Prime numbers are numbers which has only two factors, thus only two numbers can divide without a remainder (1 and itself).

Note: 1 is not a prime number because 1 can only be divided by one number, 1 itself.

Expansion simply means multiply each of the terms outside the bracket by each of the terms inside the bracket.

3a(a – 4b) = 3a x a - 3a x 4b
3a(a – 4b) = 3a² - 12ab

Note: Every number is raised to the power 1 but it is not written.

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

a x a = a¹ x a¹
a x a = a¹⁺¹
a x a = a²

3 x 4 = 12

Expressing one number as a % of another number

You calculate one number as a percentage of another by dividing the numbers and multiplying by 100.

The percentage = $\frac{\mathrm{The\; number}}{\mathrm{The\; other\; number}}$ x 100%

$\frac{5}{4}$ x 100% = 5 x 25%

$\frac{5}{4}$ x 100% = 125%

Expressing a number as a product of prime numbers

1. Divide the number with the smallest prime factor until it cannot be divided any further

2. Move to the next prime factor. Keep on dividing until it cannot be divided any further.

3. Repeat step 2 until you get 1 as a remainder.

4. Write down the product (multiplication) of each of the prime factors in the division.

5. Simplify the indices using the multiplication law of indices.

Prime products of 2700

$$\begin{array}{cc}\mathrm{}& 2700\\ 2& 1350\\ 2& 675\\ 3& 225\\ 3& 75\\ 3& 25\\ 5& 5\\ 5& 1\end{array}$$

Count each of the prime numbers to get their exponents/power.

2700 = 2² x 3³ x 5²

The sum of the ratios correspond to the total items. Each ratio correspond to its number of items.

Ratio of mangoes = 3
Ratio of oranges = 2

If ratio 3 (mangoes) = 36

then ratio 2 (oranges) = $\frac{\mathrm{2\; x\; 36}}{3}$ = 24 oranges

Note: 3 divides itself one and 36, 12 times.

Change the decimal to fraction and simplify.

Changing decimal to fraction

1. Every number is divisible by 1, hence divide the decimal by 1

0.125 = $\frac{0.125}{1}$

2. Count how many digits are after the decimal point (.)

0.125 = 0.⌒1⌒2⌒5

There are 3 digits after the decimal point. Or the decimal point has to be moved 3 times to obtain a whole number.

3. Multiply both the numerator and the denominater of the fraction obtained in step 1 by 1n0s, where n0s is how many zeros (0) to join 1. The number of zeros (0) to join the 1 is how many count you obtained in step 2.

Since we obtained 3 counts, we have to multiply both the numerator and the denominator of the fraction obtained in step 1 by 1000.3 zeros (0) is joined to the 1 to obtain 1000 because we obtained three counts in step 2.

Multiplying the decimal by 1n0s changes it to a whole number.

$\frac{0.125}{1}$ = $\frac{\mathrm{0.125\; x\; 1000}}{\mathrm{1\; x\; 1000}}$

$\frac{0.125}{1}$ = $\frac{125}{1000}$

Simplifying the fraction obtained

125 divides itself once and 1000, 8 times.

Alternatively, 25 divides 125, 5 times and 1000, 40 times. 5 divides itself 1 time and 40, 8 times.

$\frac{0.125}{1}$ = $\frac{125}{1000}$ = $\frac{1}{8}$

222_five = 2 x 5² + 2 x 5¹ + 2 x 5⁰
222_five = 2 x 25 + 2 x 5 + 2 x 1

Note: Every number raised to the power 0 is 1.

Every number raised to the power 1 is the same number.

5² = 5 x 5 = 25

222_five = 50 + 10 + 2
222_five = 62

∩ is an intersection and is the list of elements which can be found in both set A and B.

There is no element that can be found in both set A and B. Hence A∩B is an empty set, {}.

Law of division of indices

$\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

The law states that the power of the indices dividing is subtrated from the powers of the indices of the numerator so far as the bases of the indices are the same.

$\frac{3^9}{3^3}$ = 3^{9 - 3}

$\frac{3^9}{3^3}$ = 3⁶

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

Thus profit over the cost price x 100%

Profit % = $\frac{\mathrm{35\; -\; 25}}{25}$ x 100%

Profit % = $\frac{10}{25}$ x 100% = 10 x 4 = 40%

Note: 25 divides itself once and 100, 4 times.

