Cambridge International A Level Mathematics · 9709
Probability & Statistics 2 is Paper 6 in the current syllabus. Use this page to choose a matching question paper and mark scheme, practise the methods behind the questions, and turn errors into a focused revision plan.
The preserved archive below contains 82 question-paper / scheme rows from 2010–2022, including the 2020 specimen. One mismatched scheme and one unavailable specimen link are flagged and explained below. It is a historical collection, not a complete list of the latest papers. Read the current-syllabus notes before using older Paper 7 questions.
Choose the right Statistics 2 paper
For examinations in 2026–2027, Paper 6 lasts 1 hour 15 minutes, carries 50 marks, and contributes 20% of the full A Level. It belongs to the pure-mathematics-and-statistics route. It is available only as part of the full A Level, not at AS Level. Check your entered combination with your teacher rather than choosing components from the title alone.
The archive labels “Series 1”, “Series 2” and “Series 3” mean paper variants. For example, 62 is Paper 6, variant 2. A mark scheme for 61 does not match a question paper for 62. The session, year and component must all agree.
Historical-number warning: S2 was Paper 7 before the 2020 syllabus change. Older papers are useful topic practice, but their numbering and syllabus coverage must be checked. Current Paper 5 is Statistics 1; it is not interchangeable with S2.
Build the prerequisite skills with our Statistics 1 past-paper collection and Pure Mathematics 3 collection. S2 assumes S1 knowledge and the relevant Pure 3 calculus.
Five methods to recognise before calculating
| Topic | First decision | Check before finishing |
|---|---|---|
| Poisson counts | Define the count and its time or area interval; rescale the mean to that interval. | Count is a non-negative integer; tail direction matches “at least”, “more than” or “at most”. |
| Linear combinations | Write the new variable, including every coefficient and constant. | Square coefficients in the variance; state independence when adding variances. |
| Continuous variables | Write the support and check the density integrates to 1. | Integrate over the requested interval; a density height is not a probability. |
| Sampling and estimation | Separate a population spread from the standard error of a sample mean. | Use the appropriate variance estimate and conditions; interpret a confidence interval correctly. |
| Hypothesis tests | State hypotheses for a population parameter and choose the tail from the claim. | Compare like quantities: p-value with significance level, or statistic with critical value; conclude in context. |
Small errors that cost method marks
- Subtracting variances: for independent variables, uncertainty adds even in a difference. A negative variance signals an error.
- Using the old Poisson mean: a rate per hour must be converted before modelling a 15-minute count.
- Forgetting continuity correction: replace a discrete cutoff with the correct half-unit boundary when using a normal approximation.
- Dividing by n instead of n − 1: the unbiased sample-variance estimate uses n − 1 after centring on the sample mean.
- Writing “accept H₀” as proof: a non-significant result means insufficient evidence against the null at the chosen level, not proof that it is true.
- Dropping units: means and standard deviations have the original units; variances have squared units. Probabilities are unitless.
Six original worked examples
These questions are original HeLovesMath practice, not reproduced Cambridge examination questions. State a model and its assumptions before substituting numbers.
Notation: N(μ, σ²) gives a mean and variance; Po(λ) uses the mean count in the stated interval.
1. Poisson probability: scale the rate before finding a tail
Question. Messages arrive according to a Poisson process at an average rate of 12 per hour. Find the probability that at least 4 arrive in a 15-minute interval.
Model check. The arrival rate is constant, counts in disjoint intervals are independent, and simultaneous arrivals can be neglected.
- Let X count arrivals in 15 minutes. Match the mean to this interval: λ = 12 × (15/60) = 3, so X ~ Po(3).
- “At least 4” means X ≥ 4. Its complement is X ≤ 3, so P(X ≥ 4) = 1 − P(X ≤ 3).
- Using P(X = r) = e^(−λ)λʳ/r!, P(X ≥ 4) = 1 − e^(−3)[1 + 3 + 3²/2! + 3³/3!].
