The commutative property states that we can swap numbers over and still get the same answer.

The associative property states that it doesn't matter how we group the numbers (i.e. which we calculate first).

The distributive property states that multiplying the sum of two or more addends by a number is the same as multiplying each addend individually by the number and then adding the products together.

Change the decimal 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.

0.1 = 0.⌒1. = 1 x 10^-1

0.02 = 0.⌒0⌒2. = 2 x 10^-2

0.003 = 0.⌒0⌒0⌒3. = 3 x 10^-3

0.1 x 0.02 x 0.003 = 1 x 10^-1 x 2 x 10^-2 x 3 x 10^-3

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

0.1 x 0.02 x 0.003 = 1 x 2 x 3 x 10^-1 x 10^-2 x 10^-3

0.1 x 0.02 x 0.003 = 6 x 10^{-1 + -2 + -3}

0.1 x 0.02 x 0.003 = 6 x 10^{-1 - 2 - 3}

0.1 x 0.02 x 0.003 = 6 x 10^-6

The total percentage of the candidates is 100%

Percentage of candidates who passed = 100% - Percentage who failed

Percentage of candidates who passed = 100% - 25%
Percentage of candidates who passed = 75%

If 75% = 150

25% = $\frac{\mathrm{150\; x\; 25\%}}{\mathrm{75\%}}$ = 2 x 25 = 50

Note: 75 divides itself 1 and 150, 2 times.

Faces of a cube

Law of division of indices

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

2⁸ ÷ 2³ = 2^{8 - 3}

2⁸ ÷ 2³ = 2⁵

The intersection (where the two circle join) represents pupils who speak both Twi and Ga

The region outside the two circles represent pupils who don't speak any of the two languages (neither Twi nor Ga)

The easiest way of finding the L.C.M is using the prime factors method.

Prime factors of 10

$$\begin{array}{cc}\mathrm{}& 10\\ 2& 5\\ 5& 1\end{array}$$

Prime factors of 10 = 2 x 5

Prime factors of 15

$$\begin{array}{cc}\mathrm{}& 15\\ 3& 5\\ 5& 1\end{array}$$

Prime factors of 15 = 3 x 5

Prime factors of 25

$$\begin{array}{cc}\mathrm{}& 25\\ 5& 5\\ 5& 1\end{array}$$

Prime factors of 25 = 5 x 5

Prime factors of 25 = 5²

The L.C.M is the product/multiplication of each of the prime factors in their highest power.

L.C.M = 2 x 3 x 5² = 6 x 5 x 5 = 150

< means less than

≤ means less than or equal to

> means greater than

≥ means greater than or equal to

When the solid is shaded, it is either ≤ or ≥ . The solid above is not shaded so is neither of them.

If the direction of the arrow faces the right, then > and the left < . You would realize that the direction of the arrow and the inequality sign both faces the same direction.

From the above number line the arrow is facing the right. Hence x > -3

$\frac{\mathrm{2a}}{3}$ - $\frac{\mathrm{(a\; -\; b)}}{2}$ = $\frac{\mathrm{2a}}{3}$ - $\frac{\mathrm{(a\; -\; b)}}{2}$

$\frac{\mathrm{2a}}{3}$ - $\frac{\mathrm{(a\; -\; b)}}{2}$ = $\frac{\mathrm{2\; x\; 2a\; -\; 3(a-b)}}{6}$

Note: - x - = +. - 3 x -b = +3b

$\frac{\mathrm{2a}}{3}$ - $\frac{\mathrm{(a\; -\; b)}}{2}$ = $\frac{\mathrm{4a\; -\; 3a\; +\; 3b}}{6}$

$\frac{\mathrm{2a}}{3}$ - $\frac{\mathrm{(a\; -\; b)}}{2}$ = $\frac{\mathrm{a\; +\; 3b}}{6}$

Significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact.

The significant figures in 0.00025 are 2 and 5. The number next to the first significant figure is more than 4 hence 1 is added to the first significant figure.

0.00025 = 0.0003

y : 28 = 5 : 7

Change the ratios to fraction format.

