GCSE maths
Every key GCSE formula, when to use it, and one worked example set out step by step.
Use it whenyou need, say, 15% of something.
Example
Find 15% of £80.
Times by the percentage
80 × 15 = 1200
Divide by 100
1200 ÷ 100 = £12
Use it whensomething goes up or down by a percentage. Increase by 25%: multiplier 1.25. Decrease by 25%: multiplier 0.75.
Example 1 · increase
Increase £80 by 25%.
Turn the percentage into a multiplier
100% + 25% = 125% = 1.25
Times the original by it
80 × 1.25 = £100
Example 2 · decrease
Decrease £80 by 15%.
Turn the percentage into a multiplier. Decrease, so take away
100% − 15% = 85% = 0.85
Times the original by it
80 × 0.85 = £68
Use it whenyou're asked "what is the percentage increase/decrease?"
Example 1 · increase
A price goes from £40 to £50. Find the percentage increase.
Find the change
50 − 40 = 10
Divide by the ORIGINAL
1040 = 0.25
Times by 100
0.25 × 100 = 25%
Example 2 · decrease
A price goes from £50 to £35. Find the percentage decrease.
Find the change (bigger take away smaller)
50 − 35 = 15
Divide by the ORIGINAL, £50
1550 = 0.3
Times by 100
0.3 × 100 = 30% decrease
Always divide by the original (starting) number, never the new one.
Use it whenyou're given the price AFTER a change and asked for the price before.
Example 1 · after a decrease
After a 20% discount a coat costs £72. Find the original price.
Find the multiplier
100% − 20% = 80% = 0.8
Divide the new price by it
72 ÷ 0.8 = £90
Example 2 · after an increase
After a 25% increase a price is £60. Find the original price.
Find the multiplier. Increase, so add
100% + 25% = 125% = 1.25
Divide the new price by it
60 ÷ 1.25 = £48
Don't add 20% back on to £72. That gives £86.40, which is wrong.
Use it whena percentage is added every year on top of last year's total (savings, depreciation).
Example
£500 earns 4% compound interest a year. How much after 3 years?
Find the multiplier
100% + 4% = 1.04
Raise it to the number of years
1.04³ = 1.124864
Times by the original
500 × 1.124864 = £562.43
Use it whenan amount is shared in a ratio like 2 : 3.
Example
Share £60 in the ratio 2 : 3.
Add the parts
2 + 3 = 5 parts
Find one part
60 ÷ 5 = £12
Times each side of the ratio
2 × 12 = £24 and 3 × 12 = £36
Use it whena question mentions km/h, mph or m/s. Density = mass ÷ volume and pressure = force ÷ area work the same way.
Example 1 · finding the speed
A car travels 150 km in 2.5 hours. Find its speed.
Speed = distance ÷ time
150 ÷ 2.5 = 60 km/h
Example 2 · finding the time
How long does it take to travel 135 km at 60 km/h?
You want time, so time = distance ÷ speed
135 ÷ 60 = 2.25 hours
If they want hours and minutes, times the decimal part by 60
0.25 × 60 = 15, so 2 hours 15 minutes
Check the units match. 30 minutes is 0.5 hours, not 0.3.
Use it whennumbers are very big or very small. Times: multiply the numbers, add the powers. Divide: divide the numbers, take away the powers.
Example
Work out (3 × 10⁴) × (4 × 10³).
Times the front numbers
3 × 4 = 12
Add the powers
10⁴ × 10³ = 10⁷
Fix the front number so it's between 1 and 10
12 × 10⁷ = 1.2 × 10⁸
Use it whenthere's one letter and no x². Undo things in reverse order: add/take away first, then times/divide.
Example 1 · all plus
Solve 3x + 5 = 20.
Take away 5 from both sides
3x = 15
Divide both sides by 3
x = 5
Example 2 · a minus and a negative answer
Solve 4x − 7 = −19.
Minus 7, so add 7 to both sides
4x = −12
Divide both sides by 4
x = −3
x on both sides (5x − 2 = 2x + 7)? Move the x's to the side with more x first: 3x − 2 = 7, so 3x = 9, so x = 3.
Use it whena number or letter sits outside a bracket. Times it by everything inside.
