problem solving maths ks4

Visual maths worksheets, each maths worksheet is differentiated and visual.

KS4 Maths Worksheets

Maths Worksheets / KS4 Maths Worksheets

Cazoom Maths offers an outstanding range of maths worksheets for students in KS4 (Key stage 4). We are a trusted provider of maths worksheets for secondary school children and this set of maths worksheets is ideal for students in Years 10 and 11. Our KS4 maths worksheets cover all topics in the curriculum and are perfect for use in the classroom or for additional home learning. Our maths worksheets are used by over 30,000 teachers, parents and schools around the world and we are a Times Educational Supplement recommended resource for helping key stage 3 and key stage 4 students learn mathematics.

KS4 Maths Worksheets for Students in Years 10 and 11

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Try some free sample ks4 maths worksheets

Indices KS4 Maths Worksheet

Outstanding KS4 Maths Worksheets

  • Separate answers are included to make marking easy and quick.
  • Over 500 pages of the highest quality KS3 maths worksheets. Each worksheet is differentiated, including a progressive level of difficulty as the worksheet continues.
  • Single user licence for parents or teachers. Separate school licences are also available.
  • Single digital pdf download, with worksheets organised into high level chapters of Algebra, Statistics, Number and Geometry, and further by subtopics. See below for the extensive range of sheets included.
  • Free Scheme of Work included, showing all worksheets included in the download and the relevant GCSE grade and GCSE Tier.

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List of Topics

Our Year 10 and Year 11 printable maths worksheets cover the full range of topics. See below the list of topics covered. All our maths worksheets can be accessed here .

  • Expanding Brackets
  • Factorising
  • Inequalities
  • Linear Functions
  • Quadratic and Cubic Functions
  • Real Life Graphs
  • Rearranging Equations
  • Simplification
  • Solving Equations
  • Substitution
  • Fractions Decimals Percentages
  • Percentages
  • Types of Number
  • Area and Perimeter
  • Bearings Scale and Loci
  • Compound Measures
  • Constructions
  • Coordinates
  • Lines and Angles
  • Similarity and Congruence
  • Transformations
  • Trigonometry
  • Volume and Surface Area
  • Cumulative Frequency and Box Plots
  • Mean Median Mode
  • Pie Charts and Bar Charts
  • Probability
  • Stem-and-Leaf Diagrams
  • Surveys and Sampling
  • Two-Way Tables and Pictograms

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30 Problem Solving Maths Questions And Answers For GCSE

Martin Noon

Problem solving maths questions can be challenging for GCSE students as there is no ‘one size fits all’ approach. In this article, we’ve compiled tips for problem solving, example questions, solutions and problem solving strategies for GCSE students. 

Since the current GCSE specification began, there have been many maths problem solving exam questions which take elements of different areas of maths and combine them to form new maths problems which haven’t been seen before. 

While learners can be taught to approach simply structured problems by following a process, questions often require students to make sense of lots of new information before they even move on to trying to solve the problem. This is where many learners get stuck.

How to teach problem solving

6 tips to tackling problem solving maths questions, 10 problem solving maths questions (foundation tier), 10 problem solving maths questions (foundation & higher tier crossover), 10 problem solving maths questions (higher tier).

In the Ofsted maths review , published in May 2021, Ofsted set out their findings from the research literature regarding the sort of curriculum and teaching that best supports all pupils to make good progress in maths throughout their time in school.

Regarding the teaching of problem solving skills, these were their recommendations:

  • Teachers could use a curricular approach that better engineers success in problem-solving by teaching the useful combinations of facts and methods, how to recognise the problem types and the deep structures that these strategies pair to.
  • Strategies for problem-solving should be topic specific and can therefore be planned into the sequence of lessons as part of the wider curriculum. Pupils who are already confident with the foundational skills may benefit from a more generalised process involving identifying relationships and weighing up features of the problem to process the information. 
  • Worked examples, careful questioning and constructing visual representations can help pupils to convert information embedded in a problem into mathematical notation.
  • Open-ended problem solving tasks do not necessarily mean that the activity is the ‘ideal means of acquiring proficiency’. While enjoyable, open ended problem-solving activities may not necessarily lead to improved results.  

