Math Formulas for Classes 3 to 12

Geometry & measurement

Perimeter of a rectangle

P = 2 × (l + b)

In words: Perimeter equals two times the sum of length and breadth.

Perimeter is the distance all the way around a shape. A rectangle has two long sides and two short sides, so add one long and one short side, then double it.

What the letters mean

P

perimeter, in cm, m and so on

l

length (the longer side)

b

breadth (the shorter side)

Worked examples

1. Example 1. A garden is 12 m long and 5 m wide. How much fencing is needed to go around it?

   1. P = 2 × (l + b)
   2. P = 2 × (12 + 5) = 2 × 17

   Answer: 34 m of fencing

2. Example 2. A photo frame is 30 cm by 20 cm. How long is the ribbon that goes around its edge?

   1. P = 2 × (30 + 20) = 2 × 50

   Answer: 100 cm

Watch out: Perimeter is a length, so the unit is cm or m, not cm² or m².

Geometry & measurement

Perimeter of a square

P = 4 × a

In words: Perimeter equals four times the side.

All four sides of a square are equal, so the distance around is four times one side. To find the side from the perimeter, divide by 4.

What the letters mean

P

perimeter

a

length of one side

Worked examples

1. Example 1. A square card has sides of 9 cm. What is its perimeter?

   1. P = 4 × a = 4 × 9

   Answer: 36 cm

2. Example 2. A square playground has a perimeter of 80 m. How long is each side?

   1. a = P ÷ 4 = 80 ÷ 4

   Answer: 20 m

Geometry & measurement

Area of a rectangle

A = l × b

In words: Area equals length times breadth.

Area is the amount of flat space inside a shape, counted in square units. Imagine covering the rectangle with small squares: there are l squares in each row and b rows.

What the letters mean

A

area, in square units such as cm² or m²

l

length

b

breadth

Worked examples

1. Example 1. A floor is 8 m long and 5 m wide. Square tiles of 1 m² cover it. How many tiles are needed?

   1. A = l × b = 8 × 5

   Answer: 40 m², so 40 tiles

2. Example 2. A notebook cover is 25 cm long and 20 cm wide. What is its area?

   1. A = 25 × 20

   Answer: 500 cm²

Watch out: Both sides must be in the same unit before you multiply.

Geometry & measurement

Area of a square

A = a × a = a²

In words: Area equals side times side, or side squared.

A square is a rectangle whose length and breadth are the same, so multiply the side by itself.

What the letters mean

A

area

a

length of one side

Worked examples

1. Example 1. A square stamp has a side of 7 cm. What is its area?

   1. A = a × a = 7 × 7

   Answer: 49 cm²

2. Example 2. A square garden has a side of 12 m. One packet of seeds covers 1 m². How many packets are needed?

   1. A = 12 × 12 = 144 m²

   Answer: 144 packets

Geometry & measurement

Volume of a cuboid

V = l × b × h

In words: Volume equals length times breadth times height.

Volume is the space inside a solid, counted in cubic units. A cuboid is a box: its base area is l × b, and h layers of that base fill the box.

What the letters mean

V

volume, in cm³ or m³

l

length

b

breadth

h

height

Worked examples

1. Example 1. A box is 5 cm long, 4 cm wide and 3 cm high. What is its volume?

   1. V = l × b × h = 5 × 4 × 3

   Answer: 60 cm³

2. Example 2. An aquarium is 50 cm × 30 cm × 40 cm. How many litres of water can it hold? (1 litre = 1000 cm³)

   1. V = 50 × 30 × 40 = 60 000 cm³
   2. Litres = 60 000 ÷ 1000

   Answer: 60 litres

Statistics & probability

Average (mean)

Average = (sum of all values) ÷ (number of values)

In words: Average equals the sum of all the values divided by how many values there are.

The average shows what each value would be if everything were shared out equally. Add everything up, then divide by how many things you added.

What the letters mean

sum

all the values added together

number of values

how many values there are

Worked examples

1. Example 1. Asha scored 70, 80 and 90 in three tests. What is her average score?

   1. Sum = 70 + 80 + 90 = 240
   2. Average = 240 ÷ 3

   Answer: 80

2. Example 2. Rainfall on four days was 12, 15, 9 and 14 mm. What was the average daily rainfall?

   1. Sum = 12 + 15 + 9 + 14 = 50
   2. Average = 50 ÷ 4

   Answer: 12.5 mm

Watch out: Divide by the number of values, not by the biggest value.

Number & arithmetic

Division and its check

Dividend = Divisor × Quotient + Remainder

In words: Dividend equals divisor times quotient plus remainder.

Division and multiplication undo each other. This rule lets you check any division answer, and it also works backwards to find a missing number.

What the letters mean

Dividend

the number being divided

Divisor

the number you divide by

Quotient

how many whole times it fits

Remainder

what is left over, smaller than the divisor

Worked examples

1. Example 1. Divide 47 by 5 and check the answer.

   1. 47 ÷ 5 gives quotient 9 and remainder 2
   2. Check: 5 × 9 + 2 = 45 + 2 = 47

   Answer: Correct: it equals the dividend, 47

2. Example 2. Check this: 125 ÷ 8 = 15 remainder 5.

   1. Divisor × Quotient + Remainder = 8 × 15 + 5 = 120 + 5

   Answer: 125, so the division is correct

Number & arithmetic

Equivalent fractions

a/b = (a × n)/(b × n) = (a ÷ n)/(b ÷ n)

In words: A fraction stays the same when you multiply or divide its top and bottom by the same number.

Two fractions can look different but show the same amount. Multiply or divide the numerator and the denominator by the same non-zero number and the value does not change. Dividing both by their common factor simplifies the fraction.

What the letters mean

a

numerator (top number)

b

denominator (bottom number)

n

any whole number except zero

Worked examples

1. Example 1. Write a fraction equal to 2/3 with denominator 6.

   1. 6 = 3 × 2, so multiply top and bottom by 2
   2. 2/3 = (2 × 2)/(3 × 2)

   Answer: 4/6

2. Example 2. Simplify 12/18.

   1. The biggest number that divides both 12 and 18 is 6
   2. 12/18 = (12 ÷ 6)/(18 ÷ 6)

   Answer: 2/3

Money & commercial maths

Unitary method

Cost of 1 = total cost ÷ number of items; cost of many = cost of 1 × number wanted

In words: First find the cost of one, then multiply by the number you want.

When you know the price of several things and want the price of a different number, go through "one". Divide to find the value of one, then multiply to find the value of many.

What the letters mean

cost of 1

price of a single item

total cost

price of the group you know

Worked examples

1. Example 1. 6 pens cost ₹48. What do 10 pens cost?

   1. Cost of 1 pen = 48 ÷ 6 = ₹8
   2. Cost of 10 pens = 8 × 10

   Answer: ₹80

2. Example 2. A cyclist rides 60 km in 4 hours at a steady speed. How far does the cyclist ride in 1 hour, and in 7 hours?

   1. In 1 hour: 60 ÷ 4 = 15 km
   2. In 7 hours: 15 × 7

   Answer: 15 km in 1 hour, and 105 km in 7 hours

Geometry & measurement

Area of a triangle

A = ½ × b × h

In words: Area equals one half times base times height.

Any triangle fills exactly half of a rectangle (or parallelogram) with the same base and height. The height is the straight-up distance from the base to the opposite corner, not the slanted side.

What the letters mean

A

area

b

base

h

perpendicular height

Worked examples

1. Example 1. A triangle has a base of 10 cm and a height of 6 cm. Find its area.

   1. A = ½ × 10 × 6 = ½ × 60

   Answer: 30 cm²

2. Example 2. A triangular sail has a base of 4 m and a height of 3 m. How much cloth does it need?

   1. A = ½ × 4 × 3

   Answer: 6 m²

Watch out: Use the perpendicular height. A slanted side is longer than the height.

Geometry & measurement

Area of a parallelogram

A = b × h

In words: Area equals base times height.

Cut a triangle off one slanted end of a parallelogram and fit it on the other end: you get a rectangle with the same base and height.

