WJEC GCSE Mathematics Revision
Explore WJEC GCSE Mathematics revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
Assessment Legacy GCSE Mathematics: both units at the same tier. Unit 1 is non-calculator; Unit 2 allows a calculator. Each is 50%. Foundation: 90 minutes, 65 marks per paper; Intermediate/Higher: 105 minutes, 80 marks. Final Maths resits are November 2026 and June 2027, subject to demand and eligibility. This is not the new Double Award.
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Reviewed course guidance
Revision points, with the explanation beside them
Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.
Options and source
Legacy WJEC GCSE tier
- Foundation tier
- Intermediate tier
- Higher tier
Summary of assessment: candidates sit both units at the same Foundation, Intermediate or Higher tier
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Topic group
Number
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 6, 16, 28: Reading and writing whole numbers of any magnitude expressed in figures or words; Rounding whole numbers to the nearest 10, 100, 1000, etc; Understanding place value and decimal places; Rounding decimals to the nearest whole number or a given number of decimal places
Open revision guidance
What to know
- Whole numbers use base-10 place value: units, tens, hundreds, thousands, and so on.
- Rounding rule: examine the digit to the right of the target position; if it is 5 or greater, round up; otherwise, round down.
- The first three decimal positions are tenths, hundredths and thousandths: one part in 10, 100 and 1,000 respectively.
What to do in an assessment
- Convert numerical figures to words: 1,045,002 is 'one million, forty-five thousand and two'.
- Round 5,823 to the nearest 100: the tens digit is 2, so round down to 5,800.
- Round 0.478 to two decimal places: the third digit is 8, so round up to 0.48.
In plain language
Place value defines a digit's worth (e.g., tens vs. hundreds). Rounding simplifies numbers by looking at the next digit right to decide whether to increment the target digit, making complex figures easier to estimate.
Self-check
Round 5,823 to the nearest 100 and 0.478 to two decimal places.
Show the answer
5,800 and 0.48.
View diagrams
Place-value table for 50.063
Key and description
50.063 = 5 tens + 0 ones + 0 tenths + 6 hundredths + 3 thousandths.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 6, 16, 28: Using the equivalences between decimals, fractions, ratios and percentages; Converting numbers from one form into another; Ordering and comparing whole numbers, decimals, fractions and percentages; Understanding and using directed numbers, including ordering directed numbers
Open revision guidance
What to know
- Fractions, decimals, and percentages are interchangeable representations of ratios.
- Part:part ratios describe separate groups; part:whole fractions describe a proportion of the total.
- Directed numbers are integers (positive, negative, and zero) ordered on a number line.
What to do in an assessment
- Convert 0.2 to a percentage: multiply by 100 to get 20%.
- Compare negative numbers: -8 is to the left of -3 on a number line, so -8 < -3.
- Convert ratio 1:3 to a fraction of the whole: 1 part out of 4 total parts equals 1/4.
- To order mixed forms, convert to a common form. 80% = 0.8, 3/4 = 0.75, 0.7 stays 0.7, and 1 = 1.0; therefore 0.7 < 3/4 < 80% < 1.
In plain language
Equivalences convert between forms to simplify operations. Ratios compare parts to parts (e.g., 1 red, 3 blue), while fractions compare parts to the whole (1 red out of 4 total). Directed numbers extend beyond zero, where further left on the number line indicates a smaller value.
Self-check
Convert 0.2 to a percentage and compare -8 and -3.
Show the answer
0.2 is 20%; -8 < -3 because -8 is smaller.
View diagrams
Rounding and integer number lines
Key and description
56 lies between 50 and 60, so to the nearest ten it rounds to 60. The second line marks the integers −5, −2, 0 and 3.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 6, 16, 28: Using the common properties of numbers, including odd, even, multiples, factors, primes; Expressing numbers as the product of their prime factors; Using the terms square, square root and cube; The use of index notation for positive integral indices; Interpreting numbers written in standard form in the context of a calculator display
Open revision guidance
What to know
- A prime number has exactly two factors: 1 and itself.
- Prime factorization expresses a number as a unique product of primes using index notation (e.g., 2² × 3).
- Standard form represents numbers as a × 10ⁿ where 1 ≤ a < 10.
What to do in an assessment
- Calculate prime factors of 18: 18 = 2 × 9 = 2 × 3 × 3 = 2 × 3².
- Calculate 2⁴: 2 × 2 × 2 × 2 = 16.
- Interpret calculator '1.2 E 4' as 1.2 × 10⁴ = 12,000.
- A square multiplies a number by itself: 7² = 7 × 7 = 49. The positive square root reverses this, √49 = 7. A cube multiplies three copies: 3³ = 3 × 3 × 3 = 27. For a negative base use brackets: (−3)² = 9, but (−3)³ = −27.
In plain language
Prime numbers are building blocks. Index notation (aⁿ) is shorthand for repeated multiplication. Standard form uses powers of ten to handle very large or small numbers, often seen as 'E' notation on calculators.
Self-check
Express 18 as a product of prime factors and calculate 2⁴.
Show the answer
18 = 2 × 3²; 2⁴ = 16.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 7, 17, 29: Using the facilities of a calculator to plan a calculation and evaluate expressions; Using addition, subtraction, multiplication, division, square and square root; Knowing how a calculator orders its operations; (Candidates will not be expected to list the key depressions that they have made.) Using calculators effectively and efficiently; Reading a calculator display correct to a specified number of decimal places
Open revision guidance
What to know
- Calculators follow BODMAS/PEMDAS: Brackets, Orders (Indices/Roots), Division/Multiplication, Addition/Subtraction.
- Basic operations include +, -, ×, ÷, x², and √.
- Results must be rounded based on the specific required decimal places.
What to do in an assessment
- Calculate (5.2 + 3.1)²: 1. Add inside parentheses (8.3). 2. Square the result (8.3² = 68.89).
- Plan steps: identify priority operations before entering into the calculator.
- Round 68.889 to 2 decimal places: look at the third decimal (9), round up to 68.89.
- You are not required to list every calculator key pressed. Show the mathematical expression and sufficient working: for √(81 + 19), write √100 = 10 rather than a list of button names.
In plain language
Calculators are literal machines; they follow strict rules for which operation happens first. To ensure accuracy, plan your equation in stages, use parentheses if you need to override the default order, and verify rounding.
Self-check
Calculate (5.2 + 3.1)² and round to 2 decimal places.
Show the answer
68.89.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 7, 17, 29: Understanding and using number operations and the relationships between them, including inverse operations and the hierarchy of operations; Addition, subtraction, multiplication and division of whole numbers, decimals, fractions and negative numbers; Finding a fraction or percentage of a quantity; Expressing one number as a fraction or percentage of another; Calculating fractional and percentage changes (increase and decrease); Calculating using ratios in a variety of situations; proportional division; The use of a non-calculator method to multiply and divide whole numbers up to and including the case of multiplication and division of a three-digit number by a two-digit number
Open revision guidance
What to know
- Inverse operations check accuracy: subtraction reverses addition; division reverses multiplication.
- Non-calculator division/multiplication: break 3-digit by 2-digit problems into smaller partial products or steps.
- Fractional/percentage change: (change / original) × 100.
What to do in an assessment
- Divide £60 in ratio 2:3: 1. Add parts (2+3=5). 2. Find value of one part (£60/5 = £12). 3. Allocate: (2×£12=£24) and (3×£12=£36).
- Add fractions: find common denominators, then sum the numerators.
- Calculate percentage change: if value goes 80 to 100, increase is 20. 20/80 = 0.25 = 25%.
- Without a calculator, 347 × 26 = 347 × 20 + 347 × 6 = 6940 + 2082 = 9022. In long multiplication, align the units and the tens partial products before adding.
- For 938 ÷ 14, 14 goes into 93 six times (84), leaving 9; bring down 8 to get 98. Seven further groups of 14 give 98, so the quotient is 67. Check 14 × 67 = 938.
- Line up decimal points for addition and subtraction: 4.70 + 0.85 = 5.55 and 4.70 − 0.85 = 3.85. For 1.2 × 0.3, calculate 12 × 3 = 36 then restore two decimal places to give 0.36. For 4.8 ÷ 0.6, multiply both numbers by 10: 48 ÷ 6 = 8.
- Negative-number operations: −5 + 8 = 3, −5 − 8 = −13, (−5) × (−2) = 10 and (−10) ÷ 2 = −5. Multiplying or dividing two numbers with the same sign gives a positive result; different signs give a negative result.
- For fractions, 1/3 + 1/4 = 4/12 + 3/12 = 7/12; 3/4 − 1/6 = 9/12 − 2/12 = 7/12; (2/3) × (3/5) = 2/5; (2/3) ÷ (4/5) = (2/3) × (5/4) = 5/6. Divide by a fraction by multiplying by its reciprocal.
- Find 3/5 of 40: divide by 5 then multiply by 3, giving 24. Find 15% of 80: 10% is 8 and 5% is 4, so the answer is 12.
- To express 18 as a fraction of 60, use 18/60 = 3/10. As a percentage, 18 ÷ 60 × 100 = 30%. The comparison quantity belongs in the denominator.
In plain language
Math operations are reversible, which helps in verification. Proportional division is just sharing a total into equal-sized parts defined by a ratio. For percentages, always divide the 'change' by the 'starting' number to get the relative increase.
Self-check
Divide £60 in the ratio 2:3.
Show the answer
£24 and £36.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 7, 17, 30: Recognising that recurring decimals are exact fractions, and that some exact fractions are recurring decimals
Open revision guidance
What to know
- A recurring decimal has a digit or sequence that repeats infinitely.
- Recurring decimals are exact fractions (e.g., 0.666... = 2/3).
- Algebraic conversion involves creating two equations with the same repeating tail and subtracting them.
What to do in an assessment
- Convert 0.666... to a fraction: 1. Let x = 0.666... 2. 10x = 6.666... 3. 10x - x = 6.666... - 0.666... 4. 9x = 6, so x = 6/9 = 2/3.
- Recognise that terminating decimals are also fractions.
In plain language
A repeating decimal is not just an approximation; it is an exact fraction. By multiplying the decimal by 10 or 100 to shift the decimal point and subtracting the original number, you cancel out the repeating 'tail' to reveal the fraction.
Self-check
Convert 0.666... to a fraction.
Show the answer
2/3.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 7, 17, 29: Estimating and approximating solutions to numerical calculations; Using estimation in multiplication and division problems with whole numbers to obtain approximate answers, e.g. by first rounding the numbers involved to 1 significant figure; Candidates must show sufficient working in order to demonstrate how they have obtained their estimate
Open revision guidance
What to know
- Estimation uses rounding to 1 significant figure to simplify arithmetic.
- It provides a check for reasonableness; it is not the exact value.
- Working steps must be shown to justify the estimate.
What to do in an assessment
- Estimate 712 ÷ 18: 1. Round to 1 s.f.: 700 ÷ 20. 2. Calculate: 35.
- Multiply 482 × 29: 1. Round to 500 × 30. 2. 15,000.
In plain language
Estimation rounds complex numbers to the nearest easy-to-use power of ten. This lets you quickly approximate the answer to verify that your precise calculator output is in the right 'ballpark'.
Self-check
Estimate 712 ÷ 18.
Show the answer
700 ÷ 20 = 35.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 8, 18, 30: Interpretation and use of mathematical information presented in written or visual form when solving problems, e.g. TV programme schedules, bus/rail timetables, distance charts, holiday booking information; Money: The basic principles of personal and household finance, including fuel and other bills, hire purchase, discount, VAT, taxation, best buys, wages and salaries, loan repayments, mortgages, budgeting, exchange rates and commissions; Simple interest; Profit and loss; Foreign currencies and exchange rates; Carrying out calculations relating to enterprise, saving and borrowing, investing, appreciation and depreciation
Open revision guidance
What to know
- Simple interest formula: I = P × r × t.
- Profit = Selling Price - Cost Price.
- Best buy: compare price per unit (e.g., £ per kg) rather than total price.
- All rates in these worked examples are hypothetical teaching values, not current tax rules, tariff quotes or financial advice. Read which charges are fixed and which depend on use, time or the number of people.
What to do in an assessment
- Calculate simple interest: For £200 at 5% for 3 years: 200 × 0.05 × 3 = £30.
- Convert currency: Multiply amount by exchange rate.
- Determine best buy: Divide total price by quantity to find unit cost.
- Under a fictional tax rule, the first £12,000 is tax-free and the remaining income is taxed at 10%. For £16,000, taxable income is £4000 and tax is 0.10 × £4000 = £400. These figures illustrate a stated rule, not current UK tax bands.
- A mortgage is a loan secured on a property. In a simplified repayment example, a £500 payment contains £100 interest and £400 capital repayment. A £100,000 outstanding balance becomes £99,600: interest is a charge, not a reduction in the capital owed.
- For a timetable, compare arrival and departure times on the same clock. A train departs at 09:47 and arrives at 11:12: 13 minutes to 10:00, one hour to 11:00 and 12 more minutes gives 85 minutes. If a connection leaves at 11:20, the transfer time is 8 minutes; compare that with the minimum transfer time before choosing it.
- In a distance chart, find the row for the starting town and the column for the destination. If A–B is 24 km and B–C is 31 km, travel via B is 55 km. A direct A–C entry of 40 km is a different route, not an instruction to add all three distances.
- For a holiday advertised as £420 per person plus a £35 booking fee per booking, two people pay 2 × £420 + £35 = £875, not 2 × (£420 + £35). Read whether each charge is per person, per night or per booking.
- For an energy bill using 250 kWh at 28p per kWh plus a 50p daily standing charge for 30 days, usage is £70 and the standing charge £15, giving £85 before any other stated charges.
- Hire purchase with a £150 deposit and 18 monthly payments of £40 costs £150 + 18 × £40 = £870. If cash price is £800, the extra cost is £70. Compare total costs, not only the monthly payment.
- With a hypothetical 20% VAT rate, a £40 pre-VAT price becomes £40 + £8 = £48. A 15% discount on £80 takes off £12, giving £68.
- For wages, 8 hours at £12 and 3 overtime hours at £18 give £96 + £54 = £150 gross. A £25,200 annual salary paid in 12 equal instalments is £2100 gross per month; net pay requires the stated deductions.
- A £1200 loan with a stated fixed total interest charge of £96 repaid in 12 equal monthly instalments has total repayment £1296 and instalments £108. This example has no additional fees.
- A monthly budget with income £1400 and expenses £920, £150 and £80 has total expenses £1150 and £250 remaining. Saving £50 each month reaches a £600 target in 12 months, before any interest.
- An enterprise buys goods for £180 and sells them for £240: profit is £60. Selling for £150 instead gives a £30 loss. A £1200 investment earning simple interest of 3% for 2 years adds £72, making £1272.
- A one-off 10% appreciation changes £1000 to £1100; a one-off 10% depreciation changes £2200 to £1980. Repeated compound changes are studied at Intermediate tier.
- If a bureau deducts a 2.5% commission from £200 before converting at €1.20 per £1, commission is £5, leaving £195, which converts to €234. If the fee is charged at another stage, follow that wording instead.
In plain language
Financial math calculates costs and growth. Simple interest is interest paid only on the initial amount invested, not on the interest added over time. 'Best buy' calculations normalise prices so you can compare goods of different sizes.
Self-check
If you invest £200 at 5% simple interest per year, how much interest do you earn after 3 years?
Show the answer
£30.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 8, 18, 30: Giving solutions in the context of a problem, interpreting the display on a calculator; Interpreting the display on a calculator; Knowing whether to round up or down as appropriate
Open revision guidance
What to know
- Calculators provide mathematical results; you must provide the logical answer.
- Context dictates rounding: currency needs 2 decimal places; items like people must be whole numbers.
- Rounding up or down depends on whether you have enough to cover a total.
What to do in an assessment
- Interpret money: if a display shows 45.666, round to £45.67.
- Interpret physical quantities: if you need 12.1 buses to move people, round up to 13, because 12 is insufficient.
In plain language
A calculator shows digits, but it doesn't know you are buying items or counting people. If the answer is 12.6, you must round up because you can't have a fraction of a bus. Context is the final step in the calculation.
Self-check
If a calculator shows 45.66666 for a money calculation, what is the answer?
Show the answer
£45.67.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 8, 18, 31: Understanding and using Venn diagrams to solve problems
Open revision guidance
What to know
- Venn diagrams use overlapping circles to show intersection (A ∩ B) and union (A ∪ B).
- Place items in the intersection first to avoid double-counting.
- Total = (A only) + (B only) + (Intersection) + (Neither).
What to do in an assessment
- Problem: 20 students total, 12 French, 8 German, 5 both. 1. Intersection is 5. 2. French only: 12 - 5 = 7. 3. German only: 8 - 5 = 3.
- Check sum: 7 (French only) + 5 (Both) + 3 (German only) = 15. Neither = 20 - 15 = 5.
In plain language
Venn diagrams are maps of sets. The overlap is the most important part because those items belong to both groups. By filling this section first, you ensure you don't count the same items twice when adding up categories.
Self-check
In a class of 20 students, 12 study French and 8 study German. 5 study both. How many study only French?
Show the answer
7 students.
View diagrams
Venn diagram for two events
Key and description
Constructed group of 30 pupils: n(A) = 18, n(B) = 14, n(A ∩ B) = 8. A only = 18 − 8 = 10; B only = 14 − 8 = 6; neither = 30 − (10 + 8 + 6) = 6. P(A ∪ B) = 24/30 = 4/5; P(A ∩ B) = 8/30 = 4/15.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 16, 28: Rounding numbers to a given number of significant figures
Open revision guidance
What to know
- The first significant figure is the first non-zero digit in a number.
- Subsequent digits (including zeros between non-zero digits) are significant.
- Rounding determines the precision of a number by comparing the digit to the right of the target position to the threshold of 5.
What to do in an assessment
- To round to n significant figures, locate the nth digit from the left (starting at the first non-zero digit).
- If the digit to the right is 0-4, keep the nth digit unchanged; if 5-9, add 1 to the nth digit.
- Replace remaining digits to the left of the decimal point with zeros; remove digits to the right of the decimal point.
- Worked example: Round 0.005672 to 2 significant figures. The first significant figure is 5, the second is 6. The next digit is 7, so round up: 0.0057.
In plain language
Rounding to significant figures keeps the most important parts of a number. It is like looking at a value's 'importance' from left to right. By rounding 0.005672 to 0.0057, we represent the number's magnitude using only the two most vital digits.
Self-check
Round 0.04085 to 3 significant figures.
Show the answer
The first significant figure is 4, the second is 0, the third is 8. The next digit is 5, so round up: 0.0409.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 16, 28: Least common multiple and highest common factor; Finding the LCM and HCF of numbers written as the product of their prime factors; Using the terms cube root and reciprocal; The use of index notation for zero and negative integral indices; The use of index notation for positive unit fractional indices; Writing whole numbers in index form; Using the rules of indices; Expressing and using numbers in standard form with positive and negative powers of 10
Open revision guidance
What to know
- Highest Common Factor (HCF) is the largest integer dividing two numbers. Least Common Multiple (LCM) is the smallest common multiple.
- Prime factors: HCF uses the lowest power of shared factors; LCM uses the highest power of all factors.
- Index Laws: x^a × x^b = x^(a+b), x^a / x^b = x^(a-b), (x^a)^b = x^(ab).
- Special indices: x^0 = 1 (for x ≠ 0), x^(-n) = 1/x^n, x^(1/n) = n-th root of x.
- Standard Form: A × 10^n where 1 ≤ A < 10.
What to do in an assessment
- To find HCF of 24 (2³×3¹) and 36 (2²×3²): Take 2² and 3¹, resulting in 4 × 3 = 12.
- To find LCM of 24 and 36: Take 2³ and 3², resulting in 8 × 9 = 72.
- Standard form example: 0.0045 = 4.5 × 10^(-3).
- Calculate 2³ × 2^(-1) = 2^(3-1) = 2² = 4.
- Write whole numbers in index form by factorising: 32 = 2 × 2 × 2 × 2 × 2 = 2⁵. A mixed-prime example is 72 = 2³ × 3²; its factors multiply back to 8 × 9 = 72.
- A cube root reverses cubing: ∛27 = 3 because 3³ = 27. The reciprocal of a non-zero number multiplies with it to make 1: the reciprocal of 2/3 is 3/2, since (2/3)(3/2) = 1. Zero has no reciprocal.
- A positive unit fractional index 1/n means the nth root. For positive bases, 8^(1/3) = ∛8 = 2 and 16^(1/4) = 2 because 2⁴ = 16. This is not division of the base by n.
In plain language
Prime factors are the DNA of a number. Standard form is a way to handle very large or tiny numbers by moving the decimal point and tracking the shift with a power of 10, making calculations much cleaner.
Self-check
Find the HCF and LCM of 12 and 18 using prime factors.
Show the answer
12 = 2² × 3¹, 18 = 2¹ × 3². HCF = 2¹ × 3¹ = 6. LCM = 2² × 3² = 4 × 9 = 36.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 17, 29: Using calculator facilities including the constant function, memory and brackets; Using power, root, constant, memory, brackets and appropriate statistical functions; Reading a calculator display correct to a specified number of significant figures; Using appropriate trigonometric functions on a calculator
Open revision guidance
What to know
- Calculators process operations using the hierarchy of operations (BIDMAS).
- Brackets must be used for denominators or multi-step operations to enforce order.
- Trigonometric functions (sin, cos, tan) require DEG mode.
- Memory (M+) stores values for reuse.
What to do in an assessment
- Calculate 5 / (2 + 3) by typing 5 ÷ ( 2 + 3 ) = 1.
