P = {2,3,5,7} and Q = {2,4,6,8}

Intersection (∩) is the set of elements that can be found in both sets.

Only 2 can be found in both sets P and Q, hence P∩Q = {2}

An integer is a whole number (not a fractional number) that can be positive, negative, or zero.

The Lowest Common Multiple (LCM) is the multiplication of each of the prime factors in their highest powers.

The prime factors are 2, 3 and 5

The highest power for the prime number 2 is 3
The highest power for the prime number 5 is 2
The highest power for the prime number 3 is 2

Lowest Common Multiple (LCM) = 2³ x 3² x 5²

3-5 = -2
7+4 = 11

-4(3-5)+10-3(7+4)+30 = -4(-2)+10-3(11)+30

The ones in the bracket multiplies the numbers in front.

-4(3-5)+10-3(7+4)+30 = -4 x -2 +10 -3 x 11 + 30

- x - = +

-4(3-5)+10-3(7+4)+30 = 8 + 10 - 33 + 30

-4(3-5)+10-3(7+4)+30 = 18 - 3

-4(3-5)+10-3(7+4)+30 = 15

Divide the 15 m by 12 to get each part

Write the remainders from the bottom to the top

42 = 101010_two

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

Law of division of indices

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

$\frac{5^{7}x{5}^4}{5^2}$ = $\frac{5^{\mathrm{7+4}}}{5^2}$

$\frac{5^{7}x{5}^4}{5^2}$ = $\frac{5^{11}}{5^2}$

$\frac{5^{7}x{5}^4}{5^2}$ = 5^11-2

$\frac{5^{7}x{5}^4}{5^2}$ = 5⁹

Method I

The complete is 100%

400 litres = 100%
100 litres = $\frac{\mathrm{100\; litres\; x\; 100\%}}{\mathrm{400\; litres}}$ = 25% used

Percentage left = 100% - 25% = 75%

Method II

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Water left = 400 litres - 100 litres = 300 litres

Percentage Left = $\frac{300}{400}$ x 100% = 3 x 25% = 75%

Note: The two zeros (00) in 300 and 400 cancels each other and 4 divides itself, once and 100, 25 times.

Apply Pythagoras theorem

The side facing the right-angled is the hypothenus (c)

AC² = 3² + 4²

a² = a x a

AC² = 3 x 3 + 4 x 4

AC² = 9 + 16

AC² = 25

A number that can multiply itself and give 25.

25 = 5 x 5 = 5²

AC² = 5²

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

AC = 5 cm

In order to compare the fractions, we have to make sure they have the same denominator by finding the equivalent fraction for each of the fractions.

To find the equivalent fractions, follow the steps below:

1. Find the L.C.M of the denominators of the fraction (3,7,5,2).

The L.C.M is 3 x 7 x 5 x 2 = 210

2. Multiply both the numerator and denominator of each of the fractions by the number of times the denominator of that fraction divides the L.C.M

Equivalent Fractions

$\frac{2}{3}$ = $\frac{\mathrm{2\; x\; 7\; x\; 5\; x\; 2}}{\mathrm{3\; x\; 7\; x\; 5\; x\; 2}}$ = $\frac{140}{210}$

$\frac{5}{7}$ = $\frac{\mathrm{5\; x\; 3\; x\; 5\; x\; 2}}{\mathrm{7\; x\; 3\; x\; 5\; x\; 2}}$ = $\frac{150}{210}$

$\frac{2}{5}$ = $\frac{\mathrm{2\; x\; 3\; x\; 7\; x\; 2}}{\mathrm{5\; x\; 3\; x\; 7\; x\; 2}}$ = $\frac{84}{210}$

$\frac{1}{2}$ = $\frac{\mathrm{1\; x\; 3\; x\; 7\; x\; 5}}{\mathrm{2\; x\; 3\; x\; 7\; x\; 5}}$ = $\frac{105}{210}$

Once the denominators are the same for the equivalent fractions, you can compare their numerators in the order requested.

In this question, the arrangement is in descending (highest to lowest)

150 > 140 > 105 > 84

Replace the comparism by their corresponding original fractions.

