Vector

The Complete Student Workbook

Student Name:

Week 1Factors, Multiples & PrimesNumbers
Week 2Ratio & ProportionNumbers
Week 3Signed Numbers & CoordinatesNumbers
Week 4Powers, Roots & Scientific NotationNumbers
Week 5Expressions & FactoringAlgebra
Week 6Linear Equations & InequalitiesAlgebra
Week 7Quadratics & the FormulaAlgebra
Week 8Cartesian Plane & Linear GraphsAlgebra
Week 9Arithmetic Sequences & SeriesAlgebra
Week 10Functions & Rate of ChangeAlgebra
Week 11Geometric Angle TheoremsSpace
Week 12Pythagorean Theorem & TrigonometrySpace
Week 13Non-Right Trigonometry Pt 1: Sine RuleSpace
Week 14Non-Right Trigonometry Pt 2: Cosine RuleSpace
Week 15Surface Area & Volume of SolidsSpace
Week 16Similarity & CongruenceSpace
Week 17Probability FundamentalsData
Week 18Combinatorics & CountingData
Week 19Measures of SpreadData
Week 20Bivariate Data & Sampling BiasData
Week 21Applied Real-World PercentagesFinance
Week 22Simple vs. Compound InterestFinance
Week 23Personal Budgeting & Cash FlowFinance
Answer Key & Step-by-Step SolutionsAppendix

Week 1 · Module 1.1

01 / Numbers

Factors, Multiples & Primes

The Fundamental Theorem of Arithmetic, canonical factorisation, HCF and LCM

Concept Introduction

Every integer greater than 1 is either prime or can be written as a product of primes in exactly one way. This is the Fundamental Theorem of Arithmetic, and it is the reason prime factorisation is a reliable structural fingerprint of a number.

A prime factor tree splits a number repeatedly until every branch terminates in a prime. Collecting the terminal primes and writing repeated primes as powers gives the canonical form, e.g. 72 = 2³ × 3².

The Highest Common Factor (HCF) is built from the primes the numbers share, each taken to its lowest power. The Lowest Common Multiple (LCM) takes every prime that appears in either number, each to its highest power.

Because HCF collects the shared core and LCM collects the full union, HCF(a, b) × LCM(a, b) = a × b for any two positive integers.

Prime factor tree with intersecting Venn factor sets

Key Relationships

N = p₁^a × p₂^b × p₃^c … (canonical prime form)
HCF: shared primes to the lowest power
LCM: all primes to the highest power
HCF(a, b) × LCM(a, b) = a × b

Worked Examples

Worked Example 1 — Canonical form of 360

Express 360 as a product of primes in index form.

1Divide by the smallest prime repeatedly: 360 ÷ 2 = 180, 180 ÷ 2 = 90, 90 ÷ 2 = 45.
245 is odd, so move to the next prime: 45 ÷ 3 = 15, 15 ÷ 3 = 5.
35 is prime, so the tree terminates.
4Collect terminal primes: 2 × 2 × 2 × 3 × 3 × 5.
5Write in index form: 360 = 2³ × 3² × 5.

Core Heuristic: always divide by the smallest prime that works, and only move up when the current prime no longer divides. This forces the canonical order automatically.

Worked Example 2 — HCF and LCM of 84 and 120

Find HCF(84, 120) and LCM(84, 120).

1Factorise: 84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5.
2Shared primes are 2 and 3. Lowest powers: 2² and 3¹.
3HCF = 2² × 3 = 12.
4All primes present: 2, 3, 5, 7. Highest powers: 2³, 3¹, 5¹, 7¹.
5LCM = 8 × 3 × 5 × 7 = 840.
6Check: 12 × 840 = 10 080 = 84 × 120. ✓

Core Heuristic: lay both factorisations in a column aligned by prime. HCF reads the minimum of each row, LCM reads the maximum.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
List all factors of 48.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Write 90 as a product of primes in index form.
Bronze
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Final Answer:
3.
State whether 91 is prime, with justification.
Bronze
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4.
Find HCF(126, 210) using prime factorisation.
Silver
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5.
Find LCM(18, 24, 30).
Silver
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6.
Two lighthouses flash every 24 s and every 36 s. They flashed together at 21:00, when will they next flash together?
Silver
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7.
The HCF of two numbers is 8 and their LCM is 96. One number is 24. Find the other.
Silver
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8.
A rectangular floor measures 315 cm by 450 cm. It is to be covered completely with identical square tiles, laid in rows and columns with no tiles cut. Find the side length of the largest possible square tile, and the number of tiles needed.
Gold
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Final Answer:
9.
N = 2³ × 3² × 5. How many factors does N have, and how many are even?
Gold
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10.
Find the smallest positive integer k such that 1 176 × k is a perfect cube.
Gold
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Self-Audit Checkpoint

I can build a prime factor tree and write the result in index form.
I can extract HCF and LCM from aligned factorisations without guessing.
I can decide when a word problem needs HCF (splitting/grouping) versus LCM (repeating cycles).
I can use HCF × LCM = a × b as a checking tool.

Week 2 · Module 1.2

01 / Numbers

Ratio & Proportion

Part-to-part, part-to-whole, the unitary method, direct and inverse proportion, scale maps

Concept Introduction

A ratio a : b compares parts. Converting to part-to-whole fractions (a/(a+b)) is the single most useful move when a total is given.

The unitary method finds the value of one part or one unit first, then scales. It converts every ratio problem into a multiplication.

Direct proportion: y = kx, so y/x is constant. Inverse proportion: y = k/x, so xy is constant. Identifying which invariant holds is the whole decision.

Scale on a map, 1 : n, means one unit on the map represents n identical units in reality. Convert units before scaling, not after.

Direct proportion (y = kx, straight line through origin) vs inverse proportion (xy = k)

Key Relationships

share = (part / total parts) × total amount
Direct: y = kx, y₁/x₁ = y₂/x₂
Inverse: y = k/x, x₁y₁ = x₂y₂
Real distance = map distance × scale factor

Worked Examples

Worked Example 1 — Sharing in a ratio

Share $1 620 between A, B and C in the ratio 4 : 5 : 9.

1Total parts = 4 + 5 + 9 = 18.
2One part = 1 620 ÷ 18 = $90.
3A = 4 × 90 = $360, B = 5 × 90 = $450, C = 9 × 90 = $810.
4Check: 360 + 450 + 810 = 1 620. ✓

Core Heuristic: find the value of one part before touching any individual share.

Worked Example 2 — Inverse proportion

6 pumps drain a tank in 14 hours. How long would 21 pumps take, working at the same rate?

1More pumps means less time, so this is inverse proportion.
2The invariant is pumps × hours = 6 × 14 = 84 pump-hours.
3With 21 pumps: hours = 84 ÷ 21 = 4.
4Answer: 4 hours.

Core Heuristic: ask 'does the answer grow or shrink as the input grows?' Growth means constant quotient; shrinkage means constant product.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Simplify the ratio 72 : 120.
Bronze
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Final Answer:
2.
Share 84 sweets in the ratio 3 : 4.
Bronze
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3.
8 identical books cost $52. Find the cost of 14 books.
Bronze
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4.
In a ratio 2 : 7, the larger share is 133. Find the total.
Silver
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5.
A map has scale 1 : 25 000. Two towns are 8.4 cm apart on the map. Find the real distance in km.
Silver
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6.
y is directly proportional to x, and y = 45 when x = 12. Find y when x = 20.
Silver
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7.
A recipe for 6 people uses 750 g of flour. How much for 15 people?
Silver
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8.
Ratio of red to blue counters is 5 : 3. After adding 16 blue counters the ratio becomes 5 : 7. Find the original number of red counters.
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9.
The time t for a journey varies inversely with speed v. At 80 km/h it takes 45 min. Find the speed needed for 30 min.
Gold
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10.
A and B share profit in ratio 7 : 5. B receives $2 400 less than A. Find the total profit.
Gold
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Self-Audit Checkpoint

I convert ratios to part-to-whole fractions when a total is known.
I can decide between direct and inverse proportion from the context.
I convert units before applying a map scale.
I can set up ratio equations with a parameter k when shares change.

Week 3 · Module 1.3

01 / Numbers

Signed Numbers & Coordinates

Four-quadrant operations, double negatives, absolute value and the coordinate plane

Concept Introduction

Subtracting a negative reverses direction twice, so a − (−b) = a + b. Multiplying or dividing two numbers of the same sign gives a positive result; opposite signs give a negative result.

Absolute value |x| is the undirected distance of x from zero, so |−7| = 7. Distance on a number line is |a − b|, which never depends on the order.

The coordinate plane is divided into four quadrants by the axes. Signs are (+,+), (−,+), (−,−), (+,−) reading anticlockwise from the top right.

Horizontal distance between points is |x₂ − x₁| and vertical distance is |y₂ − y₁|; the midpoint averages coordinates.

The four quadrants: signed coordinates on a unit grid

Key Relationships

a − (−b) = a + b
(−)(−) = +, (−)(+) = −
|x| = x if x ≥ 0, −x if x < 0
Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Worked Examples

Worked Example 1 — Signed arithmetic chain

Evaluate −6 − (−9) + (−4) × (−3).

1Handle multiplication first: (−4) × (−3) = +12.
2Rewrite the double negative: −6 − (−9) = −6 + 9 = 3.
3Combine: 3 + 12 = 15.

Core Heuristic: convert every double negative into an addition before you start counting.

Worked Example 2 — Distance across quadrants

Find the horizontal and vertical distance between A(−5, 4) and B(3, −6), and the midpoint.

1Horizontal: |3 − (−5)| = |8| = 8 units.
2Vertical: |−6 − 4| = |−10| = 10 units.
3Midpoint: ((−5+3)/2, (4+(−6))/2) = (−1, −1).

Core Heuristic: use absolute value for distances so the answer is positive regardless of the order you subtract.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Evaluate −13 + 21.
Bronze
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Final Answer:
2.
Evaluate −8 − (−15).
Bronze
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3.
Evaluate |−12| + |5 − 9|.
Bronze
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4.
Evaluate (−48) ÷ (−6) × (−2).
Silver
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5.
State the quadrant of each point: (−3, 7), (4, −2), (−6, −1).
Silver
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6.
The temperature falls from 6 °C to −11 °C. Find the drop.
Silver
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7.
Find the midpoint of P(−7, 2) and Q(5, −8).
Silver
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8.
Solve |x − 4| = 9.
Gold
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9.
A(−4, 3), B(6, 3) and C(6, −5) form a right triangle. Find its area.
Gold
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10.
A submarine at −180 m rises 4 times in steps of 35 m, then dives 60 m. Find its final depth.
Gold
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Self-Audit Checkpoint

I never leave a double negative unsimplified.
I apply the sign rules for products and quotients confidently.
I can name a quadrant from a coordinate pair instantly.
I use absolute value for distances and can solve simple |x − a| = b equations.

Week 4 · Module 1.4

01 / Numbers

Powers, Roots & Scientific Notation

Index laws, fractional and negative indices, roots and standard form A × 10^k

Concept Introduction

Indices count repeated multiplication, so multiplying powers of the same base adds indices and dividing subtracts them: aᵐ × aⁿ = aᵐ⁺ⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ.

A power of a power multiplies indices: (aᵐ)ⁿ = aᵐⁿ. Any non-zero base to the power 0 equals 1, and a negative index is a reciprocal: a⁻ⁿ = 1/aⁿ.

Fractional indices are roots: a^(1/n) = ⁿ√a and a^(m/n) = (ⁿ√a)ᵐ. Taking the root first usually keeps the numbers small.

Scientific notation writes a number as A × 10^k where 1 ≤ A < 10. Multiplying adds the exponents; dividing subtracts them; always renormalise A back into range afterwards.

Geometric power expansion: square and cube growth

Key Relationships

aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ
a⁰ = 1, a⁻ⁿ = 1/aⁿ
a^(m/n) = (ⁿ√a)ᵐ
A × 10^k with 1 ≤ A < 10

Worked Examples

Worked Example 1 — Simplifying an index expression

Simplify (3x²y³)³ ÷ (9x⁴y²).

