proportion problem solving examples

• The Basic math blog
• Member Login
• Pre-Algebra Lessons
• Pre-Algebra Word Problems
• Pre-Algebra Calculators
• Algebra Lessons
• Algebra Word Problems
• Algebra Proofs
• Advanced Algebra
• Algebra Calculators
• Geometry Lessons
• Geometry Word Problems
• Geometry Proofs
• Geometry Calculators
• Trigonometry Lessons
• Numeration System
• Basic Concepts of Set Theory
• Consumer Math
• Baseball Math
• Math for Nurses
• Statistics Made Easy
• High School Physics
• Basic Mathematics Store
• Math Vocabulary Quizzes
• SAT Math Prep
• Math Skills by Grade Level
• Ask an Expert
• Other Websites
• K-12 Worksheets
• Worksheets Generator
• Algebra Worksheets
• Geometry Worksheets
• Fun Online Math Games
• Pre-Algebra Games
• Math Puzzles
• Math Tricks

Proportion word problems

It is very important to notice that if the ratio on the left is a ratio of number of liters of water to number of lemons, you have to do the same ratio on the right before you set them equal. 

More interesting proportion word problems

Check this site if you want to solve more proportion word problems.

Ratio word problems

Special Math Topics

Applied math.

About me :: Privacy policy :: Disclaimer :: Donate   Careers in mathematics  

Copyright © 2008-2021. Basic-mathematics.com. All right reserved

Ratio and Proportion Word Problems — Examples & Practice - Expii

Ratio and proportion word problems — examples & practice, explanations (4).

Ratio and Proportion Word Problems

Ratio and proportion word problems are really applicable to real life when you need to calculate costs, determine time limits, or even convert between measurement systems ! Here's a breakdown of what a ratio and proportion word problem might look like. We'll also see how to understand what the problem is asking, and how to solve it!

Image by Caroline Kulczycky

It's important to note that in Step Two, when you plug the appropriate values into the proportion, the units in both the numerators must be the same, and the units in both the denominators must be the same. For example, in the graphic above, the units in the numerators are books. The units in the denominators are hours. If the units given to you in the word problem are not consistent, you must somehow make them match. This next problem will show you how. For certain units we will use conversion factors to help make the units the same. We don't always have conversion factors, but it's good to know what ones we do have. Here are some examples:

• Units of length: convert between feet and meters
• Units of mass: convert between grams and kilograms
• Units of time: convert between minutes and days

Once that's done, the numerators will have the same units as each other. Likewise for the denominators. Then we'll be able to solve the proportion just using algebra. The equation will have a variable times a number on one side, and a number on the other side. Thus we can solve the equation in one step using division .

Practice Problem

Trisha can make 5 pairs of earrings in 1 week. Assuming she works at this constant rate, how many complete pairs of earrings can she make in 10 days?

Related Lessons

Math in Music: Rhythm

Now that the basic concepts of ratios and proportions have been covered, we are going to dive a little deeper into working with them through the subject of music.

Ratios are found in music within two main concepts: Rhythm/Polyrhythm and Pitch Intervals . These may seem a little foreign right now, but I promise they won't be too hard to understand. This explanation will focus on Rhythm and Polyrhythm, but the explanation on Pitch Intervals has been linked above.

Rhythm/Polyrhythm

All rhythms are referred to by names based off of fractions/ratios. There are whole notes (44 of a measure), half notes (24 of a measure), quarter notes (14 of a measure), eighth notes (18 of a measure), sixteenth notes (116 of a measure), etc. and the name denotes how much of a 4 beat measure the note takes up, as shown below. For our purposes, we are always going to use a 4 beat measure to describe our ratios.

In order, Whole Note, Half Note, Quarter Note, Eighth Note, Sixteenth Note, all taking up the same amount of time

Not all notes and rhythms are based on groups of 4, notes can be based on any division! Triplets can be used to split a four beat measure into groups of 3 or 6 instead of groups of 2 or 4.

Three half note triplets take up a measure, 6 quarter note triplets, 12 eighth note triplets, etc.

The brackets over the notes tell how many notes are played in the span of the bracket. You can see the 3 placed above all of the notes. This is because each group of 3 takes a set amount of time. The half note triplets take up 4 beats (one full measure) with 3 notes, the quarter note triplets take up 2 beats (one half of a measure) with 3 notes, and the eighth note triplets take up 1 beat (one quarter of a measure) with 3 notes. As ratios, these could be written as 3:4, 3:2, and 3:1. If we wanted to write these notes as fractions, they could be called 13, 16, and 112 notes.

