1202 Question Bank Mathematics Form 4

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Contents






Must Know iii - x


Chapter 1 Quadratic Functions and Equations in One Variable 1 – 11
NOTES 1
Paper 1 3
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Paper 2 6
Chapter 2 Number Bases 12 – 20
NOTES 12
Paper 1 14
Paper 2 16

Chapter 3 Logical Reasoning 21 – 29
NOTES 21
Paper 1 22
Paper 2 24
Chapter 4 Operations on Sets 30 – 40
NOTES 30
Paper 1 31
Paper 2 35

Chapter 5 Network in Graph Theory 41 – 49
NOTES 41
Paper 1 42
Paper 2 46

Chapter 6 Linear Inequalities in Two Variables 50 – 60
NOTES 50
Paper 1 51
Paper 2 54

Chapter 7 Graphs of Motion 61 – 73
NOTES 61
Paper 1 62
Paper 2 66

Chapter 8 Measures of Dispersion for Ungrouped Data 74 – 86
NOTES 74
Paper 1 76
Paper 2 79
Chapter 9 Probability of Combined Events 87 – 95
NOTES 87
Paper 1 88
Paper 2 90
Chapter 10 Consumer Mathematics : Financial Management 96 – 102
NOTES 96
Paper 1 97
Paper 2 98

Form 4 Assessment 103 – 114

Answers 115 – 134

ii




Content 1202QB Maths Form 4.indd 2 23/02/2022 6:09 PM

MUST


KNOW Important Facts






Quadratic Expressions and Functions Sketching the Graph of a Quadratic Functions

1. The general form of the quadratic expression is ax + bx + c For example, f(x) = 2x + 5x – 12
2
2
with a, b and c are constants and a ≠ 0. Step 1: When a = 2 . 0, the shape
2. The highest power of the quadratic expression is 2 and Step 2: When c = –12, y-intercept = –12
involved only one variable. Step 3: When f(x) = 0
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3. The general form of the quadratic functions is 2x + 5x – 12 = 0
2
f(x) = ax + bx + c.
2
4. For a graph of quadratic function f(x) = ax + bx + c, (2x – 3)(x + 4) = 0
2
3
When a . 0, When a , 0, x = or x = – 4
2
f(x) f(x)
x = m x = m f(x)
2
y = ax + bx + c (m, n)
n
c
x x x
c 0 0 0 3
n (–4, 0) (–, 0)
2
(m, n) y = ax + bx + c 2

• Curved graph: . • Curved graph: . (0, –12)
• Minimum point: (m, n) • Maximum point, (m, n)
Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 7 @ Pan Asia Publications Sdn. Bhd.
Values of a, b and c on Quadratic Functions Number Bases
When a . 0, 1. Digits used in base two up to base ten:
The value of a y y y Number base Digit
determines the Base 2 0, 1
shape of graph 0 x 0 x 0 x Base 3 0, 1, 2
b < 0 b = 0 b > 0 Base 4 0, 1, 2, 3
Base 5 0, 1, 2, 3, 4
When a , 0,
The value of b y y y Base 6 0, 1, 2, 3, 4, 5
determines the Base 7 0, 1, 2, 3, 4, 5, 6
position of the 0 x 0 x 0 x Base 8 0, 1, 2, 3, 4, 5, 6, 7
Base 9
axis of symmetry b < 0 b = 0 b > 0 Base 10 0, 1, 2, 3, 4, 5, 6, 7, 8
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
2
When a . 0 When a , 0 2. For 2431 → Place value = 5 = 5 × 5 = 25
5
The value of Digit value = 4 × 5 = 4 × 25 = 100
2
c determines y y Number 2 4 3 1
the position of c c Place value 5 3 5 2 5 1 5 0
y-intercept. x x Digit value 250 100 15 1
0 0
Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd.
Roots of a Quadratic Equation Conversion of Number Bases
1. The roots of a quadratic y 1. To convert a number in a certain base to base ten
equation ax + bx + c = 0 are 172 = (1 × 8 )+ (7 × 8 ) + (2 × 8 )
2
1
2
0
the points of intersection of Root 4 Root 8 = 64 + 56 + 2
the graph and the x-axis. 2 x = 122
2. The roots of the quadratic –2 –1 0 1 2 3 4 10
–2
equation can be determined by: –4 2. To convert a number in a base ten to certain base
(b) 234 to base eight
(a) 123 to base five
(a) Factorisation method: 10 10
x + 5x + 6 = 0 5 123 8 234
2
(x + 3)(x + 2) = 0 5 24 –3 8 29 –2
Thus, the roots are –3 and –2. 5 4 – 4 8 3 –5
(b) Graphical method: 0 – 4 0 –3
Step 1: Determine the shape of the graph by identifying º 123 = 443 5 º 234 = 352 8
10
10
the value of a. 3. To convert a number in a certain base (not base ten) to
Step 2: Determine the y-intercept another base (not base ten)
Step 3: Determine the x-intercept
Base p Base 10 Base q
Important Facts (Chapter 1) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd.
Must Know 1202QB Maths Form 4.indd 1 21/02/2022 10:48 AM

MUST


KNOW Common Mistakes







Number Bases Quadratic Expression
1
State the value of the underlined digit. Determine whether 7k + k is a quadratic expression in one
2
2
549
5 variable.
Correct Wrong Correct Wrong
= 1132 ©PAN ASIA PUBLICATIONS
Place 5 2 5 1 5 0 Place 5 3 5 2 5 1 Not a quadratic expression in A quadratic expression in
value value one variable because there is a one variable.
variable with a power which is
Number 4 Number 4 not a whole number.
4 × 5 = 100
2
Ensure the power of place Determine whether 2x + 5 is a quadratic expression in one
value is start with 0. variable.
4 × 5 = 20 Correct Wrong
1
Not a quadratic expression in A quadratic expression in
one variable because the highest one variable.
power of the variable is not two.

Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd.


Conversion of Number Bases Factorisation Method
2
Calculate the sum of value of digit 3 in base 10 for the number Solve the quadratic equation (3x – 2) = 9.
1302 and 2356 . Correct Wrong
6 9
2
Correct Wrong (3x – 2) = 9 3x – 2 = 3
2
For 1302 ˜ Digit value Sum of the value for digit 3 9x – 12x + 4 − 9 = 0 3x = 5
2
6 9x – 12x – 5 = 0 5
= 3 × 6 2 = 6 + 9 2 (3x – 5)(3x + 1) = 0 x = 3
2
= 108 10 = 117 10 5 1
For 2356 ˜ Digit value x = or x = –
3
3
9
= 3 × 9 2
2
= 243 10 Determine the roots of x + 7x + 6 = 14 by factorisation method.
Sum of the value for digit 3 Correct Wrong
= 108 + 243 10 x + 7x + 6 = 14 x + 7x + 6 = 14
2
2
10
= 351 10 x + 7x – 8 = 0 (x + 6)(x + 1) = 14
2
(x + 8)(x – 1) = 0 x + 6 = 14 or x + 1 = 14
x = –8 or x = 1 x = 8 x = 13
Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd.
Conversion of Number Bases Quadratic Functions
It is given 1m32 = 247 , find the place value for m. Diagram shows the graph of a y
8
5
Correct Wrong quadratic function. Determine (–2, 6)
the quadratic function in the
247 8 5 167 Remainder form of y = ax + bx + c.
2
= 2 × 8 + 4 × 8 × 7 × 8 0 5 33 – 2
2
1
= 167 10 5 6 – 3 –3 0 1 x
5 5 1 – 1 Correct Wrong
0 1 y = a(x + 3)(x – 1) y = (x + 3)(x – 1)
º 247 = 1132 5 y = a(x + 3x – x – 3) y = x + 2x – 3
2
2
8
2
Place value for m Place value for m = 1 y = a(x + 2x – 3)…a
= 5 2 Substitute (–2, 6) into a:
2
= 25 6 = a[(–2) + 2(–2) – 3]
a = –2
Thus, y = –2(x + 2x – 3)
2
y = –2x – 4x + 6
2
Common Mistakes (Chapter 2) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 6 @ Pan Asia Publications Sdn. Bhd.
Must Know 1202QB Maths Form 4.indd 2 21/02/2022 10:48 AM

Quadratic Functions and
Chapter 1 Equations in One Variable




NOTes



1.1 Quadratic Functions and 6. Effect of changing the values of a, b and c on graphs
Equations of quadratic function, f(x) = ax + bx + c:
2
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(a) Changing the value of a
1. A quadratic expression is an expression in the form • Affects the shape and width of the graph.
of ax + bx + c, where a, b and c are constants and y-intercept remain unchanged.
2
a ≠ 0 while x is the variable. • The width of the graph is decreasing when the
2. A quadratic expression in one variable is an value of a is increasing and vice versa.
2
expression such as y = ax , a > 1 f(x) y = x 2
(a) involves one variable
(b) the highest power for the variable is 2 2
2
y = ax , 0 < a < 1
3. General form for:
2
(a) Quadratic expression: ax + bx + c
x
2
(b) Quadratic function: f(x) = ax + bx + c –2 0 2
2
(c) Quadratic equation : ax + bx + c = 0
y = ax , a < 0
2
4. The type of relation of a quadratic function is –2
many-to-one relation.
5. Characteristics of quadratic functions:
(b) Changing the value of b
(a) When a . 0
• Affects the position of the axis of symmetry.
• Has minimum point.
• The shape and y-intercept remain unchanged.
• The axis of symmetry of the graph is parallel
For example, if a . 0:
to the y-axis and passes through the minimum
f(x)
point. y = (x + b ) 2 y = (x – b ) 2
• The shape:
f(x) y = x 2
Axis of symmetry 2
y = f(x) x
–4 –2 0 2 4
x (c) Changing the value of c
0
• Affects the position of the y-intercept.
Minimum point
• Shape of the graph unchanged.
(b) When a , 0 For example, if a . 0.
• Has maximum point. f(x)
• The axis of symmetry of the graph is parallel
to the y-axis and passes through the maximum
point.
c y-intercept
• The shape:
x
f(x)
Maximum point
For example, if a , 0.
x
0 f(x)
y = f(x)
x
0
y-intercept c
Axis of symmetry


1




C01 1202QB Maths Form 4.indd 1 21/02/2022 3:00 PM

PAPER 1


Each question has four answer choices A, B, C and D. Choose one answer for each question.

1.1 Quadratic Functions and Which of the following is true?
Equations A Maximum point: (–5, 10)
B Minimum point: (–5, 10)
1. Which of the following is a quadratic expression in C Maximum point: (0, 1)
one variable?
D Minimum point: (0, 1)
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A 2x + 2y C x + y + 5
2