- Use square root for hypotenuse/side problems: √(3² + 4²) = 5.
- Display rounding: If 2.523555 is calculated, rounding to 3 decimal places is 2.524.
- Worked example: Find sin(30°). In DEG mode, enter sin, then 30, equals 0.5.
- For a calculator memory task, calculate 18.5 × 4 = 74 and store it in memory. Calculate 12.5 × 2 = 25, add it to memory, then recall 99. Clear the memory before a new task. Key names differ between models; check your own device instructions.
- A constant multiplication function can generate 3, 6, 12, 24 by repeatedly multiplying by 2. Check each result rather than assuming repeated equals behaves identically on every calculator.
- For the data 2, 4, 9, enter the three values in the appropriate statistics mode and obtain mean (2 + 4 + 9)/3 = 5. Check that the data count is 3 and clear old data first.
In plain language
The calculator is a literal machine; it does exactly what you command. If you do not use brackets for a fraction like 1 / 2+3, it calculates (1/2)+3=3.5 instead of 1/5=0.2. Brackets tell the machine which part to calculate first.
Self-check
Calculate √(10² - 6²) / 2.
Show the answer
√(100 - 36) = √64 = 8. Then 8 / 2 = 4.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 17, 29: Calculating fractional and percentage changes using multipliers; Repeated proportional changes; appreciation and depreciation; Direct and inverse proportion
Open revision guidance
What to know
- A multiplier for a percentage increase of x% is (1 + x/100).
- Direct proportion: y = kx; Inverse proportion: xy = k.
- Ratio a:b means a total of (a+b) parts.
What to do in an assessment
- Increase 50 by 20%: 50 × 1.20 = 60.
- Decrease 50 by 20%: 50 × 0.80 = 40.
- Compound change: Increase 100 by 10% twice: 100 × 1.1 × 1.1 = 121.
- Inverse proportion calculation: If x=2 when y=10 (xy=20), find y when x=4: 4y=20, so y=5.
In plain language
Multipliers turn percentage problems into simple multiplication. Instead of finding the percent and adding it on in two steps, multiplying by 1.10 does both at once. It is the most efficient way to scale values.
Self-check
Decrease 80 by 10% and then by 10% again.
Show the answer
80 × 0.9 = 72; 72 × 0.9 = 64.8.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 18, 30: Compound interest and efficient calculation methods; Finding the original quantity given the result of a proportional change
Open revision guidance
What to know
- Simple interest: Interest = Principal × Rate × Time.
- Compound interest formula: A = P(1 + r)^n.
- Reverse percentage: To find the original price before a 20% increase, divide by 1.20.
What to do in an assessment
- Compound interest calculation: £100 at 5% for 3 years: 100 × 1.05³ = 115.76.
- Reverse percentage example: A price is £120 after a 20% increase. Original: 120 / 1.20 = 100.
- Profit calculation: (Selling Price - Cost Price) / Cost Price × 100.
In plain language
Compound interest creates more money over time because you earn interest on your interest, not just the original amount. Using the formula P(1+r)^n is faster than doing the interest year-by-year.
Self-check
Calculate the final amount of £200 at 5% compound interest for 2 years.
Show the answer
200 × 1.05² = 200 × 1.1025 = 220.50.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 18, 30, 31: Selecting an appropriate degree of accuracy and recognising limitations on the accuracy of data and measurements; Rounding an answer to a reasonable degree of accuracy in the light of the context; Recognising that measurement is approximate and that a measurement expressed to a given unit is in possible error of half a unit; The upper and lower bounds of numbers expressed to a given degree of accuracy; Calculating the upper and lower bounds in the addition and subtraction of numbers expressed to a given degree of accuracy
Open revision guidance
What to know
- For a value rounded to the nearest unit, the bounds lie half a unit below and above the recorded value; the upper endpoint is excluded under ordinary rounding.
- Lower bound: x - 0.5(unit). Upper bound: x + 0.5(unit).
- Bounds for sum/difference: use the upper/lower bounds of individual components.
What to do in an assessment
- Find bounds of 10cm (nearest cm): Lower 9.5, Upper 10.5.
- Find bounds of 10.0 (nearest 0.1): Lower 9.95, Upper 10.05.
- If a rounds to 10 and b to 5 to the nearest whole number, 9.5 ≤ a < 10.5 and 4.5 ≤ b < 5.5. The sum has 14 ≤ a + b < 16: its lower endpoint is attainable, its upper endpoint is not.
- For the same independent rounded measurements, 9.5 − 5.5 = 4 and 10.5 − 4.5 = 6 are the bounds of a − b, but neither is attainable because it requires an excluded upper endpoint: 4 < a − b < 6.
In plain language
Measurements are never exact. If you round to the nearest cm, the actual value could be anything between 0.5 below and 0.5 above your reading. Bounds help us understand the range of possible true values.
Self-check
A mass is 20kg to the nearest kg. What are the upper and lower bounds?
Show the answer
Lower bound: 19.5kg. Upper bound: 20.5kg.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 28: The use of index notation for other fractional indices
Open revision guidance
What to know
- For positive bases, xᵃ × xᵇ = xᵃ⁺ᵇ, xᵃ/xᵇ = xᵃ⁻ᵇ and (xᵃ)ᵇ = xᵃᵇ. The rules x⁰ = 1 and x⁻ᵃ = 1/xᵃ require x ≠ 0. Real roots also need the appropriate domain, such as non-negative input for an even root.
- Fractional index: x^(1/n) is the nth root of x.
- Standard form: A × 10^n (1 ≤ A < 10).
What to do in an assessment
- Simplify expressions: 8^(2/3) = (cube root of 8)^2 = 2^2 = 4.
- A negative fractional index also takes a reciprocal: 16^(−3/4) = 1/16^(3/4) = 1/8.
In plain language
Index notation is a shorthand for repeated multiplication. Fractional indices combine powers and roots; for example, the denominator of the fraction tells you which root to take, and the numerator tells you which power to raise it to.
Self-check
Evaluate 16^(3/4).
Show the answer
(16^(1/4))^3 = 2^3 = 8.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 30: Understanding annual rates, e.g. AER, APR
Open revision guidance
What to know
- Simple Interest: I = P × r × t.
- Compound Interest: A = P(1 + r)^n.
- APR and AER represent the annual cost/return of interest rates.
- Reverse percentages: Original = New Amount / Multiplier.
What to do in an assessment
- For 5% increase, multiply by 1.05. For compound interest, use (1 + rate)^time.
- Example: £1000 at 2% for 3 years: 1000 × 1.02^3 = £1061.21.
- If interest is 0.5% per month and compounded monthly, the annual growth factor is (1.005)^12 ≈ 1.0616778, so AER ≈ 6.17%. The nominal annual rate is 12 × 0.5% = 6%, not the same as AER. In the provided formula use i = 0.06 and n = 12.
- APR describes an annualised borrowing cost taking account of relevant charges and timing. For a simplified one-year £1000 loan with £100 interest and a £20 fee, the stated total charges are £120, or 12% of the initial principal. This is a simple cost comparison, not a calculation of regulated APR; repayment timing can change the actual APR.
In plain language
Compound interest creates more money than simple interest because you earn interest on your previous interest payments. APR is useful because it standardises different loan products, showing the 'real' yearly cost inclusive of fees.
Self-check
A price is £90 after a 10% increase. What was the original price?
Show the answer
90 / 1.10 = £81.82.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 31: Calculating the upper and lower bounds in calculations involving multiplication and division of numbers expressed to given degrees of accuracy
Open revision guidance
What to know
- Measurement error is half the unit of accuracy: ±(0.5 × unit).
- Upper bound = value + error; Lower bound = value - error.
- For division A/B: Max = Max(A) / Min(B), Min = Min(A) / Max(B).
What to do in an assessment
- Calculate max and min of values. If X=10 (to nearest 1), bounds are 9.5 and 10.5.
- Use these to find extremes of calculations.
- For positive a and b recorded as 8 to the nearest unit and 2 to the nearest 0.1, 7.5 ≤ a < 8.5 and 1.95 ≤ b < 2.05. Product bounds are 7.5×1.95 = 14.625 and 8.5×2.05 = 17.425, so 14.625 ≤ ab < 17.425.
- For the corresponding ratio, divide the smallest numerator by the largest denominator for the lower bound, and the largest numerator by the smallest denominator for the upper: 7.5/2.05 < a/b < 8.5/1.95. These bounds are approximately 3.6585 and 4.3590. Neither endpoint is attained; keep exact ratios when stating strict inequalities.
In plain language
Because measurements are rounded, the true value sits within a range. When we perform calculations, we use these ranges to find the largest and smallest possible results, which ensures we understand the potential error in our final answer.
Self-check
A distance is 8 cm to the nearest cm and a time 2 s to the nearest 0.1 s. Calculate the upper bound for speed, to two decimal places.
Show the answer
Upper bound = 8.5/1.95 ≈ 4.36 cm/s. It is a bound, not an attained maximum, because distance is less than 8.5 cm.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 30: Converting recurring decimals to fractional form; Distinguishing between rational and irrational numbers; Manipulating surds; using surds and π in exact calculations; Simplifying numerical expressions involving surds, excluding rationalisation of a denominator such as 2 − √3
Open revision guidance
What to know
- Recurring decimals can be written as fractions by creating equations that cancel out the repeating part.
- Surds are irrational roots like √2 or √3.
- Simplified surd rule: √ab = √a * √b.
- Exact form means keeping values as fractions or roots (no rounding).
- For non-negative a and b, √(ab) = √a × √b. Rational numbers can be written as ratios of integers with a non-zero denominator; irrational numbers cannot. For example 0.5 = 1/2 is rational, while √2 and π are irrational.
What to do in an assessment
- Convert 0.555…: let x = 0.555… . Then 10x = 5.555… and subtracting gives 9x = 5, so x = 5/9.
- Simplify √72: √72 = √(36 * 2) = √36 * √2 = 6√2.
- Add surds: 2√3 + 5√3 = 7√3.
- To rationalise a single-term denominator, 3/√2 = (3√2)/2, multiplying numerator and denominator by √2. Rationalisation of a two-term denominator such as 2 − √3 is excluded by this legacy specification.
In plain language
We use exact forms like surds and fractions because decimals often require rounding, which introduces error. By keeping √2, we maintain perfect accuracy throughout the calculation.
Self-check
Convert 0.444... to a fraction and simplify √20.
Show the answer
x = 4/9. √20 = √(4 * 5) = 2√5.
Topic group
Algebra
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 9, 19, 32: Recognition, description and continuation of patterns in number; Description, in words, of the rule for the next term of a sequence
Open revision guidance
What to know
- A sequence is an ordered set of numbers following a logical rule.
- An arithmetic sequence is one where the difference between consecutive terms is constant.
- Pattern recognition requires finding the difference between term n+1 and term n.
What to do in an assessment
- Identify the difference between terms: 4, 7, 10, 13 has differences 7-4=3, 10-7=3, 13-10=3.
- State the rule in words: Add 3 to the previous term.
- To find the next term, perform the rule on the last given term: 13 + 3 = 16.
In plain language
If you look at the sequence 4, 7, 10, 13, you see the gap between each number is always 3. Because this gap is constant, we can predict the next number by simply adding 3 to the last one. The rule is 'add 3 to the previous term'.
Self-check
Describe the rule and find the next term for the sequence: 5, 9, 13, 17.
Show the answer
The rule is 'add 4 to the previous term'. The next term is 17 + 4 = 21.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 9, 19, 32: Generating linear sequences given the nth term rule
Open revision guidance
What to know
- The nth term rule is a formula using n to denote the position of a term.
- In 3n + 2, '3' is the multiplier for the position n and '2' is the constant adjustment.
- This is a linear sequence because the change between each step is constant (the coefficient of n).
What to do in an assessment
- To find the 1st term, substitute n=1: 3(1) + 2 = 5.
- To find the 2nd term, substitute n=2: 3(2) + 2 = 8.
- To find the 3rd term, substitute n=3: 3(3) + 2 = 11.
- Sequence: 5, 8, 11.
In plain language
The letter n stands for the position of a number in a list. If your rule is 2n + 5, you calculate the 1st number by multiplying 1 by 2 and then adding 5. Using the formula allows you to jump directly to any position, like the 10th term, without calculating all the numbers before it.
Self-check
Generate the first three terms for the rule 4n - 1.
Show the answer
n=1: 4(1)-1=3; n=2: 4(2)-1=7; n=3: 4(3)-1=11. Terms are 3, 7, 11.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 9, 19, 32: Construction and interpretation of conversion graphs; Construction and interpretation of travel graphs; Construction and interpretation of graphs that describe real-life situations; Interpretation of graphical representation used in the media, recognising that some graphs may be misleading
Open revision guidance
What to know
- Conversion graphs allow for graphical estimation between two units by plotting linear data points.
- Travel graphs plot distance (y-axis) against time (x-axis); the gradient indicates speed.
- A horizontal line on a travel graph indicates zero speed (the object is stationary).
- Misleading graphs often use non-zero origins or inconsistent scale increments to skew perception.
What to do in an assessment
- To construct a conversion graph: plot two conversion pairs (e.g., 0 miles=0 km, 5 miles=8 km) and connect them with a straight line.
- To interpret a travel graph: calculate speed as vertical change (distance) divided by horizontal change (time).
- To evaluate media graphs: always verify if the y-axis starts at 0 and if all intervals represent the same quantity.
In plain language
Graphs represent relationships visually. In a conversion graph, if you know that 5 miles equals 8 kilometers, you can mark those coordinates and draw a line to estimate any other value. However, you must be careful: if a graph starts at 100 instead of 0, it might make a small change look like a huge jump.
Self-check
If a conversion graph for currency shows £1 = €1.20, what would be the coordinates for £10?
Show the answer
If £1 = €1.20, then £10 = €12. The coordinate is (10, 12).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 9, 19, 32: Using coordinates in 4 quadrants; Drawing and interpreting the graphs of x = a, y = b, y = ax + b
Open revision guidance
What to know
- The coordinate plane uses four quadrants defined by x (horizontal) and y (vertical) axes.
- x = a represents a vertical line passing through the x-axis at value a.
- y = b represents a horizontal line passing through the y-axis at value b.
- y = ax + b is a straight line with gradient a and y-intercept b. It slopes when a ≠ 0 and is horizontal when a = 0.
What to do in an assessment
- Plot (x, y): move along x, then move vertically by y.
- Graph y = 2x + 3: find two points. If x=0, y=3. If x=1, y=5. Plot (0,3) and (1,5) and draw the line.
- Identify y = -4 as a horizontal line passing through -4 on the y-axis.
In plain language
Coordinates act like an address for a point. When graphing equations, y = ax + b tells us two things: where to start on the middle line (the intercept b) and how steeply to tilt the line (the gradient a). A positive 'a' means the line goes up from left to right.
Self-check
For the line y = 3x - 2, find the coordinates when x = 0 and x = 2.
Show the answer
When x=0, y=3(0)-2 = -2. Coordinate is (0, -2). When x=2, y=3(2)-2 = 4. Coordinate is (2, 4).
View diagrams
Vertical, horizontal and sloping straight lines
Key and description
x = 1 is vertical, y = −2 is horizontal, and y = 2x + 1 has gradient 2 and y-intercept 1. Points (−1, −1), (0, 1) and (1, 3) lie on the sloping line.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 9, 20, 33: Substitution of positive and negative whole numbers, fractions and decimals into simple formulae expressed in words or in symbols; Understanding the basic conventions of algebra; Formation and simplification of expressions involving sums, differences and products; Collection of like terms; Expansion of a(bx + c), where a, b and c are integers; Formation, manipulation and solution of linear equations
Open revision guidance
What to know
- Algebraic expressions are formed using variables and constants.
- Like terms are terms containing the same variables raised to the same powers.
- Substitution involves replacing variables with given numerical values.
- Expanding brackets follows the distributive property: a(b+c) = ab + ac.
- Linear equations are solved by performing inverse operations to isolate the variable.
- Algebraic conventions: 3x means 3 × x, xy means x × y, and x² means x × x. They do not mean joining numerical digits.
What to do in an assessment
- Simplify: 5x + 3y + 2x - y = (5+2)x + (3-1)y = 7x + 2y.
- Expand: 4(2x - 3) = 8x - 12.
- Solve 3x - 4 = 8: Add 4 (3x = 12), then divide by 3 (x = 4).
- Substitute into 3x + 2: x = −2 gives 3(−2) + 2 = −4; x = 1/2 gives 3/2 + 2 = 3.5; x = 0.4 gives 1.2 + 2 = 3.2. Keep brackets around negative substitutions.
- A delivery charge has a fixed £4 plus £3 per parcel, so C = 4 + 3n. For n = 5, C = £19. Given C = £19, solve 4 + 3n = 19: subtract 4 and divide by 3 to get n = 5.
- For products of simple terms, (2x)(3y) = 6xy. For sums and differences, 2a + 3b − a = a + 3b; unlike terms cannot be collected into one term.
In plain language
Think of an equation like a balanced scale. To find the unknown, you must keep the scale balanced. If you add 4 to one side, you must add 4 to the other. Collecting like terms is simply grouping similar items, like sorting apples and oranges into separate baskets.
Self-check
Solve for x: 5x + 2 = 17.
Show the answer
Subtract 2: 5x = 15. Divide by 5: x = 3.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 19, 32: Description of the rule for the next term of a sequence in symbols
Open revision guidance
What to know
- Linear sequences change by a constant difference between terms.
- The nth term formula (dn + c) describes any term without listing them all.
What to do in an assessment
- Identify difference: For 3, 7, 11, 15, the difference is +4.
- Describe in words: Start at 3 and add 4 each time.
- Find n=1 term: 3. If formula is 4n + c, 4(1) + c = 3, so c = -1. Formula: 4n - 1.
- For 3,7,11,15, write the next-term rule in symbols as uₙ₊₁ = uₙ + 4, with u₁ = 3. Thus after u₄ = 15 comes u₅ = 19. This recurrence uses the previous term; the nth-term rule 4n − 1 uses the term position directly.
In plain language
A sequence is a set of numbers following a rule. By finding the difference between numbers, we find how to 'generate' the next one. Algebra lets us skip straight to the 100th term without counting.
Self-check
Describe the pattern 2, 5, 8, 11 in words and as an nth term.
Show the answer
Words: Start at 2 and add 3. Formula: 3n - 1.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 19, 32: Finding the nth term of a sequence where the rule is linear or quadratic; Generating non-linear sequences given the nth term rule
Open revision guidance
What to know
- Linear sequences have a constant first difference.
- Quadratic sequences have a constant second difference.
- Quadratic nth term formula: an² + bn + c.
What to do in an assessment
- Linear nth term: For 5, 8, 11, difference is 3, start is 2, so 3n + 2.
- Quadratic nth term: For 1, 4, 9, 16... second difference is 2. a = 2/2 = 1. Rule: n².
- Generating: Given 2n² + 1, find 3rd term: 2(3²) + 1 = 18 + 1 = 19.
In plain language
Quadratic sequences involve n². While linear sequences go up by a set amount, quadratic sequences accelerate. If the second difference is constant, the rule must involve n².
Self-check
Find the nth term for 4, 7, 12, 19.
Show the answer
Differences: 3, 5, 7. Second difference: 2. n² term. n² gives 1, 4, 9, 16. The sequence is n² + 3.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 19, 32: Recognising and interpreting graphs that illustrate direct and inverse proportion
Open revision guidance
What to know
- Direct proportion: Graph is a straight line through (0,0).
- Inverse proportion: Graph is a curve y = k/x.
- Speed = distance / time (gradient of distance-time graph).
What to do in an assessment
- For direct proportion y = 3x, plot (0,0),(1,3),(2,6),(3,9); a straight line through the origin has constant y/x = 3 for x ≠ 0. Doubling x doubles y.
- For inverse proportion y = 12/x, plot (1,12),(2,6),(3,4),(4,3). The graph is curved and xy stays 12: doubling x halves y. x = 0 is excluded. Negative x values give a separate branch, such as (−3,−4).
- A straight line y = 3x + 2 is not direct proportion because it does not pass through the origin. A decreasing straight line is not automatically inverse proportion; check the constant-product rule.
In plain language
Direct proportion has a constant ratio y/x and a straight graph through the origin. Inverse proportion has a constant product xy and a reciprocal curve. Recognise the relationship from the graph and verify it with coordinates rather than assuming every rising or falling graph is proportional.
Self-check
A graph contains (2,6) and (4,3) and follows y = 12/x. Is it direct or inverse proportion, and what happens if x doubles?
Show the answer
Inverse proportion: xy = 12. Doubling x halves y. The curve excludes x = 0 and is not a straight line through the origin.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 19, 32: Recognising and sketching the graphs of x = a, y = b, y = ax + b; The gradients of parallel lines
Open revision guidance
What to know
- x = a is a vertical line; y = b is a horizontal line.
- y = ax + b: 'a' is the gradient (steepness), 'b' is the y-intercept.
- Parallel lines have the same gradient.
What to do in an assessment
- Sketch y = 2x + 1: Intercept is 1, gradient is 2 (for every 1 unit right, go 2 units up).
- Identify parallel lines: y = 3x + 2 and y = 3x - 5 have the same gradient (3).
- Plotting points: (x, y) = (2, 3) is 2 units right, 3 units up.
In plain language
The equation y = ax + b tells you exactly where a line lives on a grid. 'a' tells you how tilted it is; if two lines have the same 'a', they will never touch, which is why they are parallel.
Self-check
What is the gradient of a line parallel to y = -2x + 4?
