B = {b, d, i, o, u}

B has 5 members.

A factor of a number is a number which can divide the number without a remainder.

3 divides 18, 6 times
6 divides 18, 3 times
9 divides 18, 2 times

4 divides 18, 4 times (4 x 4 = 16) with a remainder of 2.

Change the decimals 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.

0.014 = 0.⌒0⌒1⌒4. = 14 x 10^-3

0.2 = 0.⌒2. = 2 x 10^-1

0.014 x 0.2 = 14 x 10^-3 x 2 x 10^-1

0.014 x 0.2 = 14 x 2 x 10^-3 x 10^-1

Apply the law of multiplication of indices for the base 10

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

0.014 x 0.2 = 28 x 10^{-3 + -1} = 28 x 10^-4

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

Every whole number has a decimal point at the end which is not written.

28 = 28.0

In 28 x 10^-4, you are to move the decimal point 4 times to the left.

28 x 10^-4 = 0.⌒0⌒0⌒2⌒8.0 = 0.0028

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

The element(s) that can be found in both the sets A and B is/are {i}

Method I

2² = 2 x 2 = 4

a = 64 and b = 4

$\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{64}{4}$ = 16

Method II

Using indices.

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

a = 2⁶

$\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2^6}{2^2}$

Apply the law of division of indices.

Law of division of indices

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

$\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2^6}{2^2}$ = 2^{6 - 2} = 2⁴

2⁴ = 2 x 2 x 2 x 2 = 16

$\frac{\mathrm{a}}{\mathrm{b}}$ = 16

Use the number line to arrange them. Numbers decrease as you move to the left of the number line. Thus 0 is greater than -1 and - 1 is greater than -2 and so forth.

$\frac{2}{3}$ = 0.67 which is greater than 0 (is located between 0 and 1 on the number line) and 0 is also greater than -7.

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

2² × 3² × 2³ × 3⁴ = 2² × 2³ × 3² × 3⁴

2² × 3² × 2³ × 3⁴ = 2^{2 + 3} × 3^{2 + 4}

2² × 3² × 2³ × 3⁴ = 2⁵ × 3⁶

The total percentage of the students is 100%

Percentage that passed = 100% - Percentage that failed

Percentage that passed = 100% - 40% = 60%

Since you know the percentage that passed and the number, you can use proportionality to find those who failed.

If 60% = 180

40% = $\frac{\mathrm{180\; x\; 40\%}}{\mathrm{60\%}}$ = 3 x 40 = 120

Note: 60 divides itself 1 and 180, 3 times.

$\frac{1}{3}$[(5 – 1) – (2 – 7)] = $\frac{1}{3}$[(5 – 1) – (2 – 7)]

Simplify the brackets first.

$\frac{1}{3}$[(5 – 1) – (2 – 7)] = $\frac{1}{3}$[(4) – (-5)]

- - = +

$\frac{1}{3}$[(5 – 1) – (2 – 7)] = $\frac{1}{3}$[4 + 5]

$\frac{1}{3}$[(5 – 1) – (2 – 7)] = $\frac{1}{3}$ x 9 = 1 x 3 = 3

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

6n + 4 = 16
6n = 16 - 4
6n = 12

Divide both sides by 6

n = $\frac{12}{6}$ = 2

The next term is obtained by multiplying the preceding term by 2 and adding 1 to the result

For:

term 2 = 1 x 2 + 1 = 3
term 3 = 2 x 3 + 1 = 7
term 4 = 2 x 7 + 1 = 15
term 5 = 2 x 15 + 1 = 31
term 6 = 2 x 31 + 1 = 62 + 1 = 63

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

Mean = $\frac{\mathrm{10\; +\; 12\; +\; 14\; +\; 16}}{4}$ = $\frac{52}{4}$ = 13

3a² × 2ab × 4bc

a = a¹

b = b¹

Apply the law of multiplication of indices.

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

3a² × 2ab × 4bc = 3 x 2 x 4 x a² x a x b x b x c

3a² × 2ab × 4bc = 24 x a^{2 + 1} x b^{1 + 1} x c

3a² × 2ab × 4bc = 24 x a³ x b² x c

3a² × 2ab × 4bc = 24 x a³b²c

3 (2a + 3b) = 3 (2a + 3b)
3 (2a + 3b) = 3 x 2a + 3 x 3b
3 (2a + 3b) = 6a + 9b

Area of a circle = πr²

Where r = radius of the circle.

r² = r x r

Radius is half the diameter

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{28\; cm}}{2}$ = 14 cm

Area of the circle = $\frac{22}{7}$ x 14 cm x 14 cm = 22 x 2 cm x 14 cm = 616 cm ²

Note: the 7 dividing the 22 divides itself 1 and 14, 2 times.