What the letters mean

A

area

b

base

h

perpendicular height between the two parallel sides

Worked examples

1. Example 1. A parallelogram has a base of 8 cm and a height of 5 cm. Find its area.

   1. A = b × h = 8 × 5

   Answer: 40 cm²

2. Example 2. A field shaped like a parallelogram has a base of 25 m and a height of 12 m. What is its area?

   1. A = 25 × 12

   Answer: 300 m²

Geometry & measurement

Area of a trapezium

A = ½ × (a + b) × h

In words: Area equals one half times the sum of the parallel sides times the height.

A trapezium has one pair of parallel sides, a and b. Average those two sides and multiply by the distance h between them.

What the letters mean

a, b

the two parallel sides

h

perpendicular distance between them

Worked examples

1. Example 1. A trapezium has parallel sides of 6 cm and 10 cm, and the distance between them is 4 cm. Find its area.

   1. A = ½ × (6 + 10) × 4 = ½ × 16 × 4

   Answer: 32 cm²

2. Example 2. A plot has parallel sides of 20 m and 30 m, 12 m apart. What is its area?

   1. A = ½ × (20 + 30) × 12 = ½ × 50 × 12

   Answer: 300 m²

Geometry & measurement

Circumference of a circle

C = 2πr = πd

In words: Circumference equals two pi r, which is pi times the diameter.

The circumference is the distance around a circle. For every circle, the circumference is about 3.14 times the diameter. That number is π (pi). Many school problems say "take π = 22/7".

What the letters mean

C

circumference

r

radius (centre to edge)

d

diameter = 2r

π

≈ 3.14 or 22/7

Worked examples

1. Example 1. A circle has a radius of 7 cm. Find its circumference. (π = 22/7)

   1. C = 2πr = 2 × 22/7 × 7

   Answer: 44 cm

2. Example 2. A bicycle wheel has a diameter of 70 cm. How far does it travel in one full turn? (π = 22/7)

   1. C = πd = 22/7 × 70

   Answer: 220 cm

Geometry & measurement

Area of a circle

A = πr²

In words: Area equals pi times the radius squared.

The area inside a circle is π times the radius multiplied by itself. Square the radius first, then multiply by π.

What the letters mean

A

area

r

radius

π

≈ 3.14 or 22/7

Worked examples

1. Example 1. Find the area of a circle of radius 7 cm. (π = 22/7)

   1. A = πr² = 22/7 × 7 × 7

   Answer: 154 cm²

2. Example 2. A pizza has a diameter of 28 cm. What is its area? (π = 22/7)

   1. Radius = 28 ÷ 2 = 14 cm
   2. A = 22/7 × 14 × 14

   Answer: 616 cm²

Watch out: If you are given the diameter, halve it to get the radius before squaring.

Money & commercial maths

Percentage

Percentage = (part ÷ whole) × 100

In words: Percentage equals part divided by whole, times one hundred.

"Per cent" means "out of 100". To turn a part of a whole into a percentage, divide the part by the whole and multiply by 100.

What the letters mean

part

the amount you are looking at

whole

the total amount

Worked examples

1. Example 1. Riya answered 18 out of 24 questions correctly. What percentage is that?

   1. Percentage = 18 ÷ 24 × 100 = 0.75 × 100

   Answer: 75%

2. Example 2. A shirt costs ₹800 and its price is reduced by ₹120. What percentage of the price is the reduction?

   1. Percentage = 120 ÷ 800 × 100

   Answer: 15%

Money & commercial maths

Profit and loss

Profit = SP − CP; Profit % = (Profit ÷ CP) × 100; Loss % = (Loss ÷ CP) × 100

In words: Profit equals selling price minus cost price. Profit percent is profit over cost price times one hundred.

CP is what the seller paid, SP is what the seller received. If SP is more than CP there is a profit, otherwise a loss. Percentages are always worked out on the cost price.

What the letters mean

CP

cost price

SP

selling price

Worked examples

1. Example 1. A shopkeeper buys a bag for ₹400 and sells it for ₹500. Find the profit percent.

   1. Profit = 500 − 400 = ₹100
   2. Profit % = 100 ÷ 400 × 100

   Answer: 25%

2. Example 2. A toy bought for ₹250 is sold for ₹200. Find the loss percent.

   1. Loss = 250 − 200 = ₹50
   2. Loss % = 50 ÷ 250 × 100

   Answer: 20%

Watch out: Profit and loss percentages use the cost price as the base, never the selling price.

Money & commercial maths

Discount

Discount = MP − SP; SP = MP × (1 − d/100)

In words: Selling price equals marked price times one minus discount percent over a hundred.

The marked price (MP) is the price on the tag. A discount of d% takes d% off it. Multiplying by (1 − d/100) gives the selling price in one step.

What the letters mean

MP

marked price

SP

selling price after discount

d

discount percent

Worked examples

1. Example 1. A jacket is marked ₹1000 with a 20% discount. What is the selling price?

   1. SP = 1000 × (1 − 20/100) = 1000 × 0.8

   Answer: ₹800

2. Example 2. A bag marked ₹1500 is sold at a 15% discount. What do you pay?

   1. SP = 1500 × (1 − 15/100) = 1500 × 0.85

   Answer: ₹1275

Money & commercial maths

Simple interest

SI = (P × R × T) ÷ 100; Amount = P + SI

In words: Simple interest equals principal times rate times time, over one hundred.

Interest is the extra money paid for borrowing. With simple interest, the same amount is added every year, calculated on the original sum only.

What the letters mean

P

principal (the sum borrowed or saved)

R

rate of interest, % per year

T

time, in years

SI

simple interest

Worked examples

1. Example 1. Find the simple interest on ₹5000 at 8% per year for 2 years, and the amount.

   1. SI = 5000 × 8 × 2 ÷ 100 = ₹800
   2. Amount = 5000 + 800

   Answer: Interest ₹800, amount ₹5800

2. Example 2. Find the simple interest on ₹12 000 at 5% per year for 3 years.

   1. SI = 12000 × 5 × 3 ÷ 100

   Answer: ₹1800

Number & arithmetic

Speed, distance and time

Speed = Distance ÷ Time; Distance = Speed × Time; Time = Distance ÷ Speed

In words: Speed equals distance divided by time.

These three rules are one relationship written three ways. Cover the quantity you want in the triangle D over S × T, and what is left tells you what to do.

What the letters mean

Speed

for example km per hour

Distance

for example km

Time

for example hours

Worked examples

1. Example 1. A car covers 150 km in 3 hours. What is its speed?

   1. Speed = 150 ÷ 3

   Answer: 50 km/h

2. Example 2. A train travels at 60 km/h for 2.5 hours. How far does it go?

   1. Distance = 60 × 2.5

   Answer: 150 km

3. Example 3. How long does a 90 km journey take at 45 km/h?

   1. Time = 90 ÷ 45

   Answer: 2 hours

Watch out: Keep units consistent: if speed is in km/h, time must be in hours.

Geometry & measurement

Pythagoras theorem

c² = a² + b²

In words: The square of the hypotenuse equals the sum of the squares of the other two sides.

In a right-angled triangle, the longest side (the hypotenuse, opposite the right angle) squared equals the sum of the squares of the two shorter sides. It lets you find a missing side.

What the letters mean

c

hypotenuse (longest side)

a, b

the two shorter sides

Worked examples

1. Example 1. A right triangle has shorter sides of 6 cm and 8 cm. Find the hypotenuse.

   1. c² = 6² + 8² = 36 + 64 = 100
   2. c = √100

   Answer: 10 cm

2. Example 2. A ladder reaches 12 m up a wall and its foot is 5 m from the wall. How long is the ladder?

   1. c² = 12² + 5² = 144 + 25 = 169
   2. c = √169

   Answer: 13 m

Watch out: It only works for right-angled triangles. To find a shorter side, subtract: a² = c² − b².

Algebra

Algebraic identities

(a + b)² = a² + 2ab + b²; (a − b)² = a² − 2ab + b²; (a + b)(a − b) = a² − b²

In words: a plus b all squared equals a squared plus two a b plus b squared.

These patterns are always true, whatever the numbers are, so they work as shortcuts for squaring and for multiplying numbers that sit just above and below a round number.

