OCR Mathematics Paper 2

• Simplifies (2x³)² to 4x⁶ (1m)
• Multiplies to get 16x¹¹ (1m)
• Divides to get 16x⁷ (1m)

This 3-mark challenge question requires three index laws applied in sequence. Step 1 — power of a power: (2x³)² = 2² × (x³)² = 4x⁶. A common mistake here is forgetting to square the coefficient, giving x⁶ instead of 4x⁶. Step 2 — multiplication: 4x⁶ × 4x⁵: multiply coefficients (4 × 4 = 16) and add indices (6 + 5 = 11) to get 16x¹¹. Step 3 — division: 16x¹¹ ÷ x⁴: keep coefficient, subtract indices (11 − 4 = 7) to get 16x⁷. Work through brackets first, then multiplication, then division — this order prevents errors. Each step earns one mark.

• Multiplies coefficients: 2 × 5 = 10 (1m)
• Adds indices correctly: x³⁺⁴ = x⁷ (1m)

When multiplying algebraic terms, handle the numbers and the letters separately. Multiply the coefficients (the numbers in front): 2 × 5 = 10. Then apply the multiplication law to the letters: x³ × x⁴ = x³⁺⁴ = x⁷. Combine to get 10x⁷. A common error is to only add the indices and forget to multiply the coefficients, giving x⁷ instead of 10x⁷. Another mistake is adding the coefficients (2 + 5 = 7) rather than multiplying them. The 2 marks reflect these two separate processes: M1 for the coefficient, A1 for the correct index.

• Divides coefficients: 12 ÷ 4 = 3 (1m)
• Subtracts indices correctly: x⁵⁻² = x³ (1m)

When dividing algebraic terms, handle numbers and letters separately. Divide the coefficients: 12 ÷ 4 = 3. Then apply the division law to the letters: x⁵ ÷ x² = x⁵⁻² = x³. Combine to get 3x³. A common mistake is to subtract the coefficients (12 − 4 = 8) rather than dividing them. Another error is adding the indices instead of subtracting. The 2 marks reflect the two separate processes: M1 for the coefficient calculation, A1 for the correct power of x.

• Raises coefficient to power: 3⁴ = 81 (1m)
• Multiplies indices correctly: (x²)⁴ = x⁸ (1m)

When the entire bracket (3x²)⁴ is raised to a power, everything inside the bracket gets raised to that power. Work through two separate operations. First, raise the coefficient to the power: 3⁴ = 81. Then, apply the power-of-a-power law to the letter: (x²)⁴ = x²ˣ⁴ = x⁸. Combine: 81x⁸. A common mistake is only raising the letter and forgetting the coefficient (giving x⁸ instead of 81x⁸). Another is writing 3 × 4 = 12 as the coefficient rather than computing 3⁴ = 81. The 2 marks reflect: M1 for the coefficient (81) and A1 for the correct power (x⁸).

• Finds cube root of 8 = 2 (1m)
• Squares to get 4 (1m)

For a^(m/n), the rule is: the denominator gives the root, the numerator gives the power. Always do the root first, then the power. Here, 8^(2/3) = (³√8)² = 2² = 4. Step 1: ³√8 = 2 (because 2³ = 8). Step 2: 2² = 4. A common misconception is thinking you can divide 8 by the index (8 ÷ 3 × 2 ≈ 5.3) — this is wrong. The 2 marks reflect: M1 for finding the cube root (2), A1 for squaring it (4). Always work with the root before the power.

• Calculates 4^(3/2) = 8 (1m)
• Takes reciprocal: 1/8 = 0.125 (1m)

A negative fractional index combines two rules. Work in stages: first handle the fractional part (ignoring the negative), then apply the negative (reciprocal). For 4^(-3/2): Step 1 — fractional part: 4^(3/2) = (√4)³ = 2³ = 8. Step 2 — negative index: 4^(-3/2) = 1/4^(3/2) = 1/8. A common error is getting 8 and forgetting to take the reciprocal. The marks are: M1 for correctly calculating 4^(3/2) = 8, A1 for applying the reciprocal to get 1/8 = 0.125.

• Simplifies to a⁸ (1m)

The multiplication law states: when multiplying powers with the same base, add the indices. Here, a⁵ × a³ = a⁵⁺³ = a⁸. Do NOT multiply the indices together (which would give a¹⁵) — that is the power-of-a-power law, not the multiplication law. You only add indices when the base letters are identical. The answer must stay as a letter with a power — do not try to work out a numerical value.

• Simplifies to y³ (1m)

The division law states: when dividing powers with the same base, subtract the indices. Here, y⁸ ÷ y⁵ = y⁸⁻⁵ = y³. Crucially, you subtract (not divide) the indices — y⁸ ÷ y⁵ does not become y⁸/⁵ or y^1.6. Also note you subtract the bottom index from the top index, not the other way around (8 − 5 = 3, not 5 − 8 = −3). This law is the opposite of the multiplication law and is essential for simplifying expressions in algebra and standard form.

• Simplifies to m¹² (1m)

The power-of-a-power law states: when raising a power to another power, multiply the indices. Here, (m³)⁴ = m³ˣ⁴ = m¹². Do NOT add the indices (which would give m⁷) — that mistake comes from confusing this with the multiplication law. The key difference: a⁵ × a³ means you ADD (multiplication law), but (a³)⁴ means you MULTIPLY (power-of-a-power law). You can verify this: (m³)⁴ means m³ written four times, i.e. m³ × m³ × m³ × m³ = m³⁺³⁺³⁺³ = m¹².

• States 1 (1m)

The zero index law: any non-zero number raised to the power 0 equals 1. So 5⁰ = 1. This is not obvious from first principles, but it follows from the division law: aⁿ ÷ aⁿ = aⁿ⁻ⁿ = a⁰, and any number divided by itself is 1, so a⁰ = 1. Two common wrong answers are 0 (confusing the zero index with a zero result) and 5 (ignoring the index). This law holds for any non-zero base: 100⁰ = 1, (2x)⁰ = 1, even 999999⁰ = 1.

• Evaluates to 0.125 or 1/8 (1m)

The negative index law: a⁻ⁿ = 1/aⁿ. Apply this in two steps. First, apply the rule: 2⁻³ = 1/2³. Then evaluate: 2³ = 8, so 2⁻³ = 1/8 = 0.125. The key understanding is that the negative index makes the number a fraction (reciprocal), NOT a negative number. Writing -8 or -0.125 is a very common mistake. Equally common is writing just 8 (calculating 2³ and ignoring the negative index). Always think: 'negative index → flip to denominator, make index positive'.

• Writes x⁻² (1m)

This question applies the negative index law in reverse: instead of calculating a⁻ⁿ as a fraction, you convert a fraction back to a negative index. The rule is 1/aⁿ = a⁻ⁿ. Here, 1/x² = x⁻². Think of it this way: the denominator 'moves up' to the top, and the index becomes negative. This is the same law used in standard form and simplifying algebraic fractions. Accept x^-2 or x^(-2) as alternative written forms.

The correct answer is a⁸. The multiplication law says: same base, multiplying, so ADD the indices: a³ × a⁵ = a³⁺⁵ = a⁸. The distractors test common errors: a¹⁵ comes from multiplying indices (3 × 5 = 15 — that is the power-of-a-power law, not multiplication). a² comes from subtracting indices (5 − 3 = 2 — that is the division law). 2a⁸ adds a spurious coefficient of 2. When multiplying two identical letter terms, the coefficient stays at 1 unless numbers are present — there is no '2 lots of a⁸' here.

• Evaluates to 3 (1m)

The fractional index law: a^(1/n) = ⁿ√a. The denominator of the fraction tells you which root to take. Here, 27^(1/3) = ³√27 (cube root of 27). Ask yourself: what number cubed equals 27? Since 3³ = 27, the answer is 3. A common mistake is dividing by 3 (giving 9), which confuses the fractional index with ordinary division. The calculator allows you to use the fraction index directly — look for the x^y or ^ button. This law is the basis for all fractional index questions at GCSE.

Calculate the discriminant b² - 4ac = 25 - 40 = -15, which is negative. A negative discriminant means the equation has NO real solutions (the parabola doesn't cross the x-axis).

We need factors of -12 that add to -1. The factors 3 and -4 work: 3 × (-4) = -12 and 3 + (-4) = -1. So the factorised form is (x + 3)(x - 4).

(x + 3)² is always ≥ 0, with minimum value 0 when x = -3. So x² + 6x + 13 = (x+3)² + k has minimum value k. Since (x+3)² = x² + 6x + 9, we have x² + 6x + 13 = (x+3)² + 4, so k = 4 and the minimum value is 4.

• States that denominators must be the same (1m)
• Explains that you can only add the numerators when denominators match (1m)

Fractions can only be added when they have the same denominator because the denominator tells you the size of the parts. You can only combine parts that are the same size.

Cancel common factors: 6÷3 = 2, and x²÷x = x. So 6x²/3x = 2x.

When the denominators are the same, keep the denominator and add the numerators: 3/(x+1) + 5/(x+1) = (3+5)/(x+1) = 8/(x+1).

(x+2) appears in numerator and denominator, so it cancels. Left with 6/3 = 2.

• Expands both brackets correctly: n²+2n+1 and n²−2n+1 (1m)
• Simplifies to 4n (1m)
• Writes n=2k (even number) (1m)
• Substitutes to get 4(2k)=8k (1m)
• States that 8k is divisible by 8 (1m)

Write all quantities in algebraic form, expand and simplify systematically, and arrive at the required conclusion through clearly stated steps. In formal proofs, each line of working follows logically from the previous. The final line should be a clear concluding statement. Omitting steps or jumping to the conclusion without showing the algebra risks losing method marks even if the final statement is correct.

• Expands (n+1)² to n²+2n+1 (1m)
• Subtracts n² to get 2n+1 (1m)
• States that n² and (n+1)² are consecutive squares (1m)
• Explains that 2n+1 is odd (even+1 form) (1m)

Represent consecutive integers as n and n+1 (or n−1, n, n+1 for three). Expand the given product or expression, collect like terms, and show the result has the form required by the question. Every step of algebra must be shown. The conclusion should explicitly state what has been proved: 'Since [expression] = [factorised form], this proves that [the result is always...]'.

• Uses n and n+1 to represent consecutive integers (1m)
• Adds to get 2n+1 (1m)
• States that 2n+1 is odd (even+1 form) (1m)

In algebraic proof, represent the quantities using general forms: even numbers as 2n, odd numbers as 2n+1, consecutive integers as n and n+1. Manipulate these algebraically and show that the result has the required form. For example, proving the sum of two odd numbers is even: (2m+1)+(2n+1) = 2(m+n+1), which has factor 2 so is even. The proof must work for all integers, not just specific examples.

• Factorises to n(n+1) (1m)
• States that n and n+1 are consecutive, so one is even (1m)
• Concludes that n(n+1) is even (1m)

Represent consecutive odd numbers as 2n+1 and 2n+3. Multiply: (2n+1)(2n+3) = 4n²+8n+3. Analyse the structure of this result to prove the stated property. All algebraic steps must be shown. Specific numerical examples are not proofs — the algebraic manipulation must hold for all integer values of n. State a clear conclusion at the end.

• Represents two odd numbers as 2n+1 and 2m+1 (or equivalent with different variables) (1m)
• Adds to get 2n+2m+2 (or equivalent) (1m)
• Factorises to 2(n+m+1) and states it's even (1m)

Expand the left-hand side step by step and show it equals the right-hand side. Do not work backwards from the answer. Start from the left-hand side, expand all brackets, collect like terms, and simplify until the right-hand side is reached. Each algebraic manipulation earns marks. The proof is invalid if you assume the result and work backwards.

• Uses n, n+1, n+2 to represent three consecutive integers (1m)
• Adds to get 3n+3 (1m)
• Factorises to 3(n+1) and states it's a multiple of 3 (1m)

Let the starting integer be n and express related integers using n. Expand the product or sum of these expressions, collect like terms, and show the result has the required property (e.g. has factor 2, or factor 3). State the conclusion explicitly: 'This shows the result is always [divisible by 3 / even / etc.]'. Without the conclusion, the proof is incomplete.

• Represents odd number as 2n+1 (1m)
• Squares and expands to 4n²+4n+1 (1m)
• Factorises and states it's in the form even+1, so odd (1m)

A disproof by counter-example needs one specific case where the claim fails. Choose a value, compute the result, and clearly explain why the claim is false for this value. State your conclusion explicitly: 'This shows the claim is false because...' You only need one counter-example. Verify your chosen example actually fails the claim before stating it as your answer.

• Correctly expands both brackets: n²+6n+9 and n²−2n+1 (1m)
• Simplifies to 8n+8 or 8(n+1) (1m)

Expand both brackets: (n+3)² = n²+6n+9 and (n−1)² = n²−2n+1. Subtracting: (n²+6n+9) − (n²−2n+1) = 8n+8 = 8(n+1). The n² terms cancel, leaving 8n+8. Factorising gives 8(n+1), confirming the identity. Each expansion step must be shown clearly — markers need to see the full expansion, the subtraction (including the cancellation of n²), and the factorisation.

• Tests the prime number 2 (or shows working with 2) (1m)
• States 2 as the counter-example OR shows 2²-1=3 (odd) (1m)

The counter-example is the prime number 2 — the only even prime. All odd primes squared give an odd number, and odd minus 1 is even, so they appear to support Sarah's claim. But 2² − 1 = 4 − 1 = 3, which is odd, disproving it. When testing primes, always try 2 first since it is uniquely even. The answer required is the prime number 2, not the result 3.

Test all 9 single-digit positive integers: 1÷4=0 r1, 2÷4=0 r2, 3÷4=0 r3, 4÷4=1 r0, 5÷4=1 r1, 6÷4=1 r2, 7÷4=1 r3, 8÷4=2 r0, 9÷4=2 r1. All remainders are 0, 1, 2, or 3. Therefore, when dividing any single-digit positive integer by 4, the remainder is always 0, 1, 2, or 3.

• Tests all 9 cases (1 through 9) and identifies remainders (1m)
• States conclusion that all remainders are 0, 1, 2, or 3 (1m)

Expand the expression fully, collect all terms, and factorise to show the required factor is present. For example, to show divisibility by 6, show the expression equals 6k for some integer k. Write the expression in factorised form with the required factor explicit. The factorisation step is the key — if the factor is not shown as a clear multiplier, the proof is incomplete.

An even number is always a multiple of 2. The expression 2n represents 2 times any integer, which always gives an even number. For example: if n=3, then 2n=6 (even); if n=10, then 2n=20 (even).

A conjecture claims something is ALWAYS true. To disprove it, you only need ONE example where it's false. That single counter-example proves the statement isn't always true, which disproves the conjecture.

Consecutive integers follow one after another with a difference of 1. If the first integer is n, the next is n+1, and the third is n+2. For example: 5, 6, 7 or 20, 21, 22.

A counter-example must show the claim is FALSE. Tom claims the result is always prime. Option B gives 4×2+1=9, which is NOT prime (9=3×3), so this disproves Tom's claim.

• Finds total parts (15) or one part (120) (1m)
• Finds largest share (840) (1m)
• Calculates 20% of 840 (1m)
• States £168 (1m)

This two-step problem first requires ratio sharing, then a percentage calculation. Step 1: find total parts (5 + 7 + 3 = 15) and one part (£1800 ÷ 15 = £120). Step 2: find the largest share — 7 parts = 7 × £120 = £840. Step 3: calculate 20% of the largest share: 0.2 × £840 = £168. Common mistakes include taking 20% of the total donation (£1800) rather than the largest share, or identifying the wrong charity as the largest. Always identify which share is largest before proceeding.

• Recognises smallest + largest = 9 parts (1m)
• Finds one part (11) (1m)
• Uses middle = 5 parts (1m)
• States 55 (1m)

The sum of smallest and largest corresponds to 2 + 7 = 9 parts (not all 14). So 9 parts = 99, giving one part = 99 ÷ 9 = 11. The middle number is 5 parts = 5 × 11 = 55. A common mistake is dividing 99 by 14 (all parts), which is wrong because the middle number is not included in the 99. Always identify precisely which ratio parts are represented by the given sum.

