This page includes Mathematics Grade 12 past papers and memos.
Due to the demanding and challenging nature of Grade 12 Mathematics, we have collected more papers for this subject, more than for any other subject on this website.
Paper 1 Mathematics Grade 12 Exam Question Paper Topics and Structure:
| Topic | Focus | Marks |
| Algebra, Equations and Inequalities | Equations and expressions | 25 |
| Number Patterns | Patterns and sequences | 25 |
| Functions and Graphs | Graphs and functions | 35 |
| Finance, Growth and Decay | Financial calculations | 15 |
| Differential Calculus | Calculus applications | 35 |
| Counting Principle and Probability | Probability theory | 15 |
| TOTAL (3 Hours) | All Topics Combined | 150 |
Paper 2 Mathematics Grade 12 Exam Question Paper Topics and Structure:
| Topic | Focus | Marks |
| Statistics and Regression | Statistics and regression | 20 |
| Analytical Geometry | Coordinate geometry | 40 |
| Trigonometry | Trigonometric ratios and graphs | 50 |
| Euclidean Geometry | Circle geometry and theorems | 40 |
| TOTAL (3 Hours) | All Topics Combined | 150 |
Exam facts from the DBE November 2025 question papers. See the full exam guide.
Mathematics Grade 12 Papers
For each of the Mathematics Grade 12 exam paper Google Drive folders below, we have included Paper 1 and Paper 2 question papers, in both English and Afrikaans, together with their memos. The papers are organised by years and terms for easy access:
Exam PracticeMathematics: Most Common Exam Questions Quiz
Real questions from DBE NSC November Mathematics papers (2023 to 2025). Pick an answer, check the official memo steps and see where marks are won and lost.
37
Exam Questions
9
Topics
2
Papers
3
NSC Years (2023 to 2025)
NSC November 2024 · Paper 1 · Question 1.1.2
Memo: step-by-step solution (4 marks)
Answer: \(x = 1{,}71\) or \(x = 0{,}29\)
| \[2x^2 - 4x + 1 = 0\] | ✓ standard form |
| \[x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}\] | ✓ substitution into correct formula |
| \[x = 1{,}71 \quad \text{or} \quad x = 0{,}29\] | ✓✓ answers |
Exam tip: Most learners lose a mark by not moving 4x to the left first, so b is taken as +4 instead of -4.
NSC November 2024 · Paper 1 · Question 1.1.3
Memo: step-by-step solution (4 marks)
Answer: \(x < -1\) or \(x > 3\)
| \[(x - 3)(x + 1) > 0\] | ✓ factors |
| \[\text{Critical values: } x = -1 \text{ and } x = 3\] | ✓ critical values |
| \[x < -1 \quad \text{or} \quad x > 3\] | ✓✓ notation and answer |
Exam tip: "Greater than zero" means the two outside regions of a happy (positive) parabola.
NSC November 2024 · Paper 1 · Question 1.1.4
Memo: step-by-step solution (5 marks)
Answer: \(x = 3\)
| \[(2^x)^2 - 4 \cdot 2^x - 32 = 0\] | ✓ \(2^{x+2} = 4 \cdot 2^x\) |
| \[(2^x - 8)(2^x + 4) = 0\] | ✓ factors |
| \[2^x = 8 = 2^3 \;\Rightarrow\; x = 3\] | ✓✓ |
| \[2^x = -4 \;\text{ has no solution}\] | ✓ rejecting |
Exam tip: Always reject the negative value: an exponential like 2^x can never be negative.
NSC November 2025 · Paper 1 · Question 1.1.4
Memo: step-by-step solution (4 marks)
Answer: \(x = -1\) or \(x = 2\)
| \[(2 \cdot 2^x - 1)(2^x - 4) = 0\] | ✓ factors |
| \[2^x = \tfrac{1}{2} = 2^{-1} \;\Rightarrow\; x = -1\] | ✓✓ |
| \[2^x = 4 = 2^{2} \;\Rightarrow\; x = 2\] | ✓ |
Exam tip: Option B is the classic trap: those are the values of 2^x, not of x.
