Chapter 1

Numbers: Standard Form, Percentages & Proportion

Work with very large and very small numbers, solve advanced percentage problems, and master direct proportion.

10
Objectives
3
Sections
20+
Questions

1.1Standard Form

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📖 What is Standard Form?

Standard form writes numbers as A × 10ⁿ where 1 ≤ A < 10 and n is an integer.

Large numbers: n is positive. Small numbers: n is negative.

💡 Examples

  • 345,000,000 = 3.45 × 10⁸
  • 0.00072 = 7.2 × 10⁻⁴
  • On calculators: 2.5E8 means 2.5 × 10⁸
📐 Converting to Standard Form
(a) Write 47,000,000 in standard form. (b) Write 0.000036 in standard form.
a

Move decimal 7 places left: 4.7 × 10⁷

47,000,000 = 4.7 × 10⁷
b

Move decimal 5 places right: 3.6 × 10⁻⁵

0.000036 = 3.6 × 10⁻⁵

✏️ Practice

0/4
Q1
Write 5,600,000 in standard form.
Q2
Write 0.0089 in standard form.
Q3
Write 3.2 × 10⁴ as an ordinary number.
Q4
Calculate (2 × 10³) × (4 × 10⁵). Give in standard form.

1.2Advanced Percentages

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💡 Key Formulas

  • Percentage change = (actual change ÷ original) × 100
  • Reverse percentage: If result = original × multiplier, then original = result ÷ multiplier
📐 Percentage Change
A computer bought for $720 was originally priced higher but reduced by 20%. What was the original price?
1

After 20% reduction, you pay 80% of original.

2

80% = $720. So 100% = 720 ÷ 0.8 = $900

✅

Original price = $900

✏️ Practice

0/3
Q1
A shirt costs £45 after a 10% increase. What was the original price?
Q2
A house was bought for £200,000 and sold for £230,000. Find the percentage profit.
Q3
A computer sold for $540 was bought for $720. Find the percentage loss.

1.3Direct Proportion

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📖 y is directly proportional to x

Written as y ∝ x, this means y = kx where k is a constant. The graph is a straight line through the origin.

📐 Finding k
y is proportional to x. When x = 4, y = 20. Find y when x = 7.
1

y = kx. Substitute: 20 = k × 4. So k = 5.

2

When x = 7: y = 5 × 7 = 35

✅

y = 35

✏️ Practice

0/2
Q1
y ∝ x. When x=3, y=12. Find y when x=10.
Q2
5 books cost £35. How much do 8 books cost?

📝 Chapter 1 Exam

25 marks | 30 min

[3]1.(a) Write 6.52 × 10⁸ as an ordinary number. (b) Write 0.00047 in standard form.
(a) 652,000,000 (b) 4.7 × 10⁻⁴
[3]2.A car loses 15% value per year. It's now worth £12,750. What was it worth a year ago?
85%=£12750. Original=12750÷0.85=£15,000
[3]3.y ∝ x. When x=5, y=30. Find (a) k (b) y when x=12
(a) k=30÷5=6 (b) y=6×12=72
Chapter 2

Algebra: Quadratics, Simultaneous Equations & Inequalities

Expand double brackets, factorise quadratics, solve simultaneous equations, work with inequalities, and explore graphs.

2.1Expanding & Factorising Quadratics

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📖 Expanding Double Brackets

Use FOIL: First, Outside, Inside, Last.

📐 Expand (x + 3)(x − 2)
F

x × x = x²

O

x × (−2) = −2x

I

3 × x = 3x

L

3 × (−2) = −6

✅ Collect like terms:
x² − 2x + 3x − 6 = x² + x − 6

📖 Factorising Quadratics x² + bx + c

Find two numbers that multiply to give c and add to give b.

📐 Factorise x² − 6x + 8
1

Need two numbers that multiply to +8 and add to −6.

2

−2 × −4 = 8 ✓    −2 + −4 = −6 ✓

✅
x² − 6x + 8 = (x − 2)(x − 4)

✏️ Practice

0/4
Q1
Expand: (x + 5)(x + 2)
Q2
Expand: (x − 3)²
Q3
Factorise: x² + 5x + 6
Q4
Solve: x² − 6x + 8 = 0

2.2Simultaneous Equations

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📖 Solving Two Equations Together

Find the values of x and y that satisfy both equations at the same time. Use elimination (add or subtract equations to remove one variable).