What the letters mean

a, b

any numbers or expressions

Worked examples

1. Example 1. Find 103² without a calculator.

   1. 103 = 100 + 3
   2. (100 + 3)² = 100² + 2 × 100 × 3 + 3² = 10 000 + 600 + 9

   Answer: 10 609

2. Example 2. Find 98 × 102.

   1. 98 = 100 − 2 and 102 = 100 + 2
   2. (100 − 2)(100 + 2) = 100² − 2² = 10 000 − 4

   Answer: 9996

Algebra

Laws of exponents

a^m × aⁿ = a^m+n; a^m ÷ aⁿ = a^m−n; (a^m)ⁿ = a^mn; a⁰ = 1

In words: To multiply powers of the same base, add the exponents.

An exponent says how many times to multiply a number by itself. Multiplying powers with the same base just counts more copies, so you add the exponents. Dividing takes copies away, so you subtract.

What the letters mean

a

the base (not zero)

m, n

exponents (whole numbers here)

Worked examples

1. Example 1. Simplify 2³ × 2⁴ and find its value.

   1. 2³ × 2⁴ = 2^(3 + 4) = 2⁷
   2. 2⁷ = 128

   Answer: 128

2. Example 2. Find (3²)³.

   1. (3²)³ = 3^(2 × 3) = 3⁶

   Answer: 729

Watch out: These laws need the same base. 2³ × 3⁴ cannot be combined this way.

Geometry & measurement

Total surface area of a cuboid

TSA = 2(lb + bh + hl)

In words: Total surface area equals two times the sum of the three different face areas.

A cuboid has six faces in three matching pairs: l × b, b × h and h × l. Add the three areas and double them to cover all six faces. This is the amount of paper needed to wrap the box.

What the letters mean

l, b, h

length, breadth and height

TSA

total surface area

Worked examples

1. Example 1. A box is 5 cm × 4 cm × 3 cm. How much paper is needed to cover it completely?

   1. TSA = 2(5 × 4 + 4 × 3 + 3 × 5) = 2(20 + 12 + 15)
   2. = 2 × 47

   Answer: 94 cm²

2. Example 2. A room is 6 m long, 4 m wide and 3 m high. How much area do the four walls and ceiling add up to? (Use TSA − floor)

   1. TSA = 2(24 + 12 + 18) = 108 m²
   2. Minus the floor, 6 × 4 = 24 m²

   Answer: 84 m²

Geometry & measurement

Cube: volume and surface area

V = a³; TSA = 6a²

In words: Volume of a cube equals side cubed; total surface area equals six times side squared.

A cube has equal length, breadth and height, so volume is a × a × a. It has six identical square faces, each of area a², so the total surface is six of them.

What the letters mean

a

edge length

V

volume

TSA

total surface area

Worked examples

1. Example 1. A cube has an edge of 5 cm. Find its volume.

   1. V = a³ = 5 × 5 × 5

   Answer: 125 cm³

2. Example 2. Find the total surface area of the same cube.

   1. TSA = 6a² = 6 × 25

   Answer: 150 cm²

Geometry & measurement

Cylinder: volume and surface area

V = πr²h; CSA = 2πrh; TSA = 2πr(r + h)

In words: Volume of a cylinder equals pi r squared times height.

A cylinder is a stack of circles: the area of one circle (πr²) times the height h gives the volume. The curved surface unrolls into a rectangle of width 2πr and height h. Add two circular ends for the total surface.

What the letters mean

r

radius of the circular base

h

height

CSA

curved surface area

TSA

total surface area

Worked examples

1. Example 1. A cylinder has a radius of 7 cm and a height of 10 cm. Find its volume. (π = 22/7)

   1. V = πr²h = 22/7 × 7 × 7 × 10

   Answer: 1540 cm³

2. Example 2. A cylindrical tank has a radius of 7 m and a height of 5 m. How many litres does it hold? (π = 22/7; 1 m³ = 1000 litres)

   1. V = 22/7 × 7 × 7 × 5 = 770 m³
   2. Litres = 770 × 1000

   Answer: 770 000 litres

Geometry & measurement

Angle sum of a polygon

Sum of interior angles = (n − 2) × 180°

In words: The interior angles of a polygon with n sides add up to n minus two, times one hundred eighty degrees.

Draw lines from one corner to split any polygon into triangles. A polygon with n sides makes n − 2 triangles, and each triangle has angles adding to 180°. For a regular polygon, divide by n to get each angle.

What the letters mean

n

number of sides

Worked examples

1. Example 1. Find the sum of the interior angles of a hexagon.

   1. n = 6
   2. Sum = (6 − 2) × 180° = 4 × 180°

   Answer: 720°

2. Example 2. Find each interior angle of a regular octagon.

   1. Sum = (8 − 2) × 180° = 1080°
   2. Each angle = 1080° ÷ 8

   Answer: 135°

Number & arithmetic

Ratio and proportion

a : b = c : d ⇒ a × d = b × c

In words: In a proportion, the product of the extremes equals the product of the means.

Two ratios are in proportion when they describe the same relationship. Cross-multiplying turns the proportion into a simple equation for the missing number.

What the letters mean

a, b, c, d

four numbers, where one may be unknown

Worked examples

1. Example 1. If 3 : 4 = 9 : x, find x.

   1. 3 × x = 4 × 9
   2. x = 36 ÷ 3

   Answer: x = 12

2. Example 2. A recipe uses 2 cups of rice for 3 people. How much rice is needed for 9 people?

   1. 2 : 3 = x : 9
   2. 3 × x = 2 × 9
   3. x = 18 ÷ 3

   Answer: 6 cups

Money & commercial maths

Compound interest

A = P(1 + R/100)ⁿ; CI = A − P

In words: Amount equals principal times one plus rate over a hundred, all to the power of the number of years.

With compound interest, each year's interest is added to the principal, so the next year's interest is calculated on a bigger sum. This formula assumes interest is added once a year.

What the letters mean

P

principal

R

rate, % per year

n

number of years

A

final amount

CI

compound interest

Worked examples

1. Example 1. Find the amount and compound interest on ₹10 000 at 10% per year for 2 years.

   1. A = 10 000 × (1 + 10/100)² = 10 000 × 1.21 = ₹12 100
   2. CI = 12 100 − 10 000

   Answer: Amount ₹12 100; interest ₹2100

2. Example 2. Find the compound interest on ₹5000 at 20% per year for 2 years.

   1. A = 5000 × 1.2² = 5000 × 1.44 = ₹7200
   2. CI = 7200 − 5000

   Answer: ₹2200

Watch out: Compare with simple interest on the same sum: compound interest is a little higher because interest earns interest.

Geometry & measurement

Heron's formula (area from three sides)

A = √[ s(s − a)(s − b)(s − c) ], where s = (a + b + c) ÷ 2

In words: Area equals the square root of s times s minus a times s minus b times s minus c, where s is half the perimeter.

When you know all three sides of a triangle but not its height, first find the semi-perimeter s (half the perimeter), then put it into the formula.

What the letters mean

a, b, c

the three sides

s

semi-perimeter

Worked examples

1. Example 1. Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

   1. s = (13 + 14 + 15) ÷ 2 = 21
   2. A = √(21 × 8 × 7 × 6) = √7056

   Answer: 84 cm²

2. Example 2. Find the area of a triangle with sides 5, 12 and 13.

   1. s = 15
   2. A = √(15 × 10 × 3 × 2) = √900

   Answer: 30 square units (a right triangle: ½ × 5 × 12 = 30 agrees)

Algebra

Sum and difference of cubes

a³ + b³ = (a + b)(a² − ab + b²); a³ − b³ = (a − b)(a² + ab + b²)

In words: a cubed minus b cubed equals a minus b times a squared plus a b plus b squared.

These factorisations turn a cubic expression into a product of simpler ones. They are handy for factorising and for simplifying fractions with cubes.

