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35 formulas and key ideas

maths · Number foundations

Place value

4,728 = 4 × 1,000 + 7 × 100 + 2 × 10 + 8

Each digit is multiplied by its place value. No physical units are required.

Worked example

What is the value of the 7 in 4,728?

  1. Read from the right: 8 is in the ones column and 2 is in the tens column.
  2. The 7 is in the hundreds column, so it represents 7 × 100 = 700.
  3. Check by expanding the whole number: 4,000 + 700 + 20 + 8 = 4,728.
Rearrangement or equivalent form

10 ones = 1 ten; 10 tens = 1 hundred

Watch for: A digit and its value are different. The digit is 7; its value here is 700. Commas group digits for readability and do not change their values.

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maths · Number foundations

Positive and negative numbers

a − b = a + (−b)

a and b are signed numbers; use the same units when adding quantities.

Worked example

The temperature is −3 °C and rises by 8 °C. What is the new temperature?

  1. Start at −3 on the number line. A rise means add a positive change.
  2. Move 3 steps right to reach zero, leaving 5 of the 8 steps.
  3. Move 5 more steps right: −3 + 8 = 5 °C.
Rearrangement or equivalent form

a + (−b) = a − b

Watch for: −8 is less than −2, even though 8 is larger than 2. Compare positions on the number line rather than ignoring the signs.

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maths · Number foundations

Order of operations

Brackets → powers → × and ÷ → + and −

Operations on each shared-priority level are performed from left to right.

Worked example

Calculate 18 − 2 × (3 + 4).

  1. Brackets first: 3 + 4 = 7, leaving 18 − 2 × 7.
  2. Multiplication next: 2 × 7 = 14, leaving 18 − 14.
  3. Subtract last: 18 − 14 = 4.
Rearrangement or equivalent form

Use brackets to specify a different grouping.

Watch for: Do not simply work left to right through the entire expression. Also, division is not always before multiplication: 12 ÷ 3 × 2 = 8.

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maths · Number foundations

Fractions

a/c + b/c = (a + b)/c

a and b count parts; c is a common non-zero denominator.

Worked example

Add 1/2 and 1/4 of the same cake.

  1. Divide the half into two quarter-sized pieces: 1/2 = 2/4.
  2. Now both fractions count quarters, so add their numerators: 2 + 1 = 3.
  3. Keep the common denominator: 2/4 + 1/4 = 3/4.
Rearrangement or equivalent form

a/b = (a × k)/(b × k), for non-zero b and k

Watch for: Do not add denominators. 1/2 + 1/4 is not 2/6. Parts can only be added directly when their sizes match.

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maths · Number foundations

Decimals

0.01 = 1/100

Each decimal place to the right is one tenth of the previous place.

Worked example

Add 2.35 m and 0.8 m.

  1. Write 0.8 as 0.80 so the hundredths columns line up.
  2. Add hundredths: 5 + 0 = 5. Add tenths: 3 + 8 = 11 tenths.
  3. Regroup 10 tenths as 1 whole: 2.35 + 0.80 = 3.15 m.
Rearrangement or equivalent form

Decimal = numerator ÷ denominator

Watch for: More digits do not always mean a larger value. 0.9 is larger than 0.12 because 90 hundredths is greater than 12 hundredths.

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maths · Number foundations

Percentages

part = whole × percentage / 100

Part and whole use the same units; percentage is a number such as 15.

Worked example

What is 15% of $80?

  1. Convert 15% into a decimal: 15 ÷ 100 = 0.15.
  2. Multiply the whole amount by that fraction: 80 × 0.15 = 12.
  3. The portion is $12. Check mentally: 10% is $8 and 5% is $4.
Rearrangement or equivalent form

percentage = 100 × part / whole, with whole ≠ 0

Watch for: 15% is 0.15, not 15. A percentage change also depends on its starting value: a $10 rise is 10% of $100 but 20% of $50.

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maths · Number foundations

Ratios and proportions

share = total × selected parts / total parts

Total and share use matching units. Parts are relative counts.

Worked example

A drink uses concentrate and water in the ratio 1:4. How much of each makes 1 litre?

  1. Count the total parts: 1 + 4 = 5. Convert the total to 1,000 mL.
  2. One part is 1,000 ÷ 5 = 200 mL.
  3. Use 200 mL concentrate and 4 × 200 = 800 mL water. Their total is 1,000 mL.
Rearrangement or equivalent form

scale factor = new quantity / original quantity

Watch for: A 1:4 concentrate-to-water ratio means concentrate is one fifth of the final mixture, not one quarter.