📐 Solve: x + y = 5 and x − y = 1
1
Add the equations (y terms cancel)

(x + y) + (x − y) = 5 + 1 → 2x = 6 → x = 3

2
Substitute x = 3 into equation 1

3 + y = 5 → y = 2

✅

x = 3, y = 2. Check: 3+2=5 ✓, 3−2=1 ✓

✏️ Practice

0/2
Q1
Solve: 2x + y = 7 and x − y = 2
Q2
Solve: 3x + 2y = 12 and x + 2y = 8

2.3Inequalities

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💡 Inequality Symbols

  • < means "less than"   |   > means "greater than"
  • ≤ means "less than or equal to"   |   ≥ means "greater than or equal to"
  • On number lines: filled circle ● = included (≤, ≥). Open circle ○ = not included (<, >).
📐 Solve: 4(x + 1) ≥ 7
1

Expand: 4x + 4 ≥ 7

2

Subtract 4: 4x ≥ 3

3

Divide by 4: x ≥ 0.75

✏️ Practice

0/2
Q1
Solve: 3x + 5 > 20
Q2
Solve: 2x − 1 ≤ 9

2.4Sequences: nth Term

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💡 Arithmetic Sequences

nth term = an + b where a = common difference.

Quadratic sequences have second differences that are constant. nth term includes an n² term.

✏️ Practice

0/2
Q1
Find the nth term of: 4, 9, 14, 19, 24...
Q2
The nth term is 2n². Find the first 4 terms.

2.5Linear & Quadratic Graphs

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💡 Key Graph Facts

  • y = mx + c is a straight line. Parallel lines have the same gradient.
  • y = ax² gives a parabola (U-shape if a>0, ∩-shape if a<0).
  • Simultaneous equations can be solved by finding where two lines intersect.
🧪 Quadratic Graph Plotter

📝 Chapter 2 Exam

30 marks | 35 min

[3]1.Expand and simplify: (x + 4)(x − 3)
x²−3x+4x−12 = x² + x − 12
[3]2.Factorise and solve: x² + 3x − 4 = 0
(x+4)(x−1)=0. x=−4 or x=1
[4]3.Solve simultaneously: 2x + 3y = −4 and 4x − y = 6
From eq2: y=4x−6. Sub into eq1: 2x+3(4x−6)=−4 → 2x+12x−18=−4 → 14x=14 → x=1. y=4(1)−6=−2. x=1, y=−2
[2]4.Solve: 3 ≤ 2x + 5 < 11
Subtract 5: −2 ≤ 2x < 6. Divide by 2: −1 ≤ x < 3
Chapter 3

Geometry: Circles, Prisms, Pythagoras & Trigonometry

Calculate with circles, find volumes of prisms and cylinders, use Pythagoras' theorem, and discover trigonometry.

3.1Circles

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💡 Circle Formulae

  • Circumference = π × d = 2πr
  • Area = πr²

✏️ Practice

0/2
Q1
Circle radius 5cm. Find circumference (to 1 d.p.).
Q2
Circle radius 7cm. Find area (to 1 d.p.).

3.2Prisms & Cylinders

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💡 Volume of a Prism

Volume = cross-section area × length

Cylinder volume = πr²h   |   Cylinder surface area = 2πrh + 2πr²

✏️ Practice

0/2
Q1
Cylinder: radius 3cm, height 10cm. Volume? (to 1 d.p.)
Q2
Triangular prism: triangle area 24cm², length 15cm. Volume?

3.3Pythagoras' Theorem

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💡 a² + b² = c²

In a right-angled triangle, the square of the hypotenuse (longest side, opposite the right angle) equals the sum of the squares of the other two sides.

📐 Find the hypotenuse
A right triangle has shorter sides 6cm and 8cm. Find the hypotenuse.
1

c² = 6² + 8² = 36 + 64 = 100

2

c = √100 = 10

✅

Hypotenuse = 10cm

📐 Find a shorter side
Hypotenuse = 13cm, one side = 5cm. Find the other.
1

b² = 13² − 5² = 169 − 25 = 144

2

b = √144 = 12

✅

Missing side = 12cm

🧪 Pythagoras Calculator

Enter two sides to find the third. Leave one field blank.