Total oranges = 95

Fofo kept 45 of them

Remaining oranges to share = 95 - 45 = 50

She shared the rest equally with Dede. This means each got half of the 50 oranges remaining.

Dede's share = $\frac{50}{2}$ = 25 oranges

c = $\frac{{\mathrm{b}}^2+\; 4r}{\mathrm{2ar}}$

b = 3, r = 4 and a = 5

c = $\frac{3^2+\; 4\; x\; 4}{2\; x\; 5\; x\; 4}$

c = $\frac{\mathrm{3\; x\; 3\; +\; 16}}{40}$

c = $\frac{\mathrm{9\; +\; 16}}{40}$

c = $\frac{25}{40}$ = $\frac{5}{8}$

Note: 5 divides 25, 5 times and 40, 8 times.

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

Profit = Selling Price - Cost Price

Profit = ₵920,000 - ₵800,000 = ₵120,000

Profit % = $\frac{\mathrm{₵120,000}}{\mathrm{₵800,000}}$ x 100% = 3 x 5% = 15%

Note:
1. The zeros (0) at the end of ₵120,000 and ₵800,000 cancel each other
2. The zeros (0) at the end of 80 and 100 cancel each other
3. 2 divides 10, 5 times and 8, 4 times
4. 4 divides itself 1 and 12, 3 times

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 value of y when x is 1

U₁ = 9

y = 9 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 9 - 5 = 13 - 9 = 17 - 13 = 21 -17 = 4

We can substitute our d and simplify the expression for y

y = 9 + (x-1)4

Simplify the relation.

y = 9 + 4 x x - 1 x 4
y = 9 + 4x - 4
y = 4x + 9 - 4
y = 4x + 5

Volume of a cuboid = Length x Width x Height

Volume of the cuboid = 15 m x 12 m x 3 m = 540 m³

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

Total number of balls = 4 + 8 = 12

Number of blue balls = 4

P(Blue ball) = $\frac{4}{12}$ = $\frac{1}{3}$

Note: 4 divides itself 1 and 12, 3 times.

Sum of angles on a straight line is 180°

The ∟ is 90°.

y + 90° + 35° = 180°

y + 125° = 180°
y = 180° - 125°
y = 55°

Perimeter is the sum of all the sides of a shape.

The perimeter of the equilateral triangle = 16 cm + 16 cm + 16 cm = 48 cm

A square has the same perimeter.

Square has 4 equal sides.

Let L = side of the square.

The perimeter of the square = L + L + L + L = 4L

4L = 48

Divide both sides by 4.

L = $\frac{48}{4}$ = 12

The length of the square = 12 cm

Area of a square = Length x Length
Area of the square = 12 cm x 12 cm = 144 cm²

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

Principal = ₵500,000
Rate = 10
Time = 2

Simple Interest = $\frac{\mathrm{₵500,000\; x\; 2\; x\; 10}}{100}$ = ₵5000 x 2 x 10 = ₵100,000

Note: the zeros (0) at the end of ₵500,000 and 100 cancel each other.

l² = $\frac{4{\mathrm{\pi }}^2\mathrm{T}}{\mathrm{g}}$

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

l² x g = 4π²T

Divide both sides by 4π²

T = $\frac{\mathrm{g}{\mathrm{l}}^2}{4{\mathrm{\pi }}^2}$

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$

Change the decimals 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.

0.54 = 0.⌒5⌒4. = 54 x 10^-2

0.7 = 0.⌒7. = 7 x 10^-1

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$ = $\frac{\mathrm{54\; x}{10}^{-2}x7x{10}^{-1}}{9}$

Note: 9 divides 54, 6 times.

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$ = 6 x 10^-2 x 7 x 10^-1

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$ = 6 x 7 x 10^{-2 + -1}

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$ = 42 x 10^{-2 - 1}

$\frac{\mathrm{0.54\; x\; 0.7}}{9}$ = 42 x 10^-3

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

Every whole number has a decimal point at the end which is not written.

42 = 42.0

In 42 x 10^-3, you are to move the decimal point 3 times to the left.

42 x 10^-3 = 0.⌒0⌒4⌒2.0 = 0.042

Sum of ratios represents the amount shared

Sum of ratios = 5 + 3 = 8

Ebo's ratio is 3

If 8 = ₵800,000

3 = $\frac{\mathrm{₵800,000\; x\; 3}}{8}$ = ₵100,000 x 3 = ₵300,000

Note: 8 divides itself 1 and ₵800,000, 100000 times.

Perimeter is the sum of all the sides of a shape.

A rectangle has two equal lengths and two equal breadths.

