ratio problem solving questions year 8

Find here an unlimited supply of worksheets with simple word problems involving ratios, meant for 6th-8th grade math. In , the problems ask for a specific ratio (such as, " "). In , the problems are the same but the ratios are supposed to be simplified.

contains varied word problems, similar to these:

Options include choosing the number of problems, the amount of workspace, font size, a border around each problem, and more. The worksheets can be generated as PDF or html files.

Each worksheet is randomly generated and thus unique. The and is placed on the second page of the file.

You can generate the worksheets — both are easy to print. To get the PDF worksheet, simply push the button titled " " or " ". To get the worksheet in html format, push the button " " or " ". This has the advantage that you can save the worksheet directly from your browser (choose File → Save) and then in Word or other word processing program.

Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options:

Use the generator to make customized ratio worksheets. Experiment with the options to see what their effect is.

Primary Grade Challenge Math by Edward Zaccaro

A good book on problem solving with very varied word problems and strategies on how to solve problems. Includes chapters on: Sequences, Problem-solving, Money, Percents, Algebraic Thinking, Negative Numbers, Logic, Ratios, Probability, Measurements, Fractions, Division. Each chapter’s questions are broken down into four levels: easy, somewhat challenging, challenging, and very challenging.

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Yr 8 Rates and Ratio Quizzes (A, B, C, D)

Resource Type: Quiz, ABQuiz

Four similar quizzes (quizubes), A, B, C and D.

Contents: simplifying ratios, writing ratios, simplifying ratios involving mixed units of measurement, completing equivalent ratios, solving problems involving rates, simplifying rates, scale.

Includes original file that can be modified.

8 August 2013 Edit: 28 February 2014

Shared by Simon Borgert     Bellingen, New South Wales

Click thumbnails for selected previews.

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Key Stage 3 Maths - Lesson Objectives, Keywords and Resources - Year 8 - Number

• Cut-the-knot
• Curriculum Online
• Starter of the Day

Lesson Objectives

To be able to:

• Reduce a ratio to its simplest form, including a ratio expressed in different units, recognising links with fraction notation;
• Divide a quantity into two or more parts in a given ratio
• Use the unitary method to solve simple word problems involving ratio and direct proportion.
• Compare two ratios.

ratio, unitary method, percentage, fraction, conversion

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Welcome to our Ratio Word Problems page. Here you will find our range of 6th Grade Ratio Problem worksheets which will help your child apply and practice their Math skills to solve a range of ratio problems.

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Here you will find a range of problem solving worksheets about ratio.

The sheets involve using and applying knowledge to ratios to solve problems.

The sheets have been put in order of difficulty, with the easiest first. They are aimed at students in 6th grade.

Each problem sheet comes complete with an answer sheet.

Using these sheets will help your child to:

• apply their ratio skills;
• apply their knowledge of fractions;
• solve a range of word problems.
• Ratio Problems 1
• PDF version
• Ratio Problems 2
• Ratio Problems 3
• Ratio Problems 4

Ratio and Probability Problems

• Ration and Probability Problems 1
• Sheet 1 Answers
• Ration and Probability Problems 2
• Sheet 2 Answers

More Recommended Math Worksheets

Take a look at some more of our worksheets similar to these.

More Ratio & Unit Rate Worksheets

These sheets are a great way to introduce ratio of one object to another using visual aids.

The sheets in this section are at a more basic level than those on this page.

We also have some ratio and proportion worksheets to help learn these interrelated concepts.

• Ratio Part to Part Worksheets
• Ratio and Proportion Worksheets
• Unit Rate Problems 6th Grade

6th Grade Percentage Worksheets

Take a look at our percentage worksheets for finding the percentage of a number or money amount.

We have a range of percentage sheets from quite a basic level to much harder.

• Percentage of Numbers Worksheets
• Money Percentage Worksheets
• 6th Grade Percent Word Problems

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Part 5: Rates and Ratios

Guide Chapters

• 2. Introduction to linear relationships
• 3. Properties of linear relations
• 4. Cartesian Plane
• 5. Rates and ratio
• 6. Planar transformations
• 7. Volume and capacity
• 8. Similarity

Are you confident with your Rates and Ratios skills? In this guide, we will show you how to approach different problem questions involving rates and ratios and help you investigate, interpret and analyse graphs.

NSW Syllabus outcome

In this article, we’ll discuss:

Assumed knowledge for rates and ratios.

• Applying the unitary method
• Real-life problems: speed
• Converting rates between different units
• Distance-time (travel) graph
• Checkpoint Questions

Students should understand how to write, interpret and simplify simple ratios.

This guide aims to enrich these skills with techniques and advice that Matrix students learn.

Applying the Unitary Method

A rate is a comparison of quantities which are measured in different units.

I.e. a rate can be thought of as:

First Unit / Second Unit

The unitary method is often used to apply rates in problems and involves writing the first unit as per ONE of the second unit.

This ‘ simplified rate’ can then be then applied as required by the question.

The steps for using this method is as follows:

Step 1: Remove any fractions or decimals

In order to simplify the rate, it is important to first work in integers.

If the rate involves decimals, multiply both parts by an appropriate power of \( 10 \).

For example, the rate is given as:

\begin{align*} 0.333m /  \ 4s \end{align*}

We identify that only the \(LHS\) is a decimal.

To change this decimal into an integer, we must multiply by \(1000\) (i.e. \(10^3\)).

However, it is important to remember that we must do the required operations to both sides of the ratio .

Then we will have:

\begin{align*} 333 \ m /  \ 4000s \end{align*}

Now we are working with integers.

If we were given the rate in fractions, we would instead multiply both sides by the  LCM of the denominators.

For example:

\begin{align*} \frac{2}{3} L \ / \ \frac{3}{8}s \end{align*}

Here we can see that the LCM of the denominators is \(24\).

Therefore, we multiply both sides of the fraction by \(24\) to get:

\begin{align*} 16L \ / \ 9s \end{align*}

Again, we are now working with integers.

Step 2. Divide both sides by the second unit

The next step is to divide both sides by the second unit.