NSC November 2024 · Paper 1 · Question 1.1.5
Memo: step-by-step solution (4 marks)
Answer: \(x = 0\)
| \[\sqrt{-2x + 4} = x + 2\] | ✓ isolating the surd |
| \[-2x + 4 = x^2 + 4x + 4\] | ✓ squaring both sides |
| \[x^2 + 6x = 0 \;\Rightarrow\; x(x + 6) = 0\] | ✓ factors |
| \[x = 0 \quad (x = -6 \text{ is rejected: } \sqrt{16} + 6 = 10 \neq 2)\] | ✓ answer with checking |
Exam tip: Squaring creates extra answers. Substitute back, or you will lose the final mark.
NSC November 2024 · Paper 1 · Question 1.2
Memo: step-by-step solution (5 marks)
Answer: \(x = 1,\ y = 1\) or \(x = \tfrac{7}{2},\ y = -4\)
| \[y = 3 - 2x\] | ✓ eq 1 |
| \[(3 - 2x)^2 + (3 - 2x)x = 2\] | ✓ substitution |
| \[2x^2 - 9x + 7 = 0 \;\Rightarrow\; (2x - 7)(x - 1) = 0\] | ✓ standard form |
| \[x = \tfrac{7}{2} \text{ or } x = 1\] | ✓ both x-values |
| \[y = -4 \text{ or } y = 1\] | ✓ both y-values |
NSC November 2023 · Paper 1 · Question 2.1.1
Memo: step-by-step solution (3 marks)
Answer: \(457\)
| \[T_n = a + (n-1)d\] | ✓ formula |
| \[T_{91} = 7 + (91 - 1)(5)\] | ✓ substitution |
| \[T_{91} = 457\] | ✓ answer |
Exam tip: The most common error is using n instead of (n - 1), which gives 462.
NSC November 2024 · Paper 1 · Question 2.1.1
Memo: step-by-step solution (2 marks)
Answer: \(1\,090\)
| \[S_{20} = \dfrac{20}{2}\left[2(7) + (20 - 1)(5)\right]\] | ✓ substitution |
| \[S_{20} = 1\,090\] | ✓ answer |
NSC November 2023 · Paper 1 · Question 3.1.2
Memo: step-by-step solution (4 marks)
Answer: \(k = 15\)
| \[a = 3, \quad r = 2\] | ✓ a and r |
| \[\dfrac{3\left(2^k - 1\right)}{2 - 1} = 98\,301\] | ✓ substitution into S_n |
| \[2^k - 1 = 32\,767 \;\Rightarrow\; 2^k = 32\,768 = 2^{15}\] | ✓ simplification |
| \[k = 15\] | ✓ answer |
NSC November 2025 · Paper 1 · Question 2.1.3
Memo: step-by-step solution (2 marks)
Answer: \(\dfrac{16}{3}\)
| \[a = 8, \quad r = \dfrac{-4}{8} = -\dfrac{1}{2}\] | ✓ a and r |
| \[S_\infty = \dfrac{8}{1 - \left(-\frac{1}{2}\right)} = \dfrac{16}{3}\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 2.2.1
Memo: step-by-step solution (3 marks)
Answer: \(9\,437\)
| \[\text{First differences: } T_n = 2n - 1\] | ✓ general term of first differences |
| \[T_{99} - T_{98} = 2(98) - 1 = 195\] | ✓ 98th first difference |
| \[T_{98} = 9\,632 - 195 = 9\,437\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 3.3
Memo: step-by-step solution (4 marks)
Answer: The \(10^{\text{th}}\) circle
| \[T_n = 12\left(\tfrac{1}{2}\right)^{n-1}\] | ✓ general term of diameters |
| \[12\left(\tfrac{1}{2}\right)^{n-1} = \dfrac{3}{128}\] | ✓ equating |
| \[\left(\tfrac{1}{2}\right)^{n-1} = \dfrac{1}{512} = \left(\tfrac{1}{2}\right)^{9}\] | ✓ simplification |
| \[n - 1 = 9 \;\Rightarrow\; n = 10\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 4.1
Memo: step-by-step solution (2 marks)