Perimeter = 2L + 2B

Perimeter = 24 and breadth = 4 cm

24 = 2L + 2 x 4
24 = 2L + 8
24 - 8 = 2L
16 = 2L

Divide both sides by 2

L = $\frac{16}{2}$ = 8 cm

Area of a rectangle = Length x Breadth

Area of the rectangle = 4 cm x 8 cm = 32 cm²

4ab² – 20ba² = 4ab² – 20ba²
4ab² – 20ba² = 4ab(b - 5a)

$\frac{2}{3}$(x - 3) = $\frac{5}{6}$(x + 6)

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (3 and 6). The L.C.M is 6

6 x $\frac{2}{3}$(x - 3) = 6 x $\frac{5}{6}$(x + 6)

2 x 2(x - 3) = 5(x + 6)
2 x 2x - 2 x 2 x 3 = 5x + 30
4x - 12 = 5x + 30
4x - 5x = 30 + 12
-x = 42

Divide both sides by -1

x = $\frac{42}{-1}$ = -42

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

Number of oranges left = 100 - 80 = 20

Total oranges = 100

Percentage = $\frac{20}{100}$ x 100% = 20%

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.

2, 3, 4, 6, 10,12

There are two numbers at the middle, 4 and 6

Median = $\frac{\mathrm{4\; +\; 6}}{2}$ = $\frac{10}{2}$ = 5

Let U = pupils in the class
M = pupils who read Mathematics
E = pupils who read English

Number of pupils who read English only = 13 - 3 = 10

Let U = pupils in the class
M = pupils who read Mathematics
E = pupils who read English

Number of pupils who read English only = 13 - 3 = 10

Number of pupils who read Mathematics only = 8 - 3 = 5

Number of pupils who do not read either Mathematics or English is the number outside the circles.

When you sum everything in the rectangle, you are suppose to get the number in the universal set (U).

Let x = number of pupils who do not read either Mathematics or English

10 + 3 + 5 + x = 20
18 + x = 20
x = 20 - 18
x = 2

Change the decimals 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.

82.5 = 82.⌒5. = 825 x 10^-1

0.25 = 0.⌒2⌒5. = 25 x 10^-2

82.5 ÷ 0.25 = $\frac{825x{10}^{-1}}{25x{10}^{-2}}$

Note:
1. 25 divides itself 1 and 825, 33 times
2. Use the law of division of indices to simplify the base 10s

Law of division of indices

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

82.5 ÷ 0.25 = 33 x 10^-1-(-2)

Note: - - = +

82.5 ÷ 0.25 = 33 x 10^{-1 + 2}

82.5 ÷ 0.25 = 33 x 10¹

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 which is not written.

33 = 33.0

33 = 3.⌒3.0 = 3.3 x 10¹

Note: the power is positive 1 because we had to move the decimal point to the left once.

82.5 ÷ 0.25 = 3.3 x 10¹ x 10¹

Apply the law of multiplication of indices to simplify the base 10.

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

82.5 ÷ 0.25 = 3.3 x 10^{1 + 1}

82.5 ÷ 0.25 = 3.3 x 10²

Method I

Using proportionality

The percentage for the original price is 100%

When it is increased by 10% it became 100% + 10% = 110%

We know the price when it is at 100% hence we can use proportionality to find when it is 110%

If 100% = ₵450,000

110% = $\frac{\mathrm{₵450,000\; x\; 110\%}}{\mathrm{100\%}}$ = ₵4500 x 110 = ₵495,000

Note: the zeros (0) at the end of ₵450,000 and 100 cancel each other.

Method II

Finding the amount of increment and adding to the original price.

Percentage means over 100. 10% means 10 over 100

Increased amount = $\frac{10}{100}$ x ₵450,000 = 10 x ₵4500 = ₵45,000

Price price = Original price + Increased amount

New price = ₵450,000 + ₵45,000 = ₵495,000

A die is labelled from 1 to 6.

Possible outcomes = {1, 2, 3, 4, 5, 6}

Angle RSQ + 120° = 180° (Sum of angles on a straight line is 180°)

Angle RSQ = 180° -120°
Angle RSQ = 60°

Angle RSQ + 50° + RQS = 180°(sum of interior angles of a triangle is 180°)

60° + 50° + RQS = 180°
110° + RQS = 180°
RQS = 180° - 110°
RQS = 70°

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

r + 2s = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ + 2$\left(\begin{array}{c}-4\\ -1\end{array}\right)$

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

r + 2s = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ + $\left(\begin{array}{c}-8\\ -2\end{array}\right)$

r + 2s = $\left(\begin{array}{c}\mathrm{-2\; +\; -8}\\ \mathrm{3\; +\; -2}\end{array}\right)$

r + 2s = $\left(\begin{array}{c}\mathrm{-2\; -\; 8}\\ \mathrm{3\; -\; 2}\end{array}\right)$

r + 2s = $\left(\begin{array}{c}-10\\ 1\end{array}\right)$