This will then give us the first unit per ONE of the second unit.

Here the second unit has a value of \( 9 \).

Therefore, both sides should be divided by \( 9 \):

\begin{align*} \frac{16}{9} L \ / \ 1s \end{align*}

Step 3: Apply the simplified rate

The last step is to then apply this simplified rate to the question.

This will often involve multiplying both sides by a certain number in order to obtain a scaled rate which directly applies to the question.

1. How many litres of water will flow into a tank after \( 32 \) seconds, given that the water flows in at a rate of \( 16 L \ / \ 9 s \)?

To solve this question, we would use our simplified rate from before:

To find out how much water flows in after \(32 \ seconds\), we would need multiply both \(LHS\) and \(RHS\) by \(32 \ seconds\).

\begin{align*} \frac{512}{9}L \ / \ 32s \end{align*}

Therefore, we can easily see that \( \frac{512}{9} L \) of water would have flowed into the tank after \(32\) seconds.

Common mistakes

Common mistakes students make with travel graphs are:

• Not multiplying across both \(LHS\) and \(RHS\)
• Incorrectly identifying the second unit.

Real-life Problems: Speed

Speed is one of the most common rates encountered in real life.

Speed compares distance with time through the following formula:

\begin{align*} Speed \ = \ \frac{Distance}{Time} \end{align*}

Speed can be expressed through many different units.

However the most common ones are \(m/s\) ( metres per second ) or \(km/h\) ( kilometres per hour ).

The above formula can also be rearranged to obtain expression for distance and time:

\begin{align*} Distance \ = \ Speed \times \ Time Time \ = \ \frac{Distance}{Speed} \end{align*}

This triangle/formulae are extremely important when applying rates to solve questions involving speed.

Below are some techniques/rules on how to apply these formulas for worded questions:

Step 1: Write down the given quantities

The first step in solving any speed/distance/time questions which involve rates is to write down the known quantities.

1. Amanda drove from Newcastle to Sydney. She left Newcastle at \( 1:30 pm \) and arrived at Sydney at \(4:30 am\) the next day. If she drove a total distance of \(5427 \ kilometres\), what was her average speed?

In the question above we can clearly discern two quantities:

\begin{align*} \text{Distance} &= 5427 km \\ \text{Time} &= \text{Time between} \ 1:30 pm \ \text{and} \ 4:30 am \ \text{the next day} \\ &= 9 \ hours \end{align*}

Step 2: Choose the relevant equation using the known quantities

Continuing on from the question above, we are given distance and time and we can see that we want to solve for speed.

Hence we would use the formula:

Applying the given quantities we get:

\begin{align*} Speed \ &= \ \frac{5427 \ km }{9 \ hours} \\ &= 603 \ km \ / \ h \end{align*}

Common mistakes:

• Confusing the different formulas.
• Not ensuring that all the quantities are in the same units before beginning calculations.

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Converting Rates between Different Units

Often questions will ask to convert rates between different units.

For these questions, it is important to be wary of when to divide and when to multiply the rate.

Below is a generic step-by-step guide for converting rates:

Step 1: Determine whether the first unit, second unit or both need to be converted.

It is important to determine which part of your rates need to be converted.

a) If we are trying to convert \(9 km/h \ into \ m/s\)

We need to change both our first unit \((km)\) and our second unit \((h)\).

b) If we are trying to convert \(12 km/h \ into \ km/s\)

We need to only change the second unit \((h)\).

Step 2: Systematically work through converting each unit one at a time

Always convert each unit one at a time.

The order is not consequential, but in this guide, we start by converting the second unit into the required format and then work on converting the first unit.

E.g. To convert \(9 km/h\) into \(m/s\)

• First convert the rate to \(x \ km/s\)
• Then convert to \(y \ m/s\).

Converting the second unit

When converting the second unit from larger  measurements (e.g.  \(hours\)) into smaller measurements (e.g. \(seconds\)):

• You always divide by an appropriate number.

When converting the second unit from smaller measurements (e.g. \(seconds\)) into larger measurements (e.g. \(hours\)):

• You always multiply by an appropriate number.

The following diagram illustrates how to convert time, the most common second unit:

Therefore, for above example, we would divide \(9\) by \(360\) to convert from \(9 \ km/h\) to \(0.025 \ km/s\).

Converting the first unit

The next step is to convert the first unit. The rules for converting the first unit are the opposite.

When converting the first unit from larger measurements (e.g. \(kilometres\) ) into smaller measurements (e.g. \(metres\) )

Similarly, when converting the first unit from  smaller  measurements (e.g. \(metres\) ) into  larger  measurements (e.g. \(kilometres\) )

The following diagram illustrates how to convert length/distance, the most common first unit :

Hence, continuing on from the first example…

To convert \(0.025km/s\) into \(y \ m/s\), we multiply \(0.025\) by \(1000\).

This leaves us with an answer of \(25 m/s\).

1. Riley walks at a speed of \(10 \ 342 \ 080 \ mm/day\). Convert this rate into \(km/s\).

First convert \(10 \ 342 \ 080 \ mm/day\) into \(x \ mm/s\).

To do this, divide by \(8640\):

\(1197 mm/s\)

Then convert into \(y \ km/s\).

To do this, divide by \(1 \ 000 \ 000\):

\(0.001197 km/s\)

Common mistakes students make when converting rates are:

• Confusing the difference between changing the first unit and second unit
• Forgetting the value of key terms (i.e. how many zeros in a tonne, etc.).

Distance-Time (Travel) Graphs

Travel Graphs are a staple of the Rates topic.

It can be used to obtain fairly in-depth information about a subject’s journey.

The \(x axis\) always represents time whereas the \(y axis\) always represents distance.

An example graph is shown below:

From the graph above it is important to note certain things:

1. The label on the \(y \ axis\) in this graph is distance from home, not total distance

• Thus, this distance will fluctuate sporadically depending on the subject’s journey
• If the distance from home decreases, it means the subject is getting closer to home and vice versa

2. There are many different gradients on the travel graph

• Note that this graph represents distance on the \(y \ axis\), time on the x-axis and recall that \( Speed = \frac{Average \ distance}{ Total \ time}\). Thus, it can be surmised that the gradient  \( \big{[} \frac{distance \ (y)}{time \ (x)} \big{]}\) represents speed .
• Hence, straight-line segments represent constant speeds.
• This also means that the steeper  the gradient, the faster the speed and vice versa.