1Expand the bracket: (3x²y³)³ = 27x⁶y⁹.
2Divide coefficients: 27 ÷ 9 = 3.
3Subtract indices: x⁶⁻⁴ = x², y⁹⁻² = y⁷.
4Result: 3x²y⁷.

Core Heuristic: expand every bracket fully before dividing — never cancel across an unexpanded power.

Worked Example 2 — Standard form calculation

Evaluate (4.8 × 10⁷) ÷ (1.6 × 10⁻³), giving the answer in standard form.

1Divide the decimal parts: 4.8 ÷ 1.6 = 3.
2Subtract exponents: 10⁷⁻(⁻³) = 10¹⁰.
3Answer: 3 × 10¹⁰ (already normalised).

Core Heuristic: split the calculation into 'digits' and 'powers of ten', then renormalise A into 1 ≤ A < 10 at the very end.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Simplify 2⁵ × 2³.
Bronze
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Final Answer:
2.
Evaluate 7⁰ + 3⁻².
Bronze
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3.
Write 0.00046 in standard form.
Bronze
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4.
Simplify (2a³b)⁴.
Silver
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5.
Evaluate 64^(2/3).
Silver
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6.
Simplify (x⁷y⁻²) ÷ (x³y⁻⁵).
Silver
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7.
Compute (3 × 10⁵) × (7 × 10⁻²) in standard form.
Silver
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8.
Solve 2^(x+1) = 32.
Gold
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9.
Simplify (√50 + √18) ÷ √2.
Gold
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10.
A cell has mass 2.5 × 10⁻¹¹ g. Find the mass of 6 × 10⁸ cells in standard form.
Gold
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Self-Audit Checkpoint

I apply all three index laws without confusing addition and multiplication of indices.
I can convert between negative/fractional indices and roots or reciprocals.
I can multiply and divide numbers in standard form and renormalise.
I can solve simple exponential equations by matching bases.

Week 5 · Module 2.1

02 / Algebra

Expressions & Factoring

Double bracket expansion, common monomial factors, difference of two squares and trinomials

Concept Introduction

Expanding (x + a)(x + b) means every term in the first bracket multiplies every term in the second — the FOIL order (First, Outer, Inner, Last) is just a bookkeeping device for the four products.

The area model shows why: a rectangle of side (x + a) by (x + b) splits into four tiles of areas x², bx, ax and ab.

Factoring reverses expansion. Always look for a common monomial factor first, then check for a difference of two squares a² − b² = (a − b)(a + b), then attempt a trinomial split.

For x² + bx + c, look for two numbers with product c and sum b. For ax² + bx + c, split the middle term using two numbers with product ac and sum b, then factor by grouping.

2×2 area tile grid modelling double-bracket expansion

Key Relationships

(x + a)(x + b) = x² + (a + b)x + ab
(a + b)² = a² + 2ab + b²
a² − b² = (a − b)(a + b)
ax² + bx + c: split with product ac, sum b

Worked Examples

Worked Example 1 — Expand and simplify

Expand and simplify (2x − 5)(3x + 4).

1First: 2x × 3x = 6x².
2Outer: 2x × 4 = 8x.
3Inner: −5 × 3x = −15x.
4Last: −5 × 4 = −20.
5Collect like terms: 6x² + 8x − 15x − 20 = 6x² − 7x − 20.

Core Heuristic: write all four products before collecting. Skipping straight to a collected answer is where sign errors live.

Worked Example 2 — Factor a hard trinomial

Factorise 6x² + 7x − 20.

1ac = 6 × (−20) = −120 and b = 7.
2Find two numbers with product −120 and sum 7: 15 and −8.
3Split: 6x² + 15x − 8x − 20.
4Group: 3x(2x + 5) − 4(2x + 5).
5Factor the common bracket: (2x + 5)(3x − 4).

Core Heuristic: the ac-split turns every trinomial into a grouping problem, so you never have to guess bracket pairs.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Expand 4(3x − 7).
Bronze
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Final Answer:
2.
Factorise 12x² + 18x.
Bronze
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3.
Expand (x + 6)(x + 3).
Bronze
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4.
Expand and simplify (3x − 4)².
Silver
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5.
Factorise 49x² − 81.
Silver
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6.
Factorise x² − 11x + 24.
Silver
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7.
Expand and simplify (x + 5)(x − 2) − (x − 3)².
Silver
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8.
Factorise 10x² − 29x + 21.
Gold
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9.
Factorise fully 3x³ − 27x.
Gold
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10.
A rectangle has area 2x² + 11x + 12. Find expressions for its sides.
Gold
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Self-Audit Checkpoint

I expand double brackets by writing all four products.
I always check for a common factor before anything else.
I recognise a difference of two squares instantly.
I can use the ac-split and grouping for trinomials with a leading coefficient.

Week 6 · Module 2.2

02 / Algebra

Linear Equations & Inequalities

Brackets, variables on both sides, fractional terms and inequalities on number lines

Concept Introduction

An equation is a balance: whatever you do to one side you must do to the other. Expand brackets first, gather variables on one side, constants on the other, then divide.

Fractional equations are cleared by multiplying every term by the LCM of the denominators — every term, including the ones without a denominator.

Inequalities follow the same rules with one exception: multiplying or dividing by a negative number reverses the inequality sign.

On a number line, a solid dot means the endpoint is included (≤, ≥) and an open circle means it is excluded (<, >).

Interval ray on a number line and an operational balance scale

Key Relationships

Do the same operation to both sides
Multiply through by the LCM to clear fractions
Dividing by a negative reverses < to >
Solid dot = included, open circle = excluded

Worked Examples

Worked Example 1 — Fractional linear equation

Solve (2x − 1)/3 + (x + 4)/4 = 5.

1LCM of 3 and 4 is 12; multiply every term by 12.
24(2x − 1) + 3(x + 4) = 60.
38x − 4 + 3x + 12 = 60 → 11x + 8 = 60.
411x = 52 → x = 52/11.

Core Heuristic: clear fractions in one multiplication step; never cancel term by term.

Worked Example 2 — Inequality with a negative coefficient

Solve 7 − 3x ≥ 22 and show the solution on a number line.

1Subtract 7: −3x ≥ 15.
2Divide by −3 and reverse the sign: x ≤ −5.
3Number line: solid dot at −5 with the ray pointing left.

Core Heuristic: the moment you divide by a negative, flip the sign — then sanity-check with one test value.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Solve 5x − 8 = 27.
Bronze
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Final Answer:
2.
Solve 3(x + 4) = 21.
Bronze
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3.
Solve x/4 = 9.
Bronze
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4.
Solve 6x − 5 = 2x + 19.
Silver
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5.
Solve 4(2x − 3) = 3(x + 6).
Silver
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6.
Solve (x + 2)/5 − (x − 1)/3 = 1.
Silver
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7.
Solve 2x + 9 < 4 and show on a number line.
Silver
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8.
Solve −4 ≤ 3x − 1 < 11.
Gold
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9.
Solve 5 − 2(x − 3) > 4x + 1.
Gold
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10.
The perimeter of a rectangle is 54 cm and the length is 3 cm more than twice the width. Find the width.
Gold
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Self-Audit Checkpoint

I expand brackets before rearranging.
I clear fractions by multiplying every term by the LCM.
I reverse the inequality when dividing by a negative.
I can represent an inequality accurately on a number line.

Week 7 · Module 2.3

02 / Algebra

Quadratics & the Formula

Completing the square, completed square form, deriving the quadratic formula and the discriminant

Concept Introduction

Completing the square rewrites x² + bx + c as (x + b/2)² + (c − b²/4). The bracket captures the symmetry of the parabola, so the vertex can be read off directly.

In completed square form y = (x + p)² + q, the turning point is (−p, q) and the axis of symmetry is x = −p.

Completing the square on ax² + bx + c = 0 produces the quadratic formula x = (−b ± √(b² − 4ac)) / 2a — the formula is not a separate rule, it is the general result of the same procedure.

The discriminant Δ = b² − 4ac decides the root structure: Δ > 0 gives two distinct real roots, Δ = 0 gives one repeated root, Δ < 0 gives no real roots.

Parabola with discriminant-determined roots and vertex

Key Relationships

x² + bx + c = (x + b/2)² + c − b²/4
x = (−b ± √(b² − 4ac)) / 2a
Δ = b² − 4ac
Turning point of y = (x + p)² + q is (−p, q)

Worked Examples

Worked Example 1 — Complete the square and state the turning point

Write y = x² − 10x + 32 in completed square form and state the minimum point.

1Half of −10 is −5, so start with (x − 5)².
2(x − 5)² = x² − 10x + 25, which is 7 short of 32.
3So y = (x − 5)² + 7.
4Minimum point is (5, 7), axis of symmetry x = 5.

Core Heuristic: halve the x-coefficient for the bracket, then correct the constant by subtracting what the bracket over-produced.

Worked Example 2 — Formula and discriminant

Use Δ to describe the roots of 2x² − 7x − 3 = 0, then solve it to 2 decimal places.

1a = 2, b = −7, c = −3.
2Δ = (−7)² − 4(2)(−3) = 49 + 24 = 73 > 0, so two distinct real roots.
3x = (7 ± √73) / 4.
4√73 ≈ 8.544, so x ≈ 3.89 or x ≈ −0.39.

Core Heuristic: compute Δ first. It tells you what kind of answer to expect before you commit to arithmetic.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Solve x² − 9x + 20 = 0 by factorising.
Bronze
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Final Answer:
2.
Write x² + 6x + 1 in the form (x + p)² + q.
Bronze
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Final Answer:
3.
Find Δ for x² + 4x + 4 = 0 and state the number of real roots.
Bronze
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4.
Solve x² + 5x − 14 = 0.
Silver
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5.
Solve 3x² + 2x − 8 = 0 by factorising.
Silver
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6.
Solve x² − 6x + 2 = 0, leaving answers in surd form.
Silver
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7.
State the vertex of y = 2(x − 4)² − 5.
Silver
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8.
Find k so that x² + kx + 9 = 0 has exactly one root.
Gold
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9.
Solve 2x² − 5x − 6 = 0 to 2 dp using the formula.
Gold
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10.
A rectangle has perimeter 26 cm and area 40 cm². Find its dimensions.
Gold
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Self-Audit Checkpoint

I can complete the square and read off the vertex.
I can derive and apply the quadratic formula with correct signs.
I use Δ to predict the number of real roots before solving.
I can translate a geometry or word problem into a quadratic equation.

Week 8 · Module 2.4

02 / Algebra

Cartesian Plane & Linear Graphs

Slope-intercept form, gradients, intercepts, parallel and perpendicular lines

Concept Introduction

Gradient measures steepness: m = Δy/Δx = (y₂ − y₁)/(x₂ − x₁). A positive gradient rises left to right, a negative one falls.

In y = mx + c, m is the gradient and c is the y-intercept. The x-intercept is found by setting y = 0.

Parallel lines have equal gradients (m₁ = m₂). Perpendicular lines have gradients whose product is −1 (m₁m₂ = −1), so each is the negative reciprocal of the other.

Given a point and a gradient, y − y₁ = m(x − x₁) constructs the equation without solving for c separately.

y = ½x + 1 with intercepts, rise-over-run, and a perpendicular line (m₁m₂ = −1)

Key Relationships

m = (y₂ − y₁)/(x₂ − x₁)
y = mx + c
y − y₁ = m(x − x₁)
Parallel: m₁ = m₂; Perpendicular: m₁m₂ = −1

Worked Examples

Worked Example 1 — Equation through two points

Find the equation of the line through A(−2, 7) and B(4, −5).

1m = (−5 − 7)/(4 − (−2)) = −12/6 = −2.
2Use point A: y − 7 = −2(x + 2).
3y − 7 = −2x − 4 → y = −2x + 3.
4Check with B: −2(4) + 3 = −5 ✓

Core Heuristic: compute the gradient first, then substitute the point that makes the arithmetic easiest.

Worked Example 2 — Perpendicular line

Find the line perpendicular to y = ¾x − 1 passing through (6, 2).

