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Matric Mathematics Past Papers & Step-by-Step Worked Solutions

Stop getting stuck on marking memoranda. Here are comprehensive, step-by-step worked solutions for core NSC Paper 1 (Algebra, Calculus, Finance) and Paper 2 (Analytical Geometry, Trigonometry, Euclidean Geometry) with full marking rubric criteria.

Curriculum: South African CAPS Grade 12
Exam Papers: Paper 1 (150 Marks) & Paper 2 (150 Marks)
Reviewed: October 2026
Immediate Summary: NSC Mathematics Structure

The Grade 12 Mathematics examination consists of two 3-hour papers worth 150 marks each (300 total). Paper 1 tests Algebra, Equations & Inequalities (25±3 marks), Patterns & Sequences (25±3 marks), Functions & Inverses (35±3 marks), Financial Mathematics (15±3 marks), Differential Calculus (35±3 marks), and Probability (15±3 marks). Paper 2 tests Statistics (20±3 marks), Analytical Geometry (40±3 marks), Trigonometry (40±3 marks), and Euclidean Geometry & Proportions (50±3 marks).

How DBE Marks Mathematics (Understanding CA & Method Marks)

Before studying the worked examples, you must understand how DBE markers grade your script:

Worked Example 1: Calculus from First Principles (Paper 1 · 5 Marks)

Question: First Principles Derivative 5 Marks (Paper 1)
Exam Question: Given \( f(x) = 2x^2 - 3x \), determine \( f'(x) \) from first principles.
Step 1: State the official first principles definition formula

Write the limit definition clearly. This secures the first method mark.

f'(x) = lim (h → 0) [ f(x + h) - f(x) ] / h
Step 2: Determine and expand f(x + h)

Substitute \((x + h)\) wherever \(x\) appears in the original function and expand carefully:

f(x + h) = 2(x + h)² - 3(x + h)
f(x + h) = 2(x² + 2xh + h²) - 3x - 3h
f(x + h) = 2x² + 4xh + 2h² - 3x - 3h
Step 3: Substitute into the numerator [f(x+h) - f(x)]

Subtract the original function \( (2x^2 - 3x) \). Note how the non-\(h\) terms cancel out completely:

f(x + h) - f(x) = (2x² + 4xh + 2h² - 3x - 3h) - (2x² - 3x)
= 4xh + 2h² - 3h
Step 4: Factor out common factor h and cancel denominator

Crucial examiner rule: You must keep writing "lim (h → 0)" until you actually evaluate the limit!

f'(x) = lim (h → 0) [ h(4x + 2h - 3) ] / h
f'(x) = lim (h → 0) (4x + 2h - 3)
Step 5: Evaluate limit by setting h = 0
f'(x) = 4x + 2(0) - 3 = 4x - 3
Marking Memorandum Allocation:
• Mark 1 (M): Correct formula stated.
• Mark 2 (M): Correct substitution of \(f(x+h)\).
• Mark 3 (A): Correct simplification of numerator (\(4xh + 2h^2 - 3h\)).
• Mark 4 (M): Factoring out \(h\) and cancelling.
• Mark 5 (A): Final answer \(4x - 3\). (Penalty -1 if "lim h→0" was omitted prematurely).

Worked Example 2: Financial Maths — Loan Outstanding Balance (Paper 1 · 6 Marks)

Question: Present Value Annuity & Balance of Loan 6 Marks (Paper 1)
Exam Question: Thabo buys a vehicle for R350,000. He pays a 15% cash deposit and secures a bank loan for the remaining balance at an interest rate of 11.5% p.a. compounded monthly over 5 years. Monthly payments start one month after the loan is granted. Calculate:
(a) His monthly repayment \(x\).
(b) The outstanding balance of the loan immediately after making the 36th payment.
Part (a): Calculate Loan Amount & Monthly Payment x

1. Deposit = 0.15 × 350,000 = R52,500.
2. Present Value (Loan P) = R350,000 - R52,500 = R297,500.
3. Interest rate \( i = \frac{0.115}{12} \). Total payments \( n = 5 \times 12 = 60 \).

P = x[ 1 - (1 + i)^(-n) ] / i
297,500 = x[ 1 - (1 + 0.115/12)^(-60) ] / (0.115/12)
297,500 = x × 45.41407...
x = 297,500 / 45.41407... = R6,550.83
Part (b): Outstanding Balance after 36th Payment (Present Value Method)

The fastest and most reliable method in the NSC memorandum is to calculate the present value of the remaining unmade payments. Remaining payments = \( 60 - 36 = 24 \) payments left.

Balance = x[ 1 - (1 + i)^(-n_remaining) ] / i
Balance = 6550.83[ 1 - (1 + 0.115/12)^(-24) ] / (0.115/12)
Balance = 6550.83 × 21.3653...
Outstanding Balance = R139,960.52
Marking Memorandum Allocation:
• Deposit deduction & P = R297,500 (1 Mark)
• Correct formula selection and \(i/12\) substitution (1 Mark)
• \(x = R6,550.83\) (1 Mark)
• Recognizing \(n_{remaining} = 24\) (1 Mark)
• Substitution into Present Value formula (1 Mark)
• Final balance \(R139,960.52\) (1 Mark)

Worked Example 3: Euclidean Geometry Circle Theorem Proof (Paper 2 · 6 Marks)

Euclidean Geometry: Tan-Chord Theorem Proof 6 Marks (Paper 2 Theory)
Bookwork Proof (Examinable Every Year): Prove the theorem which states that the angle between the tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment.
Step 1: Construction (Mandatory)

Draw circle with centre O and tangent ABC touching at B. Draw chord BD subtending angle \( \angle BED \) at the circumference. Construction: Draw diameter BF through O and connect FD.

Step 2: Mathematical Proof Statements & Reasons
1. ∠FBC = 90° [Reason: Radius ⊥ tangent]
2. ∠B&sub1; + ∠B&sub2; = 90° [where ∠B&sub1; = ∠FBD and ∠B&sub2; = ∠DBC]
3. ∠BDF = 90° [Reason: Angle in semi-circle]
4. In ΔBDF: ∠F + ∠B&sub1; + ∠BDF = 180° [Reason: Sum of angles in Δ]
∴ ∠F + ∠B&sub1; + 90° = 180°
∴ ∠F + ∠B&sub1; = 90°
5. Since both (∠B&sub2; + ∠B&sub1; = 90°) and (∠F + ∠B&sub1; = 90°):
∴ ∠B&sub2; = ∠F
6. But ∠BED = ∠F [Reason: Angles in same segment / subtended by same chord BD]
7. ∴ ∠DBC = ∠BED [Tan-chord theorem proven]
Examiner Tip on Geometry Reasons: Every single geometric statement in Paper 2 must have an official DBE CAPS accepted reason abbreviation (e.g. "radius ⊥ tangent", "∠ in semi-circle", "∠s in same seg"). Omitting the reason forfeits the mark completely, even if the statement is correct.

Official Past Papers & Memoranda Links

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