3. There is also a portion of this graph which is a flat horizontal line (\(2:00 \ pm \ to \ 3:00 \ pm\))

• In this portion, the distance from home remains constant
•  Applying the formula above:

\begin{align*} Gradient \ &= \frac{Distance}{Time} \\ &= \frac{3.5 \ – \ 3.5}{1} \\ &= 0 \\ &= Speed \end{align*}

• Here we can see that the speed is \(0\) along this flat line. If the speed is \(0\), it means that the subject is at rest/not moving during this period.
• Therefore, flat horizontal lines  represents rest .
• The subject is stationary .

The graph below shows the journey of two different people, A & B.

• When was Person A was stationary?
• When was Person B was stationary?
• Person A’s fastest speed
• Person B’s fastest speed
• Did A or B return home?

1.  When was Person A was stationary?

To determine when Person A was stationary, we must look for flat horizontal line segments in A’s journey.

From the graph above, we can see that there are flat horizontal lines from \(3:00 \ pm\) to \(4:00 \ pm\) and from \(5:00 \ pm\) to \(6:00 \ pm\).

Therefore, Person A was stationary from \(3:00 \ pm\) to \(4:00 \ pm\) and from \(5:00 \ pm\) to \(6:00 \ pm\).

2. When was Person B was stationary?

To determine when Person B was stationary, we must look for flat horizontal line segments in B’s journey.

From the graph above, we can see that there are flat horizontal lines from \(1:00 \ pm\) to \(2:00 \ pm\) and from \(5:00 \ pm\) to \(6:00 \ pm\).

Therefore, Person B was stationary from \(1:00 \ pm\) to \(2:00 \ pm\) and from \(5:00 \ pm\) to \(6:00 \ pm\).

3. Person A’s fastest speed

To determine Person A’s fastest speed, we must look for the steepest straight line in A’s journey.

From the graph above, we can see that the steepest straight line segment is from \(1:00 \ pm\) to \(2:00 \ pm\).

Since \(gradient \ = \ speed\), we must determine the gradient of this interval:

\begin{align*} Gradient &= \frac{Distance}{Time} \\ &= \frac{3.0 – 1.0}{1} \\ &= 2 \ km/h \\ &= Speed \end{align*}

Therefore, Person A’s fastest speed was \(2 \ km/h\)

4. Person B’s fastest speed

To determine Person B’s fastest speed, we must look for the steepest straight line in B’s journey.

From the graph above, we can see that the steepest straight line segment is from \(12:00 \ pm\) to \(1:00 \ pm\).

\begin{align*} Gradient &= \frac{Distance}{Time} \\ &= \frac{5.0 – 0.0}{1} \\ &= 5 \ km/h \\ &= Speed \end{align*}

Therefore, Person B’s fastest speed was \(5 \ km/h\)

5. Did A or B return home at the end of the journey?

Person A or B would have returned home if there distance from home was \(0 \ km\) at the end of the graph.

From observing the graph, it is clear that only Person B had a distance of \(0 \ km\) from home at \(6:00 \ pm\).

Therefore, only Person B returned home at the end of the journey.

• Incorrectly calculating the gradient.
• Not being able to correctly identify steeper gradients.

Checkpoint questions

1.  It takes Tim and Taylor \(24 \ hours\) to paint \(8 \ fences\). How long will it take them to paint \(17 \ fences\)?

2.  Samantha bought \(5 \ kg\) of apples for \($24.50\) at Woolworths, \(3 \ kg\) of apples for \($23.42\) at Coles and \(4 \ kg\) of apples for \($22.10\) at the local supermarket.

Determine which purchase was the most value and how much it would have cost to buy \(12 \ kg\) of apples from that place.

3.  A recipe which reserves \(4\) people requires \(770 g\) of chicken. If I want to feed \(7\) people using this recipe, how much chicken will I have leftover if I buy \(2 kg\)?

4. Steve can walk \(1300 \ metres\) in \(35 \ minutes\). How long will it take him to walk a \(6 \ km\) marathon?

5. Miranda walked at a speed of \(3 \ km/h\) for \( 23 \ minutes\) and then drove a car at a speed of \(70 \ km/h\) for \(10 \ minutes\). How far did she travel?

6. A train travelled \(40 km/h\) for \(10 minutes\) and then \(80 km/h\) for \(14 minutes\). What was the average speed of the train?

7. A human body requires \(0.32 \ mg/L \) of iron every \(24 \ hours\). Express this rate in \(kg/mL\).

8. A plane is travelling at \(30 \ m/s\). Convert its speed into kilometres per day.

9. Steph had a heart rate of \(178 \ bpm\). How many beats would his heartbeat per day?

10. Alicia rode her bike from her house to the local train station. There she waited \(1 \ \text{hour}\) before boarding a train.

However, the train broke down halfway and Alicia was forced to walk the rest of the way to her destination.

After meeting with some friends, she rode a bike caught an Uber, walked back home, but in no particular order.

Using the graph below, determine Alicia’s mode of transport for each straight line segment.

\begin{align*} 24 \ hours \ &/ \ fence \\ 3 \ hours \ &/ \ 1 \ fence \ ( \text{Divided  by} \ 8 ) \\ 51 \ hours \ &/ \ 17 \ fence \ (\text{Multiplied by} \ 17) \end{align*}

Therefore, it will take them \(51 \ hours\) to paint \(17 \ fences\).

\begin{align*} $24.50 \ &/ \ 5kg \ (W) \\ $4.90 \ &/ \ 1kg \ (\text{Divided by} \ 5) \end{align*}

\begin{align*} $23.42 \ &/ \ 3kg \ (C) \\ $7.81 \ &/ \ 1kg \ (\text{ Divided by} \ 3) \end{align*}

\begin{align*} $22.10 \ &/ \ 4kg \ (LS) \\ $5.53 \ &/ \ 1kg \ (\text{Divided by} 4) \end{align*}

Therefore, Woolworths is the most value.

