Ascending order means from the lowest to the highest

The higher the negative, the smaller the number, thus -64 is lesser than -25

Hence the ascending order is -64, -25, 4, 17

Even numbers are numbers divisible by 2

Examples of even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26 etc, thus simply adding 2 to preceeding number to get succeeding number

x:x is read as x is such that x is ....

Since x is greater than two, means 2 is not inclusive

Since x is less than or equal to twelve, 12 is inclusive. Or means can be that or that. So either less than twelve or equal to twelve

Hence the list of x is {4, 6, 8, 10, 12}

Finite sets are sets having a finite or countable number of elements. It is also known as countable sets as the elements present in them can be counted.

Infinite sets in set theory are defined as sets that are not finite. The number of elements in an infinite set goes to infinity, that is, we cannot determine the exact number of elements.

Infinite sets have ... either at the beginning or the ending of the list to indicate the set is uncountable and goes on forever.

H.C.F. means Highest Common Factor

Finding H.C.F. using Prime Factors

Step I:

Find the prime factorization of each of the given numbers.

Step II:

The product of all common prime factors is the HCF of the given numbers.

Prime number is a number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11,13,17 etc.)

Applying the above to find the H.C.F of 18, 36 and 60

Step I:

18 = 2 x 9 = 2 x 3 x 3 = 2 x 3²

NOTE: a x a = a¹ x a¹

From the law of indices, a^m x aⁿ = a^{m + n}

Hence 3 x 3 = 3¹ x 3¹ = 3^{1 + 1} = 3²

36 = 18 x 2 = 2 x 3 x 3 x 2 = 2 x 2 x 3 x 3 = 2² x 3²

60 = 2 x 30 = 2 x 2 x 15 = 2 x 2 x 3 x 5 = 2² x 3 x 5

Step II:

The product of all common prime factors is 2 x 3

The least number plus the 207 must be divisible by 17

You have to find the least number which is greater than 207 and is divisible by 17

Thus simply write down the multiples of 17 and select the first number greater than 207

Then find out what number when added to 207, will give the least multiple of 17 greater than 207

Multiples of 17: 17, 34, 51, 68, 85, 102, 119, 136, 153, 170, 187, 204, 221, 238,...

The least number greater than 207 is 221

Least number to add = 221 - 207 = 14

Factors are numbers which can go into the number whose factors is being determined.

Factors of 36 are numbers that can go into 36

Factors of 36, P = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Multiples of 4 less than 40, Q= {4, 8, 12, 16, 20, 24, 28, 32, 36}

Intersection ∩ means element you can find in both sets

P∩Q = {4, 12, 36}

If a set contains 'n' elements, then the number of subsets of the set is 2ⁿ.

The number of elements in P∩Q is 3, hence the number of subsets could be calculated as:

Subsets = 2³ = 2 x 2 x 2 = 4 x 2 = 8

List the multiples of each and select the least number which appears in all

Multiples of 10 = 10, 10+10 = 20, 20+10 = 30, 30+10 = 40, 50,60,70,80,90,100,110,120,130,140,150,160,...

Multiples of 15 = 15, 15+15 = 30, 30+15 = 45, 45+15 = 60, 75, 90, 105, 120,135,150,165,...

Multiples of 25 = 25, 25+25 = 50, 50+25 = 75, 75+25 = 100,125, 150, 175, 200...

Alternatively

HOW TO: FIND THE L.C.M USING THE PRIME FACTORS METHOD

Step 1. Find the prime factorization of each number.

Step 2. Write each number as a product of primes.

Step 3. The L.C.M is the product (multiplication) of each of the prime numbers in its highest power.

Step 1 & 2

10 = 2 x 5

15 = 3 x 5

25 = 5 x 5 = 5²

Step 3

L.C.M = 2 x 3 x 5² = 6 x 25 = 150

Applying BODMAS, you have to first simplify the ones in the bracket

Divide by the L.C.M of the denominators (3 and 4).

