ratio and proportion problem solving questions

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Ratio Problem Solving

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

There are also ratio problem solving 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 ratio problem solving?

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

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

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

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

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

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

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

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

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

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

Then we multiply each part of the ratio by 5.

3 x 5:5 x 5 = 15:25

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

• Dividing ratios

Step-by-step guide: Dividing ratios (coming soon)

Ratios and fractions (proportion problems)

We also need to consider problems involving fractions. These are usually proportion questions where we are stating the proportion of the total amount as a fraction.

Simplifying and equivalent ratios

• Simplifying ratios

Equivalent ratios

Units and conversions ratio questions

Units and conversions are usually equivalent ratio problems (see above).

• If £1:\$1.37 and we wanted to convert £10 into dollars, we would multiply both sides of the ratio by 10 to get £10 is equivalent to \$13.70.
• The scale on a map is 1:25,000. I measure 12cm on the map. How far is this in real life, in kilometres? After multiplying both parts of the ratio by 12 you must then convert 12 \times 25000=300000 \ cm to km by dividing the solution by 100 \ 000 to get 3km.

Notice that for all three of these examples, the units are important. For example if we write the mapping example as the ratio 4cm:1km, this means that 4cm on the map is 1km in real life.

Top tip: if you are converting units, always write the units in your ratio.

Usually with ratio problem solving questions, the problems are quite wordy . They can involve missing values , calculating ratios , graphs , equivalent fractions , negative numbers , decimals and percentages .

Highlight the important pieces of information from the question, know what you are trying to find or calculate , and use the steps above to help you start practising how to solve problems involving ratios.

How to do ratio problem solving

In order to solve problems including ratios:

Identify key information within the question.

Know what you are trying to calculate.

Use prior knowledge to structure a solution.

Explain how to do ratio problem solving

Ratio problem solving worksheet

Get your free ratio problem solving worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on ratio

Ratio problem solving is part of our series of lessons to support revision on ratio . You may find it helpful to start with the main ratio lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• How to work out ratio  
• Ratio to fraction
• Ratio scale
• Ratio to percentage

Ratio problem solving examples

Example 1: part:part ratio.

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

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

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

We could write this as

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

We know that 465 students have school dinner.

2 Know what you are trying to calculate.

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

We need to find the value of p.

3 Use prior knowledge to structure a solution.

We are looking for an equivalent ratio to 5:7. So we need to calculate the multiplier. We do this by dividing the known values on the same side of the ratio by each other.

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

There are 651 students that have a packed lunch.

Example 2: unit conversions

The table below shows the currency conversions on one day.

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

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

The two values in the table that are important are GBP and EUR. Writing this as a ratio, we can state

We know that we have £520.

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

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

1.17 \times 520=608.4

So £520 = €608.40.

Example 3: writing a ratio 1:n

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

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

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

2l = 2000ml

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

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

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

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

Example 4: forming and solving an equation

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

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

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

We also know that Luke receives £8.

We want to calculate the total amount of pocket money for the three siblings.

We need to find the value of x first. As Luke receives £8, we can state the equation 2x=8 and so x=4.

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

The total amount of pocket money is therefore 4+7+8=£19.

Example 5: simplifying ratios

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

Express this data as a ratio in its simplest form.

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

We need to simplify this ratio.

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

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

HCF (12,16,10) = 2

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

Our solution is 6:8:5 .

Example 6: combining two ratios

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

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

We know the two ratios

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

Using equivalent ratios we can say that the ratio of silica:soda is equivalent to 15:3 by multiplying the ratio by 3.

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

Example 7: using bar modelling

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

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

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

We want to find the original quantity.

Drawing a bar model of this problem, we have

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

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

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

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

We can fill in each share to be 25g.

Adding up each share, we get

India = 5 \times 25=125g

Beau = 2 \times 25=50g

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

Common misconceptions

• Mixing units

Make sure that all the units in the ratio are the same. For example, in example 6 , all the units in the ratio were in millilitres. We did not mix ml and l in the ratio.

• Ratio written in the wrong order

For example the number of dogs to cats is given as the ratio 12:13 but the solution is written as 13:12.

• Ratios and fractions confusion

Take care when writing ratios as fractions and vice-versa. Most ratios we come across are part:part. The ratio here of red:yellow is 1:2. So the fraction which is red is \frac{1}{3} (not \frac{1}{2} ).

• Counting the number of parts in the ratio, not the total number of shares

For example, the ratio 5:4 has 9 shares, and 2 parts. This is because the ratio contains 2 numbers but the sum of these parts (the number of shares) is 5+4=9. You need to find the value per share, so you need to use the 9 shares in your next line of working.

• Ratios of the form \bf{1:n}

The assumption can be incorrectly made that n must be greater than 1 , but n can be any number, including a decimal.

Practice ratio problem solving questions

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

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

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

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

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

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

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

Ratio problem solving GCSE questions

1. One mole of water weighs 18 grams and contains 6.02 \times 10^{23} water molecules.

Write this in the form 1gram:n where n represents the number of water molecules in standard form.

2. A plank of wood is sawn into three pieces in the ratio 3:2:5. The first piece is 36cm shorter than the third piece.

Calculate the length of the plank of wood.

5-3=2 \ parts = 36cm so 1 \ part = 18cm

3. (a) Jenny is x years old. Sally is 4 years older than Jenny. Kim is twice Jenny’s age. Write their ages in a ratio J:S:K.

(b) Sally is 16 years younger than Kim. Calculate the sum of their ages.

