find an equivalent ratio where the first part of the ratio is equal to 1. You can only do this by dividing both parts of the ratio by 500 (as 500 \div 500=1 ).

So the ratio of plant food to water in the form 1 : n is 1 : 4.

Example 4: forming and solving an equation

Three siblings, Josh, Kieran and Luke, receive an allowance each week proportional to their age. Kieran is 3 years older than Josh. Luke is twice Josh’s age. If Josh receives \$ 8 allowance, how much money do the three siblings receive in total?

Three siblings, Josh, Kieran and Luke, receive an allowance each week proportional to their ages. Kieran is \bf{3} years older than Josh. Luke is twice Josh’s age. If Luke receives \bf{\$ 8} allowance, how much money do the three siblings receive in total?

You can represent the ages of the three siblings as a ratio. Taking Josh as x years old, Kieran would therefore be x+3 years old, and Luke would be 2x years old. As a ratio, you have:

You also know that Luke receives \$ 8.

You want to calculate the total amount of allowance for the three siblings.

You need to find the value of x first. As Luke receives \$ 8, you can state the equation 2x=8 and so x=4.

Now you know the value of x, you can substitute this value into the other parts of the ratio to obtain how much money the siblings each receive.

The total amount of allowance is therefore 4+7+8=\$ 19.

Example 5: simplifying ratios

Below is a bar chart showing the results for the colors of counters in a bag.

Express this data as a ratio in its simplest form.

From the bar chart, you can read the frequencies to create the ratio.

You need to simplify this ratio.

To simplify a ratio, you need to find the highest common factor of all the parts of the ratio. By listing the factors of each number, you can quickly see that the highest common factor is 2.

\begin{aligned} & 12 = 1, {\color{red}2}, 3, 4, 6, 12 \\\\ & 16 = 1, {\color{red}2}, 4, 8, 16 \\\\ & 10 = 1, {\color{red}2}, 5, 10 \end{aligned}

HCF(12,16,10) = 2

Dividing all the parts of the ratio by 2, you get

Our solution is 6 : 8 : 5.

Example 6: combining two ratios

Glass is made from silica, lime and soda. The ratio of silica to lime is 15 : 2. The ratio of silica to soda is 5 : 1. State the ratio of silica:lime:soda.

Glass is made from silica, lime and soda. The ratio of silica to lime is \bf{15 : 2}. The ratio of silica to soda is \bf{5 : 1}. State the ratio of silica:lime:soda.

You know the two ratios

You are trying to find the ratio of all 3 components: silica, lime and soda.

Using equivalent ratios you can say that the ratio of Silica:Soda is equivalent to 15 : 3 by multiplying the ratio by 3.

You now have the same amount of silica in both ratios and so you can now combine them to get the ratio 15 : 2 : 3.

Example 7: using bar modeling

India and Beau share some popcorn in the ratio of 5 : 2. If India has 75 \, g more popcorn than Beau, what was the original quantity?

India and Beau share some popcorn in the ratio of \bf{5 : 2} . If India has \bf{75 \, g} more popcorn than Beau, what was the original quantity?

You know that the initial ratio is 5 : 2 and that India has three more parts than Beau.

You want to find the original quantity.

Drawing a bar model of this problem, you have:

Where India has 5 equal shares, and Beau has 2 equal shares.

Each share is the same value and so if you can find out this value, you can then find the total quantity.

From the question, India’s share is 75 \, g more than Beau’s share so you can write this on the bar model.

You can find the value of one share by working out 75 \div 3=25 \, g.

You can fill in each share to be 25 \, g.

Adding up each share, you get

India=5 \times 25=125 \, g

Beau=2 \times 25=50 \, g

The total amount of popcorn was 125+50=175 \, g.

