All the sets except {Seth, Mary, Jacob, Evelyn} contains different categories of items. Only the set {Seth, Mary, Jacob, Evelyn} contains only one category of item, names of people.

All members of a subset can be found in the set it is subset to.

The easiest method is the prime factor method. If you decide to list the multiples of each and pick the least, its going to be cumbersome. Thus you have to list so many before reaching the common multiple in each.

Prime factors of 16

$$\begin{array}{cc}\mathrm{}& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ 2& 1\end{array}$$

Prime factors of 16 = 2 x 2 x 2 x 2

Prime factors of 16 = 2⁴

Prime factors of 30

$$\begin{array}{cc}\mathrm{}& 30\\ 2& 15\\ 3& 5\\ 5& 1\end{array}$$

Prime factors of 30 = 2 x 3 x 5

Prime factors of 36

$$\begin{array}{cc}\mathrm{}& 36\\ 2& 18\\ 2& 9\\ 3& 3\\ 3& 1\end{array}$$

Prime factors of 16 = 2 x 2 x 3 x 3

Prime factors of 16 = 2² x 3²

The list common multiple is the product (multiplication) of each of the prime factors in their highest powers

L.C.M of 16, 30 and 36 = 2⁴ x 3² x 5

L.C.M of 16, 30 and 36 = 16 x 9 x 5
L.C.M of 16, 30 and 36 = 720

Sum means addition.

$\frac{\mathrm{5\; +\; x}}{4}$ = 3.25

3.25 = $\frac{325}{100}$

$\frac{\mathrm{5\; +\; x}}{4}$ = $\frac{325}{100}$

Cross multiply and solve for x

100 (5 + x) = 4 x 325

100 x 5 + 100 x x = 1300
500 + 100x = 1300
100x = 1300 - 500
100x = 800

Divide both sides by 100.

x = $\frac{800}{100}$

x = 8

Convert the numbers from base 5 to base 10 so you can know the order of the numbers to find the missing numbers in base ten and change back to base 5

32_{five = 3 x 5¹ + 2 x 5⁰}

Any number/letter raised to the power 0 is 1

32_{five = 3 x 5 + 2 x 1}

32_{five = 15 + 2}

32_{five = 17}

33_{five = 3x 5¹ + 3 x 5⁰}

33_{five = 3 x 5 + 3 x 1}

33_{five = 15 + 3}

33_{five = 18}

34_{five = 3x 5¹ + 4 x 5⁰}

34_{five = 3 x 5 + 4 x 1}

34_{five = 15 + 4}

34_{five = 19}

The next sequence order would be 20 and 21 so you can convert the 20 and 21 to base 5

20 base 10 to base 5

$$\begin{array}{cc}\mathrm{}& 20\\ 5& \mathrm{4\; r\; 0}\\ 5& \mathrm{0\; r\; 4}\end{array}$$

You list the remainders from the bottom to the top.

20 = 40_five

21 base 10 to base 5

$$\begin{array}{cc}\mathrm{}& 21\\ 5& \mathrm{4\; r\; 1}\\ 5& \mathrm{0\; r\; 4}\end{array}$$

You list the remainders from the bottom to the top.

21 = 41_five

The missing numbers are 40 and 41.

Note: There are no numbers ≥ 5 in base 5

Alternatively, when the number gets to 35 since there is no 5 in base 5, 5 is 10 in base 5 so you write the 0 and add the 1 to 3 making 40, then you continue adding 1 so the next sequence would be 41.

An integer is a number with no decimal or fractional part and it includes negative and positive numbers, including zero.

Total costs = 25 x GH₵ 0.20 + 75 x GH₵ 0.30

Note: The zeros at the end of a decimal number are ignored.

0.20, 0.200, 0.2000 = 0.2
0.30, 0.300, 0.3000 = 0.3

0.2 = $\frac{2}{10}$

0.3 = $\frac{3}{10}$

Total costs = 25 x $\frac{\mathrm{GH₵\; 2}}{10}$ + 75 x $\frac{\mathrm{GH₵\; 3}}{10}$

Total costs = $\frac{\mathrm{25\; x\; GH₵\; 2\; +\; 75\; x\; GH₵\; 3}}{10}$

Total costs = $\frac{\mathrm{GH₵\; 50\; +\; GH₵\; 225}}{10}$

Total costs = $\frac{\mathrm{GH₵\; 275}}{10}$

Total costs = GH₵ 27.5

- - = +

The negative (-) infront of a bracket affects each number in the bracket.