Answer: \(a = \dfrac{2}{3}\)
| \[-\dfrac{5}{9} = a^2 - 1\] | ✓ substitution |
| \[a^2 = \dfrac{4}{9} \;\Rightarrow\; a = \dfrac{2}{3} \quad (a > 0)\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 5.2 – 5.3
Memo: step-by-step solution (4 marks)
Answer: \(y = -2\) and \(a = -3\)
| \[p = -1 \;(\text{vertical asymptote } x = 1)\] | ✓ p |
| \[q = 1 - 3 = -2 \;\Rightarrow\; y = -2\] | ✓✓ horizontal asymptote |
| \[1 = \dfrac{a}{0 - 1} - 2 \;\Rightarrow\; a = -3\] | ✓ substitution ✓ answer |
NSC November 2024 · Paper 1 · Question 6.1
Memo: step-by-step solution (3 marks)
Answer: \(B(2\,;\,9)\)
| \[x = -\dfrac{4}{2(-1)} = 2\] | ✓ x-value |
| \[f(2) = -(2)^2 + 4(2) + 5 = 9\] | ✓ substitution |
| \[B(2\,;\,9)\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 7.1
Memo: step-by-step solution (3 marks)
Answer: R14 706,56
| \[A = 5\,000\left(1 + \dfrac{0{,}068}{4}\right)^{64}\] | ✓ i ✓ n |
| \[A = \text{R}14\,706{,}56\] | ✓ answer |
Exam tip: R14 325,51 is the answer if you forget to convert to quarters, and R10 440 uses simple interest.
NSC November 2024 · Paper 1 · Question 7.2
Memo: step-by-step solution (2 marks)
Answer: \(12{,}5\%\) p.a.
| \[\tfrac{1}{2}P = P\left(1 - 4i\right)\] | ✓ substitution |
| \[i = 0{,}125 \;\Rightarrow\; 12{,}5\% \text{ p.a.}\] | ✓ answer |
Exam tip: 15,91% is the reducing-balance answer. Read which method the question asks for.
NSC November 2024 · Paper 1 · Question 7.3.1
Memo: step-by-step solution (2 marks)
Answer: R38 058,80
| \[\text{Total paid} = 60 \times \text{R}2\,300{,}98 = \text{R}138\,058{,}80\] | ✓ |
| \[\text{Interest} = \text{R}138\,058{,}80 - \text{R}100\,000 = \text{R}38\,058{,}80\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 8.1.2
Memo: step-by-step solution (4 marks)
Answer: \(g'(x) = -4x^{-3} - \dfrac{7}{3}x^{\frac{4}{3}}\)
| \[g(x) = 2x^{-2} - x^{\frac{7}{3}}\] | ✓ \(2x^{-2}\) ✓ \(-x^{7/3}\) |
| \[g'(x) = -4x^{-3} - \dfrac{7}{3}x^{\frac{4}{3}}\] | ✓ \(-4x^{-3}\) ✓ \(-\tfrac{7}{3}x^{4/3}\) |
NSC November 2024 · Paper 1 · Question 8.2
Memo: step-by-step solution (3 marks)
Answer: \(y = -2x + 3\)
| \[f'(x) = 3x^2 - 8x + 2 \;\Rightarrow\; m = f'(2) = -2\] | ✓ m = -2 |
| \[f(2) = (2)^3 - 4(2)^2 + 2(2) + 3 = -1\] | ✓ y = -1 |
| \[y + 1 = -2(x - 2) \;\Rightarrow\; y = -2x + 3\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 8.3.1
Memo: step-by-step solution (5 marks)
Answer: \(\displaystyle \lim_{h \to 0}\left(-12x - 6h\right) = -12x\)
| \[f(x+h) = -6x^2 - 12xh - 6h^2\] | ✓ |
| \[f'(x) = \lim_{h \to 0}\dfrac{-6x^2 - 12xh - 6h^2 + 6x^2}{h}\] | ✓ substitution |
| \[= \lim_{h \to 0}\dfrac{h(-12x - 6h)}{h}\] | ✓ simplification ✓ common factor |
| \[= \lim_{h \to 0}(-12x - 6h) = -12x\] | ✓ answer |
Exam tip: Marks are lost for dropping the lim notation before the last line, or for writing = -12x - 6h.