3 goes into 12, 4 times, then you use it to multiply the numerator (2)

4 goes into 12, 3 times, then you use it to multiply the numerator (1)

Simplify the fraction

You can now remove the bracket since its been simplified

Apply the fraction division rule

Thus you turn the fraction at the right and change the division to multiplication

Now cancel out. 5 cancels each other and 6 goes into 12, 2 times

Apply the multiplication of fraction rule

Make sure the denominators are the same for each fraction by multiplying both the numerator and denominators by the number of times the denominator goes into the L.C.M

The L.C.M of 3, 9 and 7 is 63

Thus:

3 = 3

9 = 3 x 3 = 3²

7 = 7

To get the L.C.M, multiply each of the prime numbers with each having its highest power

L.C.M = 3² x 7 = 9 x 7 = 63

Now express each fraction to have 63 as the denominator

Since the denominators are the same, you can compare them by the numerators

The question said in ascending order, hence lowest to highest

Simple interest (I) is calculated by the formula below:

Principal (P) = GH₵ 600.00

Convert the months to years.

There are 12 months in a year

8 months = 8/12 = 2/3 years

Total years (T) = 2 + 2/3 = 8/3 years

Rate (R) = 5

Cancel out

Simple Interest = 2 x 8 x 5 = 16 x 5 = 80

A number for a particular ratio can be calculated as:

Let x = total number of students

Ratio of boys : girls = 3:2

Ratio for girls = 2

Sum of ratio = 3+2 = 5

Number of girls is equal to the ratio of girls over the sum of the ratios times the total number of students

Every number is being divided by 1

420 = 420/1

Rewrite the equation

Cross multiply to solve for the value of x

2x x 1 = 5 x 420

Divide both sides by 2

Cancel out

x = 5 x 210 = 1050

Total number of students = 1050

ALTERNATIVELY

The ratio for the total students is the sum of the ratios

Ratio for the total students = 3+2 = 5

Let x = total students

We know the ratio of girls and the number of girls

Equate the ratio to the known numbers

Ratio of girls = Number of girls

Total ratios = x (Total number of students)

Cross multiply to solve for the value of x

2 x x = 5 x 420

2x = 5 x 420

Divide both sides by 2

Cancel out

x = 5 x 210 = 1050

Total number of students = 1050

Since selection is from 1 to 20, number of sample space being selected from = 20

Probability is calculated by the formula below:

We have to list all the numbers divisible by 3 (multiples of 3) from 1 to 20 to know how many numbers divisible by 3 in the sample

Numbers divisible by 3 from 1 to 20 = 3,6,9,12,15 and 18

Hence 6 of the numbers are divisible by 3

Substite the number of numbers divisible by 3 and the total sample space into the probability formula

Cancel out

2 goes into 6, 3 times and 20, 10 times

Percentage profit is calculated by the formula below:

Profilt = Selling Price - Cost Price

Selling Price = 24

Cost Price = 20

Profit = 24 - 20 = 4

Substiting the profit and the cost price in the percentage profit formula

Cancel out

Profit Percentage = 4 x 5 = 20%

Determine if the proportion is direct or indirect

For direct proportion, if one quantity is increased or decreased then the other quantity increases or decreases as well, respectively.

But in indirect or inverse proportion, if one quantity increases then other quantity decreases and vice-versa.

In the question, you will realize that if more men work (workers increases) the work will be done earlier (days decreases), hence its an indirect or inverse proportion

Thus number of men is inversely proportional to the days for the job

You can write the relation and solve for the constant (k) using the known information

Inversely proportional is 1 over what's its inversely proportional to

Remove the proportionality sign by replacing with an equal to sign and a constant (k)

Simplify

Make k the subject

k = men x days

Use the known values to solve for the constant (k)

When 8 men, the days taken was 12

k = 8 x 12, for the sake of future cancellations, we can leave the answer this way

Substite the value of k into the equation

Since we are asked to find the days, you can make days the subject and subtitute the number of men into it to get the days.