B 3y 2 D 2 + 4 8. Diagram below shows a graph of quadratic functions.
x 2
f(x)
2. Which of the following is a not quadratic expression
in one variable?
7
2
A x + 1 C 3x + 4 = 0
3
x 2
2
B 3x + 3x + 4 D 2 + 4 0 x
3. Which of the following is the values of a, b and c for (5, –2)
2
f(x) = x – 2x + 6?
Which of the following is true?
A a = 1, b = – 2, c = 6
A Maximum point: (5, −2)
B a = 6, b = – 2, c = 1
B Minimum point: (5, −2)
C a = 0, b = – 2, c = 6
D a = 0, b = – 2, c = 1 C Maximum point: (0, 7)
D Minimum point: (0, 7)
4. Which of the following is the values of a, b and c for
2
f(x) = 6x + x – 2? 9. Diagram below shows two quadratic functions
A a = 6, b = 0, c = –2 y = f(x) and y = g(x). HOTS Analysing
B a = –2, b = 0, c = 6 y
2
f(x) = tx – 8
C a = –2, b = 1, c = 6
D a = 6, b = 1, c = –2
2
g(x) = 2x – 8
2
5. A quadratic function f(x) = 2x + 5x + c passes
x
through point (−1, 2). What is the value of c? 0
A −1 C 5
B 2 D 9 Which of the following is the range of value of t?
A t , 0 C 0 , t , 2
2
6. A quadratic function f(x) = –3x + 7x + c passes
through point (–1, –12). What is the value of c? B t , 2 D t . 2
A –1 C –10 10. Diagram below shows a graph on a Cartesian plane.
B –2 D 2 f(x)
7. The diagram below shows a graph on Cartesian
plane. 4
f(x)
(–5, 10)
x
0 2
Which of the following is the axis of symmetry for
1
x the graph?
0
A 2 C 0
B 4 D 1
3
Question 2:
Compare with the general form, f(x) = ax + bx + c.
2
Question 10:
The axis of symmetry of the graph of a quadratic function is parallel to the y-axis and passes through the maximum or minimum point. SOS TIP






C01 1202QB Maths Form 4.indd 3 21/02/2022 3:00 PM

PAPER 2


Section A


1. Determine whether each of the following expressions 4. Determine the maximum point or minimum point
is a quadratic expression in one variable or not. and state the equation of the axis of symmetry for
2
(a) x – 3 (b) x + 3x –2 each graph of quadratic function below.
2
(c) y – x + 3 (d) –x 2 (a)
2
1
3
(e) x + x [5 marks] f(x)
©PAN ASIA PUBLICATIONS
Answer: 4
(a) 3
2
(b) 1
x
–3 –2 –1 0 1 2 3
(c)
(b)
f(x)
(d)
x
–3 –2 –1 0 1 2 3
–1
(e)
–2
–3
–4
2. Determine whether the shapes of the following
[4 marks]
graphs of quadratic functions is or .
2
(a) x – 2x (b) 2x – x 2 Answer:
2
2
(c) –2x + 2x + 4 (d) –5x + 2x – 3 (a)
[4 marks]
Answer: (b)
(a) (b)


(c) (d) 2
5. The quadratic function f(x) = 2x – 5x + c passes
through point P as given below. Find the value of c
for each of the following cases.
3. Determine the values of a, b and c for each of the (a) P(−1, 5) [2 marks]
following quadratic expressions. (b) P(3, 7) [2 marks]
2
(a) 2x – 3x + 5 (b) x + 4x Answer:
2
(c) 3x – 7 (d) 2 – 4x – 3x 2 (a)
2
(e) 3y(y – 1) [5 marks]
Answer:
(a) (b)


(b)
(c) (d)





6
Question 2:
SOS TIP Question 5:
Shape when a . 0, shape when a , 0.
Coordinate (x, f(x)). Substitute x and f(x) from coordinate to the equation. Hence, solve it by by making c as a subject.







C01 1202QB Maths Form 4.indd 6 21/02/2022 3:00 PM

Section B

9. Form a quadratic expression based on the diagram 12. Solve the following quadratic equation:
below. HOTS Analysing [2 marks] SPM – 3 = x
CLONE
(x + 7) cm 2x + 1 x – 2 [4 marks]
Answer:

(x + 2) cm Area = 66 cm 2


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Answer:



13. Diagram below shows a quadratic function
f(x) = px + 5x + q with a minimum point
2
(−2.5, −2.25). HOTS Analysing
f(x)

10. Form a quadratic expression based on the diagram
below. HOTS Analysing [2 marks] 4

r x
–4 –1 0

p
(a) Given p is an integer where –2 , p , 2, state the
value of p. [1 mark]
q
(b) Using the value of p from (a), find the value of q.
Hence, state the axis of symmetry. [2 marks]
(c) Is the quadratic function will change if the graph
Answer:
is reflected in the x-axis? If it changes, gives the
answer in the form of f(x) = ax + bx + c.
2
[2 marks]
Answer :
(a)



11. Solve the following quadratic equation: (b)
SPM x + 3x = –2(–3 – x)
2
CLONE
[4 marks]

Answer:
(c)











8
Question 10: 2 2 2
SOS TIP Question 13:
Use the formula of Phytagoras theorom, a + b = c .
The reflection at x-axis will change the value of a.







C01 1202QB Maths Form 4.indd 8 21/02/2022 3:00 PM

14. Sofea bought (8x + 14) text books with the price of 16. A box is filled with 12 same size balls as shown in
SPM RM5x each. She paid RM300 for all the text books. SPM the diagram below.
CLONE CLONE
Calculate the number of text books that she bought.
HOTS Analysing [4 marks]

Answer: 2x + 4 cm



3x + 3 cm

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Given the area of the box is 432 cm , find the
2
diameter, in cm, of one ball. HOTS Analysing
[4 marks]
Answer:


15. A box has the length of (x + 5) cm, the width of
SPM x cm and the height of 30 cm. The total volume of the
CLONE
3
box is 4 500 cm . Calculate the value of x.
HOTS Analysing [4 marks]
Answer:


















Section C


17. Razman wants to sell a rectangular cake during Answer :
Entrepreneur’s Day. The length and width of the (a)
cake are (x + 6) cm and (x + 3) cm respectively.
HOTS Analysing
(a) Form an expression for the area of the cake, (b)
2
L cm , in terms of x. [2 marks]
2
(b) Given that the area of the cake is 270 cm .
Calculate the value of x. [3 marks]
(c) Safi bought the cake and want to give to 30
friends at his birthday party. Will all his friends
2
got the cake if he cut the cake 9 cm each? (c)
Explain your answer. [2 marks]







9
Question 15:
Volume = Length × Width × Height
Question 16:
Diameter is two times the radius. SOS TIP







C01 1202QB Maths Form 4.indd 9 21/02/2022 3:00 PM

Chapter 6 Linear Inequalities in
Two Variables



NOTes



6.1 Linear Inequalities in Two Variables
©PAN ASIA PUBLICATIONS
1. Linear inequalities in two variables is the inequality that involved two variables such that the highest power of both
variables is 1. Table below shows the inequality that is suitable for certain situations.

