Choose a year for formulas, symbols and worked substitutions. Use your browser’s Print or Save as PDF after choosing a year. Earlier-year sheets are useful for prerequisite revision.
Year grouping is a revision guide; topics may be taught in a different sequence. A = area, V = volume and SA = surface area. Use consistent units: lengths in cm, areas in cm² and volumes in cm³, for example. Round only at the final step.
Year 7 formula sheet · NavGanitMate
Percentage of an amount
part = (p ÷ 100) × whole
p is the percentage. Use the same units for part and whole.
Example: 15% of $80 = 0.15 × $80 = $12.
Rectangle
A = l × w; P = 2(l + w)
l = length, w = width. A = area; P = perimeter.
Example: l = 5 cm, w = 3 cm: A = 15 cm²; P = 16 cm.
Triangle and parallelogram
triangle: A = bh ÷ 2; parallelogram: A = bh
b = base; h = perpendicular height, not a sloping side.
Example: Triangle: b = 8 m, h = 3 m gives A = 12 m².
Rectangular prism
V = lwh
V = volume; l, w, h are perpendicular dimensions.
Example: 2 cm × 3 cm × 4 cm = 24 cm³.
Mean and range
mean = sum of values ÷ number of values; range = maximum − minimum
For a non-empty numerical data set.
Example: 2, 4, 9: mean = 15 ÷ 3 = 5; range = 9 − 2 = 7.
Equally likely probability
P(E) = favourable outcomes ÷ total outcomes
E is an event; use a finite sample space with equally likely outcomes.
Example: Fair die: P(even) = 3 ÷ 6 = 1/2.
Speed
v = d ÷ t
d = distance; t = positive elapsed time. Match units.
Example: 120 km ÷ 2 h = 60 km/h.
Year 8 formula sheet · NavGanitMate
Circle
C = 2πr = πd; A = πr²
r = radius, d = diameter = 2r, C = circumference; π ≈ 3.14159.
Example: r = 6 m: C = 12π m; A = 36π m². Keep π for exact answers.
Trapezium
A = (a + b)h ÷ 2
a and b are parallel side lengths; h is perpendicular distance between them.
Example: a = 4 cm, b = 8 cm, h = 3 cm: A = 18 cm².
Prism
V = A_base × h
A_base = cross-sectional base area; h = perpendicular distance between congruent parallel bases.
Example: A_base = 12 cm², h = 5 cm: V = 60 cm³.
Index laws
aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ
Here m and n are integers; a ≠ 0 when dividing or using non-positive powers.
Example: 2³ × 2² = 2⁵ = 32; 5⁰ = 1.
Percentage change
change (%) = (new − original) ÷ original × 100
Original must be non-zero. A negative result denotes a decrease for positive starting amounts.
Example: From $80 to $100: 20 ÷ 80 × 100 = 25% increase.
Complementary events
P(not E) = 1 − P(E)
Probabilities lie between 0 and 1.
Example: P(rain) = 0.3 gives P(no rain) = 0.7.
Year 9 formula sheet · NavGanitMate
Pythagoras
a² + b² = c²
Right-angled triangles only; c is the hypotenuse opposite the right angle.
Example: a = 3 cm, b = 4 cm gives c = √25 = 5 cm.
Right-triangle trigonometry
sin θ = opposite/hypotenuse; cos θ = adjacent/hypotenuse; tan θ = opposite/adjacent
Sides are relative to acute angle θ. Set the calculator to degrees for degree angles.
Example: opposite = 3, hypotenuse = 5: sin θ = 3/5, so θ ≈ 36.87°.
Straight lines
m = (y₂ − y₁)/(x₂ − x₁); y = mx + b
m = gradient, b = y-intercept; gradient needs x₂ ≠ x₁. Vertical lines have equation x = constant.
Example: Through (1, 3), (3, 7): m = 2; y = 2x + 1.
Cylinder
V = πr²h; SA = 2πr² + 2πrh
V = volume; SA = surface area. Right circular cylinder, closed at both ends; r = radius, h = perpendicular height.
Example: r = 2 cm, h = 5 cm: V = 20π cm³; SA = 28π cm².
Simple interest
I = Prt; total = P + I
P = principal, r = annual rate as a decimal, t = years. No compounding.
Example: P = $500, r = 0.04, t = 2: I = $40; total = $540.
Each guide explains why the idea matters and gives an example beyond maths. Year bands describe possible destinations, not a promise that a topic is taught next year. Connections may support current learning or later study; school sequences vary. Advanced uses may need additional teaching.