- The bracket is 13, giving 1 − 13e^(−3) = 0.352768… ≈ 0.353.
Watch out: Use λ = 3, not 12. For X ≥ 4, subtract P(X ≤ 3), not P(X ≤ 4).
2. Independent normal variables: square every variance coefficient
Question. Independent X ~ N(8, 4) and Y ~ N(5, 9) are given. Let T = 2X − 3Y + 7. Find the distribution of T and P(T > 10).
Model check. Both variables are normal and independent. The second parameter is variance, not standard deviation.
- E(T) = 2E(X) − 3E(Y) + 7 = 2(8) − 3(5) + 7 = 8.
- Independence gives Var(T) = 2²Var(X) + (−3)²Var(Y) = 4(4) + 9(9) = 97. The constant adds no variance.
- A linear combination of independent normal variables is normal, so T ~ N(8, 97). Its standard deviation is √97.
- Standardise: P(T > 10) = P[Z > (10 − 8)/√97] = P(Z > 0.203069…).
- The probability is 0.419540… ≈ 0.420.
Watch out: The negative coefficient does not subtract variance. Also, 2X repeats the same random value: Var(2X) = 4Var(X), whereas two independent copies have Var(X₁ + X₂) = 2Var(X).
3. A probability density: normalise, then calculate moments
Question. X has density f(x) = kx(2 − x) for 0 ≤ x ≤ 2 and zero otherwise. Find k, E(X), Var(X) and P(1/2 < X < 3/2).
Model check. A density is non-negative and has total area 1. Here x(2 − x) is non-negative on the given interval.
- Normalise: 1 = ∫₀² kx(2 − x) dx = k[x² − x³/3]₀² = 4k/3. Therefore k = 3/4.
- E(X) = (3/4)∫₀² x²(2 − x) dx = (3/4)[2x³/3 − x⁴/4]₀² = 1.
- E(X²) = (3/4)∫₀² x³(2 − x) dx = (3/4)[x⁴/2 − x⁵/5]₀² = 6/5.
- Var(X) = E(X²) − [E(X)]² = 6/5 − 1 = 1/5.
- P(1/2 < X < 3/2) = (3/4)[x² − x³/3] from 1/2 to 3/2 = 11/16 = 0.6875.
Watch out: E(X²) is not the variance. A density height is not a point probability; for a continuous variable P(X = a) = 0, so including interval endpoints does not change this probability.
4. A confidence interval for a population mean
Question. Package masses are normally distributed with unknown mean μ and known standard deviation 6 g. An independent random sample of 64 packages has mean 51.2 g. Find a 95% confidence interval for μ, then the minimum sample size for a margin of error no greater than 1 g at the same confidence level.
Model check. The sample is random and independent; the population is normal with known standard deviation.
- The standard error is σ/√n = 6/√64 = 0.75 g.
- A two-sided 95% interval leaves 2.5% in each normal tail. Use z = 1.96.
- x̄ ± zσ/√n = 51.2 ± 1.96(0.75) = 51.2 ± 1.47, giving (49.73, 52.67) g.
- Interpretation: over many independent samples of the same size, about 95% of intervals constructed this way would contain the fixed population mean.
- For margin of error at most 1 g, require 1.96(6)/√n ≤ 1. Thus n ≥ (11.76)² = 138.2976. Round up: n = 139.
Watch out: This interval estimates the mean; it does not contain 95% of individual package masses. If variance is unknown, a large-sample approximate interval can use the unbiased sample variance; this justification does not automatically extend to a small non-normal sample.
5. A Poisson hypothesis test and its error probabilities
Question. A machine historically produces faults at a Poisson rate of 4 per week. After maintenance, 3 faults occur during 2 weeks. Test at the 5% level whether the weekly fault rate has decreased. Find the critical region and Type I error probability. Find the Type II error probability if the true new rate is 2 per week.
Model check. The Poisson rate is constant over the observation period. The one-sided alternative is chosen because a decrease is the claim being tested, before inspecting the result.
