IGCSE Mathematics Formula Practice
Worked Examples for Key Topics | Cambridge IGCSE Mathematics (0580)
Master calculations with step-by-step solutions
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Using the Quadratic Formula
x = (-b ± √(b²-4ac)) / 2a
How to Use This Guide
- Read the formula and understand what each variable means
- Study the example - note how values are substituted
- Try the practice problem before looking at the answer
- Show your working - Cambridge asks for all necessary working, and if a question tells you to show it, a correct answer alone will not get full marks
- Check your tier - items marked (Extended) are Extended content only; Core candidates can skip them
1. Percentages & Interest
Percentage Change
% Change = (Change ÷ Original) × 100
Worked Example:
A phone's price increased from $400 to $460. Calculate the percentage increase.
Step 1: Find the change
Change = 460 - 400 = $60
Step 2: Apply the formula
% change = (60 / 400) × 100
= 15% increase
Practice Problem:
A shop reduces a $80 jacket to $68. Calculate the percentage discount.
Show Answer
Change = 80 - 68 = $12
% = (12/80) × 100 = 15% decrease
Reverse Percentage (Extended)
Original = Final ÷ (1 ± rate)
Rate as a decimal (20% = 0.20); + for increase, - for decrease
Worked Example:
After a 20% increase, a TV costs $540. What was the original price?
20% increase means multiplied by 1.20
Original = 540 ÷ 1.20
= $450
Practice Problem:
After a 15% discount, a laptop costs $595. What was the original price?
Show Answer
15% discount = multiply by 0.85
Original = 595 ÷ 0.85 = $700
Compound Interest
A = P(1 + r/100)ⁿ
A = final amount, P = principal, r = rate %, n = years
Not given in the exam: learn it
Worked Example:
$2000 is invested at 5% compound interest for 3 years. Find the final amount.
A = P(1 + r/100)ⁿ
A = 2000 × (1 + 5/100)³
A = 2000 × (1.05)³
A = 2000 × 1.157625
A = $2315.25
Practice Problem (Extended):
A car depreciates by 15% each year. If it costs $12,000 now, what will it be worth in 2 years?
Show Answer
Depreciation = multiply by 0.85 each year
A = 12000 × (0.85)² = 12000 × 0.7225 = $8670
2. Algebra
Using the Quadratic Formula (Extended)
x = (-b ± √(b² - 4ac)) / 2a
For ax² + bx + c = 0 · Given in the Extended List of formulas
Worked Example:
Solve 2x² + 5x - 3 = 0
Identify: a = 2, b = 5, c = -3
Step 1: Work out b² - 4ac
b² - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49
Step 2: Apply formula
x = (-5 ± √49) / (2 × 2)
x = (-5 ± 7) / 4
x = 2/4 = 0.5 or x = -12/4 = -3
Practice Problem:
Solve x² - 6x + 5 = 0 using the quadratic formula.
Show Answer
a=1, b=-6, c=5
b²-4ac = 36-20 = 16
x = (6 ± 4)/2
x = 5 or x = 1
Finding the nth Term (Linear Sequence)
nth term = dn + (a - d)
d = common difference, a = first term
Worked Example:
Find the nth term of: 5, 8, 11, 14, ...
Step 1: Find common difference
d = 8 - 5 = 3
Step 2: Apply formula
nth term = 3n + (5 - 3)
nth term = 3n + 2
Check: n=1 → 3(1)+2=5 ✓, n=2 → 3(2)+2=8 ✓
Practice Problem:
Find the nth term of: 7, 12, 17, 22, ...
Show Answer
d = 12 - 7 = 5
nth term = 5n + (7-5) = 5n + 2
3. Coordinate Geometry
Finding the Gradient from Two Points (Extended)
m = (y₂ - y₁) / (x₂ - x₁)
Core finds the gradient from a grid, or reads m from y = mx + c
Worked Example:
Find the gradient of the line through A(2, 3) and B(6, 11).
m = (y₂ - y₁) / (x₂ - x₁)
m = (11 - 3) / (6 - 2)
m = 8 / 4
m = 2
Practice Problem:
Find the gradient of the line through P(-1, 5) and Q(3, -3).
Show Answer
m = (-3 - 5) / (3 - (-1)) = -8/4 = -2
Finding the Midpoint (Extended)
Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
Worked Example:
Find the midpoint of A(1, 4) and B(7, 10).
x-coordinate: (1 + 7) / 2 = 4
y-coordinate: (4 + 10) / 2 = 7
Midpoint = (4, 7)
Practice Problem:
Find the midpoint of P(-2, 6) and Q(4, -2).