When we start combining these rhythms to be played at the same time, we get what are called polyrhythms. These are normally defined by ratio of how many notes are being played in the same amount of time.

The ratios in this case are going to be written in terms of the top line against the bottom line. The first measure is a 4:3 while the second measure is a 3:2. These same polyrhythms can be written without needing a second line to compare by once again using the brackets above the notes.

The groupings on the top line both take up an entire measure, 4 beats. That means that the first measure on the top line shows a 5:4 ratio (aka five-lets), and the second shows a 7:4 (aka seven-lets) ratio. The bottom groupings take up 2 beats , so they show a 5:2 and 7:2 ratio.

We're gonna try a practice question that builds on some of these concepts.

Brandon is playing a trombone solo that has 24 evenly spaced notes. How many evenly spaced notes would Libby need to play in her counter melody to make the entire section a 4:3 polyryhthm?

(Video) Proportion Examples and Word Problems

by Math Meeting

This video by Math Meeting explains proportions and works through a few examples.

A proportion is two ratios that are set equal to one another.

An example of a proportion would be: 1 dollar10 pesos=10 dollars100 pesos As you can see, they are equivalent. The units are the same, and then we notice the fractions are equivalent . The fractions are equivalent because 10100 can be reduced to 110.

If you multiply the numerator and denominator of the first ratio by 10, you get the second ratio. If you have 10 times more dollars, you will have 10 times more pesos.

The standard proportion problem is as follows: 5x=103 You are given two proportions and you want to solve for x.

The best way to solve these types of equations is with cross multiplication .

To cross multiply, you take the product of the numerator of the first ratio and the denominator of the second ratio. Set that equal to the product of the numerator of the second ratio and the denominator of the first ratio.

Let's solve this problem now. 5x=103⇒5⋅3=x⋅1015=10x1510=10x10315210=10x1032=x

Another Example

Here's a word problem using proportions.

The currency in Spain is the euro. If I exchange $250 at the airport, how many euros will I get in return? (1 euro = 1.38 dollars)

The parentheses at the bottom give us our first ratio. 1 euro1.38 dollars

Within the question itself is our second ratio. x euros250 dollars

Set them equal to each other. Remember that the same units have to be in the top/bottom. 11.38=x250

To solve, we cross multiply. 11.38=x250⇒1⋅250=1.38⋅x250=1.38x2501.38=1.38x1.382501.38=1.38x1.38181.16=x

You will get approximately 181.16 euros!

Math in Music: Pitch

Ratios are found in music within two main concepts: Rhythm/Polyrhythm and Pitch Intervals . These may seem a little foreign right now, but I promise they won't be too hard to understand. This explanation is going to cover Pitch and Intervals, and Rhythm/Polyrhythm is linked above.

Pitch and Intervals

When two notes are being played at once, there is an Interval between the two pitches . Every pitch can be written as both a note (think A,B,C, and their placings on a staff) or as a measurement of Hertz (Hz). We are going to use A=440 Hz as a baseline to demonstrate how ratios are used in intervals and tuning in music.

As notes move in octaves, there is always a 2:1 ratio between the notes. Based on A at 440 Hz, the A one octave higher is going to be measured at 880 Hz, and the A one octave lower is going to be measured at 220 Hz.

This pattern continues in both directions (up and down) from A in the second space until eventually the note is outside of the range of human hearing.

The ratios between notes exist for every interval and are the basis for tuning many instruments as well as playing in tune with another musician. For example, were going to use the next two most common intervals, the perfect fifth and the major third. Using A=440 Hz as a basis again, we are going to use the intervals of 3:2 to find the measurement of the perfect fifth (E), and a ratio of 5:4 to find the major third (C# or C Sharp).

In this case we are going to put 440 as a denominator because we want to find the Hz of a note that is higher than our original A.

32=x4402x=1320x=660

Our perfect fifth of E will be at 660 Hz when it is perfectly in tune. Now lets calculate the pitch of the major third, C#. Once again, the 440 is going to be in the denominator as we are finding a note higher than A.

54=x4404x=2220x=550

Our major third of C# is going to be at 550 Hz when perfectly in tune. Combining all of these notes, you get a nice consonant A major chord. Just to make sure our musical brains are keeping up with the math, here is what this looks like on a staff.

Let's try a practice problem to see if everything is sticking.

Brandon and Libby are trying to practice their duet. If Brandon is playing a low A at 220 Hz, and Libby wants to play a perfectly in tune F# that is one octave and a major sixth (5:3) higher, what pitch should she play?