Are you a KS2 teacher needing more support teaching reasoning, problem solving & planning for depth ? See this article for FREE downloadable CPD

There is no ‘one size fits all’ approach to successfully tackling problem solving maths questions however, here are 6 general tips for students facing a problem solving question:

  • Read the whole question, underline important mathematical words, phrases or values.
  • Annotate any diagrams, graphs or charts with any missing information that is easy to fill in.
  • Think of what a sensible answer may look like. E.g. Will the angle be acute or obtuse? Is £30,000 likely to be the price of a coat?
  • Tick off information as you use it.
  • Draw extra diagrams if needed.
  • Look at the final sentence of the question. Make sure you refer back to that at the end to ensure you have answered the question fully.

There are many online sources of mathematical puzzles and questions that can help learners improve their problem-solving skills. Websites such as NRICH and our blog on SSDD problems have some great examples of KS2, KS3 and KS4 mathematical problems.

Read more: KS2 problem solving and KS3 maths problem solving

In this article, we’ve focussed on GCSE questions and compiled 30 problem solving maths questions and solutions suitable for Foundation and Higher tier students. Additionally, we have provided problem solving strategies to support your students for some questions to encourage critical mathematical thinking . For the full set of questions, solutions and strategies in a printable format, please download our 30 Problem Solving Maths Questions, Solutions & Strategies.

30 Problem Solving Maths Questions, Solutions & Strategies

30 Problem Solving Maths Questions, Solutions & Strategies

Help your students prepare for their maths GCSE with these free problem solving maths questions, solutions and strategies

These first 10 questions and solutions are similar to Foundation questions. For the first three, we’ve provided some additional strategies.

In our downloadable resource, you can find strategies for all 10 Foundation questions .

1) L-shape perimeter 

Here is a shape:

l-shape perimeter

Sarah says, “There is not enough information to find the perimeter.”

Is she correct? What about finding the area?

  • Try adding more information – giving some missing sides measurements that are valid. 
  • Change these measurements to see if the answer changes.
  • Imagine walking around the shape if the edges were paths. Could any of those paths be moved to another position but still give the same total distance?

The perimeter of the shape does not depend on the lengths of the unlabelled edges.

solution to finding perimeter of l-shape

Edge A and edge B can be moved to form a rectangle, meaning the perimeter will be 22 cm. Therefore, Sarah is wrong.

The area, however, will depend on those missing side length measurements, so we would need more information to be able to calculate it.

2) Find the missing point

Here is a coordinate grid with three points plotted. A fourth point is to be plotted to form a parallelogram. Find all possible coordinates of the fourth point.

coordinate grid

  • What are the properties of a parallelogram?
  • Can we count squares to see how we can get from one vertex of the parallelogram to another? Can we use this to find the fourth vertex?

There are 3 possible positions.

coordinate grid

3) That rating was a bit mean!

The vertical line graph shows the ratings a product received on an online shopping website. The vertical line for 4 stars is missing.

vertical graph

If the mean rating is 2.65, use the information to complete the vertical line graph.

Strategies 

  • Can the information be put into a different format, either a list or a table?
  • Would it help to give the missing frequency an algebraic label, x ?
  • If we had the data in a frequency table, how would we calculate the mean?
  • Is there an equation we could form?

Letting the frequency of 4 star ratings be x , we can form the equation \frac{45+4x}{18+x} =2.65

Giving x=2 

vertical graph

4) Changing angles

The diagram shows two angles around a point. The sum of the two angles around a point is 360°.

two angles around a point diagram

Peter says “If we increase the small angle by 10% and decrease the reflex angle by 10%, they will still add to 360°.”

Explain why Peter might be wrong.

Are there two angles where he would be correct?

Peter is wrong, for example, if the two angles are 40° and 320°, increasing 40° by 10% gives 44°, decreasing 320° by 10% gives 288°. These sum to 332°.

10% of the larger angle will be more than 10% of the smaller angle so the sum will only ever be 360° if the two original angles are the same, therefore, 180°.

5) Base and power

The integers 1, 2, 3, 4, 5, 6, 7, 8 and 9 can be used to fill in the boxes. 

base and power empty boxes

How many different solutions can be found so that no digit is used more than once?