To find out how much it would cost to buy \(12 \ kg\) of apples from Woolworths, multiply the simplified rate by \(12\):

\begin{align*} $58.80 \ \text{for} \ 12 \ kg \ \text{of apples from Woolworths} \end{align*}

\begin{align*} 770 \ g \ &/ \ 4 \ people \\ 192.5 \ g \ &/ \ 1 \ person \ (\text{Divided by} 4) \\ 1347.5 \ g \ &/ \ 7 \ people \ (\text{Multiplied by} 7) \end{align*}

Therefore, remainder form \( 2kg \):

\begin{align*} 2000 \ – \ 1347.5 \ = \ 652.5 \ g \end{align*}

Hence, \(652.5 \ g\) of chicken will be left over.

\begin{align*} Time &= \frac{Distance}{Speed} \\ &= \frac{6000}{ \frac{1300}{35}} \\ &= 161.54 \ minutes \end{align*}

Therefore, it will take Steve \(161.54 \ minutes\).

Using the formula:

\begin{align*} Distance \ &= \ Speed \times Times \\ &= 3 \times \frac{23}{60} + 70 \times \frac{10}{60} \\ &= 12.82 \ km \end{align*}

Hence, Miranda walked a total of \(12.82 \ km\).

Since \( Average \ Speed \ = \ Total \ Distance\) divided by \(Total \ Time\), we must first figure out the total distance.

\begin{align*} Total \ distance \ &= \ Speed \times \ Time \\ &= 40 \times \ \frac{10}{60} + 80 \times \frac{14}{60} \\ &= 25.22 \ km \\ \end{align*}

\begin{align*} Total \ Time &= \ 10 \ mins + 14 \ mins \\ &= 21 \ minutes \\ \end{align*}

\begin{align*} Avg. Speed &= \frac{Total \ Distance}{Total \ Time} \\ &= \frac{12.33}{\frac{24}{60}} \\ &= 30.83 \ km/h \end{align*}

First convert \(0.32 \ mg/L\) into \(x \ mg/mL\).

To do this, divide by \(1000\):

\begin{align*} 0.00032 \ mg/ml \end{align*}

Then convert into \(y \ kg/L\).

\begin{align*} 0.00000032 \ mg/ml \end{align*}

First convert \(30 \ m/s\) into \(x \ m/day\).

To do this, multiply by \(8640\):

\begin{align*} 259 \ 200 \ m \ / \ day \end{align*}

Then convert into \(y \ km \ / \ day\).

\begin{align*} 259.2km \ / \ day \end{align*}

To answer this question, we must convert \(178 \ bpm\) into \(x \ beats\).

To do this, multiply by \(1440\):

\begin{align*} 256 \ 320 \ beats \ / \ day \end{align*}

As the question states, the first straight line interval \([6:00 \ am \ to \ 8:00 \ am]\) must be through bike.

She was then stationary for the next flat horizontal line interval \([8:00 \ am \ to \ 9:00 \ am]\).

The next straight line interval must then represent the train journey \([9:00 \ am \ to \ 10:00 \ am]\).

The following flat horizontal line interval represents a stationary train after it broke down \([10:00 \ am \ to \ 11:00 \ am]\).

Therefore, as the question states, the next straight line interval must represent walking \([11:00 \ am \ to \ 1:00 \ pm]\).

After her meeting with friends where she is stationary \([1:00 \ pm \ to \ 3:00 \ pm]\), we are given \(3\) different straight line intervals and \(3\) different modes of transport ( car, bike walk ).

The car is clearly the fastest mode of transport and must be represented by the straight line interval with the steepest gradient \([3:00 \ pm \ to \ 4:00 \ pm]\).

Similarly, walking is the slowest mode of transport and must be represented by the straight line interval with the shallowest gradient \([4:00 \ pm \ to \ 5:00 \ pm]\).

This leaves the bike as the mode of transport for the last interval \([5:00 \ pm \ to \ 6:00 \ pm]\).

Summary of Rates and Ratios

Questions involving rates are fundamental part of junior mathematics.

It is important to develop a good foundation in this area as this is a concept which is heavily present in harder senior topics.

Always pay attention to the techniques you can apply when approaching a question as well as the variety of common mistakes you should watch out for.

Finally, we hope that you’ve learnt something new from this subject guide, so get out there and ace mathematics!

Part 6: Planar Transformations

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Sharing in a Ratio: Worksheets with Answers

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24 Ratio Word Problems for Year 6 to Year 8 With Tips On Supporting Pupils’ Progress

Emma Johnson

Ratio word problems are introduced for the first time in upper Key Stage 2. The earliest mention of ‘Ratio’ in the National Curriculum is in the Year 6 programme of study, where a whole section is dedicated to Ratio and Proportion.

At this early stage, it is essential to concentrate on the language and vocabulary of ratio. Children need to be clear on the meaning of the ratio symbol right from the start of the topic. Word problems really help children understand this concept, as they make it much more relevant and meaningful than a ratio question with no context.

Ratio in KS2

Ratio in ks3.

• Why are word problems important for children’s understanding of ratio?

How to teach ratio word problem solving in Year 6 and early secondary school 

Ratio word problems for year 6, ratio word problems for year 7, ratio word problems for year 8, more word problems.

Concrete resources and pictorial representations are key to the success of children’s early understanding of ratio. These resources are often used in word problems for year 3 , word problems for year 4 and word problems for year 5 . There is often a misconception amongst upper Key Stage 2 teachers and students, that mathematical equipment is only for children who struggle in maths. However, all students should be introduced to this new concept through resources, such as two-sided counters and visual representations, such as bar models as this can help with understanding basic mathematical concepts such as addition and subtraction word problems .

All Kinds of Word Problems Four Operations

Download this free pack of mixed word problems covering all four operations. Test your student's problem solving skills over a range of topics.

