The union of two sets is a set containing all elements that are in both sets. If an element occurs more than once in both sets, its written only once.

Hence A ∪ B = {2,4,6,8,10,12}

Standard form, or standard index form, is a system of writing numbers which can be particularly useful for working with very large or very small numbers.

It is based on using powers of 10 to express how big or small a number is.

Standard form is written in the form of a x 10ⁿ, where a is a number bigger than or equal to 1 and less than 10.

n can be any positive or negative whole number.

For example 1.2 x 10³, 3.14 x 10^-5

The rules when writing a number in standard form is that first you write down a number between 1 and 10, then you write × 10(to the power of a number).

Thus the decimal point must be behind the first non-zero digit

If you move the decimal point to the left, the power (number of times of movement) of 10 will be positive and if to the right, negative

The first non-zero digit in 0.000344 is 3, hence we have to move the decimal point (.) 4 times to the right before placing it behind the 3. Since the movement is to the right, the power (4) of 10 is negative.

⇒ 0.000344 = 3.44 x 10^-4 in standard form

The lower the negative value the larger the value and the higher the negative value the smaller the value

You can confirm this also on the number line

Starting zeros are not significant and so the number of significant digits starts from 24561. The 3 digits lands on 5 but the digit next to it is more than 5 (6) and so 1 is added to the 5 to make it 6.

NOTE
Zeros between other digits are significant. Only starting zeros are insignificant. For instance 3.052306, all the zeros are significant because they are not starting zeros after a decimal point.

Applying BODMAS, the bracket needs to be solved first

Applying a^mxaⁿ = a^m+n

⇒ 7⁵x7³ = 7⁵⁺³=7⁸

⇒ (7⁵x7³)÷7⁶ = 7⁸÷7⁶

Applying a^m÷aⁿ = a^m-n

⇒ 7⁸÷7⁶ = 7^8-6 = 7²

A line of symmetry is a line that cuts a shape exactly in half.

This means that if you were to fold the shape along the line, both halves would match exactly. Equally, if you were to place a mirror along the line, the shape would remain unchanged.

If all the sides and angles of a polygon are equal then it is said to be a regular polygon. Like the equilateral triangle, square etc. All the regular polygons are symmetrical shapes. In the regular polygon, the number of lines of symmetry is the same as the number of its sides

Regular Polygon	Number of Sides	Number of Symmetry	Image
Equilateral Triangle	3	3
Square	4	4
Regular Pentagon	5	5
Regular Hexagon	6	6

10 - 1(x+3)/2 = 8

Get rid of the fraction by multiplying each side of the equation by the LCM of the denominators

The denominators are 1,2 and 1 and their LCM is 2

NOTE:10 = 10/1 and 8 = 8/1, thus every number has one dividing it but its not written

⇒ 10 - 1(x+3)/2 = 8

2 x 10 -1x2(x+3)/2 = 8 x 2

20 -1(x+3) = 16

Applying a(b+c) = axb+axc

⇒ - 1(x+3) = -1xx-1x3 = -x-3

20 - 1(x+3) = 16

⇒ 20 -x-3 = 16

Making sure x is at one side of the equation,

-x = 16+3-20

-x = -1

Divide both sides by -1

⇒ x = 1

NOTE: You can as well move x to the right size of the equation and 16 to the left and the answer will still be the same.Thus,

20 -x-3 = 16

20 - 3 - 16 = x

20-19 = x

1 = x

kx+2xt-4k-8t

Group like terms

kx-4k+2xt-8t

Factorize like terms (factors) out

k(x-4)+2t(x-4)

(k+2t)(x-4)

Fraction = Value/Total

Fraction of boys = No. of boys in the class / Total No. of students

Total No. of students = No. of boys + No. of girls = 12+18 = 30

No. of boys = 12

⇒ fraction of boys = 12/30 = 2/5 (Thus 6 goes into 12, 2 times and 30, 5 times)

NOTE: Fraction of girls = 18/30 = 3/5 (Thus 6 goes into 18, 3 times and 30, 5 times)

The sum of all fractions should be 1

Verify: Fraction of boys + Fraction of girls = ⅖ + ⅗ = (2+3)/5 = 5/5 = 1

% means over 100

⇒ 30% = 30⁄100

The top 0 in the 30 cancels one of the 0s in the 100

⇒ 30% = 30⁄100 = 3⁄10

ky - k = y²

Factorize k out

ky = k x y

- k = -1 x k

When you factorize, you take the letter(s) common at both sides or the number that divides each numbers out and write the remain(s) in a bracket

ky - k = k x y -1 x k

Bring the k out ⇒ k x y -1 x k = k(y -1)

ky - k = y²

⇒ k(y -1) = y²

Since we are making k the subject (let k stand alone), we divide both sides of the equation by (y -1) so that the (y -1) cancels each other, leaving the k alone.

