Cost for buying 5 units of brand S = 5 x GH₵ 3,000.00
Cost for buying 5 units of brand S = GH₵ 15,000.00

Cost for 5 units of brand T = 5 x GH₵ 4,000.00
Cost for 5 units of brand T = GH₵ 20,000.00

Amount saved = Cost for buying 5 units of brand T - Cost for buying 5 units of brand S
Amount saved = GH₵ 20,000 - GH₵ 15,000
Amount saved = GH₵ 5,000.00

Method I

The sum of ratios represent the total amount shared

The difference (larger ratio - smaller ratio) represents how much more

Zalia : Amina = 2 : 5

Sum of ratios = 2 + 5 = 7

Difference of ratios = 5 - 2 = 3

Amina had 150 more than Zalia.

If 3 = 150

7 = $\frac{\mathrm{7\; x\; 150}}{3}$ = 7 x 50 = 350

Total amount shared = GH₵ 350

Method II

Let x = Zalia's share

Amina had GH₵ 150.00 more than Zalia

Amina's share = Zalia's share + GH₵ 150

Amina's share = x + 150

Zalia share : Amina share = 2 : 5

x : x + 150 = 2 : 5

Change the ratios to fractions.

$\frac{\mathrm{x}}{\mathrm{x\; +\; 150}}$ = $\frac{2}{5}$

Cross multiply

5 x x = 2 x (x + 150)

5x = 2x + 300

5x - 2x = 300

3x = 300

Divide both sides by 3

x = $\frac{300}{3}$

x = 100

∴ Zalia's share = GH₵ 100.00

Amina's share = Zalia's share + GH₵ 150.00

Amina's share = GH₵ 100.00 + GH₵ 150.00

Amina's share = GH₵ 250.00

Total amount shared = Zalia's share + Amina's share

Total amount shared = GH₵ 100.00 + GH₵ 250.00

Total amount shared = GH₵ 350

Quantitative data is made up of numerical values has numerical properties, and can easily undergo math operations like addition and subtraction.

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

P ∩ Q = {c, d, e}

Elements/members of set Q are the ones inside the circle Q.

Method I

Change the kilogram to gram and calculate for the weight of rice and change it back to kilogram.

If 1 kg = 1000 g

2 kg = $\frac{\mathrm{2\; kg\; x\; 1000\; g}}{\mathrm{1\; kg}}$ = 2000 g

Weight of rice = bag of rice - empty bag

Weight of rice = 2000 g - 150 g

Weight of rice = 1850 g

Change the gram to kilogram

If 1000 g = 1 kg

1850 g = $\frac{\mathrm{1850\; g\; x\; 1\; kg}}{\mathrm{1000\; g}}$ = 1.85 kg

Note: 1.85 kg is the same as 1.850 kg

Method II

Change the gram to kilogram

If 1000 g = 1 kg

150 g = $\frac{\mathrm{150\; g\; x\; 1\; kg}}{\mathrm{1000\; g}}$ = 0.15 kg

Weight of rice = bag of rice - empty bag

Weight of rice = 2 kg - 0.15 kg
Weight of rice = 1.85 kg

First length = 10 cm

The second is 15 cm longer than the first one

Second length = First length + 15 cm

Second length = 10 cm + 15 cm

Second length = 25 cm

The third is 9 cm less than the second

Third length = Second length - 9 cm

Third length = 25 cm - 9 cm

Third length = 16 cm

2x - 4 < 6 + 3x
2x - 3x < 6 + 4
-x < 10

Divide both sides by - 1. When dividing by a negative, the inequality changes direction.

x > $\frac{10}{-1}$

x > -10

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$

Note:
When simplifying surds, express the surd as a product of a perfect square and a number
2² = 4 and $\sqrt{4}$ = 2
3² = 9 and $\sqrt{9}$ = 3
4² = 16 and $\sqrt{\mathrm{16}}$ = 4
5² = 25 and $\sqrt{\mathrm{25}}$ = 5
6² = 36 and $\sqrt{\mathrm{36}}$ = 6
7² = 49 and $\sqrt{\mathrm{49}}$ = 7
8² = 64 and $\sqrt{\mathrm{64}}$ = 8
9² = 81 and $\sqrt{\mathrm{81}}$ = 9
10² = 100 and $\sqrt{\mathrm{100}}$ = 10
11² = 121 and $\sqrt{\mathrm{121}}$ = 11
12² = 144 and $\sqrt{\mathrm{144}}$ = 12
13² = 169 and $\sqrt{\mathrm{169}}$ = 13
14² = 196 and $\sqrt{\mathrm{196}}$ = 14
15² = 225 and $\sqrt{\mathrm{225}}$ = 15