Situation Linear Inequality Situation Linear Inequality
y is more than x y . x y is at most h times of x y < hx
y is less than x y , x The maximum value of y is h. y < h
The minimum value of y is h. y > h
y is not more than x y < x
The sum of x and y is at least k. x + y > k
y is not less than x y > x
3
3
y is at least h times of x. y > hx The ratio of x to y is not less than . y < x
2
2

2. Dotted line ( ) is used for the inequalities that 4. All the points located on the straight line y = mx + c
has the sign . or , and solid line ( ) is used for satisfied the equation y = mx + c.
the inequalities that has the sign > or <. 5. All the points in the region above the straight line
3. Linear inequalitis for the straight line on the graph y = mx + c satisfy the inequality y . mx + c.
can be determined using the general form equation 6. All the points in the region below the straight line
of the straight line, y = mx + c such that m is the y = mx + c satisfy the inequality y , mx + c.
gradient of the straight line and c is the y-intercept.
7. Diagram below shows a few regions that satisfy the
certain inequalities.
Linear Inequality
y > mx + c, the region is y , mx + c, the region is y < h, the region is x , k, the region is on the
above the solid line below the dashed line below the solid line y = h left side of the dashed line
y = mx + c y = mx + c x = k

y y = mx + c y y y x = k
y > h
y < mx + c y = h x < k x > k
y > mx + c y > mx + c
y < mx + c
x x x x
0 0 0 0
y < h
y = mx + c


6.2 Systems of Linear Inequalities in Two Variables
1. A combination of two or more linear inequalities is known as a system of linear inequalities.

2. The region that satisfy a system of linear inequalities can be determined by using y x = 2
the following steps:
I Determine and mark the region that is represented by each linear inequalities.
II Determine the common region that satisfies all the linear inequalities.
III Shade the region and make sure the region is being bounded by all the linear
R
inequalities.
2
3. For example, the shaded R in the diagram on the left satisfies all the inequalities 2 0 2 x
of y < 3x + 2, y > –x + 2 and x , 2. – – 3 y = –x + 2
y = 3x + 2

50




C06 1202QB Maths Form 4.indd 50 21/02/2022 10:45 AM

PAPER 1


Each question has four answer choices A, B, C and D. Choose one answer for each question.

6.1 Linear Inequalities in 5. Diagram below shows the shaded region that satisfy
Two Variables the linear inequality Q.
y
1. Which of the following inequalities is the linear
inequality in two variables? 4
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A y . 3 2
B x + y < 4 x
–6 –4 –2 0 2
C 2x – 3 , x – 2
–2
D 3x + 5y , z – 2
Determine the linear inequality Q. HOTS Applying
2.
Mr. Mark prepares x unit of tables and y unit of A 4y , 4 + x
1
chairs for class 5A such that the number of tables B y < x + 1
is more than the number of chairs. 4
C 4y – x , 1
Represents the situation above in the form of linear D y – 4x + 1 , 0
inequality.
6. Diagram below shows a shaded region which satisfy
A x . y C y + x . 0
linear inequality P.
B y , x D y – x . 0
y
3. Diagram below shows a shaded region that satisfies
x
the linear inequality P. –3 0
y
–9
4
2
Determine the linear inequality P. HOTS Applying
x
–4 –2 0 2 A y , –3x – 9 C y + 3x > –9
–2
1
B 3y + x . –9 D y < – x – 9
Determine the linear inequality P. 3
3 3 7. Which of the following region represents the
A y . x + 3 C y , x + 3
2 2 inequality 4y , x – 8. HOTS Analysing
3 3 A C
B y > x + 3 D y < x + 3
2 2 y y
4. Diagram below shows the shaded region that satisfy
x
the linear inequality R. 0 2
y 2
x
4 –8 0 8

2
x B D
–2 0 2 4
–2 y y

x x
Determine the linear inequality R. HOTS Applying 0 8 –2 0
–2
1 1
A y . – x + 2 C y < – x + 2
2 2
–8
1 1
B y , – x + 2 D y > – x + 2
2 2
51
Question 1:
A linear inequality in two variables is formed when symbols ,, <, >, or . used to relate two expressions involving two variables and the highest power of a
variable is 1.
Question 5: SOS TIP
Equation of the straight line can be determined by using the equation y = mx + c such that m = y – y 1 and c is the y-intercept.
2
x – x
2 1


C06 1202QB Maths Form 4.indd 51 21/02/2022 10:45 AM

PAPER 2


Section A


1. Diagram below shows a shaded region that satisfy 3. (a) Determine whether the point (5, 6) satisfy
SPM the three inequalities. SPM 2y = x + 10, 2y . x + 10 or 2y , x + 10.
CLONE CLONE
y [1 mark]
(b) In the graph below, shade the region that satisfy
6 1
3y + x . 9, y > x + 5 and x . –5. [3 marks]
©PAN ASIA PUBLICATIONS
4 2
2 Answer:
x (a)
–8 –6 –4 –2 0 2 4 6 8
–2
–4
–6
–8
–10
(b)
State all the linear inequalities. [4 marks] y
Answer:
8
6

4

2

x
–4 –2 0 2 4


2. In the graph in the diagram below, shade the region 4. In the graph below, shade the region that satisfy
SPM that satisfy the three inequalities y > –x + 6, SPM the three inequalities 2y < x + 12, y + x > – 4 and
CLONE CLONE
y < x + 6 and x < 6. [3 marks] y . 2x. [3 marks]
Answer: Answer:
y y
y = x + 6 10
2y = x + 12
8

6
6
4
2

x
x –6 –4 –2 0 2 4 6
0
6 y = –x + 6
–2
–4
y + x = –4
54
Question 1: y-intercept
SOS TIP Equation of the straight line can be found using the equation y = mx + c where m = – x-intercept and c is y-intercept.
Question 3 (a):
Point (5, 6) can be determined whether that satisfy the linear equation or the linear inequalities by the substituting the point into the equation of the straight
line.





