Formula Bank

Showing: CBE Grades 6-9 ▾

System Grade

Quick reference for mathematical rules and equations.

Algebraic Expressions

Algebra Grade 7

1. Structure of an Algebraic Term

An algebraic term is built from a numerical multiplier (coefficient) and an unknown value (variable) raised to a power.

\text{Term} = \text{coefficient} \times \text{variable}^{\text{power}}

Example: In

5x^3

5

is the coefficient,

x

is the variable,

3

is the power.
Note:

x

alone means

1x^1

2. Expanding Brackets (Distributive Property)

To expand brackets, multiply the outer number by every term inside the brackets.

a(b + c) = ab + ac

Example:

3(x + 4) = (3 \times x) + (3 \times 4) = 3x + 12

3. Combining Like Terms

Add or subtract terms with the exact same variable and power. Combine the coefficients and keep the variable part unchanged.

k_1x + k_2x = (k_1 + k_2)x

Example:

4x + 7x = (4 + 7)x = 11x

💡 Tip: Only like terms combine:

3x + 2y

cannot be simplified.

4. Multiplying Variables (Index Law 1)

When multiplying terms with the same base, keep the base and add the exponents.

x^m \times x^n = x^{m+n}

Example:

x^3 \times x^2 = x^{3+2} = x^5

5. Substitution and Evaluation

Replace each variable with its numeric value, then compute following BODMAS order.

Example: Evaluate

2x + 5

when

x = 3

2(3) + 5 = 6 + 5 = 11

💡 Tip: Always substitute with brackets:

2x

x = -3

means

2(-3)

, not

2-3

View Full Lesson

Linear Equations

Algebra Grade 7

1. Standard Form of a Linear Equation

A linear equation in one variable balances an unknown value

x

with given numbers.

ax + b = c

Example: In

2x + 4 = 10

2

is the coefficient

a

4

is the constant

b

, and

10

is the result

c

2. Isolating the Variable Term

Move the constant term to the opposite side by performing the inverse operation (subtracting

b

ax = c - b

Example: For

2x + 4 = 10

:
Subtract

4

from both sides:

2x = 10 - 4 \implies 2x = 6

3. Solving for $x$

Divide both sides by the coefficient

a

to isolate

x

completely.

x = \frac{c - b}{a}

Example: From

2x = 6

:
Divide by

2

x = \frac{6}{2} = 3

4. Variables on Both Sides

When

x

appears on both sides, collect all variable terms on one side and constant terms on the other.

ax + b = cx + d \implies (a - c)x = d - b

Example: Solve

5x + 2 = 2x + 11

:
Subtract

2x

from both sides:

3x + 2 = 11

Subtract

2

3x = 9 \implies x = 3

💡 Tip: Move the smaller variable term to avoid negative coefficients.

5. Verifying Your Solution

Substitute your answer back into the original equation to confirm both sides balance.

\text{Check: } a(\text{solution}) + b = c

Example: Check

x = 3

2x + 4 = 10

2(3) + 4 = 6 + 4 = 10

. The solution is correct!

💡 Tip: If Left Hand Side equals Right Hand Side, your answer is guaranteed correct.

View Full Lesson

Linear Inequalities

Algebra Grade 7

1. Solving a Linear Inequality

Isolate

x

using the same operations as a normal equation.

ax + b < c \implies x < \frac{c - b}{a}

Example:

3x - 2 < 10

Add

2

3x < 12

Divide by

3

x < 4

2. The Negative Division Rule

Multiplying or dividing both sides by a negative number flips the inequality direction.

\text{If } a < 0: \quad ax < b \implies x > \frac{b}{a}

Example: Solve

-2x < 8

:
Divide by

-2

and flip the sign:

x > -4

💡 Tip: This is the most-tested trap in inequalities. ALWAYS ask: did I divide by a negative?

3. Compound Inequalities

When

x

is bounded between two numbers, perform the exact same operation on all three parts at once.

p < ax + b \le q \implies \frac{p - b}{a} < x \le \frac{q - b}{a}

Example: Solve

5 < 2x + 1 \le 9

:
Subtract

1

everywhere:

4 < 2x \le 8

Divide everywhere by

2

2 < x \le 4

4. Number Line and Interval Notation

Solution sets are drawn on a number line or written as intervals. The circle type tells you whether the end value is included.

Strict (

<, >

)Open circle

\circ

, interval

(k)

Inclusive (

\le, \ge

)Closed circle

\bullet

, interval

[k]

Example:

x < 3

in interval notation is

(-\infty, 3)

x \ge 5

in interval notation is

[5, \infty)

View Full Lesson

Algebraic Expressions

Algebra Grade 8

1. Expanding Brackets

To expand an expression, multiply the single outer factor by every term inside the bracket.

a(b + c) = ab + ac

Example:

4(2x + 3) = (4 \times 2x) + (4 \times 3) = 8x + 12

💡 Tip: Pay close attention to negative signs when expanding:

-2(x - 3) = -2x + 6

2. Factorising Expressions

Factorising is the reverse of expanding. Identify and pull out the highest common factor shared by all terms.

ab + ac = a(b + c)

Example: Factorise

6x + 9

:
The highest common factor of

6x

and

9

3

6x + 9 = 3(2x + 3)

💡 Tip: Test your factorised answer by expanding it back out in your head.

3. Simplifying Expressions with Common Variables

When terms share the same variable part, simplify by adding or subtracting their coefficients.

ax + bx = (a + b)x

Example: Simplify

5x + 3x - 2x

(5 + 3 - 2)x = 6x

4. Multiplying Scalar and Binomial Terms

A constant multiplier outside the bracket affects both the variable term and the constant term inside.

k(ax + b) = kax + kb

Example: Expand

5(3x - 4)

(5 \times 3x) + (5 \times -4) = 15x - 20

View Full Lesson

Linear Equations

Algebra Grade 8

1. Equations with Brackets

When an equation contains a bracketed expression multiplied by a number, expand the bracket first or divide both sides by the multiplier.

a(x + b) = c \implies ax + ab = c

Example: Solve

3(x + 2) = 15

:
Expand the bracket:

3x + 6 = 15

Subtract

6

3x = 9

Divide by

3

x = 3

💡 Tip: Alternatively, divide by

a

first:

x + b = \frac{c}{a}

c

is divisible by

a

2. Isolating the Variable Term

Collect all constant terms on the opposite side of the variable by applying inverse operations.

ax = c - ab

Example: From

3x + 6 = 15

:
Subtract

6

from both sides:

3x = 15 - 6 = 9

3. Solving for the Unknown

Divide the remaining total by the variable's coefficient to find the final value.

x = \frac{c - ab}{a}

Example: From

3x = 9

x = \frac{9}{3} = 3

4. Clearing Fractional Equations

To eliminate a denominator, multiply every term on both sides by that denominator.

\frac{ax + b}{c} = d \implies ax + b = cd

Example: Solve

\frac{2x + 4}{3} = 6

:
Multiply both sides by

3

2x + 4 = 18

Subtract

4

2x = 14

Divide by

2

x = 7

💡 Tip: Multiplying by the denominator clears the fraction in one single step.

View Full Lesson

Equation of a Straight Line

Algebra Grade 9

1. Gradient (Slope) of a Line

The gradient (

m

) measures steepness, calculated as the vertical change (rise) divided by the horizontal change (run) between two points.

m = \frac{y_2 - y_1}{x_2 - x_1}

Example: Find the gradient passing through

(1, 2)

and

(3, 8)

m = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3

💡 Tip: Always subtract coordinates in the same order:

(y_2 - y_1)

over

(x_2 - x_1)

2. Slope-Intercept Form

A linear equation written in slope-intercept form directly reveals its gradient (

m

) and

y

-intercept (

b

y = mx + b

Example: For the line

y = 2x + 5

:
The gradient

m = 2

, and the line crosses the

y

-axis at

(0, 5)

💡 Tip: The

y

-intercept

b

is the point where

x = 0

3. Point-Slope Form

Use this formula to find the equation of a straight line when you know one point

(x_1, y_1)

and the gradient

m

y - y_1 = m(x - x_1)

Example: Find the line with gradient

m = 4

through point

(2, 3)

y - 3 = 4(x - 2)

y - 3 = 4x - 8 \implies y = 4x - 5

4. General Form of a Line

Rearrange all terms to one side so the equation equals zero, keeping coefficients as whole numbers.

Ax + By + C = 0

Example: Convert

y = 3x - 4

into general form:
Subtract

y

from both sides:

3x - y - 4 = 0

💡 Tip: Conventionally,

A

should be positive.

5. Finding the $y$-Intercept from a Point

If you know the gradient

m

and any point

(x_1, y_1)

, solve for

b

by rearranging

y = mx + b

b = y_1 - mx_1

Example: A line has

m = 3

and passes through

(2, 7)

b = 7 - 3(2) = 7 - 6 = 1

.
Equation:

y = 3x + 1

View Full Lesson

Linear Inequalities

Algebra Grade 9

1. One-Step Inequalities

Divide or multiply both sides by a positive number without changing the inequality sign.

ax < b \implies x < \frac{b}{a} \quad (a > 0)

Example: Solve

4x < 12

:
Divide by

4

x < 3

2. Two-Step Inequalities

Undo additions or subtractions first, then divide by the coefficient.

ax + c < d \implies x < \frac{d - c}{a}

Example: Solve

2x + 5 < 13

:
Subtract

5

2x < 8

Divide by

2

x < 4

3. Reversing the Sign (Negative Rule)

Multiplying or dividing both sides by a negative number reverses the inequality symbol.

-kx < m \implies x > -\frac{m}{k}

Example: Solve

-3x < 12

:
Divide by

-3

and reverse sign:

x > -4

💡 Tip: Always reverse

< \to >

\le \to \ge

when dividing by a negative number.

4. Compound Intervals

Express solutions bounded between two endpoints using interval notation.

a < x \le b \implies (a, b]

Example: Convert

2 < x \le 7

to interval notation:
Written as

(2, 7]

💡 Tip: Parentheses

( )

exclude the endpoint; square brackets

[ ]

include it.

5. Graphing Two-Variable Inequalities

Graph the line

y = mx + c

, then shade the region containing satisfying coordinate pairs.

Strict (

<, >

)Use a dashed boundary line

Inclusive (

\le, \ge

)Use a solid boundary line

Example: Graph

y \ge 2x + 1

:
1. Draw solid line

y = 2x + 1

.
2. Test

(0,0)

0 \ge 2(0) + 1 \implies 0 \ge 1

(False).
3. Shade the half-plane above the line.

💡 Tip: Test point

(0,0)

is the fastest way to confirm which side of the line to shade.

View Full Lesson

Matrices

Algebra Grade 9

1. Matrix Dimensions (Order)

A matrix is a rectangular grid of numbers described by its number of rows (

m

) and columns (

n

A_{m \times n}

Example: A matrix with

2

rows and

3

columns has dimensions

2 \times 3

\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}_{2 \times 3}

💡 Tip: Remember: Rows first, Columns second (RC = 'Remote Control').

2. Matrix Addition

Add corresponding elements in matrices that have the exact same dimensions.

C = A + B \implies c_{ij} = a_{ij} + b_{ij}

Example:

\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1+5 & 3+1 \\ 2+0 & 4+2 \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 2 & 6 \end{pmatrix}

💡 Tip: Addition is impossible if dimensions do not match.

3. Matrix Subtraction

Subtract corresponding elements in matrices with matching dimensions.

C = A - B \implies c_{ij} = a_{ij} - b_{ij}

Example:

\begin{pmatrix} 5 & 8 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 5-2 & 8-3 \\ 3-1 & 4-4 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 2 & 0 \end{pmatrix}

💡 Tip: Subtract element by element in the exact same positions.

