How to tackle GCSE Maths word problems

Look at any GCSE Foundation or Higher paper and you'll find that a large share of the total marks sit inside worded, multi-step problems rather than short pure-technique questions. Students who are perfectly competent at the underlying skills — percentages, ratio, algebra, area — regularly lose marks on these questions simply because they don't know how to get started, or they answer a slightly different question to the one asked.

The general method

1. Read the whole question once before writing anything, to understand what is actually being asked.
2. Underline or note every number given, and what it represents.
3. Identify exactly what you are being asked to find — the final quantity, in the right units.
4. Decide which topic or technique connects the given information to that final quantity.
5. Write down your method as you go, even before you're sure it's right — method marks are awarded independently of the final answer.
6. Check your answer makes sense in context: a shop discount shouldn't turn a bag more expensive; a probability shouldn't be above 1; a length shouldn't be negative.

Worked example: ratio and proportion

A recipe for 6 people uses 480g of flour and 3 eggs. Sam wants to make the recipe for 15 people. How much flour and how many eggs does he need?

1. Find the scale factor: 15 ÷ 6 = 2.5.
2. Scale the flour: 480 × 2.5 = 1200g.
3. Scale the eggs: 3 × 2.5 = 7.5, so Sam needs 8 eggs (you cannot buy half an egg, so round up in context).

Worked example: percentage change and reverse percentages

A coat is on sale at £68 after a 15% discount. What was the original price?

This is a reverse percentage question, and the most common error is calculating 15% of £68 and adding it back on — that gives the wrong answer because 15% was taken off the original price, not off £68.

1. £68 represents 100% − 15% = 85% of the original price.
2. Set up the equation: 0.85 × original = 68.
3. Solve: original = 68 ÷ 0.85 = £80.

Worked example: a multi-step geometry problem

A rectangular garden measures 12m by 8m. A circular pond of radius 2m sits inside it. Turf costs £5.50 per m². What is the total cost of turfing the garden, excluding the area of the pond?

1. Find the area of the rectangle: 12 × 8 = 96 m².
2. Find the area of the circle: π × 2² = 4π ≈ 12.57 m².
3. Subtract to find the turfed area: 96 − 12.57 = 83.43 m².
4. Multiply by the cost per m²: 83.43 × 5.50 = £458.87 (or £458.86 depending on rounding at the intermediate step).

Turning words into algebra

Many worded problems, particularly on Higher papers, want you to form and solve an equation. The skill is translating phrases into algebraic expressions consistently.

Once you have the equation, the algebra itself is usually the easy part — GCSE examiners weight worded algebra questions heavily towards the translation step rather than the solving step, which is exactly why practising translation specifically pays off.

Why students lose marks even with the right final answer

• No working shown — a correct answer with no method can score less than full marks on multi-step questions, since most of the marks are for method.
• Units missing or wrong from the final answer.
• Correct maths, wrong quantity — answering 'total cost' when 'cost per item' was asked.
• Not rounding sensibly for the context (people, buses, tins of paint).
• Giving an answer to more decimal places than the question's context supports, such as £458.865 for money.

A practice routine that works

• Practise identifying the topic hidden inside a worded question before solving it — many students can do the maths but not spot which maths a paragraph is asking for.
• After every worded question, write one sentence describing what made it hard to start, and review those sentences after a set of practice.
• Deliberately practise questions across different topics mixed together, since real papers never group worded problems by topic.

Word problems are ultimately testing whether maths is useful to you, not just whether you can execute a method in isolation. Reading carefully, translating deliberately, and checking your answer against the real-world context closes almost all of the gap between knowing the maths and scoring the marks for it.

Worked example: compound measures

A car travels 165 miles in 3 hours. It then travels a further 120 miles at an average speed of 40 mph. What is the average speed for the whole journey?

The trap in this question is assuming average speed for the whole journey is the average of the two individual speeds — it is not. Average speed for the whole journey is always total distance divided by total time.

1. First leg: distance 165 miles, time 3 hours (speed is not needed directly here).
2. Second leg: time = distance ÷ speed = 120 ÷ 40 = 3 hours.
3. Total distance: 165 + 120 = 285 miles. Total time: 3 + 3 = 6 hours.
4. Average speed = total distance ÷ total time = 285 ÷ 6 = 47.5 mph.

Worked example: probability in context

A bag contains only red and blue counters. The probability of picking a red counter at random is 0.35. If there are 20 blue counters in the bag, how many counters are there in total?

1. Probability of blue = 1 − 0.35 = 0.65.
2. 0.65 of the total number of counters is 20, so set up the equation: 0.65 × total = 20.
3. Solve: total = 20 ÷ 0.65 ≈ 30.8 — since the number of counters must be a whole number, and this doesn't give one exactly, double-check the given probability was meant to divide the total evenly (a well-posed version of this question would use a probability like 0.6 or 0.4 that gives a whole-number answer).

This example illustrates a useful exam habit: if a worded probability or ratio question gives you a non-whole-number answer for something that must be a whole number (people, counters, tickets), it's worth re-reading the question, since it usually signals a misread number rather than an error in your method.

Worked example: compound interest and growth

£2,000 is invested at 3% compound interest per year. Find the value of the investment after 4 years, to the nearest penny.

1. Identify the multiplier for a 3% increase: 1.03.
2. Apply it once per year, which means raising it to the power of the number of years: 2000 × 1.03⁴.
3. Calculate: 2000 × 1.12550881 ≈ £2,251.02.

A bank of sentence starters for getting unstuck

When a word problem feels impossible to start, working through a short list of prompts often reveals the hidden structure of the question.

• What quantity is changing, and by how much, in each sentence?
• Is anything being compared — 'more than', 'less than', 'twice as many' — that could become an equation?
• Could I draw this? Many geometry and ratio problems become obvious once sketched, even roughly.
• What are the units of the final answer, and do my intermediate steps use consistent units throughout?
• Have I used every piece of given information — an unused number is a strong sign a step has been missed.

Common contexts and their standard structure