A factor of a number is a number that divides the given number evenly or exactly, leaving no remainder.

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Note: 1 is not a prime number.

Union (∪) is simply listing each of the elements from both sets. If an element appeared in both sets, it is listed only once.

Q U T is the region of the two circles

n(Q U T) = 20 + 10 + 15 = 45

Change the mixed fraction to improper fraction.

$\sqrt{6\frac{1}{4}}$ = $\sqrt{6\frac{1}{4}}$

6$\frac{1}{4}$ = $\frac{\mathrm{6\; x\; 4\; +\; 1}}{4}$ = $\frac{\mathrm{24\; +\; 1}}{4}$ = $\frac{25}{4}$

$\sqrt{6\frac{1}{4}}$ = $\sqrt{\frac{25}{4}}$

From the law of indices,

${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{c}}$ = $\frac{{\mathrm{a}}^{\mathrm{c}}}{{\mathrm{b}}^{\mathrm{c}}}$

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

$\sqrt{6\frac{1}{4}}$ = $\frac{\sqrt{25}}{\sqrt{4}}$

25 = 5 x 5 = 5²

4 = 2 x 2 = 2²

$\sqrt{6\frac{1}{4}}$ = $\frac{\sqrt{5^2}}{\sqrt{2^2}}$

$\sqrt{6\frac{1}{4}}$ = $\frac{5}{2}$

$\sqrt{6\frac{1}{4}}$ = 2.5

39 base 10 to base 5

$$\begin{array}{cc}\mathrm{}& 39\\ 5& \mathrm{7\; r\; 4}\\ 5& \mathrm{1\; r\; 2}\\ 5& \mathrm{0\; r\; 1}\end{array}$$

You list the remainders from the bottom to the top.

39_ten = 124_five

The least number greater than 207 and divisible by 17 is 221 (17 x 13)

Let x = the least whole number

207 + x = 221

x = 221 - 207

x = 14

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

Method I

Simplify the whole numbers and the fractions differently.

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x [(3 + 1)($\frac{1}{3}$ + $\frac{1}{6}$)]

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x [(4)($\frac{\mathrm{2\; x\; 1\; +\; 1\; x\; 1}}{6}$)]

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x [(4)($\frac{\mathrm{2\; +\; 1}}{6}$)]

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

Note: 3 divides itself 1 time and 6, 2 times.

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

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x (4$\frac{1}{2}$)

Change the mixed fraction to improper fraction and multiply by 2

4$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 4\; +\; 1}}{2}$ = $\frac{\mathrm{8\; +\; 1}}{2}$ = $\frac{9}{2}$

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x $\frac{9}{2}$

Note: the 2s cancel each other 1 time.

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 9

Method II

Change the mixed fractions to improper fractions and simplify.

3$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 3\; +\; 1}}{3}$ = $\frac{\mathrm{9\; +\; 1}}{3}$ = $\frac{10}{3}$

1$\frac{1}{6}$ = $\frac{\mathrm{6\; x\; 1\; +\; 1}}{6}$ = $\frac{\mathrm{6\; +\; 1}}{6}$ = $\frac{7}{6}$

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x ($\frac{10}{3}$ + $\frac{7}{6}$ )

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x ($\frac{\mathrm{2\; x\; 10\; +\; 1\; x\; 7}}{6}$)

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x ($\frac{\mathrm{20\; +\; 7}}{6}$)

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x ($\frac{27}{6}$)

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 2 x $\frac{27}{6}$

Note:
1. 2 divides itself 1 time and 6, 3 times
2. 3 divides itself 1 time and 27, 9 times

2 x (3$\frac{1}{3}$ + 1$\frac{1}{6}$) = 9

8.51 ÷ 2.3 = 3.7

Express 85.1 ÷ 2.3 to be in the form 8.51 ÷ 2.3

85.1 = 8.⌒5.1 = 8.51 x 10¹

Note: when you move the decimal point to the left, the power of the base 10 is positive.

85.1 ÷ 2.3 = $\frac{85.1}{2.3}$

85.1 ÷ 2.3 = $\frac{8.51x{10}^1}{2.3}$

85.1 ÷ 2.3 = $\frac{8.51}{2.3}$ x 10¹

8.51 ÷ 2.3 = $\frac{8.51}{2.3}$ = 3.7

85.1 ÷ 2.3 = 3.7 x 10¹

Any number/variable raised to the power 1 is the same number/variable.

85.1 ÷ 2.3 = 3.7 x 10

85.1 ÷ 2.3 = 3⌒7. = 37

Express a decimal as a fraction in its lowest term

1. Count (n) how many digits are after the decimal points

2. Divide the number without the decimal point by 1n0s

n0s is the number of zeros to join the 1 as counted in step 1.