Show the answer
-2.
View diagrams
Vertical, horizontal and sloping straight lines
Key and description
x = 1 is vertical, y = −2 is horizontal, and y = 2x + 1 has gradient 2 and y-intercept 1. Points (−1, −1), (0, 1) and (1, 3) lie on the sloping line.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 19, 32: Identifying the equation of lines parallel or perpendicular to a given line, to satisfy given conditions; Knowledge and use of the form y = mx + c to represent a straight line where m is the gradient of the line, and c is the value of the y-intercept; Drawing, interpretation, recognition and sketching the graphs of y = ax² + b; Drawing and interpretation of graphs of y = ax² + bx + c; Drawing and interpreting graphs when y is given implicitly in terms of x
Open revision guidance
What to know
- The form y = mx + c represents a straight line, where m is the gradient (change in y divided by change in x) and c is the y-axis intercept.
- Parallel lines share the same gradient m; perpendicular lines have gradients m₁ and m₂ such that m₁ × m₂ = -1.
- The graph y = ax² + b is a parabola with a y-intercept at b; y = ax² + bx + c is a parabolic curve whose vertex and intercepts depend on a, b, and c.
- Implicit equations like 2x + y = 6 must be rearranged to y = -2x + 6 to identify m and c.
What to do in an assessment
- Identify the gradient and intercept: For y = 3x - 4, the gradient is 3 and the intercept is -4.
- Find a perpendicular line: For y = 2x + 5, the perpendicular gradient is -1/2. Using point (0, 2), the equation is y = -0.5x + 2.
- Sketch y = x² + 2: The vertex is at (0, 2) and the curve is a 'U' shape opening upwards.
- Rearrange to y = mx + c: For x + 2y = 8, subtract x to get 2y = -x + 8, then divide by 2 to get y = -0.5x + 4.
In plain language
The equation y = mx + c acts as a roadmap for a line: 'm' dictates the steepness and 'c' tells you where it crosses the vertical axis. If a line is perpendicular, its steepness is the negative reciprocal of the original; this is why 2 becomes -0.5, creating a 'flip' in slope. Parabolas are curves formed by quadratic expressions; the 'x²' term forces the shape to turn around, creating the signature 'U' shape.
Self-check
Find the equation of a line perpendicular to y = 0.5x + 3 that passes through (0, 0).
Show the answer
The original gradient is 0.5, so the perpendicular gradient is -1 / 0.5 = -2. The line passes through the origin (0, 0), so the intercept c is 0. The equation is y = -2x.
View diagrams
Roots and turning point of a quadratic
Key and description
y = x² − 2x − 3 = (x − 3)(x + 1). Roots: (−1, 0) and (3, 0). Completing the square gives (x − 1)² − 4, so the turning point is (1, −4).
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 20, 33: Formation and simplification of expressions involving powers; Changing the subject of a formula when the subject appears in one term
Open revision guidance
What to know
- Algebraic expressions involve variables, coefficients, and exponents, such as 3x² or 4a + 3b.
- Substitution requires replacing variables with specific numerical values following the order of operations.
- Changing the subject of a formula uses inverse operations to isolate a target variable.
- Expressions are simplified by combining like terms and applying the laws of indices, such as x² × x³ = x⁵.
What to do in an assessment
- Substitute a = -3, b = 2 into 2a² + b: 2(-3)² + 2 = 2(9) + 2 = 18 + 2 = 20.
- Simplify expressions: 5x + 2y - 3x + 4y = 2x + 6y.
- Change the subject to x: Given y = 4x - 7. Add 7: y + 7 = 4x. Divide by 4: x = (y + 7) / 4.
In plain language
Think of an algebraic expression as a puzzle where letters are placeholders for numbers. To change the subject of a formula, you are essentially performing a 'rescue mission' for a specific variable. By applying inverse operations—like adding to cancel subtraction or dividing to cancel multiplication—you move everything else to the other side of the equals sign to set that variable free.
Self-check
Change the subject of the formula P = 3x - 5 to x.
Show the answer
Add 5 to both sides to get P + 5 = 3x, then divide by 3 to get x = (P + 5) / 3.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 20, 33: Extraction of common factors; Formation and manipulation of simple linear inequalities; Multiplication of two linear expressions; expansion of (ax + by)(cx + dy) and (ax + by)2, where a, b, c, d are integers; Factorisation of quadratic expressions of the form x² + ax + b
Open revision guidance
What to know
- Common factors are the highest values or variables that divide evenly into all terms in an expression.
- Expansion of (ax + by)(cx + dy) is achieved by multiplying each term (FOIL method).
- Factorising quadratics of the form x² + ax + b requires finding two numbers that multiply to b and sum to a.
- Linear inequalities describe ranges of values rather than a single point.
What to do in an assessment
- Extract the common factor: 10x² + 15x = 5x(2x + 3).
- Expand brackets: (2x + 1)(x + 4) = 2x² + 8x + x + 4 = 2x² + 9x + 4.
- Square a binomial: (x + 5)² = (x + 5)(x + 5) = x² + 10x + 25.
- Factorise x² + 6x + 8: Numbers that multiply to 8 and add to 6 are 2 and 4. Result is (x + 2)(x + 4).
- If a budget is £20, a fixed fee is £5 and each item costs £3, form 3n + 5 ≤ 20. Subtract 5: 3n ≤ 15; divide by 3: n ≤ 5. Non-negative whole-number item counts are 0,1,2,3,4,5.
- For two-variable expansion, (2x + 3y)(x − y) = 2x² − 2xy + 3xy − 3y² = 2x² + xy − 3y². Squaring (2x + 3y) gives 4x² + 12xy + 9y², including the middle term.
In plain language
Expanding is the process of breaking brackets down into a full polynomial, while factorisation is the reverse: putting it back into bracket form. Imagine you are unravelling a knot (expanding) or sorting items back into their original packages (factorising). When you factorise a quadratic, you are hunting for the 'building block' factors that, when multiplied, create the original structure.
Self-check
Factorise x² + 7x + 12.
Show the answer
We look for two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4. The result is (x + 3)(x + 4).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 20, 34: The solution of linear equations with whole number coefficients in solving problems set in real-life contexts
Open revision guidance
What to know
- Linear equations translate real-world scenarios into mathematical statements where the unknown is represented by a letter.
- The equality sign signifies a balanced relationship between two quantities.
- Coefficients represent the rate or quantity per unit.
What to do in an assessment
- Construct an equation: If an item costs £5 plus £2 per hour (h), the total cost (C) is C = 2h + 5.
- Solve for an unknown: Given C = 15, solve 2h + 5 = 15. Subtract 5: 2h = 10. Divide by 2: h = 5 hours.
- Verify: 2(5) + 5 = 10 + 5 = 15. The calculation matches the requirement.
In plain language
Solving a word problem is about translation: converting English sentences into a mathematical language. Once you have an equation, you treat it like a scale. Whatever you do to one side, you must do to the other to keep the scale balanced while isolating the 'unknown' until it is revealed.
Self-check
A gym charges a £10 registration fee and £5 per visit (v). If a member spent £30, how many visits did they make?
Show the answer
The equation is 5v + 10 = 30. Subtract 10 from both sides: 5v = 20. Divide by 5: v = 4 visits.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 20, 34: Solution of linear equations and linear inequalities with whole number and fractional coefficients; The formation and solution of two simultaneous linear equations with whole number coefficients by graphical and algebraic methods in solving problems set in real-life contexts; Solution by factorisation and graphical methods of quadratic equations of the form x² + ax + b = 0; Solution of a range of cubic equations by trial and improvement methods, justifying the accuracy of the solution
Open revision guidance
What to know
- Linear equations with fractional coefficients are solved by multiplying the entire equation by the common denominator.
- Simultaneous equations are two equations solved together to find values for two variables that satisfy both.
- Quadratic equations x² + ax + b = 0 are solved by factorising and setting each bracket equal to zero.
- Cubic equations can be solved by testing values (trial and improvement) to narrow the window where the root lies.
What to do in an assessment
- Solve (x/2) + 3 = 5: Multiply all by 2 to get x + 6 = 10, so x = 4.
- Solve x² + 4x + 3 = 0: Factorise to (x + 1)(x + 3) = 0. Solutions are x = -1, x = -3.
- Trial and improvement for x³ + x = 10: Try x=2 (8+2=10). The solution is x = 2.
- For a fractional inequality, x/2 + 3 ≤ 5 gives x/2 ≤ 2 and x ≤ 4. For −3x > 6, divide by −3 and reverse the sign: x < −2. Test x = −3: 9 > 6 is true.
- For four tickets costing £16, with adult tickets £5 and pupil tickets £3, let counts be x and y. Then x + y = 4 and 5x + 3y = 16. Multiply the first by 5: 5x + 5y = 20. Subtract the second to get 2y = 4, so y = 2 and x = 2. Check both equations. Graph y = 4 − x and y = (16 − 5x)/3: their intersection (2, 2) gives the same ticket counts.
- To solve x² + 4x + 3 = 0 graphically, plot y = x² + 4x + 3 at x = −4, −3, −2, −1, 0, giving y = 3, 0, −1, 0, 3. Draw a smooth parabola; the x-axis crossings are x = −3 and −1, agreeing with the factors (x + 3)(x + 1).
- For x³ + x = 12, try x = 2.1: 2.1³ + 2.1 = 11.361, below 12. At 2.2, the value is 12.848, above 12. To justify the answer to one decimal place test the rounding midpoint 2.15: 2.15³ + 2.15 = 12.088375, above 12. Since x³ + x increases on this positive interval, the root lies between 2.1 and 2.15, so it rounds to 2.1. Merely testing 2.1 and 2.2 would not decide which is the nearest tenth.
In plain language
Use inverse operations to isolate a variable. For two simultaneous equations, the solution must satisfy both; elimination and the graph intersection are two ways to locate it. Factorisation and x-axis crossings solve suitable quadratics. Trial and improvement narrows an interval for a cubic root, then a midpoint test justifies the requested rounding.
Self-check
Solve the quadratic equation x² - 9 = 0.
Show the answer
This is a difference of two squares: (x - 3)(x + 3) = 0. The solutions are x = 3 and x = -3.
View diagrams
Two simultaneous linear equations
Key and description
Constructed example: x is the number of £1 items and y is the number of £2 items. There are 5 items, so x + y = 5 and y = 5 − x. There is one more £2 item than £1 items, so y = x + 1. The lines meet at (2, 3): 2 one-pound items and 3 two-pound items. The cost relation is C = x + 2y, giving C = 2 + 6 = £8.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 20, 34: Distinguishing in meaning between equations, formulae and expressions
Open revision guidance
What to know
- An expression describes a relationship without an equals sign, e.g., 5x + 2.
- An equation uses an equals sign to state that two expressions have the same value, requiring a solution for the variable.
- A formula is a specific type of equation that describes a rule, such as Area = length × width.
What to do in an assessment
- Identify: 3x² + 2y is an expression.
- Identify: 2x = 10 is an equation.
- Identify: V = lwh is a formula for volume.
In plain language
Think of an expression as a phrase, an equation as a statement of fact, and a formula as a tool or a recipe. An expression represents a quantity; an equation tests whether two quantities are the same; and a formula provides a consistent rule to calculate an outcome from specific inputs.
Self-check
Is A = πr² an expression, equation, or formula?
Show the answer
It is a formula, as it provides a rule for calculating area based on the variable radius.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 33: Understanding and using function notation; Interpreting and applying the transformation of functions in the context of their graphical representation, including y = f(x + a), y = f(kx), y = kf(x) and y = f(x) + a, applied to y = f(x)
Open revision guidance
What to know
- Function notation f(x) represents a rule where x is the input. The output is y = f(x).
- y = f(x) + a represents a vertical translation by the vector (0, a).
- y = f(x + a) represents a horizontal translation by the vector (-a, 0).
- y = kf(x) represents a vertical stretch by a scale factor of k, parallel to the y-axis.
- For k > 0, y = f(kx) scales horizontal coordinates by 1/k. For k < 0, it also reflects in the y-axis; k = 0 gives the constant y = f(0), when defined.
What to do in an assessment
- To translate f(x) up by 3 units: y = f(x) + 3.
- To translate f(x) left by 2 units: y = f(x + 2).
- Example: If f(x) = x², then y = 2f(x) transforms the parabola by stretching it vertically by a factor of 2, resulting in y = 2x².
- Example: If f(x) = x², then y = f(2x) transforms the parabola by compressing it horizontally by a factor of 0.5, resulting in y = (2x)² = 4x².
- If (p,q) lies on y = f(x), the matching point is (p − a,q) on f(x + a), (p/k,q) on f(kx) for k ≠ 0, (p,kq) on kf(x), and (p,q + a) on f(x) + a. For example, (2,3) maps to (−2,3) under f(−x), or (2,−3) under −f(x).
- When k is negative, y = kf(x) reflects in the x-axis as well as multiplying vertical distances by |k|. If k = 0, the transformed graph is y = 0 on the original function’s domain.
In plain language
Think of f(x) as a machine that processes an input x. Transformations change the graph's position or shape. Vertical changes (f(x) + a, kf(x)) affect the output directly. Horizontal changes (f(x + a), f(kx)) occur inside the input bracket and often act in the 'opposite' direction to what is expected, such as f(x + 2) moving the graph left, not right.
Self-check
If f(x) = x², describe the transformation y = f(x) - 5 and the result.
Show the answer
This is a vertical translation by the vector (0, -5), resulting in the graph y = x² - 5.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 33: Constructing and using tangents to curves to estimate rates of change for non- linear functions, and using appropriate compound measures to express results, including finding velocity in distance-time graphs and acceleration in velocity- time graphs; Interpreting the meaning of the area under a graph, including the area under velocity-time graphs and graphs in other practical and financial contexts; Using the trapezium rule to estimate the area under a curve
Open revision guidance
What to know
- The gradient of a tangent at any point on a curve represents the instantaneous rate of change.
- On a distance–time graph the gradient gives speed; on a signed displacement–time graph it gives velocity. On a velocity–time graph the tangent gradient gives acceleration.
- The area under a velocity-time graph represents displacement.
- The trapezium rule for area is: Area ≈ (h/2) * (y0 + 2y1 + 2y2 + ... + yn), where h is the strip width.
What to do in an assessment
- To find gradient: Pick two points on the tangent line (x1, y1) and (x2, y2), calculate (y2 - y1) / (x2 - x1).
- To estimate area: Divide the region into n strips of width h. The formula uses the height of the ordinates (y) at each boundary.
- Example: Estimate area under y = x² from x = 0 to 2 using 2 strips (h = 1). Ordinates at x=0, 1, 2 are y=0, 1, 4. Area ≈ (1/2) * (0 + 2(1) + 4) = 3.
- For a constructed distance graph d = t² metres, a drawn tangent at t = 2 s passes through (1,0) and (3,8). Its gradient is (8 − 0)/(3 − 1) = 4 m/s, the estimated instantaneous speed. These points are on the tangent, not both on the curve. If the vertical axis were velocity in m/s instead, the same tangent gradient would be 4 m/s² acceleration.
- A cost rate of £40 per week for 3 weeks gives rectangular area 40 × 3 = £120. For a changing rate with values £10, £20, £40 per week at weeks 0,1,2, two trapezia estimate total cost as ½(10 + 20) × 1 + ½(20 + 40) × 1 = £45. The area has units of money because rate is multiplied by time.
In plain language
Gradients measure how steep a slope is at a single point, revealing the 'speed' of change. Accumulation is found by calculating area. Since curves are not rectangles, the trapezium rule approximates the area by splitting the space into trapezoids, which are easier to calculate than curved shapes.
Self-check
Velocity is v = 2t m/s with time t in seconds. Find displacement from t = 0 to t = 2 using two trapezium strips of width 1 s.
Show the answer
Ordinates are 0,2,4 m/s. Area = ½ × 1 × (0 + 2×2 + 4) = 4 m. Here the graph is a straight line, so the trapezium result is exact.
View diagrams
Area under a velocity-time graph
Key and description
Velocity rises uniformly from 0 to 10 m/s over 5 s. The shaded triangle has base 5 s and height 10 m/s: displacement = ½ × 5 × 10 = 25 m. Because velocity is non-negative, distance travelled is also 25 m.
Four trapezium strips under y = x²
Key and description
From x=0 to 2 the width is h=0.5 and ordinates are 0, 0.25, 1, 2.25, 4. The straight tops of the four strips approximate the curve. Total estimated area = ½ × 0.5[0 + 2(0.25+1+2.25) + 4] = 2.75 square units.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 33: Changing the subject of a formula when the subject appears in more than one term; Factorisation of quadratic expressions of the form ax² + bx + c, including the difference of two squares; Formation and manipulation of quadratic equations; Constructing and using equations that describe direct and inverse proportion; Simplifying algebraic fractions
Open revision guidance
What to know
- Factorization of ax² + bx + c involves finding two numbers that multiply to ac and add to b.
- Difference of two squares: a² - b² = (a - b)(a + b).
- Direct proportion: y = kx. Inverse proportion: y = k/x.
- Algebraic fractions are simplified by cancelling common factors present in both the numerator and denominator.
What to do in an assessment
- Factorize x² + 5x + 6: Find two numbers that multiply to 6 and add to 5 (2 and 3). Result: (x + 2)(x + 3).
- Change subject of y = mx + c to x: y - c = mx, then x = (y - c) / m.
- Simplify (x² − 9)/(x − 3): factor the numerator as (x − 3)(x + 3), cancel the common factor and obtain x + 3, retaining x ≠ 3 because the original expression is undefined there.
- Solve for k: If y is directly proportional to x and y=10 when x=2, then 10 = k(2), so k=5.
- For a difference of two squares, 4x² − 9 = (2x)² − 3² = (2x − 3)(2x + 3). Expanding gives 4x² + 6x − 6x − 9, so the middle terms cancel.
- When the subject appears twice, collect its terms and factor. From y = ax + bx + c, subtract c: y − c = ax + bx = x(a + b), so x = (y − c)/(a + b), provided a + b ≠ 0. With y = 17, a = 2, b = 3, c = 2, x = 15/5 = 3; substitution gives 2×3 + 3×3 + 2 = 17. If a + b = 0, division is invalid and the original equation must be considered separately.
- Factorise 2x² + 7x + 3: ac = 6 and the required sum is 7, so split 7x into 6x + x. Then 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Expand to check.
- If y is directly proportional to x² and y = 12 at x = 2, write y = kx², giving 12 = 4k, k = 3 and y = 3x². At x = 4, y = 48. If instead y is inversely proportional to x², write y = k/x²: 12 = k/4 gives k = 48, and x = 4 gives y = 3. State the quantity being squared when constructing the model.
In plain language
Algebraic manipulation is like rearranging furniture to see the room better. Factoring breaks polynomials into building blocks. Proportion describes scaling relationships; if you double x in a direct proportion, you double y, but in an inverse proportion, you halve y.
Self-check
Factorize x² - 25 and make x the subject of y = x/a + b.
Show the answer
x² - 25 = (x - 5)(x + 5). From y = x/a + b, subtract b: y - b = x/a, then multiply by a: x = a(y - b). The rearrangement assumes a ≠ 0, as required by the original term x/a.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 34: The use of straight line graphs to locate regions given by linear inequalities; Solution by factorisation, graphical methods and formula, of quadratic equations of the form ax² + bx + c = 0, selecting the most appropriate method for the problem concerned; Solution of equations involving linear denominators leading to quadratic or linear equations
Open revision guidance
What to know
- Quadratic equations (ax² + bx + c = 0) are solved by factorization or the quadratic formula x = (-b ± √(b² - 4ac)) / 2a.
- Simultaneous equations represent the intersection of lines on a coordinate plane.
- Inequalities are represented by regions bounded by lines, often shaded to show valid sets of values.
- Iterative methods like trial and improvement help solve equations where exact algebraic solutions are impractical.
What to do in an assessment
- Solve x² - 6x + 8 = 0: (x - 2)(x - 4) = 0. Solutions are x = 2, x = 4.
- Solve 2x + y = 5 and x - y = 1 by adding the equations: 3x = 6, x = 2. Then y = 1.
- Solve 1/x + 1/2 = 3: Multiply by 2x to get 2 + x = 6x, so 5x = 2, x = 0.4.
- Linear inequalities: Draw the boundary line (use dashed for <, >), then test a point to decide which side to shade.
- Solve 1/x + 1/(x + 1) = 3/2 with x ≠ 0, −1. Multiply by 2x(x + 1): 2(x + 1) + 2x = 3x(x + 1). Rearrange to 3x² − x − 2 = 0 = (3x + 2)(x − 1). Thus x = 1 or −2/3. Both satisfy the domain restriction; substitution gives 3/2 in each case.
- For 2x² + 3x − 1 = 0, a = 2, b = 3, c = −1 and b² − 4ac = 9 + 8 = 17. Thus x = (−3 ± √17)/4, approximately 0.281 or −1.781. A zero discriminant gives one repeated real root; a negative discriminant gives no real roots.
- A rectangle with sides x + 2 and x + 3 metres has area 30 m². Form (x + 2)(x + 3) = 30, hence x² + 5x − 24 = 0 = (x + 8)(x − 3). The algebraic roots are −8 and 3, but −8 gives negative side lengths, so x = 3 and the sides are 5 m and 6 m.
- For x + y ≤ 4 with x ≥ 0 and y ≥ 0, draw solid boundaries x + y = 4, x = 0 and y = 0. The point (0,0) satisfies 0 ≤ 4, so keep the triangular region with vertices (0,0),(4,0),(0,4), including its boundary. If the first inequality is strict, use a dashed sloping boundary and exclude that edge; the non-strict axis boundaries stay solid.