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maths · Number foundations

Estimation and rounding

19.8 × 6 ≈ 20 × 6 = 120

The ≈ symbol means approximately equal. Keep the original quantity’s units.

Worked example

Round 3.476 m to two decimal places.

  1. Two decimal places means retain the hundredths digit, which is 7.
  2. Look at the next digit: 6 is at least 5, so round the 7 up to 8.
  3. The rounded result is 3.48 m. It lies within 0.005 m of the original.
Rearrangement or equivalent form

Round at the end when possible.

Watch for: Rounding and truncating differ. Truncating 3.476 to two decimal places gives 3.47; rounding gives 3.48.

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maths · Everyday calculations

Discounts

final price = original × (1 − discount/100)

Prices use the same currency; discount is a percentage from 0 to 100.

Worked example

A $120 jacket is discounted by 25%. What do you pay?

  1. Find the saving: 120 × 25/100 = $30.
  2. Subtract the saving from the original: 120 − 30 = $90.
  3. Check another way: 75% remains, and 120 × 0.75 = $90.
Rearrangement or equivalent form

original = final price / (1 − discount/100), for discount < 100

Watch for: Two successive 20% discounts are not a 40% discount. $100 becomes $80 and then $64, a total saving of 36%.

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maths · Algebra

Variables and expressions

C = 5 + 4h

C is cost in dollars; h is time in hours; 4 is a rate in dollars per hour.

Worked example

A bike hire costs $5 to start and $4 for each hour h. Write its cost and evaluate it for 3 hours.

  1. The fixed part is 5, regardless of hours. The changing part is 4h.
  2. Combine them into C = 5 + 4h, where C is the cost in dollars.
  3. For h = 3, C = 5 + 4 × 3 = 5 + 12 = $17.
Rearrangement or equivalent form

h = (C − 5)/4

Watch for: 4h means multiply by h; it does not mean 4 + h. Also, the letter is a number that varies, not an abbreviation for an object.

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maths · Algebra

Substitution

y = 2x² + 3

x is the input and y is the output; this example uses dimensionless numbers.

Worked example

Evaluate y = 2x² + 3 when x = −2.

  1. Replace x with (−2): y = 2 × (−2)² + 3.
  2. Square the whole negative value: (−2) × (−2) = 4.
  3. Multiply and add: y = 2 × 4 + 3 = 11.
Rearrangement or equivalent form

Substitute with brackets: y = 2 × (input)² + 3

Watch for: Without brackets, −2² conventionally means −(2²) = −4. The substituted value squared is (−2)² = 4.

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maths · Algebra

Solving linear equations

ax + b = c → x = (c − b)/a

a is a non-zero coefficient; b and c are constants; x is the unknown.

Worked example

Solve 3x + 5 = 20.

  1. Subtract 5 from both sides to remove the added constant: 3x = 15.
  2. Divide both sides by 3 to undo multiplication: x = 5.
  3. Check: 3 × 5 + 5 = 20. The left side matches the right side.
Rearrangement or equivalent form

c = ax + b

Watch for: Changing only one side breaks equality. “Move it across and change the sign” hides the real reason: you applied an inverse operation to both sides.

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maths · Algebra

Rearranging formulas

v = d/t ↔ t = d/v

v is constant speed in m/s; d is distance in m; t is time in s.

Worked example

Rearrange v = d/t to make t the subject, then find the time for 150 m at 5 m/s.

  1. Multiply both sides by t: vt = d. This removes t from the denominator.
  2. Divide both sides by v: t = d/v, provided v is not zero.
  3. Substitute: t = 150 m ÷ 5 m/s = 30 s.
Rearrangement or equivalent form

d = vt; t = d/v for v ≠ 0

Watch for: Do not swap symbols casually. From A = lw, solving for w gives A/l, not l/A. The inverse operation must affect the whole side.

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maths · Measurement

Metric units and conversion

1 km = 1,000 m; 1 m = 100 cm

km: kilometres; m: metres; cm: centimetres.

Worked example

Convert 2.4 km to metres.

  1. Use the relationship 1 km = 1,000 m.
  2. Multiply 2.4 by 1,000 because metres are smaller units: 2.4 × 1,000 = 2,400.
  3. Write the new unit: 2,400 m. The distance itself has not changed.
Rearrangement or equivalent form

metres = kilometres × 1,000; kilometres = metres / 1,000

Watch for: 1 m² is 10,000 cm², not 100 cm²: a square metre is 100 cm wide and 100 cm long. Do not use a length factor unchanged for area.