3. Simplify the fraction.

There are three (3) digits after the decimal point in 0.625.

Hence 625 is divided by 1000 and simplified.

0.625 = $\frac{625}{1000}$ = $\frac{5}{8}$

Note: 125 divides 625, 5 times and 1000, 8 times.

Ama's ratio = 5

Kofi's ratio = 3

If 5 = ₵4650

3 = $\frac{\mathrm{₵4650\; x\; 3}}{5}$ = ₵930 x 3 = ₵2,790

6.5 = $\frac{65}{10}$

If $1.00 = ₵340

$\frac{65}{10}$$ = ₵340 x $\frac{65}{10}$$ ÷ $1

Note:
1. $ cancel each other
2. the zeros at the end of 340 and 10 cancel each other
3. any number/variable divided by 1 is the same number/variable

$\frac{65}{10}$$ = ₵65 x 34 = ₵2,210

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

Principal = ₵28000

Note
1. time is in year so you have to convert the 6 months to year
2. there are 12 months in a year hence 6 months is half a year

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

Rate = 3$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 3\; +\; 1}}{2}$ = $\frac{\mathrm{6\; +\; 1}}{2}$ = $\frac{7}{2}$

Simple Interest = $\frac{\mathrm{₵28000}x\frac{1}{2}x\frac{7}{2}}{100}$

Note:
1. the zeros (0) at the end of 28000 and 100 cancel each other
2. 2 divides itself 1 time and 28, 14 times
3. 2 divides itself 1 time and 14, 7 times

Simple Interest = ₵70 x 7 = ₵490

Q = 10 km

R = 5 km

Much farther = 10 km - 5 km = 5 km

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

Number of samples = 20

Let x = number of red beads.

P(red) = $\frac{1}{4}$

$\frac{\mathrm{x}}{20}$ = $\frac{1}{4}$

Cross multiply

4 x x = 1 x 20

Divide both sides by 4

x = $\frac{20}{4}$ = 5

Method I

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

Number of samples = 12

Number of ripe mangoes = 12 - number of unripe mangoes

Number of unripe mangoes = 4

Number of ripe mangoes = 12 - 4

Number of ripe mangoes = 8

P(ripe) = $\frac{8}{12}$ = $\frac{2}{3}$

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

Method II

P(ripe) + P(not ripe) = 1

P(not ripe) = $\frac{4}{12}$ = $\frac{1}{3}$

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

P(ripe) + $\frac{1}{3}$ = 1

P(ripe) = 1 - $\frac{1}{3}$

Every whole number is being divided by 1

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

P(ripe) = $\frac{1}{1}$ - $\frac{1}{3}$

P(ripe) = $\frac{\mathrm{3\; x\; 1\; -\; 1\; x\; 1}}{3}$

P(ripe) = $\frac{\mathrm{3\; -\; 1}}{3}$

P(ripe) = $\frac{2}{3}$

p² – 6p + 9

p² = p x p

p² – 6p + 9 = p x p – 6p + 9

When p = -2

p² – 6p + 9 = -2 x -2 - 6 x -2 + 9

Note: - x - = +. -2 x -2 = + 4. - 6 x -2 = + 12

p² – 6p + 9 = 4 + 12 + 9

p² – 6p + 9 = 25

3x – 4 = x + q

x = 2

3 x 2 - 4 = 2 + q

6 - 4 = 2 + q

2 = 2 + q

2 - 2 = q

0 = q

q = 0

For ≥ and ≤ , the solids are shaded.

On the number line, the solid is shaded and is pointing to the left hence the appropriate inequality sign is ≤

If you observe the mapping, you will realize that the rule is 2x + 1

x maps onto 21

2x + 1 = 21

2x = 21 - 1

2x = 20

Divide both sides by 2

x = $\frac{20}{2}$ = 10

If you observe the mapping, you will realize that the rule is 2x + 1

-2 maps onto y

-2 → 2 x -2 + 1

-2 → -4 + 1

-2 → -3

Bearing starts from the North clockwise.

Bearing of P from Q is ∠ NQP

Since triangle PRQ has two equal sides (2 km), the triangle is isosceles triangle and the base angles are the same.

∠RPQ = ∠PQR

∠PRQ is right-angle (90°)

∠RPQ + ∠PQR + ∠PRQ = 180°(Sum of interior angles of a triangle)

∠RPQ + ∠PQR + 90° = 180°

∠RPQ + ∠PQR = 180° - 90°

∠RPQ + ∠PQR = 90°

∠RPQ = ∠PQR = $\frac{\mathrm{90°}}{2}$ = 45°

∠ NQP + ∠PQR = 90° (The sum is right-angle)

∠ NQP + 45° = 90°

∠ NQP = 90° - 45°

∠ NQP = 45°

The Bearing of P from Q is therefore 045°

The formula for finding the sum of interior angles of a regular polygon is (n - 2) x 180°

Where n is the number of sides of the regular polygon.