For every circle, circumference divided by diameter is π. The symbol π is exact; 3.14 is only an approximation.
A circle of diameter 1 unit has circumference π units. The straight line represents the unrolled circumference: a little more than three diameters. Schematic.
A useful approach
Identify the diameter
Multiply by π for circumference
Retain π if an exact result is requested.
Worked example
A circle of diameter 5 cm has circumference 5π cm. Keep π when an exact answer is requested.
Common mistake
Neither 3.14 nor 22/7 is exactly π.
Quick check
Which is the exact value of π?
3.14
22/7
π
Check answer and reasoning
π. The symbol π represents the exact constant; both numerical alternatives are approximations.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Area measures the surface inside a circle. A = πr² uses the radius, squared. To recover the radius from an area, first divide by π, then take the square root.
For radius 3 cm, area = π × 3² = 9π cm². Schematic.
A useful approach
Identify the radius
Square it
Multiply by π and attach square units. Reverse these operations if the area is given.
Worked example
If A = 16π cm², then r² = 16, so r = 4 cm. Its circumference would be 2π × 4 = 8π cm.
Common mistake
Do not use a diameter as the radius, or forget to square the radius.
Quick check
A circle has radius 3 cm. What is its area?
6π cm²
9π cm²
3π cm²
Check answer and reasoning
9π cm². A = π × 3² = 9π cm².
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
For addition and subtraction, use a common denominator. For multiplication, multiply numerators and denominators. Divide by a nonzero fraction by multiplying by its reciprocal.
The first strip selects 4/12 = 1/3; the second selects 3/12 = 1/4. The result selects 7/12. Every strip represents the same whole.
A useful approach
Identify the operation
For addition use a common denominator; for division use the reciprocal
Simplify and check the size.
Worked example
Add 1/3 + 1/4
Use denominator 12: 4/12 + 3/12
The sum is 7/12.
Common mistake
Do not add denominators. They describe the size of each part.
Quick check
What is 1/2 + 1/3?
2/5
5/6
1/6
Check answer and reasoning
5/6. 1/2 = 3/6 and 1/3 = 2/6, so the sum is 5/6.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M7N05, VC2M7N06, VC2M8N04.
A decimal can be written as a fraction using place value. A reduced fraction terminates in base ten exactly when its denominator has only factors 2 and 5.
1/2 = 0.5 = 50%. The whole is divided into two equal parts.
A useful approach
Write the decimal using place value
Simplify the fraction
Inspect the prime factors of the reduced denominator.
Worked example
0.75 = 75/100
Divide top and bottom by 25
0.75 = 3/4
For a recurring example, let x = 0.333…
10x = 3.333…
Subtract x from 10x: 9x = 3
X = 1/3 exactly.
Common mistake
A recurring decimal is exact as a fraction; cutting it off gives only an approximation.
Quick check
Which reduced fraction has a terminating decimal?
1/3
3/8
2/7
Check answer and reasoning
3/8. 8 = 2³ has only the prime factor 2, and 3/8 = 0.375.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A prime number is an integer greater than 1 with exactly two positive factors: 1 and itself. Prime factorisation expresses a positive integer greater than 1 as a product of primes.
24 splits into 2 × 12; 12 into 2 × 6; 6 into 2 × 3. The prime leaves give 24 = 2³ × 3.
A useful approach
Try a small prime divisor
Divide and continue with the quotient
Stop when every factor is prime and multiply to check.
Worked example
24 = 2 × 12
12 = 2 × 6 and 6 = 2 × 3
24 = 2³ × 3
Compare with 18 = 2 × 3²
The shared prime factors with the smaller powers give HCF(24,18) = 2 × 3 = 6.
Common mistake
1 is not prime, and composite factors must be broken down further.
Quick check
Which is the prime factorisation of 18?
2 × 9
2 × 3²
3 × 6
Check answer and reasoning
2 × 3². 18 = 2 × 9 = 2 × 3 × 3.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
For a nonzero base, multiplying like bases adds exponents, dividing subtracts exponents, and raising a power to a power multiplies exponents. Also a⁰ = 1 and a⁻ⁿ = 1/aⁿ.
x³ × x² has three x-factors followed by two more. That is five factors of x, so x³ × x² = x⁵.
A useful approach
Check that the bases match
Select the exponent law for the operation
Simplify without changing the base.
Worked example
Simplify x³ × x²
Keep base x and add 3 + 2
X⁵.
Common mistake
The laws combine exponents only when the bases match. Do not use them across a sum.
To round to a place, inspect the next digit. Digits 0–4 keep it; 5–9 increase it under the usual school rounding convention. An estimate is an approximation, not an exact answer.