Example 1 · plus outside
Expand 4(2x − 3).
Times the 4 by each term
4 × 2x = 8x and 4 × −3 = −12
Write it out
8x − 12
Example 2 · minus outside
Expand −3(2x − 5).
Times the −3 by each term. Minus × minus = plus
−3 × 2x = −6x and −3 × −5 = +15
Write it out
−6x + 15
Use it whenevery term shares a number or letter. Take out the BIGGEST one you can.
Example 1 · number and letter
Factorise 6x² + 9x completely.
Find the biggest common factor
numbers: 3 · letters: x · so 3x
Divide each term by it
6x² ÷ 3x = 2x and 9x ÷ 3x = 3
Write it out
3x(2x + 3)
Example 2 · just a number, with a minus
Factorise 10x − 15.
Find the biggest common factor. Only 10x has an x, so it's just a number
5
Divide each term by it. The minus stays with its term
10x ÷ 5 = 2x and −15 ÷ 5 = −3
Write it out
5(2x − 3)
Use it whenasked to "make x the subject".
Example 1 · one step
Make u the subject of v = u + at.
Take away at from both sides
v − at = u
Flip it round
u = v − at
Example 2 · two steps
Make x the subject of y = mx + c.
Undo the + c first. Take c from both sides
y − c = mx
Undo the times m. Divide both sides by m
x = y − cm
Use it whenthere are two equations with two letters. Look at the sign in front of the matching letter: same sign (both + or both −), take away. Different signs (one + and one −), add. If no letter matches yet, times one equation first.
Example 1 · same sign, take away
Solve 2x + y = 7 and x + y = 4.
The y's are +y and +y, the same sign, so take away
(2x + y) − (x + y) = 7 − 4
Simplify
x = 3
Put x back into an equation
3 + y = 4, so y = 1
Example 2 · different signs, add
Solve 3x + 2y = 16 and x − 2y = 0.
The y's are +2y and −2y, different signs, so add
(3x + 2y) + (x − 2y) = 16 + 0
The y's cancel out
4x = 16
Divide by 4
x = 4
Put x back into an equation
4 − 2y = 0, so y = 2
Example 3 · nothing matches yet
Solve 2x + 3y = 12 and x + y = 5.
Times the second equation by 3 so the y's match
3x + 3y = 15
The y's are +3y and +3y, the same sign, so take away
(3x + 3y) − (2x + 3y) = 15 − 12
Simplify
x = 3
Put x back into an equation
3 + y = 5, so y = 2
Use it whenyou see <, >, ≤ or ≥.
Example
Solve 5 − 2x > 11.
Take away 5
−2x > 6
Divide by −2 and flip the sign
x < −3
Use it whentwo brackets are multiplied. Every term in the first times every term in the second (FOIL: first, outside, inside, last).
Example 1 · all plus
Expand (x + 3)(x + 5).
Times each pair
x² + 5x + 3x + 15
Collect the x terms
x² + 8x + 15
Example 2 · with a minus
Expand (x + 2)(x − 6).
Times each pair. Keep each sign with its number
x² − 6x + 2x − 12
Collect the x terms (−6x + 2x = −4x)
x² − 4x − 12
Use it whena quadratic starts with plain x².
Example 1 · plus on the end
Factorise x² − 9x + 18.
Two numbers that times to 18 and add to −9
−3 and −6
Put them in brackets
(x − 3)(x − 6)
Example 2 · minus on the end
Factorise x² + 2x − 15.
Two numbers that times to −15 and add to 2. One must be minus
5 and −3
Put them in brackets
(x + 5)(x − 3)
c plus: both numbers take the sign of b. c minus: one plus, one minus, and the bigger one takes the sign of b.
Use it whenthere's a number in front of x².
Example 1 · all plus
Factorise 2x² + 7x + 6.
a × c
2 × 6 = 12
Two numbers that times to 12 and add to 7
3 and 4
Split the middle term
2x² + 3x + 4x + 6
Factorise each pair
x(2x + 3) + 2(2x + 3)
Write the brackets
(x + 2)(2x + 3)
Example 2 · with minus signs
Factorise 3x² − 10x + 8.
a × c
3 × 8 = 24
Two numbers that times to 24 and add to −10
−6 and −4
Split the middle term
3x² − 6x − 4x + 8
Factorise each pair. Take −4 out of the second pair, so +8 becomes −2
3x(x − 2) − 4(x − 2)
Write the brackets
(3x − 4)(x − 2)
Use it whenit's one square take away another, with no middle term.