What the letters mean

a, b

any numbers or expressions

Worked examples

1. Example 1. Check the identity for a = 3 and b = 2: find 3³ − 2³ both ways.

   1. Directly: 27 − 8 = 19
   2. By the identity: (3 − 2)(9 + 6 + 4) = 1 × 19

   Answer: 19 both ways

2. Example 2. Factorise x³ + 8.

   1. 8 = 2³, so a = x and b = 2
   2. x³ + 2³ = (x + 2)(x² − 2x + 4)
   3. Check at x = 1: 1 + 8 = 9 and (3)(1 − 2 + 4) = 9

   Answer: (x + 2)(x² − 2x + 4)

Geometry & measurement

Cone: volume and surface area

V = ⅓πr²h; CSA = πrl; l = √(r² + h²)

In words: Volume of a cone equals one third pi r squared times height.

A cone holds exactly one third of the cylinder with the same base and height. The slant height l runs along the side from the tip to the edge of the base, and the Pythagoras theorem links it to r and h.

What the letters mean

r

base radius

h

vertical height

l

slant height

CSA

curved surface area

Worked examples

1. Example 1. A cone has r = 3 cm and h = 4 cm. Find the slant height and the volume. (π = 3.14159)

   1. l = √(3² + 4²) = √25 = 5 cm
   2. V = ⅓ × π × 9 × 4 = 12π

   Answer: l = 5 cm; V ≈ 37.70 cm³

2. Example 2. A conical tent has a base radius of 7 m and a slant height of 25 m. How much canvas covers its curved surface? (π = 22/7)

   1. CSA = πrl = 22/7 × 7 × 25

   Answer: 550 m²

Geometry & measurement

Sphere and hemisphere

V = (4/3)πr³; SA = 4πr²; hemisphere: CSA = 2πr², TSA = 3πr²

In words: Volume of a sphere equals four thirds pi r cubed.

A sphere's surface area is exactly four times the area of its great circle (πr²). A hemisphere is half a sphere: its curved surface is half of 4πr², and its total surface adds the flat circular top.

What the letters mean

r

radius

V

volume

SA

surface area

Worked examples

1. Example 1. Find the volume of a sphere of radius 3 cm. (π = 3.14159)

   1. V = (4/3) × π × 27 = 36π

   Answer: ≈ 113.10 cm³

2. Example 2. A ball has a radius of 7 cm. Find its surface area. (π = 22/7)

   1. SA = 4πr² = 4 × 22/7 × 49

   Answer: 616 cm²

Coordinate geometry

Distance between two points

d = √[ (x₂ − x₁)² + (y₂ − y₁)² ]

In words: Distance equals the square root of the change in x squared plus the change in y squared.

Draw a right triangle between two points: the horizontal side is x₂ − x₁ and the vertical side is y₂ − y₁. The distance is its hypotenuse, so this is the Pythagoras theorem on a grid.

What the letters mean

(x₁, y₁)

first point

(x₂, y₂)

second point

Worked examples

1. Example 1. Find the distance between A(1, 2) and B(4, 6).

   1. d = √[(4 − 1)² + (6 − 2)²] = √(9 + 16) = √25

   Answer: 5 units

2. Example 2. Find the distance between P(−2, 3) and Q(2, 0).

   1. d = √[(2 − (−2))² + (0 − 3)²] = √(16 + 9)

   Answer: 5 units

Coordinate geometry

Section formula and midpoint

P = ( (m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n) ); midpoint = ( (x₁ + x₂)/2, (y₁ + y₂)/2 )

In words: The point that divides a line in the ratio m to n.

The point P divides the segment from (x₁, y₁) to (x₂, y₂) in the ratio m : n, counted from the first point. The midpoint is the special case m = n = 1, which simply averages the coordinates.

What the letters mean

m : n

the ratio in which P divides the line

(x₁, y₁), (x₂, y₂)

the end points

Worked examples

1. Example 1. Find the point that divides the line from (2, 3) to (8, 9) in the ratio 1 : 2.

   1. x = (1 × 8 + 2 × 2) ÷ 3 = 12 ÷ 3 = 4
   2. y = (1 × 9 + 2 × 3) ÷ 3 = 15 ÷ 3 = 5

   Answer: (4, 5)

2. Example 2. Find the midpoint of the line joining (−2, 4) and (6, −2).

   1. x = (−2 + 6) ÷ 2 = 2
   2. y = (4 + (−2)) ÷ 2 = 1

   Answer: (2, 1)

Coordinate geometry

Area of a triangle from coordinates

A = ½ | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

In words: Area is half the absolute value of this sum.

When you know the three corners of a triangle on a grid, you can find its area without finding a height. The bars | | mean "take the positive value", because area cannot be negative. If the answer is zero, the three points lie on one straight line.

What the letters mean

(x₁, y₁), (x₂, y₂), (x₃, y₃)

the three corners

Worked examples

1. Example 1. Find the area of the triangle with corners (0, 0), (4, 0) and (0, 3).

   1. A = ½ |0(0 − 3) + 4(3 − 0) + 0(0 − 0)| = ½ × 12

   Answer: 6 square units

2. Example 2. Find the area of the triangle with corners (1, 1), (5, 1) and (3, 5).

   1. A = ½ |1(1 − 5) + 5(5 − 1) + 3(1 − 1)| = ½ |−4 + 20 + 0|

   Answer: 8 square units (base 4 × height 4 ÷ 2 agrees)

Algebra

Quadratic formula

x = [ −b ± √(b² − 4ac) ] ÷ 2a

In words: x equals minus b, plus or minus the square root of b squared minus four a c, all over two a.

It solves any quadratic equation ax² + bx + c = 0 (with a not zero). Put in a, b and c, and the ± gives the two roots.

What the letters mean

a, b, c

the coefficients in ax² + bx + c = 0

x

the roots (solutions)

Worked examples

1. Example 1. Solve x² − 5x + 6 = 0.

   1. a = 1, b = −5, c = 6
   2. x = [5 ± √(25 − 24)] ÷ 2 = (5 ± 1) ÷ 2

   Answer: x = 3 or x = 2

2. Example 2. Solve 2x² + 3x − 2 = 0.

   1. a = 2, b = 3, c = −2
   2. x = [−3 ± √(9 + 16)] ÷ 4 = (−3 ± 5) ÷ 4

   Answer: x = ½ or x = −2

Watch out: Check your roots by putting them back into the equation.

Algebra

Discriminant: the nature of the roots

D = b² − 4ac

In words: The discriminant equals b squared minus four a c.

The part under the square root in the quadratic formula tells you what kind of roots to expect, before you solve anything. D > 0: two different real roots. D = 0: two equal real roots. D < 0: no real roots.

What the letters mean

D

the discriminant

a, b, c

the coefficients of ax² + bx + c

Worked examples

1. Example 1. Find the nature of the roots of x² − 4x + 4 = 0.

   1. D = (−4)² − 4 × 1 × 4 = 16 − 16

   Answer: D = 0, so the roots are equal (x = 2 twice)

2. Example 2. Does x² + x + 1 = 0 have real roots?

   1. D = 1² − 4 × 1 × 1 = 1 − 4

   Answer: D = −3 is negative, so there are no real roots

Algebra

Sum and product of the roots

sum of roots = −b/a; product of roots = c/a

In words: For a quadratic, the roots add up to minus b over a and multiply to c over a.

You can know the sum and the product of the roots without solving the equation. This is also how you build an equation from two given roots.

What the letters mean

a, b, c

coefficients of ax² + bx + c = 0

Worked examples

1. Example 1. For x² − 5x + 6 = 0, find the sum and product of the roots.

   1. Sum = −(−5) ÷ 1 = 5
   2. Product = 6 ÷ 1 = 6

   Answer: Sum 5, product 6 (the roots are 2 and 3)

2. Example 2. For 2x² + 3x − 2 = 0, find the sum and product of the roots.

   1. Sum = −3 ÷ 2
   2. Product = −2 ÷ 2

   Answer: Sum −3/2, product −1

Sequences & series

Arithmetic progression: the nth term

a_n = a + (n − 1)d

In words: The nth term equals the first term plus n minus one times the common difference.

An arithmetic progression (AP) goes up or down by the same amount d each time. To reach the nth term, start at the first term a and add d exactly n − 1 times.