View Full Lesson

Bar Graphs

Data Handling Grade 6

1. Scale Factor for Drawing Bars

The scale factor determines how many scale units or centimeters on paper represent one data unit.

s = \frac{H}{v_{\text{max}}}

Example: If the maximum value is

50

students and your graph height is

10\text{ cm}

s = \frac{10}{50} = 0.2\text{ cm per student}

2. Calculating Bar Heights

To find the height of a specific bar, multiply its category value by the scale factor or divide by the maximum value and scale to total height.

h_i = \frac{v_i}{v_{\text{max}}} \times H

Example: For a category value of

20

students on a

10\text{ cm}

chart where

v_{\text{max}} = 50

h_i = \frac{20}{50} \times 10 = 4\text{ cm}

3. Finding Total Frequency

Add up the values of every category to find the grand total of data collected.

V_{\text{total}} = \sum_{i=1}^{n} v_i

Example: If three bars have values

12

15

, and

13

V_{\text{total}} = 12 + 15 + 13 = 40

4. Percentage Share of a Category

Calculate the percentage share of a single category relative to the entire dataset.

p_i = \frac{v_i}{V_{\text{total}}} \times 100

Example: If a category value is

10

out of a total of

40

p_i = \frac{10}{40} \times 100 = 25\%

5. Comparing Categories as a Ratio

Compare two categories directly by writing their values as a simplified fraction or ratio.

\text{Ratio } a : b = \frac{v_a}{v_b}

Example: Compare Category A (

15

) to Category B (

5

\frac{15}{5} = \frac{3}{1} \implies 3 : 1

💡 Tip: Always simplify ratios to their lowest terms.

View Full Lesson

Data Handling

Data Handling Grade 7

1. Understanding Frequency

The frequency is the total count of observations for a specific item, recorded by counting tally marks.

\text{Frequency} = \text{Total tally marks}

Example: Tally marks

\cancel{||||}\ ||

equal a frequency count of

7

2. Total Number of Observations

Sum all individual frequencies together to determine the total number of data points collected.

N = \sum f

Example: In a table with frequencies

4

8

, and

6

:
Total data points

N = 4 + 8 + 6 = 18

3. Bar Height Proportionality

In a bar chart, the height of each bar is directly proportional to its frequency value.

\text{Bar Height} \propto f

Example: If a frequency of

5

is drawn as

2\text{ cm}

tall, a frequency of

10

must be drawn twice as tall (

4\text{ cm}

💡 Tip: Always ensure the frequency scale starts at zero to avoid misleading bars.

View Full Lesson

Data Presentation and Interpretation

Data Handling Grade 8

1. The Mean (Average)

Add up all individual values and divide by the total number of values.

\bar{x} = \frac{\sum x}{n}

Example: Find the mean of

4, 6, 8, 10

\bar{x} = \frac{4 + 6 + 8 + 10}{4} = \frac{28}{4} = 7

2. The Median (Middle Value)

Sort data from smallest to largest. The median is the exact middle value (or the average of the two middle values).

Odd count (

n = 2k - 1

)Middle value at position

\frac{n+1}{2}

Even count (

n = 2k

)Average of values at positions

\frac{n}{2}

and

\frac{n}{2} + 1

Example: For

3, 5, 7, 9, 11

(

n=5

):
Median is

7

.
For

2, 4, 6, 8

(

n=4

):
Median is

\frac{4 + 6}{2} = 5

💡 Tip: Always arrange the numbers in order from smallest to largest first!

3. The Mode & Range

The mode is the value that appears most often. The range measures spread by subtracting the smallest value from the largest.

\text{Range} = x_{\text{max}} - x_{\text{min}}

Example: For dataset

2, 5, 5, 8, 12

:
Mode =

5

(appears most often).
Range =

12 - 2 = 10

💡 Tip: A dataset can have no mode, one mode, or more than one mode.

4. Pie Chart Angles

Convert a category frequency into a sector angle out of

360^\circ

for pie chart construction.

\theta = \frac{f}{n} \times 360^\circ

Example: If a category frequency is

15

out of

60

total points:

\theta = \frac{15}{60} \times 360^\circ = 0.25 \times 360^\circ = 90^\circ

💡 Tip: Check that all sector angles add up to

360^\circ

5. Category Relative Frequency (Percentage)

Calculate the percentage representation of a frequency out of the total count.

p = \frac{f}{n} \times 100\%

Example: If

f = 12

and total

n = 50

p = \frac{12}{50} \times 100\% = 24\%

View Full Lesson

Probability

Data Handling Grade 8

1. Basic Theoretical Probability

The probability of an event is the number of favourable outcomes divided by the total number of equally likely outcomes.

P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}}

Example: Probability of rolling a

4

on a standard

6

-sided die:

P(4) = \frac{1}{6}

💡 Tip: Probability values are always between

0

(impossible) and

1

(certain).

2. The Complement Rule

The chance that an event does not happen is equal to

1

minus the chance that it does.

P(E^c) = 1 - P(E)

Example: If the chance of rain is

P(\text{Rain}) = 0.3

P(\text{No Rain}) = 1 - 0.3 = 0.7

💡 Tip: Remember:

P(\text{Happens}) + P(\text{Does not happen}) = 1

3. Mutually Exclusive Events (Addition Rule)

For events that cannot happen at the same time, add their individual probabilities together.

P(A \cup B) = P(A) + P(B)

Example: Picking a Red card (

P = 0.4

) OR a Blue card (

P = 0.35

P(\text{Red or Blue}) = 0.4 + 0.35 = 0.75

4. Independent Events (Multiplication Rule)

For events where one outcome does not affect the other, multiply their probabilities.

P(A \cap B) = P(A) \times P(B)

Example: Flipping Heads (

0.5

) AND rolling a

6

(

1/6

P(\text{Heads and } 6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

💡 Tip: Keywords: 'OR' usually means add; 'AND' usually means multiply.

5. Converting Probability Forms

Convert probability fractions to decimals by dividing, and to percentages by multiplying by

100\%

Decimal Form

\text{Decimal} = \frac{\text{numerator}}{\text{denominator}}

Percentage Form

\text{Percentage} = \text{Decimal} \times 100\%

Example: Convert fraction

\frac{2}{5}

:
Decimal:

2 \div 5 = 0.4

Percentage:

0.4 \times 100\% = 40\%

View Full Lesson

Data Interpretation (Grouped Data)

Data Handling Grade 9

1. Class Width and Midpoint

For grouped data, the class width is the span of a single interval, while the midpoint (

x_i

) represents the central value of that class.

x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}

Example: For class interval

10 - 20

:
Width

= 20 - 10 = 10

Midpoint

x_i = \frac{10 + 20}{2} = 15

2. Estimated Mean for Grouped Data

Multiply each class midpoint by its frequency, sum all products, and divide by the total frequency.

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Example: Interval

10-20

(midpoint

15

f=3

) and

20-30

(midpoint

25

f=7

\sum fx = (15 \times 3) + (25 \times 7) = 45 + 175 = 220

Total

f = 3 + 7 = 10 \implies \bar{x} = \frac{220}{10} = 22

3. Cumulative Frequency

Keep a running total of frequencies from the first interval up to the current interval.

F_i = \sum_{k=1}^{i} f_k

Example: Frequencies

4

6

5

F_1 = 4

F_2 = 4 + 6 = 10

F_3 = 10 + 5 = 15

💡 Tip: The final cumulative frequency value must equal the total number of data points.

4. Estimated Median by Linear Interpolation

Estimate the median inside the median class using proportional positioning relative to the lower boundary.

\text{Median} = L + \left( \frac{\frac{N}{2} - F_b}{f_m} \right) \times w

Example: If median class is

20-30

, lower boundary

L=20

, width

w=10

, frequency

f_m=8

, total

N=40

, and cumulative frequency before class

F_b=14

\text{Median} = 20 + \left( \frac{20 - 14}{8} \right) \times 10 = 20 + \left( \frac{6}{8} \right) \times 10 = 20 + 7.5 = 27.5

View Full Lesson

Probability

Data Handling Grade 9

1. Independent Events (Multiplication Rule)

When two events are independent, the outcome of one does not change the other. Multiply their probabilities to find the combined chance.

P(A \cap B) = P(A) \times P(B)

Example: If

P(A) = 0.5

and

P(B) = 0.4

P(A \text{ and } B) = 0.5 \times 0.4 = 0.2

2. General Addition Rule (Combined Events)

To find the chance that event

A

or event

B

occurs, add their probabilities and subtract the overlapping intersection.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Example: If

P(A) = 0.6

P(B) = 0.5

, and

P(A \cap B) = 0.2

P(A \text{ or } B) = 0.6 + 0.5 - 0.2 = 0.9

💡 Tip: Always subtract

P(A \cap B)

so the overlap is not counted twice!

3. Dependent Events & Conditional Probability

When the outcome of the first event alters the probability of the second, multiply by the conditional probability

P(B \mid A)

P(A \cap B) = P(A) \times P(B \mid A)

Example: Drawing two red marbles from a bag with

3

red and

2

blue without replacement:

P(R_1) = \frac{3}{5}

P(R_2 \mid R_1) = \frac{2}{4} \implies P(\text{Both Red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = 0.3

💡 Tip: Remember to reduce both the numerator and denominator by

1

on the second draw when selecting without replacement.

4. Complementary Events

The probability that an event does not occur is equal to

1

minus the probability that it does.

P(A^c) = 1 - P(A)

Example: If

P(\text{Rain}) = 0.25

P(\text{No Rain}) = 1 - 0.25 = 0.75

View Full Lesson

3-D Objects

Geometry Grade 6

1. Euler's Formula for 3D Polyhedra

For any solid polyhedral shape, the number of faces (

F

) plus vertices (

V

) minus edges (

E

) always equals

2

F + V - E = 2

Example: A cube has

6

faces,

8

vertices, and

12

edges:

6 + 8 - 12 = 14 - 12 = 2

. Formula holds!

2. Properties of 3D Shapes

Key terms used to classify and measure three-dimensional solid objects.

Faces (

F

)Flat or curved individual surfaces

Edges (

E

)Straight line segments where two faces meet

Vertices (

V

)Corner points where three or more edges intersect

Example: A triangular pyramid (tetrahedron) has

F = 4

flat faces,

V = 4

corner vertices, and

E = 6

straight edges.

View Full Lesson

Angles

Geometry Grade 6

1. Classifying Angle Types by Size

Angles measure rotational turn in degrees (

^\circ

), categorized by their size compared to right angles and straight lines.

AcuteBetween

0^\circ

and

90^\circ

Right AngleExactly

90^\circ

ObtuseBetween

90^\circ

and

180^\circ

Straight LineExactly

180^\circ

ReflexBetween

180^\circ

and

360^\circ

Example: An angle of

45^\circ

is Acute;

120^\circ

is Obtuse;

210^\circ

is Reflex.

2. Angles on a Straight Line

Adjacent angles along a straight line always sum to a half-turn (

180^\circ

\alpha + \beta = 180^\circ

Example: If two angles on a line are

\alpha = 110^\circ

and

\beta

\beta = 180^\circ - 110^\circ = 70^\circ

3. Complementary Angles

Two angles are complementary if their sum equals a right angle (

90^\circ

\alpha + \beta = 90^\circ

Example: Find the complement of

35^\circ

\beta = 90^\circ - 35^\circ = 55^\circ

💡 Tip: Complementary = Corner (

90^\circ

); Supplementary = Straight line (

180^\circ

View Full Lesson

Lines

Geometry Grade 6

1. Line Segments, Rays, and Lines

Geometric lines differ based on whether they have fixed endpoints or extend infinitely.

