168° = Interior angle of a regular polygon = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

Note: Sum of the interior angles = (n - 2) x 180°

Make n the subject of the formula:

Get rid of the fraction by multiplying both sides by the denominator, n

168° x n = Interior angle of a regular polygon = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$ x n

168n = 180n - 360

360 = 180n - 168n

360 = 12n

Divide both sides by 12

n = $\frac{360}{12}$

n = 30

Note:12 divides itself 1 time and 360, 3 times

Let the number be x added to both the numerator and the denominator of the fraction $\frac{1}{8}$

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

Cross-multiply to get rid of the fractions:

2(1 + x) = 1(8 + x)

2 + 2x = 8 + x

2x - x = 8 - 2

x = 6

Sum of ratios represents the total oranges shared

Sum of ratios = 5 + 3 + 7 = 15

Note: respectively means the ratios are in the order the names were mentioned.

Justina's ratio = 3

If 15 = 60

3 = $\frac{\mathrm{3\; x\; 60}}{15}$ = 12

Justina's share = 12 oranges

$\frac{\mathrm{x\; -\; 3}}{4}$ + $\frac{\mathrm{x\; +\; 1}}{8}$ ≥ 2

Get rid of the fractions by multiplying both sides by the LCM of 4 and 8, which is 8

8 x $\frac{\mathrm{x\; -\; 3}}{4}$ + 8 x $\frac{\mathrm{x\; +\; 1}}{8}$ ≥ 2 x 8

2(x - 3) + (x + 1) ≥ 16

2x - 6 + x + 1 ≥ 16

3x - 5 ≥ 16

3x ≥ 21

Divide both sides by 3

$\frac{\mathrm{3x}}{3}$ ≥ $\frac{21}{3}$

x ≥ 7

Y₂ = y
Y₁ = -8
X₂ = 1
X₁ = 2

Gradient = $\frac{\mathrm{y-(-8)}}{\mathrm{1-2}}$ = $\frac{\mathrm{y+8}}{-1}$

Notes:

1. - (-) = -- = +

2 -(-8) = + 8

But gradient of the line is -4

Therefore, $\frac{\mathrm{y+8}}{-1}$ = -4

Get ride of the fraction by multiplying both sides by the denominator, -1

-1 x $\frac{\mathrm{y+8}}{-1}$ = -4 x -1

Notes:

1. - x - = +

2. - 4 x - 1 = + 4

y + 8 = 4

y = 4 - 8

y = -4

∠WYZ = ∠TWX = 44° (Alternate angles)

∠TWX + ∠WXT + ∠WTX = 180° (Sum of angles in a triangle)

∠WXT = 50°

44° + 50° + ∠WTX = 180°

∠WTX = 180° - 94° = 86°

Let the leangth = L

Width is 7 m less than the length means when you subtract 7 m from the length, you will get the width

Width = L - 7

2 x Length + 2 x Width = Perimeter.

Perimeter = 90 m

2 x L + 2 x (L - 7) = 90

2L + 2L - 14 = 90

4L - 14 = 90

4L = 90 + 14

4L = 104

Divide both sides by 4 to solve for L:

L = $\frac{104}{4}$ = 26 m

Notes:

1. Both lengths of a rectangle are equal and both widths of the rectangle are equal

2. The two opposite sides could either be a length or a width

Assuming the two opposite sides are lengthd

The two opposite sides are equal

5x + 3 = 2x + 9

Solve for x

5x - 2x = 9 - 3

3x = 6

Divide both sides by 3

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

x = 2

Substitute the value of x into the given equations.

Length = 5x + 3 = 5(2) + 3 = 10 + 3 = 13 m

OR

Length = 2x + 9 = 2(2) + 9 = 4 + 9 = 13 m

Width = 6x - 7 = 6(2) - 7 = 12 - 7 = 5 m

Area of rectangle = Length x Width

Area = 13 m x 5 m = 65 m²

A die is numbered 1 to 6.

