Intersection (∩) are elements that can be found in both sets

P ∩ Q = {2, 4}

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(4 – 1) – (5 – 6)]

Simplify the brackets first

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[(3) – (-1)]

Note: - (-) = - x - = +. -(-1) = + 1

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[3 + 1]

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$[4]

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = $\frac{1}{2}$ x 4

Note: 2 divides itself 1 time and 4, 2 times.

$\frac{1}{2}$[(4 – 1) – (5 – 6)] = 2

The easiest way of finding the highest common factor (H.C.F) is using the prime factors method.

Prime factors of 35

$$\begin{array}{cc}\mathrm{}& 35\\ 5& 7\\ 7& 1\end{array}$$

Prime factors of 35 = 5 x 7

Prime factors of 70

$$\begin{array}{cc}\mathrm{}& 70\\ 2& 35\\ 5& 7\\ 7& 1\end{array}$$

Prime factors of 70 = 2 X 5 x 7

The highest (greatest) common factor is the product of the prime factors common in all in their lowest power.

The common prime factors are 5 and 7. Hence the H.C.F = 5 x 7 = 35

1101101_two = 1 x 2⁶ + 1 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰

Any number raised to the power 0 is 1

Any number raised to the power 1 is the same number

2² = 2 x 2 = 4

2³ = 2 x 2 x 2 = 8

2⁴ = 2 x 2 x 2 x 2 = 16

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64

1101101_two = 1 x 64 + 1 x 32 + 0 x 16 + 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1

1101101_two = 64 + 32 + 0 + 8 + 4 + 0 + 1

1101101_two = 109

The commutative property states that we can swap numbers over and still get the same answer.

The associative property states that it doesn't matter how we group the numbers (i.e. which we calculate first).

The distributive property states that multiplying the sum of two or more addends by a number is the same as multiplying each addend individually by the number and then adding the products together.

n² + 1 = 50

n² = 50 - 1

n² = 49

49 = 7 x 7 = 7²

n² = 7²

Since the powers are the same, the bases are also the same.

n = 7

Zeros(0) at the end of a decimal number is ignored. 0.70 = 0.7 and 9.00 = 9

Change the decimal to whole number x 10^-n.

Where n is the number of times you have to move the decimal point to the right to obtain a whole number.

0.63 = 0.⌒6⌒3. = 63 x 10^-2

0.7 = 0.⌒7. = 7 x 10^-1

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = $\frac{63x{10}^{-2}x7x{10}^{-1}}{9}$

Note: 9 divides itself 1 time and 63, 7 times.

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = 7 x 10^-2 x 7 x 10^-1

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = 7 x 7 x 10^-2 x 10^-1

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = 49 x 10^-2 x 10^-1

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = 49 x 10^{-2 + -1}

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = 49 x 10^-3

Change the whole number x 10^-n to decimal by moving the decimal point n times to the left.

Every whole number has .0 at the end but it is no written.

49 = 49.0

$\frac{\mathrm{0.63\; x\; 0.70}}{9.00}$ = .⌒0⌒4⌒9.0 = 0.049

$\frac{1}{\mathrm{R}}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$

R₁ = 1 and R₂ = 3

$\frac{1}{\mathrm{R}}$ = $\frac{1}{1}$ + $\frac{1}{3}$

$\frac{1}{\mathrm{R}}$ = $\frac{\mathrm{3\; x\; 1\; +\; 1\; x\; 1}}{3}$

$\frac{1}{\mathrm{R}}$ = $\frac{\mathrm{3\; +\; 1}}{3}$

$\frac{1}{\mathrm{R}}$ = $\frac{4}{3}$

Cross multiply

4 x R = 3 x 1

Divide both sides by 4

R = $\frac{\mathrm{3\; x\; 1}}{4}$

R = $\frac{3}{4}$

Numbers decrease from the right to the left (<) on the number line. Negative numbers are lesser than 0 and positive fractions are greater than 0.

So 0 is greater than - 9.

Now compare the fractions.

Arranging/Ordering Fractions

Before you can compare fractions, their denominators must be the same. Once they are the same, you can arrnage the fractions in the magnitude of their numerators.

To make the denominators the same, you must multiply both the numerator and the denominators of each fraction by the product (multiplication) of the denominators of the other fractions.

$\frac{2}{3}$ = $\frac{\mathrm{2\; x\; 5}}{\mathrm{3\; x\; 5}}$ = $\frac{10}{15}$

$\frac{3}{5}$ = $\frac{\mathrm{3\; x\; 3}}{\mathrm{5\; x\; 3}}$ = $\frac{9}{15}$

The magnitude of the numerators from the highest to lowest is 10, 9 hence the order of the fraction from highest to lowest is:

$\frac{2}{3}$, $\frac{3}{5}$

Hence the complete order is $\frac{2}{3}$, $\frac{3}{5}$, 0, -9

The mode is the one which occurs the most/ the one with the highest frequency/number of students

The mode is 4 because it occurred the most (3 times).