There are 8 solutions.

6) Just an average problem 

Place six single digit numbers into the boxes to satisfy the rules.

boxes

The mean in maths is 5  \frac{1}{3}

The median is 5

The mode is 3.

How many different solutions are possible?

There are 4 solutions.

2, 3, 3, 7, 8, 9

3, 3, 4, 6, 7, 9

3, 3, 3, 7, 7, 9

3, 3, 3, 7, 8, 8

7) Square and rectangle  

The square has an area of 81 cm 2 . The rectangle has the same perimeter as the square.

Its length and width are in the ratio 2:1.

square and rectangle areas

Find the area of the rectangle.

The sides of the square are 9 cm giving a perimeter of 36 cm. 

We can then either form an equation using a length 2x and width x . 

Or, we could use the fact that the length and width add to half of the perimeter and share 18 in the ratio 2:1. 

The length is 12 cm and the width is 6 cm, giving an area of 72 cm 2 .

8) It’s all prime

The sum of three prime numbers is equal to another prime number.

empty number sequence

If the sum is less than 30, how many different solutions are possible?

There are 6 solutions. 

2 can never be used as it would force two more odd primes into the sum to make the total even.

9) Unequal share

Bob and Jane have £10 altogether. Jane has £1.60 more than Bob. Bob spends one third of his money. How much money have Bob and Jane now got in total?

Initially Bob has £4.20 and Jane has £5.80. Bob spends £1.40, meaning the total £10 has been reduced by £1.40, leaving £8.60 after the subtraction.

10) Somewhere between

Fred says, “An easy way to find any fraction which is between two other fractions is to just add the numerators and add the denominators.” Is Fred correct?

Solution 

Fred is correct. His method does work and can be shown algebraically which could be a good problem for higher tier learners to try.

If we use these two fractions \frac{3}{8} and \frac{5}{12} , Fred’s method gives us \frac{8}{20} = \frac{2}{5}

\frac{3}{8} = \frac{45}{120} , \frac{2}{5} = \frac{48}{120} , \frac{5}{12} = \frac{50}{120} . So \frac{3}{8} < \frac{2}{5} < \frac{5}{12}

The next 10 questions are crossover questions which could appear on both Foundation and Higher tier exam papers. We have provided solutions for each and, for the first three questions, problem solving strategies to support learners.

11) What’s the difference?

An arithmetic sequence has an nth term in the form an+b .

4 is in the sequence.

16 is in the sequence.

8 is not in the sequence.

-2 is the first term of the sequence.

What are the possible values of a and b ?

  • We know that the first number in the sequence is -2 and 4 is in the sequence. Can we try making a sequence to fit? Would using a number line help?
  • Try looking at the difference between the numbers we know are in the sequence.

If we try forming a sequence from the information, we get this:

Sequence

We can now try to fill in the missing numbers, making sure 8 is not in the sequence. Going up by 2 would give us 8, so that won’t work.

Number sequence

The only solutions are 6 n -8 and 3 n -5.

12) Equation of the hypotenuse

The diagram shows a straight line passing through the axes at point P and Q .

Q has coordinate (8, 0). M is the midpoint of PQ and MQ has a length of 5 units.

diagram with points p m and q

Find the equation of the line PQ .

  • We know MQ is 5 units, what is PQ and OQ ?
  • What type of triangle is OPQ ?
  • Can we find OP if we know PQ and OQ ?
  • A line has an equation in the form y=mx+c . How can we find m ? Do we already know c ?

PQ is 10 units. Using Pythagoras’ Theorem OP = 6

The gradient of the line will be \frac{-6}{8} = -\frac{3}{4} and P gives the intercept as 6.

13) What a waste

Harry wants to cut a sector of radius 30 cm from a piece of paper measuring 30 cm by 20 cm. 

section of a radius

What percentage of the paper will be wasted?

  • What information do we need to calculate the area of a sector? Do we have it all?
  • Would drawing another line on the diagram help find the angle of the sector?

The angle of the sector can be found using right angle triangle trigonometry.

The angle is 41.81°.

This gives us the area of the sector as 328.37 cm 2 .