6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.
A useful approach
Identify the place requested
Inspect the next digit
Round once and retain the requested precision.
Worked example
Round 6.347 to two decimal places
Inspect the third decimal, 7
6.35.
Common mistake
Keep exact values through intermediate steps unless the task specifically asks for rounding.
Quick check
Round 4.26 to one decimal place.
4.2
4.3
4.26
Check answer and reasoning
4.3. The hundredths digit is 6, so the tenths digit increases from 2 to 3.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A logarithm answers an exponent question: log_b(a) = c means bᶜ = a, for a > 0, b > 0 and b ≠ 1. Equal steps on a logarithmic scale represent equal multiplicative factors.
On this base-ten logarithmic scale, 10, 100 and 1000 are equally spaced because each step multiplies the value by 10. Their logarithms are 1, 2 and 3.
A useful approach
Identify the base and positive argument
Rewrite the logarithm as an exponent equation
Check the resulting power.
Worked example
Since 10³ = 1000
Log₁₀(1000) = 3
Moving from 10 to 100 to 1000 adds one each time to the base-ten logarithm.
Common mistake
A logarithmic scale does not have equal differences in the original quantities.
Quick check
What is log₁₀(100)?
2
10
50
Check answer and reasoning
2. 10² = 100, so the required exponent is 2.
Year-level reference
Level 10 covers applied logarithmic scales. The guide’s general logarithm definition provides supporting background.
Angles on a straight line total 180°; angles around a point total 360°. Vertically opposite angles are equal. With parallel lines, corresponding and alternate angles are equal.
The straight angle is 180°. The ray splits it into 65° and 115°, so the two adjacent angles sum to 180°. Drawing is schematic.
A useful approach
Mark stated parallel lines and angle facts
Choose the straight-line, full-turn or parallel-line relationship
Calculate and name the reason.
Worked example
Two angles share a straight line and one is 65°
Subtract from 180°
The other is 115°.
Common mistake
Equal-looking angles in a sketch are not enough; use the stated line or shape properties.
Quick check
What is the missing angle on a straight line next to 72°?
18°
108°
288°
Check answer and reasoning
108°. 180° − 72° = 108°.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A triangle has interior-angle sum 180°. A simple n-sided polygon has interior-angle sum (n − 2) × 180°. A regular polygon has equal sides and equal angles.
Two diagonals from one vertex split this pentagon into three triangles. Its interior-angle sum is 3 × 180° = 540°.
A useful approach
Count sides or identify triangles
Use the correct total-angle relationship
Divide equally only when regularity is stated.
Worked example
A pentagon has 5 sides
Its interior-angle sum is (5 − 2) × 180°
The total is 540°.
Common mistake
The angle-sum formula gives the total, not one angle unless you also use regularity.
Quick check
What is the interior-angle sum of a quadrilateral?
180°
360°
540°
Check answer and reasoning
360°. A quadrilateral splits into two triangles, giving 2 × 180° = 360°.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Use stated properties to classify shapes. A parallelogram has two pairs of parallel opposite sides. A rectangle has four right angles; a rhombus has four equal sides. A square has both properties.
A square is both a rectangle and a rhombus. The arrows mean “is a type of”, not a step in a calculation.
A useful approach
List the properties actually given
Compare side, angle and parallel-line conditions
Use a definition or theorem to justify the classification.
Worked example
A parallelogram has one right angle
Adjacent angles sum to 180°
All its angles are right angles, so it is a rectangle.
Common mistake
A diagram alone does not establish equal lengths, right angles or parallel lines.
Quick check
Which property is always true of a rectangle?
All four sides are equal
All four angles are 90°
Its diagonals are perpendicular
Check answer and reasoning
All four angles are 90°. A rectangle is defined by its four right angles; its adjacent side lengths need not match.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Congruent figures have the same shape and size. Similar figures have equal corresponding angles and proportional corresponding side lengths. A length scale factor k produces area factor k² and volume factor k³.
The right triangles have side lengths 3,4,5 and 9,12,15. The length scale factor is 3 and the area scale factor is 9. Both use the same drawing scale.
A useful approach
Match corresponding angles and sides
Find the length scale factor
Use its square for areas or cube for volumes.
Worked example
A similar triangle has scale factor 3
A 4 cm corresponding side becomes 12 cm
Its area is multiplied by 9.
Common mistake
Match corresponding sides; area does not scale by the same factor as length.
Quick check
Lengths are enlarged by factor 2. By what factor does area change?