Example 1 · plain x²
Factorise x² − 49.
Spot the squares
x² and 7²
One bracket plus, one minus
(x + 7)(x − 7)
Example 2 · a number in front of x²
Factorise 4x² − 25.
Spot the squares. 2x × 2x = 4x²
(2x)² and 5²
One bracket plus, one minus
(2x + 5)(2x − 5)
Use it whenthe quadratic factorises easily. Always get = 0 first.
Example
Solve x² + x − 12 = 0.
Factorise (times to −12, add to 1: 4 and −3)
(x + 4)(x − 3) = 0
Set each bracket to 0
x + 4 = 0 or x − 3 = 0
Solve each
x = −4 or x = 3
Use it whenit won't factorise, or the question says "give your answer to 2 decimal places". Works on any ax² + bx + c = 0.
Example 1 · b plus, a = 1
Solve x² + 4x − 6 = 0 to 2 d.p.
Write down a, b and c
a = 1, b = 4, c = −6
Work out b² − 4ac first
16 − 4(1)(−6) = 16 + 24 = 40
Put it in the formula
x = −4 ± √402
Do the + and the − separately
x = 1.16 or x = −5.16
Example 2 · b minus, a not 1
Solve 2x² − 5x + 1 = 0 to 2 d.p.
Write down a, b and c
a = 2, b = −5, c = 1
Work out b² − 4ac. (−5)² is +25
25 − 4(2)(1) = 25 − 8 = 17
Put it in the formula. −b = −(−5) = 5, and 2a = 4
x = 5 ± √174
Do the + and the − separately
x = 2.28 or x = 0.22
Watch the signs: c = −6, so −4ac = −4 × 1 × −6 = +24.
Use it whenthe question says "write in the form (x + p)² + q", or asks for the turning point of a graph.
Example 1 · plus in the middle
Write x² + 6x + 2 in the form (x + p)² + q.
Halve the number in front of x
6 ÷ 2 = 3
Write the bracket and take away 3²
(x + 3)² − 9 + 2
Tidy the numbers
(x + 3)² − 7
Example 2 · minus in the middle
Write x² − 8x + 5 in the form (x + p)² + q.
Halve the number in front of x, keeping the minus
−8 ÷ 2 = −4
Write the bracket and take away (−4)² = 16. Always take away
(x − 4)² − 16 + 5
Tidy the numbers
(x − 4)² − 11
The turning point of y = (x + 3)² − 7 is (−3, −7). The x-coordinate flips sign.
Use it whenpowers of the SAME number are multiplied, divided or raised to a power.
Example
Simplify (x³)² × x⁴.
Power of a power: times the powers
(x³)² = x⁶
Multiplying: add the powers
x⁶ × x⁴ = x¹⁰
Use it whenthe power is 0, negative or a fraction.
Example
Work out 8−2/3.
The bottom of the fraction is a root
∛8 = 2
The top is a power
2² = 4
The minus means flip it
14
Use it whenthe number under the root has a square number factor (4, 9, 16, 25, 36...).
Example
Simplify √72.
Find the biggest square factor
72 = 36 × 2
Split the root
√36 × √2
Square root the square
6√2
Use it whenasked to "rationalise the denominator" (get the surd off the bottom).
Example 1 · bracket on the bottom
Rationalise 13 + √2.
Swap the sign in the bracket
3 + √2 → 3 − √2
Times top and bottom by it
3 − √2(3 + √2)(3 − √2)
Bottom = first number² take away the number under the root
9 − 2 = 7
Answer
3 − √27
The bottom is always a take away, whether the bracket had a plus or a minus.
Example 2 · root on the bottom
Rationalise 10√5 and simplify.
Times top and bottom by √5. √5 × √5 = 5
10√55
Divide the whole numbers (10 ÷ 5)
2√5
The bottom of a bracket one can come out negative, and that's fine: for 1/(2 + √7) it's 4 − 7 = −3.