Line Segment

AB

Fixed part of a line between endpoints

A

and

B

Ray

AB

Starts at

A

and extends infinitely through

B

Line

AB

Extends infinitely in both directions through

A

and

B

Example: A

5\text{ cm}

ruler measurement represents a line segment because it has two fixed endpoints.

2. Parallel and Perpendicular Lines

Parallel lines never cross and have equal gradients. Perpendicular lines cross at

90^\circ

, and their gradients multiply to

-1

Parallel Lines

m_1 = m_2

Perpendicular Lines

m_1 \times m_2 = -1

Example: If Line 1 has gradient

m_1 = 3

:
• A parallel line has

m_2 = 3

.
• A perpendicular line has

m_2 = -\frac{1}{3}

3. Distance Between Two Points

Calculate the straight-line length of segment

AB

between coordinates

(x_1, y_1)

and

(x_2, y_2)

using Pythagoras' theorem.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Example: Find length of segment between

(1, 2)

and

(4, 6)

d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

💡 Tip: Always subtract

x

-values together and

y

-values together before squaring.

View Full Lesson

Angles

Geometry Grade 7

1. Angles on a Line and Around a Point

Angles along a straight line sum to

180^\circ

(half-turn), while angles meeting at a single point sum to

360^\circ

(full turn).

\theta_1 + \theta_2 + \dots + \theta_n = 360^\circ

Example: If three angles at a point are

100^\circ

120^\circ

, and

x

x = 360^\circ - (100^\circ + 120^\circ) = 360^\circ - 220^\circ = 140^\circ

2. Vertically Opposite Angles

When two straight lines intersect, the angles directly opposite each other are equal.

\theta_{\text{vert}} = \theta_{\text{vert}'}

Example: If two lines cross and one angle is

55^\circ

, the angle directly opposite it is also

55^\circ

3. Parallel Line Angle Relationships

A transversal line crossing parallel lines forms predictable angle patterns.

Corresponding ('F' shape)Angles are equal:

\theta_{\text{corr}} = \theta_{\text{corr}'}

Alternate ('Z' shape)Angles are equal:

\theta_{\text{alt}} = \theta_{\text{alt}'}

Co-interior ('C' shape)Angles sum to

180^\circ

\theta_{\text{co-int}} + \theta_{\text{co-int}'} = 180^\circ

Example: If two co-interior angles are

x

and

70^\circ

x = 180^\circ - 70^\circ = 110^\circ

💡 Tip: Look for F, Z, and C letter patterns along parallel lines to spot angle rules fast!

View Full Lesson

Geometrical Constructions

Geometry Grade 7

1. Midpoint of a Segment

The midpoint

M

is the center point of a line segment, calculated by averaging the endpoints.

M = \frac{A + B}{2}

Example: Find the midpoint between endpoints

x = 4

and

x = 10

M = \frac{4 + 10}{2} = \frac{14}{2} = 7

2. Perpendicular Bisector

A perpendicular bisector cuts a line segment into two equal halves at a right angle (

90^\circ

\ell_{\perp} \perp AB \text{ at } M

Example: If segment

AB = 12\text{ cm}

, a perpendicular bisector crosses

AB

M = 6\text{ cm}

from

A

at an angle of

90^\circ

💡 Tip: When constructing with a compass, keep your arc radius greater than half the segment length.

3. Angle Bisector Theorem

An angle bisector cuts an angle into two equal parts and divides the opposite side proportionally to adjacent sides.

\frac{a}{b} = \frac{c}{d}

Example: In a triangle with adjacent sides

a = 6

and

b = 8

, if one segment of the divided side is

c = 3

\frac{6}{8} = \frac{3}{d} \implies 6d = 24 \implies d = 4

4. Copying a Segment Length

To copy a distance

d

accurately using geometric tools, lock your compass width at radius

d

\text{Arc radius} = d

Example: To construct segment

PQ

equal to

AB = 5\text{ cm}

, set compass point at

A

, open pencil to

B

, and strike an arc from point

P

View Full Lesson

Common Solids

Geometry Grade 8

1. Euler's Polyhedron Formula

For any 3D solid with flat faces, Vertices (

V

) minus Edges (

E

) plus Faces (

F

) always equals

2

V - E + F = 2

Example: For a triangular prism (

V = 6

E = 9

F = 5

6 - 9 + 5 = 2

. Formula holds!

💡 Tip: Remember 'VEF' order:

V - E + F = 2

2. Properties of Common 3D Solids

Three-dimensional solids are classified by their face count, edge count, and corner vertices.

Cube

F = 6

square faces,

E = 12

V = 8

Cylinder

F = 3

(

2

flat circles,

1

curved face),

E = 2

V = 0

Example: A cylinder has

V = 0

vertices because its curved sides do not form sharp corner points.

View Full Lesson

Coordinates and Graphs

Geometry Grade 8

1. Coordinates on the Cartesian Plane

An ordered pair

(x, y)

gives the precise location of a point, where

x

measures horizontal distance and

y

measures vertical distance from the origin

(0,0)

\text{Position} = (x, y)

Example: To plot

(3, -4)

: start at the origin, move

3

units right, then

4

units down.

💡 Tip: Always go along the corridor (

x

-axis) before going up or down the stairs (

y

-axis).

2. The Four Quadrants

The axes intersect at the origin to divide the Cartesian plane into four distinct regions called quadrants.

Quadrant IRight & Up:

(+, +)

Quadrant IILeft & Up:

(-, +)

Quadrant IIILeft & Down:

(-, -)

Quadrant IVRight & Down:

(+, -)

Example: The point

(-5, 2)

lies in Quadrant II because

x

is negative and

y

is positive.

3. Plotting Linear Graphs

To plot the straight line

y = mx + c

, choose several

x

-values, calculate their matching

y

-values, plot the coordinate points, and join them with a straight line.

y = mx + c

Example: Plot

y = 2x + 1

using

x = 0, 1, 2

:
When

x = 0 \implies y = 1 \implies (0,1)

When

x = 1 \implies y = 3 \implies (1,3)

When

x = 2 \implies y = 5 \implies (2,5)

View Full Lesson

Geometrical Constructions

Geometry Grade 8

1. Locus of a Perpendicular Bisector

Every point

P

on the perpendicular bisector of segment

AB

is equidistant from endpoints

A

and

B

PA = PB

Example: If point

P

lies on the perpendicular bisector of line

AB

and

PA = 7\text{ cm}

, then

PB

must also equal

7\text{ cm}

2. Locus of a Circle

A circle is defined as the locus of all points

P

that remain at a fixed distance

r

(radius) from a central point

O

OP = r

Example: All points

P

that are exactly

4\text{ cm}

away from center

O

form a circle with radius

r = 4\text{ cm}

3. Angle Bisector Property

An angle bisector inside a triangle divides the opposite side into segments proportional to the adjacent sides.

\frac{AD}{DB} = \frac{AE}{EB}

Example: In triangle

ABC

, if adjacent sides are

AE = 6

and

EB = 9

, and

AD = 4

\frac{4}{DB} = \frac{6}{9} \implies 6(DB) = 36 \implies DB = 6

4. Regular Hexagon Inscribed in a Circle

When constructing a regular hexagon inside a circle, the side length of the hexagon is exactly equal to the radius of the circle.

\text{Side length} = r

Example: To draw a regular hexagon inside a circle of radius

5\text{ cm}

, keep your compass set to

5\text{ cm}

and step off six equal arcs around the perimeter.

💡 Tip: Never change your compass width once you set the initial radius!

View Full Lesson

Scale Drawing

Geometry Grade 8

1. Linear Scale Factor

The scale factor is the ratio comparing a distance on the drawing to the actual real-world distance.

\text{Scale} = \frac{\text{drawing length}}{\text{real length}}

Example: If a

2\text{ cm}

line on a map represents a real distance of

100\text{ cm}

(

1\text{ m}

\text{Scale} = \frac{2}{100} = \frac{1}{50}

1 : 50

💡 Tip: Ensure both drawing length and real length are in the same units before writing as a ratio!

2. Calculating Real Lengths and Drawing Lengths

Multiply real dimensions by the scale factor to find map lengths, or divide map lengths by the scale factor to find real dimensions.

Drawing Length

\text{real length} \times \text{Scale}

Real Length

\frac{\text{drawing length}}{\text{Scale}}

Example: On a

1 : 200

map, find the real distance for a

4\text{ cm}

line:

\text{Real length} = \frac{4}{\frac{1}{200}} = 4 \times 200 = 800\text{ cm} = 8\text{ m}

3. Area Scale Factor

When lengths are scaled by a factor

k

, the resulting area changes by the square of the scale factor (

k^2

\text{Area Scale} = \text{Scale}^2

Example: A map has a linear scale of

1 : 100

(

k = \frac{1}{100}

). If a field on the map has an area of

5\text{ cm}^2

\text{Real area} = 5 \times 100^2 = 5 \times 10,000 = 50,000\text{ cm}^2 = 5\text{ m}^2

💡 Tip: Remember: Linear scale

= k

, Area scale

= k^2

, Volume scale

= k^3

View Full Lesson

Coordinates and Graphs

Geometry Grade 9

1. Horizontal and Vertical Separations

The horizontal change (

\Delta x

) and vertical change (

\Delta y

) form the perpendicular legs of a right triangle between two points.

\Delta x = x_2 - x_1, \quad \Delta y = y_2 - y_1

Example: Between points

(2, 3)

and

(6, 8)

\Delta x = 6 - 2 = 4

\Delta y = 8 - 3 = 5

2. Distance Between Two Points

Calculate the exact straight-line distance

d

between points

(x_1, y_1)

and

(x_2, y_2)

using Pythagoras' theorem.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Example: Find distance between

(1, 2)

and

(4, 6)

d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

💡 Tip: Squaring turns negative differences positive:

(-3)^2 = 9

3. Midpoint of a Segment

Find the point

M

exactly halfway between two coordinates by taking the average of their

x

-values and

y

-values.

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Example: Find the midpoint between

(2, 8)

and

(6, 4)

M = \left( \frac{2 + 6}{2}, \frac{8 + 4}{2} \right) = \left( \frac{8}{2}, \frac{12}{2} \right) = (4, 6)

💡 Tip: Add the coordinates together before dividing by 2!

View Full Lesson

Scale Drawing

Geometry Grade 9

1. Scale Factor & Dimensions

The scale factor

s

links real-world lengths to scale drawing lengths.

s = \frac{\text{drawing length}}{\text{real length}}

Example: If a

10\text{ m}

(

1000\text{ cm}

) real wall is drawn as

5\text{ cm}

s = \frac{5}{1000} = \frac{1}{200}

(or scale

1 : 200

💡 Tip: Always convert drawing and real measurements to the same units before calculating the scale ratio.

2. Converting Drawing and Real Lengths

Multiply real lengths by scale factor

s

to draw them, or multiply drawing lengths by

n

(from scale

1 : n

) to find real distances.

Drawing Length

L_{\text{draw}} = s \times L_{\text{real}}

Real Length

L_{\text{real}} = L_{\text{draw}} \times n

Example: On a

1 : 50

scale plan (

n = 50

), a room length drawn as

8\text{ cm}

has a real length of:

L_{\text{real}} = 8 \times 50 = 400\text{ cm} = 4\text{ m}

3. Three-Digit Bearings

A bearing is an angle measured clockwise starting from the North line, expressed using three digits.