C06 1202QB Maths Form 4.indd 54 21/02/2022 10:45 AM

9. Diagram below shows the shaded region defined by a Answer:
system of linear inequalities. (a)
y


16
12

8
©PAN ASIA PUBLICATIONS
4
(b)
x
0 2 4 6 8 10
(a) State all the inequalities that satisfy the shaded
region in the diagram above.
(b) Hence, determine the maximum value of x if
y = 12. HOTS Applying
[5 marks]

Section B

10. Mr. Zakry has allocated RM4 800 to purchase (b)
x units of calculator and y units of geometrical
set for school cooperation. The price of a unit of
calculator and a unit of geometrical set is RM20 and
RM16 respectively. The total of the calculator and
geometrical set is at least 200 units and the number
of unit of calculator purchased is not more than twice
the number of unit of geometrical set. HOTS Applying
(a) Write three inequalities, other than x > 0 and
y > 0, that satisfy the situation above. [3 marks]
(b) Using the scale of 1 cm to 50 units on both axes,
construct and shade the region that satisfies the
system of linear inequalities. [3 marks]
(c) Determine the number minimum and maximum
units of calculator that can be purchased by
Mr. Zakry if he purchased 100 units of
geometrical set. [3 marks]
Answer:
(a)






(c)










56
Question 9: y – y 1
SOS TIP (a) The equation of straight line can be determined by the equation y = mx + c such that m = x – x 1 and c is y-intercept.
2
2
(b) Draw the line y = 12 and determine the maximum value for x.






C06 1202QB Maths Form 4.indd 56 21/02/2022 10:45 AM

13. Ramesh opens a restaurant that needs x waiters and (c)
y cooks. Total waiters and cooks employed by y
Ramesh is not more than 100 person and the number
of waiters is not more than twice the number of 100
cooks employed. HOTS Applying
(a) Write two inequalities, other than x > 0 dan 90
y > 0, that satisfy the situation above. [3 marks]
(b) The third inequality is represented by the shaded 80
region in the graph in Diagram (a). Write, in
words, for the third inequality. [3 marks] 70
(c) Using Diagram (a), construct and shade the region
that satisfy the system of linear inequalities. 60
[3 marks]
(d) It is given that the monthly salary for each waiter 50
and cook employed is RM2 000 and RM4 000
respectively. Calculate, in RM, the maximum 40
salary allocation that need to be prepared by
30
Ramesh if he employed 60 cooks employed.
[3 marks]
20
Answer:
(a) 10
x
0 10 20 30 40 50 60

Diagram (a)
(d)


(b) ©PAN ASIA PUBLICATIONS












Section C


14. Diagram below shows two types of hotel room that is Type A room and Type B room can accommodate
available in a hotel in Kuala Lumpur. 2 and 3 person respectively. The hotel rental will be
charged with service tax of 6%.
(a) Mr. Rahman paid RM439.90 for a unit of
Type A room and a unit of Type B room
for a night stay. Madam Mariah paid
RM1 293.20 for 4 units of Type A room and
2 units of Type B for a night stay. Calculate, in
RM, the total payment payable by Mr. Siva if he

Type A room Type B room would like to rent 3 units of Type A room and
2 units of Type B room for 2-night stay.
[5 marks]
58 Question 13(d):
SOS TIP Draw the straight line y = 60 on the graph to get maximum value of x. Hence, determine the maximum salary of the allocation.
Question 14 (a):
The room rate of Type A room and Type B room can be obtained using simultaneous equation method.







C06 1202QB Maths Form 4.indd 58 21/02/2022 10:45 AM

Form 4 Assessment





PAPER 1

Time: 1 hour 30 minutes
This question paper contains 40 questions. Answer all the questions.
B 840 ©PAN ASIA PUBLICATIONS
1. Factorise 2x(2x – 7) – 3x – 15. 8. Determine the value of digit 3 in the number 13007 .
8
A (4x + 3)(x – 5) A 512 C 192
B (4x – 3)(x – 5) B 12288 D 1536
C (4x + 3)(x + 5)
D (4x – 3)(x + 5) 9. 321 + 112 =
5 5
A 434 C 443
5
5
2
2
2
2. Given x = z + , express y in terms of x and z. B 433 D 334 5
5
y
1
A y = 10. Diagram 1 shows a regular hexagon ABCDEF and
2(x + z)(x – z)
an equilateral triangle AGB. FAGH is a straight line.
1
B y =
2(x – z) 2 D C
2
C y =
(x + z)(x – z)
2 E B
D y =
(x – z) 2
x y
3. Round off 0.08095 correctly to 3 significant figures.
A 0.0810 F A G H
B 0.0800 Diagram 1
C 0.0890 Find the value of x + y.
D 0.0809 A 60° C 180°
B 120° D 240°
5
6