You can as well substite the men into the equation and solve for days. However you do it, you will arrive at the same answer

Make days the subject

Every number is being divided by 1, hence men = men/1

Rewrite the equation

Cross multiply

men x days = 1 x 8 x 12

Divide both sides by men

men cancels each other at the left side of the equation.

Now calculate for the days for the 6 men

Cancel out

6 goes into itself, 1 and 12, 2 times

days = 8 x 2 = 16 days

PARALLEL LINES ANGLE PROPERTIES

VERTICALLY OPPOSITE ANGLES

Vertically opposite angles are the same.

From the above diagram, angle TRS = angle PRQ = 40^o

ANGLES IN A TRIANGLE

Angles in a triangle add up to 180

angle QPR + angle PRQ + angle PQR = 180

angle QPR = 68^o

angle PRQ = 40^o

68 + 40 + PQR = 180

108 + PQR = 180

PQR = 180 - 108

PQR = 72^o

PARALLEL LINES ANGLE PROPERTIES

Alternate angles are the same

From the above diagram, angle TSR and angle QPR are the same = 68^o

Speed is calculated by the formula below:

Distance can also be calculated as speed x time

Distance = Speed x Time

From the question, the speed = 60 km/h

10.45 am to 12.15 pm is 1 hours 30 minutes

NOTE: 10.45 am - 11.45 am = 1 hour

15 minutes time will be 12 pm

Another 15 minutes time will be 12.15 pm

Hence time taken is 1 hour 30 minutes

Since the speed is in hours, convert the minutes (30 minutes) to hours

60 minutes = 1 hour

30 goes into 60, 2 times

Now add the 1/2 hours to the other 1 hour to get the total hours

Every number is being divided by 1

Rewrite the fractions

Express it in single fraction

Divide by the L.C.M of the denominators (1 and 2)

The L.C.M of 1 and 2 is 2

Substite the time into the distance equation

Cancel out

2 goes into 60, 30 times

Distance = 30 x 3 km = 90 km

Length = 10 cm

Breadth is unknown

Since we are given the perimeter, we have to use the concept of perimeter to solve for the breadth

Perimeter is the sum of all the dimensions of the shape

Perimeter = 26 cm

Perimeter = 2 x Length + 2 x Breadth

Let breadth = B

26 = 2 x 10 + 2 x B

26 = 20 + 2B

26 - 20 = 2B

6 = 2B

Divide both sides by 2

Cancel out

B = 3

Since we know the length and breadth of the rectangle, we can calculate the area

Area = Length x Breadth

Length = 10 cm

Breadth = 3 cm

Area = 10 cm x 3 cm

Area = 30 cm²

The general equation of a line is given by:

As you can see, the gradient or slope is the number multiplying the x variable

In order to find the slope, we must also express the equation in a similar way, where y must be alone just as it is in the general equation

Let's then make y the subject of the equation

3x - 6y = 33

- 6y = - 3x + 33

Divide both sides of the equation by -6

Cancel out

ASSIGNMENT

What's the y - intercept of the equation of line obtained?

Answer

y = c + bx²

Substitute the values into the equation.

y = $\frac{14}{5}$ + $\frac{4}{5}$ x 2²

a² = a x a

y = $\frac{14}{5}$ + $\frac{4}{5}$ x 2 x 2

y = $\frac{14}{5}$ + $\frac{4}{5}$ x 4

y = $\frac{14}{5}$ + $\frac{\mathrm{4\; x\; 4}}{5}$

y = $\frac{14}{5}$ + $\frac{16}{5}$

y = $\frac{\mathrm{14\; +\; 16}}{5}$

y = $\frac{30}{5}$

y = 6

Volume of a cylinder = πr²h

Volume = 20 π cm³
Height = 5 cm

Substitute volume and height into the equation and solve for the value of r, radius

20 π = π x r² x 5

Divide both sides of the equation by π x 5 to make r the subject.