In plain language
A quadratic can have two, one or no real solutions, depending on its discriminant. Factorisation works when suitable factors can be found; the quadratic formula works for any quadratic with non-zero leading coefficient. Simultaneous equations require both statements to be true. A two-variable linear inequality gives a half-plane rather than one point.
Self-check
Solve x² + 2x - 8 = 0.
Show the answer
(x + 4)(x - 2) = 0, so x = -4 or x = 2.
View diagrams
Roots and turning point of a quadratic
Key and description
y = x² − 2x − 3 = (x − 3)(x + 1). Roots: (−1, 0) and (3, 0). Completing the square gives (x − 1)² − 4, so the turning point is (1, −4).
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 34: Distinguishing identities from equations, formulae and expressions
Open revision guidance
What to know
- An expression (e.g., 3x + 2) has no equals sign.
- An equation states an equality. It may hold for particular values, no values, or every permitted value. An identity is an equation true for every value in its stated domain.
- A formula (e.g., E = mc²) defines a relationship between quantities.
- An identity (e.g., (x + 1)² ≡ x² + 2x + 1) is true for all values of x, denoted by ≡.
What to do in an assessment
- Distinguish: If you can choose any x and the sides remain equal, it is an identity.
- Solve an equation to determine the values for which it holds. For 2x + 1 = 5, x = 2; for x + 1 = x there is no solution; 2(x + 1) = 2x + 2 is an identity.
- Rearrange: When changing a formula like v = u + at, use inverse operations: v - u = at, then t = (v - u) / a.
- Verify identity: Expand (x + 2)(x - 2) = x² - 4. Since the expansion equals the expression, it is an identity.
In plain language
Algebra uses different labels based on purpose. We evaluate expressions, solve equations to find secrets, use formulas to model the real world, and prove identities to show two different ways of writing the same thing are actually identical.
Self-check
Is 4(x + 3) = 4x + 12 an equation or an identity? Why?
Show the answer
It is an identity, because it is true for all values of x.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 32: Drawing, interpretation, recognition and sketching the graphs of y = a/x; Drawing, interpretation, recognition and sketching the graphs of y = ax³; Drawing and interpretation of graphs of y = ax³ + b; Drawing and interpretation of graphs of y = ax + b + a/x, with x not equal to 0; Drawing and interpretation of graphs of y = ax³ + bx² + cx + d; Drawing and interpretation of graphs of y = k^x for integer x and simple positive k
Open revision guidance
What to know
- For a ≠ 0, y = a/x is a reciprocal curve with asymptotes x = 0 and y = 0. For a > 0 it lies in quadrants one and three; for a < 0 it lies in quadrants two and four.
- For a ≠ 0, y = ax³ is a cubic through the origin; negative a reverses its vertical orientation. Adding b shifts it vertically.
- For k > 1, y = k^x grows as x increases. For 0 < k < 1 it decays; for k = 1 it is the horizontal line y = 1. In each case it passes through (0,1).
- A cubic y = ax³ + bx² + cx + d has a ≠ 0 and can cross the x-axis up to three times.
What to do in an assessment
- For y = 1/x, plot (−2,−0.5),(−1,−1),(1,1),(2,0.5) on two separate branches; never join across x = 0. For y = −2/x, points (−2,1),(−1,2),(1,−2),(2,−1) show the two different quadrants.
- For y = 2x³ + 1, values at x = −2,−1,0,1,2 are −15,−1,1,3,17. The standard cubic is scaled vertically by 2 and moved up 1; plot and join smoothly.
- For y = 2^x, values at x = −2,−1,0,1,2 are 1/4,1/2,1,2,4. For y = (1/2)^x they are 4,2,1,1/2,1/4. Both approach y = 0 on one side without reaching it; one grows and the other decays.
- For y = x³ − x, make a table: at x = −2, −1, 0, 1, 2, y = −6, 0, 0, 0, 6. The roots are −1, 0 and 1 since x³ − x = x(x − 1)(x + 1). Plot extra points between roots, such as (−0.5, 0.375) and (0.5, −0.375), to reveal the two turns, then join smoothly. Calculus is not required.
- For y = x + 1/x, exclude x = 0. Values at x = −2, −1, −0.5, 0.5, 1, 2 are −2.5, −2, −2.5, 2.5, 2, 2.5. Keep the two branches separate. Near x = 0 the reciprocal term dominates; for large |x| the graph approaches the line y = x. The y-axis is a vertical asymptote and the graph has no x-intercept because x² + 1 = 0 has no real solution.
In plain language
These graphs represent non-linear relationships. Exponential growth (k^x) describes things like compound interest, while reciprocals (a/x) describe inverse relationships, such as how time taken decreases as speed increases.
Self-check
Sketch the graph of y = 1/x. What are the asymptotes?
Show the answer
The curves approach the axes, so the asymptotes are the x-axis (y=0) and the y-axis (x=0).
Topic group
Geometry and Measure
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 21, 35: The geometrical terms: point, line, plane, parallel, right angle, clockwise and anticlockwise turns, perpendicular, horizontal, vertical, acute, obtuse and reflex angles, face, edge and vertex; Vocabulary of triangles, quadrilaterals and circles: isosceles, equilateral, scalene, exterior/interior angle, diagonal, square, rectangle, parallelogram, rhombus, kite, trapezium, polygon, pentagon, hexagon, radius, diameter, tangent, circumference, chord, arc, sector, segment; Simple solid figures: cube, cuboid, cylinder, cone and sphere; Interpretation and drawing of nets; Using and drawing 2-D representations of 3-D shapes, including the use of isometric paper
Open revision guidance
What to know
- Polygons are 2D closed shapes; circles have unique properties like diameter (2 × radius).
- 3D shapes are defined by faces, edges, and vertices.
- Nets are 2D patterns that fold into 3D shapes.
- Interior angles of a polygon can be calculated using the sum formula: 180(n-2).
- A point marks a position; a line is straight and extends in both directions; a plane is a flat surface. Parallel lines in the same plane never meet. Perpendicular lines meet at a right angle, 90°. Horizontal means side-to-side; vertical means up-and-down. Clockwise follows clock hands and anticlockwise is the opposite turn.
- Acute angles lie between 0° and 90°, obtuse angles between 90° and 180°, and reflex angles between 180° and 360°. A face is a flat surface of a polyhedron, an edge is where two faces meet, and a vertex is a corner.
- An equilateral triangle has three equal sides, an isosceles triangle at least two, and a scalene triangle none. A square has four equal sides and four right angles; a rectangle has four right angles and equal opposite sides; a parallelogram has parallel opposite sides; a rhombus has four equal sides; a kite has two pairs of adjacent equal sides; a trapezium has a pair of parallel sides. A pentagon has five sides and a hexagon six. A diagonal joins non-adjacent vertices. An interior angle is inside the polygon; its adjacent exterior angle on an extended side adds with it to 180°.
- A circle’s radius joins its centre to its circumference; a diameter is a chord through the centre and is twice the radius. A chord joins two circumference points. A tangent touches at one point and is perpendicular to the radius there. An arc is part of the circumference, a sector is bounded by two radii and an arc, and a segment by a chord and an arc.
- A cube has six equal square faces; a cuboid has six rectangular faces. A cylinder has two matching circular ends and a curved wall. A cone has a circular base and a curved surface ending at a vertex. A sphere consists of points a fixed distance from its centre.
What to do in an assessment
- Classify a shape with 5 sides as a pentagon.
- Calculate one interior angle of a square: 180(4-2)/4 = 90°.
- Construct a cube net from six equal squares: draw four in a row, then attach one above and one below the second square. Fold along the common edges and check that the six squares form the six faces without overlap. A sphere has no exact flat net.
- On isometric dot paper, choose three grid directions for the width, depth and vertical edges of a cuboid. Draw a 4-unit width, 3-unit depth and 2-unit height, then add parallel matching edges to close the faces. Mark the dimensions 4, 3 and 2: the oblique drawing is not an ordinary scale drawing from which diagonal lengths can be measured. Omit or dash hidden edges consistently.
In plain language
Geometry helps us describe the physical world. A 'net' is like a blueprint for a cardboard box. If you know the properties of a shape, like how many sides it has, you can calculate missing angles because the sum of internal angles is always predictable.
Self-check
How many faces, edges, and vertices does a cube have?
Show the answer
A cube has 6 faces, 12 edges, and 8 vertices.
View diagrams
Cube net and isometric cuboid
Key and description
Six equal squares form a cube net: four in a row, with one above and one below the second square. The separate isometric cuboid has width 4, depth 3 and height 2 units. Parallel edges remain parallel; the drawing is not an ordinary scale drawing.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 21, 35: Accurate use of ruler, pair of compasses and protractor; (Lengths accurate to 2mm and angles accurate to 2°.) Bisecting a given line, bisecting a given angle; Constructing 2-D shapes from given information
Open revision guidance
What to know
- Constructions use only a compass and straightedge (ruler).
- A perpendicular bisector cuts a line segment into two equal halves at a 90° angle.
- An angle bisector cuts an angle into two equal halves.
- Accuracy is defined by a tolerance of ±2mm for lengths and ±2° for angles.
What to do in an assessment
- Perpendicular bisector: Place compass at one end, draw an arc, repeat from other end, join the intersection points.
- Angle bisector: Place compass on vertex, draw arc cutting both lines, place compass on each intersection and draw new arcs, join vertex to new intersection.
- Use a protractor to ensure the final angle is exactly half the original.
- For an 8 cm segment AB, set the compass radius to 5 cm, greater than half AB. Draw arcs above and below AB from A, then from B without changing the radius. Join the two arc intersections. This line crosses AB at its midpoint, 4 cm from either endpoint; both intersections are equidistant from A and B, so their joining line is the perpendicular bisector.
- To bisect angle ABC, centre the compass at B and draw an arc cutting BA and BC at D and E. Without changing the radius, draw arcs from D and E meeting inside the angle at F. Draw BF: equal distances BD = BE and DF = EF make the two sides of the construction match, so BF splits the angle into two equal parts. Check a 60° angle becomes two 30° angles.
In plain language
Constructions are the 'pure' way of drawing. By using a compass, you ensure every point on your arc is the exact same distance from the centre, which allows you to find perfect midpoints without ever needing to read a number on a ruler.
Self-check
If you bisect a 100° angle, what is the size of each new angle?
Show the answer
100° / 2 = 50°.
View diagrams
Compass-and-straightedge constructions
Key and description
Perpendicular bisector: equal-radius arcs from A and B meet at C and D; CD crosses AB at its midpoint M at 90°. Angle bisector: an arc from O cuts both arms at P and Q; equal-radius arcs from P and Q meet at R; OR bisects the angle.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 21, 35: Use of ruler and pair of compasses to do constructions; Construction of triangles, quadrilaterals and circles
Open revision guidance
What to know
- Triangles are constructed using given side lengths (SSS) or angles (ASA/SAS).
- Circles are defined by a centre and a constant radius.
- Quadrilaterals can be constructed by treating them as two triangles joined together.
What to do in an assessment
- SSS construction: Draw the base, use compasses set to the lengths of the other two sides to find the meeting point, join the vertices.
- Circle: Mark the centre, set compass to radius length, draw the full rotation.
- Always label vertices and check lengths with a ruler.
- For an isosceles triangle with base AB = 6 cm and other sides 5 cm, draw AB, then draw compass arcs of radius 5 cm from A and B. Choose an intersection C and join AC and BC. Check the two 5 cm sides; the other intersection gives a congruent triangle below AB.
- Construct a 6 cm by 4 cm rectangle: draw AB = 6 cm, construct perpendicular rays at A and B on the same side of AB, mark AD = BC = 4 cm, then join D to C. Check CD = 6 cm and all four angles are 90°.
In plain language
When you need to draw a specific triangle, measuring each side and trying to force it together is difficult. By using a compass to 'swing' arcs from the endpoints of the base, you guarantee that the other two sides connect at exactly the right point.
Self-check
What is the radius required for a circle with a diameter of 8cm?
Show the answer
Radius = diameter / 2. Radius = 8 / 2 = 4cm.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 21, 35: Simple description of symmetry in terms of reflection in a line/plane or rotation about a point; Order of rotational symmetry
Open revision guidance
What to know
- Line symmetry refers to reflectional balance across a line.
- Rotational symmetry refers to appearance after turning around a central point.
- Order of rotational symmetry is how many times a shape matches its starting position during one 360° rotation.
What to do in an assessment
- Test line symmetry: fold the image mentally along lines of symmetry.
- Test rotation: turn the shape 360° and count how many times it looks identical to the original orientation.
- For a square, it matches at 90°, 180°, 270°, and 360°, so the order is 4.
In plain language
Symmetry describes how a shape stays the same when you move it. A butterfly has line symmetry because the left side matches the right. A star has rotational symmetry because if you twist it, it still looks like the same star.
Self-check
What is the order of rotational symmetry for a rectangle?
Show the answer
A rectangle looks identical when turned 180° and 360°. The order is 2.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 22, 36: Angles at a point; Angles at a point on a straight line; Opposite angles at a vertex; Parallel lines; Corresponding, alternate and interior angles; Angle properties of triangles; Using the fact that the angle sum of a triangle is 180°
Open revision guidance
What to know
- Angles at a point: 360°.
- Angles on a line: 180°.
- Vertically opposite angles: Equal.
- Parallel lines: Corresponding angles are equal (F-shape), Alternate angles are equal (Z-shape), Interior angles sum to 180°.
- Triangle sum: 180°.
What to do in an assessment
- Find a missing angle on a line: 180° - known angle.
- Find a missing angle in a triangle: 180° - sum of two other angles.
- Use 'Z' and 'F' patterns to identify equal angles between parallel lines.
In plain language
Think of parallel lines like train tracks. If you draw a line across them, the angles created follow strict rules. If you know one angle, you can use the 'F' or 'Z' shape rules to find all the others around the intersection.
Self-check
In a triangle, if the interior angles are 40° and 70°, what is the third angle?
Show the answer
180° - 40° - 70° = 70°.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 10, 22, 36: Using the fact that the exterior angle of a triangle is equal to the sum of the interior angles at the other two vertices; Using angle properties of equilateral, isosceles and right-angled triangles; understand congruence; explain why the angle sum of any quadrilateral is 360°; Regular and irregular polygons; Sum of the interior and sum of the exterior angles of a polygon
Open revision guidance
What to know
- The exterior angle of a triangle equals the sum of the two opposite interior angles.
- Equilateral triangles have three equal sides and three 60° angles; isosceles triangles have two equal sides and base angles; right-angled triangles contain a 90° angle.
- Congruence means two shapes have identical size and shape (all corresponding sides and angles match).
- The sum of interior angles of a quadrilateral is 360°.
- The sum of the interior angles of an n-sided polygon is (n-2) × 180°.
- The sum of the exterior angles of any convex polygon is always 360°.
- Regular polygons have equal sides and equal interior angles; irregular polygons do not.
What to do in an assessment
- Find the exterior angle: If a triangle has interior angles of 40° and 60°, the exterior angle is 40° + 60° = 100°.
- Find a missing angle in a quadrilateral: If three angles are 80°, 100°, and 90°, the fourth is 360° - (80° + 100° + 90°) = 90°.
- Calculate the interior angle of a regular octagon (8 sides): Sum = (8-2) × 180° = 1080°. One angle = 1080° / 8 = 135°.
- Only for a regular polygon are the exterior angles equal: exterior angle = 360°/n and interior angle = 180° − 360°/n. For a regular pentagon, these are 72° and 108°. An irregular polygon need not have equal individual angles, even though its angle sums still obey the appropriate polygon rules.
In plain language
An exterior angle of a triangle is formed by extending a side and equals the sum of the two remote interior angles. A quadrilateral can be split into two triangles, proving its interior angle sum is 360°. For a convex n-sided polygon, triangulation gives (n − 2) × 180° for the interior sum and one full turn gives 360° for the exterior sum. Divide by n to find an individual angle only when the polygon is regular.
Self-check
A regular pentagon has 5 sides. What is the measure of one interior angle?
Show the answer
Sum of interior angles = (5-2) × 180° = 540°. One interior angle = 540° / 5 = 108°.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 11, 23, 37: Finding the coordinates of points identified by geometrical information, for example, finding the coordinates of the mid-point of the line segment AB, given points A and B; finding the coordinates of the fourth vertex of a parallelogram, given the coordinates of the other three vertices; Location determined by distance from a given point and angle made with a given line
Open revision guidance
What to know
- The mid-point of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is given by ((x₁ + x₂) / 2, (y₁ + y₂) / 2).
- In a parallelogram, diagonals bisect each other, meaning the mid-point of one diagonal is the same as the mid-point of the other.
- Location can be determined by a distance (r) from a point and an angle (θ) relative to a line.
- Parallelogram property: The sum of the x-coordinates of opposite vertices is equal, and the sum of the y-coordinates of opposite vertices is equal.
What to do in an assessment
- Find mid-point: For A(2, 4) and B(6, 10), the mid-point is ((2+6)/2, (4+10)/2) = (4, 7).
- Find vertex D of parallelogram ABCD: Given A(0, 0), B(3, 0), and C(4, 2), find D. The diagonal AC has mid-point ((0+4)/2, (0+2)/2) = (2, 1). The diagonal BD must have the same mid-point: (3+xD)/2 = 2 → xD = 1; (0+yD)/2 = 1 → yD = 2. So, D is (1, 2).
- To locate P 5 cm from A at 30° anticlockwise from a baseline AB, place the protractor centre at A and align its zero with AB. Mark 30°, draw the ray, then use a ruler to mark P exactly 5 cm along it. Check both the distance AP and the angle BAP.
In plain language
The mid-point is simply the average of the coordinates. For a parallelogram, because the diagonals cross exactly in the middle, you can find the centre point using two known opposite corners and then use that centre to calculate the missing fourth corner.
Self-check
Find the mid-point of the line connecting A(2, 3) and B(8, 7).
Show the answer
Mid-point = ((2+8)/2, (3+7)/2) = (5, 5).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 11, 23, 37: Transformations, including: • Reflection • Rotation through 90°, 180°, 270°; Clockwise or anticlockwise rotations; centre of rotation • Enlargement with positive scale factors • Translation; Candidates will be expected to draw the image of a shape under transformation
Open revision guidance
What to know
- Reflection: A transformation across a mirror line where points are equidistant from the line.
- Rotation: A transformation about a centre point by a specified angle (90°, 180°, 270°) clockwise or anticlockwise.
- Enlargement at Foundation changes size using a positive whole-number scale factor k; each point’s distance from the centre is multiplied by k.
- Translation slides every point the same distance in the same direction.
- Foundation tier enlargements use positive whole-number scale factors.
What to do in an assessment
- Rotate point (1, 0) 90° clockwise about origin (0, 0): Result is (0, -1).
- Translate a shape with vertex (2, 2) three units right and one unit down: its new vertex is (2 + 3, 2 − 1) = (5, 1).
- Enlarge a shape with vertices (0,0), (1,0), (1,1) by scale factor 2: New vertices are (0,0), (2,0), (2,2).
In plain language
Reflections, rotations and translations preserve side lengths and angles. Enlargement preserves angles and multiplies lengths. To translate, move each vertex by the same horizontal and vertical amounts. Fractional scale factors and column-vector descriptions are Intermediate; negative enlargement is Higher.
Self-check
Rotate point (0, 1) 180° about the origin (0, 0). What is the new coordinate?
Show the answer
(0, -1).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 11, 23, 37: Solving problems in the context of tiling patterns and tessellation
Open revision guidance
What to know
- Tessellation occurs when shapes cover a plane perfectly with no gaps or overlaps.
- A regular polygon tessellates alone if its interior angle is a factor of 360°.
- Squares (90°) and hexagons (120°) tessellate because 360 is divisible by 90 and 120.
What to do in an assessment
- Test if a regular octagon (135°) tessellates: 360 / 135 ≈ 2.66. Since it is not an integer, regular octagons cannot tessellate alone.
In plain language
Tessellation is about geometry fitting perfectly at a point. Around any vertex where shapes meet, the angles must sum to exactly 360°. If the interior angle of the shape does not divide into 360 evenly, gaps will appear.
Self-check
Does a regular triangle (60°) tessellate? Explain.
Show the answer
Yes, because 360 / 60 = 6, which is an integer. Six triangles can meet at a single vertex.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 11, 23, 37: Using and interpreting maps; Interpretation and construction of scale drawings; Scales may be written in the form 1 cm represents 5 m, or 1:500; Use of bearings; (Three figure bearings will be used e.g. 065°, 237°.)
Open revision guidance
What to know
- Scale factors are ratios: 1 cm : 500 cm is the same as 1:500.
- Bearings are measured clockwise from North and must be expressed as three figures (e.g., 045°).
- Actual distance = Map distance × Scale factor.
What to do in an assessment
- Calculate distance: Scale 1:1000. Map distance 4 cm. Actual distance = 4 × 1000 = 4000 cm = 40 m.
- Write bearing: A direction 30 degrees clockwise from North is 030°.
- Read a map key before measuring: symbols can identify a station, footpath or bridge, and an arrow gives north. On a map with 1 cm representing 2 km, a 3.5 cm marked route represents 7 km. Measure along the route, splitting a bent path into pieces; a straight-line distance is not automatically the road distance.
In plain language
Scale drawings shrink reality to fit on paper. A 1:100 scale means every measurement on paper is 100 times larger in the real world. Bearings ensure consistency by always starting from North and going clockwise.