Teaching tips for ratio problem solving

• Continue to remind students that when solving ratio word problems, it’s important to identify the quantities being compared and express the ratio in its simplest form.
• Create practice problems for students using the information in your classroom. For example, ask students to find the ratio of boys to the ratio of girls using the total number of students in your classroom, then the school.
• To find more practice questions, utilize educational websites and apps instead of worksheets. Some of these may also provide tutorials for struggling students. These can also be helpful for test prep as they are more engaging for students.
• Use a variety of numbers in your ratio word problems – whole numbers, fractions, decimals, and mixed numbers – to give students a variety of practice.
• Provide students with a step-by-step process for problem solving, like the one shown above, that can be applied to every ratio word problem.

Easy mistakes to make

• Mixing units Make sure that all the units in the ratio are the same. For example, in example 6, all the units in the ratio were in milliliters. You did not mix ml and l in the ratio.
• Writing ratios in the wrong order For example, the number of dogs to cats is given as the ratio 12 : 13 but the solution is written as 13 : 12.

• Counting the number of parts in the ratio, not the total number of shares For example, the ratio 5 : 4 has 9 shares, and 2 parts. This is because the ratio contains 2 numbers but the sum of these parts (the number of shares) is 5+4=9. You need to find the value per share, so you need to use the 9 shares in your next line of working.
• Ratios of the form \bf{1 : \textbf{n}} The assumption can be incorrectly made that n must be greater than 1, but n can be any number, including a decimal.

Related ratio lessons

• Unit rate math
• Simplifying ratios
• Ratio to fraction
• How to calculate exchange rates
• Ratio to percent
• How to write a ratio

Practice ratio problem solving questions

1. An online shop sells board games and computer games. The ratio of board games to the total number of games sold in one month is 3 : 8. What is the ratio of board games to computer games?

8-3=5 computer games sold for every 3 board games.

2. The ratio of prime numbers to non-prime numbers from 1-200 is 45 : 155. Express this as a ratio in the form 1 : n.

You need to simplify the ratio so that the first number is 1. That means you need to divide each number in the ratio by 45.

45 \div 45=1

155\div{45}=3\cfrac{4}{9}

3. During one month, the weather was recorded into 3 categories: sunshine, cloud and rain. The ratio of sunshine to cloud was 2 : 3 and the ratio of cloud to rain was 9 : 11. State the ratio that compares sunshine:cloud:rain for the month.

3 \times S : C=6 : 9

4. The angles in a triangle are written as the ratio x : 2x : 3x. Calculate the size of each angle.

You should know that the 3 angles in a triangle always equal 180^{\circ}.

\begin{aligned} & x+2 x+3 x=180 \\\\ & 6 x=180 \\\\ & x=30^{\circ} \\\\ & 2 x=60^{\circ} \\\\ & 3 x=90^{\circ} \end{aligned}

5. A clothing company has a sale on tops, dresses and shoes. \cfrac{1}{3} of sales were for tops, \cfrac{1}{5} of sales were for dresses, and the rest were for shoes. Write a ratio of tops to dresses to shoes sold in its simplest form.

\cfrac{1}{3}+\cfrac{1}{5}=\cfrac{5+3}{15}=\cfrac{8}{15}

1-\cfrac{8}{15}=\cfrac{7}{15}

6. The volume of gas is directly proportional to the temperature (in degrees Kelvin). A balloon contains 2.75 \, l of gas and has a temperature of 18^{\circ}K. What is the volume of gas if the temperature increases to 45^{\circ}K?

The given ratio in the word problem is 2. 75 \mathrm{~L}: 18^{\circ} \mathrm{K}

Divide 45 by 18 to see the relationship between the two temperatures.

45 \div 18=2.5

45 is 2.5 times greater than 18. So we multiply 2.75 by 2.5 to get the amount of gas.

2.75 \times 2.5=6.875 \mathrm{~l}

Ratio problem solving FAQs

A ratio is a comparison of two or more quantities. It shows how much one quantity is related to another.

