You can substitute the value of x into the rule and see which one gets the correct coressponding y in all cases

x → x - 5

When x = 0

x → x - 5

0 → 0 - 5
0 → -5

This is false since 0 rather maps to 5 from the table

Hence the rule of the mapping is incorrect

x → 5 - x

When x = 0

0 → 5 - 0

0 → 5

When x = 2

2 → 5 - 2

2 → 3

When x = 4

4 → 5 - 4

4 → 1

When x = 6

6 → 5 - 6

6 → -1

When x = 8

8 → 5 - 8

8 → -3

Each of the x values resulted in the correct corresponding y, hence the rule of the mapping is x → 5 - x

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

Note: The hypotenuse sides are |PR| and |PQ|, the sides facing the right angle (90°)

sin 30° = $\frac{\mathrm{PS}}{8}$

Multiply both sides by 8

8 x sin 30° = |PS|

|PS| = 8 sin 30°

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan 50°= $\frac{\mathrm{PS}}{\mathrm{QS}}$

But |PS| = 8 sin 30°

tan 50°= $\frac{\mathrm{8\; sin\; 30°}}{\mathrm{QS}}$

Multiply both sides by QS

QS x tan 50° = 8 sin 30°

Divide both sides by tan 50°

|QS| = $\frac{\mathrm{8\; sin\; 30°}}{\mathrm{tan\; 50°}}$

The significant figures are the non-zero digits. In 75882, there are 5 significant figures. We are asked to express to three significant figures.

From the left (7) to the right, the third significant figure is 8 and the next digit to it is 8 which is more than 4, hence 1 is added to the third significant figure (8 + 1 = 9) and the other significant figures are replaced by zeros (0s)

Point + Translation vector = Image

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

Point = (2, 4) = $\left(\begin{array}{c}2\\ 4\end{array}\right)$

Image = (5, 3) = $\left(\begin{array}{c}5\\ 3\end{array}\right)$

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

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

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

2 + x = 5
x = 5 - 2
x = 3

4 + y = 3
y = 3 - 4
y = -1

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

V = $\frac{1}{3}$πr²h

Get rid of the fraction by multiplying both sides by 3

3 x V = 1 x πr²h

3V = πr²h

Divide both sides by πr²

h = $\frac{\mathrm{3V}}{\mathrm{\pi }{\mathrm{r}}^2}$

Let l = length of the side of the rhombus

From Pythagorean theorem,

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

l² = 4² + 3²

l² = 16 + 9

Note:
1. 4² = 4 x 4 = 16
2. 3² = 3 x 3 = 9

l² = 25

A number when you multiply by itself gives 25, the number is 5

5 x 5 = 5² = 25

l² = 5²

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

l = 5

length of the side of the rhombus = 5 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.

0.000344 = 0.⌒0⌒0⌒0⌒3.44

0.000344 = 3.44 x 10^-4

Note: the power is negative (-4) because the decimal point was moved 4 times to the right.

Sum of angles in a circle is 360°

∠ VARIETY + 45° + 60° + 75° + 60° = 360°

∠ VARIETY + 240° = 360°

∠ VARIETY = 360° - 240°

∠ VARIETY = 120°

The entire angle (360°) represents the 6 hours of telecast.

The angle for DOCUMENTARY is 75°

If 360° = 6 hours

75° = $\frac{\mathrm{75°\; x\; 6}}{\mathrm{360°}}$ = $\frac{5}{4}$ = 1$\frac{1}{4}$

Note:
1. 6 divides itself, 1 time and 360, 60 times
2. 5 divides, 60, 12 times and 75, 15 times
3. 3 divides 15, 5 times and 12, 4 times

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$

500 = 5 x 100 (in order to obtain a perfect square)
125 = 5 x 25 (in order to obtain a perfect square)

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3$\sqrt{\mathrm{5\; x\; 100}}$ - 2$\sqrt{\mathrm{5\; x\; 25}}$

From the rule of surd, $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3$\sqrt{5}$ x $\sqrt{\mathrm{100}}$ - 2$\sqrt{5}$ x $\sqrt{\mathrm{25}}$