1Gradient of the given line is 3/4.
2Perpendicular gradient is the negative reciprocal: −4/3.
3y − 2 = −4/3 (x − 6).
4y = −4/3 x + 8 + 2 = −4/3 x + 10.

Core Heuristic: flip the fraction and change the sign — then verify m₁m₂ = −1 numerically.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
State the gradient and y-intercept of y = 5x − 3.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the gradient through (1, 2) and (5, 14).
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find the x-intercept of y = 2x − 8.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find the equation of the line with gradient −3 through (2, 5).
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Are y = 2x + 1 and 4x − 2y = 9 parallel? Justify.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Find the gradient of a line perpendicular to y = −2/5 x + 7.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Rewrite 3x + 4y = 24 in slope-intercept form and state both intercepts.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
Find the equation of the line through (−3, 4) and (5, 4), and describe it.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
Line L passes through (0, −2) and is perpendicular to 2x + y = 5. Find its equation.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
Find the intersection point of y = 3x − 4 and y = −x + 8.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can compute a gradient from two points with correct signs.
I can build an equation from a point and a gradient.
I can test for parallel and perpendicular using gradients.
I can find intercepts and intersections algebraically.

Week 9 · Module 2.5

02 / Algebra

Arithmetic Sequences & Series

Common difference, the general term and the sum of an arithmetic series

Concept Introduction

An arithmetic sequence adds a fixed common difference d each step. Because the step is constant, the nth term is linear in n: Tₙ = a + (n − 1)d.

The common difference is found by subtracting consecutive terms: d = T₂ − T₁ = T₃ − T₂.

Gauss pairing gives the sum: pairing the first and last terms repeatedly shows Sₙ = n/2 (a + l) where l is the last term, equivalently Sₙ = n/2 [2a + (n − 1)d].

Sequences model any situation with a constant increment — savings with a fixed deposit, seat rows increasing by a fixed number, staircase patterns.

Gauss pairing staircase for an arithmetic series

Key Relationships

d = Tₙ₊₁ − Tₙ
Tₙ = a + (n − 1)d
Sₙ = n/2 (a + l)
Sₙ = n/2 [2a + (n − 1)d]

Worked Examples

Worked Example 1 — Which term equals a value

For the sequence 7, 11, 15, 19, …, which term equals 143?

1a = 7 and d = 4.
2Tₙ = 7 + (n − 1)4 = 4n + 3.
3Set 4n + 3 = 143 → 4n = 140 → n = 35.
4So 143 is the 35th term.

Core Heuristic: simplify Tₙ into the linear form dn + (a − d) first; then any 'which term' question is a one-line equation.

Worked Example 2 — Sum of a series

Find the sum of the first 30 terms of 5, 9, 13, ….

1a = 5, d = 4, n = 30.
2S₃₀ = 30/2 [2(5) + 29(4)] = 15[10 + 116].
3= 15 × 126 = 1 890.

Core Heuristic: if you already know the last term, use Sₙ = n/2 (a + l) — it is faster and less error-prone.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find d and the next term of 3, 10, 17, 24, …
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find T₁₀ for a = 4, d = 6.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find the sum 1 + 2 + 3 + … + 50.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find a formula for Tₙ of 20, 16, 12, 8, …
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Which term of the sequence 6, 13, 20, … equals 202?
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Find S₂₀ for the sequence 11, 15, 19, …
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
T₅ = 23 and T₁₂ = 65. Find a and d.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
How many terms of 4, 9, 14, … are needed for a sum of 1 030?
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A theatre has 25 rows of seats. The first row has 18 seats, and each row after that has 3 more seats than the row in front of it. Find the total number of seats in the theatre.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
The sum of the first n terms of an arithmetic sequence is Sₙ = 2n² + 3n. Find a and d.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can find a and d from any two given terms.
I can write Tₙ in simplified linear form.
I choose the appropriate sum formula for the given information.
I can model real contexts with an arithmetic series.

Week 10 · Module 2.6

02 / Algebra

Functions & Rate of Change

Function notation, domain and range, rational expressions and average rate of change

Concept Introduction

f(x) names a rule; f(3) means substitute 3 for every x. The domain is the set of permitted inputs, and the range is the set of resulting outputs.

Rational expressions are undefined where the denominator is zero, so those values are excluded from the domain. Simplify by factorising numerator and denominator and cancelling common factors.

The average rate of change of f between x = a and x = b is (f(b) − f(a))/(b − a) — the gradient of the secant line joining the two points on the curve.

For a straight line the average rate of change is constant; for a curve it depends on the interval chosen, which is the first step towards calculus.

Curved function with a secant line and rise-over-run triangle

Key Relationships

f(a) = value of the rule at x = a
Domain excludes values making a denominator 0
Average rate = (f(b) − f(a))/(b − a)
Secant gradient = Δy/Δx

Worked Examples

Worked Example 1 — Evaluate and simplify

For f(x) = (x² − 9)/(x − 3), find f(5) and state the domain.

1f(5) = (25 − 9)/(5 − 3) = 16/2 = 8.
2Factorise: (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3.
3The denominator vanishes at x = 3, so the domain is all real x except 3.

Core Heuristic: read the domain from the original expression, not the simplified one — cancelling hides the excluded value.

Worked Example 2 — Average rate of change

For f(x) = x² + 2x, find the average rate of change from x = 1 to x = 4.

1f(1) = 1 + 2 = 3 and f(4) = 16 + 8 = 24.
2Average rate = (24 − 3)/(4 − 1) = 21/3 = 7.
3This is the gradient of the secant joining (1, 3) and (4, 24).

Core Heuristic: average rate of change is always 'output difference over input difference' — the units are output per input.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
For f(x) = 3x − 5, find f(4).
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
For g(x) = x² + 1, find g(−3).
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
State the domain of f(x) = 1/(x − 2).
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
For f(x) = 2x² − x, find the average rate of change from x = 0 to x = 3.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Simplify (x² − 4)/(x² + 5x + 6).
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
For f(x) = 5 − 2x, solve f(x) = 17.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
State the range of f(x) = x² − 4 for real x.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
For f(x) = x², find the average rate of change from x = a to x = a + h, where h ≠ 0.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A ball's height is h(t) = 20t − 5t² metres. Find the average velocity from t = 1 to t = 3 s.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
For f(x) = (2x + 1)/(x − 4), find f(6) and state the domain and the range.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can substitute into function notation reliably.
I state domains from the original expression.
I can simplify rational expressions by factorising.
I can compute and interpret an average rate of change with units.

Week 11 · Module 3.1

03 / Space

Geometric Angle Theorems

Parallel line angle pairs, polygon interior sums and exterior angles

Concept Introduction

When a transversal cuts parallel lines, corresponding angles are equal (F-shape), alternate angles are equal (Z-shape) and co-interior angles sum to 180° (C-shape).

Angles on a straight line sum to 180°, angles at a point sum to 360°, and vertically opposite angles are equal.

The interior angles of an n-sided polygon sum to (n − 2) × 180°, because the polygon splits into n − 2 triangles.

The exterior angles of any convex polygon always sum to 360°, so a regular n-gon has exterior angle 360°/n.

Transversal cutting parallel lines with highlighted angle pairs

Key Relationships

Interior sum = (n − 2) × 180°
Exterior sum = 360°
Regular polygon exterior angle = 360°/n
Co-interior angles sum to 180°

Worked Examples

Worked Example 1 — Chain of angle reasons

Two parallel lines are cut by a transversal. One angle is 118°. Find the co-interior and alternate angles, with reasons.

1The alternate angle equals 118° (alternate angles, parallel lines).
2The co-interior angle = 180° − 118° = 62° (co-interior angles are supplementary).
3The vertically opposite angle to 118° is also 118° (vertically opposite angles).

Core Heuristic: state a reason for every step. In deduction questions the reason is worth as much as the number.

Worked Example 2 — Regular polygon

A regular polygon has an interior angle of 156°. How many sides does it have?

1Exterior angle = 180° − 156° = 24°.
2Number of sides = 360° ÷ 24° = 15.
3Check: interior sum = 13 × 180° = 2 340°, and 2 340 ÷ 15 = 156° ✓

Core Heuristic: go through the exterior angle. Dividing 360° by the exterior angle is always faster than solving the interior formula.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find x if angles on a straight line are x and 47°.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the interior angle sum of a hexagon.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find the exterior angle of a regular decagon.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Parallel lines cut by a transversal give co-interior angles 3x and 2x. Find x.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
A regular polygon has exterior angle 15°. Find n and the interior angle.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
In a triangle the angles are 2x, 3x and 4x. Find each angle.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Find the interior angle of a regular pentagon.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
A polygon's interior angles sum to 1 800°. Find n.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
Two parallel lines are cut by a transversal; one angle is (4x + 10)° and its corresponding angle is (6x − 30)°. Find x and the angle.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
An irregular pentagon has four angles 100°, 115°, 90° and 130°. Find the fifth.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can name the correct angle pair (alternate, corresponding, co-interior).
I give a reason for every deduction step.
I use the exterior angle route for regular polygons.
I can solve for an unknown angle expressed algebraically.

Week 12 · Module 3.2

03 / Space

Pythagorean Theorem & Trigonometry

2D and 3D Pythagoras, SOH-CAH-TOA and angles of elevation and depression

Concept Introduction

In a right triangle, a² + b² = c² where c is the hypotenuse (always opposite the right angle). Rearranged, a shorter side is c² − b².

In three dimensions, the space diagonal of a cuboid follows d² = l² + w² + h², which is Pythagoras applied twice.

SOH-CAH-TOA: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Label the triangle relative to the angle you are using before choosing a ratio.

An angle of elevation is measured upward from the horizontal; an angle of depression is measured downward. They are equal when measured between the same two points.

Right triangle SOH-CAH-TOA reference with 3D diagonal

Key Relationships

a² + b² = c²
d² = l² + w² + h² (cuboid diagonal)
sin θ = O/H, cos θ = A/H, tan θ = O/A
Elevation from A to B = depression from B to A

Worked Examples

Worked Example 1 — 3D Pythagoras

A box measures 8 cm × 6 cm × 5 cm. Find the longest straight rod that fits inside, to 2 dp.

1Base diagonal: √(8² + 6²) = √100 = 10 cm.
2Space diagonal: √(10² + 5²) = √125.
3√125 ≈ 11.18 cm.

Core Heuristic: solve the base right triangle first and keep the exact value — rounding early corrupts the second step.

Worked Example 2 — Angle of elevation

From 40 m away, the angle of elevation to a tower top is 32°. The observer's eye is 1.6 m above ground. Find the tower height, to 1 dp.

1Opposite = height above eye level, adjacent = 40 m → use tan.
2tan 32° = h/40 → h = 40 tan 32° ≈ 24.998 m.
3Add eye height: 24.998 + 1.6 ≈ 26.6 m.

Core Heuristic: identify which two sides you have or want, then pick the single ratio containing both.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find the hypotenuse of a right triangle with legs 9 cm and 12 cm.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the missing length if c = 26 and a = 10.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find x if sin 30° = x/18.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find θ if tan θ = 7/5, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
A 6.5 m ladder rests against a wall with its foot 2.5 m out. Find the height reached.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Find the side adjacent to a 41° angle if the hypotenuse is 15 cm, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Find the space diagonal of a cube of side 7 cm, to 2 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
From a cliff 85 m high, the angle of depression to a boat is 24°. Find the horizontal distance, to nearest metre.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A cuboid is 12 × 9 × 8 cm. Find the angle the space diagonal makes with the base, to 1 dp.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
An isosceles triangle has equal sides 13 cm and base 10 cm. Find its height and area.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I always identify the hypotenuse before applying Pythagoras.
I can apply Pythagoras twice for 3D diagonals.
I label O, A, H relative to the working angle before choosing a ratio.
I distinguish elevation from depression correctly in word problems.

Week 13 · Module 3.3a

03 / Space

Non-Right Trigonometry Pt 1: Sine Rule

The Sine Rule, the ambiguous SSA case and the area formula ½ab sin C

Concept Introduction

In any triangle, a/sin A = b/sin B = c/sin C. Use it whenever you have a matched side-and-opposite-angle pair.