75 = 25 x 3
18 = 9 x 2
27 = 9 x 3

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$ = $\sqrt{\mathrm{25\; x\; 3}}$ + $\sqrt{\mathrm{9\; x\; 2}}$ - $\sqrt{\mathrm{9\; x\; 3}}$

$\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$ = $\sqrt{\mathrm{25}}$ x $\sqrt{3}$ + $\sqrt{9}$ x $\sqrt{2}$ - $\sqrt{9}$ x $\sqrt{3}$

$\sqrt{\mathrm{25}}$ = 5

$\sqrt{9}$ = 3

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$ = 5$\sqrt{3}$ + 3$\sqrt{2}$ - 3$\sqrt{3}$

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$ = 5$\sqrt{3}$ - 3$\sqrt{3}$ + 3$\sqrt{2}$

$\sqrt{\mathrm{75}}$ + $\sqrt{\mathrm{18}}$ - $\sqrt{\mathrm{27}}$ = 2$\sqrt{3}$ + 3$\sqrt{2}$

Draw a third parallel line

a° = 35° (Vertically opposite angles are the same)

a° = b° = 35° (Alternate angles are the same)

140° + d° = 180° (Sum of angles on a straight line is 180°)

d° = 180° - 140°

d° = 40°

c° = d° = 40° (Alternate angles are the same)

b° + c° + x° = 360° (Sum of angles in a circle is 360°)

35° + 40° + x° = 360°

75° + x° = 360°

x° = 360° - 75°

x° = 285°

Method I

The sum of all fractions should be 1

Fraction of oranges + fraction of mangoes + fraction of pears = 1

Fraction of pears = 1 - (fraction of oranges + fraction of mangoes)

Fraction of pears = 1 - ($\frac{2}{5}$ + $\frac{6}{25}$)

Fraction of pears = 1 - ($\frac{\mathrm{5\; x\; 2\; +\; 1\; x\; 6}}{25}$)

Fraction of pears = 1 - ($\frac{\mathrm{10\; +\; 6}}{25}$)

Fraction of pears = 1 - ($\frac{16}{25}$)

Every whole number is being divided by 1

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

Fraction of pears = $\frac{1}{1}$ - $\frac{16}{25}$

Fraction of pears = $\frac{\mathrm{25\; x\; 1\; -\; 1\; x\; 16}}{25}$

Fraction of pears = $\frac{\mathrm{25\; -\; 16}}{25}$

Fraction of pears = $\frac{9}{25}$

The whole fraction is 100%

If 1 = 100%

$\frac{9}{25}$ = $\frac{9}{25}$ x 100% ÷ 1

Any number/fraction divided by 1 is the same number/fraction

$\frac{9}{25}$ = $\frac{9}{25}$ x 100%

Note:
1. 25 divides itself 1 time and 100, 4 times
2. 9 x 4 = 36

$\frac{9}{25}$ = 36%

∴ Percentage of pears = 36%

Method II

Find the percentage of oranges and mangoes and subtract the sum from 100%

Percentage of oranges

The whole fraction is 1 which represents 100%

If 1 = 100%

$\frac{2}{5}$ = $\frac{2}{5}$ x 100% ÷ 1

Note:
1. Any number/fraction divided by 1 is the same number/fraction
2. 5 divides itself 1 time and 100, 20 times
3. 2 x 20 = 40

$\frac{2}{5}$ = 40%

∴ Percentage of the fruits which are oranges = 40%

Percentage of mangoes

If 1 = 100%

$\frac{6}{25}$ = $\frac{6}{25}$ x 100% ÷ 1

Note:
1. Any number/fraction divided by 1 is the same number/fraction
2. 25 divides itself 1 time and 100, 4 times
3. 6 x 4 = 24

$\frac{6}{25}$ = 24%

∴ Percentage of the fruits which are mangoes = 24%

Percentage of fruits which are pears = 100% - (40% + 24%)
Percentage of fruits which are pears = 100% - 64%
Percentage of fruits which are pears = 36%

Slope/Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points on the straight line to calculate the slope.