$\sqrt{\mathrm{100}}$ = 10 (10 x 10 = 100)

$\sqrt{\mathrm{25}}$ = 5 (5 x 5 = 25)

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3$\sqrt{5}$ x 10 - 2$\sqrt{5}$ x 5

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3 x 10$\sqrt{5}$ - 2 x 5$\sqrt{5}$

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 30$\sqrt{5}$ - 10$\sqrt{5}$

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 20$\sqrt{5}$

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

Total number of balls = 7 + 5 + 8 = 20

Number of white balls = 8

P(white) = $\frac{8}{20}$ = $\frac{2}{5}$

Note: 4 divides 8, 2 times and 20, 5 times.

R directly proportional to l

R ∝ l

R inversely proportional to the square of d

Square means to the power 2

Square of d = d²

Inversely proportional means 1 over the variable/number

R ∝ $\frac{1}{{\mathrm{d}}^2}$

When you combine the two relations,

R ∝ $\frac{\mathrm{l}}{{\mathrm{d}}^2}$

Introduce the constant (k) to get rid of the proportionality symbol

R = k x $\frac{\mathrm{l}}{{\mathrm{d}}^2}$

R = $\frac{\mathrm{kl}}{{\mathrm{d}}^2}$

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$

Change the mixed fractions to improper fractions.

$1\frac{2}{3}$ = $\frac{\mathrm{1\; x\; 3\; +\; 2}}{3}$ = $\frac{\mathrm{3\; +\; 2}}{3}$ = $\frac{5}{3}$

$1\frac{3}{4}$ = $\frac{\mathrm{1\; x\; 4\; +\; 3}}{4}$ = $\frac{\mathrm{4\; +\; 3}}{4}$ = $\frac{7}{4}$

$2\frac{5}{8}$ = $\frac{\mathrm{2\; x\; 8\; +\; 5}}{8}$ = $\frac{\mathrm{16\; +\; 5}}{8}$ = $\frac{21}{8}$

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{5}{3}$ - $(\frac{7}{4}÷\frac{21}{8})$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{5}{3}$ - $(\frac{7}{4}x\frac{8}{21})$

Note:
1. 4 divides itself 1 time and 8, 2 times
2. 7 divides itself 1 time and 21, 3 times

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{5}{3}$ - $\left(\frac{2}{3}\right)$

Remove the bracket

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{5}{3}$ - $\frac{2}{3}$

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{\mathrm{5\; -\; 2}}{3}$

$1\frac{2}{3}$ - $(1\frac{3}{4}÷2\frac{5}{8})$ = $\frac{3}{3}$ = 1

6x : 64 = 18 : 32

Change the ratios to fractions.

$\frac{\mathrm{6x}}{64}$ = $\frac{18}{32}$

Note: 2 divides 6, 3 times and 64, 32 times

$\frac{\mathrm{3x}}{32}$ = $\frac{18}{32}$

Since the denominators at the left and right are the same, the numerators at the left and right should be the same.

3x = 18

Divide both sides by 3

x = $\frac{18}{3}$ = 6

A = {1 ≤ x ≤ 4} and B = {1,4,10}

1 ≤ x ≤ 4 is read as 1 is less than or equal to x and x is also less than or equal to 4

The set A is therefore the set of numbers greater than or equal to and less than or equal to 4

A = {1, 2, 3, 4}

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

A ∩ B = {1, 4}

a = $\left(\begin{array}{c}2\\ 4\end{array}\right)$ and b = $\left(\begin{array}{c}-2\\ 6\end{array}\right)$

2a + b = 2$\left(\begin{array}{c}2\\ 4\end{array}\right)$ + $\left(\begin{array}{c}-2\\ 6\end{array}\right)$

2a + b = $\left(\begin{array}{c}\mathrm{2\; x\; 2}\\ \mathrm{2\; x\; 4}\end{array}\right)$ + $\left(\begin{array}{c}-2\\ 6\end{array}\right)$

2a + b = $\left(\begin{array}{c}4\\ 8\end{array}\right)$ + $\left(\begin{array}{c}-2\\ 6\end{array}\right)$

2a + b = $\left(\begin{array}{c}\mathrm{4\; +\; -2}\\ \mathrm{8\; +\; 6}\end{array}\right)$

2a + b = $\left(\begin{array}{c}\mathrm{4\; -\; 2}\\ \mathrm{8\; +\; 6}\end{array}\right)$

2a + b = $\left(\begin{array}{c}2\\ 14\end{array}\right)$

4^x = 2^{3x - 3}

In order to solve for indices, make sure the bases are the same.