A recipe calls for 2 cups of flour and 1 cup of sugar. What is the ratio of flour to sugar? (2 : 1)

In middle school ( 7 th grade and 8 th grade), students transition from understanding basic ratios to working with more complex and real-life applications of ratios and proportions. They gain a deeper understanding of how ratios relate to different mathematical concepts, making them more prepared for higher-level math topics in high school.

The next lessons are

• Properties of equality
• Multiplication and division

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Privacy Overview

• 5.4 Ratios and Proportions
• Introduction
• 1.1 Basic Set Concepts
• 1.2 Subsets
• 1.3 Understanding Venn Diagrams
• 1.4 Set Operations with Two Sets
• 1.5 Set Operations with Three Sets
• Key Concepts
• Formula Review
• Chapter Review
• Chapter Test
• 2.1 Statements and Quantifiers
• 2.2 Compound Statements
• 2.3 Constructing Truth Tables
• 2.4 Truth Tables for the Conditional and Biconditional
• 2.5 Equivalent Statements
• 2.6 De Morgan’s Laws
• 2.7 Logical Arguments
• 3.1 Prime and Composite Numbers
• 3.2 The Integers
• 3.3 Order of Operations
• 3.4 Rational Numbers
• 3.5 Irrational Numbers
• 3.6 Real Numbers
• 3.7 Clock Arithmetic
• 3.8 Exponents
• 3.9 Scientific Notation
• 3.10 Arithmetic Sequences
• 3.11 Geometric Sequences
• 4.1 Hindu-Arabic Positional System
• 4.2 Early Numeration Systems
• 4.3 Converting with Base Systems
• 4.4 Addition and Subtraction in Base Systems
• 4.5 Multiplication and Division in Base Systems
• 5.1 Algebraic Expressions
• 5.2 Linear Equations in One Variable with Applications
• 5.3 Linear Inequalities in One Variable with Applications
• 5.5 Graphing Linear Equations and Inequalities
• 5.6 Quadratic Equations with Two Variables with Applications
• 5.7 Functions
• 5.8 Graphing Functions
• 5.9 Systems of Linear Equations in Two Variables
• 5.10 Systems of Linear Inequalities in Two Variables
• 5.11 Linear Programming
• 6.1 Understanding Percent
• 6.2 Discounts, Markups, and Sales Tax
• 6.3 Simple Interest
• 6.4 Compound Interest
• 6.5 Making a Personal Budget
• 6.6 Methods of Savings
• 6.7 Investments
• 6.8 The Basics of Loans
• 6.9 Understanding Student Loans
• 6.10 Credit Cards
• 6.11 Buying or Leasing a Car
• 6.12 Renting and Homeownership
• 6.13 Income Tax
• 7.1 The Multiplication Rule for Counting
• 7.2 Permutations
• 7.3 Combinations
• 7.4 Tree Diagrams, Tables, and Outcomes
• 7.5 Basic Concepts of Probability
• 7.6 Probability with Permutations and Combinations
• 7.7 What Are the Odds?
• 7.8 The Addition Rule for Probability
• 7.9 Conditional Probability and the Multiplication Rule
• 7.10 The Binomial Distribution
• 7.11 Expected Value
• 8.1 Gathering and Organizing Data
• 8.2 Visualizing Data
• 8.3 Mean, Median and Mode
• 8.4 Range and Standard Deviation
• 8.5 Percentiles
• 8.6 The Normal Distribution
• 8.7 Applications of the Normal Distribution
• 8.8 Scatter Plots, Correlation, and Regression Lines
• 9.1 The Metric System
• 9.2 Measuring Area
• 9.3 Measuring Volume
• 9.4 Measuring Weight
• 9.5 Measuring Temperature
• 10.1 Points, Lines, and Planes
• 10.2 Angles
• 10.3 Triangles
• 10.4 Polygons, Perimeter, and Circumference
• 10.5 Tessellations
• 10.7 Volume and Surface Area
• 10.8 Right Triangle Trigonometry
• 11.1 Voting Methods
• 11.2 Fairness in Voting Methods
• 11.3 Standard Divisors, Standard Quotas, and the Apportionment Problem
• 11.4 Apportionment Methods
• 11.5 Fairness in Apportionment Methods
• 12.1 Graph Basics
• 12.2 Graph Structures
• 12.3 Comparing Graphs
• 12.4 Navigating Graphs
• 12.5 Euler Circuits
• 12.6 Euler Trails
• 12.7 Hamilton Cycles
• 12.8 Hamilton Paths
• 12.9 Traveling Salesperson Problem
• 12.10 Trees
• 13.1 Math and Art
• 13.2 Math and the Environment
• 13.3 Math and Medicine
• 13.4 Math and Music
• 13.5 Math and Sports
• A | Co-Req Appendix: Integer Powers of 10