The polygon has 10 sides hence n = 10

Sum of interior angles = (10 - 2) x 180°

Sum of interior angles = 8 x 180°

Project from the 4 miles to meet the line and project to meet the distance in kilometres axis and read the point.

The interval is 1 and the small grids are 10. Each grid is therefore = $\frac{1}{10}$ = 0.1

The point touches 4 grids after the 6 kilometres. The value of the 4 grids is 0.1 x 4 = 0.4

Total mark = 6 + 0.4 = 6.4

Project from the 4 kilometres to meet the line and project to meet the distance in miles axis and read the point.

The interval is 1 and the small grids are 10. Each grid is therefore = $\frac{1}{10}$ = 0.1

The point touches 5 grids after the 2 miles. The value of the 5 grids is 0.1 x 5 = 0.5

Total mark = 2 + 0.5 = 2.5

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

|XZ|² = 5² + 12²

5² = 5 x 5 = 25

12² = 12 x 12 = 144

|XZ|² = 25 + 144

|XZ|² = 169

169 = 13 x 13 = 13²

|XZ|² = 13²

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

|XZ| = 13 cm

The opposite side of ∠YXZ is 12 cm (|YZ|)

The adjacent side of ∠YXZ is 5 cm (|YX|)

Note: |XZ| is the hypothenuse of the triangle. The side facing the right-angle.

tan ∠YXZ = $\frac{\mathrm{12\; cm}}{\mathrm{5\; cm}}$ = $\frac{12}{5}$

PQ→ = OQ→ - OP→

PQ→ = $\left(\begin{array}{c}3\\ -2\end{array}\right)$ - $\left(\begin{array}{c}-3\\ 2\end{array}\right)$

PQ→ = $\left(\begin{array}{c}\mathrm{3\; -\; -3}\\ \mathrm{-2\; -\; 2}\end{array}\right)$

Note: - - = +. - -3 = + 3

PQ→ = $\left(\begin{array}{c}\mathrm{3\; +\; 3}\\ \mathrm{-2\; -\; 2}\end{array}\right)$

PQ→ = $\left(\begin{array}{c}6\\ -4\end{array}\right)$

Length of a vector, $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ = $\sqrt{a^2+b^2}$

Length = $\sqrt{{\mathrm{(-5)}}^2+{12}^2}$

(-5)² = -5 x -5 = 25

Note: - x - = +

12² = 12 x 12 = 144

Length = $\sqrt{25+144}$

Length = $\sqrt{169}$

169 = 13 x 13 = 13²

Length = $\sqrt{{13}^2}$ = 13

New Area = 9

Original Area = 4

9 = f² x 4

Divide both sides by 4

f² = $\frac{9}{4}$

9 = 3 x 3 = 3²

4 = 2 x 2 = 2²

f² = $\frac{3^2}{2^2}$

From the law of indices,

${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{c}}$ = $\frac{{\mathrm{a}}^{\mathrm{c}}}{{\mathrm{b}}^{\mathrm{c}}}$

f² = ${\left(\frac{3}{2}\right)}^2$

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

f = $\frac{3}{2}$ = 1$\frac{1}{2}$

A cube has all dimensions equal.

Length = Width = Height

Volume of a cube = Length x Length x Length = L³

Area of a cube = Length x Length = L²

The volume = 27

L³ = 27

27 = 3 x 3 x 3 = 3³

L³ = 3³

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

L = 3

Length = 3 cm

Area of a cube = Length x Length

Area of the cube = 3 cm x 3 cm = 9 cm²

The curved surface is made from a rectangle.

The circular base (circumference) is the breadth/length of the rectangle and the height is the breadth/length of the rectangle.

Circumference of a circle = 2 x π x radius.

π = 3.142

radius = 2 cm

Circumference of the circular base = 2 x 3.142 x 2 cm

Circumference of the circular base = 12.568 cm

Length of the rectangle that formed the circular base = 12.568 cm

Breadth = the height of the cylinder = 3 cm

Area of a rectangle = Length x Breadth

Area of the curved surface = 12.568 cm x 3 cm

Area of the curved surface = 37.704 cm² ≈ 38 cm²

Method I

Using the formula

Base = 12 cm

Height = 4 cm

Area of the parallelogram PQRS = 12 cm x 4 cm

Area of the parallelogram PQRS = 48 cm²

Method II

Solving it as two triangles and a rectangle.

Area of the parallelogram PQRS = Area A + Area B + Area C

Area of the triangle A and C are the same.