NSC November 2023 · Paper 1 · Question 7.3
Memo: step-by-step solution (3 marks)
Answer: \(-\dfrac{2}{\sqrt{3}} < x < \dfrac{2}{\sqrt{3}}\)
| \[f'(x) = -6x^2 + 8 > 0\] | ✓ f'(x) > 0 |
| \[x^2 < \dfrac{4}{3}\] | ✓ |
| \[-\dfrac{2}{\sqrt{3}} < x < \dfrac{2}{\sqrt{3}} \quad (\approx -1{,}15 < x < 1{,}15)\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 9.4
Memo: step-by-step solution (2 marks)
Answer: \(-9 < k < -8\)
| \[9 + k > 0 \;\Rightarrow\; k > -9\] | ✓ |
| \[8 + k < 0 \;\Rightarrow\; k < -8 \quad\therefore\; -9 < k < -8\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 10.1
Memo: step-by-step solution (3 marks)
Answer: \(27\) km/h
| \[s''(t) = -6t + 18 = 0 \;\Rightarrow\; t = 3\] | ✓ |
| \[s'(3) = -3(3)^2 + 18(3)\] | ✓ substitution |
| \[= 27 \text{ km/h}\] | ✓ answer |
NSC November 2024 · Paper 1 · Question 11.2
Memo: step-by-step solution (2 marks)
Answer: \(\dfrac{7}{20}\)
| \[x = 2 \;(\text{all three}), \quad M\text{ only} = 11, \quad G\text{ only} = 9\] | ✓ Venn diagram |
| \[\text{Total} = 11 + 6 + 9 + 5 + 4 + 3 + 2 = 40\] | |
| \[P(\text{at least two}) = \dfrac{4 + 2 + 5 + 3}{40} = \dfrac{14}{40} = \dfrac{7}{20}\] | ✓ numerator ✓ denominator |
NSC November 2024 · Paper 1 · Question 12.2
Memo: step-by-step solution (4 marks)
Answer: \(15\,390\)
| \[\text{1st letter: } 20 - 2 = 18 \text{ choices}\] | ✓ 18 |
| \[\text{last digit (odd): } 5 \text{ choices; } \text{2nd position digit: } 9\] | ✓ 5 ✓ 9 |
| \[\text{3rd position letter: } 19\] | ✓ 19 |
| \[18 \times 9 \times 19 \times 5 = 15\,390\] |
NSC November 2024 · Paper 2 · Question 1.1
x: 34, 32, 40, 27, 33, 28, 27, 55, 39, 44, 30, 57, 28, 32, 31 | y: 26, 21, 6, 20, 16, 26, 23, 7, 18, 14, 17, 8, 28, 25, 20
Determine the equation of the least squares regression line.
Memo: step-by-step solution (3 marks)
Answer: \(\hat{y} = 39{,}46 - 0{,}59x\)
| \[a = 39{,}46\] | ✓ a |
| \[b = -0{,}59\] | ✓ b |
| \[\hat{y} = 39{,}46 - 0{,}59x\] | ✓ equation |
Exam tip: The memo gives full marks for the correct answer only, so practise your calculator steps.