Learning Objectives

After completing this section, you should be able to:

• Construct ratios to express comparison of two quantities.
• Use and apply proportional relationships to solve problems.
• Determine and apply a constant of proportionality.
• Use proportions to solve scaling problems.

Ratios and proportions are used in a wide variety of situations to make comparisons. For example, using the information from Figure 5.15 , we can see that the number of Facebook users compared to the number of Twitter users is 2,006 M to 328 M. Note that the "M" stands for million, so 2,006 million is actually 2,006,000,000 and 328 million is 328,000,000. Similarly, the number of Qzone users compared to the number of Pinterest users is in a ratio of 632 million to 175 million. These types of comparisons are ratios.

Constructing Ratios to Express Comparison of Two Quantities

Note there are three different ways to write a ratio , which is a comparison of two numbers that can be written as: a a to b b OR a : b a : b OR the fraction a / b a / b . Which method you use often depends upon the situation. For the most part, we will want to write our ratios using the fraction notation. Note that, while all ratios are fractions, not all fractions are ratios. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only.

Example 5.28

Expressing the relationship between two currencies as a ratio.

The Euro (€) is the most common currency used in Europe. Twenty-two nations, including Italy, France, Germany, Spain, Portugal, and the Netherlands use it. On June 9, 2021, 1 U.S. dollar was worth 0.82 Euros. Write this comparison as a ratio.

Using the definition of ratio, let a = 1 a = 1 U.S. dollar and let b = 0.82 b = 0.82 Euros. Then the ratio can be written as either 1 to 0.82; or 1:0.82; or 1 0.82 . 1 0.82 .

Your Turn 5.28

Example 5.29, expressing the relationship between two weights as a ratio.

The gravitational pull on various planetary bodies in our solar system varies. Because weight is the force of gravity acting upon a mass, the weights of objects is different on various planetary bodies than they are on Earth. For example, a person who weighs 200 pounds on Earth would weigh only 33 pounds on the moon! Write this comparison as a ratio.

Using the definition of ratio, let a = 200 a = 200 pounds on Earth and let b = 33 b = 33 pounds on the moon. Then the ratio can be written as either 200 to 33; or 200:33; or 200 33 . 200 33 .

Your Turn 5.29

Using and applying proportional relationships to solve problems.

Using proportions to solve problems is a very useful method. It is usually used when you know three parts of the proportion, and one part is unknown. Proportions are often solved by setting up like ratios. If a b a b and c d c d are two ratios such that a b = c d , a b = c d , then the fractions are said to be proportional . Also, two fractions a b a b and c d c d are proportional ( a b = c d ) ( a b = c d ) if and only if a × d = b × c a × d = b × c .

Example 5.30

Solving a proportion involving two currencies.

You are going to take a trip to France. You have $520 U.S. dollars that you wish to convert to Euros. You know that 1 U.S. dollar is worth 0.82 Euros. How much money in Euros can you get in exchange for $520?

Step 1: Set up the two ratios into a proportion; let x x be the variable that represents the unknown. Notice that U.S. dollar amounts are in both numerators and Euro amounts are in both denominators.