4 = 2 x 2 = 2²

From (a^m)ⁿ = a^{m x n}

(2²)^x = 2^{3x - 3}

2^{2 x x} = 2^{3x - 3}

2^2x = 2^{3x - 3}

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

2x = 3x - 3

3 = 3x - 2x
3 = x

∴ x = 3

Note:

1. $\sqrt{a}$ x $\sqrt{a}$ = a

2. $\frac{\sqrt{a}}{\sqrt{a}}$ = 1

$\frac{1}{\sqrt{2}}(\sqrt{2}+\frac{1}{\sqrt{2}})$ = $\frac{1}{\sqrt{2}}(\sqrt{2}+\frac{1}{\sqrt{2}})$

Expand the bracket

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

$\frac{1}{\sqrt{2}}(\sqrt{2}+\frac{1}{\sqrt{2}})$ = $\frac{\sqrt{2}}{\sqrt{2}}$ + $\frac{1x1}{\sqrt{2}x\sqrt{2}}$

Note: $\sqrt{2}$ x $\sqrt{2}$ = 2

$\frac{1}{\sqrt{2}}(\sqrt{2}+\frac{1}{\sqrt{2}})$ = 1 + $\frac{1}{2}$

$\frac{1}{\sqrt{2}}(\sqrt{2}+\frac{1}{\sqrt{2}})$ = $1\frac{1}{2}$

Note: respectively means the ratio are in the order the names where mentioned.

Kwaku's profit : Attah's profit = 4 : 5

Let x = Kwaku's profit

Attah received GH₵ 500.00 more than Kwaku. This means when you add GH₵ 500.00 to Kwaku's profit, you will get Attah's profit.

Attah's profit = Kwaku's profit + GH₵ 500.00

But Kwaku's profit = x

Attah's profit = x + GH₵ 500.00

Kwaku's profit : Attah's profit = 4 : 5

x : x + 500 = 4 : 5

Change the ratios to fractions and solve for x

$\frac{\mathrm{x}}{\mathrm{x\; +\; 500}}$ = $\frac{4}{5}$

Cross multiply

5 x x = 4 x (x + 500)
5x = 4 x x + 4 x 500
5x = 4x + 2000
5x - 4x = 2000
x = 2000

∴ Kwaku's share = 2000

Attah's share = Kwaku's share + 500
Attah's share = 2000 + 500
Attah's share = 2500

Total profit the company made = Kwaku's share + Attah's share
Total profit the company made = 2000 + 2500
Total profit the company made = 4500

∴ the company made a profit of GH₵ 4,500.00

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

Multiply both sides by Time

Time x Speed = Distance

Divide both sides by Speed

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

Distance = 12 km

Speed = 8 kmh^-1

Time = $\frac{12}{8}$

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

Time = $\frac{3}{2}$ = $1\frac{1}{2}$ hour

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

Mean of the seven numbers = $\frac{\mathrm{Sum\; of\; the\; seven\; numbers}}{7}$

Mean of the seven numbers is 15

$\frac{\mathrm{Sum\; of\; the\; seven\; numbers}}{7}$ = 15

Multiply both sides by 7

Sum of the seven numbers = 15 x 7
Sum of the seven numbers = 105

Mean of the ten numbers = $\frac{\mathrm{Sum\; of\; the\; ten\; numbers}}{10}$

Let x = sum of the three numbers added

Sum of the ten numbers = Sum of the seven numbers + sum of the three numbers

But sum of the seven numbers = 105

Sum of the ten numbers = 105 + x

Mean of the ten numbers = $\frac{\mathrm{105\; +\; x}}{10}$

But mean of the ten numbers is 12

$\frac{\mathrm{105\; +\; x}}{10}$ = 12

Multiply both sides by 10

105 + x = 12 x 10
105 + x = 120
x = 120 - 105
x = 15

∴ sum of the three numbers = 15

Mean of the three numbers = $\frac{\mathrm{Sum\; of\; the\; three\; numbers}}{3}$

