3-chapter 1

Chapter 1: Basic Mathematics

CHAPTER 1

BASIC MATHEMATICS

Chapter Objectives

At the end of this section, you should be able to:

• apply basic laws of operations.
• discuss the relationship between improper fractions and mixed numbers and

practice on simplifying problems.
• write down the equation of the straight line.
• solve simple simultaneous equations using graphically and algebraically.

1.0 Introduction

Whole numbers are simply the numbers 0,1,2,3,! but no fractions. There are also called
natural numbers and counting numbers. Counting numbers are whole numbers, but without
the zero, because we can’t “count” zero. So they are 1,2,3,4,5,6,…(and so on). Natural
numbers can mean either counting numbers {1,2,3,…}or whole numbers {0,1,2,3,4,5,…}
depending on subject.

Table 1.1 Set of Numbers

Name Description Examples
Whole numbers
Integers Natural numbers

Rational numbers Counting numbers
(Fractions)
Negative integers
Irrational numbers
(Fractions) Positive integers

Non-negative integers

• Represent as a , where a & b are
b
integers, b ≠ 0 .

• Decimal repeating
Decimal not repeating

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Chapter 1: Basic Mathematics

1.1 Integers

Integers are like whole numbers, they also include negative numbers but still no fractions
allowed. Integers can be negative {-1,-2,-3,-4,-5,…}, positive {1,2,3,4,5,…} or zero {0}. We
can put that all together like this:
Integers ={…,-3,-2,-1,0,1,2,3,…}

The Basic Laws of Operations

There are some rules that might you to solve the problem, which are:

1) Adding two numbers can be reversed:

a+b = b+a
2) Multiplying two numbers can be reserved:

ab = ba
3) Adding three numbers:

(a + b)+ c = a + (b + c)

4) When we have three numbers a, b, c :

a) a(b + c) = ab+ ac

b) a(bc) = (ab)c

c) If a = b , a + c = b + c
d) If a = b , ac = bc

e) If a =b, a b
c =c

5) i) Products of two like signs = positive value:

Eg: a) 4 × 20 = 80

b) (− 25)×(− 2) = 50

ii) Products of two unlike signs = negative values:

Eg: (−12)×3 = −36

6) i) Quotient of two like signs = positive value:

Eg: a) 8 ÷ 2 = 4

b) (− 72)÷ (− 4) =18

ii) Quotient of two unlike signs = negative value:

Eg: 18÷ (− 3) = −6

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Example 1 : Chapter 1: Basic Mathematics

a) 25 − 3× 8 + 18 ÷ 2 b) 12 × 30 ÷ 6 − 9
= (360 ÷ 6) − 9
= 25 − (3× 8)+ (18 ÷ 2) = 60 − 9
= 51
= 25 − 24 + 9

= 10

Exercise 1:

Solve all.
1) 8 × 5 + 10 − 2

2) 26 − 2 + 4 × 6 − 4 ÷ 2

3) {54+ (7× 2)}÷ (8 − 6)

4) 9 + (8× 20)÷ 2 − 5 Page | 3
 


 

Chapter 1: Basic Mathematics

1.2 Fractions

A fraction, also called a proper fraction, is a portion of a whole number, expressed as one

number over the other, such as: 2 . The top number is called the numerator; the bottom
5

number is the denominator. To convert fraction into decimal, we can simply divide the

numerator with denominator.

An improper fraction has a numerator that’s larger than the denominator, such as 7 .
4

Improper fraction also can be convert into decimal, for example: 7 ÷ 4 = 1.75 . An improper

fraction is considered bad form; instead you should always convert an improper fraction to a

decimal or to a mixed fraction.

A mixed fraction is a fraction that mixes a whole number and a fraction, for example: 1 1 .
2

Just as you want to convert an improper fraction to a mixed fraction, you also need to

simplify all fractions. A simplified fraction reduced any further and still remains a fraction.