2
4
8
Check answer and reasoning
4. Area uses two length dimensions, so its factor is 2² = 4.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Coordinates state positions in order. In 2D use (x,y); in 3D use (x,y,z). A translation adds the same displacement to every point; rotations and reflections preserve lengths and angles.
Illustration: the point (2, 2) lies on y = x. Read x across, then y up.
A useful approach
Label the axes and original point
Apply the stated displacement, reflection or rotation
Check distances and coordinate order.
Worked example
Translate (2,3) by 4 right and 1 down
Add 4 to x and subtract 1 from y
The image is (6,2).
Common mistake
Reflection in the y-axis changes x, not y. Always identify the axis or rotation centre.
Quick check
Reflect (3,2) in the y-axis. What is the image?
(−3,2)
(3,−2)
(−3,−2)
Check answer and reasoning
(−3,2). The horizontal coordinate changes sign while the vertical coordinate stays 2.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A three-figure bearing is measured clockwise from north. North is 000°, east 090°, south 180° and west 270°. Reverse bearings differ by 180°, reduced to the interval from 000° to less than 360°.
The route makes a 060° clockwise turn from north. The opposite route has bearing 240°. Compass directions are shown; the diagram is schematic.
A useful approach
Draw north at the starting point
Turn clockwise through the bearing
Check three-figure notation and reverse direction if needed.
Worked example
A route has bearing 060°
The reverse direction adds 180°
Its reverse bearing is 240°.
Common mistake
Bearings start at north and turn clockwise, rather than starting at east.
Quick check
What is the bearing of due west?
090°
180°
270°
Check answer and reasoning
270°. A clockwise turn from north through east and south to west is 270°.
Year-level reference
Bearing notation supports direction problems in Years 9–10; these codes do not specify a first teaching year for the notation.
Relative to a chosen acute angle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse and tan = opposite/adjacent. These ratios apply to right triangles; identify the angle first.
For angle θ at the lower-right vertex, opposite = 3, adjacent = 4 and hypotenuse = 5. Therefore sin θ = 3/5. The triangle is right-angled.
A useful approach
Choose the reference angle
Label opposite, adjacent and hypotenuse
Select the ratio containing the known and required quantities.
Worked example
An angle has opposite side 3 and hypotenuse 5
Use sin θ = 3/5
Sin θ = 0.6.
Common mistake
Opposite and adjacent depend on the chosen angle. Keep your calculator in the required angle mode.
Quick check
A right triangle has opposite side 4 and adjacent side 3. What is tan θ?
3/4
4/3
4/5
Check answer and reasoning
4/3. Tangent compares opposite to adjacent, so tan θ = 4/3.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M9SP01, VC2M9M03, VC2M10M03.
Split an area into shapes with known formulas, or subtract cut-outs. For perimeter, trace the exposed boundary and include each exposed edge once.
The outer rectangle is 10 cm by 6 cm. Removing a 2 cm by 3 cm top-right corner leaves 60 − 6 = 54 cm². All dimensions are supplied; drawing is to one consistent scale.
A useful approach
Sketch the stated parts
Add non-overlapping areas or subtract cut-outs
Trace exposed edges separately for perimeter.
Worked example
An outer rectangle is 10 cm by 6 cm
Remove a 2 cm by 3 cm corner
Remaining area = 60 − 6 = 54 cm².
Common mistake
Adding perimeters of touching pieces counts internal joins that are not exposed boundary.
Quick check
A 6 cm by 5 cm rectangle has a 2 cm by 2 cm square removed. What area remains?
26 cm²
28 cm²
34 cm²
Check answer and reasoning
26 cm². The outer area is 30 cm² and the removed area is 4 cm², giving 26 cm².
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A measurement combines a number and a unit. Convert related units using their scale factor. Area and volume conversions require squared or cubed length factors.
A 1 m by 1 m square is 100 cm by 100 cm. Its area is 10,000 cm². The 10-by-10 grid shows 100 squares, each 10 cm by 10 cm.
A useful approach
Write the starting and required units
Apply the conversion factor with the correct length/area/volume power
Check whether the numerical value should grow or shrink.
Worked example
1 m = 100 cm
1 m² = 100 × 100 cm²
1 m² = 10,000 cm².
Common mistake
Do not use a linear conversion factor directly for square or cubic units.
Quick check
How many centimetres are in 2.5 metres?
25
250
2500
Check answer and reasoning
250. Each metre contains 100 cm, so 2.5 × 100 = 250.
Year-level reference
Unit conversion is supporting knowledge for the listed measurement content; this is not its first teaching year.
Related Victorian content: VC2M7M02, VC2M8M02, VC2M9M01.