Numer of times of sales = $\frac{\mathrm{90\; oranges}}{\mathrm{3\; oranges}}$ = 30

Total sales = 30 x ₵50 = ₵1,500

Profit % = $\frac{\mathrm{Profit}}{\mathrm{Cost\; Price}}$ x 100%

Profit = Selling Price - Cost Price

Profit = ₵6.5 million - ₵5.5 million

Profit = ₵6. 5 x 1000,000 - ₵5.5 x 1000,000

Profit = ₵1 x 1000,000

Profit % = $\frac{\mathrm{₵1\; x\; 1000,000}}{\mathrm{₵5.5\; x\; 1000,000}}$ x 100%

Profit % = $\frac{\mathrm{1\; x\; 100\%}}{5.5}$

Profit % = $\frac{\mathrm{100\%}}{5.5}$

5.5 = $\frac{55}{10}$

Profit % = 100% ÷ $\frac{55}{10}$

Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.

Profit % = 100% x $\frac{10}{55}$

Profit % = $\frac{\mathrm{100\%\; x\; 10}}{55}$

Profit % = $\frac{\mathrm{1000\%}}{55}$

Note: 5 divides 10, 2 times and 55, 11 times.

Profit % = $\frac{\mathrm{200\%}}{11}$ = 18.18%

15.5 = $\frac{155}{10}$

If 1 cm = 4 km

$\frac{155}{10}$ cm = 4 km x $\frac{155}{10}$ cm ÷ 1 cm

Any number divided by 1 is the same number

$\frac{155}{10}$ cm = 4 km x $\frac{155}{10}$

$\frac{155}{10}$ cm = $\frac{\mathrm{4\; km\; x\; 155}}{10}$

$\frac{155}{10}$ cm = $\frac{\mathrm{620\; km}}{10}$ = 62 km

2(3a + 2b) = 2(3a + 2b)

2(3a + 2b) = 6a + 4b

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Percentage = $\frac{\mathrm{150\; marks}}{\mathrm{240\; marks}}$ x 100%

Note:
1. the zeros (0) at the end of 150 and 240 cancel each other
2. 3 divides 15,5 times and 24, 8 times
3. 4 divides 8, 2 and 100, 25 times
4. 5 x 25 = 125

Percentage = $\frac{125}{2}$

Mean/Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{14\; +\; 20\; +\; 25\; +\; 15\; +\; 28\; +\; 16}}{6}$

Mean = $\frac{118}{6}$

Mean = 19.67

2² × 6² = 2² × 6²

6 = 2 x 3

2² × 6² = 2² × (2 x 3)²

(a x b)^c = a^c x b^c

2² × 6² = 2² × 2² x 3²

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

2² × 6² = 2^{2 + 2} x 3²

2² × 6² = 2⁴ x 3²

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Mangoes left = 300 mangoes - 240 mangoes

Mangoes left = 60 mangoes

Percentage = $\frac{\mathrm{60\; mangoes}}{\mathrm{300\; mangoes}}$ x 100% = 20%

Note:
1. 60 divides itself 1 time and 300, 5 times
2. 5 divides itself 1 time and 100, 20 times

Probability = $\frac{\mathrm{Number\; of\; possible\; selection}}{\mathrm{Number\; of\; samples}}$

Total number of balls = 5 + 6 = 11

Number of white balls = 6

P(white ball) = $\frac{6}{11}$

15 – 2h = 6

15 - 6 = 2h

9 = 2h

Divide both sides by 2.

$\frac{9}{2}$ = h

h = 4.5

x → 3x + 7

3 → 3 x 3 + 7

3 → 9 + 7

3 → 16

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points on the straight line to calculate for the gradient

Y₂ = 8
Y₁ = 2
X₂ = 3
X₁ = 0

Gradient = $\frac{\mathrm{8\; -\; 2}}{\mathrm{3\; -\; 0}}$ = $\frac{6}{3}$ = 2

Simple Interest = $\frac{\mathrm{Principal\; x\; Time\; x\; Rate}}{100}$

Principal = ₵50,000
Time = 2
Rate = 20

Simple Interest = $\frac{\mathrm{₵50,000\; x\; 2\; x\; 20}}{100}$ = ₵500 x 40 = ₵20,000

Note:
1. the zeros (0) at the end of 50,000 and 100 cancel each other
2. 2 x 20 = 40

100% represents the total sales

If 20% = ₵800

100% = $\frac{\mathrm{₵800\; x\; 100\%}}{\mathrm{20\%}}$ = ₵800 x 5 = ₵4,000

Bearing starts from the north.