$\frac{\mathrm{y}}{28}$ = $\frac{5}{7}$

Cross multiply.

y x 7 = 28 x 5

Divide both sides by 7

y = $\frac{\mathrm{28\; x\; 5}}{7}$

Note: 7 divides itself 1 and 28, 4 times.

y = 4 x 5
y = 20

121_five = 1 x 5² + 2 x 5¹ + 1 x 5⁰

Note:
1. Any number raised to the power 0 is 1
2. Any number raised to the power 1 is the same number
3. 5² = 5 x 5

121_five = 1 x 25 + 2 x 5 + 1 x 1

121_five = 25 + 10 + 1

121_five = 36

If the x interval is 1 and the y intervals are also the same, then the mapping is linear. You can find the next image by adding the interval to the preceding number.

The y interval is 3.

k = 10 + 3 = 13

Rotation through 180^o about the origin

$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ → $\left(\begin{array}{c}\mathrm{-x}\\ \mathrm{-y}\end{array}\right)$

Thus both the x and y coordinates are negated.

$\left(\begin{array}{c}3\\ 4\end{array}\right)$ → $\left(\begin{array}{c}-3\\ -4\end{array}\right)$

K(3, 4) → K'(-3, -4)

Ordering Fractions

Before you can compare fractions, their denominators must be the same. Once they are the same, you can arrnage the fractions in the magnitude of their numerators.

To make the denominators the same, you must multiply both the numerator and the denominators of each fraction by the product (multiplication) of the denominators of the other fractions.

$\frac{1}{2}$ = $\frac{\mathrm{1\; x\; 20\; x\; 4}}{\mathrm{2\; x\; 20\; x\; 4}}$ = $\frac{80}{160}$

$\frac{17}{20}$ = $\frac{\mathrm{17\; x\; 2\; x\; 4}}{\mathrm{20\; x\; 2\; x\; 4}}$ = $\frac{136}{160}$

$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 2\; x\; 20}}{\mathrm{4\; x\; 2\; x\; 20}}$ = $\frac{120}{160}$

Descending means from the highest to the lowest.

Since the denominators are the same, you can arrange the fractions base on their numerators.

The descending order of magnitude of the numerators are 136, 120, 80

Descending order of the fractions are therefore $\frac{17}{20}$, $\frac{3}{4}$,$\frac{1}{2}$

x-axis as mirror/reflection in x-asis

$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ → $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{-y}\end{array}\right)$

Thus the y is negated.

K$\left(\begin{array}{c}3\\ 5\end{array}\right)$ → K'$\left(\begin{array}{c}3\\ -5\end{array}\right)$ = K'(3, -5)

Selling Price% = Cost Price% + Profit%

Cost Price% = 100%

Profit% = 20%

Selling Price% = 100% + 20%
Selling Price% = 120%

The selling Price = ₵300,000

If 120% = ₵300,000

100% = $\frac{\mathrm{₵300,000\; x\; 100\%}}{\mathrm{120\%}}$

Note:
1. the zeros (0) at the end of 120 and 100 cancel each other
2. 2 divides 10, 5 times and 12, 6 times
3. 6 divides itself 1 time and 30, 5 times

100% = ₵50,000 x 5

100% = ₵250,000

∴ cost price = ₵250,000

v² = u² + 2ax

v² - u² = 2ax

Divide both sides by 2a

x = $\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{\mathrm{2a}}$

Total oranges = 95 + x + 2x
Total oranges = 95 + 3x

Sum of angle of sectors of a pie chart is 360°

y + 80° + 60° + 120° = 360°

y + 260° = 360°

y = 360° - 260°
y = 100°

Method I

Using the difference in the angle of sectors.

The angle of sector for more students who offered Science subjects than Arts subjects is 120° - 80° = 40°

The entire 360° is 720 students.

If 360° = 720

40° = $\frac{\mathrm{720\; x\; 40°}}{\mathrm{360°}}$ = 2 x 40 = 80

Note: 360 divides itself 1 and 720, 2 times.

∴ 80 more students offered Science subjects than Arts subjects.