$\frac{\mathrm{20\; \pi }}{\mathrm{\pi \; x\; 5}}$ = $\frac{\pi \; x\; {\mathrm{r}}^2 \; x\; 5}{\mathrm{\pi \; x\; 5}}$

4 = r²

Take square root of both sides or r is a value when multiplied by itself will give 4, which is 2 because 2 x 2 = 4

2 = r

Thus

2 x 2 = r²

Every number is raised to the power 1

2¹ x 2¹ = r²

From the law of indices, a^m x aⁿ = a^m+n

2¹⁺¹ = r²

2² = r²

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

2 = r

From pythagoras theorem

The hypotenuse (c) is the side facing the right-angle. The longest side of the triangle.

15² = 12² + |PQ|²

225 = 144 + |PQ|²

225 - 144 = |PQ|²

81 = |PQ|²

A number when multiplied by itself, will give 81, which is 9

9 x 9 = |PQ|²

Every number is raised to the power 1

9¹ x 9¹ = |PQ|²

From the law of indices, a^m x aⁿ = a^m+n

9¹⁺¹ = |PQ|²

9² = |PQ|²

Since the powers are the same, the bases should also be the same

9 = |PQ|

Triangular Prism

q = $\frac{1}{3}$(m + n)h

Get rid of the denominator by multipliying both sides of the equation by the denominator, 3

3 x q = 3 x $\frac{1}{3}$(m + n)h

3q = 1(m + n)h

Divide both sides of the equation by h

$\frac{\mathrm{3q}}{\mathrm{h}}$ = m + n

Make m the subject, n goes to the other side of the equation.

$\frac{\mathrm{3q}}{\mathrm{h}}$ - n = m

m = $\frac{\mathrm{3q}}{\mathrm{h}}$ - n

Express 16 and 8 as base 2 to a power

16 = 2 x 2 x 2 x 2 = 2⁴
8 = 2 x 2 x 2 = 2³

16² x 8² = (2⁴)² x (2³)²

From the law of indices, (a^m)ⁿ = a^{m x n}

(2⁴)² x (2³)² = 2^{4 x 2} x 2^{3 x 2}

(2⁴)² x (2³)² = 2⁸ x 2⁶

From the law of indices, a^m x aⁿ = a^m+n

2⁸ x 2⁶ = 2⁸⁺⁶ = 2¹⁴

16² x 8² = 2¹⁴

4a - 9b -2(2a - 3b)

Expand the bracket

4a - 9b -2 x 2a -2 x (- 3b)

- x - = +

4a - 9b -4a + 2 x 3b

4a - 9b -4a + 6b

Group like terms

4a -4a - 9b + 6b

-3b

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

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

u = $\left(\begin{array}{c}3\\ 1\end{array}\right)$ and v = $\left(\begin{array}{c}-2\\ 1\end{array}\right)$

6v + 2u = 6$\left(\begin{array}{c}-2\\ 1\end{array}\right)$ + 2$\left(\begin{array}{c}3\\ 1\end{array}\right)$

6v + 2u = $\left(\begin{array}{c}\mathrm{6\; x\; -2}\\ \mathrm{6\; x\; 1}\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{2\; x\; 3}\\ \mathrm{2\; x\; 1}\end{array}\right)$

6v + 2u = $\left(\begin{array}{c}-12\\ 6\end{array}\right)$ + $\left(\begin{array}{c}6\\ 2\end{array}\right)$

6v + 2u = $\left(\begin{array}{c}\mathrm{-12+6}\\ \mathrm{6+2}\end{array}\right)$

6v + 2u = $\left(\begin{array}{c}-6\\ 8\end{array}\right)$

$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ → $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{2\; -\; y}\end{array}\right)$

$\left(\begin{array}{c}2\\ 5\end{array}\right)$ → $\left(\begin{array}{c}2\\ \mathrm{2\; -\; 5}\end{array}\right)$

$\left(\begin{array}{c}2\\ 5\end{array}\right)$ → $\left(\begin{array}{c}2\\ -3\end{array}\right)$ = (2,-3)

Anticlockwise rotation through 90^o about the origin

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

Thus the cordinates are switched and x cordinate is negated.