NSC November 2024 · Paper 2 · Question 3.1
Memo: step-by-step solution (2 marks)
Answer: \(-\dfrac{3}{4}\)
| \[m_{DC} = \dfrac{-9 - 0}{3 - (-9)}\] | ✓ substitution |
| \[m_{DC} = -\dfrac{3}{4}\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 3.4
Memo: step-by-step solution (2 marks)
Answer: \(15\) units
| \[DC = \sqrt{(3 + 9)^2 + (-9 - 0)^2}\] | ✓ substitution |
| \[DC = \sqrt{225} = 15\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 4.2
Memo: step-by-step solution (4 marks)
Answer: \(y = -\dfrac{1}{3}x + 10\)
| \[m_{MP} = \dfrac{9 - 3}{3 - 1} = 3\] | ✓ gradient of radius |
| \[m_{ST} = -\dfrac{1}{3}\] | ✓ gradient of tangent |
| \[y - 9 = -\dfrac{1}{3}(x - 3)\] | ✓ substitution |
| \[y = -\dfrac{1}{3}x + 10\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 4.4
Memo: step-by-step solution (2 marks)
Answer: \(d = 7\)
| \[d^2 + 1 - 2d - 6 - 30 = 0 \;\Rightarrow\; d^2 - 2d - 35 = 0\] | ✓ substitution |
| \[(d - 7)(d + 5) = 0 \;\Rightarrow\; d = 7 \quad (d > 0)\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 5.1.2
Memo: step-by-step solution (2 marks)
Answer: \(-\dfrac{7}{25}\)
| \[\cos 2A = 2\cos^2 A - 1\] | ✓ expansion |
| \[= 2\left(-\tfrac{3}{5}\right)^2 - 1 = -\dfrac{7}{25}\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 5.1.3
Memo: step-by-step solution (4 marks)
Answer: \(\dfrac{24}{25}\)
| \[\sin(A - B) = \sin A\cos B - \cos A\sin B\] | ✓ expansion |
| \[= \left(-\tfrac{4}{5}\right)\left(-\tfrac{3}{5}\right) - \left(-\tfrac{3}{5}\right)\left(\tfrac{4}{5}\right)\] | ✓ cos B ✓ substitution |
| \[= \dfrac{24}{25}\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 5.2
Memo: step-by-step solution (4 marks)
Answer: \(-\dfrac{p}{4}\)
| \[= \dfrac{1}{2}\cdot\dfrac{1}{2}\sin\left(2\left(\tfrac{\alpha}{2} - 45^\circ\right)\right)\] | ✓ double angle |
| \[= \dfrac{1}{4}\sin(\alpha - 90^\circ)\] | ✓ |
| \[= -\dfrac{1}{4}\cos\alpha\] | ✓ reduction |
| \[= -\dfrac{p}{4}\] | ✓ answer |
NSC November 2023 · Paper 2 · Question 5.1.2
Memo: step-by-step solution (3 marks)
Answer: \(-\dfrac{4\sqrt{2}}{9}\)
| \[\cos\beta = -\dfrac{2\sqrt{2}}{3}\] | ✓ sign ✓ value |
| \[\sin 2\beta = 2\sin\beta\cos\beta = 2\left(\tfrac{1}{3}\right)\left(-\tfrac{2\sqrt{2}}{3}\right)\] | ✓ expansion |
| \[= -\dfrac{4\sqrt{2}}{9}\] | ✓ answer |
NSC November 2024 · Paper 2 · Question 6.3.1
Memo: step-by-step solution (3 marks)
Answer: \(\dfrac{4}{3}\)
| \[g(x) = \dfrac{4(1 - 2\sin^2 x)}{3} = \dfrac{4}{3}\cos 2x\] | ✓ factor ✓ cos 2x |
| \[\text{Max of } \cos 2x = 1 \;\Rightarrow\; g_{\max} = \dfrac{4}{3}\] | ✓ answer |
NSC November 2025 · Paper 2 · Question 5.1.1
Memo: step-by-step solution (2 marks)
Answer: \(\dfrac{k}{\sqrt{1 + k^2}}\)
| \[\cos 40^\circ = \sin 50^\circ\] | ✓ co-function |
| \[\text{hyp} = \sqrt{k^2 + 1} \;\Rightarrow\; \sin 50^\circ = \dfrac{k}{\sqrt{1 + k^2}}\] | ✓ answer |
Tip: press A–D to answer multiple choice, → for the next question.