Factorization simply means bring out the common terms and simplify if possible.

b² = b x b

b² + fb – mb – fm = b x b + fb – mb – fm
b² + fb – mb – fm = b(b + f)-m(b + f)
b² + fb – mb – fm = (b + f)(b-m)

-(-) = - x - = + or -- = +

– (-3) = +3

+ - = -
+ (-10) = + - 10 = -10

-13 – (-3) + (-10) = -13 + 3 - 10
-13 – (-3) + (-10) = -10 - 10
-13 – (-3) + (-10) = -20

The highest common factor (HCF) is the product of each of the prime numbers in their lowest power.

HCF = 3² x 5²

Note: For HCF, lowest power and Least/Lowest Common Factor (LCM), highest power. So highest (HCF) is lowest and lowest (LCM) is highest.

Note: If the mapping has equal interval (difference) for the y and x interval of 1, you can use the general linear sequence formula to find the rule of the mapping.

U_n = U₁ + (n-1)d

In the mapping, U_n = y and n = x and U₁, the first value of y

U₁ = 15

y = 15 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 30-15 = 45-30 = 60 - 45 = 15

We can substitute our d and simplify the expression for y

y = 15 + (x-1)d

y = 15 + (x-1)15

y = 15 + 15 x x -1 x 15

y = 15 + 15x -15

y = 15x + 15 -15

y = 15x + 0

y = 15x

x → 15x

Get rid of the fraction by multiplying both sides of the equation by the L.C.M of the denominators. The L.C.M of 3 and 3 is 3.

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = 3 x x - 3 x $\frac{1}{3}$ ≥ 3 x $\frac{2}{3}$ - 3 x x

The denominators cancel out the ones multiplying.

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = 3x - 1 ≥ 2 - 3x

Group the x at the left and the numbers at the right. When one is negative, it becomes positive at the other side and when positive, negative.

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = 3x + 3x ≥ 2 + 1

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = 6x ≥ 3

Divide both sides by 6

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = x ≥ $\frac{3}{6}$

Note: Since we are not dividing by a negative number, the inequality sign remains the same.

x - $\frac{1}{3}$ ≥ $\frac{2}{3}$ - x = x ≥ $\frac{1}{2}$

Perimeter is the sum of all the length of each side of a shape. Square has 4 equal sides.

Perimeter of a square = Length + Length + Length + Length

Let L = the length of each sides of the square.

28 = L + L + L + L
28 = 4L

Divide both sides by 4 to get the length of the square.

L = $\frac{28}{4}$ = 7 cm

Area of a square = Length x Length
Area = 7 cm x 7 cm = 49 cm²

Simplify the ones in the bracket first.

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{1}{3}\left(\frac{\mathrm{3-2}}{6}\right)-\frac{1}{3}\left(\frac{\mathrm{2-3}}{6}\right)$

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{1}{3}\left(\frac{1}{6}\right)-\frac{1}{3}\left(\frac{-1}{6}\right)$

Expand the bracket. Thus multiply the one outside the bracket by the one in the bracket.

Note: - x - = +

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{1}{3}x\frac{1}{6}-\frac{1}{3}x\frac{-1}{6}$

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{1}{18}$ + $\frac{1}{18}$

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{\mathrm{1\; +\; 1}}{18}$

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{2}{18}$

2 divides itself once and 18, 9 times.

$\frac{1}{3}(\frac{1}{2}-\frac{1}{3})-\frac{1}{3}(\frac{1}{3}-\frac{1}{2})$ = $\frac{1}{9}$

Get rid of the fraction by multiplying both sides of the equation by the L.C.M of the denominators. The denominator is only n, hence the LCM is n. Multiply both sides of the equation by n.

θ = 180 - $\frac{360}{n}$

θ x n = 180 x n - $\frac{360}{n}$ x n

θ x n = 180 x n - 360

Group the terms with n at the right side and the number (360) at the left. When it is negative, it becomes positive at the other side and when positive, negative.

360 = 180 x n - θ x n

Factorize the n out.

360 = n(180 - θ)

Divide both sides by 180 - θ

$\frac{360}{180\; -\; \theta }$ = n

Substitute the values into the equation.

