Union (∪) is the list of elements from each sets. If an element occurs more than one, it is written only once.

Changing a decimal to percentage is simply multiplying the decimal by 100

1.25 x 100 = 1.⌒2⌒5. x 100 = 125%

Comparing/Ordering Decimals

Method I

Place the decimals in place value chart and compare them.

Examine the digit in each of the place values to order your decimals.

Ascending order

0.03,0.3,0.301,0.33

Method II

Step 1

Change the decimals to whole numbers by multiplying each decimals by 1n0s

Where n0s means number of zeros to join the 1 to obtain 10,100,1000,10000 etc.

The n0s is the highest number of digits after the decimal point of one of the decimals you are to order.

The decimal 0.301 has the highest number of digits (3) after the decimal points, hence we multiply each of the decimals by 1000

Step 2

Order the whole numbers obtained

Step 3

Replace the whole numbers by their corresponding decimal numbers.

Step 1

0.301 = 0.301 x 1000 = 0.⌒3⌒0⌒1. x 1000 = 301

0.3 = 0.3 x 1000 = 0.⌒3⌒0⌒0. x 1000 = 300

0.33 = 0.33 x 1000 = 0.⌒3⌒3⌒0. x 1000 = 330

0.03 = 0.03 x 1000 = 0.⌒0⌒3⌒0. x 1000 = 30

Step 2

Ascending order → 30,300,301,330

Step 3

0.03,0.3,0.301,0.33

-- = + or - (-) = +

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

53 - (-7) + (-15) = 53 + 7 - 15

53 - (-7) + (-15) = 60 - 15

53 - (-7) + (-15) = 45

Intersection (∩) is the list of elements that can be found in both sets (A and B)

A∩B = {}

There are no element in both sets, hence the intersection is an empty set. The number of elements in the intersection of A and B is therefore 0.

2114_five = 2 x 5³ + 1 x 5² + 1 x 5¹ + 4 x 5⁰

2114_five = 2 x 5 x 5 x 5 + 1 x 5 x 5 + 1 x 5 + 4 x 1

Note: a⁰ = 1, 5⁰ = 1, a¹ = a, 5¹ = 5

2114_five = 2 x 125 + 1 x 25 + 5 + 4

2114_five = 250 + 25 + 9

2114_five = 284

Method I

Using the law of indices

Change the 4² to base 2 so that the base 2s can simplify each other.

4 = 2 x 2

4² = (2 x 2)²

But (a^m x bⁿ)^c = a^{m x c} x x b^{n x c}

2 = 2¹

(2 x 2)² = (2¹ x 2¹)²

(2 x 2)² = 2^{1 x 2} x 2^{1 x 2}

(2 x 2)² = 2² x 2²

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

(2 x 2)² = 2²⁺²

(2 x 2)² = 2⁴

Alternate Solution

(a^m)ⁿ = a^{m x n}

4 = 2 x 2 = 2²

4² = (2²)² = 2^{2 x 2} = 2⁴

$\frac{2^{2}x{3}^2}{4^{2}x{3}^3}$ = $\frac{2^{2}x{3}^2}{2^{4}x{3}^3}$

Law of division of indices

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

$\frac{2^{2}x{3}^2}{2^{4}x{3}^3}$ = 2^2-4 x 3^2-3 = 2^-2 x 3^-1

But a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

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

2² = 2 x 2 = 4
3¹ = 3

2^-2 x 3^-1 = $\frac{1}{4}$ x $\frac{1}{3}$ = $\frac{\mathrm{1\; x\; 1}}{\mathrm{4\; x\; 3}}$ = $\frac{1}{12}$

Method II

Change the indices to multiplication, cancel out and simplify

2² = 2 x 2
4² = 4 x 4
3² = 3 x 3
3³ = 3 x 3 x 3

$\frac{2^{2}x{3}^2}{4^{2}x{3}^3}$ = $\frac{\mathrm{2\; x\; 2\; x\; 3\; x\; 3}}{\mathrm{4\; x\; 4\; x\; 3\; x\; 3\; x\; 3}}$ = $\frac{\mathrm{1\; x\; 1\; x\; 1\; x\; 1}}{\mathrm{2\; x\; 2\; x\; 1\; x\; 1\; x\; 3}}$ = $\frac{1}{\mathrm{4\; x\; 3}}$ = $\frac{1}{12}$

Change the hour to minutes and use proportionality to solve.