Q$\left(\begin{array}{c}-4\\ 5\end{array}\right)$ → Q¹$\left(\begin{array}{c}-5\\ -4\end{array}\right)$ = Q¹(-5,-4)

The marks more than the pass mark 4 are: 10, 7 and 8. Hence 3 students scored more than the pass mark.

Mean/Average = $\frac{\mathrm{Sum\; of\; the\; numbers}}{\mathrm{Number\; of\; numbers}}$

Mean/Average = $\frac{\mathrm{10\; +\; 4\; +\; 1\; +\; 4\; +\; 3\; +\; 3\; +\; 2\; +\; 1\; +\; 1\; +\; 7\; +\; 8}}{11}$

Mean/Average = $\frac{44}{11}$ = 4

Image Length = Scale factor x Original length

Scale factor = $\frac{\mathrm{Image\; Length}}{\mathrm{Original\; length}}$

Scale factor = $\frac{15}{3}$ = 5

Try each of the rules by substituting the values of x to see if you could get the corresponding y values.

The rule which gives all the corresponding values of y is the correct rule.

Rule: y → $\frac{\mathrm{x}}{2}$ - 2

When x = 0

$\frac{0}{2}$ - 2 = 0 - 2 = -2

When x = 3

$\frac{3}{2}$ - 2 = 1.5 - 2 = -0.5

Not the rule.

Rule: y → x - 2

When x = 0

0 - 2 = -2

When x = 3

3 - 2 = 1

Not the rule.

Rule: y → x² - 2

When x = 0

0² - 2 = -2

When x = 3

3² - 2 = 9 - 2 = 7

Not the rule.

Since all the three options are not, automatically the fourth is the rule but you can still verify.

Rule: y → 2x - 2

When x = 0

2(0) - 2 = 0 - 2 = -2

When x = 3

2(3) - 2 = 6 - 2 = 4

When x = 6

2(6) - 2 = 12 - 2 = 10

When x = 9

2(9) - 2 = 18 - 2 = 16

When x = 12

2(12) - 2 = 24 - 2 = 22

y → 2x - 2 is the rule.

$\frac{1}{2}$(3x - 1) + 1 ≤ 7 + 2x

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

2 x $\frac{1}{2}$(3x - 1) + 2 x 1 ≤ 2 x 7 + 2 x 2x

3x - 1 + 2 ≤ 14 + 4x

3x + 1 ≤ 14 + 4x

3x - 4x ≤ 14 - 1

-x ≤ 13

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

$\frac{\mathrm{-x}}{-1}$ ≥ $\frac{13}{-1}$

x ≥ -13

4 - x = 3(4x + 5)

Expand the bracket

4 - x = 3 x 4x + 3 x 5

4 - x = 12x + 15

4 - 15 = 12x + x

-11 = 13x

Divide both sides of the equation by 13

$\frac{-11}{13}$ = $\frac{\mathrm{13x}}{13}$

$\frac{-11}{13}$ = x

x = $\frac{-11}{13}$

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

Number of boys = 48

Total number of students = 12 + 48 = 60

Percentage of boys = $\frac{\mathrm{Number\; of\; boys}}{\mathrm{Total\; number\; of\; students}}$ x 100

Percentage of boys = $\frac{48}{60}$ x 100 = 80%

Cancellations

1. The zeros in 60 and 100 cancels each other
2. 2 divides 6, 3 times and 10, 5 times
3. 3 divides itself and 48, 16 times
4. 16 x 5 = 80

Bearing of Q from P = The alternate angle of 60^o + 180^o

Bearing of Q from P = 60^o + 180^o = 240^o

Construction of perpendicular from P to meet line AB.