For missing sides put the sides on top; for missing angles invert the ratios so the sines sit on top: sin A/a = sin B/b.

The ambiguous case arises with SSA data: because sin θ = sin(180° − θ), a second obtuse solution may exist if it keeps the angle sum below 180°.

The area of a triangle with two sides and the included angle is ½ab sin C — the included angle is essential.

Non-right triangle with ambiguous SSA double arcs

Key Relationships

a/sin A = b/sin B = c/sin C
sin A/a = sin B/b (for angles)
Ambiguous case: θ or 180° − θ
Area = ½ab sin C

Worked Examples

Worked Example 1 — Missing side

In triangle ABC, A = 52°, B = 71°, a = 14 cm. Find b, to 1 dp.

1b/sin 71° = 14/sin 52°.
2b = 14 sin 71° / sin 52°.
3= 14 × 0.9455 / 0.7880 ≈ 16.8 cm.

Core Heuristic: write the ratio with the unknown in the numerator so one multiplication finishes it.

Worked Example 2 — Ambiguous case

In triangle ABC, a = 9, b = 12, A = 38°. Find both possible values of B.

1sin B/12 = sin 38°/9 → sin B = 12 sin 38°/9 ≈ 0.8209.
2B ≈ 55.2° (acute solution).
3Or B ≈ 180° − 55.2° = 124.8°.
4Check: 38° + 124.8° = 162.8° < 180°, so both triangles exist.

Core Heuristic: with SSA, always test the obtuse partner against the 180° angle-sum limit before rejecting it.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
A = 40°, B = 65°, a = 10. Find b to 1 dp.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the area of a triangle with sides 8 cm and 11 cm and an angle of 40° in between, to 1 dp.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
A = 30°, a = 6, b = 9. Find sin B.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
In triangle ABC, C = 48°, c = 20, A = 62°. Find a to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
a = 15, A = 43°, b = 11. Find B to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Find the third angle and remaining side: A = 55°, B = 63°, a = 9.4. Give c to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
The area of a triangle is 30 cm² with sides 10 cm and 8 cm. Find the possible angles in between (it can be acute or obtuse), to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
a = 7, b = 10, A = 35°. Determine whether two triangles exist and find both B values.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
Two observers 500 m apart see a balloon between them at elevations 38° and 47°. Find the balloon height, to the nearest metre.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
In triangle ABC, a = 12, A = 100°, b = 9. Explain why only one triangle exists and find B.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I identify a matched side-angle pair before using the Sine Rule.
I invert the ratio when solving for an angle.
I always test the ambiguous obtuse solution.
I use ½ab sin C only with the included angle.

Week 14 · Module 3.3b

03 / Space

Non-Right Trigonometry Pt 2: Cosine Rule

The Cosine Rule for SAS and SSS, and three-figure navigation bearings

Concept Introduction

The Cosine Rule c² = a² + b² − 2ab cos C generalises Pythagoras; when C = 90°, cos C = 0 and it collapses to a² + b² = c².

Use the Cosine Rule when you have two sides and the included angle (SAS) to find the third side, or three sides (SSS) to find any angle.

For SSS, rearrange: cos C = (a² + b² − c²)/(2ab). A negative value means the angle is obtuse.

Bearings are measured clockwise from true north and always written with three figures, e.g. 065°. Sketch the north lines before computing anything.

True-north compass rose with three-figure bearing vector

Key Relationships

c² = a² + b² − 2ab cos C
cos C = (a² + b² − c²)/(2ab)
Bearings: 3 figures clockwise from north
Back bearing = bearing ± 180°

Worked Examples

Worked Example 1 — SAS side

In triangle ABC, a = 11, b = 14, C = 68°. Find c to 1 dp.

1c² = 11² + 14² − 2(11)(14) cos 68°.
2= 121 + 196 − 308 × 0.3746 = 317 − 115.4 = 201.6.
3c ≈ 14.2.

Core Heuristic: compute a² + b² first, then subtract the 2ab cos C correction as a single number.

Worked Example 2 — Bearings journey

A ship sails 30 km on bearing 040°, then 45 km on bearing 130°. Find its distance from the start, to 1 dp.

1The two bearings differ by 90°, so the included angle at the turn is 180° − 90° = 90°.
2Since the angle is 90°, use Pythagoras: d² = 30² + 45² = 900 + 2 025 = 2 925.
3d ≈ 54.1 km.

Core Heuristic: convert bearings into the interior angle at the turning point before applying any rule.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find c if a = 5, b = 7, C = 60°.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Write the bearing of due west in three figures.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find cos C if a = 6, b = 8, c = 10.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find the largest angle of a triangle with sides 7, 9, 12, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Find b if a = 13, c = 10, B = 105°.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
A boat sails 12 km on 070° then 18 km on 160°. Find the distance from start, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
The back bearing of 115° is what?
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
Sides 8, 15 and 20 form a triangle. Find its largest angle and its area, to 1 dp.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
Two towns: B is 40 km from A on bearing 055°; C is 65 km from A on bearing 145°. Find BC to 1 dp.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
In triangle ABC, a = 9, b = 12, c = 15. Verify it is right-angled and find its area.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I choose the Cosine Rule for SAS and SSS, not the Sine Rule.
I can rearrange for cos C and interpret a negative value as obtuse.
I write bearings with three figures and sketch north lines.
I convert bearing differences into interior triangle angles.

Week 15 · Module 3.4

03 / Space

Surface Area & Volume of Solids

Prisms, cylinders, pyramids, cones and spheres, including composite solids

Concept Introduction

Any prism or cylinder has volume V = A_base × h, because identical cross-sections stack uniformly.

Pyramids and cones taper to a point and take exactly one third of the matching prism or cylinder: V = ⅓A_base × h.

A sphere has V = 4/3 πr³ and surface area 4πr². A hemisphere has half the volume and a curved surface of 2πr², plus πr² if the flat face is included.

For composite solids, add volumes but be careful with surface area: any face hidden at a join is not part of the exterior surface.

Composite solid: cylinder capped with a hemisphere

Key Relationships

Prism/cylinder: V = A_base h; cylinder SA = 2πr² + 2πrh
Cone: V = ⅓πr²h, SA = πr² + πrl
Pyramid: V = ⅓ × base area × h
Sphere: V = 4/3 πr³, SA = 4πr²

Worked Examples

Worked Example 1 — Cylinder with hemisphere cap

A cylinder of radius 5 cm and height 12 cm is capped with a hemisphere. Find the total volume, to 1 dp.

1Cylinder: V = π(5²)(12) = 300π.
2Hemisphere: V = ⅔π(5³) = 250π/3 ≈ 83.33π.
3Total = 383.33π ≈ 1 204.3 cm³.

Core Heuristic: keep π symbolic until the final line so the arithmetic stays exact and checkable.

Worked Example 2 — Cone surface area

A cone has radius 6 cm and vertical height 8 cm. Find its total surface area, to 1 dp.

1Slant height l = √(6² + 8²) = 10 cm.
2Curved surface = πrl = π(6)(10) = 60π.
3Base = πr² = 36π.
4Total = 96π ≈ 301.6 cm².

Core Heuristic: cone surface area needs the slant height — never substitute the vertical height into πrl.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find the volume of a cuboid 6 × 4 × 9 cm.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the volume of a cylinder r = 3, h = 10, to 1 dp.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find the volume of a sphere r = 3, to 1 dp.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find the surface area of a cylinder r = 4, h = 11, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Find the volume of a square pyramid with base 9 cm and height 14 cm.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
A cone has r = 5 and h = 12. Find its volume, to 1 dp.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
A triangular prism has base area 24 cm² and length 15 cm. Find its volume.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
A hemisphere of radius 7 cm sits on a cylinder of the same radius and height 10 cm. Find the total exterior surface area (including the base), to 1 dp.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder height.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
An inverted cone-shaped container has a radius of 6 cm and a vertical height of 15 cm. Water fills the cone from its vertex up to half its vertical height. Find the volume of water, to 1 decimal place.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I use V = A_base h for prisms and one third of it for tapered solids.
I compute the slant height before a cone's curved surface area.
I exclude hidden faces from composite surface areas.
I keep π exact until the final rounding.

Week 16 · Module 3.5

03 / Space

Similarity & Congruence

Congruence tests and the scale factor ladder k, k² and k³

Concept Introduction

Congruent figures are identical in size and shape. The valid tests are SSS, SAS, ASA (and AAS) and RHS for right triangles. AAA is not a congruence test — it only proves similarity.

Similar figures have equal corresponding angles and proportional corresponding sides. The constant of proportionality is the linear scale factor k.

Areas scale by k² and volumes (and masses of the same material) scale by k³. This is the single most tested idea in the topic.

To find k from areas take the square root; from volumes take the cube root, then use k for lengths.

Scaled similar solids comparing 1 : k : k² : k³

Key Relationships

Length ratio = k
Area ratio = k²
Volume ratio = k³
Congruence: SSS, SAS, ASA/AAS, RHS

Worked Examples

Worked Example 1 — Volume scaling

Two similar jugs have heights 12 cm and 18 cm. The small one holds 640 mL. Find the large capacity.

1k = 18/12 = 1.5.
2Volume ratio = k³ = 3.375.
3Capacity = 640 × 3.375 = 2 160 mL.

Core Heuristic: identify which ladder rung the given quantity sits on (length, area or volume) before scaling.

Worked Example 2 — Working back from areas

Two similar triangles have areas 45 cm² and 125 cm². The smaller has a base of 6 cm. Find the larger base.

1Area ratio = 125/45 = 25/9.
2k = √(25/9) = 5/3.
3Larger base = 6 × 5/3 = 10 cm.

Core Heuristic: never scale a length by an area ratio — square-root down to k first.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Two similar rectangles have corresponding sides 4 cm and 10 cm. Find the scale factor k from the smaller rectangle to the larger one.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
If k = 3, state the area and volume ratios.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Name the congruence test for two right triangles with equal hypotenuse and one equal leg.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Similar triangles: sides 5, 12, 13 and the largest side of the second is 39. Find its other sides.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Two similar cones have radii 4 and 6. The smaller has volume 96π. Find the larger.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Two similar photos have areas 200 cm² and 450 cm². Find the length scale factor k from the smaller photo to the larger one.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Explain why AAA proves similarity but not congruence.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
Two similar solids have surface areas 96 cm² and 216 cm². The smaller has volume 128 cm³. Find the larger volume.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A model car scale 1 : 24 has a 6 cm² windscreen. Find the real windscreen area in cm².
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
Triangle ABC has DE parallel to BC with AD = 4, DB = 6 and DE = 5. Find BC.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can quote and apply the four congruence tests.
I know AAA gives similarity only.
I move between k, k² and k³ in the correct direction.
I can use parallel lines to justify similar triangles.

Week 17 · Module 4.1

04 / Data

Probability Fundamentals

Sample spaces, tree diagrams, independent events and mutually exclusive addition

Concept Introduction

Theoretical probability is favourable outcomes divided by equally likely total outcomes. Experimental (relative frequency) probability is observed successes divided by trials, and it approaches the theoretical value as trials increase.

A sample space lists every possible outcome. Two-way tables and tree diagrams are structured ways of enumerating it without omission.

Independent events do not influence each other, so P(A and B) = P(A) × P(B). On a tree, multiply along branches.

Mutually exclusive events cannot both happen, so P(A or B) = P(A) + P(B). On a tree, add across completed paths. In general P(A or B) = P(A) + P(B) − P(A and B).

Two-stage probability tree with outcome nodes

Key Relationships

P(E) = favourable / total
P(not E) = 1 − P(E)
Independent: P(A and B) = P(A)P(B)
P(A or B) = P(A) + P(B) − P(A and B)

Worked Examples

Worked Example 1 — Two-stage tree with replacement

A bag has 5 red and 3 blue counters. Two are drawn with replacement. Find P(exactly one red).

1P(red) = 5/8, P(blue) = 3/8 on each draw (replacement keeps them constant).
2Path RB: 5/8 × 3/8 = 15/64.
3Path BR: 3/8 × 5/8 = 15/64.
4These paths are mutually exclusive: 15/64 + 15/64 = 30/64 = 15/32.