366.67 Hz

733.33 Hz

440 Hz

550 Hz

Worked out Problems on Ratio and Proportion

Worked out problems on ratio and proportion are explained here in detailed description using step-by-step procedure. Solved examples involving different questions related to comparison of ratios in ascending order or descending order, simplification of ratios and also word problems on ratio proportion. Sample questions and answers are given below in the worked out problems on ratio and proportion to get the basic concepts of solving ratio proportion.

1. Arrange the following ratios in descending order. 

        2 : 3, 3 : 4, 5 : 6, 1 : 5  Solution:   Given ratios are 2/3, 3/4, 5/6, 1/5  The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60 

Now, 2/3 = (2 × 20)/(3 × 20) = 40/60           3/4 = (3 × 15)/(4 × 15) = 45/60           5/6 = (5 × 10)/(6 × 10) = 50/60           1/5 = (1 × 12)/(5 × 12) = 12/60  Clearly, 50/60 > 45/60 > 40/60 > 12/60  Therefore, 5/6 > 3/4 > 2/3 > 1/5  So, 5 : 6 > 3 : 4 > 2 : 3 > 1 : 5

2. Two numbers are in the ratio 3 : 4. If the sum of numbers is 63, find the numbers. Solution: Sum of the terms of the ratio = 3 + 4 = 7 Sum of numbers = 63 Therefore, first number = 3/7 × 63 = 27 Second number = 4/7 × 63 = 36 Therefore, the two numbers are 27 and 36.

3. If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y) Solution: x : y = 1 : 2 means x/y = 1/2 Now, (2x + 3y) : (x + 4y) = (2x + 3y)/(x + 4y) [Divide numerator and denominator by y.] = [(2x + 3y)/y]/[(x + 4y)/2] = [2(x/y) + 3]/[(x/y) + 4], put x/y = 1/2 We get = [2 (1/2) + 3)/(1/2 + 4) = (1 + 3)/[(1 + 8)/2] = 4/(9/2) = 4/1 × 2/9 = 8/9 Therefore the value of (2x + 3y) : (x + 4y) = 8 : 9

More solved problems on ratio and proportion are explained here with full description.

4.  A bag contains $510 in the form of 50 p, 25 p and 20 p coins in the ratio 2 : 3 : 4. Find the number of coins of each type. 

Solution:   Let the number of 50 p, 25 p and 20 p coins be 2x, 3x and 4x.  Then 2x × 50/100 + 3x × 25/100 + 4x × 20/100 = 510 x/1 + 3x/4 + 4x/5 = 510 (20x + 15x + 16x)/20 = 510  ⇒ 51x/20 = 510 x = (510 × 20)/51  x = 200 2x = 2 × 200 = 400  3x = 3 × 200 = 600  4x = 4 × 200 = 800.  Therefore, number of 50 p coins, 25 p coins and 20 p coins are 400, 600, 800 respectively. 

5. If 2A = 3B = 4C, find A : B : C Solution: Let 2A = 3B = 4C = x So, A = x/2 B = x/3 C = x/4 The L.C.M of 2, 3 and 4 is 12 Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12 = 6x : 4x : 3x = 6 : 4 : 3 Therefore, A : B : C = 6 : 4 : 3

6. What must be added to each term of the ratio 2 : 3, so that it may become equal to 4 : 5? Solution: Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5 ⇒ (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 2

7. The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now? Solution: Length of ribbon originally = 30 cm Let the original length be 5x and reduced length be 3x. But 5x = 30 cm x = 30/5 cm = 6 cm Therefore, reduced length = 3 cm = 3 × 6 cm = 18 cm

More worked out problems on ratio and proportion are explained here step-by-step. 8. Mother divided the money among Ron, Sam and Maria in the ratio 2 : 3 : 5. If Maria got $150, find the total amount and the money received by Ron and Sam. Solution: Let the money received by Ron, Sam and Maria be 2x, 3x, 5x respectively. Given that Maria has got $ 150. Therefore, 5x = 150 or, x = 150/5 or, x = 30 So, Ron got = 2x                    = $ 2 × 30 = $60 Sam got = 3x               = 3 × 60 = $90

Therefore, the total amount $(60 + 90 + 150) = $300 

9. Divide $370 into three parts such that second part is 1/4 of the third part and the ratio between the first and the third part is 3 : 5. Find each part. Solution: Let the first and the third parts be 3x and 5x. Second part = 1/4 of third part.                     = (1/4) × 5x                     = 5x/4 Therefore, 3x + (5x/4) + 5x = 370 (12x + 5x + 20x)/4 = 370 37x/4 = 370 x = (370 × 4)/37 x = 10 × 4 x = 40 Therefore, first part = 3x                                 = 3 × 40                                 = $120 Second part = 5x/4                     = 5 × 40/4                     = $50 Third part = 5x                  = 5 × 40                  = $ 200

10. The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term. Solution: Let the fourth term be x. Thus 42, 36, 35, x are in proportion. Product of extreme terms = 42 ×x Product of mean terms = 36 X 35 Since, the numbers make up a proportion Therefore, 42 × x = 36 × 35 or, x = (36 × 35)/42 or, x = 30 Therefore, the fourth term of the proportion is 30.