Samples = {1, 2, 3, 4, 5, 6}

Number of samples = 6

The prime numbers are 2, 3, and 5.

Number of possible selections = 3

Probability = $\frac{3}{6}$ = $\frac{1}{2}$

The point that lies on the line equation, when substituted should give the same value as the equation.

When the value of x and y are substituted into the equation, it should result in 11.

When the point (1, -1) is substituted into 3x - 8y, it gives 11

For the (1, -1), x = 1 and y = -1

3x - 8y = 3(1) - 8(-1) = 3 + 8 = 11

Notes:

1. - x - = -(-) = +

2. -8 x - 1 = -1(-8) = + 8

28, 29, 39, 38, 33, 37, 26, 20, 15, and 25

Largest value = 39

Least value = 15

Range = 39 - 15 = 24

Since the difference (subtraction) is 6 years, when you add 6 years to the younger age, you should get the eldest's age.

Let y = Esi's age

Amina's age = Esi's age + 6

Amina's age = y + 6

Amina's age + Esi's age = 24

(y + 6) + y = 24

y + 6 + y = 24

y + y + 6 = 24

2y = 24 - 6

2y = 18

Divide both sides by 2

y = $\frac{18}{2}$

y = 9

Esi's age is 9 years

Let |MQ| = y

Since ∆OMN is similar to ∆OPQ, the corresponding sides are proportional.

Scale factor = $\frac{\mathrm{|OP|}}{\mathrm{|ON|}}$ = $\frac{\mathrm{|OQ|}}{\mathrm{|OM|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|MN|}}$

|OM| = 10 - |MQ|

|OM| = 10 - y

$\frac{\mathrm{8\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{10\; cm}}{\mathrm{10\; -\; y}}$

$\frac{4}{3}$ = $\frac{10}{\mathrm{10\; -\; y}}$

Cross-multiply

4 (10 - y) = 3 x 10

40 - 4y = 30

-4y = 30 - 40

-4y = -10

Divide both sides by -4

y = $\frac{-10}{-4}$

y = 2.5 cm

|MQ| = 2.5 cm

3y + bx - 6 = 0

Since the line passes through (3, 4), when you substitute the x value, 3 and y value 4 into the equation, the result is supposed to be 0. Then you can find the b value.

3(4) + b(3) - 6 = 0

12 + 3b - 6 = 0

3b + 12 - 6 = 0

3b + 6 = 0

3b = -6

Divide both sides by 3

b = $\frac{-6}{3}$

b = -2

In ∆PQR, |QR| is the hypothenuse since ∠ QPR = 90°

From pythagoras theorem, p² = r² + q²

Substituting q² = 25 - r² into the equation

p² = r² + (25 - r²)

p² = r² + 25 - r²

p² = 25

Take square root of both sides

p = √25

p = 5

Notes:

25 = 5 x 5 = 5²

Let g = Gifty's current age

Gifty is 19 years younger than her mother means if you add 19 years to Gifty's age, you will get that of the mother's age.

Mother's age = g + 19

In three years time,

Gifty's age = g + 3

Mother's age = (g + 19) + 3 = g + 22

The sum of their ages in three years will be:

(g + 3) + (g + 22) = 67

Simplifying:

g + 3 + g + 22 = 67

2g + 25 = 67

2g= 42

Divide both sides by 2

$\frac{\mathrm{2g}}{2}$ = $\frac{42}{2}$

g= 21.

Gifty is currently 21 years old.

Mother's current age = g + 19 = 21 + 19 = 40

The sum of their ages now is: 21 + 40 = 61

Mean = $\frac{\mathrm{Sum\; of\; values}}{\mathrm{Number\; of\; values}}$

Given that the mean of a, b, c, d and e is 15

$\frac{\mathrm{a\; +\; b\; +\; c\; +\; d\; +\; e}}{5}$ = 15

Get rid of the denominator by multiplying both sides by the denominator, 5

5 x $\frac{\mathrm{a\; +\; b\; +\; c\; +\; d\; +\; e}}{5}$ = 15 x 5

a + b + c + d + e = 75

Mean of (a + 1), (b + 3), (c + 5), (d + 7) and (e + 9)