What the letters mean

a

first term

d

common difference

n

position of the term

aₙ

the nth term

Worked examples

1. Example 1. Find the 10th term of the AP 3, 7, 11, …

   1. a = 3, d = 4, n = 10
   2. a₁₀ = 3 + (10 − 1) × 4 = 3 + 36

   Answer: 39

2. Example 2. Which term of the AP 5, 9, 13, … is 53?

   1. 5 + (n − 1) × 4 = 53
   2. (n − 1) × 4 = 48, so n − 1 = 12

   Answer: The 13th term

Sequences & series

Arithmetic progression: sum of n terms

S_n = (n/2) × [ 2a + (n − 1)d ] = (n/2) × (a + l)

In words: The sum of n terms equals n over two times the first term plus the last term.

Pair the first term with the last, the second with the second-last, and so on: every pair has the same total. That is why the sum is "half the number of terms" times "first plus last".

What the letters mean

n

number of terms

a

first term

d

common difference

l

last term

Worked examples

1. Example 1. Find the sum of the first 10 natural numbers.

   1. a = 1, l = 10, n = 10
   2. S = (10 ÷ 2) × (1 + 10) = 5 × 11

   Answer: 55

2. Example 2. Find the sum of the first 10 terms of the AP 2, 5, 8, …

   1. a = 2, d = 3, n = 10
   2. S = (10 ÷ 2) × [4 + 9 × 3] = 5 × 31

   Answer: 155

Trigonometry

Trigonometric ratios

sin θ = opposite ÷ hypotenuse; cos θ = adjacent ÷ hypotenuse; tan θ = opposite ÷ adjacent

In words: Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent.

In a right-angled triangle, the ratios of the sides depend only on the angle θ. "Opposite" is the side facing θ, "adjacent" is the side next to it (not the hypotenuse), and the hypotenuse is the longest side. A memory aid is SOH-CAH-TOA.

What the letters mean

θ

the angle you are looking at

opposite, adjacent

the sides facing and touching θ

hypotenuse

the side opposite the right angle

Worked examples

1. Example 1. In a right triangle with sides 3, 4, 5, find sin θ, cos θ and tan θ for the angle opposite the side 3.

   1. sin θ = 3/5, cos θ = 4/5, tan θ = 3/4

   Answer: sin θ = 0.6, cos θ = 0.8, tan θ = 0.75

2. Example 2. A 10 m ladder makes an angle of 30° with the ground. How high up the wall does it reach?

   1. height = 10 × sin 30°
   2. sin 30° = ½

   Answer: 5 m

Trigonometry

Basic trigonometric identities

sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ

In words: Sine squared plus cosine squared equals one.

These are true for every angle. The first comes from the Pythagoras theorem on a triangle with hypotenuse 1. They let you find the other ratios when you know one.

What the letters mean

θ

any angle

Worked examples

1. Example 1. If sin θ = 3/5 and θ is acute, find cos θ.

   1. cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25
   2. cos θ = 4/5

   Answer: 0.8 (that is 4/5)

2. Example 2. If tan θ = 3/4, find sec θ.

   1. sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16
   2. sec θ = 5/4

   Answer: 1.25 (that is 5/4)

Trigonometry

Standard trigonometric values

sin: 0, ½, 1/√2, √3/2, 1 cos: 1, √3/2, 1/√2, ½, 0 tan: 0, 1/√3, 1, √3, not defined (for 0°, 30°, 45°, 60°, 90°)

In words: Values of sine, cosine and tangent for zero, thirty, forty-five, sixty and ninety degrees.

Five angles come up again and again, so it pays to remember them. For sine, the values for 0°, 30°, 45°, 60° and 90° are √0/2, √1/2, √2/2, √3/2 and √4/2. Cosine is the same list in reverse order.

What the letters mean

0°, 30°, 45°, 60°, 90°

the standard angles

Worked examples

1. Example 1. Find sin 30° + cos 60°.

   1. sin 30° = ½ and cos 60° = ½
   2. ½ + ½

   Answer: 1

2. Example 2. Find tan 45° × sin 90°.

   1. tan 45° = 1 and sin 90° = 1

   Answer: 1

Trigonometry

Heights and distances

tan θ = height ÷ distance

In words: Tangent of the angle of elevation equals height divided by horizontal distance.

The angle of elevation is the angle you look up from the horizontal to the top of an object. Together with the distance you stand from the base, tan θ gives the height.

What the letters mean

θ

angle of elevation

height

height of the object above your eye level

distance

horizontal distance to its base

Worked examples

1. Example 1. From a point 20 m from the foot of a tower, the angle of elevation of its top is 45°. Find the height of the tower.

   1. height = 20 × tan 45° = 20 × 1

   Answer: 20 m

2. Example 2. From a point 10 m from the foot of a pole, the angle of elevation of its top is 60°. Find the height of the pole. (√3 = 1.732)

   1. height = 10 × tan 60° = 10 × √3

   Answer: ≈ 17.32 m

Geometry & measurement

Arc length and area of a sector

Arc length = (θ/360) × 2πr; Sector area = (θ/360) × πr²

In words: Arc length is the fraction theta over three hundred sixty of the circumference.

A sector is a slice of a circle, like a slice of pizza. The angle θ at the centre tells you what fraction of the whole circle you have, so take that fraction of the circumference (for the arc) or of the area (for the sector).

What the letters mean

θ

angle at the centre, in degrees

r

radius

Worked examples

1. Example 1. A sector has a radius of 14 cm and an angle of 90°. Find the arc length and the area of the sector. (π = 22/7)

   1. Arc = 90/360 × 2 × 22/7 × 14 = ¼ × 88 = 22 cm
   2. Area = ¼ × 22/7 × 14 × 14

   Answer: Arc 22 cm; area 154 cm²

2. Example 2. Find the area of a sector with radius 21 cm and angle 60°. (π = 22/7)

   1. Area = 60/360 × 22/7 × 21 × 21 = ⅙ × 1386

   Answer: 231 cm²

Statistics & probability

Mean of grouped data

Mean = Σ(f × x) ÷ Σf

In words: The mean equals the sum of frequency times value, divided by the total frequency.

When values repeat, list each value x with its frequency f (how many times it occurs). Multiply each pair, add the results, and divide by the total number of observations.

What the letters mean

x

a value (or the middle of a class interval)

f

its frequency

Σ

means "add up all of these"

Worked examples

1. Example 1. The scores 10, 20 and 30 occur 2, 3 and 5 times. Find the mean.

   1. Σfx = 2 × 10 + 3 × 20 + 5 × 30 = 20 + 60 + 150 = 230
   2. Σf = 2 + 3 + 5 = 10
   3. Mean = 230 ÷ 10

   Answer: 23

2. Example 2. The marks 5, 6 and 7 occur 1, 2 and 3 times. Find the mean.

   1. Σfx = 5 + 12 + 21 = 38
   2. Σf = 6
   3. Mean = 38 ÷ 6

   Answer: ≈ 6.33

Statistics & probability

Mode and median of grouped data

Mode = l + [ (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) ] × h; Median = l + [ (n/2 − cf) ÷ f ] × h

In words: Formulas for the mode and the median of data grouped in class intervals.

For the mode, find the modal class (the one with the highest frequency): l is its lower limit, f₁ its frequency, f₀ the frequency before it, f₂ the frequency after it and h the class width. For the median, find the class where the running total passes n/2: cf is the total frequency before that class and f its own frequency.

What the letters mean

l

lower limit of the class

h

class width

f₁, f₀, f₂

frequency of the modal class, the class before and the class after

n

total frequency

cf

cumulative frequency before the median class

Worked examples

1. Example 1. Modal class 30–40 with f₁ = 12, f₀ = 8, f₂ = 6 and class width 10. Find the mode.

   1. Mode = 30 + [(12 − 8) ÷ (24 − 8 − 6)] × 10 = 30 + (4 ÷ 10) × 10

   Answer: 34

2. Example 2. n = 40; the median class is 20–30 with cf = 14 and f = 10. Find the median.

   1. n/2 = 20
   2. Median = 20 + [(20 − 14) ÷ 10] × 10

   Answer: 26

Statistics & probability

Probability of an event

P(E) = number of favourable outcomes ÷ total number of outcomes; P(not E) = 1 − P(E)

In words: Probability equals favourable outcomes divided by all possible outcomes.