Self-check
On a map with scale 1:50000, 2 cm represents how many kilometres in real life?
Show the answer
2 cm × 50000 = 100000 cm = 1000 m = 1 km.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 12, 24, 38: Standard metric units of length, mass and capacity; The standard units of time; the 12- and 24- hour clock; (The notation for the 12- and 24- hour clock will be 1:30 p.m. and 13:30.) Knowledge and use of the relationship between metric units of length, mass, capacity, area and volume; Making sensible estimates of measurements in everyday situations, recognising the appropriateness of units in different contexts; Conversion between the following metric and Imperial units: km - miles; cm, m - inches, feet; kg - lb; litres - pints, gallons; Candidates will be expected to know the following approximate equivalences: 8km ≈ 5 miles, 1kg ≈ 2·2 lb, 1 litre ≈ 1⋅75 pints
Open revision guidance
What to know
- Metric units: Length (m, km, cm, mm), Mass (kg, g), Capacity (l, ml).
- Time: 12-hour (a.m./p.m.) and 24-hour (e.g., 14:00) clocks.
- Approximate equivalences: 8 km ≈ 5 miles, 1 kg ≈ 2.2 lb, 1 litre ≈ 1.75 pints.
What to do in an assessment
- Convert 16 km to miles: 16 / 8 = 2. 2 × 5 = 10 miles.
- Convert 5 kg to lb: 5 × 2.2 = 11 lb.
- Convert 14:30 to 12-hour: 2:30 p.m.
- Convert metric units by the relevant power of the length factor. Because 1 m = 100 cm, 1 m² = 100 × 100 = 10,000 cm² and 1 m³ = 100³ = 1,000,000 cm³. Also 1 litre = 1000 ml = 1000 cm³, so 2.5 litres = 2500 cm³.
- Estimate sensibly and choose a unit: a pencil is about 15 cm, a doorway about 2 m high, a person’s mass tens of kilograms, and a drinking bottle about 500 ml. A pencil length of 15 km or a bottle capacity of 500 litres is not plausible.
- Use a supplied conversion when precision is specified. Given 1 inch = 2.54 cm, 10 inches = 25.4 cm. Given 1 foot = 0.3048 m, 6 feet = 1.8288 m. Given 1 imperial gallon ≈ 4.55 litres, 9.10 litres ≈ 2 gallons. Do not confuse an imperial gallon with a US gallon.
- The specification expects the approximate equivalences 8 km ≈ 5 miles, 1 kg ≈ 2.2 lb and 1 litre ≈ 1.75 pints. Thus 24 km ≈ 15 miles, 5 kg ≈ 11 lb and 2 litres ≈ 3.5 pints. These approximations are not exact equalities.
In plain language
Conversions allow us to compare measurements between different systems. 24-hour time is used to avoid the confusion of a.m./p.m. by counting continuously from midnight.
Self-check
Convert 2 litres to pints.
Show the answer
2 × 1.75 = 3.5 pints.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 12, 24, 38: Reading and interpreting scales, including decimal scales
Open revision guidance
What to know
- Scales are defined by the range between numbered markings and the number of subdivisions.
- Value per tick = (Higher number - Lower number) / Number of gaps.
What to do in an assessment
- Determine increment: Scale numbered 0 to 2 with 4 gaps between integers. Value per tick = (1 - 0) / 4 = 0.25.
In plain language
Reading a scale is a process of division. You find the total value between two labels and divide that by how many steps exist between them to know exactly what each mark represents.
Self-check
A scale is numbered 0 to 10 with 10 equal divisions. What does one tick represent?
Show the answer
(10 - 0) / 10 = 1.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 12, 24, 38: Using compound measures including speed; Using compound measures such as m/s, km/h, mph and mpg
Open revision guidance
What to know
- Speed = Distance / Time.
- Units must be consistent: If speed is in km/h, time must be in hours.
- Mpg (miles per gallon) is a measure of efficiency.
What to do in an assessment
- Calculate speed: Distance 200 km, Time 4 hours. Speed = 200 / 4 = 50 km/h.
- Calculate distance: Speed 60 mph, Time 30 minutes (0.5 hours). Distance = 60 × 0.5 = 30 miles.
- Speed = distance ÷ time. Travelling 120 m in 15 s gives 8 m/s; travelling 150 km in 3 h gives 50 km/h; travelling 90 miles in 2 h gives 45 mph. Keep the time unit consistent with the stated speed unit.
- Fuel consumption of 180 miles using 6 imperial gallons is 180 ÷ 6 = 30 mpg. For the same distance, a higher mpg uses less fuel. The unit mpg is distance per fuel volume, not a speed.
In plain language
Compound measures describe rates, like how much distance you cover per hour. If you travel for 90 minutes, convert this to 1.5 hours before using it in a formula so the units align with km/h.
Self-check
A cyclist travels 10 km in 20 minutes. What is their speed in km/h?
Show the answer
20 minutes is 1/3 of an hour. Speed = 10 / (1/3) = 10 × 3 = 30 km/h.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 12, 24, 38: Estimating of the area of an irregular shape drawn on a square grid; Calculating: - perimeter and area of a square, rectangle, triangle, parallelogram, trapezium, circle, semicircle and composite shapes. - surface area, cross-sectional area and volume of cubes and cuboids
Open revision guidance
What to know
- Perimeter: Total length around a shape (sum of sides).
- Area: Square (side²), Rectangle (l×w), Triangle (0.5×base×height), Trapezium (0.5(a+b)h), Circle (πr²).
- Volume of cuboid: length × width × height.
- Irregular area: Count full squares and estimate fractions.
What to do in an assessment
- Perimeter of rectangle 5 cm by 3 cm: 2 × (5 + 3) = 16 cm.
- Area of triangle (base 6, height 4): 0.5 × 6 × 4 = 12 cm².
- Volume of cuboid (3 × 2 × 4): 24 cm³.
- A cuboid 4 cm by 3 cm by 2 cm has opposite pairs of faces of areas 12, 8 and 6 cm², so surface area = 2(12 + 8 + 6) = 52 cm². A cross-section perpendicular to the 4 cm length has area 3 × 2 = 6 cm², and its volume is 6 × 4 = 24 cm³. Surface area, cross-sectional area and volume are different quantities.
- A parallelogram of base 8 cm and perpendicular height 3 cm has area 24 cm². If its adjacent side is 5 cm, its perimeter is 2(8 + 5) = 26 cm: do not use the sloping side as the height.
- For a semicircle of radius 4 cm, area = ½π × 4² = 8π cm². Its boundary includes the diameter: perimeter = π × 4 + 8 cm, not just 4π.
- An L-shaped region formed by removing a 3 cm by 2 cm corner from a 7 cm by 5 cm rectangle has area 35 − 6 = 29 cm². Trace its six boundary segments (7, 3, 3, 2, 4, 5 cm) to find perimeter 24 cm; the removed internal corner creates two new boundary edges.
In plain language
Perimeter is the length of the fence around a field; area is the grass inside; volume is the amount of water needed to fill the space inside a 3D box.
Self-check
What is the area of a circle with radius 3 cm? (Use π ≈ 3.14)
Show the answer
Area = π × 3² = 3.14 × 9 = 28.26 cm².
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 21, 35: Simple solid figures: prism, pyramid, tetrahedron
Open revision guidance
What to know
- 3D shapes like prisms and pyramids are defined by their base shapes and connecting faces.
- Prisms have a uniform cross-section throughout their length.
- Tetrahedrons are specific pyramids consisting of four triangular faces.
What to do in an assessment
- Recognise a right triangular prism: it has two matching triangular ends and three rectangular side faces.
- Analyse a pyramid: A square-based pyramid has one square base and four triangular sides meeting at an apex.
In plain language
A prism has matching parallel end faces, and slices parallel to those faces have the same cross-section. A pyramid has a base and triangular sides meeting at one apex. A tetrahedron is a triangular-based pyramid with four triangular faces; it need not have every edge equal.
Self-check
What is the primary difference between a prism and a pyramid?
Show the answer
A prism has a uniform cross-section throughout its length, whereas a pyramid tapers to a single apex.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 21, 35: Constructing the perpendicular from a point to a line; Essential properties of special types of quadrilateral, including square, rectangle, parallelogram, trapezium, kite and rhombus; classify quadrilaterals by their geometric properties; Drawing plans and elevations of any 3-D solid
Open revision guidance
What to know
- Quadrilaterals are classified by properties: squares (all sides/angles equal), rectangles (opposite sides equal, 90° angles), parallelograms (opposite sides/angles equal), trapeziums (at least one pair of parallel sides), kites (two pairs of adjacent equal sides), and rhombi (all sides equal).
- Construction requires compasses and a straight edge; bisecting an angle requires arcs centred on the vertex, then from the intersections.
- Orthographic projections represent 3D objects as plan (top), front, and side elevations.
What to do in an assessment
- Bisect a line: Set compass to more than half the length, draw arcs from both ends, and join the intersections.
- Plan/Elevation: For a cube, the plan, front, and side elevations are all identical squares.
- To construct the perpendicular from a point P outside a line, centre the compass at P and draw an arc cutting the line at A and B. From A and B draw equal-radius arcs meeting at Q on the other side of the line from P. Join P to Q. P and Q are both equidistant from A and B, so PQ is their perpendicular bisector and meets the original line at right angles.
In plain language
Geometric construction is the 'drawing' equivalent of using a compass and ruler to prove mathematical concepts. Quadrilateral classification is about identifying which 'DNA'—sides or angles—a shape has. Orthographic projections are how engineers communicate 3D objects on paper, viewing the object strictly from one flat side at a time.
Self-check
Which quadrilateral has all four sides equal but the angles are not necessarily 90°?
Show the answer
A rhombus.
View diagrams
Plan, front and side elevations
Key and description
For a cuboid of width 4, depth 3 and height 2 units, the plan is a 4 × 3 rectangle, the front elevation is 4 × 2 and the side elevation is 3 × 2. Each is a flat view looking straight at the corresponding face.
Compass-and-straightedge constructions
Key and description
Perpendicular bisector: equal-radius arcs from A and B meet at C and D; CD crosses AB at its midpoint M at 90°. Angle bisector: an arc from O cuts both arms at P and Q; equal-radius arcs from P and Q meet at R; OR bisects the angle.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 21, 35: Constructing angles of 60°, 30°, 90° and 45°; The identification of congruent shapes
Open revision guidance
What to know
- Standard angles like 60° are constructed using equilateral triangle principles; 30° is 60° bisected; a perpendicular bisector meets its line at 90°; 45° is 90° bisected.
- Triangle congruence criteria (SSS, SAS, ASA, RHS) verify that corresponding sides and angles match, regardless of orientation.
What to do in an assessment
- Construct 60°: Draw a line, set compass width to line length, draw intersecting arcs to form the apex.
- Check congruence: Two triangles are congruent if their three corresponding side lengths are equal (SSS).
In plain language
Construction uses the compass as a tool to maintain precise distances, effectively 'locking' an angle into place. Congruence is the geometric version of 'identical twins'—no matter how you rotate or reflect a shape, if the sides and angles match up perfectly, the shapes are mathematically the same.
Self-check
Which congruence criteria requires two angles and the included side?
Show the answer
ASA (Angle-Side-Angle).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 22, 36: Using Pythagoras’ theorem in 2-D, including reverse problems
Open revision guidance
What to know
- Pythagoras' Theorem: a² + b² = c², where c is the hypotenuse.
- The hypotenuse is always the longest side and is opposite the right angle.
- Reverse problems use b = √(c² - a²) to find a missing shorter side.
What to do in an assessment
- Find hypotenuse: Legs 5cm, 12cm. c² = 5² + 12² = 25 + 144 = 169. c = √169 = 13cm.
- Find leg: Hypotenuse 10cm, leg 6cm. a² = 10² - 6² = 100 - 36 = 64. a = √64 = 8cm.
In plain language
Pythagoras' Theorem is a bridge between the sides of a right-angled triangle. By squaring the sides, we convert lengths into areas; the theorem says the two smaller squares always add up to the area of the large square attached to the hypotenuse. When we reverse it, we are subtracting the area of a known side from the hypotenuse's area to uncover the missing side's length.
Self-check
A right-angled triangle has a hypotenuse of 10cm and one side of 8cm. Find the missing side.
Show the answer
10² - 8² = 100 - 64 = 36. √36 = 6cm.
View diagrams
Right-angled triangle with labelled sides
Key and description
For angle θ, the opposite side is 3, the adjacent side is 4 and the hypotenuse is 5.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 22, 36: Using trigonometric relationships in right-angled triangles to solve problems, including those involving bearings and angles of elevation and depression; Calculating a side or an angle of a right-angled triangle in 2-D
Open revision guidance
What to know
- In a right-angled triangle, trig ratios are defined as: sin(θ) = Opposite/Hypotenuse, cos(θ) = Adjacent/Hypotenuse, tan(θ) = Opposite/Adjacent.
- Bearings are measured clockwise from North and represent a 3-digit angle (e.g., 045°).
- Angle of elevation is measured upward from the horizontal, while angle of depression is measured downward from the horizontal.
- This Intermediate skill uses 2-D right-angled triangles. Bearings are resolved into north–south and east–west components; 3-D trigonometry is a separate Higher skill.
What to do in an assessment
- To find a side x when the hypotenuse (h) and angle (θ) are known: x = h × sin(θ) or x = h × cos(θ).
- To find an angle θ when opposite (o) and hypotenuse (h) are known: θ = arcsin(o/h).
- Example: A ship travels 10 km on a bearing of 090°. The angle between the North line and the East direction is 90°. To find the distance East, use the sin(90°) ratio: x = 10 × sin(90°) = 10 km.
- Example: A ladder 5 m long rests against a wall with an angle of elevation of 60°. The distance from the wall base to the ladder base is: x = 5 × cos(60°) = 2.5 m.
- For a 10 km journey on bearing 060°, draw north first and measure 60° clockwise. Eastward distance = 10 sin 60° ≈ 8.66 km; northward distance = 10 cos 60° = 5 km. Label which side is opposite the angle.
- From a point 4 m above level ground, the angle of depression to an object on the ground is 30°. The angle of elevation from that object is also 30° because the horizontal lines are parallel. If d is the horizontal distance, tan 30° = 4/d, giving d = 4/tan 30° ≈ 6.93 m.
In plain language
Trigonometry works because the ratio of sides in a right-angled triangle remains constant for a specific angle, regardless of the triangle's size. By using known ratios like SOH CAH TOA, we map physical distances (like ground distance) to angular directions (bearings) by creating a right-angled triangle where one axis points North and the other East.
Self-check
A ladder 6m long leans against a wall at an angle of elevation of 30°. How far up the wall does the ladder reach?
Show the answer
Using sin(30°) = Opposite/Hypotenuse: Opposite = 6 × sin(30°) = 6 × 0.5 = 3m.
View diagrams
Bearing and component distances
Key and description
Travel 10 km on a bearing of 030° (30° clockwise from north). North component = 10 cos 30° = 5√3 km (about 8.66 km); east component = 10 sin 30° = 5 km.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 22, 36: Using angle and tangent properties of circles; Understanding that the tangent at any point on a circle is perpendicular to the radius at that point; Using the facts that the angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the circumference, that the angle subtended at the circumference by a semicircle is a right angle, that angles in the same segment are equal, and that opposite angles of a cyclic quadrilateral sum to 180°; Understanding and using the fact that tangents from an external point are equal in length
Open revision guidance
What to know
- The angle subtended by an arc at the centre is twice the angle subtended at the circumference.
- The angle subtended at the circumference by a semicircle is 90°.
- Angles subtended by the same arc in the same segment are equal.
- A tangent at any point is perpendicular to the radius at that point.
- Tangents from an external point are equal in length.
- Opposite angles of a cyclic quadrilateral sum to 180°.
What to do in an assessment
- To find an angle at the center: double the angle at the circumference.
- To identify a right angle: check if the triangle's hypotenuse is a circle diameter.
- Example: If a cyclic quadrilateral has an angle of 110°, the opposite angle is 180° - 110° = 70°.
- Example: If a tangent meets a radius, the angle is 90°. Use this to solve for missing angles in triangles formed within the circle.
In plain language
Circle theorems are fixed geometric relationships. They work because the symmetry of the circle forces certain angles to be equal or supplementary based on their position relative to the centre, arcs, or tangents.
Self-check
In a circle, the angle at the centre is 140°. What is the angle at the circumference subtended by the same arc?
Show the answer
The angle at the circumference is half the angle at the center: 140° / 2 = 70°.
View diagrams
Tangent and radius at a point of contact
Key and description
The radius to the point of contact is perpendicular to the tangent.
Angle in a semicircle: labelled proof
Key and description
AB is a diameter and O is the centre. C lies on the semicircle. OA = OC = OB, so the two small triangles are isosceles. Put α at A and between CA and CO; put β at B and between CB and CO. The angle sum of ABC gives 2(α + β) = 180°, hence ∠ACB = α + β = 90°.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 23, 37: Using the knowledge that, for two similar 2-D or 3-D shapes, one is an enlargement of the other; Using the knowledge that, in similar shapes, corresponding dimensions are in the same ratio
Open revision guidance
What to know
- Similar shapes have equal corresponding angles and proportional side lengths.
- The scale factor k is the ratio of corresponding sides.
- Enlargement multiplies lengths and distances from its centre by k. Coordinates measured relative to that centre are multiplied by k; ordinary coordinates can be multiplied directly only when the centre is the origin.
What to do in an assessment
- For the enlargement from a smaller shape to a larger similar shape, find the scale factor k by dividing a known side of the larger shape by the corresponding side of the smaller shape.
- To find a missing side, multiply the smaller side by k.
- Example: Rectangle A has sides 2cm and 4cm. Rectangle B is similar with a short side of 6cm. The scale factor k = 6 / 2 = 3. The long side of B is 4cm × 3 = 12cm.
In plain language
Similarity ensures the shape's 'look' is preserved even if it is scaled up or down. Because the proportions are locked, scaling one side by a factor scales every corresponding length by that same factor.
Self-check
Two similar triangles have corresponding sides 3 cm and 9 cm. Another side of the smaller triangle is 5 cm. Find its corresponding larger side.
Show the answer
Length scale factor = 9 ÷ 3 = 3. Corresponding larger side = 5 × 3 = 15 cm. Area and volume scale factors are separate Higher-tier content.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 23, 37: Enlargement with fractional scale factors; description of translations using column vectors; Questions may involve two successive transformations
Open revision guidance
What to know
- Translations are defined by column vectors (x, y) describing horizontal and vertical movement.
- Enlargements require a scale factor k and centre of enlargement.
- Intermediate includes positive fractional scale factors, such as 1/2 or 3/2. Factors between 0 and 1 shrink a shape; positive factors greater than 1 enlarge it. Negative factors are Higher content.
- Successive transformations are performed by applying the first transformation then the second.
What to do in an assessment
- Translation by (a, b): add a to x and b to y.
- Rotation 90° clockwise about origin (0,0): (x, y) becomes (y, -x).
- Enlargement from origin by scale factor k: (x, y) becomes (kx, ky).
- Example: Translate point (2, 3) by (4, -1) to get (6, 2). Then rotate 90° clockwise: (6, 2) becomes (2, -6).
- To describe a translation from A(2, 3) to B(6, 2), subtract start coordinates from end coordinates: horizontal change 6 − 2 = 4 and vertical change 2 − 3 = −1. Write the column vector with 4 above −1; it means 4 right and 1 down.
- A positive fractional enlargement of (6, 4) from the origin by scale factor 1/2 gives (3, 2). Each distance from the centre is halved; the image stays on the same ray from the centre.
- Enlarging (2,4) from the origin by 3/2 gives (3,6). A fractional scale factor is not necessarily smaller than 1.
In plain language
Transformations are procedural movements. A translation slides every point by the same amount, while an enlargement scales distances from a fixed point by a ratio, which is why we multiply by k.
Self-check
What is the image of point (1, 2) after a translation by vector (3, 4)?
Show the answer
New point = (1+3, 2+4) = (4, 6).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 23, 37: Constructing the locus of a point which moves such that it satisfies certain conditions, for example, (i) a given distance from a fixed point or line, (ii) equidistant from two fixed points or lines; Solving problems involving intersecting loci in two dimensions; Questions on loci may involve inequalities
Open revision guidance
What to know
- A locus is a path of points meeting a specific rule.
- Distance from a fixed point: a circle.
- Equidistant from two points: perpendicular bisector.
- Equidistant from two lines: angle bisector.
- Intersecting loci represent the region where multiple conditions are satisfied.
What to do in an assessment
- Use a compass to draw arcs for fixed-distance rules.
- Construct perpendicular bisectors by finding the intersection of two equal-radius arcs from the points.
- Intersect the resulting paths to identify valid points.
- Example: A point 3cm from A and 2cm from B. Draw a circle of radius 3cm around A and 2cm around B. The intersection points are the solutions.
- For a locus within 3 cm of A, shade the inside of the circle of radius 3 cm. Distance AP ≤ 3 includes the boundary; AP < 3 excludes it. To be at least 2 cm from a straight line, take the regions outside the two parallel lines 2 cm away, including their boundaries for distance ≥ 2. Intersect the shaded regions when both conditions must hold.
In plain language
Loci translate written rules into geometric shapes. Because a compass keeps a set distance, it is the natural tool for drawing paths that must maintain a specific distance from a point or line.
Self-check
What is the locus of a point 5cm from a fixed point P?
Show the answer
A circle with radius 5cm centred at P.