Use it whenyou have two points on a straight line.
Example 1 · going up
Find the gradient through (1, 2) and (4, 11).
Change in y
11 − 2 = 9
Change in x
4 − 1 = 3
Divide
9 ÷ 3 = 3
Example 2 · going down, with negatives
Find the gradient through (−1, 5) and (2, −4).
Change in y (second y take away first y)
−4 − 5 = −9
Change in x (same order). Taking away a minus is adding
2 − (−1) = 3
Divide. Minus ÷ plus is minus
−9 ÷ 3 = −3
Take away in the same order on the top and the bottom. A line going down from left to right has a negative gradient.
Use it whenasked for the equation of a line.
Example 1 · positive gradient
Line with gradient 3 through (1, 2). Find its equation.
Put m in
y = 3x + c
Put the point in to find c
2 = 3(1) + c, so c = −1
Write it out
y = 3x − 1
Example 2 · negative gradient
Line with gradient −2 through (−3, 4). Find its equation.
Put m in
y = −2x + c
Put the point in. −2 × −3 = +6
4 = 6 + c, so c = −2
Write it out
y = −2x − 2
Parallel lines have the same gradient. Perpendicular lines: flip the gradient and change its sign (2 becomes −½).
Use it whena sequence goes up or down by the same amount each time.
Example 1 · going up
Find the nth term of 5, 8, 11, 14...
Find the difference
+3, so 3n
What do you add to 3 to get the first term, 5?
3 + 2 = 5
Write it out
3n + 2
Example 2 · going down
Find the nth term of 20, 17, 14, 11...
Find the difference. It goes down 3
−3, so −3n
What do you add to −3 to get the first term, 20?
−3 + 23 = 20
Write it out
−3n + 23
Use it whenyou have to draw a line from its equation, or find the equation of a line drawn on a grid.
How to draw it
Draw y = 2x − 1 for x from −1 to 3.
Make a table. Put each x into the equation
x: −1, 0, 1, 2, 3 → y: −3, −1, 1, 3, 5
Plot the points
(−1, −3), (0, −1), (1, 1), (2, 3), (3, 5)
Join them with a ruler, right across the grid
one straight line
How to read it
Find the equation of the line above.
Read c: where the line crosses the y-axis
c = −1
Pick two points on grid corners and draw a triangle
across 2, up 4
Gradient = up ÷ across
m = 4 ÷ 2 = 2
Put m and c into y = mx + c
y = 2x − 1
Going down from left to right means a minus gradient. Check the scale: one square is not always 1. y = 3 is a flat line. x = 2 is an upright line.
Use it whena question says parallel (never meet) or perpendicular (meet at a right angle). Higher tier.
Example 1 · parallel
Find the line parallel to y = 3x + 2 that goes through (1, 7).
Parallel, so the same gradient
m = 3
Put the point in to find c
7 = 3 × 1 + c, so c = 4
Write it out
y = 3x + 4
Example 2 · perpendicular
Find the line perpendicular to y = 2x + 1 that goes through (4, 3).
Flip the gradient and change the sign
2 → −½
Put the point in to find c
3 = −½ × 4 + c, so c = 5
Write it out
y = −½x + 5
Check: the two gradients times to −1 (2 × −½ = −1). If the equation is not y = ..., rearrange it first: 2y = 4x + 6 becomes y = 2x + 3.
Use it whenyou draw or read a curve with an x² in it. The exam asks for roots, the turning point, the y-intercept and the line of symmetry.
How to draw it
Draw y = x² − 2x − 3 for x from −2 to 4.
Make a table. Square first, then the rest
x: −2, −1, 0, 1, 2, 3, 4 → y: 5, 0, −3, −4, −3, 0, 5
Plot the points and join with a smooth curve, not a ruler
the bottom is rounded, not pointed
How to read it
Read the key points off the graph above.
Roots: where it crosses the x-axis. They solve x² − 2x − 3 = 0
x = −1 and x = 3
y-intercept: where it crosses the y-axis. It is c
−3
Turning point: the lowest point (highest for ∩)
(1, −4)
Line of symmetry: halfway between the roots
x = 1
To solve x² − 2x − 3 = 5, go across from 5 on the y-axis to the curve, then down. Completing the square gives the same turning point: (x − 1)² − 4 means (1, −4).
