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Solving problems with percentages

  • Price difference I
  • Price difference II
  • How many students?

To solve problems with percent we use the percent proportion shown in "Proportions and percent".

$$\frac{a}{b}=\frac{x}{100}$$

$$\frac{a}{{\color{red} {b}}}\cdot {\color{red} {b}}=\frac{x}{100}\cdot b$$

$$a=\frac{x}{100}\cdot b$$

x/100 is called the rate.

$$a=r\cdot b\Rightarrow Percent=Rate\cdot Base$$

Where the base is the original value and the percentage is the new value.

47% of the students in a class of 34 students has glasses or contacts. How many students in the class have either glasses or contacts?

$$a=r\cdot b$$

$$47\%=0.47a$$

$$=0.47\cdot 34$$

$$a=15.98\approx 16$$

16 of the students wear either glasses or contacts.

We often get reports about how much something has increased or decreased as a percent of change. The percent of change tells us how much something has changed in comparison to the original number. There are two different methods that we can use to find the percent of change.

The Mathplanet school has increased its student body from 150 students to 240 from last year. How big is the increase in percent?

We begin by subtracting the smaller number (the old value) from the greater number (the new value) to find the amount of change.

$$240-150=90$$

Then we find out how many percent this change corresponds to when compared to the original number of students

$$90=r\cdot 150$$

$$\frac{90}{150}=r$$

$$0.6=r= 60\%$$

We begin by finding the ratio between the old value (the original value) and the new value

$$percent\:of\:change=\frac{new\:value}{old\:value}=\frac{240}{150}=1.6$$

As you might remember 100% = 1. Since we have a percent of change that is bigger than 1 we know that we have an increase. To find out how big of an increase we've got we subtract 1 from 1.6.

$$1.6-1=0.6$$

$$0.6=60\%$$

As you can see both methods gave us the same answer which is that the student body has increased by 60%

Video lessons

A skirt cost $35 regulary in a shop. At a sale the price of the skirtreduces with 30%. How much will the skirt cost after the discount?

Solve "54 is 25% of what number?"

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Ratio and Percent Word Problems

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Solving ratio problems

ratio and percentage problem solving maths genie

The ratio of horses to donkeys at an animal sanctuary is 5 : 2. What fraction of the animals are donkeys?

Horses to donkeys. Five to two. Five plus two equals seven – all highlighted. Fractions of animals that are donkeys equals an empty highlighted box over seven.

Add the ratio parts (5 and 2) to find the denominator of the fraction. 5 + 2 = 7. The denominator is 7

The same information. Donkeys and two from the ratios are highlighted. The fraction is now two sevenths – two is highlighted.

The numerator of the fraction is the ratio part that is the focus of the question (donkeys). The numerator is 2. The fraction of the animals that are donkeys is 2⁄7

Example 2 – Given the fraction: A diagram showing a can of red paint labelled two fifths plus a can of white paint labelled with a highlighted question mark equals a can of pink paint. Written below: Two fifths red paint plus a blank fraction, with highlighted boxes, white paint equals pink paint.

To make pink paint, 2⁄5 red paint is mixed with white paint. How much white paint is needed? What is the ratio of red to white paint?

The same diagram. The can of white paint is now labelled three fifths. Written below: Two fifths red paint and three fifths white paint. Five (highlighted blue) minus two (highlighted orange) equals three. Two parts red pain and three parts white paint. Two to three.

The numerator of the fraction of red paint is 2. This is the ratio part for red paint. To find the ratio part for the white paint, subtract the numerator from the denominator. 5 – 2 = 3. The ratio part for white paint is 3. 3⁄5 of the paint is white. The ratio of red to white paint is 2 : 3

A farm has sheep and goats in the ratio 7 : 5 (sheep : goats). What fraction of the animals are sheep?

Show answer Hide answer

Add the ratio parts (7 and 5) to find the denominator of the fraction.

7 + 5 = 12. The denominator is 12

The numerator of the fraction is the ratio part that is the focus of the question (sheep). The numerator is 7

The fraction of animals that are sheep is \( \frac{7}{12} \)

Sheep to goats. Below: Seven (highlighted blue) to five. Seven plus five equals twelve – highlighted. Fraction of animals that are sheep equals seven twelfths.

A box contains toffees and truffles in the ratio 7 : 6. There are 24 truffles. How many toffees are there?

A diagram showing the sweets with bars next to them. The toffee bar is split into seven blocks. Above: A highlighted question mark with arrows pointing to each end. The truffle bar is split into six blocks. Below: Twenty-four with arrows pointing to each end.

Draw a bar model and label it to illustrate the problem. The ratio of toffees to truffles is 7 : 6. Draw 7 parts for the toffees and 6 for the truffles. Label the truffle bar with 24. To work out the total number of toffees, the value of one part needs to be found.