-27 + 18 - (10 - 14) - (-2) = - 27 + 18 - 10 - - 14 - -2
-27 + 18 - (10 - 14) - (-2) = - 27 + 18 - 10 + 14 + 2
-27 + 18 - (10 - 14) - (-2) = 18 + 14 + 2 - 27 - 10
-27 + 18 - (10 - 14) - (-2) = 34 - 37
-27 + 18 - (10 - 14) - (-2) = -3

Negative numbers are lesser than positive numbers. The higher the negative number, the lower the number. You can represent the given numbers on a number line. Numbers increases from the left to the right.

On the number line, 0.5 is between 0 and 1, hence 0.5 is greater than 0.

Number of pieces = $\frac{\mathrm{Total\; Length}}{\mathrm{Size\; to\; cut}}$

5$\frac{1}{2}$ = $\frac{\mathrm{5\; x\; 2\; +\; 1}}{2}$

5$\frac{1}{2}$ = $\frac{\mathrm{10\; +\; 1}}{2}$

5$\frac{1}{2}$ = $\frac{11}{2}$

Total length = 121 m

Size to cut = $\frac{11}{2}$ m

Number of pieces = 121 m ÷ $\frac{11}{2}$ m

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

Number of pieces = 121 x $\frac{2}{11}$

Note:
1. m cancels each other
2. 11 divides itself 1 and 121, 11 times
3. 11 x 2 = 22

Number of pieces = 22

The one decimal place is at 6 and since the number after 6 (3) is less than 4, we don't add anything to the 6.

198.635 = 198.6 (one decimal place)

Ratio of ages

Esi:Kwasi = 12:8

Note: 4 divides 12, 3 times and 8, 2 times.

Esi:Kwasi = 3:2

The sum of ratios represent the total mangoes shared.

Sum of ratios = 3 + 2
Sum of ratios = 5

Esi's ratio = 3

Use proportionality to find Esi's share.

If 5 = 60
3 = $\frac{\mathrm{3\; x\; 60}}{5}$

Note:
1. 5 divides itself 1 and 60, 12 times
2. 3 x 12 = 36

Esi got 36 mangoes

The more the students, the lesser the time.

Hence time (t) is inversely proportional to the number of students (s).

t ∝ $\frac{1}{\mathrm{s}}$

Remove the proportionality by introducing a constant (k).

t = k x $\frac{1}{\mathrm{s}}$

Use the known information to solve for k

6 students used 1 hour.

1 hour = 60 minutes.

60 minutes = k x $\frac{1}{6}$

k = 6 x 60

k = 360

The general equation is:

t = 360 x $\frac{1}{\mathrm{s}}$

t = $\frac{360}{\mathrm{s}}$

Calculate how long it will take 15 students to sweep.

s = 15

t = $\frac{360}{15}$

Note:
1. The hours was converted to minutes hence time obtained would be in minutes
2. 15 divides itself 1 and 360, 24 times

t = 24 minutes

It wiill take 15 students 24 minutes to sweep the compound.

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

Percentage commission = $\frac{\mathrm{GH₵\; 103,500}}{\mathrm{GH₵\; 690,000}}$ x 100%

Note:
1. The zeros (0) at the end of 100 and 690000 cancels each other
2. The zeros (0) at the end of 103500 and 6900 cancels each other
3. 69 divides itself 1, and 1035, 15 times

Percentage commission = 15%

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

Simple Interest = GH₵ 37,500
Principal = GH₵ 500,000
Time = 3 years

GH₵ 37,500.00 = $\frac{\mathrm{GH₵\; 500,000\; x\; 3\; x\; Rate}}{100}$

Get rid of the fraction by multiplying both sides by 100.

GH₵ 37,500 x 100 = GH₵ 500,000 x 3 x Rate

Divide both sides by GH₵ 500,000 x 3

Rate = $\frac{\mathrm{GH₵\; 37,500\; x\; 100}}{\mathrm{GH₵\; 500,000\; x\; 3}}$

Note:
1. The zeros (0) at the end of 500000 and 100 cancels each other
2. The zeros (0) at the end of 5000 and 37500 cancels each other
3. 3 divides itself 1 and 375, 125 times
4. 25 divides 125, 5 times and 50, 2 times
5 divided by 2 is 2.5

Rate = 2.5 %

Expand the bracket

(8x²y³)($\frac{3}{8}$xy⁴) = 8x²y³ x $\frac{3}{8}$xy⁴

8 cancels each other.

(8x²y³)($\frac{3}{8}$xy⁴) = x²y³ x 3xy⁴

Apply the law of indices multiplication.

a^m x aⁿ = a^{m + m}

Note: x = x¹

(8x²y³)($\frac{3}{8}$xy⁴) = 3 x x^{2 + 1}y^{3 + 4}

(8x²y³)($\frac{3}{8}$xy⁴) = 3x³y⁷

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

The distribution is not ordered and so order it first.