Recorded measurements depend on instrument precision. Absolute error is the magnitude of the difference from a reference. Percentage error compares that error with the magnitude of a nonzero reference.
Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.
A useful approach
Identify the reference and observed measurement
Calculate the magnitude of their difference
Divide by the reference magnitude only if percentage error is requested.
Worked example
A reference length is 50 cm and the measured value is 49 cm
Absolute error = 1 cm
Percentage error = 1/50 × 100% = 2%.
Common mistake
Greater displayed precision does not prove greater accuracy. Absolute error is non-negative.
Quick check
A reference value is 20 and a measured value is 21. What is the absolute error?
−1
1
5
Check answer and reasoning
1. The magnitude of 21 − 20 is 1.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Use the same time unit before calculating. For time-zone conversions, apply the stated UTC offset and check whether the date changes. Offsets may change with daylight saving.
09:40 to 10:00 is 20 minutes; 10:00 to 11:15 is 75 minutes. Total elapsed time is 95 minutes. Timeline spacing is proportional to elapsed minutes.
A useful approach
Put times on the same date and offset
Count whole hours and remaining minutes or convert to one unit
Check any midnight crossing.
Worked example
From 09:40 to 11:15
20 minutes to 10:00, then 75 minutes
Elapsed time is 95 minutes.
Common mistake
60 minutes make an hour; clock time is not decimal hours. Use the offset given for the date.
Quick check
How long is it from 14:50 to 15:20 on the same day?
20 minutes
30 minutes
70 minutes
Check answer and reasoning
30 minutes. There are 10 minutes to 15:00 and another 20 minutes to 15:20.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
An algorithm is a precise sequence of steps. Define inputs, operations, conditions and outputs. Test typical and boundary cases; a counterexample can disprove a claimed general rule.
Divide integer n by 2. If the remainder is 0, n is even; otherwise it is odd. The diagram is an algorithm, not a statement about non-integer inputs.
A useful approach
State the inputs and desired output
Write ordered steps and conditions
Test ordinary, boundary and counterexample inputs.
Worked example
To test whether a number is even
Divide by 2 and inspect the remainder
Remainder 0 means even.
Common mistake
A few successful examples do not prove that an algorithm works for every input.
Quick check
Which test best exposes a rule that incorrectly calls every positive number even?
Only test 2
Test an odd number such as 3
Only test 100
Check answer and reasoning
Test an odd number such as 3. The odd input 3 contradicts the incorrect rule.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A network consists of vertices joined by edges. A path follows edges. A network is connected when every pair of vertices can be joined by a path. The degree of a vertex counts incident edges; a loop counts twice.
This connected path has edges AB and BC. The degrees are 1 at A, 2 at B and 1 at C. Every pair of vertices is connected by a path.
A useful approach
Identify vertices and edges
Trace the required path or count incident edges
Check whether every required vertex is reachable.
Worked example
A path A—B—C has edges AB and BC
B touches two edges
B has degree 2.
Common mistake
Crossing lines are not necessarily a vertex unless a connection is marked.
Quick check
In the path A—B—C, what is the degree of A?
0
1
2
Check answer and reasoning
1. Only edge AB meets A, so its degree is 1.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Sort values to find the median. The mode is the most frequent value and need not be unique. Range is maximum minus minimum. Choose a display suited to the type of variable and comparison.
The ordered values are 2,3,3,8,9. The middle value is 3, the most frequent value is 3, and the range is 9 − 2 = 7. The row uses equal spacing, not a number-line scale.
A useful approach
Sort observations where required
Select the measure or display suited to the question
Report the statistic with context and units.
Worked example
For 2, 3, 3, 8, 9
The median and mode are 3
The range is 9 − 2 = 7.
Common mistake
With an even number of observations, average the two middle values after sorting.
Quick check
What is the median of 1, 4, 7, 10?
4
5.5
7
Check answer and reasoning
5.5. The middle pair is 4 and 7, whose mean is (4 + 7)/2 = 5.5.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A boxplot summarises median, quartiles and endpoints under a stated convention. The interquartile range is Q3 − Q1. Compare centre, spread, shape and unusual values rather than one statistic alone.
Illustrative five-number summary: minimum 1, Q1 4, median 7, Q3 10, maximum 13. IQR = 10 − 4 = 6. Whiskers use the minimum/maximum convention.
A useful approach
Identify the five-number summary and convention
Compare medians and IQRs
Describe spread and shape without inferring unsupported frequencies.
Worked example
Given Q1 = 4 and Q3 = 10
IQR = 10 − 4
IQR = 6.