The area of the paper is 600 cm 2 .

The area of paper wasted would be 600 – 328.37 = 271.62 cm 2 .

The wasted area is 45.27% of the paper.

14) Tri-polygonometry

The diagram shows part of a regular polygon and a right angled triangle. ABC is a straight line. Find the sum of the interior angles of the polygon.

Part of a regular polygon diagram

Finding the angle in the triangle at point B gives 30°. This is the exterior angle of the polygon. Dividing 360° by 30° tells us the polygon has 12 sides. Therefore, the sum of the interior angles is 1800°.

15) That’s a lot of Pi

A block of ready made pastry is a cuboid measuring 3 cm by 10 cm by 15 cm. 

cuboid with measurements

Anne is making 12 pies for a charity event. For each pie, she needs to cut a circle of pastry with a diameter of 18 cm from a sheet of pastry 0.5 cm thick.

How many blocks of pastry will Anne need to buy?

The volume of one block of pastry is 450 cm 3 . 

The volume of one cylinder of pastry is 127.23 cm 3 .

12 pies will require 1526.81 cm 3 .

Dividing the volume needed by 450 gives 3.39(…). 

Rounding this up tells us that 4 pastry blocks will be needed.

16) Is it right?

A triangle has sides of (x+4) cm, (2x+6) cm and (3x-2) cm. Its perimeter is 80 cm.

Show that the triangle is right angled and find its area.

Forming an equation gives 6x+8=80

This gives us x=12 and side lengths of 16 cm, 30 cm and 34 cm.

Using Pythagoras’ Theorem

16 2 +30 2 =1156 

Therefore, the triangle is right angled.

The area of the triangle is (16 x 30) ÷ 2 = 240 cm 2 .

17) Pie chart ratio

The pie chart shows sectors for red, blue and green. 

pie chart

The ratio of the angles of the red sector to the blue sector is 2:7. 

The ratio of the angles of the red sector to the green sector is 1:3. 

Find the angles of each sector of the pie chart.

Multiplying the ratio of red : green by 2, it can be written as 2:6. 

Now the colour each ratio has in common, red, has equal parts in each ratio.

The ratio of red:blue is 2:7, this means red:blue:green = 2:7:6.

Sharing 360° in this ratio gives red:blue:green = 48°:168°:144°.

18) DIY Simultaneously

Mr Jones buys 5 tins of paint and 4 rolls of decorating tape. The total cost was £167.

The next day he returns 1 unused tin of paint and 1 unused roll of tape. The refund amount is exactly the amount needed to buy a fan heater that has been reduced by 10% in a sale. The fan heater normally costs £37.50.

Find the cost of 1 tin of paint.    

The sale price of the fan heater is £33.75. This gives the simultaneous equations

p+t = 33.75 and 5 p +4 t = 167.

We only need the price of a tin of paint so multiplying the first equation by 4 and then subtracting from the second equation gives p =32. Therefore, 1 tin of paint costs £32. 

19) Triathlon pace

Jodie is competing in a Triathlon. 

A triathlon consists of a 5 km swim, a 40 km cycle and a 10 km run. 

Jodie wants to complete the triathlon in 5 hours. 

She knows she can swim at an average speed of 2.5 km/h and cycle at an average speed of 25 km/h. There are also two transition stages, in between events, which normally take 4 minutes each.

What speed must Jodie average on the final run to finish the triathlon in 5 hours?

Dividing the distances by the average speeds for each section gives times of 2 hours for the swim and 1.6 hours for the cycle, 216 minutes in total. Adding 8 minutes for the transition stages gives 224 minutes. To complete the triathlon in 5 hours, that would be 300 minutes. 300 – 224 = 76 minutes. Jodie needs to complete her 10 km run in 76 minutes, or \frac{19}{15} hours. This gives an average speed of 7.89 km/h.

20) Indices

a 2x × a y =a 3

(a 3 ) x ÷ a 4y =a 32

Find x and y .

Forming the simultaneous equations

Solving these gives

This final set of 10 questions would appear on the Higher tier only. Here we have just provided the solutions. Try asking your learners to discuss their strategies for each question.  