4. 1.85 × 10 + 1.49 × 10 =
A 1.999 × 10 4 11. In Diagram 2, ABC is a tangent to the circle centered
B 1.999 × 10 5 at O, at B.
C 1.999 × 10 6
D 1.999 × 10 7
y x
5. 7.89 × 10 6 = O
5 000
A 1.578 × 10 3
B 1.578 × 10 –3 80° 70°
C 1.578 × 10 2 A B C
D 1.578 × 10 –2 Diagram 2
Calculate the value of x + y.
6. Express 1507 in base 10. A 50° C 90°
8
A 6 663 C 839 B 70° D 110°
D 832
12. State the image coordinates of the point (–4, 3)
6
4
1
7. Given x = 5 + (4 × 5 ) + (4 × 5 ) + 3, state the value under the rotation 90° anticlockwise at the center
5
of x. of (0, 1).
A 6 040 010 A (–2, –3)
B 5 040 043 B (–2, 5)
C 1 400 043 C (–2, –5)
D 1 040 043 D (–2, –4)
103




APaper 1202QB Maths Form 4.indd 103 17/02/2022 2:49 PM

PAPER 2


Time: 2 hours 30 minutes
Section A
[40 marks]
Answer all questions in this section.
1. On the graph in the answer space, shade the region 3. Diagram 3 shows a solid combination of a hemisphere
which satisfies the three inequalites 2x + y > 8, and a cylinder. Given that the height of cylinder is
y > x and y , 8. [3 marks] 13 cm and the volume of the solid combination is
©PAN ASIA PUBLICATIONS
59 3
Answer: 1 282 cm .
60
y
16
14 y = x
12
13 cm
10
8
6
Diagram 3
4
By using π = 22 , calculate the radius, in cm, for
2 7
x hemisphere and cylinder. [4 marks]
–2 2 4 6 8 10 12
–2 Answer:
2x + y = 8
–4


2. The Venn diagram below shows the set P, set Q
and set R with the state of the universe set,
ξ = P  Q  R. On the given diagram, shade the set
(a) Q  R
P



Q R




4. Diagram 4 shows the OPQ and OQRS sectors. OTQ
is an right-angled triangle and TOS is a straight line.
[1 mark]

(b) P  (Q  R) R T
P
O Q

Q R 14 cm P


S
Diagram 4
Given that OS = 2OT. By using π = 22 , calculate
[2 marks] 7
(a) the perimeter, in cm, of the whole diagram.
[2 marks]
(b) the area, in cm , of the shaded region. [2 marks]
2
107




APaper 1202QB Maths Form 4.indd 107 17/02/2022 2:49 PM

Section B
[45 marks]
Answer all the questions in this section.

11. (a) Diagram 11(a) shows a set of data. 12. Table 12 shows the distance that connecting 5 cities.

21 20 31 25 26 City A – B B – C B – D C – E C – D D – E
Diagram 11(a) Distance
(i) Calculate the standard deviation of the (km) 52 38 75 55 60 28
data.
Table 12
©PAN ASIA PUBLICATIONS
(ii) Calculate the new variance if each value of (a) Complete the non-directed graph in the answer
the set of data is multiply by 3. [4 marks]
space to represent the information in Table 15.
Answer: [2 marks]
(a) (i) (b) Calculate the nearest distance from City A to
City E. [2 marks]
(c) Calculate the longest distance from City E to
City A such that all roads are only passed once.
[2 marks]
Answer:
(a)
B
A
(ii)
C



E
D
(b)
(b) Table 11(b) shows the frequency for the number
of goals scored by a team in football match. (c)

Score 0 1 2 3 4 5
Frequency 2 4 3 2 5 1
Table 11(b) 13. (a) Complete Table 13 in the answer space for the
(i) Calculate the interquartile range. equation y = 5x + 3x – 2 by writing values of
2
(ii) Hence, draw a box plot by using the y when x = –3 and x = 2.5. [2 marks]
quartiles obtained. [5 marks]
(b) For the subdivision of this question, use graph
Answer: paper. You can use a flexible ruler.
(b) (i) Using a scale of 2 cm to 1 unit on the x-axis and
2 cm to 5 units on the y-axis, draw a graph of
2
y = 5x + 3x – 2 for –3 < x < 3. [4 marks]
(c) From the graph in 13 (b), find
(i) the value of y when x = 1.8.
(ii) the values x when y = 2.2. [3 marks]
Answer:
(a)

(ii) x –3 –2.5 –1 0 1 2 3
y 34 0 –2 6 52
Table 13
(b) Refer to the graph on page 111.
(c) (i)

(ii)



110




APaper 1202QB Maths Form 4.indd 110 17/02/2022 2:49 PM

Graph for Question 15













©PAN ASIA PUBLICATIONS



































Section C
[15 marks]
Answer any one questions from this section.
16. (a) A 5-digit number is written randomly using all (b) Teacher Ngu made a questionnaire about a
digits 1, 2, 3, 4 and 5. Find the probability that hobby against one school with 100 students. It
the number is the even number. [4 marks] was found that 45 of his students’ hobby were
Answer: reading, 21 students loved reading and listening
(a) to music, 10 loved listening to music and sporting,
7 students loved reading only and 5 students had
that three hobbies. Given the ratio of the number
of students loved listening to music only to the
number of students loved sporting only is 3 : 2.
(i) Draw a Venn diagram based on the
information provided. [5 marks]
(ii) Calculate the number of pupils who are
hobby to read or listen to music only.
[2 marks]
(iii) A student is selected at random, find the
probability that the students loved to read
and listen to music only. [2 marks]
(iv) A student is randomly selected, find the
probability that the students loved all three
hobbies. [2 marks]
113