Mean = $\frac{\mathrm{(a\; +\; 1)\; +\; (b\; +\; 3)\; +\; (c\; +\; 5)\; +\; (d\; +\; 7)\; +\; (e\; +\; 9)}}{5}$

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = a + b + c + d + e + (1 + 3 + 5 + 7 + 9)

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = a + b + c + d + e + 25

But a + b + c + d + e = 75

(a + 1) + (b + 3) + (c + 5) + (d + 7) + (e + 9) = 75 + 25 = 100

Mean = $\frac{100}{5}$ = 20

The figures are 5, 4, 0, 5, 1, 4 and the third falls on 0.

The next figure after the third figure is 5.

Since 5 is greater than 4, we add 1 to the third figure (0) to get 1.

Therefore, 5.40514 rounded to three significant figures is 5.41.

Covering section for woodwork

15 + 13 - x + 6 = 30

28 - x + 6 = 30

34 - x = 30

34 - 30 = x

4 = x

Number of students who study both woodwork and metal work is 4.

Number of students who study woodwork but not metal work = 15 - x

Number of students who study woodwork but not metal work = 15 - 4 = 11

Let s = Sowah's age now

Mr Klu is 4 times as old as his son, Sowah means when you multiply Sowah's age by 4, you get Mr Klu's age.

Mr Klu's age = 4s

7 years ago, Sowah was (s - 7) years old and Mr Klu was (4s - 7) years old.

The sum of their ages 7 years ago was (s - 7) + (4s - 7) = 5s - 14.

We are told that this sum equals 76:

5s - 14 = 76

5s = 76 + 14

5s = 90

Divide both sides by 5

$\frac{\mathrm{5s}}{5}$ = $\frac{90}{5}$

s = 18

Sowah is currently 18 years old.

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

Let the two numbers be y and z

Sum of the numbers = y + z

of means multiplication

is means =

One-third of the sum of two numbers is 12 means:

$\frac{1}{3}$ x (y + z) = 12

Get rid of the fraction by multiplying both sides by the denominator, 3

3 x $\frac{1}{3}$ x (y + z) = 12 x 3

y + z = 36 ..........(1)

Difference means subtraction

Twice means 2 times

Twice their difference is 12 means:

2 x (y - z) = 12

Divide both sides by 2 to get rid of the coefficient 2

$\frac{\mathrm{2(y\; -\; z)}}{2}$ = $\frac{12}{2}$

y - z = 6 ..........(2)

Add equations (1) and (2):

(y + z) + (y - z) = 36 + 6

y + z + y - z = 42

2y = 42

Divide both sides by 2

$\frac{\mathrm{2y}}{2}$ = $\frac{42}{2}$

y = 21.

To find z, substitute y = 21 into equation (1):

21 + z = 36

z = 36 - 21

z = 15.

The two numbers are 21 and 15.

1² + h² = 2²

1 + h² = 4

h² = 4 - 1

h² = 3

Take the square root of both sides to solve for h

h = $\sqrt{3}$

Let the length of a side of the square be L

Using the Pythagorean theorem:

L² + L² = 12²

2L² = 144

Divide both sides by 2

$\frac{2L^2}{2}$ = $\frac{144}{2}$

L² = 72

Area of a square = Length x Length

Area of a square = L × L = L²

The area of the square is L², so the area is 72 cm².

4^3x = 16^{x + 1}

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

16 = 4², so we can rewrite the equation as:

4^3x = (4²)^{x + 1}

Simplify the right side:

4^3x = 4^2(x+1)

Since the bases are the same, we can equate the powers

3x = 2(x+1)

Solve for x:

3x= 2x + 2

3x - 2x = 2

x = 2

The sum of ratios represents the total amount shared.

The highest ratio represents the largest share.