21) Angles in a polygon

The diagram shows part of a regular polygon.

part of a polygon diagram

A , B and C are vertices of the polygon. 

The size of the reflex angle ABC is 360° minus the interior angle.

Show that the sum of all of these reflex angles of the polygon will be 720° more than the sum of its interior angles.

Each of the reflex angles is 180 degrees more than the exterior angle: 180 + \frac{360}{n}

The sum of all of these angles is n (180 + \frac{360}{n} ). 

This simplifies to 180 n + 360

The sum of the interior angles is 180( n – 2) = 180 n – 360

The difference is 180 n + 360 – (180 n -360) = 720°

22) Prism and force (Non-calculator)

The diagram shows a prism with an equilateral triangle cross-section.

Prism

When the prism is placed so that its triangular face touches the surface, the prism applies a force of 12 Newtons resulting in a pressure of \frac{ \sqrt{3} }{4} N/m^{2}

Given that the prism has a volume of 384 m 3 , find the length of the prism.

Pressure = \frac{Force}{Area}

Area = 12÷ \frac{ \sqrt{3} }{4} = 16\sqrt{3} m 2

Therefore, the length of the prism is 384 ÷ 16\sqrt{3} = 8\sqrt{3} m

23) Geometric sequences (Non-calculator)

A geometric sequence has a third term of 6 and a sixth term of 14 \frac{2}{9}

Find the first term of the sequence.

The third term is ar 2 = 6

The sixth term is ar 5 = \frac{128}{9}

Diving these terms gives r 3 = \frac{64}{27}

Giving r = \frac{4}{3}

Dividing the third term twice by \frac{4}{3} gives the first term a = \frac{27}{8}

24) Printing factory

A printing factory is producing exam papers. When all 10 of its printers are working, it can produce all of the exam papers in 12 days.

For the first two days of printing, 3 of the printers are broken.

At the beginning of the third day it is discovered that 2 more printers have broken down, so the factory continues to print with the reduced amount of printers for 3 days. The broken printers are repaired and now all printers are available to print the remaining exams.

How many days in total does it take the factory to produce all of the exam papers?

If we assume one printer prints 1 exam paper per day, 10 printers would print 120 exam papers in 12 days. Listing the number printed each day for the first 5 days gives:

Day 5: 5 

This is a total of 29 exam papers.

91 exam papers are remaining with 10 printers now able to produce a total of 10 exam papers each day. 10 more days would be required to complete the job.

Therefore, 15 days in total are required.

25) Circles

The diagram shows a circle with equation x^{2}+{y}^{2}=13 .

tangent and circle

A tangent touches the circle at point P when x=3 and y is negative.

The tangent intercepts the coordinate axes at A and B .

Find the length AB .

Using the equation  x^{2}+y^{2}=13 to find the y value for P gives y=-2 .

The gradient of the radius at this point is - \frac{2}{3} , giving a tangent gradient of \frac{3}{2} .

Using the point (3,-2) in y = \frac {3}{2} x+c gives the equation of the tangent as y = \frac {3}{2} x – \frac{13}{2}

Substituting x=0 and y=0 gives A and B as (0 , -\frac {13}{2}) and ( \frac{13}{3} , 0)

Using Pythagoras’ Theorem gives the length of AB as ( \frac{ 13\sqrt{13} }{6} ) = 7.812.

tangent and circle diagram

26) Circle theorems

The diagram shows a circle with centre O . Points A, B, C and D are on the circumference of the circle. 

EF is a tangent to the circle at A . 

Angle EAD = 46°

Angle FAB = 48°

Angle ADC = 78°

Find the area of ABCD to the nearest integer.

The Alternate Segment Theorem gives angle ACD as 46° and angle ACB as 48°.

Opposite angles in a cyclic quadrilateral summing to 180° gives angle ABC as 102°.

Using the sine rule to find AC will give a length of 5.899. Using the sine rule again to find BC will give a length of 3.016cm.

We can now use the area of a triangle formula to find the area of both triangles.

0.5 × 5 × 5.899 × sin (46) + 0.5 × 3.016 × 5.899 × sin (48) = 17 units 2 (to the nearest integer).

27) Quadratic function

The quadratic function f(x) = -2x^{2} + 8x +11 has a turning point at P .

