Working Mathematically - Problem Solving
KS3MA-KS3-D003
Cross-cutting domain focused on applying mathematics to routine and non-routine problems with increasing sophistication
National Curriculum context
Mathematical problem solving at KS3 requires pupils to apply and connect their mathematical knowledge in unfamiliar, multi-step contexts that do not have obvious or immediately visible solution paths. The statutory curriculum asks pupils to develop mathematical modelling skills — formulating real-world situations mathematically, working with them mathematically, and interpreting their results in context. Pupils are expected to select from a range of mathematical techniques, recognise when approximation is appropriate, and appreciate that many problems can be solved in multiple ways. Problem solving is the overarching context in which fluency and reasoning are exercised, and the curriculum expects all three to be developed together throughout KS3.
3
Concepts
1
Clusters
2
Prerequisites
3
With difficulty levels
Lesson Clusters
Apply problem-solving strategies, mathematical modelling and multiple representations
practice CuratedProblem-solving strategies, moving between representations and mathematical modelling are the three interlocking process skills in this domain. All three are process concepts that pervade KS3.
Prerequisites
Concepts from other domains that pupils should know before this domain.
Concepts (3)
Problem-solving strategies
process AI FacilitatedMA-KS3-C090
Breaking down problems into simpler steps and persevering to find solutions
Teaching guidance
Model problem-solving strategies explicitly: draw a diagram, make a simpler related problem, work backwards, look for patterns, use trial and improvement, be systematic. Present problems that require multiple steps and where the method is not immediately obvious. Encourage pupils to estimate answers before calculating, and to check answers for reasonableness afterwards. Use collaborative problem-solving where pupils discuss strategies in pairs or groups. Include problems from real-world contexts and cross-curricular applications (science, geography, economics).
Common misconceptions
Pupils often believe there is only one 'right' method for each problem and give up if their first approach does not work. Some pupils think problem-solving means applying a known procedure, not recognising that genuine problems require thinking beyond familiar methods. Others skip the estimation and checking stages, missing errors that would be caught by reasonableness checks.
Difficulty levels
Can identify what a problem is asking and attempt a first step, but struggles to plan a multi-step approach without guidance.
Example task
A cinema has 18 rows of seats. Each row has 24 seats. Tickets cost £8.50 each. If the cinema is full, how much money is collected?
Model response: First I find the total number of seats: 18 × 24 = 432 seats. Then I multiply by the ticket price: 432 × £8.50 = £3,672.
Breaks problems into smaller steps with some independence, using known mathematical techniques to work through each step.
Example task
A rectangular garden is 12 m by 8 m. A path 1 m wide runs around the outside. Find the area of the path.
Model response: The garden plus the path forms a larger rectangle: (12 + 2) by (8 + 2) = 14 m by 10 m (adding 1 m to each side for the path on both sides). Area of larger rectangle = 14 × 10 = 140 m². Area of garden = 12 × 8 = 96 m². Area of path = 140 - 96 = 44 m².
Plans and executes multi-step problem solutions independently, selecting appropriate strategies and checking the reasonableness of the answer.
Example task
A shop sells apples at 35p each or in bags of 6 for £1.80. Priya needs 20 apples. What is the cheapest way to buy them?
Model response: Option 1: Buy all individually: 20 × 35p = £7.00. Option 2: Buy 3 bags of 6 (18 apples) + 2 individual: 3 × £1.80 + 2 × £0.35 = £5.40 + £0.70 = £6.10. Option 3: Buy 4 bags of 6 (24 apples, 4 spare): 4 × £1.80 = £7.20. The cheapest option is 3 bags + 2 individual = £6.10. I checked: the bag price per apple is £1.80 ÷ 6 = 30p, which is cheaper than 35p, so buying as many bags as possible (without too many spare) gives the best price.
Solves non-routine and unfamiliar problems by selecting and combining techniques from different areas of mathematics, generalising results and evaluating the solution critically.
Example task
The first term of a sequence is 2. Each term after the first is found by squaring the previous term and subtracting 1. Which terms of this sequence are prime? Will there always be prime terms?
Model response: Term 1: 2. Term 2: 2² - 1 = 3 (prime). Term 3: 3² - 1 = 8 (not prime: 2³). Term 4: 8² - 1 = 63 = 7 × 9 (not prime). Term 5: 63² - 1 = 3968 = 3969 - 1. Note: n² - 1 = (n-1)(n+1) by difference of two squares. So from term 3 onwards, each term equals (previous - 1)(previous + 1). Since previous ≥ 3, both factors are ≥ 2, so the term is composite. Only terms 1 and 2 (which are 2 and 3) are prime. After that, every term factors as a difference of two squares and is therefore never prime.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.
Multiple representations
skill AI DirectMA-KS3-C091
Moving fluently between numerical, algebraic, graphical and diagrammatic representations
Teaching guidance
Teach pupils to translate between representations: given a table of values, draw a graph; given a graph, write an equation; given an equation, draw a bar model. Use matching activities where pupils connect equivalent representations. Emphasise that each representation has strengths: tables show exact values, graphs show trends, equations show general rules, diagrams show structure. Practise moving in all directions between representations, not just from algebra to graph. Use real-world contexts where choosing the right representation matters.
Common misconceptions
Pupils often think of representations as separate topics rather than different views of the same mathematical object. Some pupils can draw a graph from an equation but cannot write an equation from a graph. Others rely on a single preferred representation and resist using alternatives. The idea that a table, a graph and an equation can all represent the same function is a key conceptual hurdle.
Difficulty levels
Can work with one representation (e.g. a table of values or a graph) but does not yet connect it to other forms of the same relationship.