Sum of ratios = 6 + 7 + 8 = 21

Largest ratio = 8

If 21 = 105

8 = $\frac{\mathrm{8\; x\; 105}}{21}$ = 8 x 5 = 40

Notes:

The largest share = 40

Mean = $\frac{\mathrm{Sum\; of\; values}}{\mathrm{Number\; of\; values}}$

Mean of the ten numbers = $\frac{\mathrm{Sum\; of\; first\; nine\; numbers\; +\; 10th\; number}}{10}$

Mean of the first nine numbers = $\frac{\mathrm{Sum\; of\; first\; nine\; numbers}}{9}$

Given that the mean of the first nine number is 55:

$\frac{\mathrm{Sum\; of\; first\; nine\; numbers}}{9}$ = 55

Get rid of the denominator by multiplying both sides by the denominator, 9

9 x $\frac{\mathrm{Sum\; of\; first\; nine\; numbers}}{9}$ = 55 x 9

Sum of first nine numbers = 495

Mean of the ten numbers is 56:

$\frac{\mathrm{Sum\; of\; first\; nine\; numbers\; +\; 10th\; number}}{10}$ = 56

Let n = 10th number

$\frac{\mathrm{Sum\; of\; first\; nine\; numbers\; +\; n}}{10}$ = 56

But sum of the first nine numbers = 495

$\frac{\mathrm{495\; +\; n}}{10}$ = 56

Get rid of the denominator by multiplying both sides by the denominator, 10

10 x $\frac{\mathrm{495\; +\; n}}{10}$ = 56 x 10

495 + n = 560

n = 560 - 495 = 65

10th number = 65

Let h = perpendicular distance between the parallel sides

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

105 = $\frac{1}{2}$ x (9 + 12) x h

105 = $\frac{1}{2}$ x 21 x h

Get rid of the denominator by multiplying both sides by the denominator, 2

2 x $\frac{1}{2}$ x 21 x h = 105 x 2

21 x h = 210

Divide both sides by 21

$\frac{\mathrm{21h}}{21}$ = $\frac{210}{21}$

h = 10 cm

Mean = $\frac{\mathrm{Sum\; of\; values}}{\mathrm{Number\; of\; values}}$

Mean of x and y = $\frac{\mathrm{x\; +\; y}}{2}$ = 4

Get rid of the denominator by multiplying both sides by the denominator, 2

2 x $\frac{\mathrm{x\; +\; y}}{2}$ = 4 x 2

x + y = 8

Mean of x, 2x, y and 2y = $\frac{\mathrm{x\; +\; 2x\; +\; y\; +\; 2y}}{4}$

Mean of x, 2x, y and 2y = $\frac{\mathrm{3x\; +\; 3y}}{4}$

Mean of x, 2x, y and 2y = $\frac{\mathrm{3(x\; +\; y)}}{4}$

But x + y = 8

Mean of x, 2x, y and 2y = $\frac{\mathrm{3(8)}}{4}$

Mean of x, 2x, y and 2y = $\frac{24}{4}$

Mean of x, 2x, y and 2y = 6

Comparing y = mx - 4 to the general equation of a line, m is the gradient.

From the point (-4,16), x = -4 and y = 16

Substituting these values into the equation of the line:

16 = m(-4) - 4

16 = -4m - 4

16 + 4 = -4m

20 = -4m

Divide both sides by -4

$\frac{20}{-4}$ = $\frac{\mathrm{-4m}}{-4}$

m = -5

100% represents the cost price

Selling Price % = Cost Price % + Profit %

Selling Price % = 100% + 15% = 115%

Cost Price = GH₵ 120

If 100% = 120

115% = $\frac{\mathrm{115\%\; x\; 120}}{\mathrm{100\%}}$ = 138

Notes:

The store sells the product for GH₵ 138

4 - 2k = -3

-2k = -3 - 4

-2k = -7

Divide both sides by -2

$\frac{\mathrm{-2k}}{-2}$ = $\frac{-7}{-2}$

k = 3.5

Height = initial height + (growth rate × number of days)