Use it whenyou must match a graph to its equation, or sketch one. Spot the highest power of x first.
How to tell them apart
Cubic: highest power x³
S shape. A minus in front flips it
Reciprocal: x on the bottom
two separate curves that never touch the axes
Exponential: x is the power
crosses the y-axis at 1, then shoots up
Circle: x² + y² = r² (Higher)
centre (0, 0), radius r · x² + y² = 25 has radius 5
How to draw them
Make a table of values, plot, and join with a smooth curve
same as a quadratic
For y = 1/x leave out x = 0. You cannot divide by 0, so the curve never reaches the y-axis.
Use it whena graph shows distance from a start point against time.
How to draw it
Plot (time, distance) for each part of the journey. Join with straight lines
a stop is a flat line
How to read it
Describe the journey above.
Speed = distance ÷ time for that part
home to A: 30 ÷ 1 = 30 km/h
Flat line: the distance is not changing
A to B: stopped for 0.5 hours = 30 minutes
Line going down: coming back towards home
B to C: 30 ÷ 1.5 = 20 km/h
Steeper means faster. 0.5 hours is 30 minutes, not 50. Average speed = total distance ÷ total time = 60 ÷ 3 = 20 km/h.
Use it whena graph shows speed (velocity) against time.
How to read it
Use the graph above.
Acceleration = change in velocity ÷ time
0 to 4 s: 12 ÷ 4 = 3 m/s²
Flat line: steady speed, no acceleration
4 to 10 s: 12 m/s
Line going down: slowing down
10 to 14 s: 12 ÷ 4 = 3 m/s² deceleration
Distance = area. Split it into shapes
24 + 72 + 24 (triangle, rectangle, triangle)
Add them
120 m
Here a flat line means steady speed, not stopped. Quicker: it is a trapezium, ½ × (6 + 14) × 12 = 120 m.
Use it whena question says "use the graph to solve". Draw both lines, then read the crossing point.
Example
Use the graph to solve y = x + 1 and y = 7 − 2x.
Draw both lines (make a table for each)
two straight lines
Find where they cross
(2, 3)
Write the answer
x = 2, y = 3
Check in both equations
2 + 1 = 3 ✓ and 7 − 4 = 3 ✓
Parallel lines never cross, so there is no answer. The same idea works for a curve and a line: read off where they meet.
Use it whenyou are given y = f(x) and asked about a changed version. Higher tier.
How to do it
y = f(x) has its turning point at (0, 0).
Number outside the bracket: up or down, the way it looks
f(x) + 3: up 3 · (0, 0) → (0, 3)
Number inside the bracket: left or right, the OPPOSITE way
f(x − 3): right 3 · (0, 0) → (3, 0)
Minus outside: upside down, y changes sign
−f(x): (2, 5) → (2, −5)
Minus inside: left to right, x changes sign
f(−x): (2, 5) → (−2, 5)
Inside the bracket does the opposite of what it looks like: f(x + 4) moves LEFT 4.
Use it whenfinding area. The height is always measured at right angles to the base, not the slanted side.
Example
A trapezium has parallel sides 4 and 8 and height 5. Find its area.
Add the parallel sides
4 + 8 = 12
Halve it
12 ÷ 2 = 6
Times by the height
6 × 5 = 30
Use it whena question involves a circle. r = radius (centre to edge), d = diameter (all the way across) = 2r.
Example 1 · area
A circle has diameter 10 cm. Find its area to 1 d.p.
Halve the diameter to get the radius
r = 5
Square the radius
5² = 25
Times by π
25 × π = 78.5 cm²
Square the radius first, then times by π. Don't use the diameter in πr².
Example 2 · circumference
A circle has radius 4 cm. Find its circumference to 1 d.p.
Double the radius to get the diameter
d = 8
Times by π
8 × π = 25.1 cm
Use it whenyou have a slice of a circle.
Example 1 · sector area
Sector with radius 6 and angle 90°. Find its area in terms of π.