1 hour = 60 minutes

If 45 minutes = 150 litres
60 minutes = $\frac{\mathrm{150\; litres\; x\; 60\; minutes}}{\mathrm{45\; minutes}}$ = 50 x 4 = 200 litres

Note: 15 divides 45, 3 times and 60, 4 times. 3 divides itself once and 150, 50 times.

Method I

Using indices and cancellations

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{36}{27}$ x a^3-1 x b^2-3 x $\frac{\mathrm{x}}{\mathrm{y}}$

Note: 9 divides 36, 4 times and 27, 3 times.

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{4}{3}$ x a² x b^-1 x $\frac{\mathrm{x}}{\mathrm{y}}$

But a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

b^-1 = $\frac{1}{{\mathrm{b}}^1}$ = $\frac{1}{\mathrm{b}}$

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{4}{3}$ x a² x $\frac{1}{\mathrm{b}}$ x $\frac{\mathrm{x}}{\mathrm{y}}$

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{\mathrm{4\; x} {\mathrm{a}}^{2}x\; 1\; x \mathrm{x}}{\mathrm{3\; x\; b\; x\; y}}$

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{4{\mathrm{a}}^2\mathrm{x}}{\mathrm{3by}}$

Method II

Change the indices to multiplication, cancel out and simplify

a³ = a x a x a
b² = b x b
b³ = b x b x b

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{\mathrm{36\; x\; a\; x\; a\; x\; a\; x\; b\; x\; b\; x\; x}}{\mathrm{27\; x\; a\; x\; b\; x\; b\; x\; b\; x\; y}}$

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{\mathrm{4\; x\; 1\; x\; a\; x\; a\; x\; 1\; x\; 1\; x\; x}}{\mathrm{3\; x\; 1\; x\; 1\; x\; 1\; x\; b\; x\; y}}$

$\frac{36{\mathrm{a}}^3{\mathrm{b}}^{2}x}{27a{\mathrm{b}}^{3}y}$ = $\frac{4{\mathrm{a}}^2\mathrm{x}}{\mathrm{3by}}$

Finding the rule of a mapping

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 = 5

y = 5 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 8-5 = 11-8 = 14 - 11 = 17 -14 = 3

We can substitute our d and simplify the expression for y

y = 5 + (x-1)d

y = 5 + (x-1)3

y = 5 + 3 x x -1 x 3

y = 5 + 3x - 3

y = 5 - 3 + 3x

y = 2 + 3x

y = 3x + 2

y → 3x + 2

–1 = 2 – m

Take the m to the left side of the equation and -1 to the right.

Both of them are negative, hence when moved to the other side becomes positive.

m = 2 + 1
m = 3

Perimeter is the sum of each of the lengths of a shape.

Rectangle has two equal lengths and two equal widths.

Let w = width of the rectangle.

Perimeter = 2 x 14 cm + 2 x w

But perimeter = 48 cm

48 cm = 28 cm + 2w

48 cm - 28 cm = 2w

20 cm = 2w

Divide both sides of the equation by 2

w = $\frac{\mathrm{20\; cm}}{2}$ = 10 cm

n = 2d + 3

n - 3 = 2d

Divide both sides of the equation by 2

$\frac{\mathrm{n\; -\; 3}}{2}$ = d

d = $\frac{\mathrm{n\; -\; 3}}{2}$

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

Y₂ = 3
Y₁ = 5
X₂ = -2
X₁ = 3

Gradient = $\frac{\mathrm{3-5}}{\mathrm{-2-3}}$ = $\frac{-2}{-5}$ = $\frac{2}{5}$

Note: The negatives cancel each other.

The triangle above is an isosceles triangle. It has two of its sides equal.

Angles at the base of Isosceles triangle are the same.

a = 50^o

The triangle above is an isosceles triangle. It has two of its sides equal.