Using (1,1) and (0,3)

Y₂ = 3
Y₁ = 1
X₂ = 0
X₁ = 1

Slope/Gradient = $\frac{\mathrm{3\; -\; 1}}{\mathrm{0\; -\; 1}}$ = $\frac{2}{-1}$ = -2

Steps to find equation of a line
1. Find the gradient/slope of the line
2. Use the point (x, y) and any known point on the line to calculate the slope/gradient
3. Equate the expression in step 2 to the calculated gradient/slope in step 1
4. Make y the subject of the equation in step 3

Slope/Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points on the straight line to calculate the slope.

Using (1,1) and (0,3)

Y₂ = 3
Y₁ = 1
X₂ = 0
X₁ = 1

Slope/Gradient = $\frac{\mathrm{3\; -\; 1}}{\mathrm{0\; -\; 1}}$ = $\frac{2}{-1}$ = -2

Using the point (x, y) and (1,1)

$\frac{\mathrm{y\; -\; 1}}{\mathrm{x\; -\; 1}}$ = - 2

Every number is being divided by 1.

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

$\frac{\mathrm{y\; -\; 1}}{\mathrm{x\; -\; 1}}$ = $\frac{-2}{1}$

Cross multiply

1 x (y - 1) = -2 x (x - 1)

y - 1 = -2x + 2
y = -2x + 2 + 1
y = -2x + 3

Araba is 3 years younger means when you add 3 years to Araba's age, you will get the sister's age

Let x = Araba's age

The sister's age = x + 3

Sum means addition.

If the sum of their ages is 17 years means when you add their ages, you will get 17

Araba's age + Sister's age = 17

x + x + 3 = 17
2x = 17 - 3
2x = 14

Divide both sides by 2

x = $\frac{14}{2}$ = 7

∴ Araba's age = 7 years

Total oranges shared by the 50 students = 50 x number of oranges each got

Each got 15 oranges
Total oranges shared by the 50 students = 50 x 15
Total oranges shared by the 50 students = 750

Number of oranges got by each of the 30 students = $\frac{\mathrm{Total\; oranges}}{30}$

The 30 students shared the same 750 oranges.

Number of oranges got by each of the 30 students = $\frac{750}{30}$

Note:
1. the zero (0) cancel each other
2. 3 divides itself 1 time and 75, 25 times

Note:
1. Bearing starts from the north clockwise
2. Alternate angle is formed from the line drawn between the two locations (points)
3. 120° = 90° + 30°

Bearing of P from Q = 90° + 90° + 90° + 30°
Bearing of P from Q = 300°

m = $\left(\begin{array}{c}5\\ -1\end{array}\right)$ and n = $\left(\begin{array}{c}-4\\ 2\end{array}\right)$

2m + n = 2 x $\left(\begin{array}{c}5\\ -1\end{array}\right)$ + $\left(\begin{array}{c}-4\\ 2\end{array}\right)$

2m + n = $\left(\begin{array}{c}\mathrm{5\; x\; 2}\\ \mathrm{-1\; x\; 2}\end{array}\right)$ + $\left(\begin{array}{c}-4\\ 2\end{array}\right)$

2m + n = $\left(\begin{array}{c}10\\ -2\end{array}\right)$ + $\left(\begin{array}{c}-4\\ 2\end{array}\right)$

2m + n = $\left(\begin{array}{c}\mathrm{10\; +\; -4}\\ \mathrm{-2\; +\; 2}\end{array}\right)$

2m + n = $\left(\begin{array}{c}\mathrm{10\; -\; 4}\\ 0\end{array}\right)$

2m + n = $\left(\begin{array}{c}6\\ 0\end{array}\right)$

Express both the numerator and the denominator to be in the same base.