- Let λ be the population mean number of faults per week. State H₀: λ = 4 and H₁: λ < 4.
- Let X count faults during the full 2 weeks. Under H₀, X ~ Po(8). Small values support H₁.
- The lower-tail probability is P(X ≤ 3 | H₀) = e^(−8)[1 + 8 + 8²/2! + 8³/3!] = 0.0423801….
- Since 0.0423801 < 0.05, reject H₀. At the 5% significance level, there is sufficient evidence that the mean weekly fault rate has decreased after maintenance. This alone does not establish maintenance as the cause.
- P(X ≤ 3) = 0.0423801 ≤ 0.05, but P(X ≤ 4) = 0.0996324 > 0.05. Therefore the critical region is X ≤ 3.
- A Type I error is concluding that the fault rate decreased when it is still 4 per week. Its probability is P(X ≤ 3 | λ = 4) = 0.0423801 ≈ 4.24%. Because counts are discrete, this is less than the nominal 5%.
- If the true rate is 2 per week, then X ~ Po(4) over 2 weeks. A Type II error means failing to reject H₀ despite that decrease, occurring when X ≥ 4. Its probability is 1 − e^(−4)[1 + 4 + 4²/2! + 4³/3!] = 0.5665299… ≈ 0.567.
Watch out: Use a tail, not P(X = 3). A Type II probability requires a specified alternative rate. Failure to reject H₀ would mean insufficient evidence of a decrease, not proof that the rate is unchanged.
6. Normal approximation: correct the boundary first
Question. X ~ Po(25) counts arrivals in one hour. Use a normal approximation to estimate P(X ≥ 35).
Model check. The mean 25 is sufficiently large for the usual syllabus approximation guideline, approximately greater than 15. Tail probabilities can still have noticeable approximation error.
- A Poisson variable has mean and variance 25. Use Y ~ N(25, 25), with standard deviation 5.
- The discrete event includes 35, 36, 37, …. Its lower continuous boundary is 34.5: P(X ≥ 35) ≈ P(Y > 34.5).
- Standardise the corrected boundary: z = (34.5 − 25)/5 = 1.9.
- P(X ≥ 35) ≈ 1 − Φ(1.9) = 0.0287166… ≈ 0.0287.
- As a check, the exact Poisson probability is approximately 0.0338. An approximation need not agree with the exact probability to three significant figures; use the method the question requests.
Watch out: For X ≥ 35 use 34.5, not 35.5. Divide by the standard deviation 5, not the variance 25.
Eight questions to try before revealing the answers
Write your working first. The answers include the decision that matters, so you can diagnose a setup error as well as an arithmetic slip.
Practice 1: rate and tail
Arrivals follow a Poisson process at 6 per hour. Find the probability of at least 2 arrivals in 20 minutes.
Show answer and method
λ = 6 × 20/60 = 2; P(X ≥ 2) = 1 − e^(−2)(1 + 2) = 1 − 3e^(−2) ≈ 0.594. The model assumes a constant rate and independent counts in disjoint intervals.
Practice 2: binomial-to-Poisson approximation
X ~ B(200, 0.01). Use a Poisson approximation to estimate P(X = 2), and explain why it is appropriate.
Show answer and method
Use Y ~ Po(np) = Po(2). P(X = 2) ≈ e^(−2)2²/2! = 2e^(−2) ≈ 0.271. Here n = 200 is large, p = 0.01 is small, and np = 2 < 5. The binomial model itself requires independent trials with a constant success probability.
Practice 3: independent linear combinations
Independent X ~ Po(3) and Y ~ Po(5) are given. Find E(2X − Y), Var(2X − Y) and the distribution of X + Y.
Show answer and method
E(2X − Y) = 2(3) − 5 = 1; Var(2X − Y) = 2²(3) + (−1)²(5) = 17; X + Y ~ Po(8). Independence is essential for the variance sum and Poisson-sum result; 2X is not Po(6).