\text{Bearing} = \text{Angle measured clockwise from North } (000^\circ \text{ to } 360^\circ)

Example: An direction

45^\circ

East of North is written as bearing

045^\circ

. A direction due South is

180^\circ

💡 Tip: Always add a leading zero for bearings under

100^\circ

(e.g.,

065^\circ

, not

65^\circ

4. Area Scale Factor

When linear distances scale by

s

, the area changes by the square of the scale factor (

s^2

s_A = s^2

Example: A map scale is

1 : 100

(

s = \frac{1}{100}

). An area on the map of

3\text{ cm}^2

represents a real area of:

3 \times 100^2 = 3 \times 10,000 = 30,000\text{ cm}^2 = 3\text{ m}^2

View Full Lesson

Similarity and Enlargement

Geometry Grade 9

1. Scale Factor of Similar Shapes

Similar shapes have matching angles and corresponding sides that are proportional by scale factor

k

\frac{a'}{a} = \frac{b'}{b} = \frac{c'}{c} = k

Example: If side

a = 4\text{ cm}

expands to

a' = 12\text{ cm}

, the scale factor is

k = \frac{12}{4} = 3

. Side

b = 5\text{ cm}

becomes

b' = 5 \times 3 = 15\text{ cm}

💡 Tip: To find

k

, divide a new side length by its matching original side length.

2. Coordinate Enlargement About the Origin

Enlarge a point centered at the origin

(0,0)

by multiplying both coordinates by scale factor

k

(x', y') = (kx, ky)

Example: Enlarge point

(2, -3)

by scale factor

k = 3

about

(0,0)

(x', y') = (3 \times 2, 3 \times -3) = (6, -9)

3. Perimeter vs Area Scaling

Linear lengths and perimeters scale by

k

, whereas areas scale by

k^2

New Side / Perimeter

\text{Perimeter}_{\text{new}} = k \times \text{Perimeter}_{\text{orig}}

New Area

\text{Area}_{\text{new}} = k^2 \times \text{Area}_{\text{orig}}

Example: A rectangle with perimeter

12\text{ cm}

and area

8\text{ cm}^2

is enlarged with

k = 2

:
• New Perimeter

= 2 \times 12 = 24\text{ cm}

• New Area

= 2^2 \times 8 = 4 \times 8 = 32\text{ cm}^2

💡 Tip: Never multiply side lengths by addition—always multiply by scale factor

k

View Full Lesson

Trigonometry

Geometry Grade 9

1. The Primary Trigonometric Ratios

In a right-angled triangle, sine, cosine, and tangent link acute angles to ratios of side lengths.

Sine (SOH)

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}

Cosine (CAH)

\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}

Tangent (TOA)

\tan \theta = \frac{\text{opposite}}{\text{adjacent}}

Example: In a triangle with opposite

= 3

, adjacent

= 4

, and hypotenuse

= 5

\sin \theta = \frac{3}{5} = 0.6

, \quad

\cos \theta = \frac{4}{5} = 0.8

, \quad

\tan \theta = \frac{3}{4} = 0.75

💡 Tip: Remember the acronym SOH CAH TOA to pick the correct ratio quickly.

2. The Pythagorean Trigonometric Identity

For any angle

\theta

, the sum of the squares of sine and cosine always equals

1

\sin^2 \theta + \cos^2 \theta = 1

Example: If

\sin \theta = 0.6

\sin^2 \theta = 0.36 \implies \cos^2 \theta = 1 - 0.36 = 0.64 \implies \cos \theta = \sqrt{0.64} = 0.8

💡 Tip: Note:

\sin^2 \theta

means

(\sin \theta)^2

3. Tangent Ratio Identity

Tangent equals sine divided by cosine because the hypotenuse cancels out.

\tan \theta = \frac{\sin \theta}{\cos \theta}

Example: If

\sin \theta = 0.6

and

\cos \theta = 0.8

\tan \theta = \frac{0.6}{0.8} = 0.75

View Full Lesson

Area

Measurement Grade 6

1. Area of Rectangles and Squares

Find the area of a rectangle by multiplying length by width, or square the side length for a square.

A = l \times w

Example: Find area of a rectangle with length

6\text{ cm}

and width

4\text{ cm}

A = 6 \times 4 = 24\text{ cm}^2

💡 Tip: Area is always measured in square units (e.g.,

\text{cm}^2

\text{m}^2

2. Area of a Triangle

The area of a triangle is half the area of a rectangle with the same base and perpendicular height.

A = \frac{1}{2} \times b \times h

Example: Find area of a triangle with base

8\text{ cm}

and perpendicular height

5\text{ cm}

A = \frac{1}{2} \times 8 \times 5 = 4 \times 5 = 20\text{ cm}^2

💡 Tip: Height

h

MUST be perpendicular (at

90^\circ

) to the base

b

3. Composite Areas (Additive and Subtractive)

Calculate complex shapes by breaking them into simple parts (adding areas) or subtracting a cutout from an outer shape.

Additive Area

A_{\text{total}} = A_1 + A_2 + \dots + A_n

Subtractive Area

A_{\text{remaining}} = A_{\text{outer}} - A_{\text{hole}}

Example: A

10\text{ cm} \times 8\text{ cm}

card has a

3\text{ cm} \times 2\text{ cm}

hole cut out:

A_{\text{remaining}} = (10 \times 8) - (3 \times 2) = 80 - 6 = 74\text{ cm}^2

View Full Lesson

Capacity

Measurement Grade 6

1. Litres and Millilitres Relationship

Capacity measures how much liquid a container holds. The base unit is the litre (L), which equals

1000

millilitres (mL).

1\text{ L} = 1000\text{ mL}

Example: A

2\text{ L}

water bottle contains

2 \times 1000 = 2000\text{ mL}

of water.

2. Converting Between Units of Capacity

To change litres to millilitres, multiply by $1000$ . To change millilitres to litres, divide by $1000$ .

Litres to Millilitres

\text{mL} = \text{L} \times 1000

Millilitres to Litres

\text{L} = \frac{\text{mL}}{1000}

Example: Convert

3500\text{ mL}

to litres:

\text{L} = \frac{3500}{1000} = 3.5\text{ L}

💡 Tip: Dividing by

1000

moves the decimal point

3

places to the left.

View Full Lesson

Length

Measurement Grade 6

1. Standard Metric Length Equivalents

Metric length uses millimetres (mm), centimetres (cm), metres (m), and kilometres (km).

Millimetres in a Centimetre

1\text{ cm} = 10\text{ mm}

Centimetres in a Metre

1\text{ m} = 100\text{ cm}

Metres in a Kilometre

1\text{ km} = 1000\text{ m}

Example: A distance of

3\text{ km}

equals

3 \times 1000 = 3000\text{ m}

2. Converting Down the Ladder (Larger to Smaller)

When converting from a larger unit to a smaller unit, multiply by the conversion factor.

\text{km} \xrightarrow{\times 1000} \text{m} \xrightarrow{\times 100} \text{cm} \xrightarrow{\times 10} \text{mm}

Example: Convert

2.5\text{ m}

to centimetres:

2.5 \times 100 = 250\text{ cm}

💡 Tip: Going to a smaller unit gives a bigger number, so multiply!

3. Converting Up the Ladder (Smaller to Larger)

When converting from a smaller unit to a larger unit, divide by the conversion factor.

\text{mm} \xrightarrow{\div 10} \text{cm} \xrightarrow{\div 100} \text{m} \xrightarrow{\div 1000} \text{km}

Example: Convert

450\text{ cm}

to metres:

\frac{450}{100} = 4.5\text{ m}

💡 Tip: Going to a larger unit gives a smaller number, so divide!

View Full Lesson

Mass

Measurement Grade 6

1. Kilograms and Grams Relationship

Mass measures the amount of matter in an object. One kilogram (kg) equals exactly

1000

grams (g).

1\text{ kg} = 1000\text{ g}

Example: A bag of flour weighing

2.5\text{ kg}

contains

2.5 \times 1000 = 2500\text{ g}

💡 Tip: Kilo means 'one thousand'.

2. Converting Between Grams and Kilograms

Multiply kilograms by

1000

to get grams, or divide grams by

1000

to get kilograms.

Kilograms to Grams

m_{\text{g}} = 1000 \times m_{\text{kg}}

Grams to Kilograms

m_{\text{kg}} = \frac{m_{\text{g}}}{1000}

Example: Convert

750\text{ g}

to kilograms:

m_{\text{kg}} = \frac{750}{1000} = 0.75\text{ kg}

💡 Tip: Dividing by

1000

shifts the decimal point

3

places to the left.

View Full Lesson

Money

Measurement Grade 6

1. Calculating Total Cost

To find the total cost of multiple items, add up the individual prices of all items purchased.

\text{Total Cost} = \text{Price}_1 + \text{Price}_2 + \dots + \text{Price}_n

Example: Buying a book for

150

, a pen for

30

, and a ruler for

20

\text{Total Cost} = 150 + 30 + 20 = 200

2. Calculating Change

The change returned to a customer is the amount paid minus the total cost of the items.

\text{Change} = \text{Amount Paid} - \text{Total Cost}

Example: You pay with a

500

note for items costing

340

in total:

\text{Change} = 500 - 340 = 160

💡 Tip: Double-check your subtraction by adding the change back to the total cost to see if it equals the amount paid.

View Full Lesson

Time

Measurement Grade 6

1. Converting Units of Time

Convert larger time units to smaller units by multiplying by

60

, because

1\text{ hour} = 60\text{ minutes}

and

1\text{ minute} = 60\text{ seconds}

Hours to Minutes

\text{minutes} = \text{hours} \times 60

Minutes to Seconds

\text{seconds} = \text{minutes} \times 60

Example: Convert

2.5\text{ hours}

to minutes:

2.5 \times 60 = 150\text{ minutes}

2. Calculating Elapsed Time (Duration)

Convert times to total minutes past midnight first, subtract departure from arrival, then convert back to hours and minutes.

\text{duration} = \text{arrival minutes} - \text{departure minutes}

Example: Find duration from 08:15 to 10:45:
08:15

= (8 \times 60) + 15 = 495\text{ mins}

10:45

= (10 \times 60) + 45 = 645\text{ mins}

Duration

= 645 - 495 = 150\text{ mins} = 2\text{ hrs } 30\text{ mins}

💡 Tip: To convert total minutes back to hours, divide by

60

\lfloor 150 \div 60 \rfloor = 2\text{ hrs}

, remainder

30\text{ mins}

3. 12-Hour to 24-Hour Clock Conversion

For afternoon and evening times (p.m.), add

12

to the hours to get 24-hour time.

\text{24-hour time} = \text{12-hour time} + 12 \quad (\text{p.m.})

Example: Convert 3:45 p.m. to 24-hour time:

3 + 12 = 15 \implies 15:45

💡 Tip: Note: 12:00 p.m. (noon) stays 12:00, while 12:00 a.m. (midnight) becomes 00:00.

View Full Lesson

Area

Measurement Grade 7

1. Area of Rectangles, Squares, and Parallelograms

Calculate surface space using side measurements. Parallelograms use perpendicular height, not slant height.

Rectangle

A = l \times w

Square

A = s^2

Parallelogram

A = b \times h

Example: A parallelogram has base

b = 8\text{ cm}

and vertical height

h = 5\text{ cm}

A = 8 \times 5 = 40\text{ cm}^2

💡 Tip: Always use vertical height

h

90^\circ

to base

b

, never the slanted side!

2. Area of Triangles and Trapezoids

A triangle is half a rectangle. A trapezoid takes the average of its two parallel sides multiplied by height.

Triangle

A = \frac{1}{2} \times b \times h

Trapezoid

A = \frac{1}{2}(a + b)h

Example: Trapezoid with parallel sides

a = 6\text{ cm}, b = 10\text{ cm}

and height

h = 4\text{ cm}

A = \frac{1}{2}(6 + 10) \times 4 = \frac{1}{2}(16) \times 4 = 8 \times 4 = 32\text{ cm}^2

3. Area of a Circle

Multiply

\pi

by the square of the circle's radius.