More worked out problems on ratio and proportion using step-by-step explanation. 11. Set up all possible proportions from the numbers 8, 12, 20, 30. Solution: We note that 8 × 30 = 240 and 12 × 20 = 240 Thus, 8 × 30 = 12 × 20       ………..(I) Hence, 8 : 12 = 20 : 30       ……….. (i) We also note that, 8 × 30 = 20 × 12 Hence, 8 : 20 = 12 : 30       ……….. (ii) (I) can also be written as 12 × 20 = 8 × 30 Hence, 12 : 8 = 30 : 20       ……….. (iii) Last (I) can also be written as 12 : 30 = 8 : 20       ……….. (iv) Thus, the required proportions are 8 : 12 = 20 : 30 8 : 20 = 12 : 30     12 : 8 = 30 : 20     12 : 30 = 8 : 20

12. The ratio of number of boys and girls is 4 : 3. If there are 18 girls in a class, find the number of boys in the class and the total number of students in the class. Solution: Number of girls in the class = 18 Ratio of boys and girls = 4 : 3 According to the question, Boys/Girls = 4/5 Boys/18 = 4/5 Boys = (4 × 18)/3 = 24 Therefore, total number of students = 24 + 18 = 42.

13. Find the third proportional of 16 and 20. Solution: Let the third proportional of 16 and 20 be x. Then 16, 20, x are in proportion. This means 16 : 20 = 20 : x So, 16 × x = 20 × 20 x = (20 × 20)/16 = 25 Therefore, the third proportional of 16 and 20 is 25.

●   Ratio and Proportion

What is Ratio and Proportion?

Practice Test on Ratio and Proportion

●   Ratio and Proportion - Worksheets

Worksheet on Ratio and Proportion

8th Grade Math Practice From Worked out Problems on Ratio and Proportion to HOME PAGE

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

New! Comments

Share this page: What’s this?

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

2nd Grade Measurement Worksheet | Measuring Length, Mass and Volume

Nov 20, 24 12:50 AM

2nd Grade Fractions Worksheet | Basic Concept of Fractions | Answers

Nov 20, 24 12:16 AM

2nd Grade Math Practice | Second Grade Math |2nd Grade Math Worksheets

Nov 18, 24 02:23 PM

Worksheet on Addition of Length | Word Problems on Addition of Length

Nov 17, 24 10:29 PM

Worksheet on Addition of Mass | Word problems on Addition of Mass

Nov 17, 24 10:26 PM

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

Algebra: Ratio Word Problems

Related Pages Two-Term Ratio Word Problems More Ratio Word Problems Algebra Lessons

In these lessons, we will learn how to solve ratio word problems that have two-term ratios or three-term ratios.

Ratio problems are word problems that use ratios to relate the different items in the question.

The main things to be aware about for ratio problems are:

• Change the quantities to the same unit if necessary.
• Write the items in the ratio as a fraction .
• Make sure that you have the same items in the numerator and denominator.

Printable Introduction to Ratios

Ratio Word Problems using Tape diagrams Ratio Problems given One of Quantities Ratio Problems given Total Ratio Problems given Difference Ratio Problems with Change of Ratio

Online Ratio of Quantities Ratio Word Problems I Ratio Word Problems II Ratio Word Problems III Ratio Word Problems IV Ratio with Algebra

Ratio Problems: Two-Term Ratios

Example 1: In a bag of red and green sweets, the ratio of red sweets to green sweets is 3:4. If the bag contains 120 green sweets, how many red sweets are there?

Solution: Step 1: Assign variables: Let x = number of red sweets.

Step 2: Solve the equation. Cross Multiply 3 × 120 = 4 × x 360 = 4 x

Answer: There are 90 red sweets.

Example 2: John has 30 marbles, 18 of which are red and 12 of which are blue. Jane has 20 marbles, all of them either red or blue. If the ratio of the red marbles to the blue marbles is the same for both John and Jane, then John has how many more blue marbles than Jane?