Area of a triangle = $\frac{1}{2}$ x base x height

Base of triangle = 5 cm
Height of triangle = 4 cm

Area of triangle A = $\frac{1}{2}$ x 5 cm x 4 cm

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

Area of triangle A = 5 cm x 2 cm = 10 cm²

Area of triangle A = Area of triangle C = 10 cm²

Area of a rectangle = Length x Breadth

Area of rectangle B = 7 cm x 4 cm = 28 cm²

Area of the parallelogram PQRS = 10 cm² + 28 cm² + 10 cm²

Area of the parallelogram PQRS = 48 cm²

one-third = $\frac{1}{3}$

one-fifth = $\frac{1}{5}$

of means multiplication (x)

Let n = the number.

one-third of a number ⇒ $\frac{1}{3}$ x n

one-fifth of the same number ⇒ $\frac{1}{5}$ x n

one-third of a number is added to one-fifth of the same number ⇒ $\frac{1}{3}$ x n + $\frac{1}{5}$ x n

is means equal to

$\frac{1}{3}$ x n + $\frac{1}{5}$ x n = 8

Solve for n.

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators (3 and 5).

The L.C.M of 3 and 5 is 15.

15 x $\frac{1}{3}$ x n + 15 x $\frac{1}{5}$ x n = 8 x 15

5 x n + 3 x n = 8 x 15

5n + 3n = 8 x 15

8n = 8 x 15

Divide both sides by 8

n = $\frac{\mathrm{8\; x\; 15}}{8}$ = 15

Note: the 8s cancel each other 1 time.

∴ the number is 15.

Change the mixed fraction to improper fraction.

6$\frac{1}{4}$ = $\frac{\mathrm{4\; x\; 6\; +\; 1}}{4}$ = $\frac{\mathrm{24\; +\; 1}}{4}$ = $\frac{25}{4}$

Equivalent means the value is the same hence you can equate the two.

1 : x = 6$\frac{1}{4}$ : 25

1 : x = $\frac{25}{4}$ : 25

Change the ratios to fractions and solve for x.

1 : x = $\frac{1}{\mathrm{x}}$

$\frac{25}{4}$ : 25 = $\frac{25}{4}$ ÷ 25

Every whole number is being divided by 1

25 = $\frac{25}{1}$

$\frac{25}{4}$ : 25 = $\frac{25}{4}$ ÷ $\frac{25}{1}$

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

$\frac{25}{4}$ : 25 = $\frac{25}{4}$ x $\frac{1}{25}$

Note: the 25s cancel each other 1 time.

$\frac{25}{4}$ : 25 = $\frac{1}{4}$ x $\frac{1}{1}$

$\frac{25}{4}$ : 25 = $\frac{1}{4}$ x 1

$\frac{25}{4}$ : 25 = $\frac{1}{4}$

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

Since the numerators are the same (1), the denominators are also the same.

x = 4

Express a fraction as a percentage

Percentage = fraction x 100%

Percentage = $\frac{3}{8}$ x 100%

Note: 4 divides 8, 2 times and 100, 25 times.

Percentage = $\frac{3}{2}$ x 25%

Percentage = $\frac{\mathrm{3\; x\; 25\%}}{2}$

Percentage = $\frac{\mathrm{75\%}}{2}$ = 37$\frac{1}{2}$%

Profit = Selling Price - Cost Price

Profit = ₵28,000 - ₵24,000

Profit = ₵4,000

The percentage of the cost price is 100%

If ₵24000 = 100%

₵4000 = $\frac{\mathrm{100\%\; x\; ₵4000}}{\mathrm{₵24000}}$

Note:
1. the zeros (0) at the end of 4000 and 24000 cancel each other
2. 4 divides itself 1 time and 24, 6 times
3. 2 divides 6, 3 times and 100, 50 times

₵4000 = $\frac{\mathrm{50\%}}{3}$ = 16.67% ≈ 16.7%

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

The angle for the corn is right-angle (90°)

Total area = 250 hectares

Area of the corn = $\frac{90}{360}$ x 250 hectares

Note: 90 divides itself 1 time and 360, 4 times.

Area of the corn = $\frac{\mathrm{250\; hectares}}{4}$ = 62.5 ha

Fraction = $\frac{\mathrm{Angle\; of\; the\; sector}}{360}$

The sum of all the angles of the sector is 360

The angle for the corn is right-angle (90°)

80 + 90 + 70 + Angle of sector for pepper = 360

240 + Angle of sector for pepper = 360

Angle of sector for pepper = 360 - 240

Angle of sector for pepper = 120

Fraction with pepper = $\frac{120}{360}$ = $\frac{1}{3}$