As pupils progress into Key Stage 3, they continue to build on their knowledge and understanding of ratio. As students move away from the practical and visual resources, word problems continue to be a key element to any lessons involving ratio. As students move into Key Stage 4, they continue to build on this knowledge of ratio and can expect to encounter ratio and proportion word problems in their GCSE maths exams. 

Ratio word problems are an essential component of any lessons on ratio, to help children understand how ratio is used in real-life. To help you with this, we have put together a collection of 24 word problems including multi-step word problems , which can be used by pupils from Year 6 to Year 8.

Ratio word problems in the National Curriculum

Children are first introduced to ratio and ratio problems in Year 6. The National Curriculum expectations for ratio are that students will be able to:

• solve problems involving the relative sizes of 2 quantities where missing values can be found by using integer multiplication and division facts
• solve problems involving the calculation of percentages [for example, of measures and such as 15% of 360] and the use of percentages for comparison
• solve problems involving similar shapes where the scale factor is known or can be found
•  solve problems involving unequal sharing and grouping using knowledge of fractions and multiples.

Students in Key Stage 3 continue to build on their knowledge of ratio from primary. The expectations for Years 7 and 8 are that pupils will:

•  Use scale factors, scale diagrams and maps
• Express one quantity as a fraction of another, where the fraction is less than 1 and greater than 1
•  Use ratio notation, including reduction to simplest form
•  Divide a given quantity into two parts in a given part:part or part:whole ratio; express the division of a quantity into two parts as a ratio
•  Understand that a multiplicative relationship between two quantities can be expressed as a ratio or a fraction.
•  Relate the language of ratios and the associated calculations to the arithmetic of fractions and to linear functions.
• Solve problems involving percentage change, including: percentage increase, decrease and original value problems and simple interest in financial mathematics.

Why are word problems important for children’s understanding of ratio?

Solving word problems are important for helping children to develop their understanding of ratio and the different ways ratio is used in everyday life. Without this context, ratio can be quite an abstract concept, which children find difficult to understand. Word problems bring ratio to life and enable students to see how they will make use of this skill outside the classroom.

Third Space Learning’s online one-to-one tuition programme relates maths concepts to real life situations to deepen conceptual understanding. Personalised fill the gaps in each individual student’s maths knowledge, our programmes help to build skills and confidence.

It is important children learn the skills needed to solve ratio word problems. As with any maths problem, children need to make sure they have read the questions carefully and thought about exactly what is being asked and whether they have fully understood this. The next step is to identify what they will need to do to solve the problem and whether there are any concrete resources or pictorial representations which will help them. Even older pupils can benefit from drawing a quick sketch to understand what a problem is asking.

Here is an example:

 Jamie has a bag of red and yellow sweets.

For every red sweet there are 2 yellow sweets.

If the bag has 6 red sweets. How many sweets are in the whole bag?

How to solve:

What do you already know?

• We know that for every red sweet there are 2 yellow sweets.
• If there are 6 red sweets, we need to work out how many yellow sweets there must be. 
• If there are 2 yellow sweets for each 1 red sweet, then we must need to multiply 2 by 6, to work out how many yellow sweets there are with 6 red sweets.
• Once we have worked out the total number of yellow sweets (12), we then need to add this to the 6 red sweets, to work out how many sweets are in the bag altogether.
• If there are 6 red sweets and 12 yellow sweets, there must be 18 sweets in the bag altogether.

  How can this be represented pictorially?

• We can use the two-sided counters to represent the red and yellow sweets.
• If we put down 1 red counter and 2 yellow counters.
• We then need to repeat this 6 times, until there are 6 red counters and 12 yellow counters.
• We can now visually see the answer to the word problem and that there are now a total of 18 counters (18 sweets in the bag).

Word problems for year 6 often incorporate multiple skills: a ratio word problem may also include elements from multiplication word problems , division word problems , percentage word problems and fraction word problems .

Sophie was trying to calculate the number of students in her school.

She found the ratio of boys to girls across the school was 3:2

If there were 120 boys in the school

• How many girls were there?
• How many students were there altogether?

Answer: a) 80 b) 200

This can be shown as a bar model.

120 ÷ 3 = 40

40 x 2 = 80

• 200 students altogether

120 boys + 80 girls = 200

Pupils on the Eco Committee in Year 6 wanted to investigate how many worksheets were being printed each week.

They found that there were 160 maths worksheets and 80 English worksheets

What is the ratio of maths to English worksheets?

Answer: 2:1

Ratio of 160:80 

This can be simplified to 2:1 by dividing both 160 and 80 by 80

The Year 6 football club has 30 members. The ratio of boys to girls is 4:1. How many boys and girls are in the club?

Answer: 24 boys and 6 girls

The ratio of 4:1 has 5 parts

Boys: 4 x 6 = 24

Girls 1 x 6 = 6

Yasmine has a necklace with purple and blue beads.

The ratio of purple:blue beads  = 1:3

There are 24 beads on the necklace. How many purple and blue beads are there?

Answer: 6 purple beads and 18 blue beads

The ratio of 1:3 has 4 parts

24 ÷ 4 = 6 beads per part

Purple: 1 x 6 = 6

Blue 3 x 6 = 18

Maisie drives past a field of sheep and cows.

She works out that the ratio of sheep to cows is 3:1

If there are 5 cows in the field, how many sheep are there?

Answer: 15 sheep

If there are 5 cows in the field, the 1 has been multiplied by 5.

We need to also multiply the 3 by 5, which is 15

At a party there is a choice of 3 flavours of jelly – orange, blackcurrant and lemon,

The ratio of the jellies are 3:2:1 (orange: blackcurrant: lemon)

If there are 9 orange jellies. How many blackcurrant and lemon jellies are there?

Answer: 6 blackcurrant and 3 lemon jellies

To get 9 jellies, we need to multiply 3 by 3. This means we need to multiply 2 x 3 = 6 and 1 x 3 = 3

The school photocopier prints out 150 sheets in 3 minutes.

How many sheets can it print out in 15 minutes?