Core Heuristic: multiply along a branch, add between branches.

Worked Example 2 — Without replacement

From the same bag (5 red, 3 blue), two are drawn without replacement. Find P(both red).

1First draw: P(red) = 5/8.
2Second draw: one red gone, one counter gone → 4/7.
3P(both red) = 5/8 × 4/7 = 20/56 = 5/14.

Core Heuristic: without replacement, reduce both the numerator and the denominator on the second branch.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
A fair die is rolled. Find P(even).
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
P(rain) = 0.35. Find P(no rain).
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Two coins are tossed. List the sample space and find P(two heads).
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
A spinner has 8 equal sectors numbered 1–8. Find P(prime).
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Independent events with P(A) = 0.4, P(B) = 0.25. Find P(A and B) and P(A or B).
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
A die is rolled 300 times and shows a six 62 times. Find the experimental probability and compare with theory.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
A bag has 4 green and 6 yellow balls; two are drawn without replacement. Find P(both yellow).
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
Two dice are rolled. Find P(sum = 8) and P(sum > 9).
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
P(A) = 0.5, P(B) = 0.3, P(A or B) = 0.65. Are A and B independent?
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
A test is 96% accurate. Two independent tests are run. Find P(at least one error).
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can list a sample space systematically.
I multiply along branches and add across paths.
I adjust probabilities correctly for draws without replacement.
I can test independence numerically.

Week 18 · Module 4.2

04 / Data

Combinatorics & Counting

The counting principle, factorials, permutations and combinations

Concept Introduction

The fundamental counting principle: if a first choice has m options and a second has n, the pair has m × n outcomes. This extends to any number of independent stages.

n! = n × (n − 1) × … × 1 counts the arrangements of n distinct objects, with 0! defined as 1.

Permutations count ordered selections: nPr = n!/(n − r)!. Combinations count unordered selections: nCr = n!/(r!(n − r)!).

The decisive question is always: does order matter? Seating, ranking, passwords and podium places are permutations; committees, teams and hands of cards are combinations.

Permutation branching versus unordered combination grouping

Key Relationships

Counting principle: m × n × …
n! = n(n−1)…1, 0! = 1
nPr = n!/(n−r)!
nCr = n!/(r!(n−r)!)

Worked Examples

Worked Example 1 — Ordered versus unordered

From 9 athletes, how many ways can (a) gold, silver and bronze be awarded, (b) a team of 3 be chosen?

1(a) Order matters: 9P3 = 9 × 8 × 7 = 504.
2(b) Order does not matter: 9C3 = 504 ÷ 3! = 504/6 = 84.
3Note that nCr = nPr ÷ r!.

Core Heuristic: compute the permutation first, then divide by r! if order is irrelevant.

Worked Example 2 — Restricted committee

A committee of 4 is chosen from 6 women and 5 men. How many contain exactly 2 women?

1Choose 2 women from 6: 6C2 = 15.
2Choose 2 men from 5: 5C2 = 10.
3Multiply the independent stages: 15 × 10 = 150.

Core Heuristic: split a restricted selection into independent blocks, count each with nCr, then multiply.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Evaluate 6!.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
A menu has 4 starters and 5 mains. How many different two-course meals can you have?
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Evaluate 7C2.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
How many 4-letter arrangements from the 8 distinct letters of PLUMBING (using each once)?
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
How many ways can 5 people be seated in a row?
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
From 10 books, choose 4 to take on holiday. How many different selections can you have?
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
A PIN uses 4 digits 0–9 with repetition allowed. How many different possible PINs are there?
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
How many different arrangements of the letters of BANANA are there?
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
From 7 boys and 5 girls, choose a team of 5 with at least 4 girls.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
Five people sit in a row but two specific people must sit together. How many arrangements can there be?
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I decide whether order matters before choosing a formula.
I can compute factorials, nPr and nCr accurately.
I split restricted problems into independent blocks and multiply.
I handle repeated letters with the division rule.

Week 19 · Module 4.3

04 / Data

Measures of Spread

Five-number summary, interquartile range, box plots and the 1.5 × IQR outlier rule

Concept Introduction

The five-number summary is minimum, Q₁, median, Q₃, maximum. It must be computed from ordered data.

The median splits the data in half; Q₁ is the median of the lower half and Q₃ the median of the upper half. IQR = Q₃ − Q₁ measures the spread of the middle 50%.

A box-and-whisker plot draws the box from Q₁ to Q₃ with the median inside, and whiskers to the extreme values that are not outliers.

A value is an outlier if it lies below Q₁ − 1.5 × IQR or above Q₃ + 1.5 × IQR. IQR is preferred over range because it resists outliers.

Box-and-whisker plot with 1.5 × IQR outlier fences

Key Relationships

IQR = Q₃ − Q₁
Range = max − min
Lower fence = Q₁ − 1.5 IQR
Upper fence = Q₃ + 1.5 IQR

Worked Examples

Worked Example 1 — Five-number summary

Find the five-number summary of 4, 7, 8, 11, 13, 14, 18, 21, 25.

1Data is already ordered; n = 9, so the median is the 5th value = 13.
2Lower half: 4, 7, 8, 11 → Q₁ = (7 + 8)/2 = 7.5.
3Upper half: 14, 18, 21, 25 → Q₃ = (18 + 21)/2 = 19.5.
4Summary: 4, 7.5, 13, 19.5, 25 with IQR = 12.

Core Heuristic: order the data and locate the median before touching the quartiles.

Worked Example 2 — Outlier test

For a data set with Q₁ = 22, Q₃ = 34, is the value 55 an outlier?

1IQR = 34 − 22 = 12, so 1.5 × IQR = 18.
2Upper fence = 34 + 18 = 52.
355 > 52, so 55 is an outlier and is plotted as an isolated point.

Core Heuristic: compute both fences once, then test every extreme value against them.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find the median of 3, 8, 9, 12, 15.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the range of 12, 18, 5, 27, 9.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find Q₁ and Q₃ of 2, 4, 6, 8, 10, 12, 14, 16.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Find the IQR of 11, 13, 16, 18, 21, 24, 29.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Q₁ = 40, Q₃ = 58. Find both outlier fences.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Sketch the box plot for the summary 5, 9, 14, 20, 26 and state the IQR.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Two classes have equal medians but IQRs of 6 and 15. Compare consistency.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
For 8, 11, 12, 14, 15, 17, 19, 20, 22, 41, find the five-number summary and identify outliers.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A box plot has whiskers at 12 and 48, a box from 20 to 32 and a median of 26. Describe the skew.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
A data set of 20 values has IQR 14 and Q₃ 51. A new value 18 is added. Is it an outlier?
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I always order data before computing quartiles.
I can compute IQR and both outlier fences.
I can draw and read a box plot accurately.
I can compare distributions using spread, not just averages.

Week 20 · Module 4.4

04 / Data

Bivariate Data & Sampling Bias

Scatter plots, line of best fit, correlation coefficient r, causation and sampling design

Concept Introduction

A scatter plot shows paired data. Its pattern reveals the direction (positive/negative), form (linear/curved) and strength of association.

The line of best fit minimises the vertical residuals. Predicting inside the data range is interpolation (reasonably safe); outside it is extrapolation (unreliable).

Pearson's r lies between −1 and 1. Values near ±1 indicate strong linear association; near 0 indicates weak linear association — which does not rule out a curved relationship.

Correlation is not causation: a lurking third variable or reverse causation may explain the pattern. Sound conclusions require a well-designed sample: random, representative and large enough, with stratified sampling used when the population has distinct subgroups.

Scatter plot with line of best fit and residuals

Key Relationships

−1 ≤ r ≤ 1
Residual = observed − predicted
Line of best fit: ŷ = a + bx
Stratum sample size = (stratum/population) × total sample

Worked Examples

Worked Example 1 — Interpreting r and predicting

For study hours x and score y, the fit is ŷ = 42 + 5.5x with r = 0.88 (data from x = 1 to 8). Predict the score for 6 hours and comment on x = 20.

1ŷ = 42 + 5.5(6) = 42 + 33 = 75.
2r = 0.88 indicates a strong positive linear association, so the prediction is reasonable (interpolation).
3x = 20 lies far outside 1–8, so predicting there is extrapolation and unreliable — the linear trend may not continue.

Core Heuristic: always check whether the prediction point lies inside the observed data range before trusting it.

Worked Example 2 — Stratified sample

A school has 540 juniors, 360 middles and 300 seniors. Design a stratified sample of 80.

1Population = 540 + 360 + 300 = 1 200.
2Juniors: 540/1 200 × 80 = 36.
3Middles: 360/1 200 × 80 = 24.
4Seniors: 300/1 200 × 80 = 20. Total 80 ✓

Core Heuristic: each stratum's sample is its population share multiplied by the total sample size.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Describe the correlation if r = −0.91.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
For ŷ = 12 + 3x, predict y when x = 7.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
State whether ice-cream sales causing drownings is a valid causal claim.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Observed y = 46 while ŷ = 41. Find the residual.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Data ranges from x = 2 to x = 15. Classify predictions at x = 9 and x = 30.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
A survey is conducted only outside a gym about exercise habits. Name the bias.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
A town has 800 adults and 200 teens. Design a stratified sample of 50.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
For ŷ = 90 − 2.4x with r = −0.76, interpret the gradient and strength.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A scatter plot is strongly curved but r ≈ 0.05. Explain.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
Points (2, 9) and (8, 27) lie on the line of best fit. Find its equation and predict y at x = 5.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can describe direction, form and strength from a scatter plot.
I distinguish interpolation from extrapolation.
I never infer causation from correlation alone.
I can compute stratified sample sizes and name sampling biases.

Week 21 · Module 5.1

05 / Finance

Applied Real-World Percentages

Discounts, markups, profit margins, VAT/GST and reverse percentages

Concept Introduction

A percentage change is applied with a multiplier: an increase of r% multiplies by (1 + r/100) and a decrease by (1 − r/100). Chained changes multiply their multipliers.

Markup is measured on cost price; profit margin is measured on selling price. The same dollar profit gives a larger markup percentage than margin percentage.

VAT/GST is added to the pre-tax price, so tax-inclusive price = pre-tax × (1 + rate).

Reverse percentages undo a multiplier: divide, do not subtract. If a price after a 20% discount is $96, the original is 96 ÷ 0.8 = $120.

Segmented percentage bar: cost, markup and tax

Key Relationships

New = original × (1 ± r/100)
Markup % = profit / cost × 100
Margin % = profit / selling price × 100
Original = final ÷ multiplier

Worked Examples

Worked Example 1 — Chained multipliers

A jacket costs a retailer $80. It is marked up 45%, then discounted 20% in a sale, then 10% GST is added. Find the final price.

1Marked price: 80 × 1.45 = $116.
2Sale price: 116 × 0.8 = $92.80.
3With GST: 92.80 × 1.10 = $102.08.

Core Heuristic: multiply the multipliers in order — never add or subtract percentages of different bases.

Worked Example 2 — Reverse percentage and margin

An item sells for $253 including 15% tax. Find the pre-tax price, and the margin if it cost $180.

1Pre-tax = 253 ÷ 1.15 = $220.
2Profit on the pre-tax sale = 220 − 180 = $40.
3Margin = 40/220 × 100 ≈ 18.2% (markup would be 40/180 ≈ 22.2%).

Core Heuristic: to remove a percentage that was added, divide by its multiplier.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find 18% of $450.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Increase $240 by 25%.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
A $60 shirt is discounted 30%. Find the sale price.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Cost $75, selling price $96. Find the markup percentage.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
A price is $184 after a 20% discount. Find the original price.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Add 12% GST to $340.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Cost $120, selling price $150. Find the profit margin.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
A price rises 20% then falls 20%. Find the overall percentage change.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
A shop must make a 25% margin on a $180 cost item. Find the selling price.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
After a 15% pay rise, a salary is $63 250. Find the rise in dollars.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I use multipliers rather than adding percentage points.
I distinguish markup (on cost) from margin (on price).
I divide to reverse a percentage change.
I can chain several percentage changes in the right order.