• Finds total parts (10) or one part (24) (1m)
• Calculates at least two shares correctly (1m)
• All three shares correct (1m)

To share 240 sweets in the ratio 2:3:5: (1) add ratio numbers to get total parts: 2 + 3 + 5 = 10; (2) find the value of one part: 240 ÷ 10 = 24 sweets; (3) multiply by each ratio number: Tom = 2 × 24 = 48, Jerry = 3 × 24 = 72, Spike = 5 × 24 = 120. A common error is adding the part value to the ratio number instead of multiplying. Always verify by adding all shares: 48 + 72 + 120 = 240.

• Finds difference in parts (3) (1m)
• Finds one part (7) (1m)
• Calculates larger number (56) (1m)

Use the difference method: the difference between the two values corresponds to the difference in ratio parts. The ratio 5:8 has a part-difference of 8 − 5 = 3. Since 3 parts = 21, one part = 21 ÷ 3 = 7. The larger number = 8 × 7 = 56. A common mistake is dividing 21 by 8 (the larger ratio number) instead of 3 (the difference between the ratio numbers). Always use the difference in parts, not the individual parts themselves.

• Finds difference in parts (2) (1m)
• Finds one part (6) (1m)
• Calculates total (48) (1m)

Use the difference method: the difference between cats and dogs (12 animals) corresponds to the difference in their ratio parts (5 − 3 = 2 parts). So 2 parts = 12 animals, meaning one part = 12 ÷ 2 = 6 animals. The total number of parts = 5 + 3 = 8, so the total animals = 8 × 6 = 48. A common mistake is finding only the number of cats (30) and forgetting to add the dogs (18). The question asks for the total, so add both groups at the end.

• Finds total parts (10) (1m)
• Forms fraction 3/10 or decimal 0.3 (1m)
• Converts to 30% (1m)

To convert a ratio part to a percentage: (1) find the total number of parts by adding all ratio numbers: 5 + 3 + 2 = 10; (2) write blue paint as a fraction of the total: 3/10; (3) convert to a percentage: 3/10 × 100 = 30%. A common mistake is giving the answer as a decimal (0.3) without converting to a percentage. Another mistake is using 3 directly as the answer — this is the ratio number, not the percentage.

• States Sarah used ratio numbers directly / didn't divide the total (1m)
• Shows correct method: 90÷5 = 18 (1m)
• States correct shares: £36 and £54 (1m)

Sarah's error was using the ratio numbers (2 and 3) as the actual pound amounts instead of finding what one part is worth. Correct method: total parts = 2 + 3 = 5; one part = £90 ÷ 5 = £18; shares = 2 × £18 = £36 and 3 × £18 = £54. In error-explanation questions, state what the mistake was, show the correct calculation, and give the correct answer.

• Finds total parts (8) or one part (15) (1m)
• Calculates smaller share £45 (1m)

To share in a ratio: (1) add the ratio numbers to find total parts — here 3 + 5 = 8; (2) divide the total by the number of parts to find one part — £120 ÷ 8 = £15; (3) multiply one part by the smaller ratio number — 3 × £15 = £45. A common mistake is dividing by 3 or 5 directly without finding the total parts first. Always check: the shares should add back to the original amount (£45 + £75 = £120).

• Finds one part (15) (1m)
• Calculates total (120) (1m)

Working backwards from one share: Ben's £45 represents 3 parts, so one part = £45 ÷ 3 = £15. The total number of parts is 3 + 5 = 8, so the total amount = 8 × £15 = £120. A common mistake is stopping after finding Aisha's share (5 × £15 = £75) rather than the total. The question asks for the total, not Aisha's individual share. Always re-read what the question is asking for before writing your final answer.

• Finds one part (50) (1m)
• Calculates sugar (150) (1m)

When one quantity in a ratio is known, divide it by its ratio number to find the value of one part. Flour = 7 parts = 350g, so one part = 350 ÷ 7 = 50g. Sugar = 3 parts = 3 × 50 = 150g. A common mistake is multiplying 350 by 3 directly (which gives 1050g — far too much) or dividing 350 by 3. The key step is always dividing the known amount by its ratio number first to find the one-part value.

• States total parts is 5 (1m)
• Shows calculation: 50÷5=10, then 4×10=40 (1m)

In 'explain' questions you must show the working that proves the claim, not just restate it. For full marks: state the total number of parts (1 + 4 = 5); show how one part is found (50 ÷ 5 = £10); and calculate the larger share (4 × 10 = £40). The mark scheme requires the numerical calculation to verify Liam's statement — writing 'because he calculated it correctly' earns no marks. Always work through the mathematics step by step.

Total parts = 2 + 3 = 5. One part = 300 ÷ 5 = 60. Smaller share = 2 × 60 = £120.

Total parts = 4+9 = 13. One part = 52÷13 = 4. Red = 4 parts = 4×4 = 16.

Angles in a triangle sum to 180°. Total parts = 2+3+4 = 9. One part = 180÷9 = 20°. Smallest angle = 2 × 20° = 40°.

• Sets up 5x and 3x for initial adults and children (1m)
• Forms expressions 5x-15 and 3x+9 after changes (1m)
• Forms equation 5x-15 = 3x+9 from ratio 1:1 (1m)
• Solves to get x=12 (1m)
• Calculates original total as 96 (1m)

Let adults = 5x and children = 3x initially. After the changes: adults = 5x − 15 and children = 3x + 9. Since the new ratio is 1:1, the two quantities are equal: 5x − 15 = 3x + 9. Solving: 2x = 24, so x = 12. The original total = 5x + 3x = 8x = 8 × 12 = 96 people. Two common errors: (1) finding x = 12 and giving that as the answer — x is not the total; (2) calculating the final total after changes (which would be 72) rather than the original. Always re-read the question to check what you are being asked to find.

• Uses 3x and 4x for initial red and blue (1m)
• Forms equation (3x+6):4x = 3:2 (1m)
• Solves to get x=2 (1m)
• Calculates blue = 8 (1m)

This is a changing ratio problem requiring algebra. Start by letting red = 3x and blue = 4x (using the initial ratio 3:4). Blue doesn't change, so after adding 6 red counters, red = 3x + 6 and blue = 4x. The new ratio gives the equation: (3x + 6) / 4x = 3/2. Cross-multiplying: 2(3x + 6) = 3(4x), which simplifies to 6x + 12 = 12x, so 12 = 6x and x = 2. Blue counters = 4x = 4 × 2 = 8. A common error is finding x = 2 and stopping — always go back to answer the actual question being asked.

• Recognises need to make dogs equal using LCM(3,5)=15 (1m)
• Scales both ratios correctly (10:15 and 15:12) (1m)
• Combines to 10:15:12 (1m)

When combining two linked ratios, the shared quantity (dogs) must have the same value in both. Dogs is 3 in the first ratio and 5 in the second — find the LCM of 3 and 5, which is 15. Scale the first ratio (cats:dogs = 2:3) by multiplying by 5 to get 10:15. Scale the second ratio (dogs:rabbits = 5:4) by multiplying by 3 to get 15:12. Now dogs = 15 in both, so the combined ratio is cats:dogs:rabbits = 10:15:12. You cannot simply add or concatenate the two ratios — they must share a common dog value first.

• Divides £240 by 8 to get £30 per part (1m)
• Calculates Tom=£150 and Jerry=£90 (1m)
• Finds difference of £60 (1m)

First share the total: total parts = 5 + 3 = 8, so one part = £240 ÷ 8 = £30. Tom gets 5 × £30 = £150 and Jerry gets 3 × £30 = £90. The question asks how much MORE Tom gets, so find the difference: £150 − £90 = £60. A common mistake is stopping at Tom's share (£150) and missing the final step. The phrase 'how much more' always signals you need to find the difference between two amounts, not just one person's share.

• Finds LCM(5,2)=10 or equivalent method to make B equal (1m)
• Scales to 8:10 and 10:15 (1m)
• Combines to 8:10:15 (1m)

B is the linking quantity — it appears in both ratios. B is 5 in A:B and 2 in B:C. To combine these ratios, make B equal in both using the LCM of 5 and 2, which is 10. Scale A:B = 4:5 by multiplying by 2 to get 8:10. Scale B:C = 2:3 by multiplying by 5 to get 10:15. Now B = 10 in both, giving A:B:C = 8:10:15. Check: HCF(8,10,15) = 1, so this is already in its simplest form.

• Multiplies both by 10 to get 6:15 (1m)
• Simplifies to 2:5 (1m)

When a ratio contains decimals, first convert to whole numbers by multiplying both parts by 10 (or 100 if needed): 0.6 × 10 = 6, 1.5 × 10 = 15, giving 6:15. Then simplify by dividing both by the HCF of 6 and 15, which is 3: 6 ÷ 3 = 2, 15 ÷ 3 = 5, so the answer is 2:5. A common mistake is stopping at 6:15 without simplifying further — always check whether the resulting whole-number ratio can be simplified.

• Adds 3+5 to get total of 8 parts (1m)
• Writes 3/8 (1m)

The ratio 3:5 tells you there are 3 parts boys and 5 parts girls. To find the fraction of the class that are boys, first find the total: 3 + 5 = 8 parts. Boys are 3 out of 8 parts, so the fraction is 3/8. A very common error is writing 3/5 — this is the ratio of boys to girls, not boys as a fraction of the whole class. Always add the ratio parts to get the denominator.

• Identifies scale factor of 4 (1m)
• Calculates 7×4 = 28 (1m)

To find a missing value in an equivalent ratio, determine the scale factor from the part you know: 3 has become 12, so the scale factor is 12 ÷ 3 = 4. Apply the same scale factor to the other part: 7 × 4 = 28. The key principle is that equivalent ratios multiply or divide both parts by the same number. A common mistake is adding the difference (12 − 3 = 9) to 7, giving 16 — but equivalent ratios use multiplication, not addition.

• Identifies scale factor of ÷7 (1m)
• Calculates 35÷7 = 5 (1m)

Identify the scale factor from the part that's complete: 14 has become 2, so the scale factor is 14 ÷ 7 = 2, meaning we are dividing by 7. Apply the same operation to the other part: 35 ÷ 7 = 5. So the missing value is 5. This ratio scales down rather than up. A common error is subtracting the difference (14 − 2 = 12 from 35) to get 23 — equivalent ratios always use multiplication or division, never addition or subtraction.

• Identifies that total parts = 2+3 = 5 (1m)
• States apples are 2/5 of the total (or 2/3 compares apples to oranges, not to total) (1m)

The ratio 2:3 means 2 parts apples and 3 parts oranges. Total = 2+3 = 5 parts. Apples are 2 out of 5, which is 2/5, not 2/3. Sarah confused the ratio (2:3) with a fraction of the whole.

• Divides both parts by 5 to get 3:5 (1m)

To simplify a ratio, divide both parts by their highest common factor (HCF). The HCF of 15 and 25 is 5, so divide both: 15 ÷ 5 = 3, 25 ÷ 5 = 5, giving 3:5. The method is identical to simplifying fractions — find the HCF and divide. A common mistake is dividing the two numbers by different values (e.g. dividing 15 by 3 but 25 by 5), which gives an incorrect and unequal ratio.

The ratio 5:2 means 5 parts flour and 2 parts sugar. Total parts = 5+2 = 7. Flour is 5 out of 7 parts, so the fraction is 5/7.

To simplify a ratio with fractions, multiply both parts by the LCM of the denominators. LCM(2,4)=4. So (1/2)×4 : (1/4)×4 = 2:1.

The ratio 7:3 means 7 parts apples and 3 parts bananas. Total parts = 7+3 = 10. Bananas are 3 out of 10 parts, so the fraction is 3/10.

• Calculates time for first part (1.5 hours) (1m)
• Calculates remaining distance (150 km) (1m)
• Calculates time for second part (2 hours) (1m)
• Adds the two times (3.5 hours) (1m)
• Converts to hours and minutes (3h 30min) (1m)

Find the time for each stage separately using Time = Distance ÷ Speed. Stage 1: 90 ÷ 60 = 1.5 hours (1 hour 30 min). Stage 2: remaining distance = 240 − 90 = 150 km; time = 150 ÷ 75 = 2 hours. Total time = 1.5 + 2 = 3.5 hours = 3 hours 30 minutes. The key step students miss is calculating the remaining distance for stage 2 before finding its time.

• Calculates total journey time (1.5 hours) (1m)
• Calculates time for first 40 km (0.5 hours) (1m)
• Finds remaining distance (80 km) and time (1 hour) (1m)
• Divides 80 by 1 (1m)
• Correct answer of 80 km/h (to 3 sf) (1m)

Step 1: total journey time = 10:55 − 09:25 = 90 minutes = 1.5 hours. Step 2: time for first 40 km at 80 km/h = 40 ÷ 80 = 0.5 hours. Step 3: remaining time = 1.5 − 0.5 = 1 hour. Step 4: remaining distance = 120 − 40 = 80 km. Step 5: average speed for remaining part = 80 ÷ 1 = 80 km/h. This multi-step question requires careful management of both distances and times — write down each step clearly.

• Calculates time for first part (2 hours) (1m)
• Calculates time for second part (2 hours) (1m)
• Finds total distance (300) and total time (4) (1m)
• Correct answer of 75 km/h (1m)

Average speed = total distance ÷ total time. Find each time separately: Part 1: 120 ÷ 60 = 2 hours. Part 2: 180 ÷ 90 = 2 hours. Total distance = 120 + 180 = 300 km. Total time = 2 + 2 = 4 hours. Average speed = 300 ÷ 4 = 75 km/h. Note that here (75) equals the mean of the two speeds (60+90)÷2 = 75 only because the times happen to be equal. Never just average the speeds without checking this condition.

• Calculates distance for first part (9 km) (1m)
• Calculates distance for second part (30 km) (1m)
• Finds total distance (39) and total time (1.75) (1m)
• Correct answer of 22.3 km/h to 1 dp (1m)

Convert times to hours first: 30 min = 0.5 h; 1 hour 15 min = 1.25 h. Find each distance: Part 1: 18 × 0.5 = 9 km. Part 2: 24 × 1.25 = 30 km. Total distance = 9 + 30 = 39 km. Total time = 0.5 + 1.25 = 1.75 h. Average speed = 39 ÷ 1.75 ≈ 22.3 km/h. The key error to avoid is simply averaging 18 and 24 (which gives 21), because the cyclist spends more time at the higher speed.

• Converts km to m (×1000) (1m)
• Converts hours to seconds (÷3600) (1m)
• Correct answer of 20 m/s (1m)

To convert km/h to m/s, divide by 3.6 (because 1 km = 1000 m and 1 hour = 3600 s, so 1 km/h = 1000/3600 = 1/3.6 m/s). So 72 ÷ 3.6 = 20 m/s. Alternatively: convert 72 km to metres (72,000 m) and 1 hour to seconds (3600 s), then divide: 72,000 ÷ 3600 = 20 m/s. The reverse process (m/s to km/h) is to multiply by 3.6.

• Converts seconds to hours (×3600) (1m)
• Converts m to km (÷1000) (1m)
• Correct answer of 32.4 km/h (1m)

To convert m/s to km/h, multiply by 3.6 (because 1 m/s = 3600 m per hour = 3.6 km per hour). So 9 × 3.6 = 32.4 km/h. You can also think of it step by step: 9 m/s × 3600 s/hour = 32,400 m/hour = 32.4 km/hour. A common error is dividing by 3.6 instead of multiplying — remember, m/s to km/h is the larger unit so the number gets bigger.

• Converts 45 minutes to 0.75 hours (1m)
• Divides 10 by 0.75 (1m)
• Correct answer of 13.33 or 13⅓ or 40/3 (1m)

The key step is converting 45 minutes into hours: 45 ÷ 60 = 0.75 hours. Then apply Speed = Distance ÷ Time: 10 ÷ 0.75 = 13.33 km/h (to 2 decimal places). A very common mistake is to use 45 directly as the time, giving 10 ÷ 45 = 0.22, which is wrong because the speed formula requires time in hours when the distance is in km.