A = \pi r^2

Example: Circle with radius

r = 7\text{ cm}

(using

\pi \approx \frac{22}{7}

A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154\text{ cm}^2

💡 Tip: If given diameter

d

, divide by

2

first to get radius

r = \frac{d}{2}

before squaring.

View Full Lesson

Length

Measurement Grade 7

1. Metric Length Conversion Scale

Convert lengths between kilometres, metres, centimetres, and millimetres using base-10 conversion factors.

Larger to Smaller UnitMultiply by factor (

10, 100, 1000

)

Smaller to Larger UnitDivide by factor (

10, 100, 1000

)

Example: Convert

4.2\text{ m}

to centimetres:

4.2 \times 100 = 420\text{ cm}

2. Perimeter of Compound Shapes

Find total perimeter by adding together the lengths of every outer side around the shape.

P = s_1 + s_2 + s_3 + \dots + s_n

Example: An L-shaped field has outer side lengths

8\text{ m}, 6\text{ m}, 5\text{ m}, 2\text{ m}, 3\text{ m}, 4\text{ m}

P = 8 + 6 + 5 + 2 + 3 + 4 = 28\text{ m}

💡 Tip: Make sure to calculate any missing inner or notched outer side lengths before adding.

View Full Lesson

Money

Measurement Grade 7

1. Profit, Loss, and Percentage Profit

Compare selling price to cost price. Percentage profit is always calculated relative to the cost price.

Profit

\text{Selling Price} - \text{Cost Price}

Loss

\text{Cost Price} - \text{Selling Price}

Percentage Profit

\frac{\text{Profit}}{\text{Cost Price}} \times 100\%

Example: Bought an item for

400

and sold for

500

\text{Profit} = 500 - 400 = 100

\text{Percentage Profit} = \frac{100}{400} \times 100\% = 25\%

💡 Tip: Never divide by selling price when finding percentage profit—always divide by cost price!

2. Discounts and Discounted Price

A discount reduces the marked price by a percentage fraction.

\text{Discounted Price} = \text{Marked Price} - \text{Discount}

Example: A shirt marked at

1000

has a

20\%

discount:

\text{Discount} = \frac{20}{100} \times 1000 = 200

\text{Discounted Price} = 1000 - 200 = 800

3. Calculating a Simple Budget Total

Calculate overall budget cost by summing the individual item unit costs multiplied by quantities.

\text{Budget Total} = \sum (\text{Item Cost} \times \text{Quantity})

Example: Buy

3

notebooks at

150

each and

2

pens at

50

each:

\text{Budget Total} = (3 \times 150) + (2 \times 50) = 450 + 100 = 550

View Full Lesson

Pythagorean Relationship

Measurement Grade 7

1. The Pythagorean Theorem

In any right-angled triangle, the area of the square on the longest side (hypotenuse) equals the sum of the areas of the squares on the two shorter sides (legs).

a^2 + b^2 = c^2

Example: For legs

a = 3\text{ cm}

and

b = 4\text{ cm}

c^2 = 3^2 + 4^2 = 9 + 16 = 25

💡 Tip: The hypotenuse

c

is always the longest side, directly opposite the

90^\circ

right angle.

2. Finding the Hypotenuse

To calculate the unknown hypotenuse

c

, sum the squares of both legs and take the square root of the total.

c = \sqrt{a^2 + b^2}

Example: Find hypotenuse

c

when

a = 6\text{ cm}

and

b = 8\text{ cm}

c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}

3. Finding a Missing Leg

To calculate a missing leg

a

b

, subtract the square of the known leg from the square of the hypotenuse, then take the square root.

a = \sqrt{c^2 - b^2}

Example: Find missing leg

a

when hypotenuse

c = 13\text{ cm}

and leg

b = 12\text{ cm}

a = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}

💡 Tip: Always subtract the smaller squared leg from the larger squared hypotenuse to avoid negative square roots.

View Full Lesson

Temperature

Measurement Grade 7

1. Calculating Temperature Change

The change in temperature (

\Delta T

) measures how much a value increases or decreases from start to end.

\Delta T = T_{\text{final}} - T_{\text{initial}}

Example: A liquid heats from

-5^\circ\text{C}

25^\circ\text{C}

\Delta T = 25 - (-5) = 25 + 5 = 30^\circ\text{C}

rise.

💡 Tip: Subtracting a negative temperature is equivalent to adding its positive value.

2. Temperature Difference

The overall gap between the highest and lowest recorded temperatures is found using absolute difference.

\text{Difference} = |T_{\text{high}} - T_{\text{low}}|

Example: Find the difference between maximum daytime temp

18^\circ\text{C}

and minimum night temp

-4^\circ\text{C}

\text{Difference} = |18 - (-4)| = |18 + 4| = 22^\circ\text{C}

3. Finding Final Temperature

To determine the final temperature after a change, add the temperature change (

\Delta T

) to the initial starting value.

T_{\text{final}} = T_{\text{initial}} + \Delta T

Example: Initial temp is

12^\circ\text{C}

and drops by

15^\circ\text{C}

(change

\Delta T = -15^\circ\text{C}

T_{\text{final}} = 12 + (-15) = -3^\circ\text{C}

View Full Lesson

Volume and Capacity

Measurement Grade 7

1. Volume of Cuboids and Cubes

Volume measures 3D space occupied. Multiply length, width, and height for a cuboid, or cube the side length for a cube.

Cuboid Volume

V = l \times w \times h

Cube Volume

V = s^3

Example: A box has length

10\text{ cm}

, width

4\text{ cm}

, and height

5\text{ cm}

V = 10 \times 4 \times 5 = 200\text{ cm}^3

💡 Tip: Ensure all three dimension measurements share the same length unit before multiplying.

2. Volume and Capacity Conversions

Convert physical 3D space measurements into liquid container capacity values.

Litres to

\text{cm}^3

1\text{ L} = 1000\text{ cm}^3

Cubic Metres to Litres

1\text{ m}^3 = 1000\text{ L}

Example: A tank holding

2.5\text{ m}^3

of water contains:

2.5 \times 1000 = 2500\text{ L}

3. Converting Volume ($\text{cm}^3$) to Capacity (L)

Calculate liquid capacity in litres by dividing the total volume in cubic centimetres by

1000

\text{Capacity (L)} = \frac{V_{\text{cm}^3}}{1000}

Example: A fish tank has a volume of

15,000\text{ cm}^3

\text{Capacity} = \frac{15000}{1000} = 15\text{ L}

💡 Tip: Dividing by

1000

shifts the decimal point

3

places to the left.

View Full Lesson

Area

Measurement Grade 8

1. Standard Plane Figures

Core formulas used to calculate 2D area for basic geometric shapes.

Rectangle

A = l \times w

Square

A = s^2

Triangle

A = \frac{1}{2} \times b \times h

Circle

A = \pi r^2

Example: Find area of a circle with radius

r = 7\text{ cm}

(using

\pi = \frac{22}{7}

A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154\text{ cm}^2

2. Semicircle and Quarter-Circle Areas

Calculate fractional circle areas by dividing the full circle area formula by

2

4

Semicircle

A = \frac{1}{2} \pi r^2

Quarter-Circle

A = \frac{1}{4} \pi r^2

Example: Find area of a semicircle with radius

r = 10\text{ cm}

(using

\pi = 3.14

A = \frac{1}{2} \times 3.14 \times 10^2 = 0.5 \times 3.14 \times 100 = 157\text{ cm}^2

3. Composite and Subtractive Shapes

Calculate complex combined shapes by adding basic sections together or removing cutout areas from outer boundaries.

Additive Area

A_{\text{total}} = A_1 + A_2 + \dots + A_n

Subtractive Area

A_{\text{remaining}} = A_{\text{large}} - A_{\text{hole}}

Example: A square metal plate (

10\text{ cm} \times 10\text{ cm}

) has a circular hole of area

28\text{ cm}^2

punched through:

A_{\text{remaining}} = (10 \times 10) - 28 = 100 - 28 = 72\text{ cm}^2

View Full Lesson

Circles

Measurement Grade 8

1. Radius, Diameter, and Circumference

The diameter (

d

) is twice the radius (

r

), and the circumference (

C

) is the total boundary distance around the circle.

C = 2\pi r = \pi d

Example: For a circle with radius

r = 7\text{ cm}

(using

\pi = \frac{22}{7}

d = 2(7) = 14\text{ cm}

C = 2 \times \frac{22}{7} \times 7 = 44\text{ cm}

💡 Tip: Always check whether a question gives you the radius or the diameter before using the formula.

2. Area of a Circle

The total surface area inside a circle equals

\pi

multiplied by the square of the radius.

A = \pi r^2

Example: Find area of a circle with radius

r = 7\text{ cm}

(using

\pi = \frac{22}{7}

A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154\text{ cm}^2

💡 Tip: If given diameter

d

, divide by

2

first to get radius

r

before squaring.

3. Arc Length of a Sector

An arc is a curved fraction of the full circumference, proportional to its central angle

\theta

s = \frac{\theta}{360^\circ} \times 2\pi r

Example: Find arc length for a central angle

\theta = 60^\circ

and radius

r = 21\text{ cm}

(using

\pi = \frac{22}{7}

s = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22\text{ cm}

4. Area of a Sector

A sector is a slice of a circle. Its area is the fraction

\frac{\theta}{360^\circ}

of the circle's total area.

A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2

Example: Find sector area for

\theta = 90^\circ

and radius

r = 14\text{ cm}

(using

\pi = \frac{22}{7}

A_{\text{sector}} = \frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2

View Full Lesson

Money

Measurement Grade 8

1. Simple Interest

Simple interest (

I

) is calculated on the initial money borrowed or invested (principal) over time.

I = P \times \frac{r}{100} \times t

Example: Calculate interest on a principal

P = 5000

6\%

per year for

3

years:

I = 5000 \times \frac{6}{100} \times 3 = 5000 \times 0.06 \times 3 = 900

💡 Tip: Ensure time

t

is expressed in years (e.g.,

6

months

= 0.5

years).

2. Total Accumulated Amount

The total amount (

A

) owed or received is the original principal plus the calculated interest.

A = P + I = P\left(1 + \frac{r}{100} \times t\right)

Example: Find total amount for principal

P = 5000

with interest

I = 900

A = 5000 + 900 = 5900

3. Commission and Total Earnings

Commission is a performance-based percentage cut of sales. Total earnings combine base salary with commission.

Commission

\text{Commission} = \frac{\text{Rate}}{100} \times \text{Sales}

Total Earnings

\text{Total Earnings} = \text{Base Pay} + \text{Commission}

Example: A agent has a base pay of

15,000

and earns a

5\%

commission on sales of

200,000

\text{Commission} = \frac{5}{100} \times 200,000 = 10,000

\text{Total Earnings} = 15,000 + 10,000 = 25,000

View Full Lesson

Approximations and Errors

Measurement Grade 9

1. Absolute, Relative, and Percentage Errors

Absolute error is the numerical gap between true and estimated values. Relative error expresses this as a fraction of the true value.

Absolute Error

E_{\text{abs}} = |x_{\text{true}} - x_{\text{approx}}|

Relative Error

E_{\text{rel}} = \frac{E_{\text{abs}}}{|x_{\text{true}}|}

Percentage Error

E_{\%} = E_{\text{rel}} \times 100\%

Example: True length is

20\text{ cm}

, measured as

19.5\text{ cm}

E_{\text{abs}} = |20 - 19.5| = 0.5\text{ cm}

E_{\text{rel}} = \frac{0.5}{20} = 0.025

E_{\%} = 0.025 \times 100\% = 2.5\%

💡 Tip: Absolute error is always a positive number because of the absolute value brackets.

2. Rounding Limit of Error

The maximum possible error when rounding to

n

decimal places is half the value of the last retained place value.

