Multiples of 4

4 x 1 = 4
4 x 2 = 4 + 4 = 8
4 x 3 = 8 + 4 = 12

4 + 2 = 6. In base 5, its 11. The 1 is written and the other 1 is carried forward.

2 + 2 = 4 and the carried forward is added making 5. 5 in base 5 is 10

At 5:35, the farmer would had spent 1 hour

Remaining minutes to 6:00 is 25 minutes.

At 6:18 am, total minutes = 25 + 18 = 43 minutes.

Total time taken is 1 hour 43 minutes.

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.

2474.5 = 2.⌒4⌒7⌒4.5 = 2.4745 x 10³

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$

Method I

Solving the whole numbers and fractions differently.

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [1 + 2 - 3]($\frac{1}{2}$ + $\frac{1}{4}$ - $\frac{5}{8}$)

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [3 - 3]($\frac{\mathrm{4\; x\; 1\; +\; 2\; x\; 1\; -\; 1\; x\; 5}}{8}$)

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = [0]($\frac{\mathrm{4\; +\; 2\; -\; 5}}{8}$)

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{\mathrm{6\; -\; 5}}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$

Method II

Changing the mixed fractions to improper fractions.

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

2$\frac{1}{4}$ = $\frac{\mathrm{4\; x\; 2\; +\; 1}}{4}$ = $\frac{\mathrm{8\; +\; 1}}{4}$ = $\frac{9}{4}$

3$\frac{5}{8}$ = $\frac{\mathrm{8\; x\; 3\; +\; 5}}{8}$ = $\frac{\mathrm{24\; +\; 5}}{8}$ = $\frac{29}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = 1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{3}{2}$ + $\frac{9}{4}$ - $\frac{29}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{\mathrm{4\; x\; 3\; +\; 2\; x\; 9\; -\; 1\; x\; 29}}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{\mathrm{12\; +\; 18\; -\; 29}}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{\mathrm{30\; -\; 29}}{8}$

1$\frac{1}{2}$ + 2$\frac{1}{4}$ - 3$\frac{5}{8}$ = $\frac{1}{8}$

When you analyze the sequence, you will realize that the rule for he sequence is:

U_n-1 + n + 1

Where n is the number of term.

When n = 1

U₁ = U_1-1 + 1 + 1

U₁ = U₀ + 2

U₁ = 0 + 2

U₁ = 2

When n = 2

U₂ = U_2-1 + 2 + 1

U₂ = U₁ + 3

U₂ = 2 + 3

U₂ = 5

When n = 3

U₃ = U_3-1 + 3 + 1

U₃ = U₂ + 4

U₃ = 5 + 4

U₃ = 9

When n = 4

U₄ = U_4-1 + 4 + 1

U₄ = U₃ + 5

U₄ = 9 + 5

U₄ = 14

When n = 5

U₅ = U_5-1 + 5 + 1

U₅ = U₄ + 6

U₅ = 14 + 6

U₅ = 20

When n = 6

U₆ = U_6-1 + 6 + 1

U₆ = U₅ + 7

U₆ = 20 + 7

U₆ = 27

When n = 1

U₇ = U_7-1 + 7 + 1

U₇ = U₆ + 8

U₇ = 27 + 8

U₇ = 35

Let x = the third number.

Sum means addition.

x + 10,250 + 9,750 = 28,542

x + 20,000 = 28,542

x = 28,542 - 20,000

x = 8,542

∴ the third number = 8,542

Number of boxes = $\frac{\mathrm{Number\; of\; pencils}}{\mathrm{12\; pencils}}$

Number of boxes = $\frac{\mathrm{135\; pencils}}{\mathrm{12\; pencils}}$ = 11.25 ≈ 11 boxes

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 number of time the denominator divides the L.C.M of the denominators.

The L.C.M of 20, 5, 4 and 10 is 20.