View diagrams
Intersection of a line and circular locus
Key and description
A = (0, 0), B = (4, 0). The line locus x = 2 intersects the circle centre A, radius 3 at (2, √5) and (2, −√5). The shaded solution is inside the circle and to the right of x = 2: x > 2.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 24, 38: Distinguishing between formulae for length, area and volume by considering dimensions; Using compound measures including density and population density; Using compound measures such as kg/m3, g/cm3, population per km2
Open revision guidance
What to know
- Dimensions: Length (L), Area (L²), Volume (L³).
- Speed = Distance / Time.
- Density = Mass / Volume.
- Population Density = Population / Area.
- Formulae can be checked using dimensions: e.g., Volume = L × L × L = L³.
What to do in an assessment
- Calculate density by dividing mass (g or kg) by volume (cm³ or m³).
- Calculate population density by dividing population count by area (km²).
- Example: Mass 200g, Volume 50cm³. Density = 200 / 50 = 4 g/cm³.
- Example: 500 people, area 2km². Density = 500 / 2 = 250 people/km².
In plain language
Compound measures compare one quantity relative to another (like people per unit area). Dimensions verify logic: if you calculate an area, the answer must be in units squared (e.g., cm²).
Self-check
A substance has a mass of 500g and a volume of 100cm³. What is the density?
Show the answer
Density = 500 / 100 = 5 g/cm³.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 24, 38: Calculating surface area, cross-sectional area and volume of prisms, cylinders and composite solids
Open revision guidance
What to know
- Area of rectangle (lw), triangle (½bh), parallelogram (bh), trapezium (½(a+b)h), circle (πr²).
- Volume of a prism = cross-sectional area × length.
- Surface area = sum of all face areas.
- Units of area are squared (e.g., cm²); units of volume are cubed (e.g., cm³).
What to do in an assessment
- Calculate area of a trapezium with parallel sides 6cm, 4cm and height 5cm: Area = 0.5 × (6 + 4) × 5 = 25 cm².
- Calculate volume of a triangular prism with cross-section 10cm² and length 4cm: Volume = 10 × 4 = 40 cm³.
- For composite shapes, partition into simple polygons and sum their areas.
- For a closed cylinder of radius 3 cm and height 10 cm, the curved surface unrolls to a rectangle with width 2πr = 6π cm and height 10 cm, so curved area is 60π cm². Add two end circles, 18π cm², to get total surface area 78π cm². A cross-section perpendicular to its axis is a circle of area 9π cm²; one through the axis is a 6 cm by 10 cm rectangle.
- Join two cuboids of base 4 cm by 3 cm and heights 2 cm and 5 cm across that full base. Combined volume = 24 + 60 = 84 cm³. Separate surface areas are 52 and 94 cm²; the two hidden 12 cm² faces are removed, so outside area = 52 + 94 − 24 = 122 cm². The final 4 × 3 × 7 cuboid gives the same results.
In plain language
Volume measures the 3D space inside an object. By multiplying the base area (the 'slice' of the shape) by its length, you effectively stack that area through the length to fill the volume.
Self-check
A cylinder has a radius of 3cm and a height of 10cm. Calculate the volume (use π ≈ 3.14).
Show the answer
Area = π × 3² = 9π. Volume = 9π × 10 = 90π ≈ 282.6 cm³.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 35: Understanding and using SSS, SAS, ASA and RHS conditions to prove the congruence of triangles using formal arguments; Reasons may be required in the solution of problems involving congruent triangles
Open revision guidance
What to know
- Geometric constructions require a compass and straightedge for precision.
- Congruence means two triangles are identical in size and shape.
- Conditions for congruence are: SSS (three sides equal), SAS (two sides and included angle), ASA (two angles and included side), and RHS (right angle, hypotenuse, and one other side).
What to do in an assessment
- To bisect an angle: Draw an arc from the vertex, then draw two intersecting arcs from those points; draw a line from the vertex through the intersection.
- To prove congruence: State the property matched (e.g., 'side AB = side XY') and cite the rule (e.g., 'by SSS').
- Constructions: Use the compass to set the exact distance to be marked.
- For RHS, compare two right-angled triangles with hypotenuses 5 cm and one matching leg 3 cm. Their other legs are both √(25 − 9) = 4 cm, so the triangles have the same size and shape. State the right angles, equal hypotenuses and equal corresponding legs before concluding RHS congruence.
- For triangles ABC and DEF, pair A with D, B with E and C with F. If AB = DE = 5, BC = EF = 6 and AC = DF = 7, state these three corresponding side equalities and conclude congruence by SSS.
- For SAS, AB = DE = 5 and AC = DF = 7 need the included angles BAC and EDF equal, for example both 60°. For ASA, equal angles A and D of 40° and B and E of 60°, with their included sides AB = DE = 5, establish congruence. A side or angle must correspond correctly; equal lists in the wrong positions do not constitute a proof.
In plain language
Congruence is like a mirror image that fits perfectly over the original. We don't need to check every single angle and side to know two triangles are identical; if they meet specific criteria like SSS, we know the remaining parts must also be identical.
Self-check
Two triangles have angles 40°, 60° and the side between them is 5cm. Are they congruent?
Show the answer
Yes, by the ASA (Angle-Side-Angle) condition.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 36: Using Pythagoras’ theorem in 3-D, including reverse problems
Open revision guidance
What to know
- Pythagoras' theorem: a² + b² = c² (c is the hypotenuse).
- In 3D, the space diagonal d of a cuboid with sides l, w, h is d = √(l² + w² + h²).
- Reverse: If a² + b² = c², the triangle is definitely right-angled.
What to do in an assessment
- Find missing side in 2D: If legs are 3 and 4, c² = 3² + 4² = 9 + 16 = 25, c = 5.
- Find diagonal in 3D cuboid (3x4x12): d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. d = 13.
- Check: For a triangle with sides 5, 12, 13, is it right-angled? 5² + 12² = 25 + 144 = 169. Since 13² = 169, it is right-angled.
- For a cuboid with base 3 cm by 4 cm, height h and space diagonal 13 cm, first find the base diagonal √(3² + 4²) = 5 cm. Then 5² + h² = 13², so h² = 144 and h = 12 cm. This is a reverse 3-D length problem, distinct from checking whether a 2-D triangle is right-angled.
In plain language
Pythagoras' theorem is a rule about how the sides of right-angled triangles relate to the hypotenuse. In 3D, we are simply doing it in stages: first find the base diagonal, then use that to find the space diagonal.
Self-check
A cuboid has dimensions 2cm, 2cm, and 1cm. Find the space diagonal.
Show the answer
d² = 2² + 2² + 1² = 4 + 4 + 1 = 9. d = 3cm.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 36: Calculating a side or an angle of a right-angled triangle in 3-D
Open revision guidance
What to know
- SOH CAH TOA: sin(θ) = O/H, cos(θ) = A/H, tan(θ) = O/A.
- Bearings are measured clockwise from North.
- 3D Trig: Solve by finding a shared length in a 2D triangle.
What to do in an assessment
- Find side: x = 10 * sin(30) = 5.
- Find angle: θ = arcsin(5 / 10) = 30°.
- 3D Example: To find angle between a line and the floor in a room, use the floor length and wall height to form a right-angled triangle.
- In a room 3 m by 4 m with height 12 m, the floor diagonal is √(3² + 4²) = 5 m. The triangle formed by that floor diagonal, the vertical height and the space diagonal is right-angled. Its angle with the floor is θ = tan⁻¹(12/5) ≈ 67.38°. The space diagonal is √(5² + 12²) = 13 m.
- In a second cuboid with floor dimensions 6 m by 8 m, the floor diagonal is 10 m. If its space diagonal rises at 30° to the floor, height = 10 tan 30° ≈ 5.77 m and space diagonal = 10/cos 30° ≈ 11.55 m. The floor angle is measured to the diagonal’s projection on the floor, not to an arbitrary edge.
In plain language
Trigonometry connects the angles of a triangle to the ratios of its sides. If you know one angle and one side, you can find the others. In 3D, the difficulty is just finding the correct 2D triangle hidden inside the shape.
Self-check
A right triangle has an opposite side of 3cm and adjacent side of 4cm. Find the angle.
Show the answer
tan(θ) = 3/4 = 0.75. θ = arctan(0.75) ≈ 36.87°.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 36: Extending trigonometry to angles of any size; The application of trigonometric functions to the solution of problems in 2-D or 3-D, including appropriate use of the sine and cosine rules; Using the formula: area of a triangle = ½absinC
Open revision guidance
What to know
- Sine Rule: a/sinA = b/sinB = c/sinC.
- Cosine Rule: a² = b² + c² - 2bc*cosA.
- Area of triangle = (1/2)ab*sinC.
- These apply to any triangle, regardless of whether it contains a right angle.
What to do in an assessment
- Use Sine Rule when you have an angle and its opposite side (a side-angle pair).
- Use Cosine Rule when you have three sides (SSS) or two sides and the included angle (SAS).
- Find area: For a triangle with sides 6 and 7 and included angle 50°, area = 0.5 * 6 * 7 * sin(50°) ≈ 16.07.
- Extend sine and cosine using the unit circle: an angle θ measured anticlockwise from the positive horizontal axis reaches coordinates (cos θ, sin θ); tan θ = sin θ/cos θ where cos θ ≠ 0. For θ = 150°, the reference angle is 30°, so sin 150° = ½, cos 150° = −√3/2 and tan 150° = −1/√3. For −30°, sine is −½ and cosine √3/2. Adding 360° repeats sine and cosine; tangent repeats after 180°. Tangent is undefined at 90° + 180°k, not zero.
In plain language
Right-angle trigonometry fails for non-right triangles. The Sine and Cosine rules are 'upgrades' that generalize the relationships to any triangle. The area formula ½absinC is useful because it doesn't require knowing the perpendicular height.
Self-check
Use the Cosine rule to find side 'a' if b=5, c=8, A=60°.
Show the answer
a² = 5² + 8² - 2(5)(8)cos(60°) = 25 + 64 - 80(0.5) = 89 - 40 = 49. a = 7.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 36: The graphs and behaviour of trigonometric functions; Sketching of trigonometric graphs
Open revision guidance
What to know
- y = sin(x) has a period of 360°, amplitude 1, passes through (0,0).
- y = cos(x) has a period of 360°, amplitude 1, passes through (0,1).
- y = tan(x) has vertical asymptotes at 90°, 270°, etc., and is undefined at those points.
What to do in an assessment
- Sketch sin(x): Plot points at 0, 90, 180, 270, 360 and join with a smooth wave.
- Sketch cos(x): Start at 1, go to 0 at 90, -1 at 180, 0 at 270, 1 at 360.
- Identify features: Amplitude is the height of the wave from its centre, period is how far it travels before the cycle repeats.
In plain language
Trigonometric functions describe repetitive motion, like waves. Because they repeat, the graphs look like smooth waves (sin and cos) or repeating branches (tan). Asymptotes in tan(x) occur because the function tries to divide by zero.
Self-check
What is the value of cos(180°)?
Show the answer
-1.
View diagrams
Sine and cosine over a full turn
Key and description
Solid curve: y=sin x; dashed curve: y=cos x, in degrees. At 0°,90°,180°,270°,360°, sine is 0,1,0,−1,0 and cosine is 1,0,−1,0,1. Both have amplitude 1 and period 360°.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 36: Using the alternate segment theorem; Understanding and constructing geometrical proofs using circle theorems
Open revision guidance
What to know
- A tangent is perpendicular to the radius at the point of contact.
- The angle at the centre is twice the angle at the circumference for the same arc.
- Angles in a semicircle are 90°.
- Angles in the same segment are equal.
- Opposite angles in a cyclic quadrilateral sum to 180°.
- Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
- Tangents from an external point to a circle are equal in length.
What to do in an assessment
- To find an angle using the alternate segment theorem, identify the tangent and the chord, then match the angle between them to the angle in the alternate segment.
- If the tangent at A makes a 50° angle with chord AB, the angle ACB at a circumference point C in the alternate segment is 50°. The point of tangency must be an endpoint of the chord.
- To construct a proof, list each theorem used as a geometric reason for each step of your derivation.
- In cyclic quadrilateral ABCD, angle BAD subtends the arc BCD and angle BCD subtends the other arc DAB. Those two arcs together make the full circle. The corresponding central angles sum to 360°; each inscribed angle is half its corresponding central angle, including a reflex central angle when needed. Therefore angle BAD + angle BCD = 180°.
- For tangents PA and PB from external point P to a circle with centre O, OA and OB are equal radii. Radius and tangent are perpendicular, so OAP and OBP are right-angled triangles. OP is their common hypotenuse and OA = OB. RHS gives congruence, hence PA = PB.
In plain language
Circle theorems are geometric identities. For example, the alternate segment theorem works because the angle between a tangent and a chord is essentially the same as the angle formed by two lines meeting at the circumference if you trace the path of the arc. By connecting these known relationships, you can prove the size of unknown angles without direct measurement.
Self-check
A cyclic quadrilateral ABCD is inscribed in a circle. Angle ABC = 110°. What is the size of angle ADC?
Show the answer
Since opposite angles of a cyclic quadrilateral sum to 180°, angle ADC = 180° - 110° = 70°.
View diagrams
Tangent and radius at a point of contact
Key and description
The radius to the point of contact is perpendicular to the tangent.
Angle in a semicircle: labelled proof
Key and description
AB is a diameter and O is the centre. C lies on the semicircle. OA = OC = OB, so the two small triangles are isosceles. Put α at A and between CA and CO; put β at B and between CB and CO. The angle sum of ABC gives 2(α + β) = 180°, hence ∠ACB = α + β = 90°.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 37: Using the relationships between the ratios of: • lengths and areas of similar 2-D shapes, and • lengths, areas and volumes of similar 3-D shapes
Open revision guidance
What to know
- If two shapes are similar with a length scale factor k, their area ratio is k² and their volume ratio is k³.
- Length ratio is the ratio of corresponding linear dimensions.
- Dimensions include length, width, height, perimeter, and radius.
What to do in an assessment
- To find the area ratio, square the linear scale factor: (side_new / side_old)².
- To find the volume ratio, cube the linear scale factor: (side_new / side_old)³.
- Worked Example: A shape has area 10cm² and scale factor k=3. The new area is 10 * 3² = 10 * 9 = 90cm².
- Worked Example: A shape has volume 5cm³ and scale factor k=2. The new volume is 5 * 2³ = 5 * 8 = 40cm³.
In plain language
When you enlarge a shape by a factor k, every dimension is scaled by k. Because area is a product of two dimensions (e.g., base × height), it scales by k × k = k². Volume is a product of three dimensions (e.g., length × width × height), so it scales by k × k × k = k³.
Self-check
A small cube has a side length of 2cm and a large cube has a side length of 6cm. What is the ratio of their volumes?
Show the answer
The length ratio k = 6 / 2 = 3. The volume ratio is k³ = 3³ = 27, so the ratio is 27:1.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 37: Enlargement with negative scale factors
Open revision guidance
What to know
- Enlargement involves a scale factor (k) and a centre of enlargement (x, y).
- A negative scale factor means the shape is reflected through the centre of enlargement while being scaled.
- Vector displacement formula: Image_point = Centre + k * (Original_point - Centre).
What to do in an assessment
- To perform an enlargement with scale factor -2 from centre (0,0): Multiply each vertex coordinate by -2.
- Worked Example: Point (1, 2) enlarged by factor -2 about (0,0) becomes (1 * -2, 2 * -2) = (-2, -4).
- For a centre (a,b), subtract the centre, apply the factor, then add the centre back.
In plain language
A negative enlargement is like a rotation of 180 degrees combined with a positive enlargement. By treating the centre as the origin of a local coordinate system, multiplying by a negative number flips the point to the opposite side of the centre.
Self-check
Enlarge point (2, 2) by scale factor -3 about the origin (0,0). What is the image point?
Show the answer
The image is (2 * -3, 2 * -3) = (-6, -6).
View diagrams
Negative enlargement from centre O
Key and description
Original triangle: (1, 1), (3, 1), (1, 2). Enlargement centre O, scale factor −2 gives (−2, −2), (−6, −2), (−2, −4). Join O to each vertex and extend through O: the image is on the opposite side and twice as far away. This is a negative enlargement, not a reflection in a mirror line.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 38: Lengths of circular arcs; Perimeters and areas of sectors and segments of circles; Surface areas and volumes of spheres, cones, pyramids and compound solids
Open revision guidance
What to know
- Arc length = (θ/360) * 2πr.
- Sector area = (θ/360) * πr².
- Segment area = Sector area - Triangle area.
- Volume of a pyramid = 1/3 * base_area * height.
- Volume of a sphere = 4/3 * πr³.
- Surface area of a sphere = 4πr².
What to do in an assessment
- For arc length, multiply the circumference (2πr) by the fraction of the circle (θ/360).
- Worked Example: Radius 10cm, angle 90°. Arc length = (90/360) * 2 * π * 10 = 0.25 * 20π = 5π ≈ 15.71cm.
- For compound solids, sum the surface areas of visible external faces only.
- For a cone with radius 3 cm, vertical height 4 cm and slant height 5 cm, volume = (1/3)π × 3² × 4 = 12π cm³. Curved surface area = π × 3 × 5 = 15π cm²; including the circular base gives total surface area 15π + 9π = 24π cm².
- A square-based pyramid with base side 6 cm and perpendicular height 4 cm has volume (1/3) × 36 × 4 = 48 cm³. Its triangular face slant height is √(4² + 3²) = 5 cm. Total surface area = 36 + 4 × (1/2 × 6 × 5) = 96 cm².
- For two cubes of side 2 cm joined across a full face, volume = 2 × 2³ = 16 cm³. External surface area = 2 × 6 × 2² − 2 × 2² = 40 cm²: both hidden joined faces must be removed.
- For a 90° sector of radius 10 cm, area = (90/360)π × 10² = 25π cm² and arc length = 5π cm. Sector perimeter includes both radii: 5π + 20 cm. The triangle between the radii has area ½ × 10 × 10 × sin 90° = 50 cm², so the minor segment area is 25π − 50 cm². Its chord is √(10² + 10²) = 10√2 cm, so segment perimeter is 5π + 10√2 cm. Do not confuse the chord boundary with the two radii.
In plain language
Calculating circular portions is about proportions. A sector is simply a fraction of the total area. Solids require breaking complex objects into known parts (prisms, pyramids) and summing volumes or external surface areas.
Self-check
Calculate the volume of a sphere with radius 3cm.
Show the answer
V = 4/3 * π * (3³) = 4/3 * π * 27 = 36π ≈ 113.10cm³.
View diagrams
Vertical and face slant heights
Key and description
A square pyramid has base side 6 cm and vertical height 4 cm. Join the apex to the midpoint of a base side: this face slant height is √(4² + 3²) = 5 cm. Volume = ⅓ × 36 × 4 = 48 cm³; total area = 36 + 4 × ½ × 6 × 5 = 96 cm². The perspective drawing is schematic.
Labelled constructed examples. Each diagram has its own data and method.
Topic group
Statistics
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 13, 25, 39: Understanding and using the statistical problem solving process: specifying the problem/planning; collecting, processing and representing data; interpreting and discussing results
Open revision guidance
What to know
- Stage 1: Specify the problem/plan.
- Stage 2: Collect data.
- Stage 3: Process/represent data (clean, calculate, graph).
- Stage 4: Interpret and discuss results.
What to do in an assessment
- If investigating height, write a hypothesis (e.g., 'Year 11s are taller than Year 7s').
- Choose a data collection method (e.g., survey or observation).
- Choose a representation (e.g., bar chart for categorical data).
- A hypothetical height study plans to compare Year 7 and Year 11 using the same measurement method and centimetres. A tiny illustrative sample is Year 7: 140,145,150 and Year 11: 155,160,165. Processing gives means 145 cm and 160 cm. This sample supports the prediction of greater Year 11 height by 15 cm, but three learners per year are too few and may be unrepresentative. Report that limitation rather than claiming a whole-school result.
In plain language
Statistics isn't just drawing graphs; it's a cycle. You start with a question, get the data to answer it, process that data so it makes sense, and finally explain what you found in the context of your original question.
Self-check
What is the purpose of the 'processing' stage in the statistical cycle?
Show the answer
To clean, calculate, and organise raw data into a visual or numerical form that makes it easier to analyse.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 13, 25, 39: Specifying and testing hypotheses, taking account of the limitations of the data available; Designing and criticising questions for a questionnaire, including notions of fairness and bias
Open revision guidance
What to know
- A hypothesis is a prediction that can be tested using data.
- Data limitations include small sample sizes, non-representative sampling, and observer bias.
- Questionnaires must use neutral, non-leading language to be fair; leading questions influence respondents, introducing bias.
- Questions should use clear, non-overlapping categories for quantitative ranges to ensure each response fits into exactly one group.
What to do in an assessment
- To test a hypothesis: State the prediction, collect data, analyse whether the results support the claim, and identify potential flaws in the data collection process.
- Critique questionnaire design: Check for leading language (e.g., 'Don't you agree...') and replace with neutral queries (e.g., 'What is your view on...').
- Example: Given the hypothesis 'The park is too noisy', a biased question is 'Why is the park too noisy?'. A fair version is 'How would you rate the noise level in the park? (Quiet, Moderate, Noisy)'.
In plain language
A testable hypothesis needs data that could support or contradict it. Neutral questions reduce leading-question bias, but they do not by themselves guarantee representative or accurate data. Consider sample size, who was asked and when observations were taken before accepting a conclusion.
Self-check
A researcher testing the hypothesis 'The local library is too crowded' asks: 'Why do you find the library too crowded to work in?' Identify the flaw and write a better question.