Solution: Step 1: Sentence: Jane has 20 marbles, all of them either red or blue. Assign variables: Let x = number of blue marbles for Jane 20 – x = number red marbles for Jane

Step 2: Solve the equation

Cross Multiply 3 × x = 2 × (20 – x ) 3 x = 40 – 2 x

John has 12 blue marbles. So, he has 12 – 8 = 4 more blue marbles than Jane.

Answer: John has 4 more blue marbles than Jane.

How To Solve Word Problems Using Proportions?

This is another word problem that involves ratio or proportion.

Example: A recipe uses 5 cups of flour for every 2 cups of sugar. If I want to make a recipe using 8 cups of flour. How much sugar should I use?

How To Solve Proportion Word Problems?

When solving proportion word problems remember to have like units in the numerator and denominator of each ratio in the proportion.

• Biologist tagged 900 rabbits in Bryer Lake National Park. At a later date, they found 6 tagged rabbits in a sample of 2000. Estimate the total number of rabbits in Bryer Lake National Park.
• Mel fills his gas tank up with 6 gallons of premium unleaded gas for a cost of $26.58. How much would it costs to fill an 18 gallon tank? 3 If 4 US dollars can be exchanged for 1.75 Euros, how many Euros can be obtained for 144 US dollars?

Ratio problems: Three-term Ratios

Example 1: A special cereal mixture contains rice, wheat and corn in the ratio of 2:3:5. If a bag of the mixture contains 3 pounds of rice, how much corn does it contain?

Solution: Step 1: Assign variables: Let x = amount of corn

Step 2: Solve the equation Cross Multiply 2 × x = 3 × 5 2 x = 15

Answer: The mixture contains 7.5 pounds of corn.

Example 2: Clothing store A sells T-shirts in only three colors: red, blue and green. The colors are in the ratio of 3 to 4 to 5. If the store has 20 blue T-shirts, how many T-shirts does it have altogether?

Solution: Step 1: Assign variables: Let x = number of red shirts and y = number of green shirts

Step 2: Solve the equation Cross Multiply 3 × 20 = x × 4 60 = 4 x x = 15

5 × 20 = y × 4 100 = 4 y y = 25

The total number of shirts would be 15 + 25 + 20 = 60

Answer: There are 60 shirts.

Algebra And Ratios With Three Terms

Let’s study how algebra can help us think about ratios with more than two terms.

Example: There are a total of 42 computers. Each computer runs one of three operating systems: OSX, Windows, Linux. The ratio of the computers running OSX, Windows, Linux is 2:5:7. Find the number of computers that are running each of the operating systems.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Home / United States / Math Classes / 7th Grade Math / Writing and Solving Proportions

Writing and Solving Proportions

A proportion is an equation that states that two ratios are equivalent. We can perform operations on proportions, just l ike we do with normal equations. Here we will learn to perform operations on proportions and the steps involved in solving them. ...Read More Read Less

About Proportions in Math

What are Proportions?

Writing proportions as fractions, using operations to solve proportions, solved examples.

• Frequently Asked Questions

Proportions are mathematical equations that are used to relate equivalent ratios. Two ratios having different antecedents and consequents can have the same value. This relation can be expressed with the help of proportions. Let us consider the ratios \( 2:5 \)  and \( 8:20 \) . When we simplify the ratio \( 8:20 \) , we obtain \( 2:5 \) . This implies that \( 2:5 \)  and \( 8:20 \)  are equivalent ratios. 

In other words, these ratios are in proportion. If \( a:b \)  and \( c:d \) are equivalent ratios, we can express the relation as \( a:b::c:d \) , where the ‘ \( ~::~ \) ’ sign is used to express proportion. So, we can state the proportion in the example that was just observed as, \( 2:5 :: 8:20 \) .

We have learned that a ratio is a comparison of two quantities having the same unit. We also compare two quantities using fractions. Hence, a ratio can also be written as a fraction. For example, we can write the ratio \( a:b \)  as \( \frac{a}{b} \) and \( c:d \)  as \( \frac{c}{d} \) .

Similarly, we can write a proportion as a fraction. Instead of writing \( 2:5::8:20 \) , we can write the proportion as \( \frac{2}{5} = \frac{8}{20} \)   , and this makes it easier to perform operations on proportions to solve for unknown values.

Since a proportion is basically an equation, we can perform operations on them to find unknown values. We can use operations like addition, subtraction, multiplication, and division to solve a proportion. In most cases, we only need to use multiplication and division. Let’s consider a proportion in which one of the values is unknown. 