⇒ k = y² / (y -1)

Mean = Sum of Numbers/Total number of numbers(items)

When you count the numbers,5, 2x,4 and 3, you will get 4, hence total number of numbers = 4

Mean = 5

Sum means add all the numbers

⇒ 5 = (5 + 2x + 4 + 3) / 4

Solve for x

Get rid of the fraction by multiplying both sides of the equation by the LCM of the denominators (1 and 4). The LCM of the denominators is 4 hence multiply both sides by 4

5 = 5/1

⇒ 5 x 4 = 5 + 2x + 4 + 3

20 = 5 + 2x + 4 + 3

20 = 5 + 4 + 3 + 2x

20 = 12 + 2x

20 - 12 = 2x

8 = 2x

Divide both sides by 2

8/2 = 2x/2

4 = x

⇒ x = 4

Test all the available rules with the values of x to see if you get the corresponding y values of the x

Option A

y = 2x + 2

When x = 1

y = 2(1) + 2 = 2 x1 + 2 = 4

Since y ≠ 3, option A is not the answer

Option B

y = -2x + 2

When x = 1

y = -2(1) + 2 = -2 x 1 + 2 = -2 + 2 = 0

Since y ≠ 3, option B is not the answer

Option C

y = 4x

When x = 1

y = 4(1) = 4 x 1 = 4

Since y ≠ 3, option C is not the answer

So automatically the answer will be option D but you can still verify

y = -2x + 5

When x = 1

y = -2(1) + 5 = -2 x 1 + 5 = -2 + 5 = 3

When x = 2

y = -2(2) + 5 = -2 x 2 + 5 = -4 + 5 = 1

When x = 3

y = -2(3) + 5 = -2 x 3 + 5 = -6 + 5 = -1

When x = 4

y = -2(4) + 5 = -2 x 4 + 5 = -8 + 5 = -3

When x = 5

y = -2(5) + 5 = -2 x 5 + 5 = -10 + 5 = -5

Parallelogram has two sides equal. Which means the other two sides of the parallelogram are also 5.8 m and 8.2 m.

Perimeter is the sum of the length of all sides

⇒ The perimeter of the parallelogram = 2x4.8 + 2x7.2 = 9.6+14.4 = 24.0 m

Volume of a cuboid = Length x Breadth x Height

Dividing both sides by Length x Breadth to make Height the subject

⇒ Height = Volume of cuboid / (Length x Breadth)

⇒ The height of the cuboid = 36 / (6 x 4) = 3/2 = 1.5 m

NOTE: 6 goes into 36, 6 times and 2 goes into 6, 3 times and 4, 2 times

Draw the bearing of A from B and calculate the bearing of A

240 = 180 + 60, hence the location of B bearing above

Use the property of alternative angles to calculate the bearing of B from A

Bearing of B from A is 060^o

2y + 5 < 4y - 5

5 + 5 < 4y - 2y

10 < 2y

Divide both sides by 2

10/2 < 2y/2

5 < y

You can change the y position and change the direction of the sign

⇒ y > 5

Gradient of a line = Change in Y/Change in X = (y₂-y₁)/(x₂-x₁)

The general coordinates of a point is (x,y)

y₂ = -2

y₁ = 4

x₂ = 3

x₁ = -3

⇒ Gradient of the line = (-2 - 4)/(3--3) = (-6)/(3+3) = -6/6 = -1

Express the ratios in fractions

6:8 = 6/8

r:48 = r/48

⇒ 6/8 = r/48

Cross multiply them. The numerator at the left multiplies the denominator at the right and the denominator at the left also multiplies the numerator at the right

6 x 48 = 8 x r

Divide both sides by 8 to make r the subject (stand alone)

6 x 48 / 8 = r

6 x 6 = r

NOTE: 8 goes into 48, 6 times

36 = r

⇒ r = 36

Concepts

A triangle which has two of its sides equal is called an isosceles triangle

Properties of Isosceles Triangle

1. The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.

2. The side opposite the vertex angle is called the base and base angles are equal.

3. The perpendicular from the vertex angle bisects the base and the vertex angle.

Both triangle QPS and QSR are isosceles triangles (have two sides equal and hence 2 angles equal)