For example, 1 is a simplified fraction, but 2 is not since, 2 = 1.
2 4 4 2

Addition and Subtraction of Fraction
The addition and subtraction only can be done whenever they have same denominator. To
add or subtract fractions that do not have same denominator, the lowest common denominator
or the smallest possible number divisible by both denominators, need to be find first. For
example,

11 − 5 = 11× 2 − 5 × 5
15 6 15 × 2 6 × 5

= 22 − 25
30 30

=− 3
30

=− 1
10

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Chapter 1: Basic Mathematics

After finding a common denominator and multiplying the denominators to keep the values of
the fractions the same, adding and subtraction fractions becomes simple arithmetic. For
example;

213
5+5=5

Exercise 2:

Solve all,

1) 24
3+8

2) 1 14 7
9 + 20 + 12

3) 14 − 2
5 7

4) 2 3 − 11 1
8 3

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Chapter 1: Basic Mathematics

Multiplication and Division of Fraction
Multiplying and dividing fractions is actually simpler than adding and subtracting them
because, do not have to find a common denominator. To multiply fractions, multiply the
numerators and denominators, and then simplify it. For example,

3 × 2 = ⎛⎜ 3× 2 ⎞⎟
4 5 ⎝ 4×5 ⎠

6
= 20

3
= 10

Dividing fractions is just as simple as multiplying, but, first, we need to invert the fraction
and then multiply the two fractions. For example,

6 3 6 11
7 ÷ 11 = 7 × 3

= 6 ×11
7×3

66
= 21

22
=7

Exercise 3:

Solve all,

1) 1 10
5 × 15

2) 36
42 × 15

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Chapter 1: Basic Mathematics

3) 89
20 ÷ 21

4) 2 1 ÷ 4 3
2 8

1.3 Decimal

A decimal is a portion of a number, expressed as one or more digit to the right of a decimal
point. A decimal point separates the whole number from the places that are less than one.
Place values extend infinitely in two directions from a decimal point. A number containing a
decimal point is called a decimal number or simply a decimal.

Significant Figures
Significant figure is any digit of a number that is known with certainty. There are some rules
of significant figure, which are;
1. Any non-zero number counts as a significant figure (1, 2, 3, 4, 5, 6, 7, 8, and 9).
2. Zeros between any of the numbers listed above count as significant. (Ex: 40908 has 5 SF).
3. When there is no decimal, all zeros on the right are NOT significant, all others are.

(Ex:5730000 has 3 SF).
4. When there is a decimal, all zeros on the left are NOT significant, all others are. (Ex:

0.000000012 has 2 SF.

Example 2 :

1. 0.00815 has three significant figures.
2. 0.0502, 0.710and 0.600 have 3 significant figures.
3. 70.31, 703.1and 7031 have 4 significant figures.

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Chapter 1: Basic Mathematics

Exercise 4:

How many SF do the following have?
1) 234
2) 0.568
3) 72000
4) 0.0045

Rounding
Rounding decimals is the same as rounding any other whole number; the only difference is
that, there is a decimal point. In order to round up the decimals, we need to look at the digit to
the right of where you are rounding. Where;
-­‐ If the digit is 4 or less, round down.
-­‐ If the digit is 5 or more, round up.

Example 3 :

1) 0.03129 to 2 decimal places is 0.03 .
2) 7.129 to 2 decimal places is 7.13 .
3) 25128 to 3 significant figures is 25100 .
4) 0.0514 to 2 significant figures is 0.051 .

Exercise 5:

Solve;
1) 4.1254 to 2 decimal places.
2) 7.0421 to 3 decimal places.
3) 46873to 2 significant figures.
4) 2.5261to 3 significant figures.

Adding and Subtracting Decimals
Adding and subtracting decimals exactly the same way on adding and subtracting whole
numbers. The only difference is, we need to align on the decimal points.