APaper 1202QB Maths Form 4.indd 113 17/02/2022 2:49 PM

Answers Complete Answers (Paper 1)

https://bit.ly/3KypArA


CHAPTER 1 Value of c = 2, y-intercept is 2
(b) Axis of symmetry at x = 3 + (–2) Thus, the maximum point is (0, 2).
Paper 1 2 When f(x) = 0,
1 2
1. B 2. A 3. C 4. D 5. C = 2 –x + 2 = 0
6. B 7. A 8. B 9. D 10. A ( ) ( ) ( ) x = ±! 2
1
1
1 2
11. B 12. A 13. D 14. D 15. B f 2 = 2 – 2 – 6 x = 1.4 or x = –1.4
16. A 17. C 18. C 19. D 20. D = – 25 The graph:
21. C 22. A 23. D 24. A 25. B 4 f(x)
26. A 27. D Thus, minimum point is ( 1 2 , – 25 ) .
4
Paper 2 (c) From the graph on (a), the roots for 2
Section A the graph function are x = –2 or x = 3.
x
2
1. (a) Quadratic expression in one variable. 8. (a) f(x) = x – 6x + 8 –1.4 0 1.4
(b) Not a quadratic expression in one Value of a = 1 . 0, shape
variable because there is a variable Value of c = 8, y-intercept = 8
with a power which is not a whole When f(x) = 0,
2
2
number. x – 6x + 8 = 0 (e) f(x) = x – 2
(c) Not a quadratic expression in one (x – 4)(x – 2) = 0 Value of a = 1 . 0, shape
variable because there are two x = 2 or x = 4 Value of b = 0, axis of symmetry is
variables, x and y. The graph: the y-axis
(d) Quadratic expression in one variable. f(x) Value of c = –2, y-intercept is –2
(e) Not a quadratic expression in one Thus, the minimum point is (0, –2).
variable because there is a variable When f(x) = 0,
2
with a power which is not a whole 8 –x + 2 = 0
number. x = ±! 2
x = 1.4 or x = –1.4
2. (a) (b) (c) (d)
The graph:
3. (a) a = 2, b = –3, c = 5 x
(b) a = 1, b = 4, c = 0 0 2 4 f(x)
(c) a = 3, b = 0, c = –7
2
(d) a = –3, b = –4, c = 2 (b) f(x) = x – 4x + 4
(e) a = 3, b = –3, c = 0 Value of a = 1 . 0, shape
Value of c = 4, y-intercept = 4 CHAPTER 1
4. (a) Maximum point : (–1, 4) When f(x) = 0, x – 4x + 4 = 0 0 x
2
–1.4
1.4
©PAN ASIA PUBLICATIONS
Axis of symmetry, x = –1
2
(x – 2) = 0
(b) Minimum point : (0, –4) x = 2 –2
Axis of symmetry, x = 0
The graph:
5. (a) Given f(x) = 2x – 5x + c. Section B
2
Substitute the values of x = –1 and f(x) 9. Area of rectangle = 66 cm 2
f(x) = 5 into the quadratic function: (x + 2)(x + 7) = 66
2
5 = 2(–1) – 5(–1) + c x + 2x + 7x + 14 = 35
2
5 = 2 + 5 + c x + 9x + 14 – 35 = 0
2
c = –2
4 x + 9x – 21 = 0
2
(b) Given f(x) = 2x – 5x + c.
2

Substitute the values of x = 3 and 10. By using Pythagoras Theorom,
2
2
f(x) = 7 into the quadratic function: 0 2 x r = p + q 2
7 = 2(3) – 5(3) + c 11. x + 3x = –2(–3 – x)
2
2
2
7 = 18 – 15 + c (c) f(x) = x – 4 x + 3x = 6 + 2x
2
c = 7 – 3 Value of a = 1 > 0, shape x + 3x – 2x – 6 = 0
2
2
= 4 Value of b = 0, axis of symmetry is x + x – 6 = 0
the y-axis
6. Let Sufi’s age = x – 2 Value of c = –4, y-intercept is –4 (x – 2)(x + 3) = 0
x(x – 2) = 35 x = 2, x = –3
x – 2x = 35 Thus, the minimum point is (0, –4). 12. – 3 = x
2
2
x – 2x – 35 = 0 When f(x) = 0, x – 4 = 0 2x + 1 x – 2
2
(x + 2)(x – 2) = 0 –3(x – 2) = x(2x + 1)
2
7. (a) f(x) = x – x – 6 x = 2 or x = –2 –3x + 6 = 2x + x
2
Value of a = 1 > 0, shape The graph: 2x + x + 3x – 6 = 0
2
Value of c = –6, y-intercept = –6 2x + 4x – 6 = 0
2
When f(x) = 0, f(x) x + 2x – 3 = 0
2
2
x – x – 6 = 0 (x – 1)(x + 3) = 0
(x + 2)(x – 3) = 0 x = 1, x = –3
x = –2 or x = 3
The graph: x 13. (a) p = 1
2
–2 0 2 (b) f(x) = x + 5x + q
f(x)
From the graph, when x = 0, f(x) = 4.
–4 Substitute x = 0 and f(x) = 4 into
quadratic function:
x 2
2
–2 0 3 (d) f(x) = –x + 2 4 = 1(0) + 5(0) + q
Value of a = –1 . 0, shape q = 4

–6 Value of b = 0, axis of symmetry is
the y-axis
115
Answers 1202QB Maths Form 4.indd 115 21/02/2022 6:27 PM

Given minimum point is (−2.5, −2.25), Side length of triangle = 5 – 1 Paper 2
Axis of symmetry, x = –2.5. = 4 cm
2
(c) When f(x) = x + 5x + 4 is reflected 19. Area of square – Area of triangle = 40 Section A
in the x-axis, the quadratic function 1 1. (a) 4 2 (b) 6 2
]
2
change to f(x) = –x – 5x – 4 where the 2y(2y) – [ 2 (y + 2)(2y) = 40 (c) 8 1 (d) 2 5
value of a will change to −a.
4y – y – 2y – 40 = 0 2. (a) 3 × 7 = 21 (b) 4 × 5 = 500
2
2
3
1
14. Price = RM300 3y – 2y – 40 = 0 (c) 7 × 9 = 63 (d) 2 × 3 = 162
2
4
1
(8x + 14)(5x) = 300 (3y + 10)(y – 4) = 0 0 1
4
2
40x + 70x – 300 = 0 10 3. (a) 32 = 2 × 4 + 3 × 4
= 2 + 12