Find the coordinate of the turning point after the transformation -f(x-3) .

There are two methods that could be used. We could apply the transformation to the function and then complete the square, or, we could complete the square and then apply the transformation.

Here we will do the latter.

This gives a turning point for f(x) as (2,19).

Applying -f(x-3) gives the new turning point as (5,-19).

28) Probability with fruit

A fruit bowl contains only 5 grapes and n strawberries.

A fruit is taken, eaten and then another is selected.

The probability of taking two strawberries is \frac{7}{22} .

Find the probability of taking one of each fruit. 

There are n+5 fruits altogether.

P(Strawberry then strawberry)= \frac{n}{n+5} × \frac{n-1}{n+4} = \frac{7}{22}

This gives the quadratic equation 15n^{2} - 85n - 140 = 0

This can be divided through by 5 to give 3n^{2} - 17n- 28 = 0

This factorises to (n-7)(3n + 4) = 0

n must be positive so n = 7.

The probability of taking one of each fruit is therefore, \frac{5}{12} × \frac{7}{11} + \frac {7}{12} × \frac {5}{11} = \frac {70}{132}

29) Ice cream tub volume

An ice cream tub in the shape of a prism with a trapezium cross-section has the dimensions shown. These measurements are accurate to the nearest cm.

prism with a trapezium cross-section image

An ice cream scoop has a diameter of 4.5 cm to the nearest millimetre and will be used to scoop out spheres of ice cream from the tub.

Using bounds find a suitable approximation to the number of ice cream scoops that can be removed from a tub that is full.

We need to find the upper and lower bounds of the two volumes. 

Upper bound tub volume = 5665.625 cm 3

Lower bound tub volume = 4729.375 cm 3

Upper bound scoop volume = 49.32 cm 3  

Lower bound scoop volume = 46.14 cm 3  

We can divide the upper bound of the ice cream tub by the lower bound of the scoop to get the maximum possible number of scoops. 

Maximum number of scoops = 122.79

Then divide the lower bound of the ice cream tub by the upper bound of the scoop to get the minimum possible number of scoops.

Minimum number of scoops  = 95.89

These both round to 100 to 1 significant figure, Therefore, 100 scoops is a suitable approximation the the number of scoops.

30) Translating graphs

 The diagram shows the graph of y = a+tan(x-b ).

The graph goes through the points (75, 3) and Q (60, q).

Find exact values of a , b and q .

graph of y= a + tan (x-b)

The asymptote has been translated to the right by 30°. 

Therefore, b=30

So the point (45,1) has been translated to the point (75,3). 

Therefore, a=2

We hope these problem solving maths questions will support your GCSE teaching. To get all the solutions and strategies in a printable form, please download the complete resource .

Looking for additional support and resources at KS3? You are welcome to download any of the secondary maths resources from Third Space Learning’s resource library for free. There is a section devoted to GCSE maths revision with plenty of maths worksheets and GCSE maths questions . There are also maths tests for KS3, including a Year 7 maths test , a Year 8 maths test and a Year 9 maths test Other valuable maths practice and ideas particularly around reasoning and problem solving at secondary can be found in our KS3 and KS4 maths blog articles. Try these fun maths problems for KS2 and KS3, SSDD problems , KS3 maths games and 30 problem solving maths questions . For children who need more support, our maths intervention programmes for KS3 achieve outstanding results through a personalised one to one tuition approach.

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Maths Problem Solving Tasks for KS3/4

Maths Problem Solving Tasks for KS3/4

Subject: Mathematics

Age range: 14-16

Resource type: Other

Ace Maths

Last updated

5 January 2021

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PowerPoint slides. Problem solving questions. Suitable for extension, stretch or plenary. Questions that will get students engaged with thinking about mathematical concepts. Suitable for revision at KS3 or KS4 Grade c-d

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  • Ks3 Ks4 Maths Problem Solving

Maths problem solving – Why schools should prioritise this skill

Student standing in front of maths problem on board, representing maths problem solving

Problem-solving is a crucial, yet somehow underserved component of maths lessons – and that needs to change, says Dr Ems Lord…

Ems Lord

Maths teachers will be all too familiar with the three core focuses of the curriculum – fluency, reasoning and problem-solving. They might, however, find themselves struggling over that last one…

At NRICH’s many events and workshops, we’ve repeatedly heard teachers tell us that they’re struggling to consistently deliver high-quality problem-solving opportunities for their classes. Why?