Angles at the base of Isosceles triangle are the same.

a = 50^o

Sum of angles in a triangle is 180^o

a + 50 + angle PRQ = 180

a = 50

50 + 50 + angle PRQ = 180

100 + angle PRQ = 180

Angle PRQ = 180 - 100

Angle PRQ = 80

The sum of angles on a straight line is 180^o

Angle PRQ + b = 180

But angle PRQ = 80

80 + b = 180

b = 180 - 80

b = 100

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

Odd numbers are numbers which are not divisible by 2 without a remainder.

The list of odd numbers from the set S are {1,3,5,7,9}

Number of possible selection of an odd number = 5

Number of samples = 10

Probability of = $\frac{5}{10}$ = $\frac{1}{2}$

Point + Translation Vector = Translated Point

Translated Point = (3,-2) = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

Point = (4,-5) = $\left(\begin{array}{c}4\\ -5\end{array}\right)$

Let the translation vector = $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$

$\left(\begin{array}{c}4\\ -5\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

Addition of vectors

$\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{c}\\ \mathrm{d}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{a\; +\; c}\\ \mathrm{b\; +\; d}\end{array}\right)$

$\left(\begin{array}{c}4\\ -5\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

$\left(\begin{array}{c}\mathrm{4\; +\; x}\\ \mathrm{-5\; +\; y}\end{array}\right)$ = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

The top at the left is equal to the top at the right and the down at the left is equal to the down at the right

4 + x = 3

x = 3 - 4

x = -1

-5 + y = -2

y = -2 + 5

y = 3

The translation vector = $\left(\begin{array}{c}-1\\ 3\end{array}\right)$

Bring out the common factors or expressions outside

2xy – 6y + 7x – 21

2xy – 6y + 7x – 21 = 2y(x - 3) + 7(x – 3)

2xy – 6y + 7x – 21 = (x - 3)(2y + 7)

Area of a circle = πr²

Where r = radius

Diameter = 2 x radius

154 = $\frac{22}{7}$ x r²

Get rid of the fraction by multiplying both sides of the equation by the denominator (7)

154 x 7 = 22 x r²

Divide both sides of the equation by 22

$\frac{\mathrm{154\; x\; 7}}{22}$ = r²

Note: 22 divides itself once and 154, 7 times.

7 x 7 = r²

7² = r²

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

7 = r

Radius = 7 cm

Diameter = 2 x radius

Diameter = 2 x 7 cm = 14 cm

She goes to school and back everyday. She covers the distance twice (to school and back home) every day.

Total distance covered each day = 2 x 2345 m = 4690 m

Calculations

Kilo means 1000, hence divide the meters by 1000 to obtain the kilometers.

4690 m = $\frac{\mathrm{4,690\; m}}{1000}$ = 4.69 km

Calculations

1000 = 10 x 10 x 10 = 10³

4690 m = $\frac{\mathrm{4690\; m}}{1000}$ = $\frac{\mathrm{4690\; m}}{{10}^3}$ = 4690 x 10^-3

Note: From law of indices a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

$\frac{1}{{10}^3}$ = 10^-3

Change the whole number x 10^-n form to decimal by moving the decimal point by the value of the power to the left.

4690 = 4690.0 (Every whole number has .0 which isn't written)

4690 x 10^-3 = 4.⌒6⌒9⌒0.0 x 10^-3 = 4.69 km

In summary, any time you are dividing by 1n0s, you move the decimal point of the number you are dividing n times to the left.

Where n0s means number of zeros after 1

In the above example 1000 has three (3) zeros hence we have to move the decimal point of 4690.0 three times to the left.

Length:Width = 5:3

Since you know the value of the length, you can use proportionality to get the width.

If the ratio 5 = 25 m
3 = $\frac{\mathrm{25\; m\; x\; 3}}{5}$ = 5 m x 3 = 15 m

$\frac{20}{\mathrm{a}}$ - b

a = 30 and b = 1

Substitute the values into the expression and simplify.

$\frac{20}{\mathrm{a}}$ - b = $\frac{20}{30}$ - 1

Every whole number is being divided by 1 which is not written.