Probability is a number from 0 (impossible) to 1 (certain). It assumes every outcome is equally likely. The probability that something does not happen is one minus the probability that it does.

What the letters mean

E

the event you are interested in

P(E)

probability of E, between 0 and 1

Worked examples

1. Example 1. A fair die is rolled. What is the probability of an even number?

   1. Favourable: 2, 4, 6, which is 3 outcomes; total: 6
   2. P = 3 ÷ 6

   Answer: ½

2. Example 2. A bag has 3 red and 5 blue balls. What is the probability of NOT drawing a red ball?

   1. P(red) = 3/8
   2. P(not red) = 1 − 3/8

   Answer: 5/8

Geometry & measurement

Similar triangles and the basic proportionality theorem

Area ratio = (side ratio)²; if DE ∥ BC then AD/DB = AE/EC

In words: The ratio of the areas of similar triangles equals the square of the ratio of their sides.

Similar triangles have the same shape but different sizes. Doubling every side makes the area four times as big. In a triangle ABC, a line DE parallel to BC cuts the other two sides in the same ratio.

What the letters mean

side ratio

ratio of any pair of matching sides

D, E

points on AB and AC with DE parallel to BC

Worked examples

1. Example 1. Two similar triangles have corresponding sides in the ratio 2 : 3. What is the ratio of their areas?

   1. Area ratio = 2² : 3² = 4 : 9

   Answer: 4 : 9

2. Example 2. In triangle ABC, DE ∥ BC with AD = 2, DB = 3 and AE = 4. Find EC.

   1. AD/DB = AE/EC, so 2/3 = 4/EC
   2. EC = 4 × 3 ÷ 2

   Answer: EC = 6

Number & arithmetic

HCF and LCM of two numbers

HCF × LCM = product of the two numbers

In words: The highest common factor times the lowest common multiple equals the product of the two numbers.

For any two positive whole numbers, multiplying their HCF by their LCM gives the same answer as multiplying the numbers themselves. It is a quick way to find the LCM once you know the HCF.

What the letters mean

HCF

highest common factor

LCM

lowest common multiple

Worked examples

1. Example 1. The HCF of 12 and 18 is 6. Find their LCM.

   1. LCM = (12 × 18) ÷ 6 = 216 ÷ 6

   Answer: 36

2. Example 2. The HCF of 8 and 12 is 4. Find their LCM.

   1. LCM = (8 × 12) ÷ 4 = 96 ÷ 4

   Answer: 24

Watch out: This rule works for two numbers only, not for three or more.

Trigonometry

Compound angle formulas

sin(A ± B) = sinA cosB ± cosA sinB; cos(A ± B) = cosA cosB ∓ sinA sinB

In words: The sine of A plus B equals sine A cosine B plus cosine A sine B.

They give the sine or cosine of a sum or difference of two angles, which is how exact values such as sin 75° and cos 15° are found from the standard angles. Note that cosine flips the sign: the plus on the left becomes a minus on the right.

What the letters mean

A, B

any two angles

Worked examples

1. Example 1. Find sin 75° exactly.

   1. 75° = 45° + 30°
   2. sin 75° = sin45° cos30° + cos45° sin30° = (√2/2)(√3/2) + (√2/2)(1/2)
   3. = (√6 + √2) ÷ 4

   Answer: (√6 + √2)/4 ≈ 0.9659

2. Example 2. Find cos 15° exactly.

   1. 15° = 45° − 30°
   2. cos 15° = cos45° cos30° + sin45° sin30° = (√6 + √2) ÷ 4

   Answer: (√6 + √2)/4 ≈ 0.9659 (the same value as sin 75°, as expected)

Trigonometry

Double angle formulas

sin2A = 2 sinA cosA; cos2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A; tan2A = 2tanA ÷ (1 − tan²A)

In words: Sine of two A equals two sine A cosine A.

These come from the compound angle formulas with B = A. The three forms of cos 2A are all equal, so pick whichever uses the ratio you already know.

What the letters mean

A

any angle

Worked examples

1. Example 1. If sin A = 3/5 and cos A = 4/5, find sin 2A and cos 2A.

   1. sin 2A = 2 × 3/5 × 4/5 = 24/25
   2. cos 2A = 16/25 − 9/25 = 7/25

   Answer: sin 2A = 24/25 = 0.96; cos 2A = 7/25 = 0.28

2. Example 2. Find cos 60° from cos 30° = √3/2.

   1. cos 2A = 2cos²A − 1 with A = 30°
   2. = 2 × 3/4 − 1

   Answer: ½

Trigonometry

Sine rule and cosine rule

a/sinA = b/sinB = c/sinC = 2R; c² = a² + b² − 2ab cosC

In words: In any triangle, each side divided by the sine of its opposite angle gives the same number.

These work in any triangle, not only right-angled ones. Use the sine rule when you know a side and its opposite angle plus one more piece. Use the cosine rule when you know two sides and the angle between them, or all three sides.

What the letters mean

a, b, c

sides opposite angles A, B, C

R

radius of the circle through all three corners

Worked examples

1. Example 1. Two sides of a triangle are 5 and 7 with 60° between them. Find the third side.

   1. c² = 5² + 7² − 2 × 5 × 7 × cos 60° = 25 + 49 − 35
   2. c² = 39

   Answer: c ≈ 6.24

2. Example 2. In a triangle, A = 30°, B = 45° and a = 10. Find b.

   1. b = a sinB ÷ sinA = 10 × sin45° ÷ sin30°
   2. = 10 × (√2/2) ÷ (1/2) = 10√2

   Answer: ≈ 14.14

Counting & sets

Union of two sets

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

In words: The number in A or B equals the number in A plus the number in B minus the number in both.

If you add n(A) and n(B), anything in both sets is counted twice, so subtract the overlap once. This is the key to "how many like at least one of these?" problems.

What the letters mean

n(A)

number of elements in A

A ∩ B

elements in both A and B

A ∪ B

elements in A or B or both

Worked examples

1. Example 1. 30 people like tea, 20 like coffee and 10 like both. How many like at least one?

   1. 30 + 20 − 10

   Answer: 40

2. Example 2. In a class of 50, 28 play cricket, 20 play football and 8 play both. How many play neither?

   1. At least one: 28 + 20 − 8 = 40
   2. Neither: 50 − 40

   Answer: 10

Counting & sets

Permutations and combinations

ⁿP_r = n! ÷ (n − r)!; ⁿC_r = n! ÷ [ r!(n − r)! ]

In words: n P r counts arrangements where order matters; n C r counts selections where order does not matter.

Use a permutation when the order matters (arranging books on a shelf, a race podium). Use a combination when only the group matters (choosing a team). n! (n factorial) means n × (n − 1) × … × 1.

What the letters mean

n

number of things to choose from

r

number you choose or arrange

n!

n factorial

Worked examples

1. Example 1. In how many ways can 3 books out of 5 be arranged in order on a shelf?

   1. ⁵P₃ = 5! ÷ 2! = 120 ÷ 2

   Answer: 60

2. Example 2. In how many ways can a team of 2 be chosen from 5 players?

   1. ⁵C₂ = 5! ÷ (2! × 3!) = 120 ÷ (2 × 6)

   Answer: 10

Watch out: Ask yourself: "would swapping two chosen items give a different result?" If yes, use P. If no, use C.

Algebra

Binomial theorem

(a + b)ⁿ = Σ ⁿC_r a^n−r b^r; general term T_r+1 = ⁿC_r a^n−r b^r

In words: a plus b to the power n expands as a sum of n choose r times a to the n minus r times b to the r.

It expands a power of a two-term expression without multiplying it out repeatedly. The coefficients come from combinations (the numbers in Pascal's triangle), and the powers of a go down while the powers of b go up.