Method II

Finding the number of students who offered each subjects and calculate the difference (subtraction).

The entire 360° is 720 students.

Angle of sector for Science = 120°

If 360° = 720

120° = $\frac{\mathrm{720\; x\; 120°}}{\mathrm{360°}}$ = 2 x 120 = 240

Angle of sector for Arts = 80°

If 360° = 720

80° = $\frac{\mathrm{720\; x\; 80°}}{\mathrm{360°}}$ = 2 x 80 = 160

Number of more students who offered Science subjects than Arts subjects = 240 - 160 = 80

Kofi is two years older than Ama

This means if you add 2 years to Ama's age, you would get Kofi's age.

Let x = Ama's age

Kofi's age = x + 2

If the sum of their ages is 16

Sum means addition. Hence if you add their ages, you are suppose to get 16.

Ama's age + Kofi's age = 16

x + x + 2 = 16

2x + 2 = 16
2x = 16 - 2
2x = 14

Divide both sides by 2.

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

∴ Ama is 7 years old.

J = $\frac{1}{2}$mv²

v = 4 and J = 12

12 = $\frac{1}{2}$m x 4²

4² = 4 x 4 = 16

12 = $\frac{1}{2}$m x 16

Note: 2 divides itself 1 and 16, 8 times.

12 = m x 8

Divide both sides by 8

m = $\frac{12}{8}$ = $\frac{3}{2}$ = 1.5

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

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

Base of triangle = 3 cm
Height of triangle = |PQ|

Area of the triange = 6 cm²

$\frac{1}{2}$ x 3 x |PQ| = 6

Get rid of the fraction by multiplying both sides by the denominator (2)

2 x $\frac{1}{2}$ x 3 x |PQ| = 6 x 2

3 x |PQ| = 6 x 2

Divide both sides by 3.

|PQ| = $\frac{\mathrm{6\; x\; 2}}{3}$ = 2 x 2 = 4 cm

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

Apply pythagorean theorem to solve for |PR|

The hypothenuse (c) is the longest side. The side facing the 90°

|PR|² = |PQ|² + |QR|²

|PQ| = 4 and |QR| = 3

|PR|² = 4² + 3²

4² = 4 x 4 = 16

3² = 3 x 3 = 9

|PR|² = 16 + 9

|PR|² = 25

25 = 5 x 5 = 5²

|PR|² = 5²

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

|PR| = 5 cm

You can apply pythagorean theorem to solve for the length of the diagonal.

The diagonal is the side facing the 90° or the longest side of triangle ABC and BCD hence d is the hypothenuse (c).

d² = 8² + 6²

8² = 8 x 8 = 64

6² = 6 x 6 = 36

d² = 64 + 36

d² = 100

100 = 10 x 10 = 10²

d² = 10²

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

d = 10 cm

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$)

Change the mixed fraction to improper fraction.

2$\frac{2}{3}$ = $\frac{\mathrm{3\; x\; 2\; +\; 2}}{3}$ = $\frac{\mathrm{6\; +\; 2}}{3}$ = $\frac{8}{3}$

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$($\frac{8}{3}$ + $\frac{5}{6}$)

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$($\frac{\mathrm{2\; x\; 8\; +\; 1\; x\; 5}}{6}$)

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$($\frac{\mathrm{16\; +\; 5}}{6}$)

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$($\frac{21}{6}$)

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{1}{3}$ x $\frac{21}{6}$

Note: 3 divides itself 1 and 21, 7 times.

$\frac{1}{3}$(2$\frac{2}{3}$ + $\frac{5}{6}$) = $\frac{7}{6}$

The sum of ratios represent the total cattle shared and the smallest ratio represent the smallest share.

Sum of ratios = 7 + 6 + 5 = 18

Smallest ratio = 5

If 18 = 360

5 = $\frac{\mathrm{360\; x\; 5}}{18}$ = 20 x 5 = 100

Note: 18 divides itself 1 and 36, 2 times.

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 middle mark in 3, 3, 4, 5, 6, 7, 7, 7, 8 is 6 hence the median is 6.