Mean of the three numbers = $\frac{15}{3}$ = 5

Let U = {Students}
n(U) = 20
S = {Students who play soccer}
n(S) = 16
H = {Students who play hockey}
n(H) = 12

Let x = the number of students who play both games

2 + 16 - x + x + 12 - x = 20
30 - x = 20
30 - 20 = x
10 = x

∴ 10 students play both games.

16 - x = 16 - 10 = 6
12 - x = 12 - 10 = 2

Number of students who play only hockey is 2

$(3-2\sqrt{2})(3+2\sqrt{2})$ = $(3-2\sqrt{2})(3+2\sqrt{2})$

Method I

Expansion

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 3 x 3 + 3 x $2\sqrt{2}$ - $2\sqrt{2}$ x 3 - $2\sqrt{2}$ x $2\sqrt{2}$

Note: $\sqrt{a}$ x $\sqrt{a}$ = a

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 + $6\sqrt{2}$ - $6\sqrt{2}$ - 2 x 2 x $\sqrt{2}$ x $\sqrt{2}$

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 - 4 x 2

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 - 8

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

Method II

Difference of two squares

(a + b)(a - b) = a² - b²

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 3² - ${\left(2\sqrt{2}\right)}^2$

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 - 2² x ${\sqrt{2}}^2$

Note:
1. 2² = 2 x 2 = 4
2. ${\sqrt{2}}^2$ = $\sqrt{2}$ x $\sqrt{2}$ = 2

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 - 4 x 2

$(3-2\sqrt{2})(3+2\sqrt{2})$ = 9 - 8

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

a = $\frac{2}{\mathrm{b}}$ - $\frac{1}{\mathrm{c}}$

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators (b and c). The L.C.M is b x c

b x c x a = b x c x $\frac{2}{\mathrm{b}}$ - b x c x $\frac{1}{\mathrm{c}}$

abc = 2c - b

Group the cs

b = 2c - abc

Factorize c out

b = c(2 - ab)

Divide both sides by (2 - ab)

c = $\frac{\mathrm{b}}{\mathrm{(2\; -\; ab)}}$

c = $\frac{\mathrm{b}}{\mathrm{2\; -\; ab}}$

$\frac{\mathrm{x}}{2}$ + $\frac{\mathrm{2x}}{3}$ = 14

Get rid of the fractions by multiplying both sides of the equations by the L.C.M of the denominators (2 and 3). The L.C.M is 6

6 x $\frac{\mathrm{x}}{2}$ + 6 x $\frac{\mathrm{2x}}{3}$ = 14 x 6

3 x x + 2 x 2x = 84
3x + 4x = 84
7x = 84

Divide both sides by 7

x = $\frac{84}{7}$ = 12

Sum of interior angles of a regular polygon = (n - 2) x 180°

Exterior angle of a regular polygon = $\frac{\mathrm{360°}}{\mathrm{n}}$

Where n = the number of sides of the polygon.

Exterior angle = 72°

$\frac{\mathrm{360°}}{\mathrm{n}}$ = 72°

Multiply both sides by n to get rid of the fraction

360° = 72° x n

Divide both sides by 72°

n = $\frac{\mathrm{360°}}{\mathrm{72°}}$ = 5

∴ the number of sides of the regular polygon is 5

Sum of interior angles of a regular polygon = (n - 2) x 180°

But n = 5

Sum of interior angles of a regular polygon = (5 - 2) x 180°

Sum of interior angles of a regular polygon = 3 x 180°

Sum of interior angles of a regular polygon = 540°

Clockwise rotation through 180° about the origin

(x, y) → (-x, -y)

Thus both the x and y coordinates are negated.

P(-5, 1) → (-(-5), -1)

-(-) or - x - = +

P(-5, 1) → (5, -1)

Sum of interior angles of a regular polygon = (n - 2) x 180°

Where n = number of sides of the regular polygon.