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

$\frac{20}{\mathrm{a}}$ - b = $\frac{20}{30}$ - $\frac{1}{1}$

$\frac{20}{30}$ - $\frac{1}{1}$ = $\frac{\mathrm{1\; x\; 20\; -\; 1\; x\; 30}}{30}$

$\frac{20}{30}$ - $\frac{1}{1}$ = $\frac{\mathrm{20\; -\; 30}}{30}$

$\frac{20}{30}$ - $\frac{1}{1}$ = $\frac{-10}{30}$ = -$\frac{1}{3}$

Note: 10 divides itself once and 30, 3 times or the zeros cancel each other.

You have to divide the Ghana pesewas by 100 to get the cedis.

15 Gp = GH₵$\frac{15}{100}$

Number of cards that can be bought = $\frac{\mathrm{Total\; Amount}}{\mathrm{Cost\; of\; each\; card}}$

Number of cards that can be bought = GH₵18 ÷ $\frac{\mathrm{GH₵15}}{100}$

Change the division to multiplication by reciprocating the numerator and the denominator and changing the sign from division to multiplication.

Number of cards that can be bought = GH₵18.00 x $\frac{100}{\mathrm{GH₵15.00}}$

Number of cards that can be bought = 6 x 20

Number of cards that can be bought = 120

Note: 3 divides 18, 6 times and 15, 5 times. 5 divides itself once and 100, 20 times.

Addition of vectors

$\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{c}\\ \mathrm{d}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{a\; +\; c}\\ \mathrm{b\; +\; d}\end{array}\right)$

u = $\left(\begin{array}{c}6\\ 9\end{array}\right)$ and v = $\left(\begin{array}{c}4\\ -5\end{array}\right)$

u + v = $\left(\begin{array}{c}6\\ 9\end{array}\right)$ + $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{6\; +\; 4}\\ \mathrm{9\; +\; -5}\end{array}\right)$ = $\left(\begin{array}{c}10\\ 4\end{array}\right)$

Note: + - = -

Express 4956 × 25 = 123,900 in the form of 495.6 × 2.5

4956 = 4956.0 (Every whole number has .0)

25 = 25.0 (Every whole number has .0)

4956 = 495.6 x 10¹

Note: The decimal point is moved to the left once,hence the power is 1

25 = 2.5 x 10¹

495.6 x 10¹ × 2.5 x 10¹ = 123,900

495.6 x 2.5 x 10¹ × 10¹ = 123,900

a¹ = a, 10¹ = 10

495.6 x 2.5 x 10 x 10 = 123,900

10 x 10 = 100

495.6 x 2.5 x 100 = 123,900

Divide both sides by 100 so that you obtain 495.6 x 2.5 as requested by the question.

495.6 x 2.5 = $\frac{123900}{100}$

Note: The zeros (0) in 123900 and 100 cancel each other.

495.6 x 2.5 = 1239

495.6 x 2.5 = 1.239 x 10³ in standard form

Note:1239 = 1239.0

For standard form, the decimal point must be after the first non-zero digit. We must therefore move the decimal point three times to the left and multiply by 10^{number of moves (3)}.

1. When the arrow is to the left, its less than just like the arrow direction of less than and when to the right it is greater than.

2. The solid is placed on the value of the x inequality.

3. When the inequality is also equal to (Thus either ≤ or ≥), the solid is shaded.

From the image the arrow is on -2 and pointing left (less than direction) and the solid is shaded, hence less than or equal to (≤) -2.

The total ratios = 7+5 = 12

The total ratio represents the total oranges so you can use proportionality to calculate the oranges received by any of them.

Kwame:Ama = 7:5

Ama's ratio = 5

If 12 = 180 oranges
5 = $\frac{\mathrm{180\; oranges\; x\; 5}}{12}$ = 15 x 5 = 75 oranges

Note: 12 divides itself once and 180, 15 times.

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

Principal = GH₵450.00
Time = 2
Rate = 12

Simple Interest = $\frac{\mathrm{GH₵450.00\; x\; 2\; x\; 12}}{100}$ = GH₵9 x 1 x 12 = GH₵108.00

Note: 50 divides 450, 9 times and 100, 2 times. 2 divides itself once and the other 2, once.