• Uses speed = distance ÷ time formula (1m)
• Correct answer of 12 (km/h) (1m)

Use the formula Speed = Distance ÷ Time. Substitute the values: Speed = 24 ÷ 2 = 12 km/h. The units tell you the answer must be in km/h because you divided kilometres by hours. A common mistake is to multiply 24 × 2 = 48 or to divide 2 ÷ 24. Remember: to find speed, distance is always divided by time.

• Uses distance = speed × time formula (1m)
• Correct answer of 180 (km) (1m)

Use the formula Distance = Speed × Time. Substitute: Distance = 60 × 3 = 180 km. Here the units work out correctly: km/h × hours = km. A common error is dividing (60 ÷ 3 = 20) instead of multiplying. The DST triangle helps: cover D to see S × T underneath.

• Uses time = distance ÷ speed formula (1m)
• Correct answer of 3 (hours) (1m)

Use the formula Time = Distance ÷ Speed. Substitute: Time = 240 ÷ 80 = 3 hours. A common error is multiplying (240 × 80 = 19200) or inverting the division (80 ÷ 240 ≈ 0.33). The DST triangle helps: cover T to see D ÷ S underneath. Always include units in your answer — here the answer is 3 hours.

The car spent more time at 30 km/h (2 hours) than at 90 km/h (1 hour), so the average speed is weighted towards 30. Average speed = total distance / total time = 150 / 3 = 50 km/h, not 60.

• States that total distance ÷ total time gives average speed (or calculates 50 km/h) (1m)
• Explains that the times are different/unequal OR that more time was spent at one speed (1m)

Average speed is total distance divided by total time — it is NOT simply the arithmetic mean of the speeds. Here the car travels different amounts of time at each speed: 2 hours at 30 km/h and only 1 hour at 90 km/h. Total distance = (30 × 2) + (90 × 1) = 60 + 90 = 150 km. Total time = 3 hours. Average speed = 150 ÷ 3 = 50 km/h, not 60. The arithmetic mean of speeds (60 km/h) only equals the true average speed when the times spent at each speed are equal.

Speed = Distance ÷ Time. This makes sense: if you travel further in the same time, your speed is higher. If you take longer for the same distance, your speed is lower.

Total distance = (40 × 0.5) + (80 × 0.5) = 20 + 40 = 60 km. Total time = 1 hour. Average speed = 60 ÷ 1 = 60 km/h. Note that this equals the MEAN of the speeds (40+80)÷2 = 60 ONLY because the times are equal.

A person walks at about 5 km/h (or about 1.4 m/s). 5 m/s would be running very fast (18 km/h), 50 km/h is cycling speed, and 500 m/s is faster than a jet plane!

• 25 × 10 × 1.5 or 375 seen for volume in m³ (M1) (1m)
• 375 × 1000 or 375,000 seen for litres (M1) (1m)
• 375,000 ÷ 100 or 3750 seen (M1) (1m)
• 3750 × 0.12 seen (M1) (1m)
• £450 or 450 cao (A3) — accept minor rounding differences (3m)

Volume = 25 × 10 × 1.5 = 375 m³. Convert to litres: 375 × 1000 = 375,000 litres. Number of 100-litre units = 375,000 ÷ 100 = 3750. Cost = 3750 × £0.12 = £450. This is a multi-step problem requiring: volume calculation, unit conversion (m³ to litres), then a cost calculation. Work through each step clearly and check units at every stage.

• 48 seen for first rectangle area (M1) (1m)
• 15 seen for second rectangle area (M1) (1m)
• 48 + 15 or 63 seen for total cross-sectional area (M1) (1m)
• 63 × 20 seen (M1) (1m)
• 1260 cao (A2) (2m)

Split the L-shaped cross-section into two rectangles: area 1 = 12 × 4 = 48 cm², area 2 = 5 × 3 = 15 cm². Total cross-section area = 48 + 15 = 63 cm². Volume = 63 × 20 = 1260 cm³. The key is adding all parts of the compound cross-section before multiplying by the prism length. Common errors: using only one rectangle's area, or multiplying each rectangle's area by 20 and then adding (which gives the same answer, but requires extra care).

• ½(8+12) or ½ × 20 seen in area calculation (M1) (1m)
• 8 + 12 or 20 seen (M1) (1m)
• 50 for trapezoid area (M1) (1m)
• 50 × 15 seen (M1) (1m)
• 750 cao (A1) (1m)

Volume of prism = cross-section area × length. Cross-section is a trapezium: area = (1/2)(a + b)h = (1/2)(8 + 12) × 5 = (1/2) × 20 × 5 = 50 cm². Volume = 50 × 15 = 750 cm³. The critical step is correctly applying the trapezium area formula before multiplying by the prism length. A common error is forgetting the (1/2) factor, giving area = 100 and volume = 1500.

• πr² × 8 = 500 or equivalent formed (M1) (1m)
• πr² = 62.5 seen (M1) (1m)
• r² = 62.5 ÷ π or 19.89 seen (M1) (1m)
• r = √(62.5 ÷ π) or √19.89 seen (M1) (1m)
• 4.46 (A1) — must be to 2 dp (1m)

Rearrange V = πr²h for r: r² = V ÷ (πh) = 500 ÷ (π × 8) = 500 ÷ 25.133 ≈ 19.894, then r = √19.894 ≈ 4.46 cm (2 d.p.). Two key steps: first divide by (π × h) to isolate r², then take the square root. A common mistake is forgetting to take the square root at the end, leaving the answer as r² ≈ 19.89 rather than r ≈ 4.46.

• 0.6 seen for radius (M1) (1m)
• 0.6² or 0.36 seen (M1) (1m)
• π × 0.36 × 2 or 0.72π seen (M1) (1m)
• 2.26 (A1) — must be to 2 dp (1m)

The diameter is 1.2 m, so the radius = 1.2 ÷ 2 = 0.6 m. Volume = πr²h = π × 0.6² × 2 = π × 0.36 × 2 = 0.72π ≈ 2.26 m³ (2 d.p.). The most common error is using the diameter (1.2) directly in the formula instead of converting to radius first. Always halve the diameter before substituting.

• 360 = 10 × 6 × h or equivalent equation formed (M1) (1m)
• 60h or 60 × h = 360 seen (M1) (1m)
• h = 360 ÷ 60 or division seen (M1) (1m)
• 6 cao (A1) (1m)

Rearrange the cuboid volume formula: V = l × w × h, so h = V ÷ (l × w) = 360 ÷ (10 × 6) = 360 ÷ 60 = 6 cm. First calculate the product of the two known dimensions (10 × 6 = 60), then divide the total volume by that product. A common mistake is dividing by only one dimension — always multiply the two known dimensions together first.

• (x + 2)(x - 1) seen or evidence of multiplying dimensions (M1) (1m)
• x² + x - 2 seen for expanded brackets (M1) (1m)
• Multiply by 3 seen: 3(x² + x - 2) or start of distribution (M1) (1m)
• 3x² + 3x - 6 cao (A1) — must be fully expanded (1m)

V = (x + 2)(x − 1) × 3. First expand the bracket: (x + 2)(x − 1) = x² − x + 2x − 2 = x² + x − 2. Then multiply by 3: 3(x² + x − 2) = 3x² + 3x − 6. The key step is expanding the two linear brackets before multiplying by 3 — do not try to multiply all three factors simultaneously. Check at a test value, e.g. x = 2: (4)(1)(3) = 12; expression: 3(4) + 3(2) − 6 = 12 ✓.

• ½ × 6 × 4 or 12 seen for triangle area (M1) (1m)
• 12 × 10 or evidence of multiplying area by length (M1) (1m)
• 120 cao (A1) (1m)

Volume of a triangular prism = area of triangular cross-section × length. Area of triangle = (1/2) × 6 × 4 = 12 cm². Volume = 12 × 10 = 120 cm³. The most common mistake is forgetting the 1/2 in the triangle formula, which gives 6 × 4 × 10 = 240 cm³ — double the correct answer. Always find the cross-section area first, then multiply by the length.

• π × 5² or 25 seen (M1) (1m)
• π × 25 × 8 or 200π seen (M1) (1m)
• 628.3 (A1) — must be to 1 dp as requested (1m)

A cylinder is a circular prism: Volume = πr²h. Square the radius first: r² = 5² = 25. Then V = π × 25 × 8 = 200π ≈ 628.3 cm³ (1 d.p.). A critical error is forgetting to square the radius (using πrh instead), giving π × 5 × 8 ≈ 125.7. Another common mistake is using the diameter instead of the radius. Always use the π button on your calculator rather than approximating π as 3.14.

• π × 6² or 36π seen (M1) (1m)
• 36π × 10 or π × 36 × 10 seen (M1) (1m)
• 360π (A1) — must be exact form, no decimals (1m)

Volume = πr²h = π × 6² × 10 = π × 36 × 10 = 360π cm³. Leaving the answer in terms of π means writing 360π rather than computing 360 × 3.14159... ≈ 1131. Do not substitute a decimal for π — the question specifically asks for the exact form. The step to show on paper: r² = 36, then 36 × 10 = 360, so V = 360π.

• State that the cylinder has smaller volume / cuboid is larger (B1) (1m)
• Explain that the circle fits inside the square / circular base is smaller than square base (B1) (1m)
• Reference to base areas: πr² < 10² or circle area < square area (B1) (1m)

The cuboid has the greater volume. The cylinder's circular base (diameter 10 cm, so radius 5 cm) fits inside the square base of the cuboid (10 cm × 10 cm). Since the circle's area (πr² = 25π ≈ 78.5 cm²) is less than the square's area (100 cm²), and both shapes have the same height (10 cm), the cylinder's volume must be smaller.

• 8 × 5 × 4 seen or 40 × 4 or 32 × 5 (1m)
• 160 cao (condone missing/incorrect units) (1m)

Volume of a cuboid = length × width × height. Substitute l = 8, w = 5, h = 4: V = 8 × 5 × 4 = 160 cm³. A useful approach is to multiply any two dimensions first (8 × 5 = 40), then multiply by the third (40 × 4 = 160). Always write the unit as cm³ for volume — the ³ shows it is a three-dimensional measurement.

• 18 × 12 seen (M1) (1m)
• 216 cao (A1) (1m)

When the cross-sectional area is already given, the volume calculation is a single multiplication: V = cross-section area × length = 18 × 12 = 216 cm³. There is no need to work out the area from scratch. A common mistake is adding (18 + 12 = 30) or dividing (18 ÷ 12 = 1.5) instead of multiplying.

• M1: Area of cross-section = ½ × 6 × 4 = 12 cm² seen (1m)
• A1: Volume = 12 × 10 = 120 cm³ (1m)

Area of triangular cross-section = ½ × 6 × 4 = 12 cm². Volume = 12 × 10 = 120 cm³.

• M1: Rearranges to area = volume ÷ length = 360 ÷ 15 (1m)
• A1: Area of cross-section = 24 cm² (1m)

Volume = area of cross-section × length, so area of cross-section = volume ÷ length = 360 ÷ 15 = 24 cm².

A prism is defined as a 3D shape with a constant cross-section along its entire length. A cylinder has a circular cross-section that is the same all the way along its length, so it fits this definition. The volume formula volume = area of cross-section × length applies to both, where the cross-section is a circle with area πr².

• States that a prism has a constant cross-section along its length (M1) (1m)
• Applies this to a cylinder — the circular cross-section is constant, making it fit the prism definition (A1) (1m)

A prism is a 3D solid with a uniform (constant) cross-section throughout its length. A cylinder has a circular cross-section that is identical at every slice — it is simply a prism with a circular cross-section. The same volume formula (V = cross-section area × length) applies.

The volume of a cuboid is the product of all three dimensions: V = length × width × height. This gives the total space inside the 3D shape in cubic units.

Volume is always measured in cubic units. The ³ shows we're measuring 3-dimensional space: cm³ (cubic centimetres), m³ (cubic metres), etc.

1 litre = 1000 cm³. This is a key conversion for volume problems involving liquids or capacity.

The volume of any prism = area of cross-section × length. The cross-section is the shape that stays the same along the full length of the prism. For a triangular prism, the cross-section is a triangle.

• Use area = 1/2 ab sin C (1m)
• Substitute sin 60° = √3/2 (1m)
• Calculate 1/2 × 8 × 12 = 48 (1m)
• Simplify to 24√3 cm² (1m)

Using area = 1/2 ab sin C with sin 60° = √3/2: Area = 1/2 × 8 × 12 × √3/2 = 48 × √3/2 = 24√3 cm².

• Set up sin 30° = 5/h (1m)
• Substitute sin 30° = 1/2 (1m)
• Rearrange to h = 5 ÷ (1/2) or h = 2 × 5 (1m)
• h = 10 cm (1m)

Using sin 30° = opposite/hypotenuse: 1/2 = 5/h, so h = 5 ÷ (1/2) = 10 cm.

• Substitute sin 45° = √2/2 and cos 45° = √2/2 (1m)
• Multiply correctly: 2 × 2/4 or equivalent (1m)
• Final answer = 1 (1m)

Substituting sin 45° = cos 45° = √2/2 gives 2 × (√2/2) × (√2/2) = 2 × 2/4 = 1.

• Set up sin 60° = x/10 or equivalent (1m)
• Use sin 60° = √3/2 to get x = 10√3/2 (1m)
• Simplify to 5√3 cm (1m)

Using sin 60° = opposite/hypotenuse = x/10, and sin 60° = √3/2, we get x = 10 × √3/2 = 5√3 cm.

• sin²30° = 1/4 (1m)
• cos²30° = 3/4 (1m)
• Add to get 4/4 = 1 (1m)

Substituting sin 30° = 1/2 gives sin²30° = 1/4. Substituting cos 30° = √3/2 gives cos²30° = 3/4. Adding: 1/4 + 3/4 = 1 ✓

• Substitute sin 60° = √3/2 and cos 60° = 1/2 (1m)
• Simplify to (√3 + 1)/2 (1m)

sin 60° = √3/2 and cos 60° = 1/2. Adding gives (√3 + 1)/2.

• States 'Yes' or 'Correct' (1m)
• Shows both equal 1/2 using exact values (1m)

James is correct. sin 30° = 1/2 and cos 60° = 1/2, so they are equal.

• Substitute tan 45° = 1 (1m)
• Calculate 1² + 1 = 2 (1m)

tan 45° = 1, so tan²45° = 1. Therefore tan²45° + 1 = 1 + 1 = 2.

• States cos 30° is larger (1m)
• Shows cos 30° = √3/2 and sin 30° = 1/2, with comparison (1m)

cos 30° = √3/2 ≈ 0.866, while sin 30° = 1/2 = 0.5. So cos 30° is larger.

sin 30° = 1/2 is one of the key exact trig values you must memorise. It comes from the 30-60-90 triangle.

cos 60° = 1/2. Notice that cos 60° = sin 30° (complementary angle relationship).

• √2/2 or equivalent (1/√2) (1m)

sin 45° = √2/2 (or equivalently 1/√2). Both forms are rationalised versions of each other.

tan 90° is undefined because tan = sin ÷ cos, and cos 90° = 0. Division by zero is undefined.

• √3 (1m)

tan 60° = √3. Note that tan 30° = 1/√3, which is the reciprocal.

• Identifies alternate segment theorem gives angle DAB = 2x (M1) (1m)
• States opposite angles in cyclic quad sum to 180° (M1) (1m)
• Forms equation 2x + 3x + 80 = 180 (M1) (1m)
• Simplifies to 5x = 100 (M1) (1m)
• Correct answer: x = 20 (A1) (1m)

By the alternate segment theorem, angle DAB = 2x°. Since ABCD is a cyclic quadrilateral, opposite angles sum to 180°. Therefore, 2x + (3x + 80) = 180. Solving: 5x + 80 = 180, so 5x = 100, giving x = 20.

• States angle at centre = 2 × angle at circumference (M1) (2m)
• Divides 76 by 2 (M1) (1m)
• Correct answer: 38° (A1) (1m)

Using the angle at centre theorem: the angle at the centre (AOB = 76°) is twice the angle at the circumference (ACB). Therefore, angle ACB = 76 ÷ 2 = 38°.