Week 22 · Module 5.2

05 / Finance

Simple vs. Compound Interest

Linear interest I = Prt, compound growth A = P(1 + r/n)^(nt), EAR and the Rule of 72

Concept Introduction

Simple interest is earned only on the original principal, so the balance grows linearly: I = Prt and A = P(1 + rt).

Compound interest is earned on the accumulated balance, so growth is exponential: A = P(1 + r/n)^(nt), where n is the number of compounds per year.

The effective annual rate converts any compounding frequency into a single yearly figure: EAR = (1 + r/n)ⁿ − 1. More frequent compounding gives a higher EAR for the same nominal rate.

The Rule of 72 estimates doubling time: years ≈ 72 ÷ (annual percentage rate). It is an approximation, useful for sanity-checking an exponential answer.

Simple linear interest versus compounding exponential growth

Key Relationships

I = Prt, A = P(1 + rt)
A = P(1 + r/n)^(nt)
EAR = (1 + r/n)ⁿ − 1
Doubling time ≈ 72 ÷ r%

Worked Examples

Worked Example 1 — Simple versus compound

Compare $5 000 for 4 years at 6% p.a. under simple interest and annual compounding.

1Simple: I = 5 000 × 0.06 × 4 = $1 200, so A = $6 200.
2Compound: A = 5 000(1.06)⁴ = 5 000 × 1.26248 ≈ $6 312.38.
3Compounding earns about $112.38 more, entirely from interest on interest.

Core Heuristic: simple interest is a straight line, compound is a curve — the gap widens with time, not with rate alone.

Worked Example 2 — Monthly compounding and EAR

$8 000 is invested at 5% p.a. compounded monthly for 3 years. Find the balance and the EAR.

1r/n = 0.05/12 ≈ 0.0041667 and nt = 36.
2A = 8 000(1.0041667)³⁶ ≈ 8 000 × 1.16147 ≈ $9 291.79.
3EAR = (1.0041667)¹² − 1 ≈ 0.05116 = 5.12%.

Core Heuristic: divide the rate by n and multiply the years by n — never mix a monthly rate with a yearly exponent.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Find the simple interest on $2 000 at 4% p.a. for 3 years.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Find the total amount for $1 500 at 5% simple for 6 years.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Find the value of $1 000 after 2 years at 8% compounded annually.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
$4 000 at 7% p.a. compounded annually for 5 years, to the cent.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
Estimate the doubling time at 9% p.a. using the Rule of 72.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
$6 000 at 6% compounded quarterly for 2 years.
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Find the EAR for 12% p.a. compounded monthly.
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
How long for $3 000 to reach $4 500 at 8% compounded annually? Give whole years.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
Compare $10 000 for 10 years at 5%: simple versus annual compounding. Find the difference.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
What principal grows to $12 000 in 4 years at 6% compounded semi-annually?
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can distinguish and compute simple versus compound growth.
I adjust the rate and exponent for the compounding frequency.
I can compute and interpret an effective annual rate.
I use the Rule of 72 as a quick sanity check.

Week 23 · Module 5.3

05 / Finance

Personal Budgeting & Cash Flow

The 50/30/20 framework, fixed and variable costs, emergency funds and debt amortisation

Concept Introduction

A budget starts from net (after-tax) income. The 50/30/20 framework allocates 50% to needs, 30% to wants and 20% to savings and debt repayment.

Fixed costs (rent, insurance, loan repayments) stay constant each period; variable costs (food, fuel, entertainment) change with behaviour and are where adjustments are made first.

An emergency fund is usually 3–6 months of essential expenses, held liquid. It is sized from needs, not from total income.

In an amortisation schedule each payment is split into interest on the outstanding balance and principal reduction. Early payments are interest-heavy; as the balance falls, more of each payment reduces the principal.

Waterfall cash flow: income to needs, wants and savings

Key Relationships

Needs 50% / Wants 30% / Savings 20% of net income
Net cash flow = income − total expenses
Emergency fund = monthly needs × 3 to 6
Interest portion = balance × monthly rate

Worked Examples

Worked Example 1 — Building a 50/30/20 budget

Net monthly income is $3 400. Allocate it under 50/30/20 and check feasibility if needs actually cost $1 900.

1Needs target = 0.5 × 3 400 = $1 700.
2Wants = 0.3 × 3 400 = $1 020; savings = 0.2 × 3 400 = $680.
3Actual needs are $1 900, which is $200 over target.
4Cover the shortfall from wants: wants become $820, keeping savings at $680.

Core Heuristic: protect the savings allocation first and absorb overspending in the variable (wants) category.

Worked Example 2 — First month of amortisation

A $9 000 loan at 12% p.a. is repaid at $400 per month. Find the interest and principal in month 1 and the new balance.

1Monthly rate = 12%/12 = 1% = 0.01.
2Interest = 9 000 × 0.01 = $90.
3Principal reduction = 400 − 90 = $310.
4New balance = 9 000 − 310 = $8 690.

Core Heuristic: always compute interest on the current balance first; the principal reduction is whatever the payment leaves over.

Practice Set — 10 Questions

3 Bronze · 4 Silver · 3 Gold

1.
Net income is $2 800. Find the 50/30/20 allocations.
Bronze
Working Space & Algebraic Steps:
Final Answer:
2.
Classify rent and groceries as fixed or variable.
Bronze
Working Space & Algebraic Steps:
Final Answer:
3.
Income $3 100, expenses $2 740. Find the net cash flow.
Bronze
Working Space & Algebraic Steps:
Final Answer:
4.
Monthly needs are $1 450. Find a 4-month emergency fund.
Silver
Working Space & Algebraic Steps:
Final Answer:
5.
A $6 000 loan at 18% p.a. has a $250 monthly payment. Each payment first covers that month's interest; the rest reduces the principal (the amount owed). Find how much the principal is reduced by in month 1.
Silver
Working Space & Algebraic Steps:
Final Answer:
6.
Net income $4 200 with needs $2 310. What percentage goes to needs, and does it pass 50/30/20?
Silver
Working Space & Algebraic Steps:
Final Answer:
7.
Savings are $520 per month. How long to reach a $6 240 goal?
Silver
Working Space & Algebraic Steps:
Final Answer:
8.
Income $3 600; fixed $1 800; variable $1 350. Find savings, the savings rate, and the cut needed to reach 20%.
Gold
Working Space & Algebraic Steps:
Final Answer:
9.
For a $12 000 loan at 9% p.a. with $500 monthly payments, find the balance after 2 months.
Gold
Working Space & Algebraic Steps:
Final Answer:
10.
A student earns $1 900 net, spends $1 240 on needs and saves 20%. Find the wants budget and whether needs fit 50%.
Gold
Working Space & Algebraic Steps:
Final Answer:

Self-Audit Checkpoint

I can build a 50/30/20 budget from net income.
I can classify fixed versus variable costs.
I size an emergency fund from essential expenses.
I can split a loan payment into interest and principal.

Appendix

Answer Key & Step-by-Step Solutions

Complete worked solutions for all 230 practice questions.

Week 1 · Factors, Multiples & Primes

1. Answer: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
1. Test divisors in pairs: 1×48, 2×24, 3×16, 4×12, 6×8.
2. Stop once the pair values cross (√48 ≈ 6.9).
2. Answer: 2 × 3² × 5
1. 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5.
2. Collect: 2 × 3 × 3 × 5 = 2 × 3² × 5.
3. Answer: Not prime: 91 = 7 × 13
1. Test primes up to √91 ≈ 9.5: 2, 3, 5, 7.
2. 91 ÷ 7 = 13 exactly, so 91 is composite.
4. Answer: 42
1. 126 = 2 × 3² × 7; 210 = 2 × 3 × 5 × 7.
2. Shared primes at lowest powers: 2 × 3 × 7 = 42.
5. Answer: 360
1. 18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5.
2. Highest powers: 2³ × 3² × 5 = 360.
6. Answer: 21:01:12 (after 72 s)
1. The joint period is LCM(24, 36).
2. 24 = 2³×3, 36 = 2²×3² → LCM = 2³×3² = 72 s.
3. 21:00:00 + 72 s = 21:01:12.
7. Answer: 32
1. Use HCF × LCM = a × b: 8 × 96 = 768.
2. 768 ÷ 24 = 32.
8. Answer: 45 cm; 70 tiles
1. Largest square side = HCF(315, 450).
2. 315 = 3² × 5 × 7; 450 = 2 × 3² × 5² → HCF = 3² × 5 = 45.
3. Tiles = (315/45) × (450/45) = 7 × 10 = 70.
9. Answer: 24 factors; 18 even
1. Factor count = (3+1)(2+1)(1+1) = 24.
2. Odd factors come from 3² × 5: (2+1)(1+1) = 6.
3. Even factors = 24 − 6 = 18.
10. Answer: k = 63
1. 1176 = 2³ × 3 × 7².
2. Every index in a cube must be a multiple of 3: 2³ is fine, 3¹ needs two more 3s, 7² needs one more 7.
3. k = 3² × 7 = 63.
4. Check: 1176 × 63 = 74 088 = 42³. ✓

Week 2 · Ratio & Proportion

1. Answer: 3 : 5
1. HCF(72, 120) = 24.
2. 72÷24 : 120÷24 = 3 : 5.
2. Answer: 36 and 48
1. Total parts = 7, one part = 12.
2. 3×12 = 36, 4×12 = 48.
3. Answer: $91
1. One book = 52 ÷ 8 = $6.50.
2. 14 × 6.50 = $91.
4. Answer: 171
1. One part = 133 ÷ 7 = 19.
2. Total = 9 × 19 = 171.
5. Answer: 2.1 km
1. Real = 8.4 × 25 000 = 210 000 cm.
2. 210 000 cm = 2 100 m = 2.1 km.
6. Answer: 75
1. k = 45/12 = 3.75.
2. y = 3.75 × 20 = 75.
7. Answer: 1 875 g
1. Per person = 750 ÷ 6 = 125 g.
2. 15 × 125 = 1 875 g (1.875 kg).
8. Answer: 20
1. Let red = 5k and blue = 3k.
2. 5k : (3k + 16) = 5 : 7 → 7(5k) = 5(3k + 16).
3. 35k = 15k + 80 → 20k = 80 → k = 4.
4. Red = 5k = 20 (check: blue 12 + 16 = 28 and 20 : 28 = 5 : 7 ✓).
9. Answer: 120 km/h
1. Invariant: v × t = 80 × 0.75 = 60 km.
2. v = 60 ÷ 0.5 = 120 km/h.
10. Answer: $14 400
1. Difference = 2 parts = $2 400 → one part = $1 200.
2. Total = 12 parts = 12 × 1 200 = $14 400.

Week 3 · Signed Numbers & Coordinates

1. Answer: 8
1. Move 21 units right from −13.
2. 21 − 13 = 8.
2. Answer: 7
1. Double negative becomes addition: −8 + 15.
2. = 7.
3. Answer: 16
1. |−12| = 12 and |5 − 9| = |−4| = 4.
2. 12 + 4 = 16.
4. Answer: −16
1. (−48) ÷ (−6) = 8.
2. 8 × (−2) = −16.
5. Answer: Q2, Q4, Q3
1. Signs (−,+) → Q2; (+,−) → Q4; (−,−) → Q3.
6. Answer: 17 °C
1. Drop = |6 − (−11)| = 17.
7. Answer: (−1, −3)
1. x: (−7+5)/2 = −1.
2. y: (2−8)/2 = −3.
8. Answer: x = 13 or x = −5
1. Two cases: x − 4 = 9 or x − 4 = −9.
2. x = 13 or x = −5.
9. Answer: 40 square units
1. AB = |6 − (−4)| = 10; BC = |3 − (−5)| = 8.
2. Area = ½ × 10 × 8 = 40.
10. Answer: −100 m
1. Rise total = 4 × 35 = 140 → −180 + 140 = −40.
2. Dive: −40 − 60 = −100 m.