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

From the distribution, 7 appeared the most (3 times) hence 7 is the mode.

$\frac{1}{3}$, $\frac{2}{5}$, $\frac{7}{15}$, ...

If the interval between consecutive numbers are the same, then the sequence is linear. The subsequent term is obtained by adding the interval to the preceding term.

$\frac{2}{5}$ - $\frac{1}{3}$ = $\frac{7}{15}$ - $\frac{2}{5}$ = $\frac{1}{15}$

Note

$\frac{2}{5}$ - $\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 2\; -\; 5\; x\; 1}}{15}$ = $\frac{\mathrm{6\; -\; 5}}{15}$ = $\frac{1}{15}$

$\frac{7}{15}$ - $\frac{2}{5}$ = $\frac{\mathrm{1\; x\; 7\; -\; 3\; x\; 2}}{15}$ = $\frac{\mathrm{7\; -\; 6}}{15}$ = $\frac{1}{15}$

The fourth term is obtained by adding $\frac{1}{15}$ to the third term ($\frac{7}{15}$ and the fifth term is obtained by adding $\frac{1}{15}$ to the fourh term.

Fourth term = $\frac{7}{15}$ + $\frac{1}{15}$ = $\frac{\mathrm{7\; +\; 1}}{15}$ = $\frac{8}{15}$

Fifth term = $\frac{8}{15}$ + $\frac{1}{15}$ = $\frac{\mathrm{8\; +\; 1}}{15}$ = $\frac{9}{15}$ = $\frac{3}{5}$

Note: 3 divides 9, 3 times and 15, 5 times.

The more the workers, the lesser the time and the lesser the workers the more the time.

Hence time (t) is inversely proportional to the number of workers (w).

t ∝ $\frac{1}{\mathrm{w}}$

Remove the proportionality by introducing a constant (k).

t = k x $\frac{1}{\mathrm{w}}$

Use the known information to solve for k

30 workers used 21 days.

21 = k x $\frac{1}{30}$

k = 21 x 30

The general equation is:

t = 21 x 30 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{21\; x\; 30}}{\mathrm{w}}$

Calculate how days it will take the 14 workers.

w = 14.

t = $\frac{\mathrm{21\; x\; 30}}{14}$ = 3 x 15 = 45 days

Note:
1. 7 divides 21, 3 times and 14, 2 times
2. 2 divides itself 1 and 30, 15 times

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

Total balls = 6 + 8 = 14

Number of white balls = 6

Probability = $\frac{6}{14}$ = $\frac{3}{7}$

Note: 2 divides 6, 3 times and 14, 7 times.

Area of a rectangle = Length x Breadth

Area of the face PQRV = x cm x z cm = xz cm²

The shorter part is one-quarter of the length of the rod

one-quarter = $\frac{1}{4}$

of means multiplication (x).

The shorter part = $\frac{1}{4}$ x 200 cm = 50 cm

Longer part = 200 cm - 50 cm = 150 cm

The shorter part as a percentage of the longer part = $\frac{50}{150}$ x 100%

Note: 50 divides itself 1 and 150, 3 times.

The shorter part as a percentage of the longer part = $\frac{1}{3}$ x 100%

The shorter part as a percentage of the longer part = $\frac{\mathrm{100\%}}{3}$ = 33.33%

$\frac{\mathrm{x}}{4}$ = 2

Get rid of the fraction by multiplying both sides of the equation by the denominator (4).

4 x $\frac{\mathrm{x}}{4}$ = 2 x 4

x = 8

Diameter = 2 x radius

Circumference = 2 x π x radius

Circumference = 2 x radius x π

But 2 x radius = diameter

Circumference = diameter x π

Circumference = 15.4 m

π = $\frac{22}{7}$

$\frac{22}{7}$ x diameter = 15.4 m

Get rid of the fraction by multiplying both sides by the denominator (7).

7 x $\frac{22}{7}$ x diameter = 15.4 m x 7

22 x diameter = 15.4 m x 7

Divide both sides by 22.

diameter = $\frac{\mathrm{15.4\; m\; x\; 7}}{22}$ = 4.9 m