• Finds angle AOB = 140° using angles around a point (M1) (1m)
• States angle at centre = 2 × angle at circumference (M1) (1m)
• Calculates 140 ÷ 2 = 70° (M1) (1m)
• Clear logical structure with theorems stated (M1) (1m)

Angles around point O sum to 360°, so angle AOB = 360° - 140° - 80° = 140°. Using the angle at centre theorem, angle ABC (at circumference) = 140° ÷ 2 = 70°.

• States opposite angles sum to 180° (M1) (1m)
• Forms correct equation: 2x+15 + 3x-10 = 180 or equivalent (M1) (1m)
• Correct answer: x = 35 (A1) (1m)

In a cyclic quadrilateral, opposite angles sum to 180°. So (2x+15) + (3x-10) = 180. Simplifying: 5x+5 = 180, so 5x = 175, giving x = 35.

• Identifies alternate segment theorem (M1) (1m)
• States angle ACB equals the tangent-chord angle (M1) (1m)
• Correct answer: 58° (A1) (1m)

By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment. The tangent-chord angle is 58°, so angle ACB (in the alternate segment) is also 58°.

• States tangent is perpendicular to radius (or angle is 90°) (1m)
• Identifies this creates a right-angled triangle (1m)
• States Pythagoras' theorem applies to right-angled triangles (1m)

The tangent is perpendicular to the radius OT at point T, creating a 90° angle. This means triangle OTP is right-angled at T. Pythagoras' theorem can be applied to any right-angled triangle, so the student is correct.

• M1: Identifies triangle formed by two radii is isosceles (OA = OB = radius) (1m)
• M1: Uses angles in triangle sum to 180° → base angles = (180 − 124) / 2 = 28° (1m)
• A1: y = 28° (1m)

The two sides from the centre O to the circle are both radii, so the triangle is isosceles. The angle at the centre is 124°. The two base angles are equal and sum to 180° − 124° = 56°. Each base angle = 56° ÷ 2 = 28°. Therefore y = 28°.

• Uses angle at centre = 2 × angle at circumference theorem (1m)
• Correct answer: 62° (1m)

The angle at the centre (AOC = 124°) is twice the angle at the circumference (ABC). Therefore, angle ABC = 124 ÷ 2 = 62°.

• States angles in the same segment are equal (1m)
• Correct answer: 47° (1m)

Angles in the same segment are equal. Since angles BAC and BDC are both subtended by arc BC in the same segment, angle BDC = 47°.

• States tangents from external point are equal (M1) (1m)
• Correct answer: 8 cm (A1) (1m)

Tangents drawn from an external point to a circle are equal in length. Since PA = 8 cm, PB must also be 8 cm.

• M1: Applies angle at centre = 2 × angle at circumference, i.e. divides 124 by 2 (1m)
• A1: x = 62° (1m)

The angle at the centre is twice the angle at the circumference (subtended by the same arc). So the angle at the circumference = 124 ÷ 2 = 62°.

The theorem is: the angle at the centre is twice the angle at the circumference when both angles are subtended by the same arc. It applies here because both angles are formed from the same two points on the circle, with one vertex at the centre O and the other at the circumference.

• States the theorem: angle at centre = 2 × angle at circumference (subtended by the same arc) (M1) (1m)
• Explains why it applies: one vertex is at the centre O, the other at the circumference, both subtended by the same arc (A1) (1m)

The theorem states that the angle at the centre is twice the angle at the circumference when subtended by the same arc. To apply it you need to identify one angle with vertex at O (centre) and one at the circumference, both looking at the same chord or arc.

• States perpendicular from centre bisects chord or identifies M as midpoint (M1) (1m)
• Correct answer: 12 cm (A1) (1m)

The perpendicular from the centre to a chord bisects the chord. Since OM is perpendicular to AB, M is the midpoint of AB. Therefore, AM = 24 ÷ 2 = 12 cm.

• States 'angle in a semicircle' or equivalent theorem name (1m)
• States the angle is 90° or a right angle (1m)

Since AB is a diameter, angle ACB is an angle in a semicircle. The angle in a semicircle theorem states that this angle is always 90°.

The angle in a semicircle is always 90°. Since AB is a diameter, angle ACB is an angle in a semicircle, so it must be 90°.

• 90° (or states tangent is perpendicular to radius) (1m)

A tangent to a circle is always perpendicular to the radius at the point of contact. Therefore, the angle is 90°.

Opposite angles in a cyclic quadrilateral sum to 180°. Angle PQR and angle PSR are opposite, so angle PSR = 180° - 105° = 75°.

The angle at the centre is twice the angle at the circumference when both are subtended by the same arc. This is one of the fundamental circle theorems.

• Finds ⁴√16 = 2 (1m)
• Raises to power 3: 2³ (1m)
• 8 (1m)

For a fractional index x^(m/n), the rule is: the denominator (n) tells you which root, and the numerator (m) tells you which power. Always do the root first, then the power. Here: 16^(3/4) means find the fourth root of 16, then cube the result. Step 1 — fourth root: ⁴√16 = 2 (since 2⁴ = 16). Step 2 — cube: 2³ = 8. A very common mistake is stopping after step 1 and writing 2 — this earns 1 mark but not 2. Another mistake is multiplying the indices (3 × 4 = 12), which is incorrect. Always carry out both steps.

• Identifies √70 is between √64=8 and √81=9 (1m)
• 8.4 (accept 8.3 or 8.5) (1m)

When estimating a square root, first identify the two perfect square numbers either side of the target. Here, 64 < 70 < 81, so √70 lies between √64 = 8 and √81 = 9. Next, judge where 70 sits between 64 and 81: 70 − 64 = 6, 81 − 70 = 11, so 70 is closer to 64 than to 81, meaning √70 is closer to 8 than to 9. Trying 8.4: 8.4² = 70.56, which is close enough to accept. Always show the boundary squares — you earn a method mark just for identifying √64 = 8 and √81 = 9.

• Writes 1/5² or equivalent expression showing reciprocal (1m)
• 1/25 (1m)

A negative index means take the reciprocal (flip to a fraction): x⁻ⁿ = 1/xⁿ. Apply this rule in two steps. First, strip the negative sign and evaluate the positive power: 5² = 25. Then flip it to a fraction: 5⁻² = 1/25. Two very common mistakes are writing -25 (treating the negative index as making the answer negative) or writing just 25 (ignoring the negative index altogether). The question says 'give your answer as a fraction', so leave it as 1/25 rather than converting to 0.04.

• Shows 27^(1/3) = ∛27 or equivalent understanding (1m)
• 3 (1m)

A fractional index x^(1/n) means the nth root of x. The denominator of the fraction tells you which root to find. So 27^(1/3) means the cube root of 27. Ask yourself: which number, when cubed, gives 27? Since 3³ = 3 × 3 × 3 = 27, the answer is 3. A common error is dividing by 3 (giving 9), which confuses the fractional index with division. The key rule to remember: x^(1/2) = √x, x^(1/3) = ∛x, and in general x^(1/n) = the nth root of x.

• Shows 8^(2/3) = 4 or equivalent working (1m)
• 1/4 (1m)

When an index is both negative and fractional (like -2/3), tackle it in two stages. First, ignore the negative sign and evaluate the fractional part: 8^(2/3) means (∛8)² = 2² = 4. Then apply the negative index by taking the reciprocal: 8^(-2/3) = 1/(8^(2/3)) = 1/4. A common error is getting 4 (correct fractional part) but forgetting to flip it to 1/4. The question also specifies 'give your answer as a fraction', so do not convert to 0.25. Remember: the negative index only flips the result — it never makes it a negative number.

• States that negative index means reciprocal (or equivalent) (1m)
• Gives reasoning (e.g. x⁻¹×x=1, or pattern from index laws) (1m)

A negative index means 'take the reciprocal'. This comes from index laws: x⁻¹ × x¹ = x⁰ = 1, so x⁻¹ must equal 1/x.

• Shows 2⁶ = 64 or systematic working (1m)
• n = 6 (1m)

To solve 2ⁿ = 64, you need to find which power of 2 equals 64. The most reliable method is to systematically build up powers of 2: 2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32, 2⁶=64. Counting six steps tells you n = 6. A common mistake is dividing 64 by 2 and getting 32, which confuses the operation with the answer. You could also recognise that 64 = 2⁶ from memory (powers of 2 are a useful sequence to know). Show your working by listing the powers of 2 — this earns the method mark even if you make an arithmetic error.

• 49 (1m)

7² means 7 multiplied by itself — so 7 × 7 = 49. The notation 'squared' always means the number is used as a factor exactly twice. A very common mistake is to calculate 7 × 2 = 14, treating the power as a multiplier rather than a repeated multiplication. Memorising square numbers up to 15² will make these questions quick to answer in exams.

• 64 (1m)

4³ means 4 multiplied by itself three times: 4 × 4 × 4. Work in two steps: 4 × 4 = 16, then 16 × 4 = 64. The word 'cubed' always means the number is used as a factor exactly three times. A common error is calculating 4 × 3 = 12, which confuses the power with a multiplier. Knowing cube numbers up to 10³ by heart is very useful for GCSE questions.

• 12 (1m)

The square root symbol √ means 'what number multiplied by itself gives this value?'. √144 asks: which number squared equals 144? Since 12 × 12 = 144, the answer is 12. Note that the √ symbol always means the positive root only — do not write ±12 unless the question specifically asks for both roots. Knowing your square numbers up to at least 15² is essential for answering these quickly without a calculator.

A negative index means 'reciprocal'. 2⁻³ = 1/2³ = 1/8. The negative power doesn't make the number negative — it flips it to a fraction.

• 5 (1m)

The cube root symbol ∛ means 'what number multiplied by itself three times gives this value?'. ∛125 asks: which number cubed equals 125? Since 5 × 5 × 5 = 125, the answer is 5. A very common mistake is dividing by 3 (125 ÷ 3 ≈ 41.7), which gives the wrong operation entirely. Memorising cube numbers from 1³ to 10³ (1, 8, 27, 64, 125, 216, 343, 512, 729, 1000) will help you spot cube roots instantly.

• 100,000 (1m)

Powers of 10 follow a simple pattern: 10ⁿ equals 1 followed by exactly n zeros. So 10⁵ = 100,000 (the digit 1 with five zeros after it). The key shortcut to remember is that you do NOT multiply 10 by 5 — a common mistake that gives 50. Instead, think of 10⁵ as 10 multiplied by itself 5 times: 10 × 10 × 10 × 10 × 10 = 100,000. This power-of-10 pattern is the foundation of standard form, so it is worth knowing thoroughly.

• 3700 (1m)

Multiplying by 10³ means multiplying by 1000, which shifts the decimal point 3 places to the right. Starting from 3.7, move the decimal: 3.7 → 37.0 → 370.0 → 3700. Remember: multiplying by a power of 10 always makes numbers bigger, so the decimal moves right. Dividing by a power of 10 moves the decimal left. A common error is moving the decimal point the wrong way — ask yourself whether the answer should be larger or smaller to check your direction.

Calculate each: 2⁵=32, 3³=27, 4²=16, 5²=25. The largest is 32 (option A).

• Sets up (6 × 10²⁴) ÷ (7.4 × 10²²) (1m)
• Divides coefficients: 6 ÷ 7.4 (1m)
• Subtracts powers: 10²⁴⁻²² = 10² (1m)
• Calculates 0.81 × 10² or equivalent (1m)
• Final answer 81 (to nearest whole number) (1m)

The phrase 'how many times heavier' indicates division: divide Earth's mass by the Moon's mass. Set up (6 × 10^24) ÷ (7.4 × 10^22). Divide coefficients: 6 ÷ 7.4 ≈ 0.8108... Subtract powers: 10^24 ÷ 10^22 = 10^2 = 100. Multiply: 0.8108 × 100 ≈ 81. Round to the nearest whole number: 81. This 5-mark question follows the standard division method but requires a calculator for the coefficient division, and asks for an ordinary number (not standard form) as the final answer. Note that 0.81 × 10^2 is an acceptable intermediate answer, but you must convert it to 81 for the final mark.

• Multiplies coefficients: 6 × 4 = 24 (1m)
• Adds powers for multiplication: 10⁵⁺⁽⁻³⁾ = 10² (1m)
• Divides: 24 ÷ 8 = 3 and 10²⁻⁽⁻¹⁾ = 10³ (1m)
• Final answer 3 × 10³ in standard form (1m)

This multi-step calculation combines multiplication and division of standard form numbers with negative powers. Work through the operations in order: first multiply (6 × 10^5) × (4 × 10^-3): coefficients 6 × 4 = 24, indices 5 + (-3) = 2, giving 24 × 10^2. Then divide by (8 × 10^-1): coefficients 24 ÷ 8 = 3, indices 2 - (-1) = 3, giving 3 × 10^3. Note that subtracting a negative index is the same as adding it. This 4-mark question rewards each correct step. The sign of the power is the most common source of error when negative indices are involved.

• Multiplies coefficients to get 12 (1m)
• Adds powers to get 10⁷ (1m)
• Final answer 1.2 × 10⁸ in standard form (1m)

When multiplying numbers in standard form, multiply the coefficients and add the powers of 10 (index law: 10^a × 10^b = 10^(a+b)). For example, 4 × 10^3 × 3 × 10^4: multiply coefficients 4 × 3 = 12, and add indices 3 + 4 = 7, giving 12 × 10^7. However, 12 is not between 1 and 10, so adjust: 12 × 10^7 = 1.2 × 10^8. This adjustment step is a common source of errors. This 3-mark question rewards multiplying coefficients (M1), adding powers (M1), and giving the final answer in correct standard form (A1).

• Divides coefficients to get 4 (1m)
• Subtracts powers to get 10⁴ (1m)
• Final answer 4 × 10⁴ in standard form (1m)

When dividing numbers in standard form, divide the coefficients and subtract the powers (index law: 10^a ÷ 10^b = 10^(a-b)). For example, 9.6 × 10^7 ÷ 3.2 × 10^4: divide coefficients 9.6 ÷ 3.2 = 3, and subtract indices 7 - 4 = 3, giving 3 × 10^3. Check that the coefficient is between 1 and 10 — 3 satisfies this, so no further adjustment is needed. The marks follow the same pattern: coefficient division, power subtraction, final standard form answer.

• Converts to same power (e.g., 0.32 × 10⁴) (1m)
• Adds correctly to get 4.82 × 10⁴ (1m)
• Answer in standard form (1m)

You cannot add numbers in standard form by simply adding coefficients if the powers of 10 are different. First, express both numbers with the same power of 10. For example, to add 3.6 × 10^6 and 4.5 × 10^5, convert 4.5 × 10^5 to 0.45 × 10^6. Then add coefficients: 3.6 + 0.45 = 4.05, giving 4.05 × 10^6. Check the coefficient is between 1 and 10 — it is, so the answer is 4.05 × 10^6. This 3-mark question requires showing the conversion (M1), adding correctly (M1), and giving the final standard form answer (A1).

• Uses time = distance ÷ speed (1m)
• Calculates (1.5 × 10⁸) ÷ (3 × 10⁵) correctly (1m)
• Answer 5 × 10² (or 500) in standard form (1m)

This question applies the standard form division method to a real-world speed-distance-time problem. The formula needed is time = distance ÷ speed. Distance = 1.5 × 10^8 km, speed = 3 × 10^5 km/s. Divide: (1.5 × 10^8) ÷ (3 × 10^5). Coefficients: 1.5 ÷ 3 = 0.5. Powers: 10^8 ÷ 10^5 = 10^3. So 0.5 × 10^3. This is not in standard form (0.5 < 1), so adjust: 0.5 × 10^3 = 5 × 10^2. The answer is 5 × 10^2 seconds (500 seconds). The 3 marks cover identifying the formula (M1), dividing correctly (M1), and giving the standard form answer (A1).

• Coefficient 5.6 (1m)
• Power 10⁴ (1m)

To write a large number in standard form, find where to place the decimal point so the coefficient is between 1 and 10, then count how many places the decimal moved to determine the power of 10. For 56000: place the decimal after the 5 to get 5.6, and count 4 places (5.6 × 10^4). The number of places the decimal moves left equals the positive power. This is a 2-mark question — one for the correct coefficient (5.6) and one for the correct power (10^4). A common error is writing 56 × 10^3 — the coefficient must be between 1 and 10.