Week 4 · Powers, Roots & Scientific Notation

1. Answer: 2⁸ = 256
1. Add indices: 5 + 3 = 8.
2. 2⁸ = 256.
2. Answer: 1 1/9
1. 7⁰ = 1 and 3⁻² = 1/9.
2. Sum = 10/9.
3. Answer: 4.6 × 10⁻⁴
1. Move the point 4 places right to get 4.6.
2. Moving right gives a negative exponent.
4. Answer: 16a¹²b⁴
1. Raise each factor: 2⁴ = 16, (a³)⁴ = a¹², b⁴.
5. Answer: 16
1. Cube root of 64 = 4.
2. 4² = 16.
6. Answer: x⁴y³
1. Subtract indices: 7−3 = 4, −2−(−5) = 3.
7. Answer: 2.1 × 10⁴
1. 3 × 7 = 21 and 10⁵⁻² = 10³.
2. 21 × 10³ = 2.1 × 10⁴.
8. Answer: x = 4
1. 32 = 2⁵.
2. Equate indices: x + 1 = 5 → x = 4.
9. Answer: 8
1. √50 = 5√2, √18 = 3√2 → sum 8√2.
2. 8√2 ÷ √2 = 8.
10. Answer: 1.5 × 10⁻² g
1. 2.5 × 6 = 15 and 10⁻¹¹⁺⁸ = 10⁻³.
2. 15 × 10⁻³ = 1.5 × 10⁻² g.

Week 5 · Expressions & Factoring

1. Answer: 12x − 28
1. Multiply each term by 4.
2. Answer: 6x(2x + 3)
1. HCF of terms is 6x.
2. Divide each term by 6x.
3. Answer: x² + 9x + 18
1. Sum 9, product 18.
4. Answer: 9x² − 24x + 16
1. Use (a−b)² = a² − 2ab + b².
2. a = 3x, b = 4.
5. Answer: (7x − 9)(7x + 9)
1. Difference of two squares with a = 7x, b = 9.
6. Answer: (x − 3)(x − 8)
1. Product 24, sum −11 → −3 and −8.
7. Answer: 9x − 19
1. First product: x² + 3x − 10.
2. Second: (x − 3)² = x² − 6x + 9.
3. Subtract: (x² + 3x − 10) − (x² − 6x + 9) = 9x − 19.
8. Answer: (2x − 3)(5x − 7)
1. ac = 210, sum −29 → −14 and −15.
2. 10x² − 14x − 15x + 21 = 2x(5x − 7) − 3(5x − 7).
9. Answer: 3x(x − 3)(x + 3)
1. Common factor 3x → 3x(x² − 9).
2. x² − 9 is a difference of squares.
10. Answer: (2x + 3) and (x + 4)
1. ac = 24, sum 11 → 3 and 8.
2. 2x² + 3x + 8x + 12 = x(2x+3) + 4(2x+3).

Week 6 · Linear Equations & Inequalities

1. Answer: x = 7
1. Add 8: 5x = 35.
2. Divide by 5.
2. Answer: x = 3
1. Expand: 3x + 12 = 21.
2. 3x = 9.
3. Answer: x = 36
1. Multiply both sides by 4.
4. Answer: x = 6
1. Gather: 4x = 24.
5. Answer: x = 6
1. 8x − 12 = 3x + 18.
2. 5x = 30.
6. Answer: x = −2
1. Multiply every term by 15: 3(x + 2) − 5(x − 1) = 15.
2. 3x + 6 − 5x + 5 = 15 → −2x + 11 = 15.
3. −2x = 4 → x = −2. Check: 0/5 − (−3)/3 = 1 ✓
7. Answer: x < −2.5
1. 2x < −5 → x < −2.5; open circle at −2.5 pointing left.
8. Answer: −1 ≤ x < 4
1. Add 1 throughout: −3 ≤ 3x < 12.
2. Divide by 3.
9. Answer: x < 5/3
1. 5 − 2x + 6 > 4x + 1 → 11 − 2x > 4x + 1.
2. 10 > 6x → x < 5/3.
10. Answer: 8 cm
1. Let w be width, length = 2w + 3.
2. 2(w + 2w + 3) = 54 → 6w + 6 = 54.
3. 6w = 48 → w = 8 cm.

Week 7 · Quadratics & the Formula

1. Answer: x = 4 or x = 5
1. Product 20, sum −9 → −4, −5.
2. Answer: (x + 3)² − 8
1. Half of 6 is 3; (x+3)² = x²+6x+9; 1 − 9 = −8.
3. Answer: Δ = 0, one repeated root
1. 16 − 16 = 0.
4. Answer: x = 2 or x = −7
1. Product −14, sum 5 → 7 and −2.
5. Answer: x = 4/3 or x = −2
1. ac = −24, sum 2 → 6, −4.
2. 3x² + 6x − 4x − 8 = 3x(x+2) − 4(x+2) = (x+2)(3x−4).
6. Answer: x = 3 ± √7
1. Complete the square: (x−3)² − 7 = 0.
2. (x−3)² = 7.
7. Answer: (4, −5)
1. Completed square form reads directly; a = 2 > 0 so it is a minimum.
8. Answer: k = ±6
1. Δ = k² − 36 = 0.
2. k² = 36 → k = 6 or −6.
9. Answer: x ≈ 3.39 or x ≈ −0.89
1. Δ = 25 + 48 = 73.
2. x = (5 ± √73)/4 with √73 ≈ 8.544.
10. Answer: 5 cm by 8 cm
1. x + y = 13 and xy = 40.
2. x(13 − x) = 40 → x² − 13x + 40 = 0.
3. (x−5)(x−8) = 0 → 5 and 8.

Week 8 · Cartesian Plane & Linear Graphs

1. Answer: m = 5, c = −3
1. Read directly from y = mx + c.
2. Answer: 3
1. Δy/Δx = 12/4 = 3.
3. Answer: (4, 0)
1. Set y = 0 → 2x = 8.
4. Answer: y = −3x + 11
1. y − 5 = −3(x − 2).
5. Answer: Yes, both m = 2
1. Rearrange: −2y = −4x + 9 → y = 2x − 4.5.
6. Answer: 5/2
1. Negative reciprocal of −2/5.
7. Answer: y = −¾x + 6; (8, 0) and (0, 6)
1. 4y = −3x + 24.
2. Divide by 4.
8. Answer: y = 4, a horizontal line
1. Gradient = 0/8 = 0, so y is constant.
9. Answer: y = ½x − 2
1. Given line: y = −2x + 5, m = −2.
2. Perpendicular m = ½, c = −2.
10. Answer: (3, 5)
1. 3x − 4 = −x + 8 → 4x = 12 → x = 3.
2. y = 3(3) − 4 = 5.

Week 9 · Arithmetic Sequences & Series

1. Answer: d = 7, next 31
1. Subtract consecutive terms.
2. Answer: 58
1. T₁₀ = 4 + 9(6).
3. Answer: 1 275
1. S = 50/2 (1 + 50) = 25 × 51.
4. Answer: Tₙ = 24 − 4n
1. a = 20, d = −4 → 20 + (n−1)(−4).
5. Answer: the 29th
1. Tₙ = 7n − 1 = 202 → 7n = 203.
6. Answer: 980
1. S = 20/2[22 + 19(4)] = 10 × 98.
7. Answer: a = −1, d = 6
1. Difference: 7d = 42 → d = 6.
2. a = 23 − 4(6) = −1.
8. Answer: 20 terms
1. Sₙ = n/2[8 + 5(n−1)] = n(5n + 3)/2 = 1 030.
2. 5n² + 3n − 2 060 = 0 → (5n + 103)(n − 20) = 0.
3. n must be a positive integer, so n = 20.
9. Answer: 1 350
1. a = 18, d = 3, n = 25.
2. S = 25/2[36 + 72] = 25/2 × 108.
10. Answer: a = 5, d = 4
1. S₁ = 5 so a = 5.
2. S₂ = 14 → T₂ = 9 → d = 4.

Week 10 · Functions & Rate of Change

1. Answer: 7
1. 3(4) − 5 = 7.
2. Answer: 10
1. (−3)² + 1 = 10.
3. Answer: all real x except 2
1. Denominator zero at x = 2.
4. Answer: 5
1. f(0) = 0, f(3) = 15.
2. 15/3 = 5.
5. Answer: (x − 2)/(x + 3), x ≠ −2, −3
1. Factor: (x−2)(x+2) / (x+2)(x+3).
6. Answer: x = −6
1. 5 − 2x = 17 → −2x = 12.
7. Answer: y ≥ −4
1. x² ≥ 0, so minimum at x = 0.
8. Answer: 2a + h
1. ((a+h)² − a²)/h = (2ah + h²)/h.
9. Answer: 0 m/s
1. h(1) = 15, h(3) = 60 − 45 = 15.
2. (15 − 15)/2 = 0.
10. Answer: f(6) = 6.5; domain x ≠ 4; range y ≠ 2
1. f(6) = 13/2.
2. Denominator zero at 4.
3. Solving y = 2 gives no solution, so y = 2 is excluded.

Week 11 · Geometric Angle Theorems

1. Answer: 133°
1. 180 − 47 = 133.
2. Answer: 720°
1. (6 − 2) × 180.
3. Answer: 36°
1. 360 ÷ 10.
4. Answer: 36°
1. 3x + 2x = 180 → 5x = 180.
5. Answer: n = 24, interior 165°
1. 360 ÷ 15 = 24; 180 − 15 = 165.
6. Answer: 40°, 60°, 80°
1. 9x = 180 → x = 20.
7. Answer: 108°
1. Exterior = 72°, interior = 108°.
8. Answer: 12
1. (n − 2)180 = 1 800 → n − 2 = 10.
9. Answer: x = 20, angle 90°
1. Corresponding angles are equal: 4x + 10 = 6x − 30.
2. 40 = 2x → x = 20 → angle = 90°.
10. Answer: 105°
1. Sum = 540°.
2. 540 − 435 = 105°.

Week 12 · Pythagorean Theorem & Trigonometry

1. Answer: 15 cm
1. 81 + 144 = 225.
2. √225 = 15.
2. Answer: 24
1. 676 − 100 = 576.
2. √576 = 24.
3. Answer: 9
1. x = 18 × 0.5.
4. Answer: 54.5°
1. θ = tan⁻¹(1.4).
5. Answer: 6 m
1. 42.25 − 6.25 = 36.
2. √36 = 6.
6. Answer: 11.3 cm
1. cos 41° = A/15 → A = 15 cos 41°.
7. Answer: 12.12 cm
1. d = 7√3 ≈ 12.12.
8. Answer: 191 m
1. tan 24° = 85/d → d = 85/tan 24°.
2. ≈ 190.9 → 191 m.
9. Answer: 28.1°
1. Base diagonal = √(144+81) = 15.
2. tan θ = 8/15 → θ ≈ 28.07°.
10. Answer: 12 cm; 60 cm²
1. Half base = 5, so h = √(169 − 25) = 12.
2. Area = ½ × 10 × 12 = 60.

Week 13 · Non-Right Trigonometry Pt 1: Sine Rule

1. Answer: 14.1
1. b = 10 sin 65°/sin 40°.
2. Answer: 28.3
1. ½ × 8 × 11 × sin 40°.
3. Answer: 0.75
1. sin B = 9 sin 30°/6.
4. Answer: 23.8
1. a = 20 sin 62°/sin 48°.
5. Answer: 30.0°
1. sin B = 11 sin 43°/15 ≈ 0.5001.
6. Answer: C = 62°, c ≈ 10.1
1. C = 180 − 118 = 62°.
2. c = 9.4 sin 62°/sin 55°.
7. Answer: 48.6° or 131.4°
1. 30 = ½(80) sin C → sin C = 0.75.
2. C = sin⁻¹(0.75) ≈ 48.6°, or C = 180° − 48.6° = 131.4°.
8. Answer: Two: B ≈ 55.1° or 124.9°
1. sin B = 10 sin 35°/7 ≈ 0.8195.
2. Both satisfy the angle sum.
9. Answer: ≈ 226 m
1. Angle at the balloon = 180° − 38° − 47° = 95°.
2. Sine Rule: the side from the 38° observer = 500 sin 47°/sin 95° ≈ 367.0 m.
3. Height = 367.0 × sin 38° ≈ 226 m.
10. Answer: B ≈ 47.6°, unique
1. sin B = 9 sin 100°/12 ≈ 0.7386 → B ≈ 47.6°.
2. The obtuse partner 132.4° would exceed the angle sum with A = 100°.