E_{\text{round}} = \frac{1}{2} \times 10^{-n}

Example: When rounding to

1

decimal place (

n=1

E_{\text{round}} = \frac{1}{2} \times 10^{-1} = 0.5 \times 0.1 = 0.05

💡 Tip: This represents the lower and upper bounds of a rounded figure.

3. Error Combination in Addition and Subtraction

When adding or subtracting two measurements with independent errors, combine absolute errors in quadrature.

E_{\text{sum}} = \sqrt{E_1^2 + E_2^2}

Example: Adding lengths with errors

E_1 = 0.3

and

E_2 = 0.4

E_{\text{sum}} = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5

4. Error Combination in Multiplication and Division

For products or quotients, combine relative errors in quadrature to find the overall product relative error.

\frac{E_{\text{prod}}}{|P|} = \sqrt{\left(\frac{E_1}{|x_1|}\right)^2 + \left(\frac{E_2}{|x_2|}\right)^2}

Example: Multiply

x_1 = 10

(

E_1 = 0.3

, relative

0.03

) and

x_2 = 20

(

E_2 = 0.8

, relative

0.04

P = 10 \times 20 = 200

\text{Relative error} = \sqrt{0.03^2 + 0.04^2} = \sqrt{0.0025} = 0.05

E_{\text{prod}} = 200 \times 0.05 = 10

View Full Lesson

Area

Measurement Grade 9

1. Area Formulas for Common 2D Shapes

Standard mathematical formulas for evaluating regions enclosed by plane shapes.

Triangle

A = \frac{1}{2}bh

Rectangle & Square

A = \ell \times w \quad | \quad A = s^2

Parallelogram

A = bh

Trapezium

A = \frac{1}{2}(a + b)h

Example: Trapezium with parallel sides

a = 5\text{ m}, b = 9\text{ m}

and height

h = 4\text{ m}

A = \frac{1}{2}(5 + 9) \times 4 = \frac{1}{2}(14) \times 4 = 28\text{ m}^2

2. Polygon Decomposition into Triangles

Any non-overlapping

n

-sided polygon can be split into

n-2

interior triangles from a single vertex to compute total area.

A_{\text{polygon}} = A_{T_1} + A_{T_2} + \dots + A_{T_k}

Example: A

5

-sided pentagon (

n=5

) splits into

5 - 2 = 3

triangles. Sum the

3

triangle areas to find the total pentagon area.

💡 Tip: The formula

n - 2

also helps calculate total interior angles:

(n-2) \times 180^\circ

3. Land Area Conversions (Hectares)

Large land parcels and farms (shambas) are measured using hectares (ha) and square kilometres (

\text{km}^2

Hectare to Square Metres

1\text{ ha} = 10,000\text{ m}^2

Square Kilometre to Square Metres

1\text{ km}^2 = 1,000,000\text{ m}^2 = 100\text{ ha}

Example: A farm measuring

25,000\text{ m}^2

converted to hectares:

\text{Area} = \frac{25,000}{10,000} = 2.5\text{ ha}

💡 Tip: Think of

1\text{ hectare}

as a square plot measuring

100\text{ m} \times 100\text{ m}

View Full Lesson

Mass, Volume, Weight, and Density

Measurement Grade 9

1. Density, Mass, and Volume Triangle

Density (

\rho

) measures how tightly matter is packed inside a given 3D volume.

Density

\rho = \frac{m}{V}

Mass

m = \rho \times V

Volume

V = \frac{m}{\rho}

Example: A block of metal has mass

m = 400\text{ g}

and volume

V = 50\text{ cm}^3

\rho = \frac{400}{50} = 8\text{ g/cm}^3

💡 Tip: Unit check: if mass is in grams and volume in

\text{cm}^3

, density is in

\text{g/cm}^3

2. Weight and Gravity

Weight (

W

) is the downward gravitational force acting on a mass (

m

), measured in Newtons (N).

W = m \times g

Example: Find weight of a

5\text{ kg}

mass on Earth (

g \approx 9.8\text{ m/s}^2

W = 5 \times 9.8 = 49\text{ N}

💡 Tip: Mass never changes, but weight changes if gravity

g

changes (e.g., on the Moon).

3. Mass Unit Consistency

Always convert mass to standard matching units before calculating density or weight.

1\text{ kg} = 1000\text{ g}

Example: Convert

250\text{ g}

to kilograms before computing weight:

m = \frac{250}{1000} = 0.25\text{ kg} \implies W = 0.25 \times 9.8 = 2.45\text{ N}

View Full Lesson

Money

Measurement Grade 9

1. Discounts and Discounted Price

A discount reduces the original cost (

C

) by percentage rate

d\%

D = C \times \left(1 - \frac{d}{100}\right)

Example: An item priced at

2000

has a

15\%

discount (

d = 15

D = 2000 \times \left(1 - 0.15\right) = 2000 \times 0.85 = 1700

2. Value Added Tax (VAT)

VAT is a tax percentage added to the base price of goods and services.

T = C \times \left(1 + \frac{t}{100}\right)

Example: Calculate total price for

5000

base cost with

16\%

VAT (

t = 16

T = 5000 \times \left(1 + 0.16\right) = 5000 \times 1.16 = 5800

3. Combined Discount and VAT

When both a discount and VAT apply, apply the discount factor first, then multiply by the VAT factor.

\text{Final} = C \times \left(1 - \frac{d}{100}\right) \times \left(1 + \frac{t}{100}\right)

Example: Item cost

10,000

with

10\%

discount and

16\%

VAT:

\text{Final} = 10000 \times (1 - 0.10) \times (1 + 0.16) = 10000 \times 0.90 \times 1.16 = 10,440

💡 Tip: Apply the discount to the original price BEFORE adding VAT on the reduced price.

View Full Lesson

Time, Distance, and Speed

Measurement Grade 9

1. Distance, Speed, and Time Triangle

Speed (

v

) measures distance (

d

) traveled per unit of time (

t

Distance

d = v \times t

Speed

v = \frac{d}{t}

Time

t = \frac{d}{v}

Example: A bus travels

180\text{ km}

3\text{ hours}

v = \frac{180}{3} = 60\text{ km/h}

💡 Tip: Check that units match: if speed is

\text{km/h}

, distance must be

\text{km}

and time in

\text{hours}

2. Average Speed for a Whole Journey

Average speed is total distance divided by total time for the entire multi-stage trip.

v_{\text{avg}} = \frac{\text{total distance}}{\text{total time}}

Example: A car travels

100\text{ km}

2\text{ hrs}

, then

140\text{ km}

1\text{ hr}

v_{\text{avg}} = \frac{100 + 140}{2 + 1} = \frac{240}{3} = 80\text{ km/h}

💡 Tip: Never just average the two individual speeds together! Always sum distances and total time.

View Full Lesson

Volume of Solids

Measurement Grade 9

1. General Volume of Prisms

The volume of any prism equals its uniform cross-sectional base area (

B

) multiplied by height (

h

V_{\text{prism}} = B \times h

Example: A triangular prism has base area

B = 15\text{ cm}^2

and length

h = 10\text{ cm}

V = 15 \times 10 = 150\text{ cm}^3

2. Volume of a Cylinder

A cylinder is a prism with a circular base (

B = \pi r^2

). Multiply base circle area by height.

V_{\text{cylinder}} = \pi r^2 h

Example: Find volume of cylinder with radius

r = 7\text{ cm}

and height

h = 10\text{ cm}

(using

\pi = \frac{22}{7}

V = \frac{22}{7} \times 7^2 \times 10 = \frac{22}{7} \times 49 \times 10 = 1540\text{ cm}^3

💡 Tip: Be careful not to confuse cylinder volume (

\pi r^2 h

) with cylinder surface area!

View Full Lesson

Decimals

Numbers Grade 6

1. Decimal Place Value

Digits after the decimal point represent fractions of ten, hundred, or thousand. The

n^{\text{th}}

place value after the dot equals

\frac{d}{10^n}

\text{Value} = \frac{d}{10^n}

Example: In

0.35

:
Digit

3

is in tenths place

\implies \frac{3}{10}

Digit

5

is in hundredths place

\implies \frac{5}{100}

2. Multiplying & Dividing by Powers of 10

Multiplying moves the decimal point to the right to make the number larger. Dividing moves it to the left to make it smaller.

Multiply by

10^k

Shift decimal point

k

places right

Divide by

10^k

Shift decimal point

k

places left

Example:

3.45 \times 10 = 34.5

(shift

1

right)

3.45 \div 100 = 0.0345

(shift

2

left).

💡 Tip: Count the zeros in the power of 10 to know how many places to move the decimal point.

3. Adding & Subtracting Decimals

Line up the decimal points vertically first, then add or subtract column by column like whole numbers.

Example: Calculate

12.4 + 3.85

\begin{array}{rl} 12.40 \\ +\ 3.85 \\ \hline 16.25 \end{array}

Add place-holder zero to

12.4

to align columns.

💡 Tip: Always pad shorter decimal numbers with trailing zeros so all columns match.

4. Multiplying Decimals

Multiply as if there are no decimal points, then count total decimal places in both numbers and place the point that many digits from the right in your answer.

d_1 \times d_2 \implies \text{Count total decimal places}

Example: Calculate

0.3 \times 0.04

:
1. Multiply

3 \times 4 = 12

.
2. Total decimal places

= 1 + 2 = 3

.
3. Place point

3

places from right:

0.012

5. Dividing Decimals

If the divisor is a decimal, multiply both divisor and dividend by

10, 100,

1000

to turn the divisor into a whole number before dividing.

\frac{n}{d} = \frac{n \times 10^k}{d \times 10^k}

Example: Calculate

1.2 \div 0.04

:
Multiply both by

100

120 \div 4 = 30

View Full Lesson

Division

Numbers Grade 6

1. The Division Algorithm

When dividing dividend

a

by divisor

b

, the total equals divisor times quotient (

q

) plus a non-negative remainder (

r

a = b \times q + r \quad (0 \le r < b)

Example: Divide

17

5

17 = 5 \times 3 + 2

Quotient

q = 3

, Remainder

r = 2

💡 Tip: The remainder

r

must always be strictly smaller than the divisor

b

2. Quotient and Remainder Formulas

The quotient

q

is the whole-number floor of the division, and remainder

r

is the leftover value.

q = \left\lfloor \frac{a}{b} \right\rfloor, \quad r = a - b \times q

Example: For

23 \div 4

q = \left\lfloor \frac{23}{4} \right\rfloor = 5

r = 23 - (4 \times 5) = 23 - 20 = 3

3. Writing Results as Mixed Numbers

Express the exact division answer by placing the remainder over the divisor as a fraction.

\frac{a}{b} = q + \frac{r}{b}

Example: Express

17 \div 5

as a mixed number:

\frac{17}{5} = 3 + \frac{2}{5} = 3\frac{2}{5}

View Full Lesson

Fractions

Numbers Grade 6

1. Adding & Subtracting Like Fractions

When denominators are equal, add or subtract the numerators while keeping the common denominator unchanged.

\frac{a}{b} \pm \frac{c}{b} = \frac{a \pm c}{b}

Example:

\frac{3}{7} + \frac{2}{7} = \frac{3 + 2}{7} = \frac{5}{7}

💡 Tip: Never add or subtract denominators!

2. Adding & Subtracting Unlike Fractions

When denominators differ, convert to equivalent fractions with a common denominator before combining.

\frac{a}{b} \pm \frac{c}{d} = \frac{a \cdot d \pm c \cdot b}{b \cdot d}

Example:

\frac{1}{3} + \frac{2}{5} = \frac{(1 \times 5) + (2 \times 3)}{3 \times 5} = \frac{5 + 6}{15} = \frac{11}{15}

3. Equivalent Fractions

Multiply or divide both numerator and denominator by the same non-zero number to create an equivalent fraction.