R = $\frac{h}{2}$ + $\frac{d^2}{\mathrm{8h}}$

d = 8 and h = 6

R = $\frac{6}{2}$ + $\frac{8^2}{\mathrm{8\; x\; 6}}$

8² = 8 x 8

R = $\frac{6}{2}$ + $\frac{\mathrm{8\; x\; 8}}{\mathrm{8\; x\; 6}}$

2 divides itself once and 6, 3 times. 8 at the denominator cancels one of the 8s at the numerator. 2 divides 8, 4 times and 6, 3 times.

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

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

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

R = (3+1)$\frac{1}{3}$

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

Method I

If 8 = GH₵ 16.00
then 5 = $\frac{\mathrm{5\; x\; GH₵\; 16.00}}{8}$ = 5 x GH₵ 2.00 = GH₵ 10.00

Method II

Cost per book = $\frac{\mathrm{GH₵\; 16.00}}{8}$ = GH₵ 2.00

5 books = GH₵ 2.00 x 5 = GH₵ 10.00

$\frac{1}{5}$(2 + y) = $\frac{1}{2}$(y -1)

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators (2 and 5). The L.C.M of 2 and 5 is 10.

10 x $\frac{1}{5}$(2 + y) = 10 x $\frac{1}{2}$(y -1)

2 x (2 + y) = 5 x (y -1)
2 x 2 + 2 x y = 5 x y - 1 x 5
4 + 2y = 5y - 5
4 + 5 = 5y - 2y
9 = 3y

Divide both sides by 3.

$\frac{9}{3}$ = y

3 = y

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

y₂ = 8
x₂ = 4
y₁ = 2
x₁ = 3

Gradient = $\frac{\mathrm{8\; -\; 2}}{\mathrm{4\; -\; 3}}$

Gradient = $\frac{6}{1}$

Gradient = 6

Method I

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Distance = Speed x Time

Speed is the same.

2$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 2\; +\; 1}}{2}$

2$\frac{1}{2}$ = $\frac{\mathrm{4\; +\; 1}}{2}$

2$\frac{1}{2}$ = $\frac{5}{2}$

Speed = 60 km per hour

Time = $\frac{5}{2}$

Distance = 60 km x $\frac{5}{2}$

2 divides itself 1 and 60, 30.

Distance = 30 km x 5
Distance = 150 km

Method II

Using proportionality.

60 km per hour means a distance of 60 km is covered in an hour.

If 1 hour = 60 km

$\frac{5}{2}$ hour = 60 km x $\frac{5}{2}$ ÷ 1 hour

$\frac{5}{2}$ hour = 30 km x 5

$\frac{5}{2}$ hour = 150 km

Method III

The car covers 60 km an hour

$\frac{1}{2}$ an hour the car will cover $\frac{\mathrm{60\; km}}{2}$

$\frac{1}{2}$ an hour the car will cover 30 km

2$\frac{1}{2}$ = 1 hour + 1 hour + $\frac{1}{2}$ hour

Distance covered in 2$\frac{1}{2}$ hours = 60 km + 60 km + 30 km

Distance covered in 2$\frac{1}{2}$ hours = 150 km

Angle ABR = Angle BCW (Corresponding Angles) = 128^o

α + Angle ABR = 180^o(Sum of angles on a straight line is 180)

α + 128 = 180
α = 180 - 128
α = 58

a = b

$\left(\begin{array}{c}5\\ \mathrm{2n}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{2n\; -1}\\ 6\end{array}\right)$

The top at the left equals the top at the right and the down at the left equals the down at the right.

Method I

Top at left equals top at the right.

5 = 2n - 1
5 + 1 = 2n
6 = 2n

Divide both sides by 2

$\frac{6}{2}$ = n

n = 3

Method II

Down at the left equals down at the right

2n = 6

Divide both sides by 2

n = $\frac{6}{2}$

n = 3

Volume of a cube = Length x Length x Length

Volume of cube = 5 cm x 5 cm x 5 cm
Volume of cube = 125 cm³

Sum of angles on a straight line is 180

65 + y = 180
y = 180 - 65
y = 115

The sum of the ratios represent the total amount shared.

Sum of ratios = 5 + 4
Sum of ratios = 9

Kwame's ratio = 5 and amount received = GH₵ 9.00

If 5 = GH₵ 9.00
9 = $\frac{\mathrm{9\; x\; GH₵\; 9.00}}{5}$

9 = GH₵ 16.2

∴ Amount shared = GH₵ 16.2

Area = $\frac{1}{2}$(9 cm + 21 cm) x 24 cm

Note: 2 divides itself 1 and 24, 12 times.