For example, \( \frac{5}{8} = \frac{x}{40} \)

Use basic math operations to solve this equation. Begin by removing the denominator from both sides.

\( \frac{5 \times 8}{8} = \frac{x \times 8}{40} \)            [Multiply both sides by \( 8 \)]

\( 5 = \frac{x}{5} \)                    [Simplify]

\( 5 \times 5 = \frac{x\times 5}{5} \)          [Multiply both sides by \( 5 \)]

\( 25 = x \)                   [ Simplify]      

Hence, the value of \( x \) is \( 25 \). 

Similarly, we can use a combination of mathematical operations to solve proportions.

Example 1: Use math operations to find the value of \( x \) in the expression, \( \frac{3}{7} = \frac{x}{28} \) .

Solution:  

To find the value of \( x \) , simplify the equation.

\( \frac{3}{7} = \frac{x}{28} \)                    [Write the equation]

\( \frac{3 \times 7}{7} = \frac{x \times 7}{28} \)              [Multiply both sides by \( 7 \)]

\( 3 = \frac{x}{4} \)                      [Simplify]

\( 3 \times 4 = \frac{x \times 4}{4} \)            [Multiply both sides by \( 4 \)]

\( 12 = x \)                    [Simplify]

So, the value of \( x \) is \( 12 \).

Example 2: Solve the proportion to find the unknown value: \( 15:y :: 25:55 \) .

The proportion is \( 15:y :: 25:55 \) and this expression can also be written as \( \frac{15}{y} = \frac{25}{55} \)

To find the value of \( y \) , simplify the equation.

\( \frac{15}{y} = \frac{25}{55} \)                   [Write the proportion]

\( \frac{y}{15} = \frac{55}{25} \)                    [Taking reciprocal of both sides]

\( \frac{y \times 15}{15} = \frac{55 \times 15}{25} \)           [Multiplying both sides by \( 15 \)]

\( y = \frac{11 \times 15}{55} \)                  [Simplify]

\( y = 11 \times 3 \)                [Simplify]

\( y = 33 \)                      [Multiply]

Hence, the unknown value, \( y \) is \( 33 \).

Example 3: An athlete can run \( 100 \) meters in \( 11 \) seconds. If she runs at a constant pace, how long will she take to run \( 800 \) meters?

Time taken by the athlete to cover \( 100 \) meters \( = 11 \) seconds

Let us assume the time taken by the athlete to cover \( 800 \) meters \( = x \) seconds

Since her speed is constant, the ratio of distance and time in both cases is in proportion. 

\( \frac{100}{11} = \frac{800}{x} \)                      [Write the above condition in proportion]

\( \frac{11}{100} = \frac{x}{800} \)                      [Taking reciprocal of both sides]

\( \frac{11 \times 100}{100} = \frac{x \times 100}{800} \)              [Multiplying both sides by \( 100 \)]

\( 11 = \frac{x}{8} \)                          [Simplify]

\( x = 88 \)                            [Multiplying both sides by \( 8 \)]

Hence, the athlete will take \( 88 \) seconds to run \( 800 \) meters.

Are ratios related to proportions?

Yes, a proportion is an equation that states that two ratios are equivalent. So, we can say that proportions are directly related to ratios.

Can we perform mathematical operations on proportions?

Since proportions are basically mathematical equations, we can perform all the mathematical operations on them, just like we do with a normal mathematical equation.

How do we solve a proportion?

We can solve a proportion to find the unknown value by performing mathematical operations on them. The goal is to isolate the unknown value on one side of the equation. Thus, by solving the equation, we will get the value of the unknown on the other side of the equation.

Check out our other courses

Grades 1 - 12

Level 1 - 10

GCSE Tutoring Programme

Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring.

In order to access this I need to be confident with:

This topic is relevant for:

Here we will learn about proportion in maths, including direct and inverse proportion, real life examples, worded problems, and graphical representations. 

There are also proportion in maths worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is proportion?

Proportion is a type of relationship between two variables linked by a constant.

There are two types of proportion, direct proportion and inverse proportion . They can also be referred to as direct variation and inverse variation.

When solving problems involving proportion it is important to know which type of proportion that you are dealing with, direct proportion or inverse proportion.

Step-by-step guide: Direct and indirect proportion

• Direct proportion

If there is a directly proportional relationship between two variables then as one variable increases, so does the other.

For example, An apple costs 60p. As the number of apples increases, so does the cost.

The relationship between the number of apples and the cost of the apples is directly proportional.

The constant in this example is 60.