12, 19, 24, 39, 45, 51, 54, 67, 75, 81

There are two numbers at the middle, 45 and 51

Median = $\frac{\mathrm{45\; +\; 51}}{2}$

Median = $\frac{96}{2}$

Median = 48

Angle of sector = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 360^o

The total is 100 since the data is %. Thus if you sum all the %, you will get 100%.

Percentage for maize = 35%

Angle of sector for maize = $\frac{35}{100}$ x 360^o

Note:
1. The zeros (0) at the end of 100 and 360 cancels each other
2. 5 divides 10, 2 times and 35, 7 times
3. 2 divides itself 1 and 36, 18 times
4. 7 x 18 = 126

Angle of sector for maize = 126^o

Cards = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29}

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

Number of samples = 19

Number of cards which contain the digit 2 = 11

P (Contain digit 2) = $\frac{11}{19}$

$\frac{\mathrm{x}}{4}$ + 1 = 5

Get rid of the fraction by multiplying both sides by 4

x + 1 x 4 = 5 x 4
x + 4 = 20
x = 20 - 4
x = 16

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

2x + 1 < 8
2x < 8 - 1
2x < 7

Divide both sides by 2

x < $\frac{7}{2}$

x < 3.5

x = {2, 3} since they are less than 3.5

7x - (10x + 3) ≥ -9
7x - 10x - 3 ≥ -9
-3x ≥ -9 + 3
-3x ≥ -6

Divide both sides by -3. When dividing by a negative, the inequality sign changes direction. ≥ becomes ≤

x ≤ $\frac{-6}{-3}$

x ≤ 2

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 = 7

y = 7 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 11 - 7 = 15 - 11 = 19 - 15 = 23 - 19 = 4

We can substitute our d and simplify the expression for y

y = 7 + (x-1)d

y = 7 + (x-1)4

y = 7 + 4 x x -1 x 4

y = 7 + 4x -4

y = 7 - 4 + 4x

y = 3 + + 4x

y = 4x + 3

Circumference of a circle = 2πr

Area of a circle = πr²

Where r = radius of the circle.

But area = 64 π cm²

πr² = 64 π cm²

π cancels each other

r² = 64 cm²

64 cm² = 8 cm x 8 cm
64 cm² = (8 cm)²

r² = (8 cm)²

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

r = 8 cm

Circumference of a circle = 2π x 8 cm

Circumference of a circle = 16 π cm

The cube is obtained from the shape square.

Line of symmetry

A line of symmetry is a line that cuts a shape exactly in half. In symmetrical shapes, like a regular polygon, this means that if you were to fold the shape along the line of symmetry, both halves would match exactly and be mirror images of each other.

Number of fits = $\frac{\mathrm{Bigger\; Volume}}{\mathrm{Smaller\; Volume}}$

Volume of rectangular box = Length x Width x Height
Volume of the rectangular box = 20 cm x 6 cm x 4 cm
Volume of the rectangular box = 480 cm³

Volume of cube = Length x Length x Length
Volume of the cube = 2 cm x 2 cm x 2 cm
Volume of the cube = 8 cm³

Number of fits = $\frac{480}{8}$

Number of fits = 60

Interior angle = $\frac{\mathrm{(n\; -\; 2)\; x\; 180}}{\mathrm{n}}$

Where n = number of sides of the polygon.

Interior angle = 120^o.

$\frac{\mathrm{(n\; -\; 2)\; x\; 180}}{\mathrm{n}}$ = 120

Get rid of the fraction by multiplying both sides by n.

(n - 2) x 180 = 120 x n

n x 180 - 2 x 180 = 120 x n

180n - 360 = 120 x n
180n - 120 x n = 360
60n = 360

Divide both sides by 60

n = $\frac{360}{60}$

Note:
1. The zero (0) at the end of 360 and 60 cancels each other
2. 6 divides itself 1 and 36, 6 times

n = 6

Triangle SQP is isosceles triangle. The base angles of isosceles triangle are the same.

Angle SQP = Angle SPQ

Angle SQP + 112 = 180 (Sum of angles on a straight line is 180)
Angle SQP = 180 - 112
Angle SQP = 68

Angle SQP + Angle SPQ + Angle QSP = 180(Sum of angles in a triangle is 180)

Angle SQP = Angle SPQ = 68
Angle QSP = x

68 + 68 + x = 180
136 + x = 180
x = 180 - 136
x = 44^o

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypotenuse (c)

From the triangle, the side facing the right-angle is y, hence y is the hypotenuse (c)

z² + x² = y²

z² = y² - x²

Before you can express as a percentage, both must be in the same unit. Is either you change the minutes to hours or hours to minutes.

Change the seconds to minutes. 60 seconds make 1 minute. Hence 30 seconds is half ($\frac{1}{2}$) minutes.