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Example 4 : Chapter 1: Basic Mathematics
1) 4.7 + 3.84 +10 .
2) 2.18 −1.5468
4.70
+ 3.84 2.1800
10.00 − 1.5468
18.54
Exercise 6: 0.6332
Solve all;
1) 28.21 + 3.95 + 8.34

2) 10.455 + 1 + 4.5 + 23

3) 204.90 −193.80

4) 494.78 − 82.8

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Chapter 1: Basic Mathematics

1.4 Inequalities

Definition:
For any real numbers a and b ,
i) a < b means a is less than b
ii) a ≤ b means a is less than or equal to b
iii) b > a means b is greater than a
iv) b ≥ a means b is greater than or equal to a

Approximations

Technique of guessing the answer close to a correct one. For the least, we must be able to

capture the answer as between two close values. A simple example of the technique: 20 is
3

less than 7 ( i.e. 21 ), and it is more than 6 (i.e. 18 ) or 6 < 20 < 7.
3 3 3

The Interval Notation
The double inequality a < x ≤ b means x is more than a ( x > a ) and x is less than or equal to b
( x ≤ b ) or can be expressed by the interval;

(a,b]= {x | a < x ≤ b}
Other examples are: (a, b) = {x | a < x < b} - the open interval from a to b

and [a, b] = {x | a ≤ x ≤ b} - the closed interval from a to b

Interval Inequality Table 1.2 Type
Notation Notation Set Notation Line Graph

[a, b] a≤ x≤b {x | a ≤ x ≤ b}

[a, b) a≤ x<b {x | a ≤ x < b}

(a, b] a< x≤b {x | a < x ≤ b}

(a, b) a< x<b {x | a < x < b}

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Chapter 1: Basic Mathematics

[b, ∞) x≥b {x | x ≥ b}
{x | x > b}
(b, ∞) x>b {x | x ≤ a}

(− ∞, a] x≤a {x | x < a}

(−∞, a) x<a

Exercise 7:

Write each of the following in the form of interval notation and line graph.
1) − 2 < x ≤ 2

2) 3 ≥ x ≥ 0

3) {x | x >1}

Absolute values
If x is a real number, its absolute value is itself, if x ≥ 0 , and the positive of it if x < 0 . This
is denoted by | x |. For example; | 5 |= 5, | −9 |= 9.

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Chapter 1: Basic Mathematics

1.5 Percentages

Percentages can be expressed as a fraction or as a decimal, and any decimal or fraction can be
expressed as a percentages. In business and finance, statements like the following are
common:
-­‐ “Sales increased by 80%”
-­‐ “The Ringgit has depreciated 30% against the dollar”

To convert a decimal to a percentages, we need to move the decimal place two places to the
right and add a percent sign (%).

Example 5 :

Convert to percentages;
1) 0.255 becomes 25.5%
2) 0.03 becomes 3%

To change percentages to a decimal, we need to simply the process; move the decimal point
two places to the left.

Example 6 :

Convert to decimal;
1) 78% becomes 0.78
2) 82.25% becomes 0.8225

To convert a fraction to a percentage, we must first convert the fraction to a decimal (i.e.
divide the numerator by the denominator) and then use the same procedure with converting
decimal to percentages.

Example 7 :

Convert to percentages;

1) 1 = 0.02 = 2%
50

2) 2 = 0.1 = 10%
20

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Exercise 8: Chapter 1: Basic Mathematics
a) Convert to percentages; Page | 13
 

1) 0.836
2) 16

32

3) 1.29

4) 7
25

b) Convert to decimals;
1) 7.125%

2) 326.8%


 

Chapter 1: Basic Mathematics

Solving Percentages
Example 8 :

1) What is 10% of 300%?

10% × 300 = 10 × 300
100

= 30

2) What per cent is 200 of 600?

200 ×100% = 0.33 33 × 10 0%
600

= 33.33%

Exercise 9:
1) What is 35% of 4500?

2) A fruit seller had some apples, he sells 40% apples and still has 420 apples. Originally,
how many apples he had?