Week 14 · Non-Right Trigonometry Pt 2: Cosine Rule

1. Answer: √39 ≈ 6.2
1. 25 + 49 − 70(0.5) = 39.
2. Answer: 270°
1. Measured clockwise from north.
3. Answer: 0
1. (36 + 64 − 100)/96 = 0, so C = 90°.
4. Answer: 96.4°
1. Largest angle faces 12: cos C = (49+81−144)/126 = −1/9 ≈ −0.1111.
2. C = cos⁻¹(−0.1111) ≈ 96.4°.
5. Answer: ≈ 18.34
1. b² = 169 + 100 − 260 cos 105° ≈ 336.29.
2. b = √336.29 ≈ 18.34.
6. Answer: 21.6 km
1. Included angle 90° → d² = 144 + 324 = 468.
7. Answer: 295°
1. 115 + 180 = 295.
8. Answer: ≈ 117.6°; area ≈ 53.2
1. The largest angle faces the side 20: cos C = (64 + 225 − 400)/(2 × 8 × 15) = −0.4625.
2. C ≈ 117.6°.
3. Area = ½ × 8 × 15 × sin 117.6° ≈ 53.2.
9. Answer: 76.3 km
1. Included angle at A = 90°.
2. BC² = 1 600 + 4 225 = 5 825 → 76.3 km.
10. Answer: Right angle at C; area 54
1. 81 + 144 = 225 = 15², so C = 90°.
2. Area = ½ × 9 × 12 = 54.

Week 15 · Surface Area & Volume of Solids

1. Answer: 216 cm³
1. Multiply the dimensions.
2. Answer: 282.7 cm³
1. V = π(9)(10) = 90π.
3. Answer: 113.1 cm³
1. V = 4/3 π(27) = 36π.
4. Answer: 377.0 cm²
1. 2π(16) + 2π(4)(11) = 32π + 88π = 120π.
5. Answer: 378 cm³
1. ⅓ × 81 × 14.
6. Answer: 314.2 cm³
1. ⅓π(25)(12) = 100π.
7. Answer: 360 cm³
1. V = 24 × 15.
8. Answer: ≈ 901.6 cm²
1. Hemisphere curved surface = 2π(7²) = 98π.
2. Cylinder curved surface = 2π(7)(10) = 140π.
3. Circular base = π(7²) = 49π (the join is hidden).
4. Total = 287π ≈ 901.6 cm².
9. Answer: 18 cm
1. Sphere V = 288π.
2. 288π = 16πh → h = 18.
10. Answer: ≈ 70.7 cm³
1. Similar cone: r = 3, h = 7.5.
2. V = ⅓π(9)(7.5) = 22.5π ≈ 70.7.

Week 16 · Similarity & Congruence

1. Answer: 2.5
1. k = 10/4.
2. Answer: 9 and 27
1. k² = 9, k³ = 27.
3. Answer: RHS
1. Right angle, hypotenuse, side.
4. Answer: 15 and 36
1. k = 3.
5. Answer: 324π
1. k = 1.5, k³ = 3.375.
6. Answer: 1.5
1. √(450/200) = √2.25.
7. Answer: Equal angles fix shape, not size
1. Any enlargement preserves angles.
8. Answer: 432 cm³
1. k² = 216/96 = 2.25 → k = 1.5.
2. k³ = 3.375 → 128 × 3.375 = 432.
9. Answer: 3 456 cm²
1. Area factor = 24² = 576.
2. 6 × 576 = 3 456.
10. Answer: 12.5
1. Triangles ADE and ABC are similar with k = 10/4 = 2.5.
2. BC = 5 × 2.5 = 12.5.

Week 17 · Probability Fundamentals

1. Answer: 1/2
1. 3 of 6 outcomes are even.
2. Answer: 0.65
1. 1 − 0.35.
3. Answer: HH, HT, TH, TT; 1/4
1. 4 equally likely outcomes.
4. Answer: 1/2
1. Primes: 2, 3, 5, 7 → 4/8.
5. Answer: 0.1 and 0.55
1. 0.4 × 0.25 = 0.1.
2. 0.4 + 0.25 − 0.1 = 0.55.
6. Answer: ≈ 0.207 vs 0.167
1. 62/300 ≈ 0.207, slightly above 1/6.
7. Answer: 1/3
1. 6/10 × 5/9 = 30/90.
8. Answer: 5/36 and 1/6
1. Sum 8: 5 ways of 36.
2. Sums 10, 11, 12: 3 + 2 + 1 = 6 → 6/36.
9. Answer: Yes
1. P(A and B) = 0.5 + 0.3 − 0.65 = 0.15.
2. 0.5 × 0.3 = 0.15, so independent.
10. Answer: 0.0784
1. P(both correct) = 0.96² = 0.9216.
2. 1 − 0.9216 = 0.0784.

Week 18 · Combinatorics & Counting

1. Answer: 720
1. 6 × 5 × 4 × 3 × 2 × 1.
2. Answer: 20
1. 4 × 5.
3. Answer: 21
1. 7 × 6 / 2.
4. Answer: 1 680
1. 8P4 = 8 × 7 × 6 × 5.
5. Answer: 120
1. 5! = 120.
6. Answer: 210
1. 10C4 = 210.
7. Answer: 10 000
1. 10⁴.
8. Answer: 60
1. 6!/(3!2!) = 720/12.
9. Answer: 36
1. Case 4 girls + 1 boy: 5C4 × 7C1 = 5 × 7 = 35.
2. Case 5 girls: 5C5 = 1.
3. Total = 35 + 1 = 36.
10. Answer: 48
1. Treat the pair as one block: 4! = 24.
2. The pair can swap internally: × 2 = 48.

Week 19 · Measures of Spread

1. Answer: 9
1. Middle value of 5 ordered items.
2. Answer: 22
1. 27 − 5.
3. Answer: Q₁ = 5, Q₃ = 13
1. Lower half median (4+6)/2; upper (12+14)/2.
4. Answer: 11
1. Median 18; Q₁ = 13, Q₃ = 24.
5. Answer: 13 and 85
1. IQR = 18, 1.5 IQR = 27.
6. Answer: IQR = 11
1. Box 9 to 20 with median 14; whiskers to 5 and 26.
7. Answer: The IQR 6 class is more consistent
1. Smaller IQR means less spread in the middle 50%.
8. Answer: 8, 12, 16, 20, 41; 41 is an outlier
1. Median = (15+17)/2 = 16.
2. Q₁ = 12, Q₃ = 20 → IQR = 8, upper fence = 32.
3. 41 > 32.
9. Answer: Slight right (positive) skew
1. Median 26 sits mid-box, but the upper whisker (16) is longer than the lower (8).
10. Answer: No
1. Q₁ = 51 − 14 = 37.
2. Lower fence = 37 − 21 = 16, and 18 > 16.

Week 20 · Bivariate Data & Sampling Bias

1. Answer: Strong negative linear
1. Close to −1.
2. Answer: 33
1. 12 + 21.
3. Answer: No — a lurking variable (hot weather)
1. Correlation is not causation.
4. Answer: 5
1. 46 − 41.
5. Answer: Interpolation; extrapolation
1. 9 is inside the range, 30 outside.
6. Answer: Selection bias (unrepresentative sample)
1. Gym-goers over-represent active people.
7. Answer: 40 adults, 10 teens
1. Shares 0.8 and 0.2.
8. Answer: Each unit of x lowers y by 2.4; moderately strong negative
1. Gradient is the rate of change; |r| = 0.76.
9. Answer: r measures only linear association
1. A symmetric curve can have near-zero linear correlation.
10. Answer: ŷ = 3x + 3; 18
1. Gradient = 18/6 = 3.
2. 9 = 6 + c → c = 3.
3. ŷ(5) = 18.

Week 21 · Applied Real-World Percentages

1. Answer: $81
1. 0.18 × 450.
2. Answer: $300
1. 240 × 1.25.
3. Answer: $42
1. 60 × 0.7.
4. Answer: 28%
1. Profit 21; 21/75 = 0.28.
5. Answer: $230
1. 184 ÷ 0.8.
6. Answer: $380.80
1. 340 × 1.12.
7. Answer: 20%
1. 30/150 = 0.2.
8. Answer: 4% decrease
1. 1.2 × 0.8 = 0.96.
9. Answer: $240
1. Cost is 75% of price: 180 ÷ 0.75.
10. Answer: $8 250
1. Original = 63 250 ÷ 1.15 = $55 000.
2. Rise = 63 250 − 55 000.

Week 22 · Simple vs. Compound Interest

1. Answer: $240
1. 2 000 × 0.04 × 3.
2. Answer: $1 950
1. I = 450.
3. Answer: $1 166.40
1. 1 000 × 1.08².
4. Answer: $5 610.21
1. 4 000 × 1.07⁵ = 4 000 × 1.4025517 = 5 610.2069…
2. Round to the cent: $5 610.21.
5. Answer: ≈ 8 years
1. 72 ÷ 9.
6. Answer: $6 758.96
1. Quarterly rate = 6% ÷ 4 = 1.5%; periods = 2 × 4 = 8.
2. 6 000(1.015)⁸ = 6 000 × 1.126493 ≈ $6 758.96.
7. Answer: ≈ 12.68%
1. (1.01)¹² − 1.
8. Answer: 6 years
1. 1.08ⁿ = 1.5 → n = ln1.5/ln1.08 ≈ 5.27.
2. So 6 full years are needed.
9. Answer: ≈ $1 288.95
1. Simple: A = $15 000.
2. Compound: 10 000(1.05)¹⁰ ≈ $16 288.95.
10. Answer: $9 472.91
1. A = P(1.03)⁸ → P = 12 000/1.266770.
2. P ≈ $9 472.91.

Week 23 · Personal Budgeting & Cash Flow

1. Answer: $1 400 / $840 / $560
1. Multiply by 0.5, 0.3, 0.2.
2. Answer: Rent fixed; groceries variable
1. Rent is contractually constant.
3. Answer: +$360
1. 3 100 − 2 740.
4. Answer: $5 800
1. 1 450 × 4.
5. Answer: $160
1. Interest = 6 000 × 0.015 = $90.
2. 250 − 90 = 160.
6. Answer: 55% — over target
1. 2 310/4 200 = 0.55.
7. Answer: 12 months
1. 6 240 ÷ 520.
8. Answer: $450, 12.5%, cut $270
1. Savings = 3 600 − 3 150 = 450 → 12.5%.
2. 20% of 3 600 = 720, so cut variable spending by $270.
9. Answer: ≈ $11 176.93
1. Monthly rate = 0.75%.
2. Month 1: interest $90, principal $410 → balance $11 590.
3. Month 2: interest ≈ $86.93, principal ≈ $413.07 → balance ≈ $11 176.93.
10. Answer: $280 wants; needs are 65% — over
1. Savings = 380, needs = 1 240 → wants = 1 900 − 1 620 = 280.
2. 1 240/1 900 ≈ 65%.

Curriculum Architecture

01 / Numbers

Factors, Prime Trees, Ratios, Signed Coordinates & Index Laws

02 / Algebra

Quadratics, Formula Derivations, Linear Graphs, Sequences & Rates of Change

03 / Space

Angle Deductions, 3D Pythagoras, Sine/Cosine Rules & Scale Similarity

04 / Data

Combinatorics, Quartile Spreads, Box Plots & Bivariate Regression

05 / Finance

Commercial Percentages, Compound Interest Trajectories & Budgeting

Vector

Vector

Contents

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Concept Introduction

Worked Examples

Practice Set — 10 Questions

Week 1 · Factors, Multiples & Primes

Week 2 · Ratio & Proportion

Week 3 · Signed Numbers & Coordinates

Week 4 · Powers, Roots & Scientific Notation

Week 5 · Expressions & Factoring

Week 6 · Linear Equations & Inequalities

Week 7 · Quadratics & the Formula

Week 8 · Cartesian Plane & Linear Graphs

Week 9 · Arithmetic Sequences & Series