Practice 4: density and median
X has density f(x) = 2x for 0 ≤ x ≤ 1 and zero otherwise. Find P(X > 3/4), E(X), Var(X) and the median.
Show answer and method
P(X > 3/4) = [x²] from 3/4 to 1 = 7/16. E(X) = ∫₀¹2x² dx = 2/3. E(X²) = 1/2, so Var(X) = 1/2 − 4/9 = 1/18. The median satisfies ∫₀ᵐ2x dx = 1/2, giving m = 1/√2.
Practice 5: unbiased estimates and sampling
An independent random sample of size 10 has Σx = 50 and Σx² = 290. Estimate the population mean and variance without bias. Why might taking the first 10 items from a daily production run be unsuitable instead?
Show answer and method
The mean estimate is x̄ = 5. The unbiased variance estimate is s² = [290 − 50²/10]/9 = 40/9 ≈ 4.44. The first items might reflect start-up conditions and under-represent later production; randomly selecting across the target production run helps avoid this selection bias.
Practice 6: confidence interval for a proportion
In an independent random sample of 400 customers, 240 choose option A. Find an approximate 95% confidence interval for the population proportion choosing A.
Show answer and method
p̂ = 240/400 = 0.6. The interval is 0.6 ± 1.96√[0.6(0.4)/400] = (0.55199…, 0.64801…), approximately (0.552, 0.648). The success and failure counts, 240 and 160, are both large; the random sample must represent the target customer population.
Practice 7: normal mean test
Packet masses are normal with known standard deviation 10 g. An independent random sample of 25 packets has mean 95 g. Test H₀: μ = 100 against H₁: μ < 100 at the 1% level.
Show answer and method
Standard error = 10/√25 = 2 g; z = (95 − 100)/2 = −2.5. The one-sided p-value is Φ(−2.5) ≈ 0.00621 < 0.01, so reject H₀. There is sufficient evidence at the 1% level that the population mean packet mass is below 100 g. Population normality and known variance justify this calculation even though the sample is not large.
Practice 8: normal approximation to a binomial
X ~ B(100, 0.4). Use a normal approximation to estimate P(X ≤ 45), checking the conditions.
Show answer and method
np = 40 and n(1 − p) = 60 are both greater than 5. Use Y ~ N(40, 24). With continuity correction, P(X ≤ 45) ≈ P(Y < 45.5) = Φ[(45.5 − 40)/√24] = Φ(1.12268…) ≈ 0.869.
A practical paper-review routine
- Choose: match year, session and full component code with its scheme. Use a recent available paper aligned to your exam syllabus for a timed attempt; reserve older papers for targeted practice.
- Attempt: allow 75 minutes for a current-format full paper. Around 1.5 minutes per mark is a planning guide, not a rule to stop mid-method.
- Mark: distinguish missing method, incorrect model, numerical slip and incomplete contextual conclusion. Do not award yourself method marks solely because the final number matches.
- Repair: redo the setup without looking at the solution, then solve a new problem on the same skill. Record the model, the missed step and one prevention check.
- Retest: revisit the question after a gap and explain why the method applies. A copied correction is not evidence that the method has stuck.
Frequently asked questions
Why are some Statistics 2 papers numbered 7?
Paper 7 is the historical S2 component used before 2020. Current S2 is Paper 6. Treat old questions as topic practice after comparing them with your exam-year syllabus.
Are the variants different difficulty levels?
The final digit identifies a variant. It is not a Foundation/Higher label or a ranking from easy to hard. Always match the full code when choosing the mark scheme.
Does this page include every recent paper?
No. The original collection covers 2010–2022 and the 2020 specimen. The official Cambridge past-paper page in Sources & References is the route to publicly available newer material; schools may have additional authorised resources.
Will a high score here guarantee my A-level grade?
No. This is one component, and grades depend on the relevant combination, series and grade thresholds. Use the score to identify skills to improve rather than convert it into a promised final grade.