27 = 3 x 3 x 3 = 3³

Apply (a^m)ⁿ = a^{m x n}

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = $\frac{{(3^3)}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = $\frac{3^{\mathrm{3\; x\; (m\; +\; 1)}}}{3^{\mathrm{m\; +\; 2}}}$

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = $\frac{3^{\mathrm{3m\; +\; 3}}}{3^{\mathrm{m\; +\; 2}}}$

Law of division of indices

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

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = 3^{3m + 3 - (m + 2)}

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = 3^{3m + 3 - m - 2}

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = 3^{3m - m + 3 - 2}

$\frac{{27}^{\mathrm{m\; +\; 1}}}{3^{\mathrm{m\; +\; 2}}}$ = 3^{2m + 1}

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

Number of white balls = 15

Total balls = 15 + 25 = 40

P(white) = $\frac{15}{40}$ = $\frac{3}{8}$

Note: 5 divides 15, 3 times and 40, 8 times

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

A rectangle has 2 equal lengths and 2 equal widths.

2 x Length + 2 x Width = Perimeter.

Area of a rectangle = Length x Width

Area = 18
Width = 2

18 = Length x 2

Divide both sides by 2

Length = $\frac{18}{2}$ = 9 cm

Perimeter = 2 x 9 cm + 2 x 2 cm
Perimeter = 18 cm + 4 cm
Perimeter = 22 cm

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

Principal = 400
Time = 2
Rate = 10

Simple Interest = $\frac{\mathrm{400\; x\; 2\; x\; 10}}{100}$ = 4 x 2 x 10 = 80

Note: the two zeros (00) in 400 and 100 cancel each other

∴ Simple interest = GH₵ 80.00

Factorize 4m² + 12m - 8m - 24 out

4m² + 12m - 8m - 24 = 4m² + 12m - 8m - 24

4m² + 12m - 8m - 24 = 4m(m + 3) - 8)(m + 3)

4m² + 12m - 8m - 24 = (m + 3)( 4m - 8)

The other factor = m + 3

The percentage of the initial selling price (cost) is 100%

The percentage for the new selling price = 100% - 10%

The percentage for the new selling price = 90%

If 100% = GH₵ 600

90% = $\frac{\mathrm{GH₵\; 600\; x\; 90\%}}{\mathrm{100\%}}$ = GH₵ 540.00

Note:
1. The zeros (00) in 600 and 100 cancel each other
2. 6 x 90 = 540

$\frac{1}{\mathrm{m}}$ = $\frac{1}{\mathrm{p}}$ + $\frac{1}{\mathrm{r}}$

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators. The L.C.M of m, p and r is m x p x r

m x p x r x $\frac{1}{\mathrm{m}}$ = m x p x r x $\frac{1}{\mathrm{p}}$ + m x p x r x $\frac{1}{\mathrm{r}}$

Cancel out

p x r x 1 = m x r x 1 + m x p x 1

pr = mr + mp

Factorize m out

pr = m(r + p)

Divide both sides by r + p

m = $\frac{\mathrm{pr}}{\mathrm{r\; +\; p}}$

Point + Translation Vector = Image

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

Translation Vector = $\left(\begin{array}{c}-1\\ 3\end{array}\right)$

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

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

$\left(\begin{array}{c}\mathrm{-2\; -\; 1}\\ 6\end{array}\right)$ = R

$\left(\begin{array}{c}-3\\ 6\end{array}\right)$ = R

∴ The coordinates of R = (-3, 6)

Method I

The leap is a linear arithmetic sequence with 3 as interval (common difference)

The nth term of a linear arithmetic sequence is given as:

U_n = U₁ + (n - 1)d

Where U₁ is the first term
n is the nth position
d is the common difference

U₁ = 4

d = 7 - 4 = 10 - 7 = 3

Each succeeding term is obtained by adding 3 to the preceding term.

U₁₀ = 4 + (10 - 1) x 3

U₁₀ = 4 + 9 x 3

U₁₀ = 4 + 27

U₁₀ = 31

Method II

Common difference = 7 - 4 = 10 - 7 = 3

Each succeeding term is obtained by adding 3 to the preceding term.