Answer: 750 sheets in 15 minutes

We need to multiply 3 by 5 to get 15 minutes. This means we also need to multiply 150 by 5 = 750

Mason carried out a survey of the favourite sports of children in Year 6.

For every 3 students who chose football, 2 chose swimming and 1 chose basketball.

12 children chose football. How many took part in the survey altogether?

Answer: 24 children took part in the survey.

If we multiply 3 by 4, we get to the 12 students who chose football. 

We need to also multiply the 2 by 4 (8 children chose swimming) and the 1 by 4 (4 children chose swimming)

12 + 8 + 4 = 24

David has 2 grandchildren: Maisie (age 6) and Lottie (age 3)

He decides to share £60 between the 2 children in a ratio of their ages.

How much does each child get?

Answer: Maisie gets £40, Lottie gets £20

Ratio of 2:1 = 3 parts

60 ÷ 3 = £20 per part

Maisie: 2 x 20 = £40

Lottie: 1 x 20 = £20

A rectangle has the ratio of width to length 2:3. If the perimeter of the rectangle is 50cm, what’s the area?

Answer: Area: 150cm^{2}

Width: 10cm

Length: 15cm

Divide 50 by 5 to work out 1 part = 10

The 2 widths must by 2 x 10 = 20

The 2 lengths must be 3 x 10 = 30

To work out the width of 1 side, divide the 20 by 2 = 10

To work out the length of 1 side, divide the 30 by 2 = 15

Area: 10 x 15 = 150cm2

In Bethany’s class there are 20 girls and 12 boys.

Write down the ratio of girls to boys in the simplest form

Answer: 5:3

(Divide both sides by 4 = 5:3)

A piece of ribbon is 45cm long.

It has been cut into 3 smaller pieces in a ratio of 4:3:2

How long is each piece?

Answer: 20cm, 15cm and 10cm

4:3:2 =9 parts: 45 ÷ 9 = 5cm per part

4 x 5 = 20cm

3 x 5 = 15cm

2 x 5 = 10cm

Chloe is making a smoothie for her and her 3 friends.

She has the recipe for making a smoothie for 4 people: 240ml yoghurt, 120 ml milk, 300ml apple juice, 180g strawberries and 1 table spoon of sugar.

• How much yoghurt would be needed to make a smoothie for 8 people.
• How many g of strawberries are needed to make the smoothie for 2 people?
• 480 ml yoghurt 

240ml x 2 = 480

• 90g strawberries

180 ÷ 2 = 90

The ratio of cups of flour:cups of water in the recipe for making the dough for a pizza base is 7:4.

The pizza restaurant needs to make a large quantity of pizzas  and is using 42 cups of flour. How much water will be needed?

Answer: 24 cups of water

Multiply 7 by 6 to get 42 cups of flour.

We therefore need to also multiply the 4 by 6 to work out how many cups of water are needed.

Ahmed shared £56 between him and Hamza in a ratio of 3:5 (3 for Hamza and 5 for him).

How much did each get?

Answer: Ahmed got £35, his brother got £21

Ratio of 3:5 = 8 parts

56 ÷ 8 = £7 per part

3 x 7 =£ 21

5 x 7 = £35

Amber and Holly share some money in a ratio of 5:7

If Amber gets £30. How much do they have between them to share?

Answer: £72

Amber gets £30 which is 5 parts: 30 ÷ 5 = 6

Holly gets 7 x 6 = £42

£30 + £42 = £72       

Two companies are making an orange coloured paint.

Company A makes the orange paint by mixing red and yellow paint in a ratio of 5:7

Company B makes the orange paint by mixing red and yellow paint in a ratio of 3:4.

Which company uses a higher proportion of red paint to make the orange?

Answer: Company B uses more red paint.

Company A: 5:7 = \frac{5}{12} is red

Company B: 3:4 = \frac{3}{7} is red

We can compare the fractions by giving them the same denominator, to find the equivalent fractions.

Students in a school have to choose one Humanities subject – History or Geography.

The ratio of boys to girls is 5:4 and \frac{3}{5} of the girls study History.

There are 225 students in the year. How many girls study History?

Answer: 60 girls study History

The ratio of 5:4 has 9 parts. Divide 225 by 9 to work out 1 part = 25

There are 5 x 25 boys = 125  and 4 x 24 girls = 100

\frac{3}{5} of 100 = 60

The angles in a triangle are in the ratio of 3:4:5 for angles A, B and C

Calculate the size of each angle.

Angle A: 45°

Angle B: 60°

Angle C: 75°

3:4:5 = 12 parts.   180 ÷ 12 = 15 (each part is worth 15°)

3 x 15 = 45

4 x 15 = 60

5 x 15 = 75

The audience in a theatre has a ratio of 2:1 adults to children.

There are 2,250 people in the audience.

The cost of an adult ticket is £10 and a child ticket is £5

How much money did the theatre make from the ticket sales

Answer: £18,750

2250 ÷ 3 = 750 people per part

Number of adults: 2 x 750 = 1500          Number of children: 1 x 750 = 750

Cost of adult tickets = 1500 x £10 = £15,000    Cost of child tickets = 750 x £5 = £3,750

Total cost: £15,000 + £3,750 = £18,750

Sophia and Jessica collect stickers and stamps.

Altogether they have the same number of stickers as stamps.

The ratio of stickers Sophia has to the stickers Jessica has is 3:7. 

The ratio of stamps Sophia has to stamps Jessica has is 1:4

Show Jessica has more stamps than stickers.

Answer: 

Ratio of stickers to stickers has 10 parts

Ratio of stamps to stamps has 5 parts.

To make the stamps equivalent to the stickers, they need to be doubled – 1:4 becomes 2:8, compared to 3:7 stickers, therefore, Jessica has more stamps than stickers.

A drink is made by mixing pineapple and lemonade in the ratio of 1:4. 

Pineapple costs £1.50 per litre. Lemonade costs £1.30 per litre (Bottles are sold in 1 litre containers)

How much will it cost to make 4 litres of drink?