For triangle QSR

∠SQR = ∠QRS = 35^o

But angles in a triangle add up to 180^o

⇒ ∠SQR + ∠QRS + ∠RSQ = 180

⇒ 35 + 35 + ∠RSQ = 180

70 + ∠RSQ = 180

∠RSQ = 180 - 70

∠RSQ = 110^o

For triangle QPS

∠RSQ + ∠QSP = 180 (Angles on a straight line add up to 180^o

But ∠RSQ = 110^o

⇒ 110 + ∠QSP = 180

∠QSP = 180 - 110

∠QSP = 70

Angles in a triangle add up to 180^o

⇒ ∠SQP + ∠QSP + ∠QPS = 180

∠SQP = ∠QSP = 70

⇒ 70 + 70 + ∠QPS = 180

⇒ 140 + ∠QPS = 180

⇒ ∠QPS = 180 - 140

∠QPS = 40^o

Method I

Let x = the time taken to travel 12 km

8 km took 1hr ⇒ 8 km = 1hr

12 km will take xhr ⇒ 12 km = xhr

8 = 1

12 = x

Cross multiply both sides

x x 8 = 12 x 1

Divide both sides by 8

x8/8 = 12/8 = 3/2 = 1½

NOTE: 4 goes into 12, 3 times and into 8, 2 times

Method II

Speed = Distance/Time

Speed for 8 km in 1hr = 8km / 1hr = 8 km/h

Let x = time taken to cover the 12 km

Since the same speed was used to cover the 12 km, 12/x should give the same speed (8 km/h)

⇒ 12/x = 8

Multiply both sides by x

12 = 8 x x

Divide both sides by 8

12/8 = x

⇒ x = 3/2 = 1½

A prime is a number that is greater than 1 and divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11,...)

List of prime numbers to 100

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

The prime numbers from the 25 to 35 list are only 29, 31

Probability = No. of possible outcomes / Total numbers (items)

There are two possibilities (29 and 31)

Count the total numbers (25,26,27,28,29,30,31,32,33,34,35)

Total numbers(items) = 11

⇒ Probability of selecting a prime number = 2 / 11

The easiest way to express a fraction as decimal fraction is to covert to percentage and then divide by 100

12/25 to percentage is simply multiplying by 100

12/25 x 100 = 48%

NOTE: 25 goes into 100, 4 times and 4 x 12 = 48

Now change the percentage to fraction

48% = 48/100 = 0.48

NOTE: 100 = 10 x 10 = 10²

From the law of indices, 1/aⁿ = a^-n

48/100 = 48/(10²) = 48 x 1/(10²)= 48 x 10^-2 = 48.0 x 10^-2

NOTE: Every number has .0 behind. Thus 48 = 48.0

Now change the 48.0 x 10^-2 to decimals by moving the decimal point by the value of the power. If the power is negative you move to the left the value of the power times and if positive to the right. In 48.0 x 10^-2,the decimal point is moved to the left 2 times which gives 0.48

Circumference of a circle = 2πr

Diameter = 2 x radius = 2r

⇒ Circumference = π x diameter

π = 22/7

Circumference = 88

⇒ 88 = 22/7 x diameter

Multiply both sides by 7

88 x 7 = 22 x diameter

Divide both sides by 22

88 x 7 / 22 = diameter

NOTE: 22 goes into 88, 4 times

4 x 7 = diameter

28 = diameter

⇒ the diameter is 28 cm

Let x = the number

three times a certain number = 3 x x

Subtracted from three times a certain number = 3 x x - 12 = 3x - 12

The result is divided by four = (3x-12)/4

is means =, hence the answer is 18 ⇒ (3x-12)/4 = 18

Now solve for the value of x

(3x-12)/4 = 18

Multiply both sides by 4 to get rid of the fraction

3x-12 = 18 x 4

3x-12 = 72

3x = 72 + 12

3x = 84

Divide both sides by 3

3x/3 = 84/3 = 28

⇒ x = 28

Concepts

Transversal Angles

Alternate Angles

∠MOS = ∠NOT = 115^o(Transversal angles)

∠MOT = ∠OTU (Alternate angles)

∠MOT + ∠NOT = 180 (Angles on a straight line add up to 180)

But ∠NOT = ∠MOS = 115^o

∠MOT + 115 = 180

∠MOT = 180 - 115

∠MOT = 65^o

But ∠MOT = ∠OTU = 65^o

⇒ ∠OTU = 65^o

For a vector, simplify the tops by the operator and write the result at the top.