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2027
| Term | Download Links |
| Term 1 (March/April) | Papers coming soon |
| Term 2 (May/June) | Papers coming soon |
| Term 3 (August/September) | Papers coming soon |
| Term 4 (November) | Papers coming soon |
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Study Guides
- DBE Self Study Guide: Mathematics Analytical Geometry PDF, 1 MB
- DBE Self Study Guide: Mathematics Functions PDF, 956 KB
- DBE Self Study Guide: Mathematics Trigonometry PDF, 6.7 MB
- DBE Self Study Guide: Mathematics Literacy Book 1 PDF, 2.3 MB
- DBE Self Study Guide: Mathematics Literacy Book 2 PDF, 2.6 MB
- DBE Self Study Guide: Mathematics Literacy Book 3 PDF, 1.6 MB
- DBE Self Study Guide: Mathematics Teachers’ Self Study Guide Book 1 PDF, 1.2 MB
- DBE Self Study Guide: Mathematics Teachers’ Self Study Guide Book 2 PDF, 4.6 MB
- DBE Self Study Guide: Mathematics Functions and Finance PDF, 25.7 MB
- DBE Self Study Guide: Mathematics Trigonometry and Euclidean Geometry PDF, 25.5 MB
- Mind the Gap: Mathematics PDF, 30.2 MB
- Mind the Gap: Wiskunde PDF, 24.5 MB
Mathematics Grade 12 Exam Guide
How the final Mathematics exam is set out, taken from the cover and instruction pages of the DBE November 2025 question papers.
Mathematics Paper 1
- 150 marks
- 3 hours
- 11 pages
- November 2025
Comes with: 1 information sheet and an answer book of 23 pages.
Instructions on the paper
- This question paper consists of 11 questions.
- Answer ALL the questions in the SPECIAL ANSWER BOOK provided.
- Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers.
- Answers only will NOT necessarily be awarded full marks.
- You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise.
- An information sheet with formulae is included at the end of the question paper.
Mathematics Paper 2
- 150 marks
- 3 hours
- 14 pages
- November 2025
Comes with: 1 information sheet and an answer book of 27 pages.
Instructions on the paper
- This question paper consists of 11 questions.
- Answer ALL the questions in the SPECIAL ANSWER BOOK provided.
- Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers.
- Answers only will NOT necessarily be awarded full marks.
- You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise.
- An information sheet with formulae is included at the end of the question paper.
Mathematics Grade 12 Questions and Answers
How many papers do you write for Mathematics in Grade 12?
2 papers: Paper 1 (150 marks, 3 hours) and Paper 2 (150 marks, 3 hours). These facts are from the DBE November 2025 question papers.
Can you use a calculator in the Mathematics exam?
Yes. The November 2025 Paper 1 says: "You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise."
What do you get with the Mathematics question paper?
Paper 1 comes with 1 information sheet and an answer book of 23 pages. Paper 2 comes with 1 information sheet and an answer book of 27 pages.
How long is the Mathematics question paper?
In November 2025, Paper 1 had 11 pages and Paper 2 had 14 pages.
Which Mathematics past papers are on this page?
398 files from 2020 to 2026, covering the March, June, September and November exams. They are listed by year and term above, and each one opens as a PDF.
Are the Mathematics memos included?
Yes. 222 of the files are memos (marking guidelines). Each memo is listed next to its question paper.
Are the Mathematics papers available in Afrikaans?
Yes. Most papers on this page come in English and Afrikaans. Use the language filter above the papers to show only one.
Is there a Mathematics practice quiz?
Yes. The exam questions quiz on this page has 37 questions taken word for word from real DBE Mathematics papers. The memo answer shows after you try each question.