7 min. 30 sec. = 7 minutes + $\frac{1}{2}$ minutes.

7 min. 30 sec. = 7$\frac{1}{2}$ minutes

7 min. 30 sec. = $\frac{\mathrm{2\; x\; 7\; +\; 1}}{2}$ minutes

7 min. 30 sec. = $\frac{\mathrm{14\; +\; 1}}{2}$ minutes

7 min. 30 sec. = $\frac{15}{2}$ minutes

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

1 hour = 60 minutes

Percentage = $\frac{15}{2}$ minutes ÷ 60 minutes x 100%

The minutes cancel each other.

Percentage = $\frac{15}{2}$ ÷ 60 x 100%

Every number/letter is being divided by 1 but not written.

60 = $\frac{60}{1}$

Percentage = $\frac{15}{2}$ ÷ $\frac{60}{1}$ x 100%

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

Percentage = $\frac{15}{2}$ x $\frac{1}{60}$ x 100%

Note: 15 divides itself 1 and 60, 4 times.

Percentage = $\frac{1}{2}$ x $\frac{1}{4}$ x 100%

Note: 4 divides itself 1 and 100, 25 times.

Percentage = $\frac{1}{2}$ x 25%

Percentage = $\frac{\mathrm{1\; x\; 25\%}}{2}$

Percentage = $\frac{\mathrm{25\%}}{2}$

Percentage = 12.5%

Point + Translation vector = Image point

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

Image = (3, 1) = $\left(\begin{array}{c}3\\ 1\end{array}\right)$

Let v = Translation vector

$\left(\begin{array}{c}4\\ 5\end{array}\right)$ + v = $\left(\begin{array}{c}3\\ 1\end{array}\right)$

v = $\left(\begin{array}{c}3\\ 1\end{array}\right)$ - $\left(\begin{array}{c}4\\ 5\end{array}\right)$

v = $\left(\begin{array}{c}\mathrm{3\; -\; 4}\\ \mathrm{1\; -\; 5}\end{array}\right)$

v = $\left(\begin{array}{c}-1\\ -4\end{array}\right)$

The side QT is the hypotenuse side of the object triangle (QST). The image side would also be the hypotenuse.

Hypotenuse is the side facing the right-angle or the longest side. The hypotenuse of the enlargement (image) is TR

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

The scale factor is the same for each corresponding sides.

Scale factor = $\frac{\mathrm{E\text{'}F\text{'}}}{\mathrm{EF}}$ = $\frac{\mathrm{F\text{'}G\text{'}}}{\mathrm{FG}}$ = $\frac{\mathrm{G\text{'}H\text{'}}}{\mathrm{GH}}$ = $\frac{\mathrm{E\text{'}H\text{'}}}{\mathrm{EH}}$

$\frac{\mathrm{12\; cm}}{\mathrm{EF}}$ = $\frac{\mathrm{20\; cm}}{\mathrm{5\; cm}}$

Cross multiply and solve for EF

20 x EF = 12 x 5

Divide both sides by 20

EF = $\frac{\mathrm{12\; x\; 5}}{20}$

Note
1. 5 divides itself 1 and 20, 4 times
2. 4 divides itself 1 and 12, 3 times

EF = 3 cm

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 number has .0 at the end which is not written

4037 = 4037.0

4037 = 4.⌒0⌒3⌒7.0 = 4.037 x 10³

Note: We moved three times (3) to the left, hence the power is positive 3.

Sum of angles on a straight line is 180^o

The arc D bisects the 180^o into 2 producing 90^o

The arc C also bisects the 90^o into 2 producing 45^o

a = $\left(\begin{array}{c}-5\\ 12\end{array}\right)$ and b = $\left(\begin{array}{c}\mathrm{10x}\\ 12\end{array}\right)$

a = b

$\left(\begin{array}{c}-5\\ 12\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{10x}\\ 12\end{array}\right)$

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

10x = -5

Divide both sides by 10

x = $\frac{-5}{10}$

Note: 5 divides itself 1 and 10, 2 times.

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

For clarity, the diagram has been relabelled.

Line WY is parallel to line QT

Angle UVS is equal to angle RAS (Alternate angles) = 40^o

Bearing starts from the North.

The bearing of B from A is the angle from P to S

Angle from P to S = Angle of straight line PR + Angle RAS

Angle on a straight line = 180^o

Angle of straight line PR = 180^o

Angle RAS = 40^o

The bearing of B from A = 180^o + 40^o

The bearing of B from A = 220^o

Ama's age = Kwame's age + 9

Kwame's age = 18

Ama's age = 18 + 9
Ama's age = 27

Kwame's age : Ama's age = 18 : 27

Note: 9 divides 18, 2 times and 27, 3 times.

Kwame's age : Ama's age = 2 : 3