Comprehensive Practice Papers and Full Step-by-Step Marking Memorandums
A complete CAPS-aligned diagnostic preparation asset for South African Senior Phase learners.
Calculators Permitted • Full Structural Solutions Included • accessiodocs.com
1. Your Exam Map: Understanding CAPS Marks
Preparing for your final exams can feel overwhelming, but knowing exactly how your exam is built gives you a massive advantage! This study companion aligns perfectly with the South African Department of Basic Education’s CAPS framework. Your exam papers are broken down into four dynamic problem styles, ensuring that every type of learner has a chance to shine:
- Knowledge Profiles (25%): Evaluating straight recall, fundamental definition identification, and standard base formulas. Secure these quick marks by remembering your foundations!
- Routine Procedures (45%): Monitoring the execution of well-known calculations, multi-step algebraic transformations, and familiar geometric proofs. Practice often so these feel like second nature on exam day.
- Complex Procedures (20%): Testing multi-concept integration, structural financial forecasting, and advanced spatial calculations. Take a deep breath and break the question into smaller pieces.
- Problem Solving (10%) Challenging non-routine optimization, pattern extension decoding, and contextual abstraction. Don’t leave these blank! Try different methods to unlock clues.
Part 1: Your Practice Examination Papers
Exam Paper 1: Algebra, Numbers, and Money Mechanics
⏳ Time Allowed: 2 Hours | 💯 Total Score: 75 Marks
Top Tips for Success:
- Read carefully! Make sure to answer exactly what the question asks.
- Keep your non-programmable scientific calculator nearby to double-check your arithmetic.
- Show your steps! Examiners love awarding partial marks, even if your final answer is slightly off.
Question 1: Number Identification & Rules of Exponents
1.1 Work through the expressions below. Tell us whether your final solution is a Rational or Irrational number:
- a) $\sqrt{25} + \sqrt[3]{-8}$ [2 Marks]
- b) $\sqrt{18} \div \sqrt{2}$ [2 Marks]
1.2 Use your exponent laws to simplify the expression below. Remember to keep all your final exponents positive:
$$\frac{(2a^3b^{-2})^3 \cdot 3a^4b^5}{12a^7b^{-3}}$$
[4 Marks]
Study Workspace Prompt: Use a pencil to expand the brackets first!
Question 2: Expanding Expressions & Factorising
2.1 Multiply out the brackets completely, combine the like terms together, and clean up the expression:
$$(3x – 4)(2x + 7) – (x – 2)^2$$
[5 Marks]
2.2 Break these algebraic expressions down into brackets completely (Factorisation):
- a) $4x^2 – 49$ [2 Marks]
- b) $2x^2 – 12x – 32$ [3 Marks]
Study Workspace Prompt: Hint for b: Is there a highest common factor you can pull out first?
Question 3: Equations and Graphing
3.1 Solve for the missing value of $x$ in this linear fraction equation:
$$\frac{2x – 3}{4} – \frac{x + 1}{3} = 2$$
[4 Marks]
3.2 Let’s look at the straight-line equation: $y = -2x + 4$.
- a) Find the exact points where the line cuts through both the $x$-axis and the $y$-axis (the intercepts). [3 Marks]
- b) Draw a neat, clean sketch of this straight line on grid paper using your intercepts. [2 Marks]
Question 4: Real-World Money Mathematics
4.1 Thandi invests R24,000 into a savings account for 5 years. Calculate exactly how much more money she would earn if her account paid 9% Compound Interest per year instead of 9.5% Simple Interest per year. [6 Marks]
4.2 A consumer buys a refrigerator worth R11,500 using a Hire Purchase (HP) agreement. They pay a 15% deposit upfront cash, and pay off the rest over 36 months at a simple interest charge rate of 14% per year. Calculate their exact required payment amount per month. [6 Marks]
Exam Paper 2: Spatial Layouts, Shapes, and Measurement
⏳ Time Allowed: 2 Hours | 💯 Total Score: 75 Marks
Top Tips for Success:
- Always state your reasons! In geometry, an answer without a matching geometric rule reason will lose points.