Common mistake
Quartile conventions can differ for small samples. Use the convention supplied; a box width does not measure frequency.
Quick check
If Q1 = 7 and Q3 = 15, what is the IQR?
8
11
22
Check answer and reasoning
8. The interquartile range is 15 − 7 = 8.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
The population is the entire group of interest. A sample is the part observed. Random selection can reduce selection bias; a larger sample alone cannot correct a biased method.
The large outline represents a population of 24 members. Six filled members illustrate a sample. This picture does not establish that a selection process was random.
A useful approach
Define the target population
Examine how cases were selected and who could be missed
Distinguish sampling variation from systematic bias.
Worked example
To estimate a school opinion
Randomly sample students across the school
Surveying only one club may miss other views.
Common mistake
Random sampling and random assignment are different ideas; a sample can still have non-response bias.
Quick check
Which method best reduces selection bias in a school survey?
Ask only your friends
Randomly select students across the school
Ask only volunteers from one club
Check answer and reasoning
Randomly select students across the school. Selecting across the target population gives students a chance to be represented beyond one friendship or club.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M8ST01, VC2M8ST02, VC2M9ST02.
Different random samples can give different results. Increasing a well-designed sample size generally reduces sampling variability, but does not remove bias or guarantee an exact result.
Illustrative proportions from repeated samples: n=10 gives 0.2,0.4,0.5,0.6,0.8; n=100 gives 0.45,0.48,0.50,0.52,0.55. These are authored illustrations, not measured experiment results or a guarantee.
A useful approach
Compare the sampling methods first
Distinguish random variation from bias
Judge how the number of independent observations affects stability.
Worked example
Two random samples of 10 may give visibly different proportions
Samples of 100 usually vary less
The sampling method still matters.
Common mistake
A larger sample cannot fix a systematically biased question or selection process.
Quick check
What does a larger random sample generally reduce?
All bias
Sampling variability
The population size
Check answer and reasoning
Sampling variability. More independent observations generally make sample summaries more stable.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Bivariate data record two variables for each case. A scatterplot can show direction, strength and form of association. An association does not by itself establish causation.
Authored example pairs show a positive association: larger x tends to accompany larger y. A scatterplot alone cannot establish causation. Axes show arbitrary numerical units.
A useful approach
Identify both variables and units
Examine direction, strength and form
Separate association from a causal conclusion.
Worked example
As study time rises, a sample may show higher marks
The scatterplot has positive association
Other variables or study design still matter for a causal claim.
Common mistake
A trend is not proof of cause, and extrapolating far beyond the data is risky.
Quick check
Points tend downward as x increases. What association is suggested?
Positive
Negative
No possible relationship
Check answer and reasoning
Negative. A decreasing y as x increases is a negative association.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M10ST02, VC2M10ST05.
An event and its complement cover all outcomes without overlap. Their probabilities add to 1. Relative frequency is observed event count divided by trial count.
The first 3 of 10 equal parts represent rain, probability 0.3. The other 7 represent no rain, probability 0.7. Together they sum to 1.
A useful approach
Identify the event and all outcomes outside it
Use P(not A) = 1 − P(A)
Check that the two probabilities sum to 1.
Worked example
If P(rain) = 0.3
P(no rain) = 1 − 0.3
P(no rain) = 0.7.
Common mistake
A relative frequency is an observation; it need not equal the theoretical probability in a short experiment.
Quick check
If P(A) = 2/5, what is P(not A)?
2/5
3/5
5/2
Check answer and reasoning
3/5. 1 − 2/5 = 5/5 − 2/5 = 3/5.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
List ordered outcomes systematically using a table or tree. Multiply probabilities along a path. Add probabilities of disjoint paths. For dependent events, later branch probabilities must reflect earlier outcomes.
HH, HT, TH and TT are equally likely. Two outcomes have exactly one head: probability = 2/4 = 1/2.
A useful approach
List ordered outcomes or draw branches
Update probabilities after each event
Multiply along paths and add only disjoint paths.
Worked example
Two independent fair coin tosses have outcomes HH, HT, TH, TT
One outcome is HH
P(HH) = 1/2 × 1/2 = 1/4.
Common mistake
Do not assume every named event is equally likely. Independence is a condition, not a consequence of drawing a tree.
Quick check
For two independent fair coin tosses, what is P(exactly one head)?
1/4
1/2
3/4
Check answer and reasoning
1/2. HT and TH are two of four equally likely ordered outcomes.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M8P02, VC2M9P01, VC2M10P02.