Sources & References
Official syllabus and current public resources
Check the syllabus for your examination year. Cambridge’s public past-paper page provides a selection; registered schools can access additional authorised materials. From March 2026, some paper layouts change for accessibility, while assessment content, demand and question types are unchanged.
- Cambridge 9709 qualification and syllabus hub
- 2026–2027 Mathematics syllabus, version 4
- Official Mathematics 9709 past papers and specimen materials
Preserved 2010–2022 paper archive
Choose a year to see its question papers and schemes. Third-party availability varies. Check the first-page code against the row before use. These links retain the original collection; older content may differ from the current syllabus.
Link check, 5 October 2026: 22 original entries were verified as PDF resources; 141 could not be confirmed because of timeouts or access limits. The old specimen scheme returned 404, so current official specimen links are supplied. The November 2017 variant-73 row originally linked to a variant-72 scheme; that mismatched legacy link is clearly flagged. Availability checks do not validate every answer or every paper’s syllabus fit.
2022 · 4 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| June 2022 | 61 | Question PDFAvailability unverified | Mark scheme PDF |
| June 2022 | 62 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2022 | 63 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| March 2022 | 62 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2021 · 7 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| Nov 2021 | 61 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| Nov 2021 | 62 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| Nov 2021 | 63 | Question PDFAvailability unverified | Mark scheme PDF |
| June 2021 | 61 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2021 | 62 | Question PDF | Mark scheme PDF |
| June 2021 | 63 | Question PDF | Mark scheme PDFAvailability unverified |
| March 2021 | 62 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2020 · 8 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| Nov 2020 | 61 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| Nov 2020 | 62 | Question PDFAvailability unverified | Mark scheme PDF |
| Nov 2020 | 63 | Question PDF | Mark scheme PDF |
| June 2020 | 61 | Question PDF | Mark scheme PDFAvailability unverified |
| June 2020 | 62 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2020 | 63 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| March 2020 | 62 | Question PDFAvailability unverified | Mark scheme PDF |
| Specimen 2020 | Specimen 9709/06 | Current official specimen QP Original QP URL: availability unverified | Current official specimen scheme Old scheme link: unavailable (404) |
2019 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2019 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2019 | 73 | Question PDFAvailability unverified | Mark scheme PDF |
| June 2019 | 71 | Question PDF | Mark scheme PDF |
| June 2019 | 72 | Question PDF | Mark scheme PDF |
| June 2019 | 73 | Question PDF | Mark scheme PDFAvailability unverified |
| March 2019 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2018 · 7 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2018 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2018 | 72 | Question PDFAvailability unverified | Mark scheme PDF |
| November 2018 | 73 | Question PDF | Mark scheme PDF |
| June 2018 | 71 | Question PDF | Mark scheme PDF |
| June 2018 | 72 | Question PDF | Mark scheme PDFAvailability unverified |
| June 2018 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| March 2018 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2017 · 7 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2017 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2017 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2017 | 73 | Question PDFAvailability unverified | Legacy variant-72 link: does not match 73Matching variant 73 scheme is not verified here. Check an authorised source. |
| June 2017 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2017 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2017 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| March 2017 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2016 · 7 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2016 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2016 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2016 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2016 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2016 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2016 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| March 2016 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2015 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2015 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2015 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2015 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2015 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2015 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2015 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2014 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2014 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2014 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2014 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2014 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2014 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2014 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2013 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2013 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2013 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2013 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2013 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2013 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2013 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2012 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2012 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2012 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2012 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2012 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2012 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2012 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2011 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2011 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2011 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2011 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2011 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2011 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2011 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
2010 · 6 question-paper / scheme pairs
| Session | Component | Question paper | Mark scheme |
|---|---|---|---|
| November 2010 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2010 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| November 2010 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2010 | 71 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2010 | 72 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
| June 2010 | 73 | Question PDFAvailability unverified | Mark scheme PDFAvailability unverified |
This independent guide is not endorsed by Cambridge International. The worked questions, explanations and diagrams are original HeLovesMath material. Examination papers and schemes remain the work of their respective owners.