- Round your final calculated decimal numbers to exactly two decimal places.
Question 1: Parallel Lines & Missing Angles
1.1 Imagine two horizontal parallel lines, $PQ$ and $RS$ ($PQ \parallel RS$). A diagonal line vector $TV$ cuts straight through them at intersections $M$ and $N$. If angle $\hat{M}_2 = 4x – 15^\circ$ and angle $\hat{N}_1 = 2x + 45^\circ$, set up an equation to find the value of $x$. Remember to show your geometric reasons! [5 Marks]
Question 2: Proving Congruency & Similar Shapes
2.1 A diagonal line $BD$ cuts right through the middle of a four-sided shape $ABCD$. If side lengths are equal ($AB = CD$) and parallel ($AB \parallel CD$), write down a formal step-by-step geometric proof to show that $\triangle ABD$ is identical (congruent) to $\triangle CDB$. [5 Marks]
2.2 Triangles $\triangle PQR$ and $\triangle XYZ$ are mathematically similar ($\triangle PQR \sim \triangle XYZ$). If length $PQ = 5\text{ cm}$, its larger matching counterpart $XY = 15\text{ cm}$, and base length $QR = 8\text{ cm}$, calculate the missing length of segment $YZ$. [3 Marks]
Question 3: Using the Theorem of Pythagoras
3.1 A $13\text{-meter}$ long emergency rescue ladder is placed against a flat vertical wall. If the foot of the ladder is exactly $5\text{ meters}$ away from the bottom edge of the wall, find out exactly how high up the wall the ladder can reach. [4 Marks]
Question 4: Area & Volume of 3D Prisms
4.1 A 3D solid triangular prism has a right-angled triangular face at the front. The dimensions of this front triangle are: a height of $5\text{ cm}$, a base width of $12\text{ cm}$, and a long side (hypotenuse) of $13\text{ cm}$. The depth of the prism stretching backward is $20\text{ cm}$. Calculate:
- a) The total space inside the prism (Volume). [3 Marks]
- b) The total surface area if you had to unwrap and paint all 5 outer panels. [5 Marks]
Part 2: The Solution Guide (Learn Step-by-Step!)
Paper 1 Solutions Walkthrough
Question 1 Answer Key
1.1 Number Systems Demystified:
- a) $\sqrt{25} + \sqrt[3]{-8} = 5 + (-2) = 3$. Since $3$ can be written as a normal fraction ($\frac{3}{1}$), it is a Rational Number! (1 Mark calculation, 1 Mark classification)
- b) $\sqrt{18} \div \sqrt{2} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3$. Since this simplifies down to a neat, clean integer, it is also a Rational Number! (1 Mark simplification, 1 Mark classification)
1.2 Exponent Master Steps:
- Step 1 (Power-to-a-power rule first): Multiply the outer 3 into the top bracket: $\frac{8a^9b^{-6} \cdot 3a^4b^5}{12a^7b^{-3}}$ (1 Mark)
- Step 2 (Simplify the top row): Multiply coefficients, add exponents of matching bases: $\frac{24a^{13}b^{-1}}{12a^7b^{-3}}$ (1 Mark)
- Step 3 (Divide bases): Subtract bottom row exponents from top row (division law): $2a^{13-7}b^{-1-( -3)} = 2a^6b^2$ (1 Mark)
- Final Form Answer: $2a^6b^2$ (1 Mark)
Question 2 Answer Key
2.1 Bracket Expansion Breakdown:
- Step 1 (Double bracket distribution): $(3x – 4)(2x + 7) = 6x^2 + 21x – 8x – 28 = 6x^2 + 13x – 28$ (1 Mark)