Descending order = $\frac{5}{7}$,$\frac{2}{3}$,$\frac{1}{2}$,$\frac{2}{5}$

x → 10 - 2x

When x = 3

3 → 10 - 2(3)

3 → 10 - 2 x 3

3 → 10 - 6

3 → 4

$\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ = $\frac{\mathrm{1\; x\; 9\; +\; 1\; x\; 3\; +\; 1\; x\; 1}}{27}$

$\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ = $\frac{\mathrm{9\; +\; 3\; +\; 1}}{27}$

$\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ = $\frac{13}{27}$

2x = 5(x - 2) + 7

Expand the bracket

2x = 5 x x - 5 x 2 + 7

2x = 5x - 10 + 7

2x -5x = - 10 + 7

-3x = - 3

Divide both sides by -3

x = $\frac{-3}{-3}$ = 1

Average = $\frac{\mathrm{Sum\; of\; day\; temperatures}}{\mathrm{Number\; of\; day\; temperatures}}$

Average = $\frac{\mathrm{33+29+32+34+32+30+30}}{7}$

Average = $\frac{220}{7}$ = 31.4 ^oC

Division

The least temperature for the day is 29 which was on Tuesday.

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

Yellow balls = 30 - 16 = 14

Total sample = 30

Probability (Yellow) = $\frac{14}{30}$ = $\frac{7}{15}$

Note: 2 divides 14, 7 times and 30, 15 times.

$\frac{1}{4}$(x + 3) ≤ 2x - 1

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

4 x $\frac{1}{4}$(x + 3) ≤ 4 x 2x - 1 x 4

1(x + 3) ≤ 4 x 2x - 1 x 4

1 x x + 1 x 3 ≤ 8x - 4

x + 3 ≤ 8x - 4

x - 8x ≤ -4 - 3

- 7x ≤ -7

Divide both sides by -7. When dividing by a negative, the sign direction changes.

x ≥ $\frac{-7}{-7}$

x ≥ 1

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

Perimeter of the shape = 25 cm + 24 cm + half the circumference of the circular portion

Circumference = 2πr

Where r = radius.

Diameter = 2 x radius

Circumference = π x diameter

Length of the circular portion = $\frac{\mathrm{Circumference}}{2}$

Length of the circular portion = $\frac{\mathrm{\pi \; x\; diameter}}{2}$

Perimeter of the shape = 25 cm + 24 cm + $\frac{\mathrm{\pi \; x\; diameter}}{2}$

But perimeter = 71

71cm = 25 cm + 24 cm + $\frac{\mathrm{\pi \; x\; diameter}}{2}$

71 = 49 + $\frac{\mathrm{\pi \; x\; diameter}}{2}$

71 - 49 = $\frac{\mathrm{\pi \; x\; diameter}}{2}$

22 = $\frac{\mathrm{\pi \; x\; diameter}}{2}$

Multiply both sides of the equation by 2 to get rid of the denominator 2.

22 x 2 = π x diameter

But π = $\frac{22}{7}$

22 x 2 = $\frac{22}{7}$ x diameter

Get rid of the fraction by multiplying both sides of the equation by the denominator 7

22 x 2 x 7 = 22 x diameter

Divide both sides by 22 to make diameter the subject

Diameter = $\frac{\mathrm{22\; x\; 2\; x\; 7}}{22}$ = 2 x 7 = 14 cm

$\frac{\mathrm{3x}}{4}$ - $\frac{\mathrm{x\; -\; y}}{3}$ = $\frac{\mathrm{3\; x\; 3x\; -\; 4(x\; -\; y)}}{12}$

$\frac{\mathrm{3x}}{4}$ - $\frac{\mathrm{x\; -\; y}}{3}$ = $\frac{\mathrm{9x\; -\; 4\; x\; x\; -4\; x\; -\; y}}{12}$

- x - = +

$\frac{\mathrm{3x}}{4}$ - $\frac{\mathrm{x\; -\; y}}{3}$ = $\frac{\mathrm{9x\; -\; 4x\; +\; 4y}}{12}$

$\frac{\mathrm{3x}}{4}$ - $\frac{\mathrm{x\; -\; y}}{3}$ = $\frac{\mathrm{5x\; +\; 4y}}{12}$

Method I

Kojo's percentage relative to Afua = 100% + 20% = 120%

Afua's percentage = 100%

120% = 6kg
100% = $\frac{\mathrm{6\; kg\; x\; 100\%}}{\mathrm{120\%}}$ = 5.0 kg

Note: One of the zeros (0) in 100% and 120% cancels each other, 6 divides itself once and 12, 2 times. 2 divides itself once and 10, 5 times.