The same diagram. All the blocks in the truffle bar are labelled four. The first block is shaded blue. Written underneath: twenty-four divided by six.

To work the value of one part, divide the number of truffles (24) by the number of parts given (6). This gives the value of one part. 24 ÷ 6 = 4

The same diagram. All the block in the toffee bar are now labelled four. The question mark above it is now twenty-eight. Written below the bars: There are twenty-eight toffees – highlighted.

Multiply the value of one part (4) by the number of parts asked for: the toffees (7). 4 × 7 = 28. There are 28 toffees in the box.

Example 2 – Part-whole problem: Books for children to books for adults. Three to five.

A library stocks books for children and books for adults in the ratio 3 : 5. There are 450 books for children. How many books are in the library?

Two bars. The first is split in three and is labelled books for children. On top: Four-hundred and fifty with arrows pointing to each end. The second is split in five and is labelled books for adults. To the right: A vertical bracket around all the bar labelled with a highlighted question mark.

Draw a bar model and label it to illustrate the problem. The ratio of books for children to books for adults is 3 : 5. Draw 3 parts for books for children and 5 parts for books for adults. Label the books for children bar with 450. To work out the total number of books, the value of one part needs to be found.

The same bars. All of the blocks in the books for children bar labelled one-hundred and fifty. The first block is now highlighted orange with four-hundred and fifty divided by three above it.

To find the value of one part, divide the amount of books for children (450) by the number of parts given (3). 450 ÷ 3 = 150. The value of one part is 150

All of the block in the books for adults bar are now labelled one-hundred and fifty. The vertical bracket is now labelled one-thousand two-hundred – highlighted.

Multiply the value of one part (150) by the number of parts asked for: the number of books in the library. This is the total of books for children parts and the books for adults parts. (3 + 5 = 8). The calculation is 150 × 8. The total number of books in the library is 1200.

The ratio of desserts to pizzas in a supermarket freezer is 4 : 3 There are a total of 620 desserts. How many pizzas are in the freezer?

Draw a bar model to illustrate the problem. The ratio of desserts to pizzas is 4 : 3 . Draw 4 parts for desserts and 3 parts for pizzas.

Label it to illustrate the problem. Label the dessert bar with 620. To find the number of pizzas in the freezer, work out the value of one part.

To find the value of one part, divide the share for desserts (620) by the number of parts (4). 620 ÷ 4 = 155. The value of one part is 155

Multiply the value of one part (155) by the number of parts asked for: pizzas (3). 155 x 3 = 465

The number of pizzas in the freezer is 465

A diagram of two different sized bars with all blocks in both bars are labelled one-hundred and fifty-five. The first is split into four blocks and labelled desserts – with the first block highlighted. The second is three blocks long and labelled pizzas. Underneath: Four-hundred and sixty-five – highlighted with an arrow pointing to each end of it.

Everyone at a fancy dress party is dressed up as either a vampire or a wizard. The ratio of people dressed as vampires to wizards is 5 : 2. If there are 6 more vampires than wizards, how many people are at the party?

A diagram showing two bars each highlighted orange. The first is split into five blocks and labelled vampires. The second is split into two blocks and labelled wizards.

Draw a bar model to illustrate the problem. The ratio of vampires to wizards is 5 : 2. There are 5 parts for the vampires and 2 parts for the wizards.

The same diagram. On the right of the wizards bar: Six with arrows pointing right and left – filling the gap underneath the other three blocks of the vampire bar. To the right: A vertical bracket around both bars labelled with a highlighted question mark.

Label the given information. There are 6 more vampires than wizards. The diagram shows the comparison between the vampires bar and the wizards bar. To work out the total number of people at the party, the value of one part needs to be found.

The same diagram. The last three blocks of the vampire bar are labelled two. The first of them is highlighted orange with six divided by three above it.

To find the value of one part, divide the difference value (6) by the number of parts that make up the difference (3). 6 ÷ 3 = 2. The value of one part is 2

All blocks in both bars are labelled two and highlighted orange. The vertical bracket is now labelled fourteen – highlighted.

Multiply the value of one part (2) by the number of parts asked for (all the people so all the parts, 7). 2 × 7 = 14. The total number of people at the party is 14

Example 2: Tulips to daffodils. Three to seven.

The ratio of the number of tulips to daffodils in a flower display is 3 : 7. There are 96 fewer tulips than daffodils. Find the number of daffodils in the display.

A diagram showing two bars each highlighted orange. The first is split into three blocks and labelled tulips. The second is split into seven blocks and labelled daffodils.

Draw a bar model to illustrate the problem. The ratio of tulips to daffodils is 3 : 7. There are 3 parts for tulips and 7 parts for daffodils.