Step 2: Cross multiply, since the ratios a b a b and c d c d are proportional, then a × d = b × c a × d = b × c .

You should receive 426.4 426.4 Euros ( 426.4 € ) ( 426.4 € ) .

Your Turn 5.30

Example 5.31, solving a proportion involving weights on different planets.

A person who weighs 170 pounds on Earth would weigh 64 pounds on Mars. How much would a typical racehorse (1,000 pounds) weigh on Mars? Round your answer to the nearest tenth.

Step 1: Set up the two ratios into a proportion. Notice the Earth weights are both in the numerator and the Mars weights are both in the denominator.

Step 2: Cross multiply, and then divide to solve.

So the 1,000-pound horse would weigh about 376.5 pounds on Mars.

Your Turn 5.31

Example 5.32, solving a proportion involving baking.

A cookie recipe needs 2 1 4 2 1 4 cups of flour to make 60 cookies. Jackie is baking cookies for a large fundraiser; she is told she needs to bake 1,020 cookies! How many cups of flour will she need?

Step 1: Set up the two ratios into a proportion. Notice that the cups of flour are both in the numerator and the amounts of cookies are both in the denominator. To make the calculations more efficient, the cups of flour ( 2 1 4 ) ( 2 1 4 ) is converted to a decimal number (2.25).

Step 2: Cross multiply, and then simplify to solve.

Jackie will need 38.25, or 38 1 4 38 1 4 , cups of flour to bake 1,020 cookies.

Your Turn 5.32

Part of the definition of proportion states that two fractions a b a b and c d c d are proportional if a × d = b × c a × d = b × c . This is the "cross multiplication" rule that students often use (and unfortunately, often use incorrectly). The only time cross multiplication can be used is if you have two ratios (and only two ratios) set up in a proportion. For example, you cannot use cross multiplication to solve for x x in an equation such as 2 5 = x 8 + 3 x 2 5 = x 8 + 3 x because you do not have just the two ratios. Of course, you could use the rules of algebra to change it to be just two ratios and then you could use cross multiplication, but in its present form, cross multiplication cannot be used.

People in Mathematics

Eudoxus was born around 408 BCE in Cnidus (now known as Knidos) in modern-day Turkey. As a young man, he traveled to Italy to study under Archytas, one of the followers of Pythagoras. He also traveled to Athens to hear lectures by Plato and to Egypt to study astronomy. He eventually founded a school and had many students.

Eudoxus made many contributions to the field of mathematics. In mathematics, he is probably best known for his work with the idea of proportions. He created a definition of proportions that allowed for the comparison of any numbers, even irrational ones. His definition concerning the equality of ratios was similar to the idea of cross multiplying that is used today. From his work on proportions, he devised what could be described as a method of integration, roughly 2000 years before calculus (which includes integration) would be fully developed by Isaac Newton and Gottfried Leibniz. Through this technique, Eudoxus became the first person to rigorously prove various theorems involving the volumes of certain objects. He also developed a planetary theory, made a sundial still usable today, and wrote a seven volume book on geography called Tour of the Earth , in which he wrote about all the civilizations on the Earth, and their political systems, that were known at the time. While this book has been lost to history, over 100 references to it by different ancient writers attest to its usefulness and popularity.

Determining and Applying a Constant of Proportionality

In the last example, we were given that 2 1 4 2 1 4 cups of flour could make 60 cookies; we then calculated that 38 1 4 38 1 4 cups of flour would make 1,020 cookies, and 720 cookies could be made from 27 cups of flour. Each of those three ratios is written as a fraction below (with the fractions converted to decimals). What happens if you divide the numerator by the denominator in each?

The quotients in each are exactly the same! This number, determined from the ratio of cups of flour to cookies, is called the constant of proportionality . If the values a a and b b are related by the equality a b = k , a b = k , then k k is the constant of proportionality between a a and b b . Note since a b = k , a b = k , then b = a k . b = a k . and b = a k . b = a k .