1^st = 4
2^nd = 4 + 3 = 7
3^rd = 7 + 3 = 10
4^th = 10 + 3 = 13
5^th = 13 + 3 = 16
6^th = 16 + 3 = 19
7^th = 19 + 3 = 22
8^th = 22 + 3 = 25
9^th = 25 + 3 = 28
10^th = 28 + 3 = 31

If you add 3 hours to 9:15 pm, you will arrive at 00:15 am.

From 00:15 to 2:45 is 2 hours 30 minutes.

Total time for the journey = 3 hours + 2 hours 30 minutes

Total time for the journey = 5 hours 30 minutes

Note: x multiplied by x is x².

3x - 2(3 + 2x) + x(2x + 4) = 3x - 2(3 + 2x) + x(2x + 4)

3x - 2(3 + 2x) + x(2x + 4) = 3x - 2x3 - 2 x 2x + x x 2x + x x 4

3x - 2(3 + 2x) + x(2x + 4) = 3x - 6 - 4x + 2x² + 4x

Group like terms

3x - 2(3 + 2x) + x(2x + 4) = 2x² + 3x - 4x + 4x - 6

Note: - 4x + 4x = 0

3x - 2(3 + 2x) + x(2x + 4) = 2x² + 3x - 6

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

Opposite side of 30° is 5 cm and hypothenuse side is PQ

sin 30° = $\frac{\mathrm{5\; cm}}{\mathrm{|PQ|}}$

But sin 30° = $\frac{1}{2}$

$\frac{1}{2}$ = $\frac{\mathrm{5\; cm}}{\mathrm{|PQ|}}$

Cross multiply

1 x PQ = 2 x 5 cm

PQ = 10 cm

Note
1. Rolling a fair die does not influence the outcomes of tossing a fair coin
2. Neither tossing a fair coin affects the outcomes of rolling a fair die
3. Both events do not affect each other so they are independent events

Samples for coin = {Head, Tail}

Number of samples for coin = 2

Number of tail(s) = 1

P(Tail) = $\frac{1}{2}$

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

Number of samples for die = 6

Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.

Odd numbers from the die samples are {1, 3, 5}

Number of odd numbers = 3

P(Odd) = $\frac{3}{6}$ = $\frac{1}{2}$

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

P(Tail and Odd) = P(Tail) x P(Odd)

P(Tail and Odd) = $\frac{1}{2}$ x $\frac{1}{2}$

P(Tail and Odd) = $\frac{\mathrm{1\; x\; 1}}{\mathrm{2\; x\; 2}}$ = $\frac{1}{4}$

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

The gradient will be zero only when the y coordinates are the same for both points. Thus when Y₂ = Y₁

When both points have the same y coordinates, the straight line is horizontal.

A questionnaire is a set of printed or written questions with a choice of answers, devised for the purposes of a survey or statistical study.

Number of boxes = $\frac{\mathrm{Total\; number\; of\; apples}}{\mathrm{Number\; of\; apples\; per\; box}}$

Number of boxes = $\frac{1800}{120}$ = 15

Note:
1. The zero (0) in each of 1800 and 120 cancel each other
2. 12 divides 180, 15 times

Amount initially given = Cost of the three items + change

Amount initially given = GH₵ 72 + GH₵ 1,105 + GH₵ 216 + GH₵ 107

Amount initially given = GH₵ 1,500.00

The mode is the one which occurs the most/ the one with the highest frequency.

Number of times die was thrown = Sum of the frequencies

Number of times die was thrown = 4 + 3 + 3 + 2 + 3 + 5

Number of times die was thrown = 20

Rotation through 360° or 0 degree is equivalent to no rotation at all

Number of pages = 10 x time + Unread pages

Number of pages = 50
time = t
unread pages = N

50 = 10 x t + N

Make N the subject

N = -10t + 50

You can verify the relation. Assuming the student reads for 5 hours.

Number of unread pages (N) = -10 x 5 + 50

Number of unread pages (N) = -50 + 50

Number of unread pages (N) = 0

This means a student will complete the story book in 5 hours time.

Assuming the student reads for 4 hours.

Number of unread pages (N) = -10 x 4 + 50

Number of unread pages (N) = -40 + 50

Number of unread pages (N) = 10

This means after 4 hours, there will be 10 unread pages.