• Coefficient 8.7 (1m)
• Power 10⁻⁵ (1m)

For small decimals, the power of 10 is negative. To convert 0.000087: find the first non-zero digit (8), place the decimal after it (8.7), then count how many places the decimal moved to the right — it moved 5 places right to get from 0.000087 to 8.7. Moving right to make a larger number means the power is negative: 8.7 × 10^-5. The two marks are for the correct coefficient (8.7) and the correct negative power (-5). Confusing -5 with +5 and writing 8.7 × 10^5 is a common exam mistake.

• Easier to read/write/compare (any one of these) (1m)
• Avoids many zeros OR easier to calculate (any one) (1m)

Standard form is a compact way to express very large or very small numbers that would be unwieldy to write out fully. For example, the distance from Earth to the Sun (about 150,000,000,000 metres) is much cleaner as 1.5 × 10^11 m. Standard form also makes comparisons and calculations easier — comparing 3.4 × 10^11 with 5.2 × 10^9 is instant once you see the different powers, whereas comparing the full decimal representations takes much longer. For 2 marks, explain that it is easier to read or write these numbers, and that it simplifies comparisons or calculations.

• 340000 (1m)

Standard form is written as a × 10^n where a is between 1 and 10. To convert back to an ordinary number, multiply by the power of 10. Since the power is +5, multiply 3.4 by 10^5 = 100000: move the decimal point 5 places to the right. 3.4 becomes 340000. A positive power always gives a large number; a common mistake is moving the decimal in the wrong direction or by the wrong number of places.

• 0.00052 (1m)

A negative power of 10 means the number is very small. 10^-4 = 1/10000 = 0.0001. To convert 5.2 × 10^-4, move the decimal point 4 places to the LEFT: 5.2 → 0.52 → 0.052 → 0.0052 → 0.00052. The negative in the power tells you the direction (left = getting smaller). A common mistake is moving right instead of left, giving 52000 — but that would be a very large number, which contradicts the negative power.

Standard form requires the first number to be between 1 and 10 (1 ≤ a < 10). Only 4.5 × 10⁴ meets this requirement.

When comparing standard form numbers, first compare the powers. 10⁴ is larger than 10³ or 10², so 3.1 × 10⁴ = 31000 is the largest.

The student correctly calculated 2 × 5 = 10 and 10⁴⁺³ = 10⁷, giving 10 × 10⁷. However, standard form requires the coefficient to be between 1 and 10. Since 10 ≥ 10, they must write it as 1 × 10⁸.

• Finds UB(distance) = 185 km (1m)
• Finds LB(time) = 2.45 hours (1m)
• Sets up division: 185 ÷ 2.45 (1m)
• Calculates 75.51... (1m)
• Rounds to suitable accuracy (2 sf): 76 km/h (1m)

This 5-mark question combines two skills: identifying the correct bounds for a speed calculation, and giving the answer to a suitable accuracy. Speed = distance / time, so maximum speed = maximum distance / minimum time. UB(distance) = 180 + 5 = 185 km (nearest 10 km, so half-unit = 5). LB(time) = 2.5 - 0.05 = 2.45 hours (nearest 0.1 hours, so half-unit = 0.05). Maximum speed = 185 / 2.45 = 75.51... km/h. Round to 2 significant figures: 76 km/h (since distance has 2 sf). The marks are: UB of distance (M1), LB of time (M1), setting up the division (M1), calculating the result (M1), and suitable rounding (A1). A common error is using 180 / 2.45 or 185 / 2.55 — check which bound makes the result larger.

• Finds UB(x) = 7.25 and UB(y) = 4.55 (or 7.24 and 4.54) (1m)
• Multiplies 7.25 × 4.55 (= 32.9875) (1m)
• Calculates 32.9875 or 32.99 or 33.0 (1m)
• Rounds to suitable accuracy: 33 or 33.0 cm² (1m)

To find the maximum area, multiply the upper bounds of both dimensions. UB(x) = 7.2 + 0.05 = 7.25 cm and UB(y) = 4.5 + 0.05 = 4.55 cm. Maximum area = 7.25 × 4.55 = 32.9875 cm². However, the question asks for a suitable degree of accuracy. Because the original measurements are given to 1 decimal place (2 significant figures), the answer should also be given to 2 significant figures: 33 cm². Writing 32.9875 would give a false impression of precision that the original measurements don't support. This 4-mark question rewards finding the upper bounds (M1), multiplying them (M1), the numerical result (M1), and rounding to a suitable accuracy (A1).

• Finds LB(mass) = 145 g and UB(volume) = 42.5 cm³ (1m)
• Divides 145 ÷ 42.5 (1m)
• Calculates 3.41176... (1m)
• Rounds to 2 sf: 3.4 g/cm³ (1m)

For the minimum value of a quantity calculated by division (density = mass / volume), use the lower bound of the numerator and the upper bound of the denominator. LB(mass) = 150 - 5 = 145 g (mass is to nearest 10 g, so halve the rounding unit: 10/2 = 5). UB(volume) = 42 + 0.5 = 42.5 cm³. Minimum density = 145 / 42.5 = 3.411... g/cm³. Round to 2 significant figures: 3.4 g/cm³. The key insight is that the smallest density comes from dividing the smallest possible mass by the largest possible volume. A common mistake is swapping the bounds to get UB(mass) / LB(volume), which gives the maximum density instead.

• Finds upper bounds: 12.45 cm and 7.85 cm (1m)
• Calculates maximum area: 12.45 × 7.85 ≈ 97.7 cm² (1m)
• Compares to 98: 97.7 < 98 (1m)
• States Jamie is correct with justification (1m)

When asked to verify whether something is 'definitely' true, you must test the worst case — the extreme that would most likely break the claim. Jamie says the area is definitely less than 98 cm², so you need to check whether even the maximum possible area is below 98. UB(length) = 12.4 + 0.05 = 12.45 cm and UB(width) = 7.8 + 0.05 = 7.85 cm. Maximum area = 12.45 × 7.85 = 97.7325 cm². Since 97.7325 < 98, Jamie is correct. If the maximum had been above 98, Jamie would be wrong. This 4-mark question rewards finding upper bounds (M1), computing the maximum area (M1), comparing to 98 (M1), and stating a justified conclusion (M1). Never just state a conclusion without showing the calculation.

• Finds upper bounds 12.55 and 8.35 (or 12.54 and 8.34) (1m)
• Adds the upper bounds (1m)
• Gives 20.9 or 20.89 (1m)

To find the maximum possible value of a sum, use the upper bounds of both values. For a = 12.5 to 1 d.p., the upper bound is 12.55 (add 0.05, which is half of 0.1). For b = 8.3 to 1 d.p., the upper bound is 8.35. Maximum of a + b = 12.55 + 8.35 = 20.9 (or 20.89 for strict upper bounds using 12.54... + 8.34...). Remember the rule: Maximum of addition = UB + UB. A common mistake is just adding the given values (12.5 + 8.3 = 20.8), which gives the nominal sum, not the maximum possible sum.

• Finds LB(p) = 45.65 and UB(q) = 23.85 (1m)
• Subtracts LB(p) - UB(q) (1m)
• Gives 21.8 or 21.79 (1m)

For subtraction, the minimum possible value comes from using the lower bound of the first number and the upper bound of the second. For p - q: minimum = LB(p) - UB(q). LB(p) = 45.7 - 0.05 = 45.65 and UB(q) = 23.8 + 0.05 = 23.85. Minimum = 45.65 - 23.85 = 21.8. Think of it logically: if p is as small as possible and q is as large as possible, the difference p - q is at its smallest. The most common mistake is using UB(p) - LB(q), which gives the maximum difference (22.0), not the minimum. This 3-mark question awards marks for identifying the correct bounds (M1), setting up the subtraction (M1), and the final answer (A1).

• Finds UB(perimeter) = 48.5 cm (1m)
• Calculates maximum side length: 48.5 ÷ 4 = 12.125 cm (1m)
• Calculates area and rounds to 3 sf: 147 cm² (1m)

This is a multi-step bounds problem. The perimeter is 48 cm to the nearest cm, so UB(perimeter) = 48.5 cm. For a square, each side = perimeter / 4, so maximum side = 48.5 / 4 = 12.125 cm. Area of a square = side squared, so maximum area = 12.125² = 147.015625 cm². Rounded to 3 significant figures: 147 cm². The key here is recognising that 'maximum area' requires the maximum perimeter, which then gives the maximum side length. A common mistake is using 48 / 4 = 12 directly (giving 144 cm²), which ignores the bounds entirely. The three marks correspond to: upper bound of perimeter (M1), dividing by 4 to get maximum side (M1), squaring and rounding correctly (A1).

• Correct lower bound 44.5 with ≤ (1m)
• Correct upper bound 45.5 with < (1m)

An error interval expresses the full range of values a rounded measurement could take. For m = 45 kg to the nearest kilogram, any value from 44.5 up to (but not including) 45.5 would round to 45. The error interval is written as 44.5 ≤ m < 45.5. The key notation detail is that the lower bound uses ≤ (included — 44.5 itself rounds to 45) while the upper bound uses < (excluded — 45.5 would round up to 46). This 2-mark question gives one mark for each bound with its correct inequality sign.

• Shows understanding of truncation (1m)
• Writes 3.5 (1m)

Truncation is different from rounding. When a number is truncated to 1 decimal place, the digits beyond the first decimal place are simply cut off (not rounded up). So if x is truncated to give 3.4, x could be any value from 3.4 up to (but not including) 3.5 — for example, 3.4999... truncates to 3.4, not 3.5. The upper bound is therefore 3.5. A common error is confusing truncation with rounding: if x were rounded to 3.4, the lower bound would be 3.35 (not 3.4).

• States that original measurements are 1 dp / have limited precision (1m)
• Explains that 21.00 claims more precision than justified / gives false precision (1m)

This question asks you to explain a mathematical concept, which is an AO3 skill. There are two mark points to address. First, state that the original measurements are only given to 1 decimal place — this means they have limited precision of ±0.05. Second, explain that writing 21.00 implies the answer is accurate to 2 decimal places (to within ±0.005), which is far more precise than the measurements allow. This is called 'false precision'. The correct answer to give is 21.0 cm (1 decimal place, matching the inputs). Note that 21.00 and 21.0 are numerically the same, but they communicate different levels of precision to the reader. In scientific and mathematical contexts, significant figures always communicate precision.

When a number is rounded to the nearest 10, the lower bound is found by subtracting half of 10 (which is 5) from the rounded value. So 240 - 5 = 235 cm.

• Writes 8450 (1m)

The upper bound is found by adding half the degree of accuracy to the rounded value. The attendance is rounded to the nearest 100, so half of 100 is 50. Upper bound = 8400 + 50 = 8450. Note that 8450 itself would round up to 8500, so 8450 is strictly excluded — hence upper bounds always use a strict inequality (<) in the error interval. A common mistake is to add the full 100, giving 8500.

Truncation means cutting off digits, not rounding. If n is truncated to 5.47, it could be any value from 5.47 (exactly) up to (but not including) 5.48. So: 5.47 ≤ n < 5.48

When measurements are given to 1 decimal place, the answer cannot have more precision than the original measurements. 47.88 has 4 significant figures, which is too many. The answer should be given to 2 significant figures: 48 cm²

• Uses correct perimeter formula (1m)
• Adds dimensions to get 8 (1m)
• Multiplies by 2 (1m)
• Correct answer 16 (1m)

The key insight is that the length (4 + √5) and width (4 - √5) are a conjugate pair. When you add them: (4 + √5) + (4 - √5) = 8, because the +√5 and -√5 cancel each other out. Perimeter = 2(length + width) = 2 × 8 = 16 cm. What makes this interesting is that the answer is a whole number even though both dimensions contain surds — the surds eliminate each other. The 4 marks are: perimeter formula (M1), adding dimensions and getting 8 (M1), multiplying by 2 (M1), and the final answer 16 (A1). A common mistake is writing 2(4 + √5) + 2(4 - √5) = 8 + 2√5 — the student forgot that +√5 and -√5 cancel.

• Simplifies √12 to 2√3 (1m)
• Simplifies √27 to 3√3 (1m)
• Adds to get 5√3 (1m)

You cannot add √12 + √27 directly because they are unlike surds (different numbers under the root). You must simplify each one first. √12 = √(4×3) = 2√3, and √27 = √(9×3) = 3√3. Now both have the same surd part (√3), so they are like surds and can be added: 2√3 + 3√3 = 5√3. This 3-mark question rewards each step: simplifying √12 (M1), simplifying √27 (M1), and the final sum (A1). A classic mistake is writing √12 + √27 = √39 — you can NEVER add numbers under separate roots directly.

• Expands to four terms correctly (1m)
• Correctly evaluates √3 × -√3 = -3 (1m)
• Simplifies to 3 + √3 (1m)

Expanding two brackets with surds uses the FOIL method (First, Outer, Inner, Last), just like double brackets in algebra. For (2 + √3)(3 - √3): First = 2×3 = 6; Outer = 2×(-√3) = -2√3; Inner = √3×3 = 3√3; Last = √3×(-√3) = -3. Collect the integer terms: 6 + (-3) = 3. Collect the surd terms: -2√3 + 3√3 = √3. Final answer: 3 + √3. The critical step is recognising √3 × (-√3) = -3 (not -√9 or -3√3). This 3-mark question awards marks for correct expansion (M1), correct evaluation of the √3 × √3 term (M1), and collecting like terms to get the final answer (A1).

• Multiplies by conjugate (3-√5)/(3-√5) (1m)
• Expands denominator correctly to get 4 (1m)
• Simplifies to 9 - 3√5 (1m)

When the denominator contains a binomial like (3 + √5), you cannot just multiply by √5 — you need to multiply by the conjugate (3 - √5). The conjugate has the same terms but the opposite sign between them. Multiplying: numerator = 12(3 - √5) = 36 - 12√5. Denominator = (3 + √5)(3 - √5) = 3² - (√5)² = 9 - 5 = 4. This uses the difference of squares identity. Then divide by 4: (36 - 12√5)/4 = 9 - 3√5. The 3 marks are for: identifying and using the conjugate (M1), correctly expanding the denominator to 4 (M1), and simplifying to 9 - 3√5 (A1). A common mistake is forgetting to simplify the final fraction after rationalising.

• Uses difference of squares or expands brackets (1m)
• Correctly evaluates (√7)² = 7 and 2² = 4 (1m)
• Shows 7 - 4 = 3 (1m)

The expression (√7 + 2)(√7 - 2) matches the difference of squares identity: (a + b)(a - b) = a² - b², where a = √7 and b = 2. Substituting: (√7)² - 2² = 7 - 4 = 3. You can also expand fully using FOIL: (√7)(√7) + (√7)(-2) + (2)(√7) + (2)(-2) = 7 - 2√7 + 2√7 - 4 = 7 - 4 = 3. Either method works, but recognising the difference of squares is faster. For the show-that, you must show each step explicitly: state the pattern or expansion (M1), evaluate (√7)² = 7 and 2² = 4 (M1), and conclude 7 - 4 = 3 (A1). You cannot just write 3 = 3 without working.

• Uses √20 = √(4×5) or equivalent (1m)
• Simplifies to 2√5 (1m)

To simplify a surd, find the largest perfect square that is a factor of the number under the root. For √20: the square factors of 20 are 4 (since 4 × 5 = 20). Using the rule √(ab) = √a × √b, write √20 = √4 × √5 = 2√5. Always use the LARGEST square factor in one step — if you use a smaller square factor first (e.g., √(2×10)) you'll need extra steps and risk errors. The mark scheme awards M1 for splitting correctly (showing √4 × √5) and A1 for the final answer 2√5.

• Shows 5 + 2 or equivalent (1m)
• Correct answer 7√3 (1m)

Like surds work exactly like like terms in algebra. Just as 5x + 2x = 7x, we have 5√3 + 2√3 = 7√3. The surd part √3 acts like a letter variable — only the coefficients (the numbers in front) are added. Two key errors to avoid: (1) adding the numbers under the roots (√3 + √3 ≠ √6 — that's multiplication), and (2) multiplying the coefficients instead of adding them. The mark scheme gives M1 for recognising these are like terms and A1 for the answer 7√3.