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maths · Measurement

Perimeter

P = 2(l + w)

P is perimeter; l is length; w is width, all in the same length unit.

Worked example

Find the perimeter of a rectangular garden 8 m long and 3 m wide.

  1. The four boundary lengths are 8 m, 3 m, 8 m and 3 m.
  2. Add them: 8 + 3 + 8 + 3 = 22.
  3. The perimeter is 22 m. It is a length, so the unit is metres, not square metres.
Rearrangement or equivalent form

w = P/2 − l

Watch for: Multiplying length by width finds area. Adding boundary lengths finds perimeter. A square metre cannot be added to a metre.

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maths · Measurement

Area

A = lw; triangle A = bh/2

A is area in m²; l, w, b (base) and h (perpendicular height) are in m.

Worked example

Find the area of a triangular panel with base 6 m and perpendicular height 4 m.

  1. The matching rectangle has area 6 × 4 = 24 m².
  2. The triangle occupies half that rectangle: 24 ÷ 2 = 12.
  3. The area is 12 m². Each unit square is 1 m by 1 m.
Rearrangement or equivalent form

rectangle w = A/l; triangle h = 2A/b

Watch for: A triangle’s sloping side is usually not its height. Also, adding side lengths gives perimeter, not area.

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maths · Measurement

Volume

V = lwh

V is volume in m³ when l, w and h are all in metres.

Worked example

A tank’s internal dimensions are 2 m by 1 m by 0.5 m. What is its capacity?

  1. Find the base area: 2 × 1 = 2 m².
  2. Multiply by height: 2 × 0.5 = 1 m³.
  3. Convert to litres: 1 m³ = 1,000 L. The theoretical capacity is 1,000 L.
Rearrangement or equivalent form

h = V/(lw), for non-zero l and w

Watch for: Use internal dimensions for capacity. External dimensions overestimate capacity because the walls occupy space.

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maths · Geometry

Pythagoras’ theorem

a² + b² = c²

a and b are perpendicular side lengths; c is the hypotenuse, all in the same units.

Worked example

Find the diagonal of a rectangle 3 m wide and 4 m long.

  1. The diagonal forms a right triangle with the rectangle’s sides. Set a = 3 and b = 4.
  2. Calculate c² = 3² + 4² = 9 + 16 = 25.
  3. Take the positive square root: c = √25 = 5 m.
Rearrangement or equivalent form

c = √(a² + b²); a = √(c² − b²)

Watch for: Do not stop at c² = 25 and label the length 25 m. Also, if finding a shorter side, subtract the known square from the hypotenuse square.

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maths · Graphs and coordinates

Coordinates and gradient

m = (y₂ − y₁)/(x₂ − x₁)

m is gradient; x and y are coordinates. Units are y-units per x-unit.

Worked example

Find the gradient between (1, 2) and (4, 8).

  1. The horizontal change is 4 − 1 = 3.
  2. The vertical change is 8 − 2 = 6.
  3. Divide rise by run: m = 6/3 = 2. Each step right increases y by 2.
Rearrangement or equivalent form

y = mx + c; c = y − mx

Watch for: Keep subtraction in the same order in numerator and denominator. Reversing only one gives the wrong sign.

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maths · Statistics and probability

Mean, median and mode

mean = sum of values / number of values

The mean has the same units as the values; the count is unitless and must be positive.

Worked example

Find the mean, median and mode of 2, 3, 3, 4, 8.

  1. The total is 2 + 3 + 3 + 4 + 8 = 20. Divide by 5 values: mean = 4.
  2. The list is already ordered. The third value is central: median = 3.
  3. The value 3 occurs twice, more than any other: mode = 3.
Rearrangement or equivalent form

sum = mean × count

Watch for: The median is not necessarily the middle value in the original unsorted list. Sort first, and retain repeated values.

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physics · Measurement and scientific practice

SI units

1 N = 1 kg·m/s²

N: newton; kg: kilogram; m: metre; s: second.

Worked example

A cart travels 12 m in 3 s. Express its average speed in SI units.

  1. Identify the measured quantities: distance in metres and time in seconds.
  2. Divide distance by time: 12/3 = 4.
  3. Carry the units through the division: average speed = 4 m/s.
Rearrangement or equivalent form

1 m/s = 3.6 km/h

Watch for: The SI base unit of mass is the kilogram, not the gram. Also, a measurement of mass in kg is not a force in N.

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physics · Measurement and scientific practice

Scalars and vectors

Δx = xfinal − xinitial

Δx is signed displacement in m; positions are measured from one origin.