The number of sides of the above polygon is 5 hence n = 5

Sum of the interior angles = (5 - 2) x 180°

Sum of the interior angles = 3 x 180°

Sum of the interior angles = 540°

2x + 3x + x + 4x + 70° = 540°

10x = 540° - 70°

10x = 470°

Divide both sides by 10

x = $\frac{\mathrm{470°}}{10}$ = 47°

Let x = the larger number and y = the smaller number

Note:
1. Sum means addition
2. Difference means subtraction
3. Is means equal to

The sum of two numbers is 72

x + y = 72 ------ eqn. 1

The difference between them is 26

x - y = 26 ------ eqn. 2

Get rid of y, eqn 1 + eqn 2

2x + 0 = 98

2x = 98

Divide both sides by 2

x = $\frac{98}{2}$ = 49

From ---- eqn. 1

x + y = 72

But x = 49

49 + y = 72
y = 72 - 49
y = 23

∴ the smaller number is 23

Note:
1. Bearing starts from the north Clockwise
2. An alternate angle is formed between the lines drawn
3. 110° = 90° + 20°

Bearing of P from Q = 90° + 90° + 90° + 20°

Bearing of P from Q = 290°

Note:
1. The base angles of isosceles triangle are the same
2. Sum of angles on a straight line is 180°
3. Sum of interior angles of a triangle is 180°

∠PSR + 158° = 180° (Sum of angles on a straight line)

∠PSR = 180° - 158°

∠PSR = 22°

∠PSR = ∠SPR = 22°(Base angles of isosceles triangles are the same)

∠PSR + ∠SPR + ∠PRS = 180° (Sum of interior angles of a triangle)

22° + 22° + ∠PRS = 180°

44° + ∠PRS = 180°

∠PRS = 180° - 44°

∠PRS = 136°

∠PRS + ∠PRQ = 180° (Sum of angles on a straight line)

136° + ∠PRQ = 180°

∠PRQ = 180° - 136°

∠PRQ = 44°

∠PRQ = ∠PQR = 44°(Base angles of isosceles triangles are the same)

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

44° + 44° + ∠QPR = 180°

88° + ∠QPR = 180°

∠QPR = 180° - 88°

∠QPR = 92°

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

Perimeter of the garden = Radius + Radius + Length of arc of sector

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Radius = 21 m

π = $\frac{22}{7}$

Angle of the sector = 60°

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Length of the arc of the garden = $\frac{60}{360}$ x 2 x $\frac{22}{7}$ x 21m

Note:
1. 60 divides itself 1 time and 360, 6 times
2. 7 divides itself 1 time and 21, 3 times
3. 3 divides itself 1 time and 6, 2 times
4. the 2s cancel each other

Length of the arc of the garden = 22 m

Perimeter of the garden = 21 m + 21 m + 22 m

Perimeter of the garden = 64 m

Selling Price = Cost Price + Profit

Cost Price = GH₵ x

Profit = GH₵ y

Selling Price = GH₵ x + GH₵ y

Selling Price = GH₵ (x + y)

She sells 5 tins

Total amount received = 5 x each tin sold

Total amount received = 5 x GH₵ (x + y)

Total amount received = GH₵ 5(x + y)

100% represents the percentage of the amount before the discount (marked price)

Percentage paid = 100% - Discount%

Percentage paid = 100% - 10%
Percentage paid = 90%

Amount paid = GH₵ 2,700

If 90% = 2700

100% = $\frac{\mathrm{2700\; x\; 100\%}}{\mathrm{90\%}}$ = 3000

∴ The marked price = GH₵ 3,000

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)

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 will 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%

Prime factors of 72

$$\begin{array}{cc}2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}$$

Prime factors of 72 = 2 x 2 x 2 x 3 x 3 = 2³ x 3²

When their multiples are the same, they will ring simultaneously.

For the first time, that will be their least common multiples (L.C.M)

Multiples of 3 = {3,6,9,12,15,...}
Multiples of 4 = {4,8,12,16,...}

After 12 hours both bells will ring at the same time.