• Multiplies to get √48 or equivalent (1m)
• Simplifies to 4√3 (1m)

To multiply two surds, use the rule √a × √b = √(a × b): combine them under one square root sign. So √6 × √8 = √(6 × 8) = √48. However, the question asks for the simplest form, so you must then simplify √48. The largest square factor of 48 is 16 (since 16 × 3 = 48), so √48 = √(16 × 3) = √16 × √3 = 4√3. The mark scheme gives M1 for reaching √48 and A1 for simplifying to 4√3. Stopping at √48 scores only 1 mark — always check whether the result can be simplified.

• Expands to 2√3 and 3 (or √3×√3) (1m)
• Correct answer 2√3 + 3 or 3 + 2√3 (1m)

Expanding a single bracket with a surd works exactly the same way as normal algebra: multiply the term outside by each term inside. For √3(2 + √3): first, √3 × 2 = 2√3; second, √3 × √3 = 3 (because √a × √a = a). So the result is 2√3 + 3. Note that 2√3 and 3 are unlike terms and cannot be combined further. The key fact to remember is √3 × √3 = 3, not √9 or 9 — squaring and taking the square root cancel out, leaving just the number. The mark scheme gives M1 for correct expansion and A1 for the simplified answer 2√3 + 3.

• Multiplies by √3/√3 to get 6√3/3 or equivalent (1m)
• Simplifies to 2√3 (1m)

To rationalise the denominator of 6/√3, multiply both numerator and denominator by √3 (this is the same as multiplying by 1, so the value does not change). Numerator: 6 × √3 = 6√3. Denominator: √3 × √3 = 3. So you get 6√3/3. The question then asks for simplest form — divide both parts by 3: 6√3/3 = 2√3. Two marks are awarded: M1 for correctly multiplying to get 6√3/3 (rationalised form), and A1 for simplifying to 2√3. A common mistake is multiplying only the denominator and not the numerator — always multiply top AND bottom to keep the fraction equivalent.

• States it's easier to compare sizes/order fractions (1m)
• States it's easier to perform further arithmetic/calculations (1m)

Rationalized form (e.g., √2/2 instead of 1/√2) makes it easier to: (1) compare sizes of fractions, (2) perform further calculations, and (3) convert to decimal approximations. It's also the standard mathematical convention.

√48 = √(16×3) = √16 × √3 = 4√3. Find the largest square factor (16), split the surd, then simplify.

To rationalize, multiply by √5/√5 (which equals 1, so doesn't change the value). This gives (1×√5)/(√5×√5) = √5/5.

Using the rule √a × √b = √(ab), we get √3 × √5 = √(3×5) = √15.

• Sets up x = 0.3666... (1m)
• Multiplies by 10 and 100 correctly (1m)
• Subtracts to get 90x = 33 (1m)
• Simplifies 33/90 to 11/30 (1m)

This is the hardest type of recurring decimal conversion because the decimal has a non-recurring part (0.3) before the recurring part (6). The trick is to use two different powers of 10. Since there is 1 non-recurring digit and 1 recurring digit, multiply by 10 (to shift past the non-recurring part: 10x = 3.666...) and by 100 (to shift one full cycle: 100x = 36.666...). Subtracting: 100x - 10x = 36.666... - 3.666..., so 90x = 33, giving x = 33/90 = 11/30 (HCF is 3). The 4 marks are: setting up x (M1), both multiplications (M1), correct subtraction to 90x = 33 (M1), and simplifying (A1).

• Converts 0.3̇ to 1/3 (1m)
• Converts 0.1̇8̇ to 2/11 (1m)
• Finds common denominator 33 and adds correctly (1m)
• Final answer 17/33 (1m)

Adding recurring decimals is done by first converting each one to a fraction. 0.3̇ = 1/3 (single recurring digit: 3/9 = 1/3). 0.1̇8̇ = 2/11 (two recurring digits: 18/99, simplify by 9). Now add 1/3 + 2/11. LCM(3, 11) = 33 (they are coprime). Convert: 1/3 = 11/33 and 2/11 = 6/33. Sum = 17/33. Check: HCF(17, 33) = 1 (17 is prime and doesn't divide 33), so already in simplest form. The 4 marks correspond to: converting 0.3̇ (M1), converting 0.1̇8̇ (M1), finding common denominator and adding (M1), and final answer (A1). Do not try adding the decimals — always convert first.

• Sets up x = 0.272727... (1m)
• Multiplies by 100 to get 100x = 27.272727... (1m)
• Subtracts and simplifies to x = 3/11 (1m)

This is a 'prove that' version of the algebraic conversion method. Since both 2 and 7 in 0.2̇7̇ recur together as a 2-digit block (27), you multiply by 100. Let x = 0.272727..., then 100x = 27.272727... Subtracting: 100x - x = 27, so 99x = 27. Therefore x = 27/99. Simplify using HCF(27, 99) = 9: 27 ÷ 9 = 3 and 99 ÷ 9 = 11, giving x = 3/11. For show-that and prove questions, write out every step clearly and state the conclusion. The mark scheme awards B1 for setting up x correctly, M1 for multiplying by 100 and subtracting, and A1 for simplifying 27/99 to 3/11.

• Sets up x = 0.454545... (1m)
• Multiplies by 100, subtracts to get 99x = 45 (1m)
• Simplifies 45/99 to 5/11 (1m)

Converting 0.4̇5̇ to a fraction uses the algebraic method for a 2-digit recurring block. Let x = 0.454545..., then multiply by 100 (since 2 digits recur): 100x = 45.454545... Subtracting: 99x = 45. So x = 45/99. Simplify: HCF(45, 99) = 9, so 45/99 = 5/11. This 3-mark question gives M1 for setting up the equations, M1 for getting 99x = 45, and A1 for simplifying to 5/11. Quick-check using the reverse: 5/11 = 0.4545... confirming the answer. The shortcut: a 2-digit recurring block 'ab' gives ab/99.

• Converts all to decimals or recognizes values (1m)
• Identifies 0.6̇ = 2/3 (1m)
• Correct order: 0.6̇, 2/3, 0.67, 0.6̇7̇ (1m)

To compare these four numbers, convert them all to equivalent decimal expansions. 0.6̇ = 0.6666..., and 2/3 = 0.6666... — these are EQUAL (the same value). 0.67 = 0.6700... (terminates, so after 2 dp it's just zeros). 0.6̇7̇ = 0.6767... (the block '67' recurs). Comparing at the 3rd decimal place: 0.6666 < 0.6700 < 0.6767, so the order is 0.6̇, 2/3, 0.67, 0.6̇7̇. This 3-mark question awards M1 for converting to decimal form, M1 for identifying 0.6̇ = 2/3, and A1 for the correct full ordering. Students often incorrectly place 0.67 last because it looks largest — but 0.6̇7̇ is larger since it recurs above 0.67.

• Defines x = 0.545454... and sets up equation using correct power of 10 (100x = 54.545454...) (1m)
• Subtracts to get 99x = 54 (1m)
• Simplifies 54/99 to 6/11 and states conclusion (1m)

This show-that question requires a fully written algebraic proof. The 2-digit block '54' recurs, so you multiply by 100. Let x = 0.5̇4̇ = 0.545454..., then 100x = 54.545454... Subtracting: 99x = 54, so x = 54/99. Simplify: HCF(54, 99) = 9, giving x = 6/11. Conclude with 'as required' or 'therefore 0.5̇4̇ = 6/11'. The three marks are: correct setup with x and 100x (B1), subtracting to get 99x = 54 (M1), and simplifying 54/99 to 6/11 with conclusion (A1). You cannot receive full marks without stating the conclusion explicitly — in show-that questions, the final written statement is assessed.

• States 6 = 2 × 3 (prime factorisation) (1m)
• States rule about only 2 and 5 giving terminating decimals (1m)
• Concludes that factor 3 (not 2 or 5) causes recurring (1m)

This 3-mark question asks you to explain WHY a fraction gives a recurring decimal using prime factors. The rule is: a fraction in its simplest form gives a terminating decimal if and only if its denominator has ONLY prime factors of 2 and 5. For 1/6: factorise 6 as 2 × 3. The denominator has the prime factor 3, which is neither 2 nor 5. Therefore 1/6 cannot terminate and must recur. The reason this rule works is that our decimal system is base 10 = 2 × 5, so only multiples of 2 and 5 divide exactly into powers of 10. The 3 marks are for: prime factorisation of 6 (M1), stating the rule about 2s and 5s (M1), and concluding that factor 3 causes the decimal to recur (A1).

• Sets up x = 0.111... and 10x = 1.111... (1m)
• Subtracts to get 9x = 1, so x = 1/9 (1m)

The algebraic method for converting a recurring decimal to a fraction has four steps. Step 1: let x equal the recurring decimal (x = 0.1̇ = 0.111...). Step 2: multiply by 10 because one digit recurs — this shifts the decimal point one place: 10x = 1.111... Step 3: subtract the original equation from the new one: 10x - x = 1.111... - 0.111..., so 9x = 1. Step 4: divide both sides by 9: x = 1/9. The magic is that subtracting eliminates the recurring part. For two recurring digits, multiply by 100; for three, multiply by 1000. The M1 mark is awarded for setting up both equations correctly.

• 0.3̇ (1m)

Dot notation is used to show that a decimal recurs (repeats forever) without writing dots. For a single repeating digit, place a dot directly above that digit. So 0.3333... is written as 0.3̇. If a block of digits repeats (e.g. 0.181818...), place a dot above the first AND last digit of the repeating block: 0.1̇8̇. A common error is writing 0.333 — this suggests the decimal terminates after three digits, which is incorrect. The dot notation is the standard mathematical shorthand used in GCSE examinations.

• 5/9 (1m)

There is a quick rule for converting a single recurring digit to a fraction: place that digit over 9. So 0.5̇ = 5/9. This works because if x = 0.5555..., then 10x = 5.5555..., so 10x - x = 5, giving 9x = 5 and x = 5/9. Check: HCF(5, 9) = 1, so 5/9 is already in simplest form. A common mistake is confusing 0.5̇ (which recurs) with 0.5 (which terminates at 1/2). The dot above the digit is the signal that the decimal never ends.

A fraction gives a recurring decimal if its denominator (in simplest form) has prime factors OTHER than 2 or 5. Only 1/3 has denominator 3, which is not 2 or 5.

• 0.1̇8̇ (1m)

To convert a fraction to a decimal, carry out the division. 2 ÷ 11 = 0.181818... — the block '18' repeats indefinitely. In dot notation, place a dot above the first digit of the repeating block (1) and a dot above the last digit (8): 0.1̇8̇. A common mistake is placing only one dot (0.1̇8 or 0.18̇) — both digits are part of the same repeating block so both need dots. You can verify the answer: 18/99 = 2/11 (dividing both by 9), confirming 2/11 = 0.1̇8̇.

The MCQ tests whether you recognise that 0.6̇ = 2/3. Using the shortcut: a single recurring digit d gives d/9, so 0.6̇ = 6/9. Simplify by dividing numerator and denominator by HCF(6, 9) = 3: 6/9 = 2/3. The distractors test common errors — 6/10 is for the terminating decimal 0.6, and 6/99 would be used for a 2-digit recurring decimal 0.0̇6̇. Worth memorising: 0.3̇ = 1/3, 0.6̇ = 2/3, 0.1̇ = 1/9. These appear frequently in GCSE questions.

A fraction in simplest form gives a terminating decimal if and only if its denominator has ONLY factors of 2 and 5. Since 12 = 2² × 3, it has factor 3, so 7/12 is recurring.

• Identifies multiplier 0.85 (1m)
• Uses 0.85³ for 3 years (1m)
• Calculates 0.85³ = 0.614125 (1m)
• Multiplies 18000 × 0.614125 = 11054.25 (1m)
• Rounds to £11054 (1m)

This 5-mark compound depreciation problem requires the formula: final value = initial value × (multiplier)^n. A 15% decrease each year means the car retains 85% of its value annually, giving multiplier 0.85 = 1 - 0.15. Over 3 years: 18000 × 0.85³ = 18000 × 0.614125 = 11054.25, rounded to £11054. The five marks are: identifying multiplier 0.85 (M1), recognising that you apply it 3 times / use 0.85³ (M1), calculating 0.85³ = 0.614125 (M1), multiplying 18000 × 0.614125 (M1), and rounding to the nearest pound (A1). A critical mistake is adding 3 × 15% = 45% and subtracting that from the original — compound depreciation is not the same as simple depreciation.

• Identifies multiplier 1.15 or 115% (1m)
• Divides 13800 by 1.15 (1m)
• Answer £12000 (1m)

This is a reverse percentage question — you know the value AFTER a change and need to find the original. A 15% increase means the original was multiplied by 1.15 to get £13800. To reverse this: divide £13800 by 1.15. 13800 / 1.15 = 12000. The three mark points are: identifying the multiplier 1.15 (M1), dividing to reverse the operation (M1), and getting £12000 (A1). The most common mistake is taking 15% off £13800 (giving £11730), which is WRONG — 15% of £13800 is not the same as 15% of the original smaller value. Always reverse by dividing by the multiplier, not by subtracting a percentage of the given value.

• Finds profit = £30 (1m)
• Divides 30 by 60 (cost price) (1m)
• Converts to 50% (1m)

Percentage profit = (profit / cost price) × 100. First, find the profit: £90 - £60 = £30. Then, express this as a percentage of the COST PRICE (not the selling price): (30 / 60) × 100 = 50%. The three marks correspond to finding the profit (M1), dividing by the cost price (M1), and giving the final percentage (A1). A very common mistake is dividing by the selling price (90) instead of the cost price (60), which gives 33.3% — always use the cost price as the base. The reason is that profit is always expressed relative to what you paid, not what you sold for.

• Uses multiplier 1.05 for 5% increase (1m)
• Applies twice: 1.05² or multiplies by 1.05 twice (1m)
• Final answer £220500 (1m)

Repeated percentage change (compound growth) means applying the percentage each year to the NEW value, not the original. A 5% increase uses the multiplier 1.05. For 2 years: 200000 × 1.05 × 1.05 = 200000 × 1.05² = 200000 × 1.1025 = 220500. The key distinction from simple percentage is that in Year 1 you get 10000 extra, but in Year 2 you get 5% of the larger amount (210000), giving 10500 extra — hence the total is 220500, not 220000. The three marks are: correct multiplier 1.05 (M1), applying it twice or using 1.05² (M1), and the final answer 220500 (A1).

• Recognizes £750 represents 120% of original and sets up division by 1.2 (1m)
• Performs 750 ÷ 1.2 correctly (1m)
• Answer £625 (1m)

A price of £750 including 20% VAT means £750 = 120% of the original (100% original + 20% VAT). To find the original, divide by 1.2: 750 / 1.2 = 625. The most common mistake is taking 20% off £750 (= 750 × 0.8 = 600), which gives the wrong answer. Why is 600 wrong? Because the VAT was originally added to the SMALLER original price, not to £750. If the original was £625, then VAT = 20% × 625 = 125, and 625 + 125 = 750. This confirms the answer. The three marks are: recognising £750 = 120% and identifying division by 1.2 (M1), performing the division (M1), and giving the correct answer £625 (A1).

• Finds 15% of 80 = 12 (1m)
• Final answer 92 (1m)

To increase £80 by 15%, there are two equivalent methods. Method 1: find 15% of 80 (= 80 × 0.15 = 12), then add to the original: 80 + 12 = 92. Method 2 (quicker): use the multiplier 1.15 (which represents 100% + 15%) — so 80 × 1.15 = 92. The mark scheme awards M1 for finding 15% of 80 and A1 for the final total of 92. A common mistake is giving just the increase amount (12) — you must add it back to the original to get the new value. Check: does 92 seem reasonable for an 80 increased by about a sixth? Yes.

• Finds 20% of 450 = 90 (1m)
• Final answer £360 (1m)

A reduction of 20% means you pay 80% of the original price. Method 1: find 20% of 450 (= 0.2 × 450 = 90), then subtract from the original: 450 - 90 = 360. Method 2 (quicker): use the multiplier 0.8 (which represents 100% - 20%) — so 450 × 0.8 = 360. The M1 mark is for finding 20% of 450 (= 90) and A1 for the final sale price of £360. Common errors include giving just the discount (90) or adding instead of subtracting (540). Always re-read whether the question asks for the sale price or the amount of reduction.