Method I

Let x = Afua's weight

Kojo's weight more than Afua = 6 - x

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Kojo's percentage more than Afua = $\frac{\mathrm{Kojo\text{'}s\; weight\; more\; than\; Afua}}{\mathrm{Afua\text{'}s\; weight}}$ x 100%

Kojo's percentage more than Afua = $\frac{\mathrm{6-x}}{\mathrm{x}}$ x 100%

But Kojo's percentage more than Afua = 20%

$\frac{\mathrm{6-x}}{\mathrm{x}}$ x 100% = 20%

Solve for the value of x

Get rid of the fraction by multiplying both sides of the equation by the denominator x

(6 - x) x 100 = 20 x x
6 x 100 - x x 100 = 20 x x
600 - 100x = 20x
600 = 20x + 100x
600 = 120x

Divide both sides by 120 to solve for x

x = $\frac{600}{120}$ = 5

Afua's weight = 5.0 kg

Volume of a cylinder = πr²h
r² = r x r

Radius = 2 cm
Height = 3 cm

Volume of a cylinder = π x 2 cm x 2 cm x 3 cm
Volume of a cylinder = π x 12 cm³
Volume of a cylinder = 12π cm³

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

Y₂ = 2
Y₁ = -2
X₂ = 3
X₁ = 5

Gradient = $\frac{\mathrm{2-(-2)}}{\mathrm{3-5}}$ = $\frac{\mathrm{2+2}}{-2}$ = $\frac{4}{-2}$ = -2

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

Simple Interest = GH₵11,250.00
Principal = GH₵ 150,000
Rate = 2.5
Time = ?

11250 = $\frac{\mathrm{150000\; x\; Time\; x\; 2.5}}{100}$

11250 = 1500 x 2.5 x Time
11250 = 150 x 10 x 2.5 x Time
11250 = 150 x 25 x Time

Divide both sides by 150 x 25 to make Time the subject

Time = $\frac{11250}{\mathrm{150\; x\; 25}}$ = 3 years

Note: 25 divides itself once and 11250, 450 times. 150 divides itself once and 450, 3 times.

11250 divided by 25

Change the decimal to fraction.

Every number is being divided by 1 which is not written.

3.75 = $\frac{3.75}{1}$

Count how many numbers are after the decimal point (n).

3.75 has two numbers (7 and 5) after the decimal point.

Multiply both the numerator and the denominator by 1 joined by n number of zeros.

In 3.75, we have to multiply both the numerator and denominator by 100 (two zeros since there are two numbers after the decimal point).

Multiplying both the numerator and the denominator by by 1 joined by n number of zeros changes the decimal point to whole number.

$\frac{3.75}{1}$ = $\frac{\mathrm{3.75\; x\; 100}}{\mathrm{1\; x\; 100}}$ = $\frac{375}{100}$

Simplify the fraction

$\frac{375}{100}$ = $\frac{15}{4}$

Note: 25 divides 375, 15 times and 100, 4 times.

375 divided by 25

Change $\frac{15}{4}$ to mixed fraction.

$\frac{15}{4}$ = 3$\frac{3}{4}$

If 1 cm = 100,000 cm on map
5 cm = $\frac{\mathrm{5\; x\; 100000}}{1}$ = 500000 cm on map

Change the centimeters to meters

100 cm = 1 m
500000 cm = $\frac{\mathrm{1\; m\; x\; 500000cm}}{\mathrm{100\; cm}}$ = 5000 m

Kilo means 1000
5,000 m = 5 km

Multiplying a vector by a number

c$\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ = c x $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{a\; x\; c}\\ \mathrm{b\; x\; c}\end{array}\right)$

Addition of vectors

$\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{c}\\ \mathrm{d}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{a\; +\; c}\\ \mathrm{b\; +\; d}\end{array}\right)$

Subtraction of vectors

$\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ - $\left(\begin{array}{c}\mathrm{c}\\ \mathrm{d}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{a\; -\; c}\\ \mathrm{b\; -\; d}\end{array}\right)$

r = $\left(\begin{array}{c}-3\\ 4\end{array}\right)$ and s = $\left(\begin{array}{c}1\\ -3\end{array}\right)$

r - 2s = $\left(\begin{array}{c}-3\\ 4\end{array}\right)$ - 2$\left(\begin{array}{c}1\\ -3\end{array}\right)$

r - 2s = $\left(\begin{array}{c}-3\\ 4\end{array}\right)$ - $\left(\begin{array}{c}\mathrm{1\; x\; 2}\\ \mathrm{-3\; x\; 2}\end{array}\right)$

r - 2s = $\left(\begin{array}{c}-3\\ 4\end{array}\right)$ - $\left(\begin{array}{c}2\\ -6\end{array}\right)$

r - 2s = $\left(\begin{array}{c}\mathrm{-3\; -\; 2}\\ \mathrm{4\; -\; -6}\end{array}\right)$

-- = +

r - 2s = $\left(\begin{array}{c}-5\\ \mathrm{4\; +\; 6}\end{array}\right)$

r - 2s = $\left(\begin{array}{c}-5\\ 10\end{array}\right)$

500 g of meat, 850 g of fish and 900 g of eggs
Total grams = 500 g + 850 g + 900 g = 2250 g

Kilo means 1000, hence divide the grams by 1000 to obtain the kilograms.