Simplify the downs by the operator and write the result at the down

The operator is +, hence r + t = (^-3+3 _{-5 + - 5}) = (⁰ _-10)

Try r - t

⇒ r - t = (^-3-3 _{-5 - - 5}) = (^-6 ₀)

NOTE: -5 -- 5 = -5 + 5 = 0

Let x = amount for selling the 90 oranges

3 = 0.75

90 = x

Cross multiply. Left side multiplying right and right side multiplying left

3 x x = 90 x 0.75

Divide both sides by 3

3 x x/3 = 90 x 0.75/3

x = 90 x 0.75/3

Change the decimal to the format, whole number x 10^a

0.75 = 75 x 10^-2 (When you move the decimal point to the right its negative to the number of times you moved)

⇒ x = 90 x 75 x 10^-2/3

3 goes into 90, 30 times

x = 30 x 75 x 10^-2

30 x 75 = 2250

⇒ x = 2250 x 10^-2

Now change the whole number x 10^a format back to decimal

Since the power in 10^-2 is negative 2, you have to move the decimal point to the left two times

2250 = 2250.0 (Every number has .0 which is not written)

2250.0 x 10^-2 = 22.50

⇒ x = GH₵ 22.50

A prime is a number that is greater than 1 and divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11,...)

Write the 72 as a multiple of a number and further break the numbers down until you express it in the simplest factors

72 = 36 x 2

72 = 6 x 6 x 2 (36 = 6 x 6)

72 = 3 x 2 x 3 x 2 x 2(6 = 3 x 2)

Since its in the simplest form, express them as powers

Applying a^m x aⁿ = a^{m + n}

3 x 2 x 3 x 2 x 2 = 2 x 2 x 2 x 3 x 3

Every number is raised to the power 1 which is not written. Thus a = a¹

⇒ 2 = 2¹, 3 = 3¹

⇒ 2 x 2 x 2 x 3 x 3 = 2¹ x 2¹ x 2¹ x 3¹ x 3¹

2¹ x 2¹ x 2¹ x 3¹ x 3¹ = 2¹⁺¹⁺¹ x 3¹⁺¹ = 2³ x 3²

Concepts

a^m x aⁿ = a^{m + n}

a = a¹(Every number or letter is raised to the power 1 which is not written)

ab = a x b

3a x 24ab = 3 x 24 x a x a x b

3 x 24 = 72

a x a = a¹ x a¹ = a¹⁺¹ = a²

3a x 24ab = 3 x 24 x a x a x b

⇒ 3a x 24ab = 72a² x b = 72a²b

For a vector, simplify the tops by the operator and write the result at the top.

Simplify the downs by the operator and write the result at the down

The operator is +, hence the addition of the two vectors = (^-2+-1 _{3 + 5}) = (^-3 ₈)

Try subtraction (-) as an exercise

(^-2--1 _{3 - 5}) = (^-2+1 _-2) = (^-1 _-2)

Change the decimals to the form whole number x 10^a where a is the number of times you move the decimal point to get the whole number. If the decimal point is moved to the right, the sign is negative

0.07 = 7 x 10^-2

0.02 = 2 x 10^-2

⇒ (0.07 x 0.02) = 7 x 10^-2 x 2 x 10^-2

(0.07 x 0.02) ÷ 14 = 7 x 10^-2 x 2 x 10^-2 ÷ 14

7 goes into 14, 2 times and 2 goes into itself 1

7 x 10^-2 x 2 x 10^-2 ÷ 14 = 1 x 10^-2 x 1 x 10^-2/1

1 x 10^-2 x 1 x 10^-2/1 = 1 x 10^-2 x 10^-2

Applying a^m x aⁿ = a^{m + n}

⇒ 1 x 10^-2 x 10^-2 = 1 x 10^-2+-2 = 1 x 10^-4

Now change the whole number x 10^a format back to decimal

Since the power in 10^-4 is negative 4, you have to move the decimal point to the left four times

1 = 1.0 (Every number has .0 which is not written)