Fraction of the circle
90 ÷ 360 = ¼
Whole circle area
π × 6² = 36π
Take the fraction
¼ × 36π = 9π
Example 2 · arc length
Sector with radius 5 cm and angle 72°. Find the arc length to 1 d.p.
Fraction of the circle
72 ÷ 360 = 0.2
Whole circumference
2 × π × 5 = 31.42...
Take the fraction
0.2 × 31.42... = 6.3 cm
Use it whenfinding how much a 3D shape holds. Cone, sphere and pyramid formulas are usually given in the exam.
Example
A cylinder has radius 3 cm and height 10 cm. Find its volume in terms of π.
Area of the circle end
π × 3² = 9π
Times by the height
9π × 10 = 90π cm³
Use it whentwo shapes are the same shape but different sizes.
Example 1 · area
Two similar shapes have lengths 3 cm and 6 cm. The small one has area 10 cm². Find the large one's area.
Length scale factor
6 ÷ 3 = 2
Area scale factor
2² = 4
Times the area
10 × 4 = 40 cm²
Example 2 · volume
Two similar solids have lengths 2 cm and 6 cm. The small one has volume 5 cm³. Find the large one's volume.
Length scale factor
6 ÷ 2 = 3
Volume scale factor: cube it
3³ = 27
Times the volume
5 × 27 = 135 cm³
Use it whenfinding a missing angle. Parallel lines: alternate angles (Z shape) are equal, corresponding angles (F shape) are equal, co-interior (C shape) add to 180°.
Example
A triangle has angles 50° and 60°. Find the third.
Add the ones you know
50 + 60 = 110
Take from 180
180 − 110 = 70°
Use it whena shape has more than 4 sides. n = number of sides. Interior + exterior = 180°.
Example
Find each interior angle of a regular hexagon.
Exterior angle
360 ÷ 6 = 60°
Interior = 180 − exterior
180 − 60 = 120°
Use it whenit's a right-angled triangle, you know 2 sides and want the third, and there are NO angles involved. Finding the long side: add. Finding a short side: take away.
Example 1 · finding the long side
Short sides 2 and 3. Find the hypotenuse to 1 d.p.
Square both short sides
2² = 4 and 3² = 9
Add (it's the long side)
4 + 9 = 13
Square root
√13 = 3.6
Example 2 · finding a short side
Hypotenuse 10, one short side 6. Find the other short side.
Square both
10² = 100 and 6² = 36
Take away (it's a short side)
100 − 36 = 64
Square root
√64 = 8
Use it whenit's a right-angled triangle and an angle is involved (given or wanted). Label the sides from the angle: Opposite, Adjacent (next to it), Hypotenuse. Pick the ratio that uses the two sides you have.
Example 1: finding a side
Angle 45°, adjacent 13. Find the opposite.
You have A and want O, so use TOA
tan 45 = O13
Times both sides by 13
O = 13 × tan 45 = 13
Example 2: finding an angle
Opposite 3, hypotenuse 6. Find the angle.
O and H, so use SOH
sin θ = 36 = 0.5
Use inverse sin (shift sin)
θ = sin⁻¹(0.5) = 30°
Use it whenthere's NO right angle and you know a side and its opposite angle (a matching pair).
Example
Angle A = 30°, side a = 5, angle B = 45°. Find side b to 1 d.p.
Put the values in
5sin 30 = bsin 45
Times both sides by sin 45
b = 5 × sin 45sin 30 = 7.1
Use it whenthere's NO right angle and you know two sides and the angle between them, or all three sides.
Example
b = 5, c = 7, angle A = 60°. Find a to 1 d.p.
Put the values in
a² = 25 + 49 − 2 × 5 × 7 × cos 60
Work it out (cos 60 = 0.5)
a² = 74 − 35 = 39
Square root
a = 6.2
Use it whenyou know two sides and the angle between them, but not the height.
Example
Sides 8 and 10 with 30° between them. Find the area.
Put the values in
½ × 8 × 10 × sin 30
sin 30 = 0.5
40 × 0.5 = 20
QuadraticsFactorises easily: factorise. Says "2 d.p." or won't factorise: formula. Says "(x + p)² + q" or "turning point": complete the square.