Answer: £6.70

Ratio of pineapple to lemonade has 5 parts

4l of drink = 4000 ml. 1 part = 4000 ÷ 5 = 800

Pineapple = 1 x 800 = 800ml

Lemonade = 4 x 800 = 3200ml

1 Bottle of pineapple will be needed (£1.50) and 4 bottles of lemonade (4 x £1.30 = £5.20)

Total cost = £1.50 + £5.20 = £6.70

The ratio of Amber’s age to Zymal’s age is 5:7

If Zymal is 12 years older than Amber, how old are both Amber and Zymal

Answer: Amber: 30 years & Zymal: 42 years

Zymal is 12 years older. This means that the 2 parts more in the ratio  of 5:7 must be worth 12 years. Therefore, 1 part = 6 years.

Amber is 5 x 6 = 30 years

Zymal is 7 x 6 = 42 years

Hamza, Jude and Adam are collecting red and yellow leaves for an art project.

In total they collect the same number of red and yellow leaves.

The 3 boys  collect red leaves in a ratio of 4:7:9 and yellow leaves in a ratio of 3:1:5 (Hamza:Jude:Adam)

Did Hamza collect more red leaves or yellow leaves?

Answer: Hamza collected more yellow leaves (60 yellow, compared to 36 red)

Red leaves have a ratio of 20 parts, yellow leaves have a ratio of 9 parts. To make them both equivalent, they both need to have 180 parts (20 x 9 = 180) and (9 x 20 = 180)

4:7:9 = 20 parts red leaves. Multiply each part by 9 36:63:81

3:1:5= 9 parts yellow leaves. Multiply each part by 20 = 60:20:100

Hamza collected more yellow leaves (60 yellow leaves and 36 red leaves)

Looking for word problems on more topics? Take a look at our practice problems for Years 3-6 including money word problems , time word problems , addition word problems and subtraction word problems .

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• Ratio Problems

Concept of Ratio

There should be two quantities required to compare them with each other. For example, a : b, they are called terms(a and b). The first term(a) is called antecedent and the second term(b) is called the consequent. Generally, a ratio is expressed in the simplest form .

Two Quantities of Ratio.

The order of the terms in a ratio is important, we can not write a : b to b : a. They are different.

Ratio exists only between quantities of the same kind as well as in the same units.

We can convert them into fractions whenever needed.

Fraction and Ratio.

Types of Ratios

Some of the important ratios are:

Inverse Ratio- The new ratio obtained by reversing the terms of a ratio is called the inverse ratio. For example, the inverse ratio of 2 : 3 will be 3 : 2.

Compound Ratio- The new ratio formed by the product of the previous terms of two or more ratios and the product of the last terms is called a mixed ratio. For example, the mixed ratio of two ratios (a : b) and (c : d) will be (ad : bc). Similarly the mixed ratio of 2 : 3 , 4 : 5 and 6 : 7 will be 2 × 4 × 6 : 3 × 5 × 7 i.e. 48 : 105 or 16 : 35.

Duplicate Ratio- If a new ratio is made by mixing a ratio with the same, then it is called a square ratio. For example, the square ratio of 2 : 3 is 2² : 3³. That is 2 × 2 : 3 × 3 or 4 : 9.

How To Solve Ratio?

When two numbers are divided by each other, the ratio of those numbers is obtained. As a and b are two numbers then a/b will be their ratio. If we want to find the ratio of two numbers 3 and 7, then 3/7 will be their ratio. It will be written as 3:7.

Simple Ratio Problems(Illustrations)

What is the ratio of 2 to 3?

Solution : 2 + 3 = 5. We have 5 parts in the ratio of 2 : 3.

25 cm and 1 m. What is the ratio?

Solution : 1 m = 100 cm.

25 cm and 100 cm ratio of 25/100 = 1 : 4.

The sum of the two numbers is 60 and the difference is 6. What will be the ratio?

Solution: Required ratio of numbers:

(60 + 6) / (60 - 6) = 66/54 = 11/9 or 11 : 9.

Solved Questions

What will be the inverse ratio of 5 : 6?

Solution: The inverse ratio of 5 : 6 is 6 : 5.

What will be the duplicate ratio of 3 : 4?

Solution: The duplicate ratio of 3 : 4 = 3² : 4² or 9 : 16

If a quantity is divided in the ratio of 5 : 7, the larger part is 100. Find the quantity.

Solution: Let the quantity be x.

Then the two parts will be 5x / 7 + 5 and 7x / 7 + 5.

Hence, if the larger part is 100, we get 7x / 5 + 7 = 100.

7x / 12 = 100

7x = 100 × 12

x = 1200 / 7

Therefore, the quantity is 171.42.

Learning by Doing

Problems On Ratio:

1. Shilpa earns ₹150 in 4 hours and isha ₹300 in 7 hours. What will be the ratio of their earnings?

Ans: Shilpa earning: 150 X 4 = 600.

Isha earning: 300 X 7 = 2100.

Ratio= 600 : 2100

2. The ratio between the speeds of two trains is 4:5. If the second train runs 400 kms. in 6 hours. What will be the speed of the first train?

Ans: The speed of train in 1 hours : 400/6 = 66(approx)

Ratio = 4x/5x=y/66

= 4x X 66 = 5x X y

= 264 x = 5xy

= 264x/5x=y

The speed of the first train is 52.8 km/hr.

The Ratio refers to the comparison of at least two quantities of each other. If a and b are two quantities of the same kind (in the same units), then the fraction a/b is called the ratio of a to b. It is written as a : b. The quantities a and b are called the terms of the ratio, a is called the first term or antecedent and b is called the second term or consequent. The ratio compounded of the two ratios a : b and c : d is ad : bc. A ratio compounded of itself is called its duplicate ratio. a² : b² is the duplicate ratio of a : b. For any ratio a : b, the inverse ratio is b : a.

FAQs on Ratio Problems

1. Isha weighs 55 kg. If she reduces her weight in the ratio of 5: 4, find his reduced weight.

Let the previous weight be 5x.

Therefore, 4 × 11 = 44 kg. The reduced weight is 44 kg.