Bearing = 360° - 85° = 275°

The triangle is isosceles triangle. The base angles are the same.

Let the base angle = x

∠PQR = ∠PRQ = x

Sum of interior angles of a triangle is 180°

x + x + 40 = 180

2x = 180 - 40

2x = 140

Divide both sides by 2

x = $\frac{140}{2}$ = 70

∠PRQ = 70°

∠PRQ + ∠PRS = 180° (Sum of angles on a straight line is 180°)

70° + ∠PRS = 180°
∠PRS = 180° - 70°
∠PRS = 110°

2xy – 8x + 5y - 20 = 2xy – 8x + 5y - 20

2xy – 8x + 5y - 20 = 2x(y - 4) + 5(y - 4)

2xy – 8x + 5y - 20 = (2x + 5)(y - 4)

Degrees to face north = 90° + 90° + 90° = 270°

From pythagorean theorem,

The hypothenuse (c) is the longest side.

RT² = 4² + 3²

4² = 4 x 4 = 16

3² = 3 x 3 = 9

RT² = 16 + 9

RT² = 25

25 = 5 x 5 = 5²

RT² = 5²

Since the powers are the same, the bases are also the same.

RT = 5

A line of symmetry is the line that divides a shape or an object into two identical parts.

a = $\left(\begin{array}{c}-3\\ 4\end{array}\right)$ and b = $\left(\begin{array}{c}1\\ 2\end{array}\right)$

b - a = $\left(\begin{array}{c}1\\ 2\end{array}\right)$ - $\left(\begin{array}{c}-3\\ 4\end{array}\right)$

b - a = $\left(\begin{array}{c}\mathrm{1\; -\; -\; 3}\\ \mathrm{2\; -\; 4}\end{array}\right)$

Note: - - = +

b - a = $\left(\begin{array}{c}\mathrm{1\; +\; 3}\\ \mathrm{2\; -\; 4}\end{array}\right)$

b - a = $\left(\begin{array}{c}4\\ -2\end{array}\right)$

u = $\left(\begin{array}{c}3\\ -1\end{array}\right)$ and v = $\left(\begin{array}{c}6\\ 1\end{array}\right)$

2u + v = 2$\left(\begin{array}{c}3\\ -1\end{array}\right)$ + $\left(\begin{array}{c}6\\ 1\end{array}\right)$

2u + v = $\left(\begin{array}{c}\mathrm{3\; x\; 2}\\ \mathrm{-1\; x\; 2}\end{array}\right)$ + $\left(\begin{array}{c}6\\ 1\end{array}\right)$

2u + v = $\left(\begin{array}{c}6\\ -2\end{array}\right)$ + $\left(\begin{array}{c}6\\ 1\end{array}\right)$

2u + v = $\left(\begin{array}{c}\mathrm{6\; +\; 6}\\ \mathrm{-2\; +\; 1}\end{array}\right)$

2u + v = $\left(\begin{array}{c}12\\ -1\end{array}\right)$

Volume of a cuboid = Length x Breadth x Height

Length = 3 m
Breadth = 6m
Depth/Height = d

3 x 2 x d = 9

Divide both sides by 3 x 2

d = $\frac{9}{\mathrm{3\; x\; 2}}$ = $\frac{3}{2}$ = 1.5 m

Note: 3 divides itself 1 time and 9, 3 times.

Scale factor = $\frac{\mathrm{Enlarged\; Side}}{\mathrm{Corresponding\; Object\; Side}}$

Enlarged Sid = AE

Corresponding Object Side = AB

Scale factor = $\frac{\mathrm{Enlarged\; Side}}{\mathrm{Corresponding\; Object\; Side}}$

AB = 15 cm

AE = AB + BE

AE = 15 cm + 30 cm = 45 cm

Scale factor = $\frac{\mathrm{45\; cm}}{\mathrm{15\; cm}}$ = 3

Area of a sector = $\frac{\mathrm{Angle\; of\; the\; sector}}{360}$ x πr²

The entire angle is 360°

Angle for the shaded sector = 360° - 60° = 300°

Area of the sector = $\frac{300}{360}$ x πr²

Note: 60 divides 300, 5 times and 360, 6 times.

Area of the sector = $\frac{5}{6}$πr²

Area of a trapezium = $\frac{1}{2}$ x (a + b) x h

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x h

OR = 20 cm + 30 cm + 16 cm = 66 cm

Area of the trapezium = $\frac{1}{2}$ x (30 cm + 66 cm) x 20 cm

Note: 2 divides itself 1 time and 20, 10 times.

Area of the trapezium = 96 cm x 10 cm = 960 cm²