Step-by-step guide: Direct proportion

• Inverse proportion

If there is an inversely proportional relationship between two variables then as one variable increases, the other variable decreases.

For example, 1 worker takes 30 days to paint a long fence. As the number of workers increases, the time it takes to paint the fence decreases. 

The relationship between the number of workers and the time taken to paint the fence is inversely proportional.

The constant in this example is 30.

Step-by-step guide: Inverse proportion

Solve ratio and proportion problems

Sometimes a proportion problem may also involve ratio.

For example,

A cafe sells apples and bananas in the ratio 2:3. They sell 35 pieces of fruit altogether. Apples cost 30p and bananas cost 45p. Work out the total amount of money the fruit cost.

So, first you use the ratio to work out how many apples and bananas were sold. The ratio 2:3 has 5 parts, so you divide 35 by 5 to get 7. Then you multiply each part of the ratio by 7. This means that 14 apples and 21 bananas were sold.

You can then use direct proportion to find the price of the apples and the price of the bananas.

Apples: 14\times 30=420

Bananas: 21\times 45=945

We add these together to get the total price for all the fruit.

Therefore the total price is 1365p or £13.65.

Notation for proportional relationships

This symbol ‘\bf{\propto}’ means ‘is proportional to’.

If y is directly proportional to x you can write it as y \propto x.

If y is inversely proportional to x you can write it as y \propto \frac{1}{x}.

Proportion formulas and the constant of proportionality

Proportion is a type of relationship between two variables linked by a constant. This is known as the constant of proportionality and is usually denoted by the letter ‘k’. 

Equations of proportion:

• The direct proportion formula is, y=kx .

Step-by-step guide: Direct proportion formula

• The inverse proportion formula is, y=\frac{k}{x}.

Step-by-step guide: Inverse proportion formula

Graphs of proportional relationships

Proportional relationships can be represented graphically.

For direct proportion, as one variable increases the other variable increases.

Therefore, the graph will go up as you look from left to right. The graph can be a straight line or a curve. All direct proportion graphs must start at the origin. 

If y is directly proportional to x then the graph will be a straight line.

If y is directly proportional to a function of x such as x^{2} or \sqrt{x} then the graph will form a curve.

For inverse proportion , as one variable increases the other variable decreases.

Therefore, the graph will go down as you look from left to right. All inverse proportion graphs form a curved shape like in the example below.

Whether y is inversely proportional to just x or a function of x it will still form a curved graph with this general shape.

Step-by-step guide: Directly proportional graph / inversely proportional graph

Types of proportion

How to use proportion in maths

In order to answer general proportion in maths problems:

• Identify if the relationship between the variables in the question is (a) directly proportional. (b) inversely proportional.
• (a) By using division, find the constant. (b) By using multiplication, find the constant.
• (a) Multiply the constant by the required value of one variable to find the answer for the other variable. (b) Divide the constant by the required value of one variable to find the answer for the other variable.

Explain how to use proportion

Proportion worksheet

Get your free proportion worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Proportion examples

Example 1: worded problem.

If 5 biscuits cost 80p, find the cost of 7 biscuits.

As the number of biscuits increases the cost will also increase. Therefore the variables are directly proportional.

2 (a) By using division, find the constant. (b) By using multiplication, find the constant. 

1 biscuit will cost 16p.

3 (a)  Multiply the constant by the required value of one variable to find the answer for the other variable. (b)  Divide the constant by the required value of one variable to find the answer for the other variable.

7 biscuits will cost 112p or £1.12.

You can also visualise these steps in a table.

Example 2: worded problem

If 6 bags of flour weigh 15 \ kg, find the weight of 17 bags of flour.

Identify if the relationship between the variables in the question is (a) directly proportional.  (b) inversely proportional.

As the number of bags of flour increases the weight will also increase. Therefore the variables are directly proportional.

(a) By using division, find the constant.  (b) By using multiplication, find the constant.

1 bag of flour will weigh 2.5kg.

(a)  Multiply the constant by the required value of one variable to find the answer for the other variable. (b)  Divide the constant by the required value of one variable to find the answer for the other variable.

17 bags of flour will weigh 42.5kg.

Example 3: worded problem

If 4 taps pumping water at a constant rate take 9 hours to fill a swimming pool, find the time it would take 3 taps pumping water at the same rate to fill the swimming pool.

As the number of taps increases the time it takes to fill the swimming pool decreases. Therefore the variables are inversely proportional.

It would take 1 tap 36 hours to fill the pool.

3 taps would take 12 hours to fill the pool.

Example 4: worded problem

A local park has a course marked out for runners to use. One runner completes the course in 30 minutes running at a speed of 8 \ km/h.  