Half is 50% of the rope. You can use proportionality to calculate the half of the rope.

If 15% = 75 cm
50% = $\frac{\mathrm{75\; cm\; x\; 50\%}}{\mathrm{15\%}}$ = 5 x 50 = 250 cm

Note: 15 divides itself once and 75, 5 times.

Total fractions paid by Tom and Azuma = $\frac{2}{3}$ + $\frac{1}{5}$ = $\frac{\mathrm{5\; x\; 2\; +\; 3\; x\; 1}}{15}$

Total fractions paid by Tom and Azuma = $\frac{\mathrm{10\; +\; 3}}{15}$

Total fractions paid by Tom and Azuma = $\frac{13}{15}$

The complete (whole) of the bill is 1

Remaining fraction paid by Tina = 1 - $\frac{13}{15}$

Every whole number is being divided by 1

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

Remaining fraction paid by Tina = $\frac{1}{1}$ - $\frac{13}{15}$

Remaining fraction paid by Tina = $\frac{\mathrm{15\; x\; 1\; -\; 1\; x\; 13}}{15}$

Remaining fraction paid by Tina = $\frac{\mathrm{15\; -\; 13}}{15}$

Remaining fraction paid by Tina = $\frac{2}{15}$

It shows the bisection of 60^o angle which gives 30^o.

Changing decimal to fraction

Every number is being divided by 1

0.055 = $\frac{0.055}{1}$

Count how many times you have to move the decimal point to obtain a whole number.

In 0.055, we have to move three (3) times.

Multiply both the decimal and 1 by 1nOs

Where nOs means number of zeros which is the number of times you have to move the decimal point to obtain a whole number.

In 0.055 we have to multiply both 0.055 and 1 by 1000, three (3) zeros after the 1 (1000).

Multiplying the decimal by 1000 makes it a whole number.

Simplify the numerator and the denominator.

$\frac{0.055}{1}$ = $\frac{\mathrm{0.055\; x\; 1000}}{\mathrm{1\; x\; 1000}}$

$\frac{0.055}{1}$ = $\frac{55}{1000}$

$\frac{0.055}{1}$ = $\frac{11}{200}$

Note: 5 divides 55, 11 times and 1000, 200 times.

Shoe making has 300,000 workers which is the highest.

Add all the number of workers in various trades.

Selling Price = Cost Price + Profit

Cost Price = GHC 80.00
Profit = GHC 13.60

Selling Price = GHC 80.00 + GHC 13.60
Selling Price = GHC 93.60

Calculation

Since the two figures are similar, figure PQRS is an enlargement of figure ABCD.

During an enlargement, each side is being multiplied by the same scale factor.

You can use any of the given sides to find the scale factor and use it to find the side of b.

Scale factor = $\frac{\mathrm{Image\; Size\; of\; similar\; side}}{\mathrm{Object\; Size\; of\; similar\; side}}$

The scale factor is the same for all similar sides.

Scale factor = $\frac{10}{6}$ = $\frac{20}{12}$ = $\frac{15}{9}$ = $\frac{\mathrm{b}}{18}$

Using any of the fractions to the left of $\frac{\mathrm{b}}{18}$, you can solve for the value of b by equating it to any of the fractions at the left since they are all the same.

$\frac{15}{9}$ = $\frac{\mathrm{b}}{18}$

Cross multiply.

b x 9 = 15 x 18

Divide both sides by 9

b = $\frac{\mathrm{15\; x\; 18}}{9}$ = 15 x 2 = 30 cm

Note: 9 divides itself once and 18, 2 times.

Total ratios = 2+3 = 5

Esi:Ato = 2:3

Ato's ratio = 3

Since we know Ato's ratio and amount received, we can use proportionality to get the total amount shared.

If 3 = GH₵ 45
5 = $\frac{\mathrm{5\; x\; 45}}{3}$ = 5 x 15 = GH₵75

Note: 3 divides itself once and 45, 15 times.

Disjoint sets are a pair of sets which does not have any common element (no intersection).