$\frac{13}{20}$ = $\frac{\mathrm{13\; x\; 1}}{\mathrm{20\; x\; 1}}$ = $\frac{13}{20}$

$\frac{3}{5}$ = $\frac{\mathrm{3\; x\; 4}}{\mathrm{5\; x\; 4}}$ = $\frac{12}{20}$

$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 5}}{\mathrm{4\; x\; 5}}$ = $\frac{15}{20}$

$\frac{7}{10}$ = $\frac{\mathrm{7\; x\; 2}}{\mathrm{10\; x\; 2}}$ = $\frac{14}{20}$

The fraction with the greatest numerator is the greatest fraction.

The numerator 15 from the $\frac{3}{4}$ is the greatest. Hence the greatest fraction is $\frac{3}{4}$

Fraction = $\frac{\mathrm{Number}}{\mathrm{Total}}$

Fraction = $\frac{\mathrm{₵325,000}}{\mathrm{₵550,000}}$ = $\frac{13}{22}$

Note: 25 divides 325, 13 times and 550, 22 times.

Cost for the four pencils = ₵200.00 x 4
Cost for the four pencils = ₵800.00

Cost for the five pens = ₵350.00.00 x 5
Cost for the five pens = ₵1,750.00

Total cost = ₵800.00 + ₵1,750.00
Total cost = ₵2,550

100% represens the amount

If 100% = ₵120,000

5% = $\frac{\mathrm{₵120,000\; x\; 5\%}}{\mathrm{100\%}}$ = ₵1200 x 5 = ₵6,000

Note: the zeros (0) at the end of 120000 and 100 cancel each other.

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

8 workers used 10 days.

10 = k x $\frac{1}{8}$

k = 10 x 8

The general equation is:

t = 10 x 8 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{10\; x\; 8}}{\mathrm{w}}$

Calculate how days it will take the 14 workers.

w = 5.

t = $\frac{\mathrm{10\; x\; 8}}{5}$ = 2 x 8 = 16 days

Note: 5 divides itself 1 time and 10, 2 times.

x → x² – 1

When x = 3

3 → 3² – 1

3 → 3 x 3 – 1

3 → 9 – 1

3 → 8

Cost for first 5 sheets = ₵2,000.00

Remaining sheets = 20 - 5 = 15

Cost for remaining sheets = 15 x ₵600.00 = ₵9,000

Total cost = ₵2,000.00 + ₵9,000 = ₵11,000

Change the metre to kilometre and seconds to hour

To change metre to kilometre, you divide the metre by 1000.

Note: kilo means 1000.

100 m = $\frac{100}{1000}$ km = $\frac{1}{10}$ km

Note: the zeros (0) at the end of 100 and 1000 cancel each other.

To change the seconds to hours you divide the seconds by 3600 (60 x 60).

12 seconds = $\frac{12}{3600}$ h = $\frac{1}{300}$ h

Note: 12 divides itself 1 time and 36, 3 times.

Per means division.

Kilometres per hour means the hour divides the kilometre.

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

Speed = $\frac{1}{10}$ km ÷ $\frac{1}{300}$ h

Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.

Speed = $\frac{1}{10}$ km x $\frac{300}{\mathrm{1h}}$ = 30 km/h

Note: the zeros (0) at the end of 10 and 300 cancel each other.

The sum of the ratios represent the total amount shared.

Sum of ratios = 3 + 4 + 5 = 12

Amina's ratio is 5.

If 5 = ₵40,000

12 = $\frac{\mathrm{₵40,000\; x\; 12}}{5}$ = ₵8,000 x 12 = ₵96,000

Note: 5 divides itself 1 time and 40, 8 times.

Original Price % = 100%

Selling Price % = 100% - 10% = 90%

Abban paid ₵630,000 at 90%.