Learning checklist

You have now learned how to:

• Relate the language of ratios and the associated calculations to the arithmetic of fractions and to linear functions
• Develop their mathematical knowledge, in part through solving problems and evaluating the outcomes, including multi-step problems
• Make and use connections between different parts of mathematics to solve problems

The next lessons are

• Compound measures
• Best buy maths

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WORD PROBLEMS ON RATIO AND PROPORTION

Problem 1 :

The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy.

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x + 5x + 7x)/3 = 25

Age of the first boy = 3x

Age of the first boy = 5x

Age of the first boy = 7x

So, the age of the youngest boy is 15 years.

Problem 2 :

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7 : 6, find his new weight.

Given :  Original weight of John = 56.7 kg. He is going to reduce his weight in the ratio 7:6.

We can use the following hint to find his new weight, after it is reduced in the ratio 7 : 6.

His new weight is

= (6  ⋅  56.7)/7

So, John's new weight is 48.6 kg.

Problem 3 :

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2 : 3.

Sum of the terms in the given ratio is

So, no. of boys in the school is

= 720  ⋅  (3/8)

= 270 

No. of girls in the school is

= 720  ⋅  (5/8) 

Given : Number of new girls admitted in the school is 18.

Let x be the no. of new boys admitted in the school.

After the above new admissions,

No. of boys in the school = 270 + x

No. of girls in the school = 450 + 18 = 468

Given : The ratio after the new admission is 2 : 3.

Then, we have

(270 + x) : 468 = 2 : 3

Use cross product rule.

3(270 + x) = 468  ⋅  2

810 + 3x = 936

So, the number of new boys admitted in the school is 42.

Problem 4 :

The monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenditures are in the ratio 7 : 9. If each saves $50 per month, find the monthly income of the second person.

From the given ratio of incomes ( 4 : 5 ),

Income of the 1st person = 4x

Income of the 2nd person = 5x

(Expenditure  =  Income - Savings)

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

So, we have

(4x - 50) : (5x - 50) = 7 : 9

9(4x - 50) = 7(5x - 50)

36x - 450 = 35x - 350

Then, the income  of the second person is

So, income of the second person is $500.

Problem 5 :

The ratio of the prices of two houses was 16 : 23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11 : 20. Find the original price of the first house.

From the given ratio 16 : 23,

Original price of the 1st house = 16x

Original price of the 2nd house = 23x

After increment in prices,

Price of the 1st house = 16x + 10% of 16x 

= 16x + 1.6x

Price of the 2nd house = 23x + 477

After increment in prices, the ratio of prices becomes 11:20.

17.6x : (23x + 477) = 11 : 20

20(17.6x) = 11(23x + 477)

352x = 253x + 5247

Then, original price of the first house is

So, original price of the first house is $848.

Problem 6 :

Two numbers are respectively 20% and 50% are more than a third number, Find the ratio of the two numbers.

Let x be the third number.

Then, the first number is

= (100 + 20)% of x

= 120% of x

The second number is

= (100 + 50)% of x

= 150% of x

The ratio between the first number and second number is

= 1.2x : 1.5x

= 1.2 : 1.5

So, the ratio of two numbers is 4 : 5.

Problem 7 :

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water ?    

[4 : 3 ----> 4 + 3 = 7, M ----> 4/7, W ----> 3/7]

Let x be the quantity of mixture taken from vessel A to obtain a new mixture in vessel C.

Quantity of milk in x = (4/7)x = 4x/7

Quantity of water in x = (3/7)x = 3x/7

[2 : 3 ----> 2 + 3 = 5, M ---> 2/5, W ----> 3/5]

Let y be the quantity of mixture taken from vessel B to obtain a new mixture in vessel C.

Quantity of milk in y = (2/5)y = 2y/5

Quantity of water in y = (3/5)y = 3y/5

Vessel A and B :

Quantity of milk from A and B is

= 4x/7 + 2y/5

= (20x + 14y)/35

Quantity of water from A and B is

= (3x/7) + (3y/5)

= (15x + 21y)/35

According to the question, vessel C must consist half of the milk and half of the water.

That is, in vessel C, quantity of milk and water must be same.

There fore,

quantity of milk in (A + B) = quantity of water in (A + B)

(20x + 14y)/35 = (15x + 21y)/35

20x + 14y = 15x + 21y

x : y = 7 : 5

So, the required ratio is 7 : 5.

Problem 8 :

A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. From the vessel, 10 liters of the mixture is removed  and replaced with an equal quantity of pure milk. Find the ratio of milk and water in the final mixture obtained.    

[3 : 2 ----> 3 + 2 = 5, M ----> 3/5, W ----> 2/5]

In 20 liters of mixture,

no. of liters of milk = 20  ⋅  3/5 = 12

no. of liters of water = 20  ⋅  2/5 = 8

Now, 10 liters of mixture removed.

In this 10 liters of mixture, milk and water will be in the ratio 3 : 2.

no. of liters of milk in this 10 liters = 10  ⋅  3/5 = 6

no. of liters of water in this 10 liters = 10  ⋅  2/5 = 4

After removing 10 liters (1st time),

no. of liters of milk in the vessel = 12 - 6 = 6

no. of liters of water in the vessel = 8 - 4 = 4

Now,  we add 10 liters of pure milk in the vessel,

After adding 10 liters of pure milk in the vessel,

no. of liters of milk in the vessel = 6 + 10 = 16

no. of liters of water in the vessel = 4 + 0 = 4

After removing 10 liters of mixture and adding 10 liters of pure milk, the ratio of milk and water is

So, the required ratio is 4 : 1.

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