Show the answer
The flaw is that the question is leading/presumptive; it assumes the library is too crowded. A neutral version is: 'How would you describe the crowd levels in the library when you visit?'
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 13, 25, 39: Sorting, classification and tabulation of qualitative (categorical) data, discrete or continuous quantitative data; Grouping of discrete or continuous data into class intervals of equal widths; Understanding and using tallying methods
Open revision guidance
What to know
- Qualitative data consists of categories (e.g., favourite colour); quantitative data consists of numerical values.
- Discrete data consists of countable values (e.g., number of children); continuous data consists of measurable values (e.g., height, weight).
- Tallies are used to count frequency by marking groups of five: 卌.
- This Foundation skill uses equal-width class intervals, for example 10 ≤ x < 20 and 20 ≤ x < 30. Unequal-width intervals are valid too, but their use and later frequency-density diagrams are separate tier-specific skills.
What to do in an assessment
- Organise raw data into frequency tables: List categories or intervals, tally each data point, and sum the tallies for the total frequency.
- Define class intervals: Ensure no data point falls into two intervals. If using 10-19 and 20-29, the width is 10 (20 - 10 = 10).
- Example: For data {2, 5, 8, 3, 5, 9, 7} with intervals 0-4 and 5-9: 2 and 3 fall into 0-4 (Frequency 2); 5, 8, 5, 9, 7 fall into 5-9 (Frequency 5).
In plain language
Sort categories or numerical values into a tally table, then count each tally to form frequencies. Equal-width integer groups such as 0–9 and 10–19 each contain ten possible integers; the boundary form 0 ≤ x < 10 and 10 ≤ x < 20 also works for continuous measurements. Never overlap the boundaries.
Self-check
Data: 12, 15, 18, 11, 14. Create a frequency table using intervals 10-14 and 15-19.
Show the answer
Interval 10-14: 12, 11, 14 (Frequency 3). Interval 15-19: 15, 18 (Frequency 2).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 13, 25, 39: Constructing and interpreting pictograms, bar charts and pie charts for qualitative data; Constructing and interpreting vertical line diagrams for discrete data; Constructing line graphs for the values of a variable at different points in time; understanding that intermediate values in a line graph may or may not have meaning; Simple cases of constructing and interpreting grouped frequency diagrams; Temperature charts; Constructing and interpreting scatter diagrams for data on paired variables
Open revision guidance
What to know
- Pictograms use symbols to represent specific frequencies.
- Bar charts show frequency of qualitative/discrete data; vertical line diagrams are specifically for discrete data.
- Pie charts divide 360° proportionally: (Category Frequency / Total Frequency) × 360°.
- Scatter diagrams plot paired variables to show correlation: positive, negative, or none.
- Line graphs show trends over time; points between marks are estimated.
What to do in an assessment
- Construct pie charts: Calculate the angle for each category by multiplying the fraction of the total by 360°.
- Interpret scatter diagrams: Check the direction of the points. If points move upward left-to-right, it is positive correlation.
- Example: Total = 40, Red = 10. Fraction = 10/40 = 0.25. Angle = 0.25 × 360° = 90°.
- For a pictogram with one symbol representing 4 learners, frequencies 8, 12 and 6 require two, three and one-and-a-half symbols. Show the key; half a symbol means 2 learners, not half a learner. A bar chart of the same categories uses equal-width separated bars with heights 8, 12 and 6.
- For grouped data 0 ≤ x < 10 with frequency 4, 10 ≤ x < 20 with frequency 7 and 20 ≤ x < 30 with frequency 3, draw adjacent equal-width bars over those intervals, with heights 4, 7 and 3 on a frequency axis. Read the total as 14 and the modal interval as 10 ≤ x < 20. This simple equal-width diagram is not the unequal-width frequency-density method.
- For a temperature chart at 06:00, 09:00, 12:00 and 15:00 showing 3°C, 7°C, 10°C and 8°C, put time on the horizontal axis and temperature on the vertical axis, plot and join the points. The maximum recorded temperature is 10°C at noon; from 09:00 to noon the increase is 3°C. An intermediate reading from the connecting line is an estimate, not an observation.
In plain language
Statistical diagrams change numbers into pictures. A pie chart turns frequencies into 'slices' of a circle, while a scatter diagram checks if two things are related—like whether taller people tend to have larger shoe sizes. Always remember that for scatter diagrams, the line of best fit is just a guide, not an exact rule for every single point.
Self-check
A survey of 80 people shows 20 prefer Blue. Calculate the angle for Blue on a pie chart.
Show the answer
(20 / 80) × 360° = 0.25 × 360° = 90°.
View diagrams
Pictogram and temperature chart
Key and description
In the pictogram, one filled circle means 4 learners: A = 8, B = 12 and C = 6. Half a circle means 2. The temperature chart plots (06:00, 3°C), (09:00, 7°C), (12:00, 10°C) and (15:00, 8°C). Connecting lines give estimates between recorded observations.
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 14, 26, 40: Mean, median and mode for a discrete (ungrouped) frequency distribution; Comparison of two distributions using one measure of central tendency (i.e. the mean or the median); Modal category for qualitative data; Modal class for grouped data; Calculating or estimating the range applied to discrete data; Drawing 'by eye' a line of 'best fit' on a scatter diagram, understanding and interpreting what this represents
Open revision guidance
What to know
- Mean: Sum of all values divided by the total number of values (Σx / n).
- Median: The middle value when data is arranged in ascending order.
- Mode: The most frequently occurring value or category.
- Modal Class: The class interval with the highest frequency.
- Range: The difference between the highest and lowest values.
- Line of Best Fit: A straight line used to represent the trend in a scatter diagram, balanced so points are evenly spread on both sides.
What to do in an assessment
- Calculate mean from frequency: Σ(frequency × value) / Σ(frequency).
- Identify Median: If n is odd, the middle is at position (n+1)/2.
- Example: Data {2, 2, 3, 5, 8}. Mean = (2+2+3+5+8)/5 = 4. Median = 3. Mode = 2. Range = 8 - 2 = 6.
- For values 1, 2, 3 with frequencies 2, 3, 1, the total frequency is 6 and total value is 1×2 + 2×3 + 3×1 = 11. Mean = 11/6 ≈ 1.83. Ordered values are 1,1,2,2,2,3; median averages the third and fourth, giving 2, and mode is 2. Range = 3 − 1 = 2.
- Compare A = 1,2,2,2,3,5 with B = 2,3,3,3,4,6: means are 15/6 = 2.5 and 21/6 = 3.5, and medians are 2 and 3. By either measure, B has a higher centre. Name which measure supports your comparison.
- For colours red, blue, green with frequencies 4,7,3, the modal category is blue. For grouped intervals 0 ≤ x < 10, 10 ≤ x < 20, 20 ≤ x < 30 with frequencies 2,7,3, the modal class is 10 ≤ x < 20.
- On a scatter diagram of (1,2),(2,4),(3,5),(4,8),(5,9), sketch a line through the middle of the points, balancing points above and below rather than joining them in order. A line approximately y = 1.8x + 0.2 follows the positive trend.
In plain language
Averages tell us where the 'center' of our data lies. The mean is the mathematical average, the median is the middle person in a line, and the mode is the most popular item. These measures help us compare two groups—for example, comparing test scores to see which class performed better overall.
Self-check
Data: 1, 1, 2, 4, 7. Find mean, median, mode and range.
Show the answer
Mean: (1+1+2+4+7)/5 = 15/5 = 3. Median: 2. Mode: 1. Range: 7 - 1 = 6.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 14, 26, 40: Recognising that graphs may be misleading; Looking at data to find patterns and exceptions; Drawing inferences and conclusions from summary measures and data representations, relating results back to the original problem; Drawing of conclusions from scatter diagrams; using terms such as positive correlation, negative correlation, little or no correlation; Appreciating that correlation does not imply causality
Open revision guidance
What to know
- Graphs can be misleading if the axes do not start at zero or use irregular scales.
- Positive correlation: Both variables increase together. Negative correlation: As one increases, the other decreases.
- Correlation does not imply causality; a relationship between two variables does not mean one causes the other.
- Outliers are data points that do not fit the general pattern.
What to do in an assessment
- Critique representations: Check if axes are labelled and scaled correctly to avoid exaggerated trends.
- Identify patterns: Describe scatter plots using terms like 'positive correlation', 'negative correlation', or 'no correlation'.
- Example: Ice cream sales and sunscreen sales are positively correlated because they both rise in summer, but ice cream sales do not cause sunscreen sales; weather causes both.
In plain language
Just because two things happen at the same time (like wearing a scarf and drinking hot chocolate) doesn't mean one causes the other. Correlation is just noticing a trend. When reading charts, always check the labels and numbers on the axes to ensure the creator isn't trying to make a small change look like a huge one.
Self-check
Does a strong positive correlation between umbrella sales and heavy rain imply umbrellas cause rain?
Show the answer
No. Correlation does not imply causality; both are caused by the weather.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 15, 27, 41: Understanding and using the vocabulary of probability, including notions of uncertainty and risk; The terms 'fair', 'evens', 'certain', 'likely', 'unlikely ' and 'impossible'
Open revision guidance
What to know
- Probability is the measure of the likelihood of an event occurring.
- The scale goes from 0 to 1.
- Impossible: 0 (cannot happen).
- Evens: 0.5 (equal chance of happening or not).
- Certain: 1 (will definitely happen).
- Likely is > 0.5; unlikely is < 0.5.
What to do in an assessment
- Assign descriptors: Match events to terms like 'fair', 'certain', or 'impossible' based on the scenario.
- Example: On a fair 6-sided die, rolling a 1, 2, 3, 4, 5, or 6 is certain; rolling a 7 is impossible.
In plain language
Probability is the language of chance. We use a scale where 0 means there is zero chance of something happening, and 1 means it is guaranteed. Anything in the middle is just a estimate of how risky or probable an event is.
Self-check
How would you describe the probability of a coin landing on a 'tail' if it is a fair coin?
Show the answer
Evens (0.5).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 15, 27, 41: Understanding and using the probability scale from 0 to 1; Use of: the probability of an event not occurring is one minus the probability that it occurs; (Probabilities must be expressed as fractions, decimals or percentages.)
Open revision guidance
What to know
- Probabilities can be written as fractions, decimals, or percentages.
- The total probability of all possible outcomes is always 1.
- Complementary rule: The probability of an event not happening is 1 minus the probability that it does happen: P(not A) = 1 - P(A).
What to do in an assessment
- Convert between forms: 0.25 = 25/100 = 1/4 = 25%.
- Find complements: Subtract the given probability from 1.
- Example: If P(rain) = 0.3, P(no rain) = 1 - 0.3 = 0.7.
In plain language
If there is a 30% chance of something happening, there must be a 70% chance of it NOT happening, because the total must add up to 100%. This is the basic idea of complementary events: everything else is what's left over.
Self-check
The probability of picking a red marble is 0.4. What is the probability of not picking a red marble?
Show the answer
1 - 0.4 = 0.6.
View diagrams
Labelled probability scale
Key and description
The equally spaced marks are 0, ¼, ½, ¾ and 1. Zero means impossible, ½ means an even chance, and 1 means certain. Both endpoints are included. The highlighted point represents probability ¾.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 15, 27, 41: Estimating the probability of an event as the proportion of times it has occurred; Relative frequency; An understanding of the long-term stability of relative frequency is expected; Calculating theoretical probabilities based on equally likely outcomes; Estimating probabilities based on experimental evidence; Comparing an estimated probability from experimental results with a theoretical probability
Open revision guidance
What to know
- Theoretical probability: Based on equally likely outcomes (e.g., 1/6 for a fair die face).
- Experimental probability: Based on observed data (Relative Frequency = Favorable results / Total trials).
- With many fair, repeated trials, relative frequency tends to stabilise near the theoretical probability. It need not become closer after every new trial, and a large sample does not guarantee an exact value.
What to do in an assessment
- Calculate theoretical: Count successful outcomes / total possible outcomes.
- Calculate experimental: Use the tally of successful trials / total number of trials.
- Example: Toss a coin 100 times. If it lands heads 52 times, the experimental probability is 52/100 = 0.52.
In plain language
Theoretical probability comes from the model, such as equally likely faces of a fair die. Experimental probability comes from observations. In 100 fair-coin tosses, 52 heads gives 0.52 compared with theoretical 0.5, a difference of 0.02. More trials generally give a more stable estimate, not a guaranteed result or a steadily shrinking difference.
Self-check
A basketball player shoots 20 free throws and makes 14. What is the experimental probability of making a shot?
Show the answer
14 / 20 = 0.7.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 15, 27, 41: Identifying all the outcomes of a combination of two experiments, e.g. throwing two dice; use tabulation, Venn diagrams, or other diagrammatic representations of compound events
Open revision guidance
What to know
- Sample space: A list or table showing all possible outcomes of an experiment.
- Compound event: Two or more events happening, like rolling two dice.
- Tabulation: A grid used to display the outcomes of two events.
- Venn diagrams: Used to show overlapping sets of outcomes.
What to do in an assessment
- Construct a grid: For two dice, create a 6x6 table. Rows = Die 1, Columns = Die 2.
- Calculate from grid: Count the cells that satisfy your condition and divide by 36.
- Example: Sum of 3 when rolling two dice. Outcomes: (1,2) and (2,1). Probability = 2/36 = 1/18.
In plain language
When you do two things at once, there are more possible results than you might think. A grid is the best way to see them all, so you don't miss any combinations. Once you have the grid, you can just count how many outcomes match what you are looking for.
Self-check
Using a 6x6 grid for two dice, how many outcomes result in a sum of 5?
Show the answer
4 outcomes: (1,4), (2,3), (3,2), (4,1).
View diagrams
Sample space for two coin tosses
Key and description
Rows identify the first toss and columns the second. Cells are HH, HT, TH, TT (H=head, T=tail). For two independent fair tosses each cell has probability ¼. Three cells contain at least one head, giving probability ¾.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Minimum tier foundation; printed pages 15, 27, 41: Knowledge that the total probability of all the possible outcomes of an experiment is 1
Open revision guidance
What to know
- The sum of all individual probabilities in an experiment must equal exactly 1.
- An experiment includes all possible outcomes; if one outcome is missing, its probability is 1 minus the sum of all others.
What to do in an assessment
- Verify: Add all known probabilities to ensure they total 1.
- Calculate missing: If P(A)=0.2, P(B)=0.3, find P(C). P(C) = 1 - (0.2 + 0.3) = 0.5.
- Example: Three coloured balls in a bag. P(Red)=0.1, P(Blue)=0.4. P(Green) = 1 - 0.5 = 0.5.
In plain language
If you have a set of results that cover every possibility, the sum of their likelihoods has to be 100%, or 1. If you know all the probabilities except one, you can just subtract the known ones from 1 to find the answer.
Self-check
A spinner has three sections: Red, Yellow, and Green. P(Red) = 0.3, P(Yellow) = 0.4. What is P(Green)?
Show the answer
1 - (0.3 + 0.4) = 0.3.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 25, 39: Testing an hypothesis such as ‘Girls tend to do better than boys in biology tests’; Specifying the data needed and considering potential sampling methods; Sampling systematically; Considering the effect of sample size and other factors that affect the reliability of conclusions drawn
Open revision guidance
What to know
- Hypotheses must be testable and specific.
- Sampling methods: Random, systematic (e.g., every 10th), or stratified.
- Leading questions create bias; questions should be neutral and exhaustive.
- Reliability depends on sample size and minimising bias.
What to do in an assessment
- To test 'Boys do better than girls': Collect representative data from both groups, calculate the mean, and compare.
- Use systematic sampling with a random starting position and a fixed interval. Check whether patterns in the ordered list could bias the sample.
- Rewrite leading questions: Change 'Why do you like X?' to 'How do you rate X?' using a 1-5 scale.
- Example: To select 10 from 100 students, pick every 10th student on the register.
- To select 10 from 100 ordered students, randomly choose a start from 1–10. If it is 7, select positions 7,17,27,37,47,57,67,77,87,97. Taking every tenth without considering ordering can create bias if the list has a repeating pattern.
- For a hypothetical biology comparison, girls’ marks 60,70,80 have mean 70 and boys’ marks 55,65,75 have mean 65. That small sample supports a 5-mark difference, not a claim about all learners. Use comparable assessments, representative sampling and sufficient numbers; a larger biased sample can still mislead.
In plain language
Good statistics depend on the quality of data. If the sample is biased (e.g., only surveying friends) or the question is leading (e.g., 'don't you agree?'), the conclusion will be unreliable.
Self-check
Identify why 'Do you agree that math is fun?' is a poor survey question.
Show the answer
It is a leading question that encourages a specific answer ('agree'), creating bias.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 25, 39: Grouping of discrete or continuous data into class intervals of unequal widths
Open revision guidance
What to know
- Discrete data uses distinct numbers; continuous data uses measurements on a scale.
- Grouped intervals are defined as lower bound ≤ x < upper bound.
- Frequency tables summarise large datasets by counting how many items fall into each interval.
What to do in an assessment
- Create intervals for continuous data: 0 ≤ x < 10, 10 ≤ x < 20.
- Use tallies for each data entry to avoid counting errors.
- Ensure all intervals are mutually exclusive (no overlap at boundaries).
- Example: Data points 5, 12, 18, 25. Intervals 0-9 (f=1), 10-19 (f=2), 20-29 (f=1).
- Unequal intervals need explicit, non-overlapping boundaries. For measurements 1, 4, 7, 12, 19 and 24 minutes, classes 0 ≤ t < 5, 5 ≤ t < 15 and 15 ≤ t < 30 have widths 5, 10 and 15 and frequencies 2, 2 and 2. A value of 15 goes in the last class. Record the widths as well as the frequencies; equal frequency does not mean equal width.
In plain language
Grouping data into intervals is necessary when you have too many different numbers to list individually. It summarises information so you can see trends, such as where most data points are clustered.
Self-check
Given data: 2, 7, 11, 14, 18. Group into intervals 0-9 and 10-19. What are the frequencies?
Show the answer
0-9: {2, 7} (f=2); 10-19: {11, 14, 18} (f=3).
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 25, 39: Constructing and interpreting frequency polygons; Constructing and interpreting cumulative frequency tables and diagrams using the upper boundaries of the class intervals
Open revision guidance
What to know
- A frequency polygon is constructed by plotting frequency against the midpoint of each class interval.
- Cumulative frequency tracks the running total of frequencies as values increase.
- Cumulative frequency diagrams (ogives) are plotted using the upper boundary of each class interval.
What to do in an assessment
- Calculate midpoints: (Lower + Upper) / 2. Plot (midpoint, frequency) and connect with straight lines.
- For cumulative frequency: Plot (upper boundary, cumulative total). Example: 0-5 (10), 5-10 (20). Point 1: (5, 10). Point 2: (10, 30).
In plain language
A frequency polygon uses a point at the centre of each group to represent the 'average' position of that group's data. Cumulative frequency is useful for finding percentiles or the median, as it shows how many data points are below a certain value at any point along the scale.
Self-check
Data: 0-10 (5), 10-20 (15). Find the cumulative frequency coordinates for the upper boundaries.
Show the answer
(10, 5) and (20, 20).
View diagrams
Frequency polygon from grouped data
Key and description
Classes and frequencies: [0, 10): 2; [10, 20): 6; [20, 30): 8; [30, 40): 4. Plot the class midpoints: (5, 2), (15, 6), (25, 8), (35, 4), then join consecutive points.
Cumulative-frequency curve and quartiles
Key and description
Frequencies 2, 6, 8, 4 give cumulative points (0, 0), (10, 2), (20, 8), (30, 16), (40, 20). N = 20. Interpolating on the graph: Q₁ at N/4 = 5 is 15; median at N/2 = 10 is 22.5; Q₃ at 3N/4 = 15 is 28.75. Do not use (N + 1)/2.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 26, 40: Selecting and using an appropriate measure of central tendency; Estimates for the median and mean of grouped frequency distributions; Comparison of two distributions using one measure of spread; Estimating the median from a cumulative frequency diagram; Selecting appropriate measures of spread; calculating or estimating the interquartile range of discrete data, and range and interquartile range of grouped and continuous data; Producing and using box-and-whisker plots to compare distributions; Drawing the line of 'best fit' through a given, calculated or plotted mean point
Open revision guidance
What to know
- Estimated Mean = Σ(midpoint × frequency) / Σ(frequency).
- Median for grouped data is estimated via linear interpolation or from a cumulative frequency graph at n/2.
- IQR = Q3 − Q1. For a cumulative-frequency estimate, locate Q1 at N/4 and Q3 at 3N/4 on the cumulative axis. For individual discrete data, order the values and take the medians of the lower and upper halves, stating the convention used rather than applying cumulative-graph positions blindly.
- Box-and-whisker plots: whiskers represent the range, box represents Q1, Median, and Q3.
What to do in an assessment
- For mean: Multiply group midpoint (x) by frequency (f). Sum (fx) and divide by sum (f).
- Line of best fit: Draw a straight line passing through the mean point (mean x, mean y) that balances points above and below the line.
- For equal-width classes 0 ≤ x < 10, 10 ≤ x < 20, 20 ≤ x < 30 and 30 ≤ x < 40 with frequencies 2, 6, 8 and 4, use midpoints 5, 15, 25 and 35. Estimated mean = (10 + 90 + 200 + 140)/20 = 22. This is an estimate because the exact values within each class are unknown.