Worked example

Walk 6 m east, then 2 m west. Find distance and displacement.

  1. Distance adds the lengths travelled: 6 + 2 = 8 m.
  2. Choose east as positive. The displacements are +6 m and −2 m.
  3. Add signed displacements: +6 − 2 = +4 m, meaning 4 m east.
Rearrangement or equivalent form

xfinal = xinitial + Δx

Watch for: A return trip can have zero displacement but non-zero distance. Do not cancel distances travelled in opposite directions.

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physics · Motion

Speed and velocity

average speed = distance / time

Distance in m and time in s give speed in m/s. Time must be positive.

Worked example

A cyclist covers 600 m in 120 s. Find average speed.

  1. Use the total distance 600 m and total elapsed time 120 s, including any stops.
  2. Divide: average speed = 600/120 = 5 m/s.
  3. Convert if useful: 5 × 3.6 = 18 km/h.
Rearrangement or equivalent form

distance = average speed × time

Watch for: Do not average two speeds directly unless the time spent at each speed is equal. Always return to total distance divided by total time.

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physics · Motion

Acceleration

a = (v − u)/t

a: acceleration in m/s²; u: initial velocity and v: final velocity in m/s; t: elapsed seconds.

Worked example

A trolley’s velocity changes from 2 m/s to 10 m/s in 4 s along a straight line.

  1. Find the velocity change: 10 − 2 = 8 m/s.
  2. Divide by elapsed time: 8/4 = 2 m/s².
  3. If acceleration is constant, the velocity increases by 2 m/s in each second.
Rearrangement or equivalent form

v = u + at, for constant acceleration

Watch for: Dividing final velocity by time ignores the initial velocity. Use the change in velocity, unless the initial velocity is zero.

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physics · Motion

Motion graphs

displacement = signed area under velocity–time graph

Time in s and velocity in m/s give area in m. Gradient has units m/s².

Worked example

A velocity-time graph rises linearly from 0 m/s to 8 m/s over 4 s. Find acceleration and displacement.

  1. Gradient is rise/run: (8 − 0)/(4 − 0) = 2 m/s².
  2. The region under the line is a triangle with base 4 s and height 8 m/s.
  3. Its area is ½ × 4 × 8 = 16 m. This is displacement; motion stays positive, so distance is also 16 m.
Rearrangement or equivalent form

constant velocity: Δx = vt

Watch for: A downward sloping position-time graph means motion in the negative direction, not necessarily slowing down. Always read the vertical axis first.

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physics · Forces

Newton’s laws

Fnet = ma

Fnet: resultant force in N; m: mass in kg; a: acceleration in m/s².

Worked example

A 4 kg trolley experiences a resultant horizontal force of 12 N. Find its acceleration.

  1. Use the resultant force, after combining all horizontal forces: Fnet = 12 N.
  2. Rearrange Fnet = ma to a = Fnet/m.
  3. Substitute: a = 12/4 = 3 m/s², in the direction of the resultant force.
Rearrangement or equivalent form

a = Fnet/m; m = Fnet/a when a ≠ 0

Watch for: Equal and opposite third-law forces do not cancel on one object: each acts on a different object. Balanced forces on one object are a separate idea.

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physics · Forces

Mass and weight

W = mg

W: weight in N; m: mass in kg; g: gravitational field strength in N/kg.

Worked example

Find the weight of a 5 kg object near Earth’s surface.

  1. Identify m = 5 kg and g = 9.8 N/kg.
  2. Multiply: W = 5 × 9.8 = 49.
  3. The weight is 49 N, directed towards Earth’s centre.
Rearrangement or equivalent form

m = W/g, for g ≠ 0

Watch for: Mass and weight are related but not interchangeable. A scale labelled in kilograms usually infers mass from a measured support force.

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physics · Forces

Friction and resultant force

Fnet = Fapplied − Ffriction

All forces are in N; this scalar form assumes opposite forces along one line.

Worked example

A 5 kg crate is pushed right with 20 N while friction acts left with 8 N. Find acceleration.

  1. Choose right as positive: the forces are +20 N and −8 N.
  2. The resultant is 20 − 8 = 12 N to the right.
  3. Apply a = Fnet/m: 12/5 = 2.4 m/s² to the right.
Rearrangement or equivalent form

sliding model: Ffriction ≈ μN, with μ unitless

Watch for: Friction does not always point opposite an object’s overall motion. It opposes relative slipping at the contact: static friction can push a walking person forward.

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physics · Energy

Work and energy transfer

W = Fd (parallel force)

W: work in J; F: constant force in N; d: displacement in m.