One piece of information that we can derive from the constant of proportionality is a unit rate. In our example (cups of flour divided by cookies), the constant of proportionality is telling us that it takes 0.0375 cups of flour to make one cookie. What if we had performed the calculation the other way (cookies divided by cups of flour)?

In this case, the constant of proportionality ( 26.66666 … = 26 2 3 ) ( 26.66666 … = 26 2 3 ) is telling us that 26 2 3 26 2 3 cookies can be made with one cup of flour. Notice in both cases, the "one" unit is associated with the denominator. The constant of proportionality is also useful in calculations if you only know one part of the ratio and wish to find the other.

Example 5.33

Finding a constant of proportionality.

Isabelle has a part-time job. She kept track of her pay and the number of hours she worked on four different days, and recorded it in the table below. What is the constant of proportionality, or pay divided by hours? What does the constant of proportionality tell you in this situation?

To find the constant of proportionality, divide the pay by hours using the information from any of the four columns. For example, 87.5 7 = 12.5 87.5 7 = 12.5 . The constant of proportionality is 12.5, or $12.50. This tells you Isabelle's hourly pay: For every hour she works, she gets paid $12.50.

Your Turn 5.33

Example 5.34, applying a constant of proportionality: running mph.

Zac runs at a constant speed: 4 miles per hour (mph). One day, Zac left his house at exactly noon (12:00 PM) to begin running; when he returned, his clock said 4:30 PM. How many miles did he run?

The constant of proportionality in this problem is 4 miles per hour (or 4 miles in 1 hour). Since a b = k , a b = k , where k k is the constant of proportionality, we have

a miles b hours = k a miles b hours = k

a 4 .5 = 4 a 4 .5 = 4 (30 minutes is ½ ½ , or 0.5 0.5 , hours)

a = 4 ( 4.5 ) a = 4 ( 4.5 ) , since from the definition we know a = k b a = k b

a = 18 a = 18

Zac ran 18 miles.

Your Turn 5.34

Example 5.35, applying a constant of proportionality: filling buckets.

Joe had a job where every time he filled a bucket with dirt, he was paid $2.50. One day Joe was paid $337.50. How many buckets did he fill that day?

The constant of proportionality in this situation is $2.50 per bucket (or $2.50 for 1 bucket). Since a b = k , a b = k , where k k is the constant of proportionality, we have

a dollars b buckets = k 337.50 b = 2.50 a dollars b buckets = k 337.50 b = 2.50

Since we are solving for b b , and we know from the definition that b = a k : b = a k :

b = 337.50 2.50 b = 135 b = 337.50 2.50 b = 135

Joe filled 135 buckets.

Your Turn 5.35

Example 5.36, applying a constant of proportionality: miles vs. kilometers.

While driving in Canada, Mabel quickly noticed the distances on the road signs were in kilometers, not miles. She knew the constant of proportionality for converting kilometers to miles was about 0.62—that is, there are about 0.62 miles in 1 kilometer. If the last road sign she saw stated that Montreal is 104 kilometers away, about how many more miles does Mabel have to drive? Round your answer to the nearest tenth.

The constant of proportionality in this situation is 0.62 miles per 1 kilometer. Since a b = k , a b = k , where k k is the constant of proportionality, we have

a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48 a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48

Rounding the answer to the nearest tenth, Mabel has to drive 64.5 miles.

Your Turn 5.36

Using proportions to solve scaling problems.

Ratio and proportions are used to solve problems involving scale . One common place you see a scale is on a map (as represented in Figure 5.16 ). In this image, 1 inch is equal to 200 miles. This is the scale. This means that 1 inch on the map corresponds to 200 miles on the surface of Earth. Another place where scales are used is with models: model cars, trucks, airplanes, trains, and so on. A common ratio given for model cars is 1:24—that means that 1 inch in length on the model car is equal to 24 inches (2 feet) on an actual automobile. Although these are two common places that scale is used, it is used in a variety of other ways as well.