\frac{a}{b} = \frac{a \times k}{b \times k}

Example:

\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}

4. Comparing Fractions with Same Numerator

When numerators are equal, the fraction with the smaller denominator is larger because the whole is split into fewer, bigger parts.

\text{If } b < d \implies \frac{a}{b} > \frac{a}{d}

Example: Compare

\frac{3}{4}

and

\frac{3}{8}

:
Since

4 < 8

\frac{3}{4} > \frac{3}{8}

View Full Lesson

Inequalities

Numbers Grade 6

1. Solving Linear Inequalities

Isolate

x

using inverse operations. Flip the inequality sign if you multiply or divide by a negative number.

ax + b < c \implies x < \frac{c - b}{a}

Example:

2x + 3 < 11 \implies 2x < 8 \implies x < 4

2. Graphing Inequalities on a Number Line

Draw an open dot for strict inequalities (

<, >

) and a closed dot for inclusive inequalities (

\le, \ge

Strict (

x > k

x < k

)Open dot

\circ

k

, shade in inequality direction

Inclusive (

x \le k

x \ge k

)Closed dot

\bullet

k

, shade in inequality direction

Example:

x > 3

: Open dot at

3

, arrow shaded to the right.

x \le 5

: Closed dot at

5

, arrow shaded to the left.

3. Compound Inequalities

A compound inequality bounds

x

between two values. Graph by placing boundary dots at both endpoints and shading the region between them.

a < x \le b

Example: Graph

2 < x \le 6

:
Open dot at

2

, closed dot at

6

, shade line segment connecting them.

View Full Lesson

Multiplication

Numbers Grade 6

1. Multiplication as Repeated Addition

Multiplication is a shortcut for adding the same number to itself a set number of times.

a \times b = \underbrace{a + a + \dots + a}_{b \text{ times}}

Example:

4 \times 3 = 4 + 4 + 4 = 12

2. Properties of Multiplication

Core operational laws that govern how numbers interact during multiplication.

Commutative Property

a \times b = b \times a

Associative Property

(a \times b) \times c = a \times (b \times c)

Multiplicative Identity

1 \times a = a

Zero Property

0 \times a = 0

Example:

5 \times 3 = 3 \times 5 = 15

(2 \times 3) \times 4 = 2 \times (3 \times 4) = 24

3. The Distributive Property

Break a large factor into smaller parts, multiply each part separately, and add the results.

a \times (b + c) = a \times b + a \times c

Example:

6 \times 14 = 6 \times (10 + 4) = (6 \times 10) + (6 \times 4) = 60 + 24 = 84

💡 Tip: Use the distributive property to solve tough mental math problems quickly!

View Full Lesson

Whole Numbers

Numbers Grade 6

1. Place Value System

The position of a digit determines its value as a power of

10

. One million is equal to one thousand thousands.

\text{Value} = d \times 10^n

Example: In

1,000,000

1

is in the millions place (

1 \times 10^6

).
In

5,400

, the digit

5

represents

5 \times 10^3 = 5000

2. Rounding Whole Numbers

To round a number, check the decision digit to the right of your target place value.

Digit 5 or higherRound UP (add 1 to target place)

Digit 4 or lowerRound DOWN (target place stays same)

Example: Round

4,372

to the nearest hundred:
Target digit is

3

(hundreds), decision digit is

7

(tens).
Since

7 \ge 5

, round up to

4,400

3. Comparing Whole Numbers

Compare two numbers by starting at the leftmost place value and comparing digits where they first differ.

a > b

Example: Compare

452,100

and

458,000

:
First two digits (

4, 5

) match. In the thousands place,

8 > 2

, so

458,000 > 452,100

💡 Tip: Always align place values or count digits first—the number with more digits is always larger.

View Full Lesson

Decimals

Numbers Grade 7

1. Converting Fractions to Decimals

Divide the numerator

p

by denominator

q

using long division to get the decimal form.

\frac{p}{q} = p \div q

Example: Convert

\frac{3}{4}

to a decimal:

3 \div 4 = 0.75

2. Terminating Decimals and Fractions

A simplified fraction gives a terminating decimal if its denominator's only prime factors are

2

and

5

\frac{p}{q} \text{ terminates} \iff q = 2^a \cdot 5^b

Example:

\frac{3}{20} = \frac{3}{2^2 \cdot 5}

terminates as

0.15

\frac{1}{3}

recurs as

0.333\dots

because

3

is a prime factor other than

2

5

💡 Tip: Simplify the fraction to lowest terms before checking prime factors of the denominator.

3. Recurring Decimals to Fractions

Convert repeating decimals to exact fractions using algebraic elimination.

0.\bar{a} = \frac{a}{9}

Example: Convert

x = 0.\bar{3}

(

0.333\dots

) to a fraction:

10x = 3.333\dots

Subtract

x

9x = 3 \implies x = \frac{3}{9} = \frac{1}{3}

💡 Tip: Multiply by

10

for

1

repeating digit,

100

for

2

repeating digits, and

1000

for

3

repeating digits.

View Full Lesson

Factors

Numbers Grade 7

1. Prime Factorisation

Every composite number can be broken down into a unique product of prime numbers raised to powers.

n = p_1^{e_1} \times p_2^{e_2} \times \dots \times p_r^{e_r}

Example: Express

60

as a product of prime factors:

60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3^1 \times 5^1

2. Highest Common Factor (HCF) & Lowest Common Multiple (LCM)

For HCF, take the smallest exponent of shared prime factors. For LCM, take the largest exponent of all prime factors present.

HCF Formula

\text{HCF}(a,b) = \prod p_i^{\min(e_{i,a}, e_{i,b})}

LCM Formula

\text{LCM}(a,b) = \prod p_i^{\max(e_{i,a}, e_{i,b})}

Example: For

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

\text{HCF} = 2^1 \times 3^1 = 6

\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36

3. The HCF and LCM Product Rule

The product of the HCF and LCM of any two positive numbers is equal to the product of the two numbers themselves.

\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

Example: For

a = 12

and

b = 18

(

\text{HCF} = 6

\text{LCM} = 36

6 \times 36 = 216

and

12 \times 18 = 216

. Equality holds!

💡 Tip: Use this rule to find an unknown LCM quickly:

\text{LCM} = \frac{a \times b}{\text{HCF}}

View Full Lesson

Fractions

Numbers Grade 7

1. Multiplying Fractions

To multiply two fractions, multiply the top numbers (numerators) together and multiply the bottom numbers (denominators) together.

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Example:

\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}

2. Dividing Fractions

Dividing by a fraction is the same as multiplying by its reciprocal (flip the second fraction).

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{a \times d}{b \times c}

Example:

\frac{3}{4} \div \frac{2}{5} = \frac{3}{4} \times \frac{5}{2} = \frac{3 \times 5}{4 \times 2} = \frac{15}{8} = 1\frac{7}{8}

💡 Tip: Remember: Keep, Change, Flip (Keep 1st fraction, Change

\div

\times

, Flip 2nd fraction).

3. Converting Mixed Numbers to Improper Fractions

Before multiplying or dividing mixed numbers, convert them into improper fractions first.

a\frac{b}{c} = \frac{(a \times c) + b}{c}

Example: Convert

2\frac{1}{3}

\frac{(2 \times 3) + 1}{3} = \frac{6 + 1}{3} = \frac{7}{3}

4. Simplifying Fractions to Lowest Terms

Divide both numerator and denominator by their Greatest Common Divisor (GCD) to simplify the final answer.

\frac{a}{b} = \frac{a \div g}{b \div g}

Example: Simplify

\frac{12}{18}

(GCD

= 6

\frac{12 \div 6}{18 \div 6} = \frac{2}{3}

View Full Lesson

Squares and Square Roots

Numbers Grade 7

1. Squaring a Number

Squaring a number means multiplying that number by itself.

n^2 = n \times n

Example:

7^2 = 7 \times 7 = 49

2. Square Roots as Inverse Operations

The square root of a number is the value that, when multiplied by itself, gives the original number.

\sqrt{a} = b \quad \text{where } b^2 = a

Example:

\sqrt{36} = 6

because

6^2 = 36

💡 Tip: By definition, principal square roots of real numbers are non-negative.

3. Square Root of a Square

Taking the square root of a squared number returns its positive magnitude (absolute value).

\sqrt{a^2} = |a|

Example:

\sqrt{(-5)^2} = \sqrt{25} = 5 = |-5|

💡 Tip: The result of a principal square root is never negative.

4. Square and Root Properties for Products and Fractions

You can split or combine squares and square roots across multiplication and division.

Product Power Law

(ab)^2 = a^2b^2

Fraction Square Root

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

Example:

(3 \times 4)^2 = 3^2 \times 4^2 = 9 \times 16 = 144

\sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}

💡 Tip: Square roots split across multiplication and division, but NEVER across addition or subtraction!

View Full Lesson

Whole Numbers

Numbers Grade 7

1. Order of Operations (BODMAS)

Always solve math expressions in the strict BODMAS order: Brackets, Orders (powers/roots), Division/Multiplication, Addition/Subtraction.

BBrackets first

OOrders (exponents and square roots)

D / MDivision and Multiplication (left to right)

A / SAddition and Subtraction (left to right)

Example: Evaluate

3 + 4 \times (5 - 2)

:
1. Brackets:

5 - 2 = 3

2. Multiply:

4 \times 3 = 12

3. Add:

3 + 12 = 15

2. Brackets Priority

Brackets force the operation inside them to be computed before any outside operations.

a \times (b + c)

Example:

2 \times (3 + 4) = 2 \times 7 = 14

.
(Without brackets,

2 \times 3 + 4 = 6 + 4 = 10

💡 Tip: Always clear the innermost brackets first.

3. Equal Priority Operations (Left to Right)

Division and multiplication have equal priority, as do addition and subtraction. Solve them in order from left to right.

a \div b \times c

Example: Evaluate

12 \div 4 \times 3

:
Working left to right:

12 \div 4 = 3

, then

3 \times 3 = 9

💡 Tip: Do not do multiplication before division just because 'M' comes after 'D' in the acronym—they share equal priority!

View Full Lesson

Decimals

Numbers Grade 8

1. Converting Fractions to Decimals

Convert any fraction to a decimal by dividing the numerator by the denominator.

\frac{a}{b} = a \div b

Example: Convert

\frac{3}{8}

to a decimal:

3 \div 8 = 0.375

2. Shifting Decimals by Powers of 10

Multiplying by powers of

10

moves the decimal point to the right; dividing moves it to the left.

Multiply by

10^k

Shift decimal point

k

places right

Divide by

10^k

Shift decimal point

k

places left

Example:

4.56 \times 100 = 456

(shift

2

places right)

4.56 \div 1000 = 0.00456

(shift

3

places left)

3. Multiplying Decimals

Multiply as whole numbers first, then place the decimal point so the total number of decimal places equals the combined decimal places of both factors.

(a.bc) \times (d.ef) \implies \text{Count total decimal places}

Example: Calculate

1.2 \times 0.03

:
1. Whole numbers:

12 \times 3 = 36

2. Total decimal places:

1 + 2 = 3

3. Result:

0.036

💡 Tip: Add leading zeros if needed to reach the total required decimal places.

4. Rounding Decimals

To round to

n

decimal places, look at the

(n+1)^{\text{th}}

digit. Round up if it is

5

or more; leave the

n^{\text{th}}

digit unchanged if it is

4

or less.

Example: Round

3.14159

2

decimal places:

n=2

digit is

4

(n+1)=3^{\text{rd}}

digit is

1

.
Since

1 < 5

, round down:

3.14

💡 Tip: Drop all digits to the right of the rounded position.

View Full Lesson

Fractions

Numbers Grade 8

1. Adding and Subtracting Fractions

To add or subtract fractions with different denominators, find a common denominator, adjust the numerators, and combine.