Worked example

A constant 15 N force moves a box 4 m in the force’s direction. Find the work done by that force.

  1. The force and displacement are parallel, so use W = Fd.
  2. Substitute F = 15 N and d = 4 m.
  3. W = 15 × 4 = 60 J. This is the energy transferred by this force.
Rearrangement or equivalent form

F = W/d; d = W/F for non-zero divisors

Watch for: If friction also acts, the work of the applied force is not the net work. Add the work done by all forces to find the change in kinetic energy.

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physics · Energy

Kinetic energy

Ek = ½mv²

Ek: kinetic energy in J; m: mass in kg; v: speed in m/s.

Worked example

Find the kinetic energy of a 2 kg trolley moving at 3 m/s.

  1. Square the speed: 3² = 9 m²/s².
  2. Multiply by the mass: 2 × 9 = 18.
  3. Halve the result: Ek = ½ × 18 = 9 J.
Rearrangement or equivalent form

v = √(2Ek/m), for m > 0

Watch for: Do not square the mass or the entire product. The formula is ½ × m × v², not (½mv)².

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physics · Energy

Gravitational potential energy

ΔEp = mgΔh

ΔEp: change in energy in J; m: kg; g: N/kg; Δh: vertical change in m.

Worked example

Raise a 2 kg book by 1.5 m, using g = 9.8 N/kg.

  1. Identify the vertical change in height: Δh = 1.5 m.
  2. Calculate the weight: mg = 2 × 9.8 = 19.6 N.
  3. Multiply by height change: ΔEp = 19.6 × 1.5 = 29.4 J.
Rearrangement or equivalent form

Δh = ΔEp/(mg), for non-zero m and g

Watch for: A longer ramp does not increase the gravitational energy change if the final vertical height is the same. Friction can still increase the work needed along the ramp.

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physics · Electricity

Current, voltage and Ohm’s law

V = IR

V: potential difference in V; I: current in A; R: resistance in Ω.

Worked example

A 6 V supply is connected across a 3 Ω resistor in an ideal circuit. Find current.

  1. Use V = IR and isolate current: I = V/R.
  2. Substitute the voltage across this resistor and its resistance: I = 6/3.
  3. The current is 2 A. Use the same current through this single series path.
Rearrangement or equivalent form

I = V/R for R > 0; R = V/I for I ≠ 0

Watch for: Current is not “used up” by a resistor. Energy is transferred, while charge is conserved. Also, use voltage across the component being analysed.

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physics · Electricity

Electrical power and energy

P = VI; E = Pt

P: W; V: V; I: A; t: s; E: J. Alternatively use kW and h to get kWh.

Worked example

A 1.5 kW heater runs at constant power for 2 h. Find energy use.

  1. Power is already in kilowatts and time is in hours.
  2. Multiply: E = 1.5 × 2 = 3 kWh.
  3. Convert if needed: 3 × 3.6 million = 10.8 million J.
Rearrangement or equivalent form

t = E/P; I = P/V for non-zero divisors

Watch for: Do not label energy kW/h. The billing unit is kWh, a product of power and time. Also, 1,500 W is 1.5 kW, not 15 kW.

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physics · Waves

Frequency and wavelength

v = fλ

v: wave speed in m/s; f: frequency in Hz; λ (lambda): wavelength in m.

Worked example

Find the wavelength of a 686 Hz sound wave travelling at 343 m/s.

  1. Use v = fλ and rearrange: λ = v/f.
  2. Substitute 343 m/s and 686 Hz: λ = 343/686.
  3. The wavelength is 0.5 m. The assumed sound speed is typical of air near room temperature.
Rearrangement or equivalent form

λ = v/f; f = v/λ for positive divisors

Watch for: Higher pitch does not necessarily mean louder sound. Frequency relates to pitch; amplitude and other factors influence perceived loudness.

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physics · Heat

Temperature and specific heat

Q = mcΔT

Q: energy in J; m: kg; c: J/(kg·K); ΔT: temperature change in K or °C.

Worked example

Heat 0.5 kg of water by 10 °C, using c = 4,200 J/(kg·K), with no losses.

  1. Identify m = 0.5 kg and ΔT = 10 K.
  2. Multiply mass and specific heat: 0.5 × 4,200 = 2,100 J/K.
  3. Multiply by temperature rise: Q = 2,100 × 10 = 21,000 J.
Rearrangement or equivalent form

ΔT = Q/(mc), for m and c > 0

Watch for: Use the temperature change, not the final temperature. This formula does not include energy for melting or boiling during a phase change.

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