- Step 2 (Square the second block safely): $(x – 2)^2 = (x^2 – 4x + 4)$ (1 Mark)
- Step 3 (Distribute the negative sign): $6x^2 + 13x – 28 – x^2 + 4x – 4$ (1 Mark)
- Step 4 (Group like terms): $(6x^2 – x^2) + (13x + 4x) + (-28 – 4)$ (1 Mark)
- Final Answer: $5x^2 + 17x – 32$ (1 Mark)
2.2 Factorisation Patterns Unlocked:
- a) $4x^2 – 49$ $\rightarrow$ Notice it’s a difference of squares! Square root both sides to split into dynamic brackets. Final Brackets: $(2x – 7)(2x + 7)$ (2 Marks)
- b) $2x^2 – 12x – 32$ $\rightarrow$ Factor out the common term 2 first: $2(x^2 – 6x – 16)$ (1 Mark). Now find numbers that multiply to $-16$ and add to $-6$ (which are $-8$ and $+2$). Final Answer Form: $2(x – 8)(x + 2)$ (2 Marks)
Question 3 Answer Key
3.1 Clearing Fractional Denominators:
- Step 1: Find the lowest common denominator (LCD) which is $12$ (1 Mark). Multiply all terms by $12$ to clear the fractions: $3(2x – 3) – 4(x + 1) = 2(12)$
- Step 2 (Multiply out brackets): $6x – 9 – 4x – 4 = 24$ (1 Mark)
- Step 3 (Combine variables): $2x – 13 = 24 \rightarrow 2x = 37$ (1 Mark)
- Final Answer: $x = \frac{37}{2} = 18.5$ (1 Mark)
3.2 Setting Up Your Straight Line Chart Intercepts:
- Y-Intercept: Let $x = 0$ $\rightarrow y = -2(0) + 4 \rightarrow y = 4$. Coordinate anchor is (0, 4) (1.5 Marks)
- X-Intercept: Let $y = 0$ $\rightarrow 0 = -2x + 4 \rightarrow 2x = 4 \rightarrow x = 2$. Coordinate anchor is (2, 0) (1.5 Marks)
- Student Reminder: Grab your ruler, plot those two exact coordinates on your axes plane, and trace a crisp line through them to claim your final 2 graphing marks!
Question 4 Answer Key
4.1 The Savings Balance Battle:
- Simple Option: $A = P(1 + it) \rightarrow A = 24000 \cdot [1 + (0.095 \cdot 5)] = R35,400$. Interest portion $= R35,400 – R24,000 = R11,400$ (2 Marks)
- Compound Option: $A = P(1 + i)^t \rightarrow A = 24000 \cdot (1 + 0.09)^5 = R36,926.98$. Interest portion $= R36,926.98 – R24,000 = R12,926.98$ (2 Marks)
- Difference: Subtract simple from compound: $R12,926.98 – R11,400.00 = R1,526.98$
- Final Conclusion: Thandi earns R1,526.98 extra using Compound Interest! (2 Marks)
4.2 Hire Purchase Installment Calculation:
- Step 1: Calculate down-payment deposit: $R11,500 \cdot 15\% = R1,725$ (1 Mark)
- Step 2: Find the remaining loan value to finance: $R11,500 – R1,725 = R9,775$ (1 Mark)
- Step 3: Calculate simple interest total over 3 years: $A = 9775 \cdot [1 + (0.14 \cdot 3)] = R13,880.50$ (2 Marks)
- Step 4: Divide final balance by the 36-month timeline: $\frac{R13,880.50}{36}$
- Final Result: Monthly Installment Obligation = R385.57 per month (2 Marks)
Paper 2 Solutions Walkthrough
Question 1 Answer Key
1.1 Solving Parallel Interior Vectors:
Because line $PQ$ and line $RS$ are parallel, our matching co-interior angles add up to exactly $180^\circ$! Let’s arrange our calculation table:
| Math Equation Calculation Steps | The Classroom Geometric Reason Code | Mark Weight |