2. Shubham leaves $ 1000000 behind. According to his wish, the money is to be divided between his wife and sister in the ratio of 3 : 2. Find the sum received by his wife.

We know if a quantity x is divided in the ratio a: b then the two parts are ax / a + b and bx / a + b.

= 3/5 × 1000000  

= 3 × 200000

Therefore, the sum received by his wife is $600000.

3. Ishita weighs 56.7 kg. If she reduced her weight in the ratio of 7 : 6, find her new weight. 

Original weight of Ishita = 56.7 kg.

She reduces her weight by a ratio of 7 : 6.

Her new weight = (6 × 56.7)/7 = 6 × 8.1 = 48.6.    

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Expected level of development

Australian Curriculum Mathematics V9 : AC9M8M05

Numeracy Progression : Proportional thinking: P5

At this level, students recognise and use rates to solve problems including constant rates, rate of pay, cost per kilogram, recipes, simple interest and average rates. Students will also be required to make comparisons between two related quantities of different units of measurement. They will write rates in their simplest form, giving the amount for one unit of the second quantity, for example, 50km/1 hour as 50km/h.

Students are to apply their knowledge of rates to solve real-world questions determining rates in different contexts, for example, to compare running rates of athletes in sport; to compare the petrol consumption price of different vehicles; to determine the best value for money between two specials; and to determine annual income.

Students will recall on prior knowledge converting between different units of measurement. They will apply rates to solve problems, for example, converting distance from miles to kilometres. They convert between area units (mm 2 , cm 2 , m 2 , hectares and km 2 ) and volume units (mm 3 , cm 3  and m 3 ).

Teaching and learning summary:

• Demonstrate and show the conversion between different units of measurement before solving problems.
• Expand knowledge of rates to then use them in various contexts.
• Apply rates to calculate and solve problems between two quantities, recognising the relationship between them.
• Record rates in their simplest form using correct unit notation.

• will confidently convert between units
• understand the meaning of the term 'rate'
• write rates in their simplest form
• determine relationships between different quantities
• solve real-world examples of rates being used, e.g. in finance, record keeping and sports analysis.

Some students may:

• become reliant on these rules and can easily forget which operation or power of 10 to use when converting between units.
• confuse formulas and methods to determine rates.
• not make connections between units of measurement.

The Learning from home activities are designed to be used flexibly by teachers, parents and carers, as well as the students themselves. They can be used in a number of ways including to consolidate and extend learning done at school or for home schooling.

Learning intention

• I will solve problems involving the conversion between units of measurement.
• I will apply rates to calculate solutions to problems.

Why are we learning about this?

There are many everyday situations where it is useful to convert rates to the same unit for easy comparison. This is a good way to determine the best value or best buy.

• How much will 4kg of salami cost at $8/kg?
• A radio station plays songs at a rate of 9 songs per hour. How many songs would you hear in 8 hours?
• A teenage girl buys 6 pairs of shoes per year. How many pairs of shoes does she buy during her teenage years? 
• A surfer catches 9 good waves per hour. How many waves would he catch if he surfed for 2 1 2 hours?
• A ream of 500 sheets of paper is 5cm thick. How thick is 1 sheet of paper in millimetres?
• A house is 24 metres from the cliff above the sea. The cliff is eroding at 30mm per year. How many years will pass before the house starts to fall into the sea?

Melbourne Stars batting scoring card (7/119)

• Calculate the overall run rate for the match (runs per over), and the team’s overall batting strike rate (runs per 100 balls).

Success criteria

• I can write rates in their simplest form
• I can convert between units of measurement.

Please note:  This site contains links to websites not controlled by the Australian Government or ESA. More information  here .

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Teaching strategies.

A collection of evidence-based teaching strategies applicable to this topic. Note we have not included an exhaustive list and acknowledge that some strategies such as differentiation apply to all topics. The selected teaching strategies are suggested as particularly relevant, however you may decide to include other strategies as well. 

Concrete, Representational, Abstract (CRA)

The CRA model is a three-phased approach where students move from concrete or virtual manipulatives, to making visual representations and on to using symbolic notation.

Explicit teaching

Explicit teaching is about making the learning intentions and success criteria clear, with the teacher using examples and working though problems, setting relevant learning tasks and checking student understanding and providing feedback.

Culturally responsive pedagogies

Mathematics is not an exclusive western construct. Therefore, it is important to acknowledge and demonstrate the mathematics to be found in all cultures.

Teaching resources

A range of resources to support you to build your student's understanding of these concepts, their skills and procedures. The resources incorporate a variety of teaching strategies.

Athletics club

A short problem involving ratios of different age groups and types of groups.

Working with rate tool

This activity is designed to highlight the common misunderstandings students have with the concept of rate. Worksheet and rubric are included.

Relevant assessment tasks and advice related to this topic.

By the end of Year 8, students identify and apply rates to compare two related quantities of different units of measure to solve problems.

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HA ratio worksheets (YR 6)

Subject: Mathematics

Age range: 7-11

Resource type: Worksheet/Activity

Last updated

8 September 2024

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Find included HA ratio worksheets, suitable for year six SATs preparation.

NC: solve problems involving the relative sizes of two quantities where missing values can be found by using integer multiplication and division facts

Enhance your Year 6 students’ understanding of ratios with my HA Ratio Worksheets, expertly designed to prepare them for SATs. These worksheets are aligned with the National Curriculum objective to solve problems involving the relative sizes of two quantities where missing values can be found by using integer multiplication and division facts.

These high-ability (HA) worksheets challenge students with complex ratio problems that require a deep understanding of multiplication and division facts to solve. Each problem is crafted to improve students’ abilities to think critically and apply their mathematical knowledge to find relationships between quantities, preparing them for similar questions they might encounter during their SATs.

Included in this resource are detailed answer sheets that not only provide solutions but also explain the methods used. This feature is crucial for helping students understand the processes behind the answers and for enabling teachers to provide targeted, effective feedback.

Regular updates ensure that the content remains relevant and reflective of the latest SATs format and standards, making these worksheets an invaluable tool for any Year 6 classroom focused on achieving excellence in mathematics.

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