Height = 9 + $\frac{1}{8}$x

Height = $\frac{1}{8}$x + 9

$\frac{1}{8}$ = 0.125

Height = y

Height = 0.125x + 9

y = 0.125x + 9

6pq-3rs-3ps+6qr = 6pq - 3rs - 3ps + 6qr

Group like terms

6pq-3rs-3ps+6qr = 6pq - 3ps - 3rs + 6qr

6pq-3rs-3ps+6qr = 6pq - 3ps + 6qr - 3rs

Factor out common terms from each group

6pq-3rs-3ps+6qr = 3p(2q - s) + 3r(2q - s)

6pq-3rs-3ps+6qr = (2q - s)(3p + 3r)

But (3p + 3r) = 3(p + r)

6pq-3rs-3ps+6qr = 3(p + r)(2q - s)

6pq-3rs-3ps+6qr = 3(2q - s)(p + r)

Change the mixed fractions to improper fractions:

2$\frac{1}{6}$ = $\frac{\mathrm{6\; x\; 2\; +\; 1}}{6}$ = $\frac{\mathrm{12\; +\; 1}}{6}$ = $\frac{13}{6}$

2$\frac{7}{12}$ = $\frac{\mathrm{12\; x\; 2\; +\; 7}}{12}$ = $\frac{\mathrm{24\; +\; 7}}{12}$ = $\frac{31}{12}$

3$\frac{1}{4}$ = $\frac{\mathrm{4\; x\; 3\; +\; 1}}{4}$ = $\frac{\mathrm{12\; +\; 1}}{4}$ = $\frac{13}{4}$

Let n = the number to be subtracted

$\frac{13}{6}$ + $\frac{31}{12}$ - n = $\frac{13}{4}$

$\frac{13}{6}$ + $\frac{31}{12}$ - $\frac{13}{4}$ = n

Find a common denominator, which is 12

$\frac{\mathrm{2\; x\; 13\; +\; 1\; x\; 31\; -\; 3\; x\; 13}}{12}$ = n

$\frac{\mathrm{26\; +\; 31\; -\; 39}}{12}$ = n

$\frac{18}{12}$ = n

$\frac{3}{2}$ = n

n = $\frac{3}{2}$ or 1$\frac{1}{2}$

When the cyclist reaches the starting point, he completes a full circle, which is 6 m.

He has covered 23 m in total, so he has completed $\frac{23}{6}$ = 3$\frac{5}{6}$ = 3 full circles

3 full circles x circumference + additional distance = 23 m

3 x 6 + additional distance = 23 m

18 + additional distance = 23 m

additional distance = 23 - 18 = 5 m

Therefore, the cyclist is 5 m from the starting point.

Since two sides are equal, the triangle is an isosceles.

Let y = the base angles of the isosceles triangle

Sum of angles in triangle = 180°

y + y + 40° = 180°

2y + 40° = 180°

2y = 140°

Divide both sides by 2

$\frac{\mathrm{2y}}{2}$ = $\frac{\mathrm{140°}}{2}$

y = 70°

y + ∠XZT = 180° (angles on a straight line)

70° + ∠XZT = 180°

∠XZT = 180° - 70° = 110°

From the diagram, the bearing of T from Q is 045°.

Note: bearing reading starts from the north and goes clockwise.

The sum of all the angles is 360°

Angle for oranges = 360° - (80° + 60° + 100°) = 120°

The angle for apples = 80°

Number of apples = 60

If 80° = 60

120° = $\frac{\mathrm{120°\; x\; 60}}{\mathrm{80°}}$ = 30 x 3 = 90

There are 90 oranges on display.

A die is numbered from 1 to 6.

Possible outcomes

Note: at least 10 means 10 or more

Total possible outcomes when a die is tossed twice = 6 x 6 = 36

Favourable outcomes for a sum of at least 10 = (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)

Favourable outcomes for a sum of at least 10 = 6

Probability = $\frac{\mathrm{Favourable\; outcomes}}{\mathrm{Total\; possible\; outcomes}}$

Probability = $\frac{6}{36}$ = $\frac{1}{6}$