Area = 30 cm x 12 cm
Area = 360 cm²

r = $\left(\begin{array}{c}2\\ -5\end{array}\right)$ and s = $\left(\begin{array}{c}-2\\ 5\end{array}\right)$

2r - 3s = 2$\left(\begin{array}{c}2\\ -5\end{array}\right)$ - 3$\left(\begin{array}{c}-2\\ 5\end{array}\right)$

2r - 3s = $\left(\begin{array}{c}\mathrm{2\; x\; 2}\\ \mathrm{-5\; x\; 2}\end{array}\right)$ - $\left(\begin{array}{c}\mathrm{-2\; x\; 3}\\ \mathrm{5\; x\; 3}\end{array}\right)$

2r - 3s = $\left(\begin{array}{c}4\\ -10\end{array}\right)$ - $\left(\begin{array}{c}-6\\ 15\end{array}\right)$

2r - 3s = $\left(\begin{array}{c}\mathrm{4\; -\; -6}\\ \mathrm{-10\; -\; 15}\end{array}\right)$

- - = +

2r - 3s = $\left(\begin{array}{c}\mathrm{4\; +\; 6}\\ -25\end{array}\right)$

2r - 3s = $\left(\begin{array}{c}10\\ -25\end{array}\right)$

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

Total balls = 10 + 15
Total balls = 25

Red balls = 10

P(red) = $\frac{10}{25}$

Note: 5 divides 10, 2 times and 25, 5 times.

P(red) = $\frac{2}{5}$

Total number of students = 3 + 10 + 6 + 7 + 4
Total number of students = 30

Modal is the number with the highest frequency. The age with the highest number of student.

The modal age is 14 because it has the highest number of students.

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

Total number of students = 3 + 10 + 6 + 7 + 4
Total number of students = 30

Number of children who are 15 years = 6

P(15 years) = $\frac{6}{30}$

Note: 6 divides itself 1 and 30, 5

P(15 years) = $\frac{1}{5}$

Number of ribbons = $\frac{\mathrm{Length\; of\; ribbon}}{\mathrm{Length\; to\; cut}}$

Number of ribbons = $\frac{\mathrm{16.8\; m}}{\mathrm{0.36\; m}}$

Change the decimals to whole number x 10^-n

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

16.8 = 16.⌒8. = 168 x 10^-1

0.36 = 0.⌒3⌒6. = 36 x 10^-2

Number of ribbons = $\frac{168\mathrm{x}{10}^{-1}}{36\mathrm{x}{10}^{-2}}$

$\frac{168}{36}$ = 4.667

Law of division of indices

$\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

$\frac{{10}^{-1}}{{10}^{-2}}$ = 10^-1--2

$\frac{{10}^{-1}}{{10}^{-2}}$ = 10^{-1 + 2}

$\frac{{10}^{-1}}{{10}^{-2}}$ = 10¹

$\frac{{10}^{-1}}{{10}^{-2}}$ = 10

Number of ribbons = 4.667 x 10

Number of ribbons = 46.67

Number of ribbons ≈ 46

The cost price is 100%

Selling Price % = Cost Price % - Loss %
Selling Price % = 100% - 10%
Selling Price % = 90%

But selling price = GH₵ 200.00

If 90% = GH₵ 200.00

100% = $\frac{\mathrm{GH₵\; 200.00\; x\; 100\%}}{\mathrm{90\%}}$

100% = GH₵ 222.22

∴ Cost price = GH₵ 222.22

Pyramid

The number is eight hundred and thirty two thousand,seven hundred and thirteen

A standard form should be in the form a.bcde x 10ⁿ

The decimal point must be after the first non-zero digit.

The n is the number of times the decimal point has to be moved to be right after the first non-zero digit.

If it is moved to the left, the n is positive and if to the right, n is negative.

Every whole number has .0 at the end but not written.

3560 = 3560.0

3560 = 3.⌒5⌒6⌒0.0 = 3.56 x 10³

Note: the power is positive because we have to move to the left.

If the number after the three decimal places is 5 or more, you add 1 to the third place.

0.02751 = 0.028