\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm cb}{bd}

Example:

\frac{1}{3} + \frac{2}{5} = \frac{(1 \times 5) + (2 \times 3)}{3 \times 5} = \frac{5 + 6}{15} = \frac{11}{15}

2. Multiplying Fractions

Multiply straight across: numerators together and denominators together.

\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}

Example:

\frac{3}{4} \times \frac{5}{7} = \frac{3 \times 5}{4 \times 7} = \frac{15}{28}

3. Dividing Fractions

Divide by a fraction by multiplying by its reciprocal (flip the second fraction).

\frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}

Example:

\frac{2}{5} \div \frac{3}{4} = \frac{2}{5} \times \frac{4}{3} = \frac{8}{15}

💡 Tip: Remember: Keep the first fraction, Change

\div

\times

, and Flip the second fraction.

4. Mixed Numbers to Improper Fractions

Convert mixed numbers to improper fractions before performing multiplication or division.

n\frac{p}{q} = \frac{nq + p}{q}

Example: Convert

3\frac{2}{5}

\frac{(3 \times 5) + 2}{5} = \frac{15 + 2}{5} = \frac{17}{5}

View Full Lesson

Integers

Numbers Grade 8

1. Adding and Subtracting Negative Numbers

Adding a negative number is equivalent to subtracting a positive. Subtracting a negative is equivalent to adding a positive.

Add a negative

a + (-b) = a - b

Subtract a negative

a - (-b) = a + b

Example:

7 + (-3) = 7 - 3 = 4

5 - (-2) = 5 + 2 = 7

💡 Tip: Two like signs together give a plus:

-(-3) = +3

. Two unlike signs give a minus:

+(-3) = -3

2. Multiplication and Division Signs Rule

Multiplying or dividing numbers with the same signs gives a positive result. Different signs give a negative result.

Same signsPositive result

(+)

Different signsNegative result

(-)

Example:

(-4) \times (-3) = 12

(-15) \div 3 = -5

3. Zero Product Property

If the product of two numbers is zero, at least one of the individual factors must be zero.

a \times b = 0 \implies a = 0 \text{ or } b = 0

Example: If

5 \times x = 0

, then

x

must equal

0

4. Absolute Value

The absolute value

|a|

is the distance of a number from zero on a number line, always non-negative.

|a| = \begin{cases} a & \text{if } a \ge 0 \\ -a & \text{if } a < 0 \end{cases}

Example:

|5| = 5

|-8| = -(-8) = 8

💡 Tip: Absolute value strips away the negative sign.

View Full Lesson

Rates, Ratio, Proportions and Percentages

Numbers Grade 8

1. Ratios and Rates

A ratio compares two quantities of the same unit as a fraction. A rate compares two quantities measured in different units.

Ratio

\text{Ratio } a:b = \frac{a}{b}

Rate

\text{Rate} = \frac{\text{quantity}_1}{\text{quantity}_2}

Example: Ratio of

4\text{ m}

8\text{ m} = \frac{4}{8} = 1:2

.
Speed rate for

120\text{ km}

2\text{ hours} = \frac{120}{2} = 60\text{ km/h}

💡 Tip: Ratios have no units, but rates must always include units!

2. Cross-Multiplication in Proportions

Solve an unknown in two equal fractions by cross-multiplying numerators and denominators.

\frac{a}{b} = \frac{c}{d} \implies a \times d = b \times c

Example: Solve

\frac{x}{6} = \frac{5}{2}

2x = 6 \times 5 \implies 2x = 30 \implies x = 15

3. Percentage Calculations

Calculate parts of a whole or reverse the process to find the original total.

Finding Part

\text{Part} = \frac{\%}{100} \times \text{Whole}

Finding Whole

\text{Whole} = \frac{\text{Part}}{\%} \times 100

Example: Find

15\%

200

\text{Part} = \frac{15}{100} \times 200 = 30

.
If

30

15\%

of a number:

\text{Whole} = \frac{30}{15} \times 100 = 200

4. Percentage Change

Measure percentage increase or decrease relative to the original value.

\% \text{ Change} = \frac{\text{New} - \text{Old}}{\text{Old}} \times 100

Example: A price increases from

80

100

\% \text{ Change} = \frac{100 - 80}{80} \times 100 = \frac{20}{80} \times 100 = 25\%

increase.

💡 Tip: Always divide by the OLD (original) value, never the new one!

View Full Lesson

Squares and Square Roots

Numbers Grade 8

1. Squares and Square Roots Definition

Squaring a number converts a side length into a square area; taking the square root reverses this.

n^2 = n \times n, \quad \sqrt{A} = r \iff r^2 = A

Example:

9^2 = 81 \implies \sqrt{81} = 9

2. Estimating Square Roots

Trap a non-perfect square between its surrounding perfect square values to estimate its root.

a^2 < A < b^2 \implies a < \sqrt{A} < b

Example: Estimate

\sqrt{20}

:
Since

16 < 20 < 25

, we know

4^2 < 20 < 5^2

.
Therefore,

4 < \sqrt{20} < 5

(approx.

4.5

3. Linear Interpolation Estimate

Refine a square root estimate proportionally between two perfect square boundaries.

\sqrt{A} \approx a + \frac{A - a^2}{b^2 - a^2}

Example: Estimate

\sqrt{20}

using lower bound

a=4

(

a^2=16

) and upper bound

b=5

(

b^2=25

\sqrt{20} \approx 4 + \frac{20 - 16}{25 - 16} = 4 + \frac{4}{9} \approx 4.44

4. Simplifying Radical Expressions

Square roots split across multiplication and division, allowing radicals to be simplified.

\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}, \quad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

Example: Simplify

\sqrt{50}

\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}

\sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}

💡 Tip: Look for the largest perfect square factor (

4, 9, 16, 25, 36, \dots

) when simplifying radicals.

View Full Lesson

Compound Proportions and Rates of Work

Numbers Grade 9

1. Direct Proportions and Cross-Multiplication

When two quantities increase or decrease at the same ratio, they are directly proportional.

\frac{a}{b} = \frac{c}{d} \implies x = \frac{a \cdot c}{b}

Example: If

4

books cost

200

, find the cost

x

6

books:

\frac{4}{200} = \frac{6}{x} \implies 4x = 1200 \implies x = 300

2. Work, Rate, and Time Relationship

Total work output equals working speed (rate) multiplied by total time spent.

Work Done

\text{Work} = \text{Rate} \times \text{Time}

Individual Rate

R = \frac{\text{Work}}{\text{Time}}

Example: If a builder completes

1

wall in

5

hours, their work rate is

R = \frac{1}{5}

wall per hour.

3. Combined Work Rates

When people or machines work together, add their individual rates to find the combined speed.

R_{\text{total}} = R_1 + R_2 + \dots + R_n, \quad T = \frac{\text{Work}}{R_{\text{total}}}

Example: Worker A completes a job in

3

hours (

R_1 = \frac{1}{3}

) and Worker B in

6

hours (

R_2 = \frac{1}{6}

R_{\text{total}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

job/hr.

\text{Combined Time } T = \frac{1}{1/2} = 2\text{ hours}

💡 Tip: Always add the RATES together, never add the completion times directly!

View Full Lesson

Cubes and Cube Roots

Numbers Grade 9

1. Cubing a Number

Cubing a number means using it as a factor three times, representing the volume of a 3D cube.

a^3 = a \times a \times a, \quad V = s^3

Example:

4^3 = 4 \times 4 \times 4 = 64

.
A cube with edge

s = 4\text{ cm}

has volume

V = 64\text{ cm}^3

2. Cube Roots as Inverse Operations

Taking the cube root finds the side length

s

that produces a given volume

V

when cubed.

\sqrt[3]{b} = c \iff c^3 = b, \quad s = \sqrt[3]{V}

Example:

\sqrt[3]{125} = 5

because

5^3 = 125

.
A cube with volume

125\text{ cm}^3

has edge length

s = 5\text{ cm}

💡 Tip: Unlike square roots, cube roots of negative numbers are negative:

\sqrt[3]{-27} = -3

3. Scaling Edge Length vs Volume

Scaling the edge length of a cube by factor

k

increases its total volume by

k^3

V_{\text{new}} = (ks)^3 = k^3 V

Example: If a cube's edge is doubled (

k = 2

):
New Volume

= 2^3 \times V = 8V

. The volume becomes

8

times larger!

4. Product Law for Cubes

The cube of a product equals the product of the individual cubed terms.

(ab)^3 = a^3 b^3, \quad \sqrt[3]{a^3} = a

Example:

(2 \times 3)^3 = 2^3 \times 3^3 = 8 \times 27 = 216

View Full Lesson

Indices and Logarithms

Numbers Grade 9

1. Core Laws of Indices

Rules for combining exponents with identical base values.

Multiplication Law

a^m \times a^n = a^{m+n}

Division Law

\frac{a^m}{a^n} = a^{m-n}

Power of Power Law

(a^m)^n = a^{mn}

Example:

2^3 \times 2^4 = 2^{3+4} = 2^7 = 128

\frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625

(3^2)^3 = 3^{2 \times 3} = 3^6 = 729

2. Negative and Zero Exponents

Any non-zero base raised to power

0

equals

1

, while a negative power creates a reciprocal fraction.

Zero Power

a^0 = 1

Negative Power

a^{-n} = \frac{1}{a^n}

Example:

7^0 = 1

2^{-3} = \frac{1}{2^3} = \frac{1}{8}

3. Definition of Logarithms

A logarithm is the inverse of an index; it finds the exponent

x

needed for base

b

to reach total value

N

\log_b(N) = x \iff b^x = N

Example:

\log_2(8) = 3

because

2^3 = 8

\log_{10}(100) = 2

because

10^2 = 100

💡 Tip: Read

\log_b(N) = x

as: 'To what power must I raise

b

to get

N

4. Logarithm Operational Rules

Logarithms simplify multiplication into addition and powers into multipliers.

Product Rule

\log_b(MN) = \log_b M + \log_b N

Power Rule

\log_b(M^k) = k \log_b M

Identity Rule

b^{\log_b N} = N

Example:

\log_{10}(2) + \log_{10}(5) = \log_{10}(2 \times 5) = \log_{10}(10) = 1

\log_2(8^2) = 2 \log_2(8) = 2 \times 3 = 6

View Full Lesson

Integers

Numbers Grade 9

1. Adding and Subtracting Signed Integers

When adding or subtracting, combine magnitudes. Like signs sum up; opposite signs subtract and take the sign of the larger magnitude.

a - b = a + (-b)

Example:

-7 + 4 = -3

-5 - (-8) = -5 + 8 = 3

2. Multiplying and Dividing Signed Integers

Multiplying or dividing integers with identical signs produces a positive result; different signs produce a negative result.

Same Signs

(+, + \text{ or } -, -)

Positive product / quotient

(+)

Opposite Signs

(+, - \text{ or } -, +)

Negative product / quotient

(-)

Example:

(-6) \times (-4) = 24

(-20) \div 5 = -4

3. Absolute Value Definition

Absolute value

|a|

strips away the negative sign, giving the non-negative distance from zero.

|a| = \begin{cases} a & a \ge 0 \\ -a & a < 0 \end{cases}

Example:

|-12| = 12

|12| = 12

4. Order of Operations with Integers

Multi-step integer expressions must strictly follow the BODMAS priority sequence.

Example: Evaluate

-4 + 3 \times (-2)^2

:
1. Orders (power):

(-2)^2 = 4

2. Multiply:

3 \times 4 = 12

3. Add:

-4 + 12 = 8

💡 Tip: Always square negative numbers inside brackets carefully:

(-2)^2 = 4

, but

-2^2 = -4

View Full Lesson