• Divides 15 by 25 to get 0.6 (1m)
• Converts to 60% (1m)

To express one quantity as a percentage of another, use the formula: (part / whole) × 100. Here, girls = 15 (the part) and the whole class = 25 (the total). So: (15/25) × 100 = 0.6 × 100 = 60%. The M1 mark is for dividing 15 by 25 correctly (getting 0.6), and A1 for multiplying by 100 to give 60%. A common mistake is dividing the wrong way (25/15 = 166.7%) — always put the part over the total. Another error is giving the decimal answer (0.6) instead of converting to a percentage.

• Uses multiplier 1.2 or finds 20% of 24 (1m)
• Final answer £28.80 (1m)

VAT at 20% is just a percentage increase in disguise. The total cost including VAT = original price × 1.2 (since 100% + 20% = 120% = 1.2). 24 × 1.2 = 28.80. Alternatively: 20% of 24 = 4.80, then 24 + 4.80 = 28.80. The M1 mark is for using multiplier 1.2 (or calculating 20% of 24 = 4.80) and A1 for the correct total of £28.80. A very common error is giving only the VAT amount (4.80) rather than the total price — the question asks for the total cost including VAT.

• Calculates 5 ÷ 40 = 0.125 (1m)
• Converts to 12.5% (1m)

Percentage change formula: ((new value - original) / original) × 100. Here: change = 45 - 40 = 5. Percentage increase = (5 / 40) × 100 = 0.125 × 100 = 12.5%. The critical rule is to ALWAYS divide by the ORIGINAL value, not the new value. Dividing by 45 (the new value) gives approximately 11.1%, which is wrong. The M1 mark is awarded for dividing the change (5) by the original value (40), and A1 for converting 0.125 to 12.5%. If you give the answer as 0.125 (the decimal), you only get M1.

• 12 (1m)

To find a percentage of an amount, convert the percentage to a decimal by dividing by 100, then multiply. 20% = 20/100 = 0.2, so 20% of 60 = 60 × 0.2 = 12. Alternatively, use the fraction equivalent: 20% = 1/5, so 60 ÷ 5 = 12. A common mistake is dividing 60 by 20 to get 3 — this gives 1/20 of 60 (which is 5%), not 20%. Always check: 0.2 represents 20%, not 2%.

For a 30% increase, you need 100% (the original) + 30% (the increase) = 130% = 1.3.

If the price decreased by 25% to £300, then £300 represents 75% of the original price (100% - 25% = 75% = 0.75). So: original × 0.75 = 300, which means original = 300 ÷ 0.75 = 400.

Percentage change depends on the ORIGINAL value. Increase: (10÷50)×100 = 20%. Decrease: (10÷60)×100 = 16.67%. Even though the absolute change (£10) is the same, the percentages differ because they're calculated from different starting values. The increase (20%) is greater.

• Prime factorises 12 = 2² × 3 (1m)
• Prime factorises 18 = 2 × 3² (1m)
• Prime factorises 30 = 2 × 3 × 5 (1m)
• Identifies highest powers (2², 3², 5) (1m)
• Correct answer: 180 (1m)

Finding LCM of three numbers: factorise all three, then for each prime that appears in ANY factorisation, take the HIGHEST power. Here: 12 = 2² × 3, 18 = 2 × 3², 30 = 2 × 3 × 5. All primes present: 2, 3, and 5. Highest powers: 2² (from 12), 3² (from 18), 5¹ (from 30). LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180. The 5 marks are for: M1 for each of the three prime factorisations, M1 for identifying highest powers, A1 for 180. Check: 180 ÷ 12 = 15, 180 ÷ 18 = 10, 180 ÷ 30 = 6 — all whole numbers, confirming 180 is divisible by all three.

• Prime factorises 180 = 2² × 3² × 5 (1m)
• Prime factorises 252 = 2² × 3² × 7 (1m)
• Finds HCF = 2² × 3² (method shown) (1m)
• (a) Correct: 36 (1m)
• (b) Finds LCM = 2² × 3² × 5 × 7 and gives 1260 (1m)

This 5-mark question tests both HCF and LCM from prime factorisation. Factorise both: 180 = 2² × 3² × 5 and 252 = 2² × 3² × 7. For HCF (part a): identify primes COMMON to BOTH (2 and 3), take LOWEST powers (2² and 3²). HCF = 2² × 3² = 4 × 9 = 36. For LCM (part b): include ALL primes from EITHER factorisation (2, 3, 5, 7), take HIGHEST powers of each. LCM = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260. A very common error is swapping the rules: LCM uses HIGHEST powers, HCF uses LOWEST powers of COMMON primes only. The Venn diagram method can help: common factors (with lowest powers) go in the overlap; remaining factors go in the separate sections. Verify: HCF × LCM = 36 × 1260 = 45360 = 180 × 252.

• Identifies that LCM of 12 and 18 is needed (1m)
• Finds LCM = 36 (1m)
• Adds 36 minutes to start time (1m)
• Correct answer: 9:36 am (accept 09:36 or 9:36) (1m)

When two repeating cycles need to coincide, you need the LCM. The first time both buses return together is after LCM(12, 18) minutes. Factorise: 12 = 2² × 3, 18 = 2 × 3². LCM = 2² × 3² = 4 × 9 = 36. So they first meet again after 36 minutes. Add 36 minutes to 9:00 am to get 9:36 am. The question has 4 marks: M1 for recognising LCM is needed, M1 for finding LCM = 36, M1 for adding to the start time, A1 for 9:36 am. A common mistake is writing just '36' without converting to a time — always answer what the question actually asks for.

• Identifies need for HCF (1m)
• Prime factorises at least one dimension (144 = 2⁴ × 3² or 108 = 2² × 3³) (1m)
• Finds HCF = 2² × 3² (or equivalent working) (1m)
• Correct answer: 36 (accept 36 cm) (1m)

The key insight here is recognising this as an HCF problem. For the square tiles to fit with no gaps, the tile side must divide exactly into BOTH dimensions. To find the LARGEST possible tile, find the HCF of 144 and 108. Prime factorise: 144 = 2⁴ × 3², 108 = 2² × 3³. Common primes: 2 and 3. Lowest powers: 2² and 3². HCF = 2² × 3² = 4 × 9 = 36. So the largest tile is 36 cm × 36 cm. Check: 144 ÷ 36 = 4 tiles along, 108 ÷ 36 = 3 tiles across. The 4-mark structure is: M1 for recognising HCF, M1 for prime factorising, M1 for applying lowest powers, A1 for 36.

• Prime factorises both numbers (18 = 2 × 3², 24 = 2³ × 3) (1m)
• Identifies highest powers (2³ and 3²) (1m)
• Correct answer: 72 (1m)

The LCM is found by (1) prime factorising both numbers, (2) listing ALL primes that appear in EITHER factorisation, (3) taking the HIGHEST power of each prime, (4) multiplying. Here: 18 = 2 × 3² and 24 = 2³ × 3. All primes present: 2 and 3. Highest power of 2 is 2³ (from 24), highest power of 3 is 3² (from 18). LCM = 2³ × 3² = 8 × 9 = 72. The 3 marks reflect: M1 for both factorisations, M1 for identifying highest powers (2³ and 3²), A1 for the answer 72. A common mistake is multiplying both numbers (18 × 24 = 432), which gives a multiple but not the LOWEST.

• Chooses n = 40 as counter-example (1m)
• Calculates 40² + 40 + 41 = 1681 (1m)
• Shows that 1681 = 41² (or 41 × 41), so it's not prime (1m)

To disprove a 'for all values' statement, you only need to find ONE counter-example where it fails. The expression n² + n + 41 is famously almost always prime, but it breaks down at n = 40. Substituting: 40² + 40 + 41 = 1600 + 40 + 41 = 1681. Now show 1681 is not prime: 1681 = 41 × 41 = 41². Since 1681 has factors other than 1 and itself (specifically 41), it is composite, not prime. This disproves Sasha's claim. You need all three steps for 3 marks: choosing n = 40, calculating 1681, and showing it equals 41². Trying small values of n (1, 2, 3 ...) will not find a counter-example — the breakdown only occurs at n = 40.

• Prime factorises at least two numbers correctly (1m)
• Identifies common primes with lowest powers (2 × 3 × 7) (1m)
• Correct answer: 42 (1m)

Finding HCF of three numbers follows exactly the same method as two numbers — prime factorise all three, find primes common to ALL THREE (not just two), and take the lowest power of each. Here: 84 = 2² × 3 × 7, 126 = 2 × 3² × 7, 210 = 2 × 3 × 5 × 7. Common to all three: 2, 3, and 7 (note 5 only appears in 210, so it is excluded). Lowest powers: 2¹, 3¹, 7¹. HCF = 2 × 3 × 7 = 42. Check: 84 ÷ 42 = 2, 126 ÷ 42 = 3, 210 ÷ 42 = 5. All three divide exactly, so 42 is correct.

• States or uses the relationship: product = HCF × LCM (1m)
• Substitutes: product = 6 × 180 (1m)
• Completes: 6 × 180 = 1080 (1m)

This 'show that' question tests whether you know the key relationship: for any two numbers a and b, a × b = HCF(a, b) × LCM(a, b). Once you know this rule, the proof is straightforward: substitute HCF = 6 and LCM = 180, giving product = 6 × 180 = 1080. In a 'show that' question, every step of reasoning must be shown — stating the rule (B1), substituting the values (M1), and completing the multiplication (A1). You cannot write '= 1080' without showing where it comes from. The rule a × b = HCF × LCM is a fundamental number theory result worth memorising.

• Shows correct prime factorisation process (e.g., 60 = 2 × 30 = 2 × 2 × 15...) (1m)
• Gives 2² × 3 × 5 (or equivalent ordering) (1m)

To express a number as a product of prime factors, use repeated division by the smallest prime that divides exactly into the number. Start with 2 (since 60 is even): 60 ÷ 2 = 30, 30 ÷ 2 = 15. Now 15 is odd so try 3: 15 ÷ 3 = 5. Finally, 5 is already prime. Collecting all the divisors: 60 = 2 × 2 × 3 × 5. In index form this is 2² × 3 × 5. Index form means using powers for any repeated factor — here 2 appears twice so it becomes 2². You earn M1 for showing the systematic division process, A1 for the correct index form answer.

• Shows prime factorisation method OR systematic factor listing (1m)
• Correct answer: 24 (1m)

The HCF is found by (1) prime factorising both numbers, (2) identifying primes common to BOTH, (3) taking the LOWEST power of each common prime, (4) multiplying. Here: 48 = 2⁴ × 3 and 72 = 2³ × 3². Common primes are 2 and 3. Lowest power of 2 is 2³ (from 72), lowest power of 3 is 3¹ (from 48). So HCF = 2³ × 3 = 8 × 3 = 24. A very common mistake is confusing HCF with LCM (getting 144 instead). Check your answer divides into both: 48 ÷ 24 = 2 and 72 ÷ 24 = 3.

• States that 51 = 3 × 17 OR that 51 has factors 3 and 17 (1m)
• Explains that prime numbers have exactly two factors, so 51 is not prime (or equivalent: composite, more than two factors) (1m)

For 'explain why a number is not prime', you must do two things to earn both marks: (1) identify at least one factor pair other than 1 and the number, and (2) link it to the definition of prime. Here: 51 = 3 × 17, so it has four factors (1, 3, 17, 51). Since a prime number has exactly two factors, and 51 has four, it is not prime. Many students test divisibility correctly but then just write 'it has more factors' without explicitly connecting to the prime definition. A quick divisibility test: check if the digit sum (5 + 1 = 6) is divisible by 3 — it is, so 3 divides 51.

29 is prime because it has exactly two factors: 1 and 29. The other numbers have additional factors: 27 = 3³, 33 = 3 × 11, 35 = 5 × 7.

Statement B is true: 12 divides into 24 exactly (24 = 12 × 2), so every factor of 12 is also a factor of 24. The factors of 12 are {1, 2, 3, 4, 6, 12}, and all of these divide 24. Statement A is false (6, 18, 30 are multiples of 6 but not 12). Statement C is false (2 is prime and even). Statement D is false (2 is even and prime, not composite).

90 = 2 × 45 = 2 × 9 × 5 = 2 × 3 × 3 × 5 = 2 × 3² × 5. Option B includes 9 which is not prime. Option C includes 10 which is not prime. Option D has 2² instead of 2¹.

• Finds ⅔ remains after first purchase (1m)
• Calculates ¼ of ⅔ = ⅙ spent on drink (1m)
• Works out ⅔ - ⅙ = ½ remaining (1m)
• Correct answer £36 (since £18 = ½ of original) (1m)

This is a multi-step fraction problem requiring careful tracking of what fraction of the original remains at each stage. Step 1: Sarah spends 1/3 on a book, leaving 2/3. Step 2: She spends 1/4 of what remains, which is 1/4 × 2/3 = 2/12 = 1/6 of the original. Step 3: The fraction remaining = 2/3 - 1/6 = 4/6 - 1/6 = 3/6 = 1/2. Step 4: If 1/2 of the original = £18, then the whole = £18 × 2 = £36. The critical point is that 'a quarter of what remains' means a quarter of 2/3 (not a quarter of the original). This 4-mark challenge question rewards each correct step separately.

• Converts to improper fractions OR separates whole and fraction parts (1m)
• Adds fractions correctly with common denominator (1m)
• Correct answer 4⅜ as a mixed number (1m)

For adding mixed numbers, the most reliable method is to convert both to improper fractions first: 2 and 3/4 = 11/4 and 1 and 5/8 = 13/8. The LCM of 4 and 8 is 8, so convert 11/4 to 22/8. Then add: 22/8 + 13/8 = 35/8. Convert back to a mixed number: 35 ÷ 8 = 4 remainder 3, so 35/8 = 4 and 3/8. This 3-mark question rewards showing the conversion (M1), the fraction addition (M1), and the correct mixed number answer (A1). A common error is to forget to convert the fraction sum back to a mixed number.

• Converts division to multiplication by reciprocal: ⅘ × 3/2 (1m)
• Multiplies correctly to get 12/10 or 6/5 (1m)
• Correct answer 1⅕ as a mixed number (1m)

Dividing by a fraction means multiplying by its reciprocal. The reciprocal is found by flipping the fraction — for 2/3, the reciprocal is 3/2. So 4/5 ÷ 2/3 = 4/5 × 3/2. Multiply: (4 × 3)/(5 × 2) = 12/10 = 6/5. Convert to a mixed number: 6 ÷ 5 = 1 remainder 1, so 1 and 1/5. This 3-mark question awards one mark each for applying the reciprocal, for multiplying correctly (12/10 or 6/5), and for the final mixed number. The mnemonic 'Keep, Change, Flip' helps: keep the first fraction, change ÷ to ×, flip the second fraction.

• Converts both to improper fractions: 8/3 and 9/5 (1m)
• Multiplies correctly to get 72/15 or simplified 24/5 (1m)
• Correct answer 4⅘ as a mixed number (1m)

You cannot multiply mixed numbers by treating the whole parts and fraction parts separately. The correct approach is to convert both to improper fractions first: 2 and 2/3 = 8/3 (2 × 3 + 2 = 8) and 1 and 4/5 = 9/5 (1 × 5 + 4 = 9). Then multiply: 8/3 × 9/5 = 72/15. Simplify by dividing by HCF(72, 15) = 3: 24/5. Convert to a mixed number: 24 ÷ 5 = 4 remainder 4, giving 4 and 4/5. This 3-mark question follows the same M1, M1, A1 pattern — converting, multiplying, final answer.

• Multiplies ⅔ × ¾ first to get 6/12 or ½ (1m)
• Adds ½ to the result (1m)
• Correct answer 1 (1m)

BIDMAS applies to fractions exactly as it does to integers. In the expression 1/2 + 2/3 × 3/4, the multiplication must be done first. Calculate 2/3 × 3/4 = 6/12 = 1/2. This is a satisfying simplification — the 3s cancel. Then add 1/2 + 1/2 = 1. A common mistake is to work left to right and add 1/2 + 2/3 first, getting 7/6, then multiply by 3/4 to get 7/8. That is wrong because multiplication takes priority over addition. This 3-mark question gives one mark for performing the multiplication first, one for correctly adding the result, and one for the final answer of 1.