If 90% = ₵630,000

100% = $\frac{\mathrm{₵630,000\; x\; 100\%}}{\mathrm{90\%}}$ = ₵7000 x 100 = ₵700,000.00

Note:
1. the zeros (0) at the end of 630,000 and 90 cancel each other
2. 9 divides itself 1 time and 63, 7 times

The percentage in 1990 = 100%

New percentage in 1998 = 100% + 20% = 120%

The population in 1990 was 88,000

If 100% = 88,000

120% = $\frac{\mathrm{120\%\; x\; 88,000}}{\mathrm{100\%}}$ = 120 x 880 = 105,600

Note: the zeros (0) at the end of 88000 and 100 cancel each other.

a * b = 2a – b

When a = 3 and b = 2

3 * 2 = 2 x 3 - 2
3 * 2 = 6 -2
3 * 2 = 4

r = $\frac{\mathrm{m\; +\; 1}}{\mathrm{m\; -\; 2}}$

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

(m - 2) x r = $\frac{\mathrm{m\; +\; 1}}{\mathrm{m\; -\; 2}}$ x (m - 2)

r(m - 2) = m + 1

rm - 2r = m + 1
rm - m = 1 + 2r

Factorize m out.

m(r - 1) = 2r + 1

Divide both sides by r - 1

m = $\frac{\mathrm{2r\; +\; 1}}{\mathrm{r\; -\; 1}}$

The percentage for the entire class is 100%

Let x = percentage of students who speak both Ga and Twi

40% - x% + x% + 75% - x% = 100%

115% - x% = 100%

115% - 100% = x%

15% = x

∴ 15% speaks both Ga and Twi

The percentage for the entire class is 100%

If 100% = 40

75% = $\frac{\mathrm{40\; x\; 75\%}}{\mathrm{100\%}}$ = 2 x 15 = 30

Note:
1. the zeros (0) at the end of 40 and 100 cancel each other
2. 2 divides 4, 2 times and 10, 5 times
3. 5 divides itself 1 and 75, 15 times

y-axis as mirror/reflection in y-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 x is negated.

$\left(\begin{array}{c}2\\ 1\end{array}\right)$ → $\left(\begin{array}{c}-2\\ 1\end{array}\right)$ = (-2, 1)

(2a + b)(a – b) = (2a + b)(a – b)
(2a + b)(a – b) = 2a(a – b) + b(a – b)
(2a + b)(a – b) = 2a² - 2ab + ab - b²
(2a + b)(a – b) = 2a² - ab - b²

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

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

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

3s + r = $\left(\begin{array}{c}-3\\ 12\end{array}\right)$ + $\left(\begin{array}{c}3\\ 2\end{array}\right)$

3s + r = $\left(\begin{array}{c}\mathrm{-3\; +\; 3}\\ \mathrm{12\; +\; 2}\end{array}\right)$

3s + r = $\left(\begin{array}{c}0\\ 14\end{array}\right)$

$\frac{1}{3}$y + $\frac{1}{5}$y = 8

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators (3 and 5). The L.C.M of 3 and 5 is 15.

15 x $\frac{1}{3}$y + 15 x $\frac{1}{5}$y = 8 x 15

5y + 3y = 8 x 15
8y = 8 x 15

Divide both sides by 8

y = $\frac{\mathrm{8\; x\; 15}}{8}$ = 15

Scale factor = $\frac{\mathrm{Enlarged\; Side}}{\mathrm{Corresponding\; Object\; Side}}$

Scale factor = $\frac{\mathrm{|ST|}}{\mathrm{|PQ|}}$ = $\frac{\mathrm{|QR|}}{\mathrm{|RT|}}$

|RT| = |QR| + |QT|

|RT| = 4 cm + 8 cm = 12 cm

$\frac{\mathrm{|ST|}}{2.4}$ = $\frac{12}{4}$

Cross multiply.

4 x |ST| = 2.4 x 12

Divide both sides by 4

|ST| = $\frac{\mathrm{2.4\; x\; 12}}{4}$ = 2.4 x 3

Note: 4 divides itself 1 time and 12, 3 times.

2.4 = $\frac{24}{10}$

|ST| = $\frac{24}{10}$ x 3

|ST| = $\frac{\mathrm{24\; x\; 3}}{10}$

|ST| = $\frac{72}{10}$ = 7.2 cm

Sum of the angle of sectors is 360°

105° + 78° + 60° + a° = 360°

243° + a° = 360°

a° = 360° - 243°

a° = 117°

360° represents the total salary

105° represents amount spent on food.