Show Answer
x = (-2+4)/2 = 1, y = (6+(-2))/2 = 2
Midpoint = (1, 2)
Finding the Distance Between Two Points (Extended)
d = √[(x₂-x₁)² + (y₂-y₁)²]
Worked Example:
Find the distance between A(1, 2) and B(4, 6).
d = √[(4-1)² + (6-2)²]
d = √[3² + 4²]
d = √[9 + 16] = √25
d = 5 units
Practice Problem:
Find the distance between P(0, 0) and Q(5, 12).
Show Answer
d = √(5² + 12²) = √(25 + 144) = √169 = 13 units
4. Areas & Volume
Area of a Triangle Using Sine (Extended)
Area = ½ × a × b × sin(C)
Two sides and the included angle · Given in the Extended List of formulas
Worked Example:
Find the area of a triangle with sides 8 cm and 10 cm, with an included angle of 60°.
Area = ½ × a × b × sin(C)
Area = ½ × 8 × 10 × sin(60°)
Area = ½ × 80 × 0.866
Area = 34.6 cm²
Practice Problem:
Find the area of a triangle with sides 12 cm and 15 cm, with an included angle of 45°.
Show Answer
Area = ½ × 12 × 15 × sin(45°)
= ½ × 180 × 0.707 = 63.6 cm²
Volume of a Cone
V = ⅓ × π × r² × h
Given in the List of formulas (Core and Extended)
Worked Example:
Find the volume of a cone with radius 6 cm and height 10 cm.
V = ⅓ × π × r² × h
V = ⅓ × π × 6² × 10
V = ⅓ × π × 36 × 10
V = 120π
V = 377 cm³ (3 s.f.)
Practice Problem:
A cone has volume 150π cm³ and radius 5 cm. Find the height.
Show Answer
150π = ⅓ × π × 25 × h
150 = ⅓ × 25 × h
450 = 25h
h = 18 cm
Volume of a Sphere
V = (4/3) × π × r³
Given in the List of formulas (Core and Extended)
Worked Example:
Find the volume of a sphere with radius 9 cm.
V = (4/3) × π × r³
V = (4/3) × π × 9³
V = (4/3) × π × 729
V = 972π
V = 3050 cm³ (3 s.f.)
Practice Problem:
A sphere has volume 288π cm³. Find the radius.
Show Answer
288π = (4/3) × π × r³
288 = (4/3) × r³
216 = r³
r = 6 cm
5. Pythagoras & Trigonometry
Pythagoras' Theorem
a² + b² = c²
c = hypotenuse (longest side)
Worked Example:
Find the hypotenuse of a right triangle with sides 5 cm and 12 cm.
c² = a² + b²
c² = 5² + 12²
c² = 25 + 144 = 169
c = √169 = 13 cm
Practice Problem:
A right triangle has hypotenuse 17 cm and one side 8 cm. Find the other side.
Show Answer
a² = c² - b² = 17² - 8² = 289 - 64 = 225
a = 15 cm
Right-Angle Trigonometry (SOH-CAH-TOA)
Worked Example:
In a right triangle, the opposite side is 7 cm and the hypotenuse is 14 cm. Find the angle θ.
We have O and H, so use sin
sin θ = O/H = 7/14 = 0.5
θ = sin⁻¹(0.5)
θ = 30°
Practice Problem:
In a right triangle, angle θ = 40° and the adjacent side is 10 cm. Find the opposite side.
Show Answer
tan θ = O/A
O = A × tan θ = 10 × tan(40°) = 10 × 0.839
O = 8.39 cm
Sine Rule (Extended)
a/sin A = b/sin B = c/sin C
Use when: 2 angles + 1 side, or 2 sides + angle opposite one · Given in the Extended List of formulas
Worked Example:
In triangle ABC: A = 40°, B = 75°, a = 8 cm. Find side b.
a/sin A = b/sin B
8/sin(40°) = b/sin(75°)
8/0.643 = b/0.966
b = (8 × 0.966) / 0.643
b = 12.0 cm
Practice Problem:
In triangle PQR: P = 55°, q = 10 cm, Q = 80°. Find side p.