What the letters mean

n

the power

r

term number minus one, from 0 to n

ⁿCᵣ

the binomial coefficient

Worked examples

1. Example 1. Find the coefficient of x² in the expansion of (1 + x)⁵.

   1. General term: ⁵C_r x^r
   2. For x², r = 2: ⁵C₂ = 10

   Answer: 10

2. Example 2. Expand (x + 2)³.

   1. x³ + ³C₁ x² (2) + ³C₂ x (2²) + 2³
   2. = x³ + 6x² + 12x + 8
   3. Check at x = 1: 27 = 1 + 6 + 12 + 8

   Answer: x³ + 6x² + 12x + 8

Sequences & series

Geometric progression

a_n = a rⁿ⁻¹; S_n = a(rⁿ − 1) ÷ (r − 1) (r ≠ 1); S_∞ = a ÷ (1 − r) (|r| < 1)

In words: Each term of a geometric progression is the previous term times r.

A geometric progression (GP) multiplies by the same ratio r each time. The sum formula adds a finite number of terms. If the ratio is between −1 and 1, the terms shrink and the infinite sum settles at a fixed value.

What the letters mean

a

first term

r

common ratio

n

number of terms

Worked examples

1. Example 1. For the GP 2, 6, 18, …, find the 5th term and the sum of the first 5 terms.

   1. a = 2, r = 3
   2. a₅ = 2 × 3⁴ = 162
   3. S₅ = 2(3⁵ − 1) ÷ 2 = 242

   Answer: 5th term 162; sum 242

2. Example 2. Find the sum of 1 + ½ + ¼ + ⅛ + …

   1. a = 1, r = ½, so |r| < 1
   2. S∞ = 1 ÷ (1 − ½)

   Answer: 2

Algebra

Laws of logarithms

log(mn) = log m + log n; log(m/n) = log m − log n; log mⁿ = n log m; log_ba = log a ÷ log b

In words: The log of a product is the sum of the logs.

A logarithm answers "what power do I raise the base to, to get this number?". Logs turn multiplication into addition and powers into multiplication. The last rule changes base, so any calculator log will do.

What the letters mean

m, n

positive numbers

b

the base, positive and not 1

Worked examples

1. Example 1. Find log₂ 8.

   1. 2 to the power 3 is 8

   Answer: 3

2. Example 2. Simplify log 25 + log 4 (base 10).

   1. log 25 + log 4 = log(25 × 4) = log 100
   2. 10² = 100

   Answer: 2

Complex numbers

Complex numbers: modulus and De Moivre

i² = −1; |a + bi| = √(a² + b²); (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ

In words: i squared equals minus one; the modulus of a plus b i is the square root of a squared plus b squared.

The number i is defined by i² = −1, which lets us take square roots of negative numbers. A complex number a + bi can be drawn as the point (a, b); its modulus is its distance from the origin. De Moivre's theorem makes powers of complex numbers on the unit circle simple.

What the letters mean

a + bi

a complex number with real part a and imaginary part b

|z|

modulus: the distance from the origin

Worked examples

1. Example 1. Find |3 + 4i|.

   1. √(3² + 4²) = √25

   Answer: 5

2. Example 2. Find (1 + i)².

   1. (1 + i)² = 1 + 2i + i² = 1 + 2i − 1

   Answer: 2i (real part 0, imaginary part 2)

Coordinate geometry

Straight lines

m = (y₂ − y₁) ÷ (x₂ − x₁); y − y₁ = m(x − x₁); distance = |Ax₁ + By₁ + C| ÷ √(A² + B²)

In words: Slope is the rise over the run; the last formula is the distance from a point to a line.

The slope m measures how steep a line is. Given a point and the slope, the point-slope form gives the equation. The distance formula finds how far a point (x₁, y₁) is from a line Ax + By + C = 0, measured at right angles.

What the letters mean

m

slope (gradient)

(x₁, y₁)

a point

Ax + By + C = 0

the line

Worked examples

1. Example 1. Find the slope of the line through (1, 2) and (3, 8), and its equation.

   1. m = (8 − 2) ÷ (3 − 1) = 3
   2. y − 2 = 3(x − 1), so y = 3x − 1

   Answer: Slope 3; y = 3x − 1

2. Example 2. Find the distance from the point (3, 4) to the line 3x + 4y − 5 = 0.

   1. |3 × 3 + 4 × 4 − 5| ÷ √(9 + 16) = |20| ÷ 5

   Answer: 4 units

Coordinate geometry

Circle, parabola and ellipse

circle: (x − h)² + (y − k)² = r²; parabola: y² = 4ax; ellipse: x²/a² + y²/b² = 1, e = √(1 − b²/a²)

In words: Standard equations of a circle, a parabola and an ellipse.

A circle is all points a fixed distance r from the centre (h, k). A parabola y² = 4ax opens to the right with its focus at (a, 0). An ellipse with a > b has eccentricity e, a number between 0 and 1 that tells how stretched it is.

What the letters mean

(h, k)

centre of the circle

r

radius

a, b

half-lengths of the ellipse, with a > b

Worked examples

1. Example 1. Does the point (5, 7) lie on the circle with centre (2, 3) and radius 5?

   1. (5 − 2)² + (7 − 3)² = 9 + 16 = 25 = 5²

   Answer: Yes, it lies on the circle

2. Example 2. Find the eccentricity of the ellipse x²/25 + y²/9 = 1.

   1. a² = 25, b² = 9
   2. e = √(1 − 9/25) = √(16/25)

   Answer: 0.8 (that is 4/5)

Calculus

Limits: standard results

lim_x→0 (sin x)/x = 1; lim_x→a (xⁿ − aⁿ)/(x − a) = n aⁿ⁻¹; lim_x→0 (e^x − 1)/x = 1

In words: The limit of sine x over x as x tends to zero is one.

A limit describes what a value approaches as x gets very close to a number. These results come up constantly and are the building blocks for finding derivatives. In the first, x must be measured in radians.

What the letters mean

x → a

x gets closer and closer to a

Worked examples

1. Example 1. Find the limit of (x² − 1)/(x − 1) as x → 1.

   1. Use the second result with n = 2 and a = 1: n a^(n−1) = 2 × 1
   2. Or factorise: (x − 1)(x + 1) ÷ (x − 1) = x + 1 → 2

   Answer: 2

2. Example 2. Find the limit of sin(3x)/x as x → 0.

   1. sin(3x)/x = 3 × sin(3x)/(3x)
   2. As x → 0, sin(3x)/(3x) → 1

   Answer: 3

Calculus

Derivatives: the rules

d/dx(xⁿ) = n xⁿ⁻¹; (uv)′ = u′v + uv′; (u/v)′ = (u′v − uv′) ÷ v²; dy/dx = dy/du × du/dx

In words: The derivative of x to the n is n times x to the n minus one.

The derivative gives the slope of a curve at any point, or how fast a quantity changes. The power rule handles powers, the product and quotient rules handle two functions multiplied or divided, and the chain rule handles a function inside another.

What the letters mean

u, v

functions of x

u′, v′

their derivatives

Worked examples

1. Example 1. Find the slope of y = x⁴ at x = 2.

   1. dy/dx = 4x³
   2. At x = 2: 4 × 8

   Answer: 32

2. Example 2. Differentiate y = (3x + 1)⁵ and find its value at x = 0.

   1. Chain rule: dy/dx = 5(3x + 1)⁴ × 3 = 15(3x + 1)⁴
   2. At x = 0: 15 × 1

   Answer: 15

Calculus

Derivatives of standard functions

d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x; d/dx(eˣ) = eˣ; d/dx(ln x) = 1/x

In words: The derivative of sine x is cosine x.

Learn these as a short list; with the rules above they let you differentiate almost anything met in school. For trigonometric functions, x is in radians.

What the letters mean

x

the variable; in radians for sin, cos and tan

Worked examples

1. Example 1. Find the slope of y = sin x at x = 0.

   1. dy/dx = cos x
   2. cos 0 = 1

   Answer: 1

2. Example 2. Find the slope of y = ln x at x = 2.

   1. dy/dx = 1/x
   2. At x = 2: 1/2

   Answer: ½

Calculus

Maxima and minima

Find f′(x) = 0, then test with f″(x): f″ < 0 gives a maximum; f″ > 0 gives a minimum

In words: Set the derivative to zero, then use the second derivative to decide whether it is a maximum or a minimum.

At the top of a hill or the bottom of a valley, the slope is zero. Solve f′(x) = 0 to find those points. The second derivative tells you the shape there: negative means curving down (a maximum), positive means curving up (a minimum).