2250 g = $\frac{\mathrm{2250\; g}}{1000}$ = 2.⌒2⌒5⌒0.0 = 2.25 kg

Note

1. Every whole number has .0 at the end which is not written. 2250 = 2250.0

2. When dividing by 1n number of zeros, the result is the movement of the decimal point to the left n times. 1000 has 3 zeros, hence the decimal point in 2250.0 is moved to the left three times.

12 noon to 12 midnight is 12 hours

Change the mixed fraction of the minutes gained to improper fraction.

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

If 1 hour = $\frac{3}{2}$ minutes
12 hour = 12 hour x $\frac{3}{2}$ minutes ÷ 1 hour = 6 x 3 minutes = 18 minutes

If 3 hours = 600 books
5 hours = $\frac{\mathrm{5\; hours\; x\; 600\; books}}{\mathrm{3\; hours}}$ = 5 x 200 books = 1000 books

Note: 3 divides itself once and 600, 200 times.

275 = 90 + 90 + 90 + 5 at Busase

At Atoru = 90 + 5 = 95°

sol

Let x = number of pupils who study both French and English
Let E = represent English
Let F = represent French

8 + x + 10 = 24
x + 18 = 24
x = 24 - 18
x = 6

Number of pupils who study English = 8 + x
But x = 6
Number of pupils who study English = 8 + 6
Number of pupils who study English = 14

84 base ten to base five

Write the remainders from the bottom to the top

84 = 314_five

Scale factor = $\frac{\mathrm{Size\; of\; Image}}{\mathrm{Size\; of\; corresponding\; object\; side}}$

Scale factor = $\frac{\mathrm{8\; cm}}{\mathrm{4\; cm}}$ = $\frac{\mathrm{6\; cm}}{\mathrm{3\; cm}}$ = 2

The least number is a number to add to 308 to get the first multiple of 19 greater than 308.

Multiples of 19

19 x 1 = 19
19 x 2 = 19 + 19 = 38
19 x 3 = 38 + 19 = 57
19 x 4 = 57 + 19 = 76
19 x 5 = 76 + 19 = 95
19 x 6 = 95 + 19 = 114
19 x 7 = 114 + 19 = 133
19 x 8 = 133 + 19 = 152
19 x 9 = 152 + 19 = 171
19 x 10 = 171 + 19 = 190
19 x 11 = 190 + 19 = 209
19 x 12 = 209 + 19 = 228
19 x 13 = 228 + 19 = 247
19 x 14 = 247 + 19 = 266
19 x 15 = 266 + 19 = 285
19 x 16 = 285 + 19 = 304
19 x 17 = 285 + 19 = 323

The least multiple of 19 greater than 308 is 323

Let x = the number to add to 308

308 + x = 323
x = 323 - 308
x = 15

Note: You could start the multiplication of 19 from 19 x 10 since the result is simply joining zero (0) to the 19 to obtain 190. You could then continue with 19 x 11, 19 x 12, etc.

Let x = number of girls in the school
The number of girls exceeds the number of boys by 150 → number of boys = x - 150

Number of girls + Number of boys = 940
x + x - 150 = 940
2x = 940 + 150
2x = 1090

Divide both sides by 2 to solve for x

x = $\frac{1090}{2}$ = 545

There are 545 girls in the school.

1090 divided by 2

For fractions which are equivalent, the simplification of the fractions should be the same.

The fraction $\frac{12}{20}$ is equivalent to $\frac{3}{5}$ because when the fraction $\frac{12}{20}$ is simplified, the simplified fraction $\frac{3}{5}$ is obtained.

4 divides 20, 3 times and 20, 5 times giving $\frac{3}{5}$.

a + b = 180 (Sum of angles on a straight line)

b = 35^o (Alternate angles)

a + 35 = 180
a = 180 - 35
a = 145^o

Corresponding angles

– x(3 – 2x) = -x x 3 -x x – 2x

- x - = +

– x(3 – 2x) = -3x + 2x²

– x(3 – 2x) = 2x² -3x