3) Two students appeared at an examination. One of them secured 9 marks more than the
other and his marks was 56% of the sum of their marks. The marks obtained by them are:

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Chapter 1: Basic Mathematics

1.6 Solving Equation
1.6.1 Coordinate System

A Cartesian coordinate system, also known as rectangular coordinate system, can be used to
plot points and graph lines. We construct two straight lines intersecting at right angles, one
vertical and the other horizontal. The following is an example of rectangular coordinate
system

(II)
  (I)
 

(III)
  (IV)
 

Figure 1.1 Rectangular coordinate systems

The horizontal line is called x − axis , from which y is measured. The vertical line is called
y − axis , from which x is measured. A point is indicated by (x, y). It is called the
coordinates of the point. The point of intersection is called the origin of the coordinate. The
origin or intersection of the two axis is equal to (0,0).There are four quadrants shown in the
figure above. In quadrant (I), x and y are positive, in quadrant (II), x is negative, but y is
positive, in quadrant (III), x and y are negative while in quadrant (IV), x is positive, but y
is negative.

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Chapter 1: Basic Mathematics

1.6.2 Straight Lines

Straight line is a line traced by a point travelling in a constant direction. A straight line may
be defined by two properties which are slope and vertical intercept. Slope is a gradient of
the line and it is usually represented by the symbol m . While, the vertical intercept is the
point at which the line crosses the y − axis and it is usually represented by the symbol c .

A linear equation can be written in the following form:

y = mx + c (1.1)
where m and c are constant.

Example 9:

In y = 3x + 2, the line has a slope of 3. As x moves forward 1 unit, y moves up 3 units. In this
case, the line has a positive slope, and y is said to be an increasing function of x . (Draw a
graph)

y

y = 3x + 2


  2

0 x

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Chapter 1: Basic Mathematics

For the relation y = − 1 x + 2 , the slope is − 1 , as x moves forward 1 unit, y goes down by 1
2 2 2

unit. The line has a negative slope; y is a decreasing function of x .

y

y = − 1 x + 2
2


 

2

0x
4

In equation (1.1), for x = 0 , we have y = c , the point at which the line cuts the y − axis . c is
called the intercept. In the previous example, both lines have intercepts 2. Intercept is the
point (0, c) and which c is y -intercept. If we change the value of c , without changing m , the

line merely shifts up or down, but the slope remains the same.

If we are given any two points on a straight line, we can easily find its slope as follows. Let
(1,2) and (3,5) be any points on a straight line. The slope of the line can be found by using the
formula for the slope,

m= y2 − y1 (1.2)
x2 − x1

= 5−2
3−1

3
=2

Hence, the slope of the line is 3 .
2

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Chapter 1: Basic Mathematics

The equation of the line can be formed when we are given (x, y)and (1,2) on a line with a
slope 3 .

2

From (1.2),

m= y2 − y1
x2 − x1

3 = y−2
2 x −1

Simplifying,

3(x −1) = 2( y − 2)
3x − 3 = 2y − 4
2y − 3x −1 = 0

Hence, from the example above, we have two general formulas for a straight line:
a) when we are given a slope, m and an intercept, c ; the equation is

y = mx + c

b) if a line passing through two points ( p, q) and (r, s) ; the equation is

y − q = s− q (x − p)
r− p

y − q = m(x − p) (1.3)

In the business world, the straight line equation can be used as a supply and demand function
which x − axis in the graph is represent the quantity demand or quantity supply of the product
and y − axis is representing price of the product.

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Chapter 1: Basic Mathematics

Exercise 10:

Write the equation 2x − 3y = 5in the general form y = mx + c and draw it on a graph.

Solution:

Exercise 11:

Ten watches are sold when the price is RM80 and 20 watches are sold when the price is
RM60. What is the demand equation?

Solution:

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Chapter 1: Basic Mathematics

1.6.3 Intersecting Lines: Simultaneous Equations

Simultaneous equations may be solved by plotting each of the equation on the same diagram,
then finding the coordinates of the intersection. In the supply and demand equations, the point
of intersection is called equilibrium points. The x − coordinate and y − coordinate represent
the solution to the equation. If the two lines cut each other at right angles; they are said to be

perpendicular. The product of their slopes m1 • m2 = −1.

However, when the two lines being plotted have the same slope m1 = m2 , they are said to be

parallel and thus never intersect. In this case, the simultaneous equations have no solution.
Thus, there are no values of x and y that satisfy both equations.