What the letters mean

f′(x)

first derivative: the slope

f″(x)

second derivative: how the slope changes

Worked examples

1. Example 1. Find the maximum and minimum values of f(x) = x³ − 6x² + 9x.

   1. f′ = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0, so x = 1 or x = 3
   2. f″ = 6x − 12: at x = 1 it is −6 (maximum); at x = 3 it is +6 (minimum)
   3. f(1) = 1 − 6 + 9 = 4; f(3) = 27 − 54 + 27 = 0

   Answer: Maximum value 4 at x = 1; minimum value 0 at x = 3

2. Example 2. A rectangle has a perimeter of 40 cm. What dimensions give the largest area?

   1. Sides x and 20 − x, so A = x(20 − x) = 20x − x²
   2. A′ = 20 − 2x = 0, so x = 10; A″ = −2 < 0 (maximum)
   3. Area = 10 × 10

   Answer: A 10 cm by 10 cm square, area 100 cm²

Calculus

Integration: standard integrals

∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1); ∫(1/x) dx = ln|x| + C; ∫eˣ dx = eˣ + C; ∫sin x dx = −cos x + C; ∫cos x dx = sin x + C

In words: Integration reverses differentiation.

Integration finds the function whose derivative you are given. Because the derivative of a constant is zero, every answer has "+ C", the constant of integration. For powers, raise the power by one and divide by the new power.

What the letters mean

C

constant of integration

n

the power (not −1)

Worked examples

1. Example 1. Find ∫3x² dx.

   1. 3 × x³/3 = x³
   2. Check: the derivative of x³ is 3x²

   Answer: x³ + C

2. Example 2. Find ∫₀² x² dx (the area under y = x² from 0 to 2).

   1. [x³/3] from 0 to 2 = 8/3 − 0

   Answer: 8/3 ≈ 2.67

Calculus

Definite integrals and integration by parts

∫ₐᵇ f(x) dx = F(b) − F(a); ∫u dv = uv − ∫v du

In words: A definite integral equals the antiderivative at the top limit minus the antiderivative at the bottom limit.

A definite integral gives the area under a curve between x = a and x = b (where the curve is above the axis). Integration by parts handles a product of two different kinds of function, for example x times eˣ.

What the letters mean

F

an antiderivative of f

a, b

the limits

u, v

chosen parts of the product

Worked examples

1. Example 1. Find the area under y = x² from x = 0 to x = 3.

   1. F(x) = x³/3
   2. F(3) − F(0) = 27/3 − 0

   Answer: 9 square units

2. Example 2. Find ∫₀¹ x eˣ dx using integration by parts.

   1. Take u = x and dv = eˣ dx, so du = dx and v = eˣ
   2. ∫x eˣ dx = x eˣ − eˣ
   3. From 0 to 1: (e − e) − (0 − 1)

   Answer: 1

Matrices & vectors

Determinant and inverse of a 2×2 matrix

|A| = ad − bc; A⁻¹ = (1 ÷ |A|) × [ d −b ; −c a ] for A = [ a b ; c d ]

In words: The determinant of a two by two matrix is a d minus b c.

For the matrix with rows (a, b) and (c, d), the determinant ad − bc tells you whether an inverse exists (it must not be zero). To invert, swap a and d, change the signs of b and c, and divide everything by the determinant. Inverses solve systems of equations.

What the letters mean

a, b, c, d

the four entries

|A|

the determinant

Worked examples

1. Example 1. Find the inverse of A with rows (2, 1) and (5, 3).

   1. |A| = 2 × 3 − 1 × 5 = 1
   2. Inverse = 1 ÷ 1 × [3 −1 ; −5 2]

   Answer: Inverse has rows (3, −1) and (−5, 2)

2. Example 2. Solve 2x + y = 7 and 5x + 3y = 19 using the inverse above.

   1. x = 3 × 7 + (−1) × 19 = 21 − 19
   2. y = (−5) × 7 + 2 × 19 = −35 + 38
   3. Check: 2 × 2 + 3 = 7 and 5 × 2 + 3 × 3 = 19

   Answer: x = 2, y = 3

Matrices & vectors

Vectors: magnitude, dot product and cross product

|a| = √(a₁² + a₂² + a₃²); a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cosθ; |a × b| = |a||b| sinθ

In words: The dot product of two vectors adds the products of matching components.

A vector has both size and direction. The dot product gives a number: it is zero when two vectors are at right angles. The size of the cross product equals the area of the parallelogram the two vectors make.

What the letters mean

a, b

vectors with components (a₁, a₂, a₃) and (b₁, b₂, b₃)

θ

angle between the vectors

Worked examples

1. Example 1. For a = (1, 2, 3) and b = (4, −5, 6), find a·b and |a|.

   1. a·b = 1 × 4 + 2 × (−5) + 3 × 6 = 4 − 10 + 18
   2. |a| = √(1 + 4 + 9) = √14

   Answer: a·b = 12; |a| ≈ 3.74

2. Example 2. Are (2, 3, −1) and (1, −1, −1) perpendicular?

   1. Dot product = 2 × 1 + 3 × (−1) + (−1) × (−1) = 2 − 3 + 1

   Answer: The dot product is 0, so yes, they are perpendicular

Statistics & probability

Conditional probability and Bayes' theorem

P(A|B) = P(A ∩ B) ÷ P(B); P(A|B) = P(B|A) P(A) ÷ P(B)

In words: The probability of A given B equals the probability of both, divided by the probability of B.

"P(A|B)" means the probability of A when we already know B has happened. Bayes' theorem reverses the question: from P(B|A) it finds P(A|B), which is how a test result updates a belief.

What the letters mean

P(A|B)

probability of A given B

P(A ∩ B)

probability that both happen

Worked examples

1. Example 1. A die is rolled and the number is even. What is the probability it is a 6?

   1. P(6 and even) = 1/6; P(even) = 1/2
   2. P(6 | even) = (1/6) ÷ (1/2)

   Answer: 1/3

2. Example 2. Machine A makes 60% of items and 2% of its items are faulty. Machine B makes 40% and 5% of its items are faulty. A faulty item is found: what is the probability it came from A?

   1. P(faulty) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032
   2. P(A | faulty) = 0.012 ÷ 0.032

   Answer: 0.375, that is 37.5%

Statistics & probability

Binomial distribution

P(X = r) = ⁿC_r p^r q^n−r, q = 1 − p; mean = np; variance = npq

In words: The probability of exactly r successes in n trials.

Use it when you repeat the same trial n times, each trial has only two outcomes (success with probability p, or failure), and the trials do not affect each other. It counts the ways to place r successes among n trials.

What the letters mean

n

number of trials

p

probability of success in one trial

r

number of successes you want

Worked examples

1. Example 1. A fair coin is tossed 5 times. What is the probability of exactly 2 heads?

   1. n = 5, p = ½, r = 2
   2. P = ⁵C₂ × (½)² × (½)³ = 10 × 1/32

   Answer: 10/32 = 0.3125

2. Example 2. A fair coin is tossed 10 times. Find the mean and the variance of the number of heads.

   1. Mean = np = 10 × ½ = 5
   2. Variance = npq = 10 × ½ × ½

   Answer: Mean 5; variance 2.5

Statistics & probability

Variance and standard deviation

σ² = Σ(x − x̄)² ÷ n; σ = √σ²

In words: Variance is the average of the squared distances from the mean.

The mean says where the data is centred; the standard deviation says how spread out it is. Find each value's distance from the mean, square it, average the squares (that is the variance), then take the square root.

What the letters mean

x̄

the mean

n

number of values

σ

standard deviation

Worked examples

1. Example 1. Find the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.

   1. Mean = 40 ÷ 8 = 5
   2. Squared distances: 9, 1, 1, 1, 0, 0, 4, 16, total 32
   3. Variance = 32 ÷ 8 = 4; σ = √4

   Answer: 2

2. Example 2. Find the variance and the standard deviation of 1, 2, 3.

   1. Mean = 2; squared distances: 1, 0, 1
   2. Variance = 2 ÷ 3

   Answer: Variance ≈ 0.67; σ ≈ 0.82