Conditional probability restricts the sample space to cases meeting a condition. P(A given B) = P(A and B)/P(B), when P(B) > 0. In a table, use the conditioned group as the denominator.
All 12 dots are cyclists. Three filled dots also represent red clothing. Within the cyclist group, P(red given cycles) = 3/12 = 1/4.
A useful approach
Identify the condition
Restrict the denominator to that group
Count cases meeting both the event and the condition.
Worked example
Of 12 students who cycle, 3 wear red
Restrict attention to those 12 cyclists
P(red given cycles) = 3/12 = 1/4.
Common mistake
P(A given B) is generally different from P(B given A). Do not divide by the whole population when the condition specifies a subgroup.
Quick check
Of 10 swimmers, 4 wear glasses. What is P(glasses given swimmer)?
2/5
4/14
3/5
Check answer and reasoning
2/5. The condition selects 10 swimmers, and 4 of them wear glasses: 4/10 = 2/5.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A simulation uses a model to imitate a chance process. Match the model to the stated probabilities, repeat trials and compare relative frequencies with theory.
Choose 0 or 1 with equal probability: 0 means tails and 1 means heads. Repeat independent trials and count outcomes. No experiment results are asserted here.
A useful approach
Specify the theoretical chance model
Map random outcomes to the process with matching probabilities
Repeat trials and compare relative frequencies.
Worked example
To model a fair coin
Assign 0 to tails and 1 to heads with equal chance
Repeat many independent trials and count the heads.
Common mistake
A simulation approximates behaviour under its assumptions; more trials do not repair an incorrect model.
Quick check
Which is suitable for modelling a fair six-sided die?
Choose uniformly from 1 to 6
Always choose 3
Choose uniformly from 1 to 5
Check answer and reasoning
Choose uniformly from 1 to 6. Each of the six outcomes needs probability 1/6.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M7P02, VC2M8P03, VC2M9P03.
A proof links stated facts to a conclusion using valid reasons. A diagram or a collection of examples can suggest a result but does not establish it for every case. One valid counterexample disproves a universal claim.
A 3 cm by 2 cm rectangle has four right angles but unequal adjacent sides. It disproves the claim that every rectangle is a square. One counterexample is enough.
A useful approach
State the assumptions and claim
Justify each inference with a definition or theorem
Look for a counterexample when testing a universal claim.
Worked example
Claim: all rectangles are squares
A 2 cm by 3 cm rectangle has four right angles but unequal adjacent sides
The claim is false.
Common mistake
Do not assume the feature you are trying to prove. State why each equality or angle fact holds.
Quick check
What disproves a claim about every member of a group?
One valid counterexample
One supporting example
A neat sketch
Check answer and reasoning
One valid counterexample. A universal claim is false if even one allowed case contradicts it.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A net unfolds the faces of a solid into a plane. Plans and elevations show views from chosen directions. A valid net must fold without overlap into the intended solid.
Four squares in a row, with one above and one below the second square, form a valid cube net. Each face has the same side length.
A useful approach
Identify the solid and viewing direction
Match adjacent faces or projected features
Check how the representation folds or aligns.
Worked example
A cube has 6 square faces
Its net must contain 6 matching squares arranged to fold correctly
Counting squares alone is not sufficient.
Common mistake
A view can hide edges or faces; a flat sketch need not show every face separately.
Quick check
How many faces does a cube have?
4
6
8
Check answer and reasoning
6. A cube has a top, bottom, front, back, left and right face.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A point (x,y,z) describes position along three perpendicular axes. For distances in this Cartesian system, add the squares of the three coordinate changes and take the square root.
A schematic projection of perpendicular x,y,z axes. The point (2,3,6) is reached by changes 2 in x, 3 in y and 6 in z. Perspective spacing is illustrative, not an angle or distance measurement.
A useful approach
Identify all three axis changes
Square and add them when calculating Cartesian distance
Take the non-negative square root and attach length units.
Worked example
From (0,0,0) to (2,3,6), the changes are 2, 3 and 6
Distance² = 4 + 9 + 36 = 49
Distance = 7 units.
Common mistake
Keep the axis order consistent; a 2D projection may hide depth.
Quick check
Which point has x = 2, y = −1 and z = 4?
(2,−1,4)
(−1,2,4)
(4,−1,2)
Check answer and reasoning
(2,−1,4). The ordered triple lists the stated x, y and z coordinates in that order.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
A geometric construction is a sequence of justified steps producing a shape or locus. With straightedge and compass, equal-radius arcs can locate points equidistant from two endpoints.
Equal-radius circles centred at A and B meet above and below AB. The line through those intersections is the perpendicular bisector. The radius exceeds half the distance AB.