• Converts both to improper fractions: 21/5 and 5/3 (1m)
• Finds common denominator and subtracts: 63/15 - 25/15 = 38/15 (1m)
• Correct answer 2 8/15 as a mixed number (1m)

For subtracting mixed numbers, convert to improper fractions first — this is safer than trying to subtract whole and fraction parts separately. 4 and 1/5 = 21/5 (4 × 5 + 1) and 1 and 2/3 = 5/3 (1 × 3 + 2). The LCM of 5 and 3 is 15: convert 21/5 to 63/15 and 5/3 to 25/15. Subtract: 63/15 - 25/15 = 38/15. Convert back: 38 ÷ 15 = 2 remainder 8, giving 2 and 8/15. This 3-mark question follows the same three-step structure. If you tried to subtract the fraction parts separately without converting, you would get 1/5 - 2/3 = 3/15 - 10/15 = -7/15, which would require borrowing — the improper fraction method avoids this complication.

• Finds common denominator (15) and converts both fractions correctly (1m)
• Correct answer 13/15 (1m)

To add fractions with different denominators, you cannot simply add the numerators and denominators together. The key step is finding a common denominator — the smallest number both denominators divide into. For 2/3 + 1/5, the LCM of 3 and 5 is 15. Convert: 2/3 = 10/15 and 1/5 = 3/15. Now add the numerators: 10 + 3 = 13, keeping the denominator 15, giving 13/15. This is a 2-mark question — one for the correct common denominator and conversion, one for the final answer 13/15.

• Converts both fractions to common denominator 24 (1m)
• Correct simplified answer 11/24 (1m)

Subtracting fractions follows the same rule as addition — you need a common denominator first. For 5/6 - 3/8, the LCM of 6 and 8 is 24. Convert: 5/6 = 20/24 and 3/8 = 9/24. Then subtract the numerators: 20 - 9 = 11, giving 11/24. Since HCF(11, 24) = 1 (11 is prime), this is already in simplest form. A common error is to subtract both numerators and denominators: 5-3=2 and 6-8=-2, which is meaningless. Always find the common denominator first.

• Multiplies numerators and denominators: 2×5=10, 3×8=24 (1m)
• Correct simplified answer 5/12 (1m)

Unlike addition, multiplying fractions does not require a common denominator. Simply multiply the numerators together and the denominators together: (2 × 5)/(3 × 8) = 10/24. Then simplify: HCF(10, 24) = 2, so divide both by 2 to get 5/12. A useful shortcut is to cancel before multiplying — notice that 2 and 8 share a factor of 2, so simplify 2/8 to 1/4 before multiplying, giving (1 × 5)/(3 × 4) = 5/12 directly. This 2-mark question rewards the multiplication step and the correct simplified answer.

• Finds ⅕ of 85 = 17 (1m)
• Subtracts to get sale price: 85 - 17 = 68 (1m)

The question says prices are reduced BY 1/5, meaning the discount is 1/5 of £85. Find the discount: 85 ÷ 5 = £17. Then the sale price is the original minus the discount: £85 - £17 = £68. An alternative method is to recognise that paying after a 1/5 reduction means you pay 4/5 of the original: 4/5 × £85 = (4 × 85) ÷ 5 = 340 ÷ 5 = £68. Both approaches earn full marks. The most common mistake is finding £17 and stopping — that is the reduction amount, not the sale price. Always re-read the question to check what is being asked.

• Explains that addition requires same sized parts/common denominator to combine (1m)
• Explains that multiplication works by multiplying tops and bottoms separately (1m)

This question tests your understanding of why the rules differ. When adding fractions, you are combining parts of the same whole — to do this fairly, the parts must be the same size, which requires a common denominator. For example, 1/3 + 1/4 must become 4/12 + 3/12 before you can add. When multiplying fractions, you are finding a fraction of a fraction — for instance, 1/3 × 1/4 means 'one third of one quarter'. This involves scaling, not combining, so you simply multiply tops together and bottoms together. No common denominator is needed. For 2 marks you need to explain both operations with the underlying reason for each rule.

To compare fractions, convert them to the same denominator or to decimals. Using decimals: ⅔ ≈ 0.667 ¾ = 0.75 ⅗ = 0.6 ⅝ = 0.625 The largest is 0.75, which is ¾. Alternatively, using common denominator (LCM of 3,4,5,8 = 120): ⅔ = 80/120 ¾ = 90/120 (largest) ⅗ = 72/120 ⅝ = 75/120

To find the equivalent fraction, simplify 18/24 by dividing both numerator and denominator by their HCF. Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 HCF = 6 18 ÷ 6 = 3 24 ÷ 6 = 4 So 18/24 = ¾

To divide by a fraction, you multiply by its RECIPROCAL. The reciprocal of ⅘ is 5/4 (flip the fraction). So ⅗ ÷ ⅘ = ⅗ × 5/4 You can verify: ⅗ × 5/4 = 15/20 = ¾, and ⅗ ÷ ⅘ = (3÷4)/(5÷5) = 0.75/1 = ¾ ✓

• Expands first two brackets to x² + 3x + 2 (1m)
• Multiplies by third bracket (x+3) (1m)
• Expands to get all six terms (1m)
• Correct answer x³ + 6x² + 11x + 6 (1m)

Triple brackets must be tackled in two stages. Stage 1: expand any two brackets first. (x + 1)(x + 2) = x² + 3x + 2. Stage 2: multiply this quadratic by the third bracket. (x² + 3x + 2)(x + 3) — multiply each term by x: x³ + 3x² + 2x, then by 3: 3x² + 9x + 6. Collecting all terms: x³ + (3x² + 3x²) + (2x + 9x) + 6 = x³ + 6x² + 11x + 6. The four marks are: expanding the first two brackets (M1), multiplying by the third (M1), expanding all six terms (M1), and the final simplified answer (A1).

• Expands first bracket to 2x + 6 (1m)
• Expands second bracket to 3x - 3 (1m)
• Simplifies to 5x + 3 (1m)

When a question says 'expand and simplify', you must do both steps. Step 1 — expand each bracket separately: 2(x + 3) = 2x + 6 and 3(x − 1) = 3x − 3. Step 2 — collect like terms: x-terms: 2x + 3x = 5x; constants: 6 − 3 = 3. Final answer: 5x + 3. The three marks are one for each bracket expansion (M1, M1) and one for the simplified answer (A1). A common mistake is leaving the answer as 2x + 6 + 3x − 3 without collecting — that earns 2 marks but not the third.

• Correctly expands at least one squared bracket (1m)
• Subtracts the two expansions with correct sign changes (1m)
• Simplifies to 10x - 5 (1m)

This 'show that' question requires expanding two perfect squares and subtracting. Step 1: (x + 2)² = x² + 4x + 4. Step 2: (x − 3)² = x² − 6x + 9. Step 3: subtract, being careful with signs — the subtraction applies to every term in the second bracket: (x² + 4x + 4) − (x² − 6x + 9) = x² + 4x + 4 − x² + 6x − 9. Step 4: collect: x² terms cancel (x² − x² = 0); x terms: 4x + 6x = 10x; constants: 4 − 9 = −5. Result: 10x − 5. The three marks are: expanding at least one bracket correctly (M1), subtracting with correct sign changes (M1), and simplifying to 10x − 5 (A1). The most common error is writing −(x² − 6x + 9) = −x² − 6x + 9 (not changing all signs).

• Correctly multiplies to get -6x (1m)
• Correct answer -6x + 10 or 10 - 6x (1m)

Expanding −2(3x − 5) requires multiplying −2 by each term inside the bracket. First: −2 × 3x = −6x (negative times positive = negative). Second: −2 × (−5) = +10 (negative times negative = positive). Result: −6x + 10. The two marks are for getting −6x (M1) and the full correct answer −6x + 10 (A1). The most common error is writing −6x − 10, which comes from forgetting that the second multiplication involves two negatives. Always write each multiplication separately and track the signs.

• Gets x² term (1m)
• Correct answer x² + 5x + 6 (1m)

Expanding double brackets requires every term in the first bracket to multiply every term in the second. Use FOIL: First (x × x = x²), Outer (x × 3 = 3x), Inner (2 × x = 2x), Last (2 × 3 = 6). This gives x² + 3x + 2x + 6. Collect the middle terms: 3x + 2x = 5x. Final answer: x² + 5x + 6. The two marks are for getting the x² term from multiplying the first terms (M1) and the complete correct answer (A1). The classic mistake is only multiplying the first and last pairs (giving x² + 6) and missing the cross-terms.

• Correctly finds all four FOIL products (2x², 8x, x, 4) (1m)
• Correct answer 2x² + 9x + 4 (1m)

Expanding (2x + 1)(x + 4) uses FOIL with a coefficient on the first term. First: 2x × x = 2x² (note: 2x squared is 2x², not x²). Outer: 2x × 4 = 8x. Inner: 1 × x = x. Last: 1 × 4 = 4. This gives 2x² + 8x + x + 4. Collect: 8x + x = 9x. Final answer: 2x² + 9x + 4. The two marks are for finding all four products correctly (M1) and the simplified answer (A1). The key mistake is writing 2x × x = x² rather than 2x², forgetting to include the coefficient in the multiplication.

• Correctly finds all four FOIL products (x², 2x, -5x, -10) (1m)
• Correct answer x² - 3x - 10 (1m)

Expanding (x − 5)(x + 2) with FOIL: First: x × x = x². Outer: x × 2 = 2x. Inner: −5 × x = −5x. Last: −5 × 2 = −10. Collect the middle terms: 2x + (−5x) = −3x. Final answer: x² − 3x − 10. The two marks are for finding all four products correctly (M1) and the full correct answer (A1). Sign errors are the main risk — writing the middle term as +3x (forgetting −5x is larger) or writing the constant as +10 (forgetting −5 × 2 = −10, not +10).

• Expands as (x+5)(x+5) or uses formula correctly (1m)
• Correct answer x² + 10x + 25 (1m)

A perfect square (x + 5)² must be expanded as (x + 5)(x + 5) — never just squaring each term separately. Expanding: x² + 5x + 5x + 25 = x² + 10x + 25. The middle term is 10x because you get two lots of 5x. Alternatively, use the identity (a + b)² = a² + 2ab + b² with a = x and b = 5: x² + 2(x)(5) + 25 = x² + 10x + 25. The two marks are for correctly treating it as double brackets or using the formula (M1) and the final answer (A1). The critical error is writing (x + 5)² = x² + 25, which misses the middle term entirely.

• Gets 9x² and 4 correct (1m)
• Correct answer 9x² - 12x + 4 (1m)

Use the identity (a − b)² = a² − 2ab + b² with a = 3x and b = 2. a² = (3x)² = 9x² (square both the coefficient and the variable). 2ab = 2 × 3x × 2 = 12x. b² = 2² = 4. So (3x − 2)² = 9x² − 12x + 4. The two marks are for getting 9x² and +4 correct (M1) and the full answer (A1). Two common errors: (1) writing (3x)² = 3x² instead of 9x², forgetting to square the 3; (2) getting the middle term as −6x (computing only 3x × 2 = 6x, forgetting the factor of 2 in 2ab).

• Uses difference of squares or expands correctly (1m)
• Shows equals x² - 16 (1m)

This is a 'show that' question — you must provide clear working starting from the left-hand side and reaching the right. Recognise (x + 4)(x − 4) as the difference of two squares pattern (a + b)(a − b). Applying the identity: a² − b² = x² − 4² = x² − 16. Alternatively, expand fully using FOIL: x² − 4x + 4x − 16 = x² − 16 (the middle terms cancel because they are +4x and −4x). The marks are for showing the method (M1) and reaching x² − 16 (A1). In 'show that' questions, you must show the working — stating the answer alone earns no marks.

• Uses difference of squares to get (2x)² - 3² (1m)
• Correct answer 4x² - 9 (1m)

Recognise (2x + 3)(2x − 3) as the difference of two squares: (a + b)(a − b) = a² − b² where a = 2x and b = 3. Apply: (2x)² − 3² = 4x² − 9. The key step is computing (2x)² correctly: you must square both parts, giving 2² × x² = 4x². The two marks are for recognising the pattern and applying it correctly (M1) and the final answer 4x² − 9 (A1). The common mistake is writing (2x)² = 2x² (only squaring the variable, not the coefficient).

Multiply 3 by each term inside the bracket: 3 × x = 3x and 3 × 2 = 6, giving 3x + 6.

Expand: x² + (-2x) + 3x + (-6) = x² + x - 6. The middle terms are -2x + 3x = x.

The difference of two squares states that (a+b)(a-b) = a² - b². When you expand, the middle terms +ab and -ab cancel out.

• Identifies LCD = 6 and correctly multiplies to clear fractions (2(2x-1) + 3(x+4) = 30 or equivalent) (1m)
• Correct expansion: 4x - 2 + 3x + 12 = 30 or 7x + 10 = 30 seen (1m)
• 7x = 20 seen (1m)
• x = 20/7 (accept exact decimal 2.857... or fraction 20/7) (1m)

To solve (2x − 1)/3 + (x + 4)/2 = 5 with two fractions, find the lowest common denominator (LCD) of 3 and 2, which is 6. Multiply every term by 6: 2(2x − 1) + 3(x + 4) = 30. Expand: 4x − 2 + 3x + 12 = 30. Collect: 7x + 10 = 30. Subtract 10: 7x = 20. Divide by 7: x = 20/7. The four marks are: clearing fractions correctly (M1), correct expansion giving 7x + 10 = 30 (M1), 7x = 20 (M1), and x = 20/7 (A1). The critical step is multiplying every term — including the right-hand side — by the LCD. Forgetting to multiply the 5 by 6 gives 30 instead of the needed 30, so it actually works here, but always multiply both sides.

• Bracket correctly expanded: 6x + 3 seen (1m)
• 6x = 18 (or equivalent intermediate step) (1m)
• x = 3 (1m)

Solving 3(2x + 1) = 21 is a three-step process. Step 1: expand the bracket — 3 × 2x + 3 × 1 = 6x + 3, giving 6x + 3 = 21. Step 2: subtract 3 from both sides: 6x = 18. Step 3: divide by 6: x = 3. The three marks reward each step: expansion (B1), isolation of 6x (M1), and x = 3 (A1). An alternative approach is dividing 21 by 3 first to give 2x + 1 = 7, then solving the simpler equation. Both methods earn full marks.

• Begins to collect x-terms: 3x seen on one side (1m)
• Reaches 3x = 12 with both x-terms and constants collected (1m)
• x = 4 (1m)

When x appears on both sides, collect all x-terms to one side. From 5x + 3 = 2x + 15: subtract 2x from both sides to get 3x + 3 = 15. Then subtract 3 from both sides: 3x = 12. Divide by 3: x = 4. The three marks are for collecting x-terms (M1), isolating the x-term (M1), and x = 4 (A1). Check: 5(4) + 3 = 23 and 2(4) + 15 = 23. Always move x-terms before constants to keep working clean.

• Both brackets correctly expanded: 6x - 8 and 5x - 5 both seen (1m)
• Correct method to collect x-terms (e.g. x - 8 = -5 or equivalent) (1m)
• x = 3 (1m)

Solving 2(3x − 4) = 5(x − 1) with brackets on both sides. Step 1: expand both brackets — left: 6x − 8; right: 5x − 5. Equation: 6x − 8 = 5x − 5. Step 2: collect x-terms on the left — subtract 5x from both sides: x − 8 = −5. Step 3: add 8 to both sides: x = 3. The three marks reward: expanding both brackets correctly (M1), collecting x-terms to one side (M1), and x = 3 (A1). The key step is getting all x-terms to one side before isolating x.

• Collects x-terms to get -2x or 2x with correct sign on both sides (1m)
• -2x = -3 or 2x = 3 seen (1m)
• x = 3/2 or x = 1.5 (1m)