| $(4x – 15^\circ) + (2x + 45^\circ) = 180^\circ$ | Co-interior angles match; given $PQ \parallel RS$ | [2 Marks] |
| $6x + 30^\circ = 180^\circ$ | Combine variable like terms smoothly | |
| $6x = 150^\circ$ | Move $30^\circ$ across, changing signs to subtract | [1 Mark] |
| $x = 25^\circ$ | Divide by 6 to isolate your final solution | [2 Marks] |
Question 2 Answer Key
2.1 Constructing Your Congruency Proof:
Let’s compare components directly inside $\triangle ABD$ and $\triangle CDB$:
- Side segment $AB = CD$ (Reason: Given directly in question text) [1 Mark]
- Alternate interior angle $\hat{B}_1 = \hat{D}_1$ (Reason: Alternate angles match up since line $AB \parallel CD$) [1 Mark]
- Line edge $BD = DB$ (Reason: It’s a common shared center side!) [1 Mark]
- Conclusion Validation: $\triangle ABD \equiv \triangle CDB$ [1 Mark]
- Core Reason Rule: SAS (Side-Angle-Side conditions met) [1 Mark]
2.2 Similarity Scale Transformations:
Since shapes are scale matches ($\triangle PQR \sim \triangle XYZ$), compute your size multiplier:
- Scale Growth Factor $= \frac{XY}{PQ} = \frac{15\text{ cm}}{5\text{ cm}} = 3$ times larger [1 Mark]
- Apply scale factor to your base length: $YZ = 3 \cdot QR \rightarrow 3 \cdot 8\text{ cm}$ [100% Tracking Method – 1 Mark]
- Final Missing Measurement Segment $YZ = 24\text{ cm}$ [1 Mark]
Question 3 Answer Key
3.1 Climbing the Pythagoras Equation:
- Formula layout: $(\text{Hypotenuse})^2 = (\text{Base Side})^2 + (\text{Vertical Height})^2$ (Theorem of Pythagoras) [1 Mark]
- Plug in your known dimensions: $13^2 = 5^2 + h^2 \rightarrow 169 = 25 + h^2$ [1 Mark]
- Isolate your variable parameter: $h^2 = 169 – 25 \rightarrow h^2 = 144$ [1 Mark]
- Solve for absolute metric: $h = \sqrt{144} = 12$
- Conclusion: The ladder secures a vertical height reach of exactly 12 meters! [1 Mark]
Question 4 Answer Key
4.1 Measuring 3D Triangular Capacity Space Volume:
- First, identify your front triangular face surface area profile:$\text{Area Base Triangle} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 12\text{ cm} \cdot 5\text{ cm} = 30\text{ cm}^2$ [1 Mark]
- Now multiply this area by the overall backward extrusion depth: $V = 30\text{ cm}^2 \cdot 20\text{ cm}$ [1 Mark]
- Final Cubic Volume Content Capacity = $600\text{ cm}^3$ [1 Mark]
4.2 Total Surface Area Outer Mapping Matrix:
Let’s find the area of all 5 separate sides and sum them together:
- 2 matching triangle face areas $= 30\text{ cm}^2 + 30\text{ cm}^2 = 60\text{ cm}^2$ [1 Mark]
- Floor rectangle area panel $= 12\text{ cm} \cdot 20\text{ cm} = 240\text{ cm}^2$ [1 Mark]
- Back vertical spine wall panel $= 5\text{ cm} \cdot 20\text{ cm} = 100\text{ cm}^2$ [1 Mark]
- Roof slanted hypotenuse slider panel $= 13\text{ cm} \cdot 20\text{ cm} = 260\text{ cm}^2$ [1 Mark]
- Sum elements to combine totals: $\text{TSA} = 60 + 240 + 100 + 260$
- Final Combined Exterior Surface Plan Area = $660\text{ cm}^2$ [1 Mark]
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