Example 5.37

Solving a scaling problem involving maps.

Figure 5.17 is an outline map of the state of Colorado and its counties. If the distance of the southern border is 380 miles, determine the scale (i.e., 1 inch = how many miles). Then use that scale to determine the approximate lengths of the other borders of the state of Colorado.

When the southern border is measured with a ruler, the length is 4 inches. Since the length of the border in real life is 380 miles, our scale is 1 inch = 95 = 95 miles.

The eastern and western borders both measure 3 inches, so their lengths are about 285 miles. The northern border measures the same as the southern border, so it has a length of 380 miles.

Your Turn 5.37

Example 5.38, solving a scaling problem involving model cars.

Die-cast NASCAR model cars are said to be built on a scale of 1:24 when compared to the actual car. If a model car is 9 inches long, how long is a real NASCAR automobile? Write your answer in feet.

The scale tells us that 1 inch of the model car is equal to 24 inches (2 feet) on the real automobile. So set up the two ratios into a proportion. Notice that the model lengths are both in the numerator and the NASCAR automobile lengths are both in the denominator.

This amount (216) is in inches. To convert to feet, divide by 12, because there are 12 inches in a foot (this conversion from inches to feet is really another proportion!). The final answer is:

The NASCAR automobile is 18 feet long.

Your Turn 5.38

Check your understanding, section 5.4 exercises.

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Course: 7th grade   >   Unit 1

Worked example: solving proportions.

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Video transcript

A free service from Mattecentrum

Ratios and proportions and how to solve them

Let's talk about ratios and proportions. When we talk about the speed of a car or an airplane we measure it in miles per hour. This is called a rate and is a type of ratio. A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours.

A ratio can be written in three different ways and all are read as "the ratio of x to y"

$$x\: to\: y$$

$$\frac{x}{y}$$

A proportion on the other hand is an equation that says that two ratios are equivalent. For instance if one package of cookie mix results in 20 cookies than that would be the same as to say that two packages will result in 40 cookies.

$$\frac{20}{1}=\frac{40}{2}$$

A proportion is read as "x is to y as z is to w"

$$\frac{x}{y}=\frac{z}{w} \: where\: y,w\neq 0$$

If one number in a proportion is unknown you can find that number by solving the proportion.

You know that to make 20 pancakes you have to use 2 eggs. How many eggs are needed to make 100 pancakes?

$$\frac{eggs}{pancakes}=\frac{eggs}{pancakes}\: \: or\: \: \frac{pancakes}{eggs}=\frac{pancakes}{eggs}$$

If we write the unknown number in the nominator then we can solve this as any other equation

$$\frac{x}{100}=\frac{2}{20}$$

Multiply both sides with 100

$${\color{green} {100\, \cdot }}\, \frac{x}{100}={\color{green} {100\, \cdot }}\, \frac{2}{20}$$

$$x=\frac{200}{20}$$

If the unknown number is in the denominator we can use another method that involves the cross product. The cross product is the product of the numerator of one of the ratios and the denominator of the second ratio. The cross products of a proportion is always equal

If we again use the example with the cookie mix used above

$$\frac{{\color{green} {20}}}{{\color{blue} {1}}}=\frac{{\color{blue} {40}}}{{\color{green} {2}}}$$

$${\color{blue} {1}}\cdot {\color{blue} {40}}={\color{green} {2}}\cdot {\color{green} {20}}=40$$

It is said that in a proportion if

If you look at a map it always tells you in one of the corners that 1 inch of the map correspond to a much bigger distance in reality. This is called a scaling. We often use scaling in order to depict various objects. Scaling involves recreating a model of the object and sharing its proportions, but where the size differs. One may scale up (enlarge) or scale down (reduce).  For example, the scale of 1:4 represents a fourth. Thus any measurement we see in the model would be 1/4 of the real measurement. If we wish to calculate the inverse, where we have a 20ft high wall and wish to reproduce it in the scale of 1:4, we simply calculate:

$$20\cdot 1:4=20\cdot \frac{1}{4}=5$$

In a scale model of 1:X where X is a constant, all measurements become 1/X - of the real measurement. The same mathematics applies when we wish to enlarge. Depicting something in the scale of 2:1 all measurements then become twice as large as in reality. We divide by 2 when we wish to find the actual measurement.