On the right of the tulips bar: Ninety-six with arrows pointing right and left – filling the gap underneath the other four blocks of the daffodils bar. Underneath the daffodils bar: A highlighted question mark with arrows pointing to each end.

Label the given information. There are 96 fewer tulips than daffodils. The diagram shoes the comparison between the tulips bar and the daffodils bar. To work out the number of daffodils in the display, the value of one part needs to be found.

The last four blocks of the daffodils bar are each now labelled twenty-four. The first of them is highlighted orange with ninety-six divided by four above.

To find the value of one part, divide the comparison value (96) by the number of parts that make up the difference (4). 96 ÷ 4 = 24. The value of one part is 24

All blocks in the daffodils bar are labelled twenty-four and highlighted orange. The question mark below is now one-hundred and sixty-eight.

Multiply the value of one part (24) by the number of parts asked for (all the daffodils, 7 parts). 24 × 7 = 168. The number of daffodils in the display is 168

The ratio of the number of robins to sparrows to blackbirds in a survey of garden birds is 1 : 3 : 8 (robins : sparrows : blackbirds)

There were 70 fewer robins than sparrows. How many birds were observed in this survey?

Draw a bar model to illustrate the problem. The ratio of robins to sparrows to blackbirds is 1 : 3 : 8. There is 1 part for robins, 3 for sparrows and 8 for blackbirds.

Label the given information. There are 70 fewer robins than sparrows. This is the difference between the robins bar and the sparrows bar. To find the total number of birds in the survey, the value of one part needs to be found.

To find the value of one part, divide the difference value (70) by the number of parts that make up the difference (2). 70 ÷ 2 = 35. The value of one part is 35

Multiply the value of one part (35) by the number of parts asked for (all the parts, 12). 35 x 12 = 420

A market stall has 60 T-shirts. The ratio of small, medium and large T-shirts being sold on the stall is 2 : 3 : 1. The stallholder sells 26 medium and 2 large T-shirts. What is the ratio of small, medium and large T-shirts now?

A diagram showing three bars. The first is split into two block and labelled small. The second is split into three block and labelled medium. The third is one block and labelled large. To the right: A vertical bracket around all the bars labelled sixty.

Draw a bar model to illustrate the starting information. Draw 2 parts for small T-shirts, 3 for medium and 1 for large. The total number of T-shirts is 60

The first block in the bar labelled small is now labelled ten and highlighted pink – with sixty divided by six above it.

Divide 60 in the ratio 2 : 3 : 1. To find the value of one part, divide the total number (60) by the sum of the parts (2 + 3 + 1 = 6). 60 ÷ 6 = 10. The value of one part is 10

All of the blocks in each bar are labelled ten. To the right: Twenty small – highlighted pink. Thirty medium – highlighted blue. Ten large – highlighted orange.

2 parts are small, so 2 × 10 = 20 small T-shirts. 3 parts are medium, so 3 × 10 = 30 medium T-shirts. 1 part is large, so 1 × 10 = 10 large T-shirts.

A diagram showing the pink small, blue medium and orange large t-shirts. Below: A three by three table. First column: Twenty small – highlighted pink. No sales. Twenty small. Second column: Thirty medium – highlighted blue. Twenty-six sales. Four medium. Third column. Ten large – highlighted orange. Two sales. Eight large.

Adjust the shared amounts according to the given information in the question. At the start there were 20 small, 30 medium and 10 large T-shirts. 26 medium and 2 large T-shirts are sold. The remaining T-shirts are 20 small, 4 medium and 8 large T-shirts.

The same t-shirt diagram. Below: S to M to L. Twenty to four to eight.

The ratio of small to medium to large T-shirts is now 20 : 4 : 8. This ratio can be simplified.

Twenty divided by four to four divided by four eight divided by four. Five to one to two – highlighted.

To simplify the ratio, divide each part by their highest common factor (HCF). The HCF of 20, 4 and 8 is 4. Divide each part by 4. The new simplified ratio of small to medium to large T-shirts is 5 : 1 : 2

There are 55 big cats in a safari park. The ratio of lions to tigers is 3 : 2 12 lion cubs and 5 tiger cubs are born. What is the ratio of lions to tigers now?

A total amount and ratio is given. Draw a bar model to illustrate this starting information. Draw 3 parts for lions and 2 parts for tigers, with a total of 55

Divide the total number of big cats (55) in the ratio 3 : 2. To find the value of one part, divide the amount (55) by the total number of parts (5). 55 ÷ 5 = 11. The value of one part is 11. Multiply one part by the number of parts for each big cat. There are 11 × 3 = 33 lions. There are 11 × 2 = 22 tigers.

Adjust the shared amounts according to given information in the question. After the cubs have been born there are 33 + 12 = 45 lions and 22 + 5 = 27 tigers.