1.0 x 10^-4 = 0.0001

Let x = number of boys in the class

The girls were 7 more than the boys ⇒ Girls = Boys + 7 = x + 7

Class of 23 students = Number of boys + Number of girls

⇒ x + (x + 7) = 23

x + x + 7 = 23

2x = 23 - 7

2x = 16

Divide both sides by 2

2x/2 = 16/2

x = 8

⇒ there were 8 boys in the class

Before you express in the percentage, they must all be in the same units

Change the hours to minutes

1 hour = 60 minutes

3 hours = 3 x 60 minutes = 180 minutes

3 hours 20 minutes = 180 minutes + 20 minutes = 200 minutes

Percentage = Number/Total x 100

30 minutes of 3 hours 20 minutes (200 minutes) in percentage,

30 / 200 x 100 = 15

100 goes into 200, 2 times and 2 goes into 30, 15 times

Write down the multiples for each. The Least Common Multiple is the least number which occurs in all of the list of multiples

Multiples of 2

2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34

Multiples of 3

3,6,9,12,15,18,21,24,27,30,33,36

Multiples of 5

5,10,15,20,25,30,35,40

The least number which appeared in all 3 multiples is 30, hence the L.C.M for 2,3 and 5 is 30

Where S.I = Simple Interest
P = Principal amount (Amount invested or loan taken/Initial value)
R = Rate
T = Time (years)

You can memorize all the formula or memorize only the main one and learn how to make the rest the subject

R = (S.I x 100)/(P x T)
S.I = 45
P = 450
T = 4

⇒ R = (45 x 100)/(450 x 4) = 5/2 = 2.5%

NOTE: 45 goes into 450, 10 times and 10 goes into 100, 10 times and 2 goes into 10, 5 times and 4, 2 times. Then divide 5 by 2

The median is the middle score in a set of given numbers.

The mean is the arithmetic average of a set of given numbers.

The mode is the most frequently occurring score in a set of given numbers.

Median

The median is the middle score of a distribution. To calculate the median
* Arrange your numbers in numerical order.
* Count how many numbers you have.
* If you have an odd number, divide by 2 and round up to get the position of the median number.
* If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.

In the question above, there are 6 numbers. 6 is even and so to get the median position, divide 6 by 2, which falls on the 3^rd number.

The numbers are not ordered so you have to order them from the lowest to the highest before determing the 3^rd position

⇒ 46, 68, 34, 37, 76, 81 = 34, 37, 46, 68, 76, 81

The 3^rd number from the ordered numbers is 46

Since the total number count (6) is even, the median is the average of the 3^rd and its next number

Average simply means the sum of the two numbers divided by 2

So the median will be the 3^rd number + the 4^th number divided by 2

⇒ the median = (46+68)/2 = 114/2 = 57

If the total number count is odd, the median is simply the middle number in the ordered numbers

Mean

The mean, or average, is calculated by adding up the scores and dividing the total by the number of scores. Consider the following number set: 3, 4, 6, 6, 8, 9, 11.

The mean is calculated in the following manner:
3 + 4 + 6 + 6 + 8 + 9 + 11 = 47
number of scores is 7
⇒ The mean = 47 / 7 = 6.7

The mean (average) of the number set is 6.7.

Mode

Since the mode is the most frequently occurring score in a distribution, simply select the most common score as your mode.

Consider the following number distribution of 2, 3, 6, 3, 7, 5, 1, 2, 3, 9.

The mode of these numbers would be 3 since three(3) is the most frequently occurring number.

In some number sets, there may actually be two modes. This is known as bi-modal distribution and it occurs when there are two numbers that are tied in frequency.

For example, consider the following set of numbers: 13, 17, 20, 20, 21, 23, 23, 26, 29, 30.

In this set, both 20 and 23 occur twice.

If no number in a set occurs more than once, then there is no mode for that set of data.

M ∩ N = {2,7}

Intersection (∩) is where the two circles join

Try M ∪ N as exercise

M ∪ N = {1,2,3,4,5,6,7,8,9}

Union (∪) you list all the elements (members) in both sets. If a member appear in both sets, its written only once

Elements of N are {1,2,4,6,7,9}. If you count the members you will get 6

Exercises

a) Find M ∩ N
b) Find M ∪ N
c) Elements of M
d) How many members are in the set M?

Answers to exercises

a) M ∩ N = {2,7}

Intersection (∩) is where the two circles join

b) M ∪ N = {1,2,3,4,5,6,7,8,9}

Union (∪) you list all the elements (members) in both sets. If a member appear in both sets, its written only once

c) M = {2,3,5,7,8}

d) 5