Video lesson

$$\frac{x}{x + 20} = \frac{24}{54}$$

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How do you solve ratio and proportion problems?

The “Box Method” For Teaching Ratio/Proportion Problems

Download Paper | Permalink

2009 Annual Conference & Exposition

Austin, Texas

June 14, 2009

June 17, 2009

Bridging the Gap and Freshman Programs

Mathematics

14.1266.1 - 14.1266.11

10.18260/1-2--5803

https://strategy.asee.org/5803

Paper Authors

James sullivan dallas independent school district, abstract note: the first page of text has been automatically extracted and included below in lieu of an abstract.

The “Box Method” for Teaching Ratio/Proportion Problems Abstract This paper details a systematic method for teaching high school students how to set up and solve ratio and/or proportion problems. Such problems frequently occur in a wide variety of engineering applications. The author, while teaching high school algebra courses, noticed a remarkable fact: Students were able to solve such problems correctly once the problems had been set up properly. In other words, their major difficulty was not in the arithmetic required to solve these types of problems, but simply in setting the problems up. After examining several textbooks, the problem became clearer: this important aspect of solving ratio/proportion problems has been neglected for many years. The author theorized that this learning, as most other learning, takes place in very small “micro- steps”. Teachers are familiar with the solution to such problems and tend to gloss over the essential phase of setting up such problems. As indicated above, algebra textbooks also neglect this important aspect of solving such problems. Students need structure while they are learning this type of process, and this fact has been overlooked for too long in the pedagogy of such problems. The author then developed a highly-structured, systematic means of setting up such problems. Students quickly began to set up the problems correctly. Once students had the problems set up correctly, completing the arithmetic details of the solution was easy for them. The result was an almost perfect success rate for students working on ratio/proportion problems. A few simple math errors remain, but student success rates have been dramatically improved using this method. This paper details the “box method” and how it should be taught. Several examples are provided to illustrate the use of this method. Introduction Ratio/proportion problems are a key area of mathematics often used in science, engineering, and business. Conversions of any type of linear direct variation, such as feet to inches, pounds to ounces, etc., are essentially ratio/proportion problems. Very often, similar triangles provide opportunities for using ratio/proportion analysis to determine the unknown lengths of the sides of these triangles. And maps, blueprints, and photographs often serve as the basis for ratio/proportion problems. The arithmetic involved is usually minimal. In its simplest form, only one multiplication and one division are required. Yet students frequently have serious difficulties learning how to use such a fundamental tool. In fact, a 2007 report on standardized testing of mathematics found that when attempting to master standards for 8th grade coordinate geometry, “Students who are unsuccessful have the greatest difficulty with setting up and solving proportions from real-world

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Mla - latex bibitem, endnote - ris.

Sullivan, J. (2009, June), The “Box Method” For Teaching Ratio/Proportion Problems Paper presented at 2009 Annual Conference & Exposition, Austin, Texas. 10.18260/1-2--5803

ASEE holds the copyright on this document. It may be read by the public free of charge. Authors may archive their work on personal websites or in institutional repositories with the following citation: © 2009 American Society for Engineering Education. Other scholars may excerpt or quote from these materials with the same citation. When excerpting or quoting from Conference Proceedings, authors should, in addition to noting the ASEE copyright, list all the original authors and their institutions and name the host city of the conference. - Last updated April 1, 2015