Write the new shares as a ratio and simplify. The new ratio of lions to tigers is 45 : 27. This ratio can be simplified by finding the HCF of 45 and 27. This is 9. Divide each ratio share by 9. The ratio simplifies to 5 : 3

The ratio of lions to tigers is now 5 : 3

A diagram of two different sized bars with all blocks in both bars are labelled eleven. The first is split into three blocks and labelled lions – with the first block highlighted and fifty-five divided by five above it. On top: thirty-three (highlighted) with arrows pointing to each end. The second is two blocks long and labelled tigers. Underneath: Twenty-two – highlighted with an arrow pointing to each end. To the right: A vertical bracket around all the bars  labelled fifty-five. Thirty-three lions plus twelve cubs equals forty-five lions – highlighted. Twenty-two tigers plus five cubs equals twenty-seven tigers – highlighted.

Real-world maths

ratio and percentage problem solving maths genie

More on Ratio

Find out more by working through a topic

Scale drawings

  • count 4 of 5

ratio and percentage problem solving maths genie

Map scales and ratio

  • count 5 of 5

ratio and percentage problem solving maths genie

Equivalent ratios and simplifying ratios

  • count 1 of 5

ratio and percentage problem solving maths genie

Division in a given ratio

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ratio and percentage problem solving maths genie

Ratio Worksheet

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Ratio Worksheet

Help your students prepare for their Maths GCSE with this free ratio worksheet of 44 questions and answers

  • Section 1 of the ratio worksheet contains 36 skills-based ratio questions, in 3 groups to support differentiation
  • Section 2 contains 4 applied ratio questions with a mix of worded problems and deeper problem solving questions
  • Section 3 contains 4 foundation and higher level GCSE exam style ratio questions 
  • Answers and a mark scheme for all ratio questions are provided
  • Questions follow variation theory with plenty of opportunities for students to work independently at their own level
  • All questions created by fully qualified expert secondary maths teachers
  • Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards

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Ratio Worksheet

Raise maths attainment across your school with hundreds of flexible and easy to use GCSE maths worksheets and lessons designed by teachers for teachers.

Ratio at a glance

A ratio compares how much there is of one thing in relation to another. For example if the ratio of the number of dogs to the number of cats is 2:3 then for every 2 dogs there are 3 cats. 

Ratios can often be simplified by dividing each part by a common factor. Writing ratios in their simplest form makes it easier to visualise the relationship between the quantities in the ratios. It also makes it easier to use ratios in other contexts.

Ratio word problems may ask us to write a ratio, simplify a ratio, divide a quantity into a ratio or use a ratio to find quantities. We can use a bar model to represent a given ratio and this can help us visualise the ratio problem more easily.

Looking forward, students can then progress to additional ratio and proportion worksheets , for example a speed distance time worksheet or a   direct proportion worksheet.

For more teaching and learning support on Ratio and Proportion our GCSE maths lessons provide step by step support for all GCSE maths concepts.

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    Advice • Read each question carefully before you start to answer it. • Keep an eye on the time. • Try to answer every question. • Check your answers if you have time at the end mathsgenie.co.uk 1 The ratio of dogs to cats is 5:3 The ratio of fish to dogs is 6:1 Find the ratio of cats to fish. Give your answer in its simplest form.

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    Ratio, Fractions, Decimals and Percentages Back to Scheme Overview Stage 1 Stage 2 In addition to the topics in stage 1, students should be fluent in the following topics. Stage 3 In addition to the topics in stage 1 and 2, students should be fluent in the following topics. Maths Genie is a free GCSE and A Level revision site.

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    Part A The magic carpet is made with a total of 150 meters of yarn. How much silver yarn is in the magic carpet? meters Part B Jasmine buys the magic carpet on sale for $ 374 . Jasmine saved $ 66 off the regular price. What percent was the price of the magic carpet discounted? % Problem 2: Soccer Joel is training for soccer season.

  9. PDF Name: GCSE (1

    Instructions Use black ink or ball-point pen. Answer all questions. Answer the questions in the spaces provided there may be more space than you need. Diagrams are NOT accurately drawn, unless otherwise indicated. You must show all your working out. Information The marks for each question are shown in brackets

  10. PDF Name: GCSE (1

    Advice • Read each question carefully before you start to answer it. • Keep an eye on the time. • Try to answer every question. • Check your answers if you have time at the end mathsgenie.co.uk 1 In a bag there are blue sweets and red sweets. The ratio of blue sweets to red sweets is 5:3 What fraction of the sweets are blue?

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    Help your students prepare for their Maths GCSE with this free ratio worksheet of 44 questions and answers Section 1 of the ratio worksheet contains 36 skills-based ratio questions, in 3 groups to support differentiation Section 2 contains 4 applied ratio questions with a mix of worded problems and deeper problem solving questions