✏️ Practice

0/3
Q1
Sides 5cm and 12cm. Find hypotenuse.
Q2
Hypotenuse 10cm, one side 6cm. Find the other (to 1 d.p.).
Q3
A ladder 5m long leans against a wall with its base 3m from the wall. How high does it reach?

3.4Trigonometry: SOH CAH TOA

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💡 SOH CAH TOA

  • Sin θ = Opposite ÷ Hypotenuse
  • Cos θ = Adjacent ÷ Hypotenuse
  • Tan θ = Opposite ÷ Adjacent

Hypotenuse: longest side (opposite right angle). Opposite: across from the angle. Adjacent: next to the angle.

📐 Find a missing side
In a right triangle, angle = 35°, hypotenuse = 10cm. Find the opposite side.
1

We have angle, hypotenuse, want opposite → use SIN

2

sin 35° = opp ÷ 10

3

opp = 10 × sin 35° = 10 × 0.5736 = 5.7cm

✅

Opposite = 5.7cm (1 d.p.)

🧪 Trigonometry Calculator

✏️ Practice

0/2
Q1
Right triangle: angle 40°, adjacent = 8cm. Find opposite (1 d.p.).
Q2
Right triangle: opposite = 7, hypotenuse = 10. Find the angle (nearest degree).

📝 Chapter 3 Exam

30 marks | 35 min

[3]1.Circle: diameter 14cm. Find (a) circumference (b) area. (1 d.p.)
(a) π×14=44.0cm (b) π×7²=153.9cm²
[3]2.Cylinder: r=4cm, h=10cm. Find volume (1 d.p.).
π×16×10=502.7cm³
[4]3.A ladder leans against a wall. Foot is 2.5m from wall, top reaches 6m up. Find ladder length (1 d.p.).
c²=2.5²+6²=6.25+36=42.25. c=√42.25=6.5m
[4]4.Right triangle: angle 50°, hypotenuse 12cm. Find (a) opposite (b) adjacent. (1 d.p.)
(a) 12×sin50°=12×0.766=9.2cm (b) 12×cos50°=12×0.643=7.7cm
Chapter 4

Statistics & Probability

Data collection, combined probability, tree diagrams, and Venn diagrams.

4.1Data Collection & Grouped Data

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💡 Key Terms

  • Primary data: You collect it yourself
  • Secondary data: Someone else collected it
  • Bias: When results are unfairly influenced
  • Random sample: Every item has equal chance of being chosen

📖 Estimated Mean from Grouped Data

Use the midpoint of each class. Estimated mean = Σ(midpoint × frequency) ÷ Σ(frequency)

✏️ Practice

0/1
Q1
Heights: 150-160cm (freq 4, midpoint 155), 160-170cm (freq 8, mid 165), 170-180cm (freq 3, mid 175). Estimated mean?

4.2Combined & Independent Probability

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💡 Key Rules

  • Independent events: P(A and B) = P(A) × P(B)
  • Mutually exclusive: P(A or B) = P(A) + P(B)
  • Tree diagrams: Multiply along branches (AND), add between branches (OR)
📐 Tree Diagram
A coin is flipped and a dice is rolled. Find P(heads AND 6).
1

P(heads) = 1/2     P(6) = 1/6

2

Independent events, so multiply:

P(heads AND 6) = 1/2 × 1/6 = 1/12

✏️ Practice

0/3
Q1
P(rain Monday) = 0.3, P(rain Tuesday) = 0.4. P(rain both days)?
Q2
Two dice rolled. P(both show 6)?
Q3
Spinner: P(red)=0.3, P(blue)=0.5. These are mutually exclusive. P(red or blue)?

📝 Chapter 4 Exam

20 marks | 25 min

[3]1.Two spinners: Spinner A has P(red)=0.4. Spinner B has P(red)=0.3. Both are spun. Find P(both red).
Independent: 0.4 × 0.3 = 0.12
[4]2.Bag: 5 red, 3 blue socks. Two socks picked at random (without replacement). Draw a tree diagram and find P(matching pair).
P(RR) = 5/8 × 4/7 = 20/56. P(BB) = 3/8 × 2/7 = 6/56. P(match) = 26/56 = 13/28
[3]3.Give one example of bias in a survey about school lunches, and explain how to avoid it.
Example: Only asking students during lunch break could bias towards those who eat school food. To reduce bias, use a random sample from the full school register so every student has an equal chance of being selected.

🎉 Year 9 Complete!

You've covered the entire lower secondary curriculum. Time for revision!