(x – 2)(4x + 15) = 0 x = – 3 (Rejected) or y = 4
15 = 14
x = 2, x = – (Rejected) Perimeter of the combination objects
4 (b) 11101 2
1
2
3
0
Number of text books bought by Sofea = (4 + 6) + (4 + 2) + [3 × 2(4)] = 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 +
©PAN ASIA PUBLICATIONS
= 10 + 6 + 8 + 8 + 8
= 8(2) + 14 = 40 cm 1 × 2 4
= 30 books = 1 + 4 + 8 + 16
1
20. (a) L = (y)(y + 4) = 29
15. Volume of the box = 4 500 2 (c) 413 = 3 × 5 + 1 × 5 + 4 × 5 2
1
0
1
(x + 5)(x)(30) = 4 500 L = (y + 4y) 5 = 3 + 5 + 100
2
30x + 150x – 4 500 = 0 2 = 108
2
1
2
x + 5x – 150 = 0 L = y + 2y (d) 624 = 4 × 7 + 2 × 7 + 6 × 7 2
2
1
0
(x – 10)(x + 15) = 0 2 7 = 4 + 14 + 294
x = 10, x = –15 (b) Area of triangle = 48 = 312
(Rejected) 1 2 y + 2y = 48
2
Thus, the value of x is 10 cm. 1 4. (a) 534 to base two
10
2
16. Area of the box = 432 2 y + 2y – 48 = 0 2 534 Remainder
2
(2x + 4)(3x + 3) = 432 y + 4y – 96 = 0
2
6x + 12x + 6x + 12 = 432 (y + 12)(y – 8) = 0 2 267 0
CHAPTER 1 – CHAPTER 2
2
6x + 18x – 420 = 0 y = 8, y = –12 2 133 1
2
x + 3x – 70 = 0 (Rejected)
(x + 10)(x – 7) = 0 Thus, y = 8 cm. 2 66 1
x = 7, x = –10 (c) Area of the polygon = 6 × 48 2 33 0
(Rejected) = 288 cm 2 2 16 1
Substitute x = 7, Thus, the area of polygon is 288 cm 2
Diameter 4 balls = 3(7) + 3 and its name is hexagon. 2 8 0
= 21 + 3 21. (a) L = (35 + y)(65 + y) 2 4 0
2
= 24 cm L = 2 275 + 65y + 35y + y
2
So, diameter 1 ball = 24 ÷ 4 L = y + 100y + 2 275 2 2 0
= 6 cm (b) Area of the tile = 6 175 2 1 0
2
Substitute x = 7 into 2x + 4, y + 100y + 2 275 = 6 175 0 1
2
Diameter 3 balls = 2(7) + 4 y + 100y – 3 900 = 0 534 = 1000010110
= 14 + 4 (y – 30)(y + 130) = 0 10 2
= 18 cm y = 30, y = –130 (Rejected) (b) 534 to base eight
10
So, diameter 1 ball = 18 ÷ 3 Thus, y = 30 cm. 8 534 Remainder
= 6 cm (c) The smallest part of tile represented
by region DEFI. 8 66 6
Section C Area of DEFI = 0.3 m × 0.3 m
17. (a) L = (x + 6)(x + 3) = 0.09 m 2 8 8 2
2
L = x + 6x + 3x + 18 Number of tiles needed 8 1 0
2
L = x + 9x + 18 = 1.08 ÷ 0.09
(b) Area of the cake = 270 = 12 tiles 0 1
x + 9x + 18 = 270 22. (a) Area A – Area B = 12 cm 2 534 = 1026 8
2
10
2
x + 9x + 18 – 270 = 0 2x(x + 3) – x(x + 5) = 12 (c) 534 to base five
10
2
2
x + 9x – 252 = 0 2x + 6x – x – 5x – 12 = 0
2
(x – 12)(x + 21) = 0 x + x – 12 = 0 5 534 Remainder
2
x = 12 cm (x – 3)(x + 4) = 0 5 106 4
(c) Number of small cakes x = 3, x = – 4 (Rejected)
= 270 cm ÷ 9 cm 2 Thus, x = 3 cm. 5 21 1
2
= 30 cakes (b) Perimeter of new arrangement 5 4 1
Thus, the small cakes is enough to = 6 + 6 + 6 + 8 + 3 + 8 + 3
give to all Safi’s friend. = 40 cm 0 4
534 = 4114
18. (a) Area of triangle, L 10 5
1
= [2(x + 3)](x – 1) CHAPTER 2 (d) 534 to base four
2 10
= (x + 3)(x – 1) 4 534 Remainder
= x + 2x – 3 Paper 1
2
(b) Area of triangle = 32 1. A 2. D 3. D 4. A 5. C 4 133 2
2
x + 2x – 3 = 32 6. C 7. A 8. B 9. B 10. D 4 33 1
2
x + 2x – 35 = 0 11. C 12. D 13. B 14. D 15. C
(x – 5)(x + 7) = 0 16. A 17. C 18. C 19. B 20. A 4 8 1
x = 5, x = –7 21. C 22. B 23. B 24. D 25. A 4 2 0
(Rejected) 26. C 27. D 28. A 29. B 30. D
When x = 5, 31. C 32. C 33. A 34. B 35. D 0 2
Height of triangle = 5 + 3 36. D 37. B 38. C 39. D 40. C 534 = 20112 4
10
= 8 cm 41. B 42. B
116
Answers 1202QB Maths Form 4.indd 116 21/02/2022 6:27 PM

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