If 360° = ₵720,000

105° = $\frac{\mathrm{₵720,000\; x\; 105°}}{\mathrm{360°}}$ = ₵2000 x 105 = ₵210,000

Note:
1. the zeros (0) at the end of 720,000 and 360
2. 36 divides itself 1 time and 72, 2 times

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

Total angles = 360°

The angle of sector for savings is 60°

Percentage = $\frac{\mathrm{60°}}{\mathrm{360°}}$ x 100% = $\frac{\mathrm{100\%}}{6}$ = $\frac{\mathrm{50\%}}{3}$ = 16.7%

Note:
1. 60 divides itself 1 time and 360, 6 times
2. 2 divides 100, 50 times and 6, 3 times

Angle ZYX + Angle YXZ + Angle XZY = 180° (Sum of interior angles of triangle is 180°)

Angle XZY + 128° = 180°(Sum of angles on a straight line is 180°)

Angle XZY = 180° - 125°
Angle XZY = 55°

Angle ZYX + Angle YXZ + Angle XZY = 180°

52° + Angle YXZ + 55° = 180°

Angle YXZ + 52° + 55° = 180°

Angle YXZ + 107° = 180°

Angle YXZ = 180° - 107°

Angle YXZ = 73°

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

Total number of students = 18 + 22 = 40

Number of girls = 18

P(Girl) = $\frac{18}{40}$ = $\frac{9}{20}$

Note: 2 divides 18, 9 times and 40, 20.

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

Rectangle has two equal lengths and two equal widths.

Let w = the width of the triangle.

length of a rectangle is three times its width

three times is 3 x

Length = 3 x w = 3w

Perimeter = 2 x Length + 2 x Width

Perimeter = 2 x 3w + 2 x w

Perimeter = 6w + 2w

Perimeter = 8w

Perimeter is 24 cm.

8w = 24

Divide both sides by 8

w = $\frac{24}{8}$ = 3 cm

∴ width is 3 cm

Interior Angle = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

135° = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

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

135 x n = (n - 2) x 180

135n = 180n - 360

360 = 180n - 135n

360 = 45n

Divide both sides by 45

n = $\frac{360}{45}$ = 8

Scale factor = $\frac{\mathrm{Enlarged\; Side}}{\mathrm{Corresponding\; Object\; Side}}$

Corresponding Object Side = 4 cm

Scale factor = 3

3 = $\frac{\mathrm{Enlarged\; Side}}{\mathrm{4\; cm}}$

Multiply both sides by the denominator (4)

Enlarged side = 3 x 4 cm = 12 cm

Area of a square = Length x Length

Length = 12 cm

Area of the enlarged square = 12 cm x 12 cm

Area of the enlarged square = 144 cm²

Volume of water = Length x Width x Height of water

Tank is filled with water to $\frac{2}{3}$ of its capacity

Height of water in the tank = $\frac{2}{3}$ x 1.5 m

1.5 = $\frac{15}{10}$

Height of water in the tank = $\frac{2}{3}$ x $\frac{15}{10}$ m = $\frac{5}{5}$ m = 1 m

Note:
1. 2 divides itself 1 time and 10, 5 times
2. 3 divides itself 1 time and 15, 5 times

Volume of water = 3 m x 2 m x 1 m = 6 m³

Volume of water = Length x Width x Height of water

Volume of water when completely filled = 3 m x 2 m x 1.5 m

1.5 = $\frac{15}{10}$

Volume of water when completely filled = 3 m x 2 m x $\frac{15}{10}$ m = 3 m x 1 m x 3 m = 9 m³

Note:
1. 2 divides itself 1 and 10, 5 times
2. 5 divides itself 1 and 15, 3 times

Every 1 m³ cost ₵40,000

If 1 = ₵40,000

9 = $\frac{\mathrm{₵40,000\; x\; 9}}{1}$ = ₵40,000 x 9 = ₵360,000

1 metre = 1000 millimetres

1 centimetre = 10 millimetres

34 m = 34 x 1000 mm = 34,000 mm

5 cm = 5 x 10 mm = 50 mm

34m 5cm 6mm = 34,000 mm + 50 mm + 6 mm = 34,056 mm