A useful approach
State the target property
Follow the permitted construction steps
Justify the result using equal distances or angle facts.
Worked example
To bisect segment AB, draw equal-radius arcs from A and B with radius greater than half AB
Join the two arc intersections
That line is the perpendicular bisector of AB.
Common mistake
A sketch that looks correct is not a construction argument. Keep compass radii equal when the method requires it.
Quick check
What is true of every point on the perpendicular bisector of AB?
It is closer to A
It is equally distant from A and B
It must lie on AB
Check answer and reasoning
It is equally distant from A and B. The perpendicular bisector is the locus of points equidistant from the two endpoints.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Translate shape definitions into explicit tests. Ask only questions supported by the supplied measurements or properties. Classification categories can overlap: every square is also a rectangle and a rhombus.
For a quadrilateral already known to have four right angles, check whether all four sides are equal. Yes gives a square; no gives a rectangle that is not a square. A square remains a rectangle too.
A useful approach
List precise shape definitions
Turn them into ordered property tests
Check overlapping categories and boundary cases.
Worked example
For a quadrilateral, test whether all angles are right angles
If yes, it is a rectangle
If its four sides are also equal, it is a square.
Common mistake
Do not treat overlapping categories as mutually exclusive. A square meets more than one broader definition.
Quick check
A quadrilateral has four right angles and four equal sides. Which is its most specific name?
Square
Rectangle that is not a square
Kite that is not a rhombus
Check answer and reasoning
Square. Four right angles and four equal sides meet the definition of a square.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Choose a graph based on the variables and purpose. Bar charts compare categories, histograms display a grouped numerical distribution, line graphs show ordered change and scatterplots show paired numerical variables.
Illustrative category counts: A = 2, B = 4, C = 9. Separate bars compare categories; the vertical scale starts at zero.
A useful approach
Identify categorical or numerical variables
Decide whether you need comparison, distribution, change or association
Choose a display and label its scale honestly.
Worked example
To compare counts for favourite colours
The variable is categorical
Use separated bars, one per category.
Common mistake
A histogram uses intervals on a numerical scale; a categorical bar chart does not imply continuous intervals.
Quick check
Which display best shows association between two numerical variables?
Pie chart
Scatterplot
One bar for the whole sample
Check answer and reasoning
Scatterplot. A scatterplot places each paired observation at its two numerical coordinates.
Year-level reference
Numerical displays relate to Year 7; choosing and justifying displays is revisited in Year 9.
A model translates a situation into variables, relationships and assumptions. Define units and the domain, solve the mathematical problem, then check whether the answer makes sense in context.
With a fixed charge of $4 and $2 per kilometre, cost C = 4 + 2d. At d = 3 km the cost is $10. Assume this tariff applies for the trip.
A useful approach
Define variables, units and assumptions
Form a relationship and solve within its domain
Interpret and check the result against the original situation.
Worked example
A taxi charges a fixed $4 plus $2 per kilometre
For d kilometres, C = 4 + 2d
At d = 3, C = $10.
Common mistake
Do not extrapolate a model beyond its stated assumptions or domain without justification.
Quick check
A service costs $5 plus $3 per hour. Which model gives cost C for h hours?
C = 5h + 3
C = 5 + 3h
C = 15h
Check answer and reasoning
C = 5 + 3h. The fixed $5 is paid once, while $3 is multiplied by the number of hours.
Year-level reference
Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.
Related Victorian content: VC2M7N10, VC2M8A03, VC2M9A06, VC2M10A15.
A scale drawing compares a represented length to a real length. Put both lengths in the same units before using the ratio. For a 1:n scale, real length is n times drawing length.
At scale 1:100, a 3 cm drawing length represents 300 cm = 3 m. The two bars are symbolic and not shown at a 1:100 size ratio.
A useful approach
Write the scale ratio in a consistent direction
Convert corresponding lengths to matching units
Calculate and convert the final unit if necessary.
Worked example
At scale 1:100, a 3 cm drawing length represents 300 cm
Convert 300 cm to metres
The real length is 3 m.
Common mistake
Scale factors apply to corresponding lengths. Convert units after applying the ratio, or consistently before it.
Quick check
At scale 1:50, what real length does 4 cm represent?
2 m
12.5 cm
200 m
Check answer and reasoning
2 m. 4 × 50 = 200 cm, which is 2 m.
Year-level reference
Proportional measurement relates to Year 7, with scale modelling extended in Years 9–10.
Related Victorian content: VC2M7M06, VC2M9M05, VC2M10M04.