Example task
Here is a table: x = 1, 2, 3, 4 and y = 3, 5, 7, 9. Plot these points on a coordinate grid.
Model response: I plot (1,3), (2,5), (3,7), (4,9) on the grid. The points form a straight line going upwards.
Can convert between two representations of the same relationship (e.g. table to graph, or equation to table) when prompted.
Example task
The equation y = 2x + 1 describes a line. Complete a table of values for x = 0, 1, 2, 3, 4, then draw the graph.
Model response: When x = 0, y = 1. When x = 1, y = 3. When x = 2, y = 5. When x = 3, y = 7. When x = 4, y = 9. I plot these points and draw a straight line through them. The line crosses the y-axis at (0,1) and goes up by 2 for every 1 across.
Moves fluently between numerical, algebraic, graphical and diagrammatic representations, choosing the most useful form to solve a given problem.
Example task
A phone contract costs £12 per month plus 8p per text. Write a formula, draw a graph, and use whichever is most useful to find the cost of sending 150 texts in a month.
Model response: Formula: C = 12 + 0.08n where C is the monthly cost in pounds and n is the number of texts. For 150 texts: C = 12 + 0.08 × 150 = 12 + 12 = £24. The formula is the most efficient representation here because we need an exact answer for a specific value. The graph would be useful if we wanted to compare this contract visually with another, or read off how many texts we could send for a given budget.
Strategically selects and switches between representations to gain insight, using each form's strengths to illuminate different aspects of a mathematical situation.
Example task
Two tanks are being filled with water. Tank A starts with 20 litres and gains 5 litres per minute. Tank B starts empty and gains 8 litres per minute. Using both algebra and graphs, find when the tanks contain the same amount and explain why both representations are valuable.
Model response: Algebraically: Tank A = 20 + 5t, Tank B = 8t. Setting equal: 20 + 5t = 8t, so 20 = 3t, giving t = 6⅔ minutes, when both contain 53⅓ litres. Graphically: the lines intersect at (6.67, 53.3). The algebra gives the exact fractional answer. The graph shows the full picture: before t = 6⅔, Tank A has more; after, Tank B has more. The graph also shows Tank B's steeper gradient, meaning it fills faster. Neither representation alone gives the complete understanding — the algebra is precise, the graph shows the behaviour over time.
Delivery rationale
Secondary maths concept — abstract, procedural, and objectively assessable.
Mathematical modeling
process AI FacilitatedMA-KS3-C092
Translating situations into mathematical representations to solve problems
Teaching guidance
Start with simple modelling tasks: 'A swimming pool is filled at 50 litres per minute. Model the volume of water over time as a formula and a graph.' Progress to models that require simplification or assumption-making. Teach the modelling cycle: formulate (translate context to mathematics), solve (work mathematically), interpret (translate the answer back to context), validate (check the answer makes sense in context). Discuss the limitations of models — what assumptions have been made and how might they affect the answer?
Common misconceptions
Pupils often skip the interpretation and validation stages, presenting mathematical answers without relating them back to the context. Some pupils think a model must be perfectly accurate, not understanding that all models involve simplification. Others struggle to identify which variables matter and which can be ignored. The transition from a real-world situation to a mathematical representation is a significant cognitive step.
Difficulty levels
Can extract relevant numbers from a word problem and set up a simple calculation, but does not yet create a mathematical model independently.
Example task
A taxi charges £3 plus £1.50 per mile. How much does a 4-mile journey cost?
Model response: The cost is £3 + 4 × £1.50 = £3 + £6 = £9.
Can translate a real-world situation into a mathematical expression or equation with some guidance, identifying the variables and relationships involved.
Example task
A plumber charges a £40 call-out fee plus £25 per hour. Write a formula for the total cost C in terms of hours h. How many hours of work would cost £140?
Model response: C = 40 + 25h. For C = 140: 140 = 40 + 25h, so 100 = 25h, so h = 4 hours.
Independently translates complex real-world situations into mathematical models, solves them, and interprets the results back in context, including checking whether the mathematical answer makes sense.
Example task
A farmer has 200 m of fencing and wants to enclose a rectangular field against a straight wall (so only three sides need fencing). What dimensions give the maximum area?
Model response: Let the width perpendicular to the wall be w metres. The length parallel to the wall is 200 - 2w metres. Area A = w(200 - 2w) = 200w - 2w². To maximise, I find the vertex of this quadratic: w = -200 / (2 × -2) = 50 m. Length = 200 - 2(50) = 100 m. Maximum area = 50 × 100 = 5000 m². Check: the fencing is 50 + 100 + 50 = 200 m ✓.
Creates, analyses and critiques mathematical models for complex situations, recognising assumptions and limitations, and refining the model where appropriate.
Example task
A ball is thrown upwards. Its height in metres after t seconds is modelled by h = 20t - 5t². (a) Find when the ball hits the ground. (b) Find the maximum height. (c) What assumption does this model make, and when does the model break down?
Model response: (a) When h = 0: 20t - 5t² = 0, so 5t(4 - t) = 0, giving t = 0 or t = 4 seconds. The ball hits the ground after 4 seconds. (b) Maximum at t = -20/(2 × -5) = 2 seconds. h(2) = 20(2) - 5(4) = 40 - 20 = 20 m. (c) The model assumes no air resistance and constant gravitational acceleration (g = 10 m/s²). It breaks down for t < 0 (before the throw) and for t > 4 (the ball cannot go below ground height). For a real ball, air resistance would reduce the maximum height and increase the flight time slightly. The parabolic model is a simplification — it works well for short flights with dense objects but would be poor for a feather or a rocket.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.