Another runner has a pace of 12km/h. How long will it take this runner to complete the course?

As the speed increases, the time taken to complete the course decreases. Therefore the variables are inversely proportional.

The constant for the course is 240.

If your speed was 1km/hour it would take 240 minutes.

It would take the runner 20 minutes to complete the course.

Note that this question can also be answered using the formula for speed.

Step-by-step guide: Speed distance time

Calculations with proportions

In order to answer problems involving a proportion formula:

Substitute in the given values and rearrange the resulting equation to determine the value of \textbf{k} .

Substitute the value of \textbf{k} into the formula written down in step 1.

Example 5: writing a proportion formula

y is directly proportional to x. 

When y=12, \ x=3.  

Find a formula for y in terms of x.

Identify the type of proportion and write down the appropriate formula. 

Example 6: writing a proportion formula

When a=6, \ b = 5.  

Find a formula for a in terms of b.

a = \frac{k}{b}

Common misconceptions

• Assuming worded problems reflect real life

Worded proportion problems can often imply things that are not reflective of real life. For example, It takes 3 workers 2 days to build a wall, how long does it take 4 workers to build the same wall? This question makes an assumption that all workers work the same hours each day and are equally efficient. This is not reflective of real life, but it is something we assume in order to model the situation mathematically.

• Assuming the unitary method is the only/best way to solve a proportion problem

There may be several ways to get to the correct answer for proportion questions. Some ways are more efficient than others depending on the numbers involved. You can always follow the step by step guide, however it is good practice to think independently about each question and consider other methods to develop your problem solving skills.

• Misinterpreting decimal answers

If the solution to a proportion problem is a decimal, think carefully about what this represents in the context of the question before writing your final answer.  For example, 5.3 pounds should be written as £5.30. 2.5 hours could be written as 2 hours 30 minutes.

Practice proportion questions

1. If 8 batteries cost £4.80, find the cost of 5 batteries.

This is a direct proportion problem. 

The cost of one battery would be,

The cost of 5 batteries would be,

2. If 3 paving slabs weigh 18 \ kg, find the weight of 7 slabs.

The weight of one slab would be,

The weight of 7 slabs would be,

3. If a taxi journey is shared by 4 people it costs £24 each. What does each person pay if there are only 3 people sharing the taxi for the same journey?

This is an inverse proportion problem. 

The cost of the whole taxi journey would be,

4\times 24=96 pounds.

Sharing this cost between 3 people would be,

4. If it takes 5 workers 12 hours to do a job, how long would it take 9 workers to complete the same job? Give your answer in hours and minutes.

3 hours 45 minutes

21 hours 6 minutes

6 hours 66 minutes

6 hours 40 minutes

The total time needed to complete the job would be,

5\times 12=60 hours.

Sharing this time between 9 workers would be,

Two thirds of an hour is 40 minutes, 60 \times \frac{2}{3} = 40 .

The final answer would be 6 hours 40 minutes.

5. Which of these equations indicates that y is directly proportional to x?

The correct equation will be in the form y=kx, where k is the constant of proportionality.

The correct equation will be, y=5x.

6. Which of these graphs indicates that y \propto x?

y \propto x means y is proportional to x. Therefore the graph should show that as x increases, y increases.

All direct proportion graphs go through the origin (0,0). As y is proportional to x (and not x^{2} \text{or} \sqrt{x} for example) the graph will be a straight line.

7. y is inversely proportional to x. When y=20, \ x=4. Find a formula for y in terms of x.

Start with the general formula for inverse proportion y = \frac{k}{x}.

Now substitute in the values of x and y and rearrange the resulting equation to find k.

\begin{aligned} 20 &= \frac{k}{4} \\\\ 20 \times 4 &= k \\\\ 80&=k \end{aligned} Finally, substitute this value of k into the general formula to get the final answer

Proportion GCSE questions

1. A recipe to make enough garlic bread for 5 people needs 75 \ g of butter.

Calculate the amount of butter needed to make garlic bread for 12 people.

2. y is inversely proportional to x.

Complete the table.

3. The graphs show different relationships between the variables x and y.

(a) Identify the graph which shows that y\propto x .

(b) Identify the graph which shows that y\propto \frac{1}{x} .

(a) Graph D

(b) Graph A

Learning checklist

You have now learned how to:

• Solve real world problems involving direct and inverse proportion
• Recognise formula and graphs for direct and inverse proportions

The next lessons are

• Compound measures
• Best buy maths

Still stuck?

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.

Privacy Overview