Example 10:

Consider the two simultaneous equations below:

y = 3x + 2 (1)

y = − 1 x + 1 (2)
3

Solution:

There are two methods to solve the two simultaneous equations.

Method 1: Graphically method

The product of the slopes, m1 • m2 = 3• −1 = −1. Hence, line (1) is perpendicular to the line
3

(2). The graph shows that the two lines cut each other at right angles.

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Chapter 1: Basic Mathematics

The point of intersection provides the solution to the simultaneous equation (1) and (2).
Method 2: Algebraically method

y = 3x + 2 (1)
(2)
y = − 1 x + 1
3

From (2) × 3, we have

⎡ y = − 1 x + 1⎦⎤⎥ × 9
⎣⎢ 3

9y = −3x + 9 (3)

Adding (3) to (1), we have

10 y = 11

y = 11
10

y = 1.1

Substitute the value of y = 1.1into (1) or (2);
y = 3x + 2

1.1 = 3x + 2
3x = −0.9
x = −0.3

Hence, the solution of the simultaneous equation above is (−0.3,1.1) .

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Chapter 1: Basic Mathematics

Exercise 12:

Solve the simultaneous equations below:

14x − 8y +12 = 0
2x − 4y − 6 = 0

Solution:

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Chapter 1: Basic Mathematics

Exercise 13:

Find the point of equilibrium for the following demand and supply equations.

y = 10 − 2x

y = 3 x +1
2

Solution:

END OF CHAPTER

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Chapter 1: Basic Mathematics

TUTORIAL 1
BASIC MATHEMATICS

1. Solve all,

a) 3 1 + 2
3 5

b) 4 2 − 3
5 4

c) ⎛⎜ 1 × 1 ⎟⎞ + 3
⎝ 6 12 ⎠ 4

2 1 − 1
2 5
d) 1 1
2 5
2 +

e) 2 1 ÷ ⎛⎜ 3 − 1 ⎟⎞
9 ⎝ 6 ⎠

2. Simplify,

a) 2+e − f −e
2e ef

b) 1 − n−4
2n 6n 2

c) m + 3 − (4m − 3)

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Chapter 1: Basic Mathematics

3. Solve all,
a) 3 + 4 − 8 × 2

b) 9 − (4 + 1)× 7

c) (7 −1)× 5 + (7 × 3)

4. Solve,
a) Find 25% of RM125.

b) Adam bought a car for RM95,000. During the first year, its value depreciated by
20%. What was the value of the car after one year?

c) Taking 18% of the total of 360 degrees we get:

d) 47% of the students in a class of 34 students has glasses or contacts. How many
students in the class have either glasses or contacts?

e) The population of a city in the year 2015 was 500,000. Over the following decade
the population grew by 8%. What was the population of the city in 2025?

f) A class has 32 students. Approximately 75% passed their last Mathematics test.
How many students were not passed the test?

5. Convert,
a) 0.875 to a per cent

b) 3.5% to a decimal

c) 11 to decimal
20

d) 0.125 to a fraction

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Chapter 1: Basic Mathematics

e) 5 to per cent
8

f) 5% to a fraction

6. Solve,
a) 4.2245 to 2 decimal places

b) 92.0998 to 3 decimal places

c) 34981 to 2 significant figures

d) 0.1902 to 2 significant figures

7. Find the point of equilibrium for the following demand and supply equations:

a) 6x + 8y = 38
4x − 2 y = 18

b) y = 10 − 6x
y = 8x + 24

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Chapter 1: Basic Mathematics

c) − 4x + y = 10
x+ y = 4

8. Ten watches are sold when price is RM80. 20 watches are sold when the price is
RM60. What is the demand equation?

9. Find the equation of the straight line and show that whether these two straight lines
are parallel or not to one another.
(i) passes through the point (4, 6) and has gradient 4,

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Chapter 1: Basic Mathematics

(ii) passes through the points A(−2, 6) and B(4, 18).

10. Find the equation of the following straight line that passes through the point (0, 4) and
is parallel to the line 2x + 3y = 5.

END OF TUTORIAL QUESTIONS

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