- The cumulative frequencies at 0, 10, 20, 30 and 40 are 0, 2, 8, 16 and 20. From the straight joined cumulative graph, the median at cumulative 10 is 22.5; Q1 at 5 is 15 and Q3 at 15 is 28.75, so estimated IQR is 13.75. The estimated boundary range is 40 − 0 = 40, not the exact range of the unseen observations.
- For ordered discrete data 1, 2, 4, 5, 7, 9, 10, 12, median = (5 + 7)/2 = 6. The lower half has median (2 + 4)/2 = 3 and the upper half (9 + 10)/2 = 9.5, so IQR = 6.5 and range = 12 − 1 = 11. Plot whiskers at 1 and 12 and a box from 3 to 9.5, with the median line at 6.
- Compare with ordered data 3, 4, 5, 5, 7, 8, 9, 10: median is also 6, Q1 = 4.5, Q3 = 8.5, IQR = 4 and range = 7. Both have the same centre by median, but the second has less spread by IQR and range. For skewed data or strong outliers, median with IQR is often more informative than mean with range; a modal category is appropriate for categorical data.
- For paired points (1,2), (2,4), (3,5), (4,8), (5,9), the mean point is (15/5, 28/5) = (3, 5.6). Draw the line of best fit through that point while balancing the scatter above and below; y = 1.8x + 0.2 passes through it.
In plain language
When we don't have individual data points, we use the midpoint of a group to represent all items in that group to find the mean. The IQR tells us where the middle 50% of the data sits, which is more reliable than the range because it excludes extreme values or outliers.
Self-check
Given: Q1=10, Median=15, Q3=20. Calculate the IQR.
Show the answer
IQR = 20 - 10 = 10.
View diagrams
Box plot and five-number summary
Key and description
Constructed summary: minimum 2, lower quartile 5, median 8, upper quartile 12, maximum 16. Range = 16 − 2 = 14; interquartile range = 12 − 5 = 7. The box spans the middle half of the distribution, not half its numerical range.
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 27, 41: Graphical representation of relative frequency against the number of trials
Open revision guidance
What to know
- Relative frequency = Number of successful trials / Total number of trials.
- Relative frequency acts as an experimental estimate for probability.
- The Law of Large Numbers states that relative frequency converges to theoretical probability as the number of trials increases.
What to do in an assessment
- Divide the occurrences by the total number of events recorded in the experiment.
- Example: If a die is rolled 50 times and a 6 appears 10 times, relative frequency is 10/50 = 0.2.
- To graph relative frequency, put cumulative number of trials on the horizontal axis and cumulative successes ÷ trials on the vertical axis. Counts 6/10, 26/50, 49/100 and 202/400 give points (10, 0.60), (50, 0.52), (100, 0.49), (400, 0.505). Join successive points to show the trend toward 0.5. Each value uses all trials so far, not just the last batch. The approach can fluctuate rather than moving steadily toward the theoretical probability.
In plain language
Theoretical probability is what we expect to happen (e.g., 1/6 for a fair die), but relative frequency is what actually happens in reality. The more trials we conduct, the more likely our results will match the theoretical prediction.
Self-check
A fair coin is flipped 200 times and lands on heads 95 times. What is the relative frequency of heads?
Show the answer
95 / 200 = 0.475.
View diagrams
Relative frequency against cumulative trials
Key and description
Cumulative successes/trials are 6/10, 26/50, 49/100 and 202/400. The plotted relative frequencies are 0.60, 0.52, 0.49 and 0.505. The dashed theoretical probability is 0.5. Fluctuations need not move steadily towards it.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 27, 41: Using tree diagrams to identify outcomes of compound events
Open revision guidance
What to know
- A tree diagram records successive stages; each leaf is a complete ordered outcome.
- Branches leaving a node list all possible next results. If probabilities are given, the probabilities leaving that node sum to 1.
What to do in an assessment
- Draw a first pair of branches for coin 1: H and T. From each endpoint draw H and T branches for coin 2. Trace the four ordered outcomes HH, HT, TH, TT.
- The second branch belongs to the first outcome already taken. For two fair, independent coins each branch probability is 1/2, so each complete path has probability 1/4. This card identifies outcomes; dependent probabilities without replacement are taught separately at Higher.
In plain language
A tree shows all the possible next outcomes after each first result. Two coin tosses give HH, HT, TH and TT; HT and TH are different ordered outcomes. Follow each path rather than counting an outcome twice.
Self-check
A spinner can land on A, B or C, followed by a coin showing H or T. List the six tree outcomes.
Show the answer
AH, AT, BH, BT, CH, CT: each first-stage branch has two second-stage branches.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Minimum tier intermediate; printed pages 27, 41: Recognising the conditions when the addition of probabilities for mutually exclusive events and the multiplication of probabilities for two independent events apply, and making the appropriate calculations; If A and B are mutually exclusive, then the probability of A or B occurring is P(A) + P(B); If A and B are independent events, the probability of A and B occurring is P(A) × P(B)
Open revision guidance
What to know
- Mutually exclusive: P(A or B) = P(A) + P(B).
- Independent: P(A and B) = P(A) × P(B).
- Sum of all exhaustive, mutually exclusive probabilities = 1.
What to do in an assessment
- Add probabilities when events cannot happen simultaneously.
- Multiply probabilities when the occurrence of one does not change the probability of the other.
- For a fair die, “roll 1” and “roll 6” are mutually exclusive, so P(1 or 6) = 1/6 + 1/6 = 1/3. The addition rule without subtraction does not apply to overlapping events.
- For a fair coin and a separate fair die, outcomes are independent: P(head and 6) = 1/2 × 1/6 = 1/12. Independence means the coin result does not change the die probability.
In plain language
Adding probabilities works when we want either of two separate things to happen. Multiplying works when we require a specific sequence or joint event, like getting heads on two different coins, because the chance of the second event is unchanged by the first.
Self-check
Events A and B are independent. P(A)=0.5, P(B)=0.2. Find P(A and B).
Show the answer
0.5 × 0.2 = 0.1.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 39: Working with stratified sampling techniques and defining a random sample
Open revision guidance
What to know
- In a simple random sample, every possible group of the required sample size has the same chance of selection; every member therefore has an equal chance of inclusion.
- Stratified sampling: population is divided into subgroups (strata), and samples are taken proportionally.
- Formula: (Number in stratum / Total population) * Required sample size.
What to do in an assessment
- Determine the total population and the size of each stratum.
- Calculate the fraction for each stratum.
- Multiply the fraction by the total sample size to get the number needed from each group.
- Worked Example: Population 200 (100 men, 100 women). Sample size 20. Men needed = (100/200) * 20 = 10.
- After allocating 20 under-18s and 30 adults in the 500-person example, number people within each group and choose 20 and 30 distinct random numbers. Do not choose the first volunteers, which would undermine the random selection. If allocations require rounding, adjust consistently so they still sum to the required total.
In plain language
Stratified sampling sets sample numbers in proportion to the sizes of the chosen population subgroups, then selects randomly within each subgroup. It ensures the planned representation of those groups, but does not guarantee an unbiased result: unsuitable strata, incomplete lists or non-response can still cause problems.
Self-check
Population of 500 contains 200 under-18s and 300 over-18s. Calculate the number of under-18s required for a stratified sample of 50.
Show the answer
(200 / 500) * 50 = 0.4 * 50 = 20.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 40: Extending skills in handling data into constructing and interpreting histograms with unequal class widths; Frequency density; Interpreting shapes of histograms representing distributions (with reference to mean and dispersion)
Open revision guidance
What to know
- Frequency density = Frequency / Class width.
- In a histogram, Area = Frequency.
- This is used for continuous data with unequal intervals.
What to do in an assessment
- Calculate frequency density for each interval.
- Plot frequency density on the y-axis against the variable on the x-axis.
- Worked Example: Class 10-20 (width 10, freq 5). Density = 5 / 10 = 0.5. Class 20-40 (width 20, freq 20). Density = 20 / 20 = 1.0.
- The second bar will be twice as tall as the first.
- Interpret a histogram using both interval positions and rectangle areas. Concentration around smaller values with a long tail to the right suggests positive skew; large values in the tail can pull the mean above the median. A long left tail can pull the mean below the median. For a roughly symmetric distribution the mean and median can be close, but the drawing alone does not give an exact mean.
- Compare spread on matching scales: data occupying 0–40 are more dispersed by range than data occupying 10–20, even if both histograms have a tallest bar of the same height. With unequal widths, frequency is area = width × density; compare frequencies using areas, not heights alone.
In plain language
Histograms use area to represent frequency. If you have an interval that is twice as wide as another, you must halve the height of the bar for that interval to ensure the area correctly represents the count of items.
Self-check
Data: Class 10-20 (freq 5), 20-30 (freq 15). Find frequency densities.
Show the answer
Class 10-20: 5 / 10 = 0.5. Class 20-30: 15 / 10 = 1.5.
View diagrams
Histogram with unequal class widths
Key and description
The vertical scale is frequency density. Bar area, not bar height alone, represents frequency.
Labelled constructed examples. Each diagram has its own data and method. - Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Minimum tier higher; printed pages 41: Understanding when and how to estimate conditional probabilities; The multiplication law for dependent events; Sampling without replacement
Open revision guidance
What to know
- P(A and B) = P(A) * P(B|A) for dependent events.
- Sampling without replacement means the denominator decreases for the second event.
- Conditional probability P(B|A) is the probability of B happening given A has already occurred.
What to do in an assessment
- Use a tree diagram to track probabilities for each stage.
- Update the fractions after the first event to reflect the removal of an item.
- Worked Example: 2 Red, 3 Blue balls. P(Red, Red) = (2/5) * (1/4) = 2/20 = 1/10.
- In a hypothetical sample of 40 learners, 20 play sport. Of those 20, 12 also play music; among the 20 who do not play sport, 8 play music. Given that a learner plays sport, estimated P(music | sport) = 12/20 = 0.6. The denominator is the 20 meeting the condition, not all 40. Estimated P(sport and music) = 12/40 = 0.3 = (20/40)(12/20). Overall P(music) = 20/40 = 0.5, different from 0.6, so these events are not independent in this sample. Estimates for a wider population depend on how representative the sample is.
In plain language
Conditional probability accounts for changing conditions. If you take an item out, there are fewer items left, which shifts the likelihood of what happens next. The multiplication law allows us to find the probability of a sequence by multiplying the likelihood of each step.
Self-check
A box has 4 green and 2 yellow balls. Two are picked without replacement. What is the probability of picking two green balls?
Show the answer
(4/6) * (3/5) = 12/30 = 0.4.
Topic group
Assessment and supplied formulas
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · Summary of assessment, printed p2; Technical information, printed p44; 2026–27 entry procedures, GCSE Mathematics notes, printed p18
Open revision guidance
What to know
- This is legacy GCSE Mathematics 3300QS, not the new Mathematics and Numeracy Double Award. Both papers must be taken at the same Foundation, Intermediate or Higher tier in the same series.
- Unit 1 is non-calculator and Unit 2 allows a calculator. Each is 50% of the qualification. At Foundation, each lasts 90 minutes and has 65 marks; at Intermediate and Higher, each lasts 105 minutes and has 80 marks.
- The legacy grade ranges are Foundation D to G, Intermediate B to E and Higher A* to C. Higher-tier content builds on Intermediate and Foundation; Intermediate builds on Foundation.
- Summer 2026 was the last full assessment. Maths resits are offered in November 2026 and June 2027, subject to demand; January 2027 is the separate Numeracy resit, not a Maths paper.
- The 2026–27 entry handbook normally restricts final certifications to post-16 candidates who previously cashed in before November 2026. It also permits post-16 adult learners, learners who sat a 9–1 GCSE, and learners unable to sit the GCSE in Year 11 or earlier without a previous entry. Confirm eligibility and availability with the centre. New Year 10 entrants from September 2025 use the new Double Award.
What to do in an assessment
- Plan practice for both units, not just the calculator paper. A Foundation practice schedule needs two separate 90-minute sittings; an Intermediate or Higher schedule needs two 105-minute sittings.
- On Unit 1, show an exact written calculation: 347 × 26 = 6940 + 2082 = 9022. On Unit 2, enter the whole expression and check it with an estimate rather than copying an unexplained display.
- Check the paper title, tier, unit and date on your entry. A January 2027 Mathematics–Numeracy entry is not a replacement for legacy Mathematics Unit 1 or 2.
In plain language
Foundation means two 90-minute papers, so total examination time is 180 minutes. Moving to Intermediate changes both papers to 105 minutes, a total of 210 minutes. You cannot choose Foundation for one paper and Higher for the other in this legacy qualification.
Self-check
A candidate plans one legacy Intermediate paper and one Higher paper. Is this valid, and what is the total time for two Intermediate papers?
Show the answer
No. Both papers must be at the same tier in the same series. Two Intermediate papers take 105 + 105 = 210 minutes.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · §3.1 Assessment objectives, printed p42; §3.3 Communicating, organising and writing accurately, printed p43
Open revision guidance
What to know
- AO1 tests recall and standard methods, with 50–60% weighting. AO2 tests selecting and applying methods, with 10–20%. AO3 tests interpreting, analysing and devising strategies, with 25–35%.
- Questions can be structured, unstructured, short or long, including multiple choice. A multi-step problem may not tell you which operation to use.
- Each paper includes two marks for communicating, organising and writing accurately: one for communication and organisation, and one for accurate writing, including mathematical notation. The questions are identified on the paper.
What to do in an assessment
- Read the problem and identify the quantities, units and required outcome. Choose a method, show the calculation, then interpret the result. For 35 pupils and coaches seating 12, 35 ÷ 12 = 2 remainder 11, so three coaches are needed, not two. At £80 each, the total is 3 × £80 = £240.
- For a numerical estimate, show the rounded values and operation: 68.2 × 29.8 is approximately 70 × 30 = 2100. The exact value is 2032.36, so the scale of the estimate is sensible.
- Write one logical step per line, use equality signs only between equal quantities, give units and explain why a choice is valid. For the coach example, explicitly state that two coaches hold only 24 pupils.
- If a result is surprising, recheck the data and method. An answer of 2.92 coaches is a calculation result, not a practical booking decision.
In plain language
Solving the problem means more than performing division. The remainder represents pupils who still need seats. Explaining that is the link between your mathematics and the final decision.
Self-check
Twenty-nine people need minibuses with 8 passenger seats each. Each minibus costs £65. Write a justified booking and total cost.
Show the answer
29 ÷ 8 = 3 remainder 5. Three minibuses seat only 24, so book four. Cost = 4 × £65 = £260. State why rounding up is necessary.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · §3.2 Formula provision, printed p43; Appendix A Foundation formula list, printed p45
Open revision guidance
What to know
- The original Foundation formula list supplies area of a trapezium = ½(a + b)h. The parallel side lengths are a and b; h is the perpendicular distance between them, not a sloping side.
- The supplied formula must be usable but does not need to be memorised. This describes the original Appendix A list, not an assumed extra 2027 formula sheet.
- Use one length unit throughout. An area from centimetre lengths is in cm²; metres give m².
What to do in an assessment
- For parallel sides 4 cm and 10 cm with perpendicular height 3 cm, area = ½(4 + 10) × 3 = 21 cm². The bracket adds the parallel lengths before multiplication.
- For area 28 cm² and parallel sides 3 cm and 11 cm, 28 = ½(3 + 11)h = 7h, so h = 4 cm. Substitute back: 7 × 4 = 28.
- Check which sides are parallel on the drawing. A sloping side labelled 5 cm cannot be used as h unless it is also the perpendicular height.
In plain language
The formula uses the average of the two parallel lengths, multiplied by the height. For 4 cm and 10 cm, the average is 7 cm, so the trapezium has the same area as a 7 cm by 3 cm rectangle.
Self-check
A trapezium has parallel sides 5 m and 9 m and perpendicular height 4 m. Find its area.
Show the answer
Area = ½(5 + 9) × 4 = 7 × 4 = 28 m².
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Intermediate tier, Higher tier · Appendix A Intermediate formula list, printed p46
Open revision guidance
What to know
- The original Intermediate list supplies the trapezium formula and volume of a prism = area of cross-section × length. The cross-section is constant along the prism.
- A cross-sectional area in cm² multiplied by length in cm gives volume in cm³. Do not multiply by the perimeter or by a sloping edge that is not the prism length.
What to do in an assessment
- For a triangular cross-section with base 6 cm and perpendicular height 4 cm, area = ½ × 6 × 4 = 12 cm². A prism length of 9 cm gives volume = 12 × 9 = 108 cm³.
- If volume is 96 cm³ and cross-sectional area is 12 cm², length = 96 ÷ 12 = 8 cm. Check by multiplying 12 × 8.
- Identify the pair of matching end faces before choosing the cross-section; then measure the perpendicular distance between them along the prism.
In plain language
A prism is like stacking identical slices. Each 1 cm thick slice has volume numerically equal to its cross-sectional area, so eight slices of area 12 cm² occupy 96 cm³.
Self-check
A prism has cross-sectional area 15 cm² and length 7 cm. Find its volume and the length of another prism with that same cross-section and volume 180 cm³.
Show the answer
Volume = 15 × 7 = 105 cm³. For the second prism, length = 180 ÷ 15 = 12 cm.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Higher tier · Appendix A Higher formula list, printed p47
Open revision guidance
What to know
- The original Higher list includes trapezium area and prism volume; sphere volume ⁴⁄₃πr³ and surface area 4πr²; cone volume ⅓πr²h and curved surface area πrl. Radius r, perpendicular height h and slant height l are different lengths.
- It also provides a/sin A = b/sin B = c/sin C, a² = b² + c² − 2bc cos A, and triangle area = ½ab sin C. Lowercase sides are opposite the corresponding capital-letter angles.
- For ax² + bx + c = 0, a ≠ 0, the supplied quadratic formula is x = (−b ± √(b² − 4ac))/(2a). The whole numerator is divided by 2a.
- The provided AER decimal is (1 + i/n)^n − 1, where i is nominal annual interest as a decimal and n is compounding periods per year. Multiply by 100 to express the resulting decimal as a percentage.
What to do in an assessment
- A sphere of radius 3 cm has volume ⁴⁄₃π × 27 = 36π cm³ and surface area 4π × 9 = 36π cm². The same numerical coefficient does not make area and volume the same quantity.
- For a cone with r = 3 cm, h = 4 cm and l = 5 cm, volume = 12π cm³ and curved surface area = 15π cm². Add the base area 9π only if total surface area is required.
- With side a = 6 cm opposite A = 60° and angle B = 30°, b = 6 sin 30°/sin 60° = 2√3 cm ≈ 3.46 cm. For sides 3 cm and 4 cm enclosing a 90° angle, cosine rule gives the opposite side √(9 + 16) = 5 cm.
- A triangle with sides 6 cm and 8 cm enclosing 30° has area ½ × 6 × 8 × sin 30° = 12 cm². Do not substitute an angle that is not between the given sides.
- For x² − 3x − 4 = 0, a = 1, b = −3, c = −4. Then x = (3 ± √(9 + 16))/2 = (3 ± 5)/2, so x = 4 or −1. Check both in the original equation.
- For nominal annual rate 12% compounded monthly, i = 0.12 and n = 12. AER = (1.01)^12 − 1 ≈ 0.126825, or 12.68%. Do not enter 12 instead of 0.12.
In plain language
First choose the formula that answers the question, then match each label to the correct measurement. The cone surface formula uses the slant height, whereas its volume formula uses the perpendicular height. A supplied formula does not choose those values for you.
Self-check
A cone has radius 2 cm and perpendicular height 3 cm. Find its volume, leaving π exact.
Show the answer
V = ⅓πr²h = ⅓π × 4 × 3 = 4π cm³. Radius is squared; perpendicular height is not.
- Core
Scope: Core · Legacy WJEC GCSE tier: any of Foundation tier, Intermediate tier, Higher tier · §3.2 Calculators, printed p43; Appendix B Use of calculators, printed p48
Open revision guidance
What to know
- A calculator is allowed in Unit 2, not Unit 1. It must be desk-sized, battery or solar powered, and free of cases or covers bearing printed instructions or formulas.
- The specification forbids language translators, symbolic algebra manipulation, symbolic differentiation or integration, and communication with other machines or the internet.
- Do not borrow another learner’s calculator during the examination. Do not keep retrievable databanks, dictionaries, mathematical formulas or text in it.
What to do in an assessment
- Before the exam, check the device with your centre, remove an instruction-covered case, clear stored information and bring a reliable battery or solar-powered device. Practise with the same model beforehand.
- For trigonometry in degrees, check DEG mode: sin 30° should give 0.5. If it does not, check the angle mode before proceeding.
- Enter compound expressions with brackets. For 5/(2 + 3), enter 5 ÷ (2 + 3) to get 1; 5 ÷ 2 + 3 gives 5.5 and represents a different expression.
- Estimate the scale of an answer and retain unrounded intermediate values. For 18.6 × 4.1, approximately 20 × 4 = 80; the calculator result 76.26 is plausible.
In plain language
A calculator can perform the arithmetic you enter, but it cannot check that your expression matches the question. Device rules, settings and brackets are separate checks, and all three matter.
Self-check
A learner plans to use a calculator with stored revision text in Unit 1. Identify two problems.
Show the answer
Unit 1 does not allow a calculator, and the specification forbids retrievable stored text in an examination calculator. Clear prohibited stored information and use an allowed device only for Unit 2.
Previous revision ticks
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Earlier topic checklist
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Mathematics topic list
Tick each item off as you revise. Ticks are saved on this device only.
These ticks keep your earlier topic checklist. They do not count towards the new reviewed requirements above. Its numbering may be internal; use the linked references above for the exact reviewed source requirements.
Foundation tier topics
Higher tier topics
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