Sometimes it’s because they lack the time to prioritise problem-solving within the timetable. They may worry that there’s a lack of consistency in approach across their school, or sometimes struggle to afford resources within their limited budget. Yet without problem-solving skills, students will be missing out on what makes maths so exciting for so many.

Why is problem-solving so important?

While exploring numbers can be very fulfilling, possessing amazing number skills isn’t always enough on its own. Learners must also be able to consider, choose and draw upon different strategies in order to successfully make decisions and solve problems.

Today’s workplaces demand employees who can solve non-routine problems in collaborative environments. To succeed, students will therefore need to develop resilience, be able to work collaboratively, think creatively and flexibly, and perhaps most importantly, be able to solve problems.

Collaborative problem-solving is becoming increasingly recognised as an important 21st century skill, as highlighted in the OECD’s PISA 2015 Collaborative Problem Solving assessment . That said, problem-solving doesn’t just empower students in the long term, but can also play a crucial role in helping them discover for themselves the joy of mathematics in the present day.

Imagine attending football practice, but only ever learning how to dribble. Or taking piano lessons, but only ever playing scales. Problem-solving can provide those real ‘lightbulb moments’ for students in maths lessons – it’s watching Messi score a goal, or listening to your favourite performer play live.

Problem-solving presents students with a mechanism for connecting the dots, as their maths skills and fluency in the subject become more purposeful and meaningful over time. They get to really see the value in what they’re doing, and once that value is better understood, they’ll become ever more engaged and motivated to learn.

Highlighted by Ofsted

Problem-solving is an area frequently picked up on by Ofsted, and was highlighted in the regulator’s July 2023 maths subject report. Said report called for schools to not just spend more time teaching specific problem-solving strategies, but for students to be given more opportunities to actually put those strategies into practice.

But where can schools turn to for support in implementing these recommendations, given their limited budgets and resources? One avenue is the nationwide Problem-Solving Schools programme recently launched by NRICH – part of the University of Cambridge, and one of the world’s largest maths outreach projects.

Problem-Solving Schools aims to ensure all schools are equipped to nurture students’ problem-solving skills through self-assessments of their existing provision, implementing whole-class problem-solving activities and rolling out dedicated teacher support – all provided entirely free of charge.

Problem-Solving Schools Charter

The programme begins with the Problem-Solving Schools Charter , which hundreds of schools have already signed up to. The Charter presents a framework to help you reflect on your current practice and identify areas for development, while highlighting five key areas:

  • Values and ethos
  • Leadership and professional development
  • Curriculum, pedagogy and assessment
  • Classroom culture
  • Problem-solving beyond the classroom

NRICH believes that every child can become a better problem-solver, and invites readers to sign up to that ambition as part of Problem-Solving Schools.

Once you’ve reflected on your current provision and practice and identified your action points, NRICH will then provide free classroom resources and support. Our low-threshold, high ceiling activities will allow you to work through maths problems with whole classes, giving students the time and support they need to consider problems in depth, and attempt different strategies for resolving them.

We can also provide wraparound support for teachers, consisting of webinars, videos and teacher notes to help build confidence. Embark on the journey towards improving problem-solving in your setting by registering as a Problem-Solving School .

Dr Ems Lord is the Director of NRICH. current Chair of The Mathematical Association , a Fellow of The Institute of Mathematics and its Applications and a Founding Fellow of the Chartered College of Teaching . Download a maths problem-solving KS2 lesson plan from NRICH.

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Problem solving with volumes of prisms

Problem solving with volumes of prisms

A problem-solving activity using volumes of prisms and cylinders, plus constructing and solving simple equations.

Students are shown prisms and a cylinder with measurements given either numerically or as unknowns x, y and z.

They must use the information to form and solve simple equations to find all the unknowns as well as the radius of the cylinder.

Students can then create their own prism with the same volume to go in the empty box.

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