Show Answer
p/sin(55°) = 10/sin(80°)
p = 10 × sin(55°) / sin(80°)
p = 8.317... = 8.32 cm (3 s.f.; keep the full sine values in your calculator)
Cosine Rule (Extended)
a² = b² + c² - 2bc × cos A
Use when: 3 sides, or 2 sides + included angle · Given in the Extended List of formulas
Worked Example:
In triangle ABC: b = 7 cm, c = 9 cm, A = 50°. Find side a.
a² = b² + c² - 2bc × cos A
a² = 7² + 9² - 2(7)(9) × cos(50°)
a² = 49 + 81 - 126 × cos(50°)
a² = 130 - 80.991... = 49.008... (keep the full value)
a = 7.00 cm (3 s.f.)
Practice Problem:
Find angle A in a triangle where a = 8 cm, b = 5 cm, c = 9 cm.
Show Answer
cos A = (b² + c² - a²) / 2bc
cos A = (25 + 81 - 64) / (2×5×9) = 42/90 = 0.467
A = 62.2°
6. Similarity & Scale Factors
Area and Volume Ratios (Extended)
Length ratio = k → Area ratio = k² → Volume ratio = k³
Worked Example:
Two similar cylinders have heights in the ratio 2:3. If the smaller has volume 40 cm³, find the volume of the larger.
Length ratio = 2:3
Volume ratio = 2³:3³ = 8:27
40/V = 8/27
V = (40 × 27) / 8 = 1080/8
V = 135 cm³
Practice Problem:
Two similar triangles have areas 12 cm² and 75 cm². If the smaller has a side of 4 cm, find the corresponding side of the larger.
Show Answer
Area ratio = 12:75 = 4:25
Length ratio = √4:√25 = 2:5
4/x = 2/5
x = 10 cm
7. Statistics & Probability
Mean from Grouped Data (Extended)
Mean = Σfx / Σf
f = frequency, x = midpoint of class
Worked Example:
Find the estimated mean:
| Height, h (cm) | 140 < h ≤ 150 | 150 < h ≤ 160 | 160 < h ≤ 170 |
| Frequency | 5 | 12 | 8 |
Midpoints: 145, 155, 165
Σfx = (5×145) + (12×155) + (8×165)
= 725 + 1860 + 1320 = 3905
Σf = 5 + 12 + 8 = 25
Mean = 3905 / 25
Mean = 156.2 cm
Practice Problem:
Age, a (years): 10 < a ≤ 20 (f=3), 20 < a ≤ 30 (f=7), 30 < a ≤ 40 (f=10). Find the estimated mean age.
Show Answer
Midpoints: 15, 25, 35
Σfx = 45 + 175 + 350 = 570
Σf = 20
Mean = 570/20 = 28.5 years
Combined Probability
P(A and B) = P(A) × P(B)
For independent events, e.g. draws with replacement (multiply along the branches of a tree diagram)
Worked Example:
A bag has 4 red and 6 blue balls. Two balls are drawn with replacement. Find P(both red).
P(red) = 4/10 = 2/5
With replacement, second draw is independent
P(red and red) = (2/5) × (2/5)
= 4/25 = 0.16
Practice Problem:
A coin is flipped and a die is rolled. Find P(heads AND a 6).
Show Answer
P(heads) = 1/2
P(6) = 1/6
P(heads AND 6) = 1/2 × 1/6 = 1/12
More Resources
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Frequently Asked Questions
How do I solve quadratic equations in IGCSE Maths?
Solving quadratics by factorising, the formula or completing the square is Extended content; Core solves them from a graph. You can solve them by: 1) Factorising if possible, 2) Using the quadratic formula x = (-b ± √(b²-4ac))/2a, or 3) Completing the square. To use the formula, write the equation as ax² + bx + c = 0 and substitute a, b and c. The formula is printed in the List of formulas on Extended Papers 2 and 4, and you may be asked to give solutions in surd form.
When do I use sine rule vs cosine rule?
The sine and cosine rules are Extended content. Use the Sine Rule (a/sinA = b/sinB) when you have: two angles and one side, or two sides and an angle opposite one of them. Use the Cosine Rule (a² = b² + c² - 2bc·cosA) when you have: three sides, or two sides and the included angle. Both rules, and area = ½ab sin C, are printed in the List of formulas on Extended papers.
How do I find compound interest?
Use the formula A = P(1 + r/100)ⁿ where P is the principal (starting amount), r is the interest rate per period, and n is the number of periods. For example, $1000 at 5% for 3 years: A = 1000(1.05)³ = $1157.63. Compound interest is in both Core and Extended, and this formula is not given in the exam, so learn it.
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