F4 ADD MATH MODULE 2021.pdf

ii
JABATAN PENDIDIKAN NEGERI SABAH
PREFACE
Understanding the concept and doing a lot of practice on
the concepts learned is a contributing factor to student
success in the SPM examination. Most students who are
weak in mastering the concept are due to lack of practice.
A compact and brief note (One Page Note) that is included
in this module is expected to help students to master the
concept of the topic. Doing intensive exercises for each
topic is also expected to help teachers and students master
the topic.
Therefore, this module is hoped to help teachers and students during teaching and
learning and during revision exercises before the SPM examination.
Sincerely from:
Lee Chiong Tee
Coordinator and Head Panels
Additional Mathematics Module 2021
MODULE OBJECTIVE
1. Make it easier for students to remember important concepts in the
form of more compact graphics.
2. Help students answer revision practice questions
before the SPM examination
3. Help teachers overcome the problems of students who do not have
any reference sources.

iii
CONTENT
TOPIC 1 FUNCTIONS 1
TOPIC 2 QUADRATICS FUNCTIONS 35
TOPIC 3 SYSTEM OF EQUATIONS 81
TOPIC 4 INDICES, SURDS AND LOGARITHMS 111
TOPIC 5 PROGRESSIONS 172
TOPIC 6 LINEAR LAW 216
TOPIC 7 COORDINATE GEOMETRY 253
TOPIC 8 VECTORS 315
TOPIC 9 SOLUTION OF TRIANGLES 386
TOPIC 10 INDEX NUMBER 430
WORKSHEET ANSWERS 464

1
FUNCTIONS
- ONE PAGE NOTE (OPN)
- WORKSHEET
Encik Dennis Chua Ah Thin

2
ONE PAGE NOTE (OPN)

3
WORKSHEET
TOPIC 1 : FUNCTIONS
[ 2 – 3 questions → 4 – 9 marks ]
==========================================================================================================================================
Revision [ RELATIONS ~ ordered pairs, arrow diagram, cartesian graph ]
==========================================================================================================================================
1 Based on the below information, the relation between P and Q is defined by the set of ordered pairs
{ ( 1, 2 ), ( 1, 4 ), ( 2, 6 ), ( 2, 8 ) }.
P = { 1, 2, 3 }
Q = { 2, 4, 6, 8, 10 }
State
(a) the image of 1,
(b) the object of 2.
[2 marks] [2003, No.1]
Answer :
(a) (b)
2 The diagram shows the relation between set P and set Q.
State
(a) the range of the relation,
(b) the type of the relation.
[2 marks] [2004, No.1]
Answer :
(a) (b)
w
d
x
y
z
e
f
Set P Set Q
types of relation
:
→ one to one
→ one to many
→ many to one
→ many to many
range  codomain
(3, 6) , (4, 8) , . . .
image of 3 = 6
object of 6 = 3
function notation, f(x) = 2x

4
3 The diagram shows the relation between set X and set Y in the graph form.
State
(a) the objects of q.
(b) the codomain of the relation.
[2 marks] [2009, No.1]
Answer :
(a) (b)
4 The diagram shows the relation between set X and set Y in the graph form.
State
(a) the relation in the form of ordered pairs.
(b) the type of the relation,
(c) the range of the relation.
[3 marks] [2010, No.1]
Answer :
(a)
(b) (c)
5 It is given that the relation between set X = { 0, 1, 4, 9, 16 } and set Y = { 0, 1, 2, 3, 4, 5, 6 } is “
square of ”
(a) Find the image of 9.
(b) Express the relation in the form of ordered pairs.
[3 marks] [2011, No.1]
Answer :
(a)
(b)
s
r
q
p
Set Y
2 4 6
Set X
Set X
Set Y
s
r
q
p
1 3 4
2

5
6 The diagram shows the relation between set M and set N.
State
(a) the object of −1,
(b) the range of the relation.
[2 marks] [2012, No.1]
Answer :
(a) (b)
7 The diagram shows the relation between Set A and Set B in the arrow diagram form.
(a) Represent the relation in the form of ordered pairs.
(b) State the domain of the relation.
[2 marks] [2014, No.1]
Answer :
(a)
(b)
1
4
−1
−3
Set M Set N
3
5
6
7
Set B
Set A
1 4
2
1
−2

6
==========================================================================================================================================
1.1 Functions
1.1.1 Explain function using graphical representations and notations.
1.1.2 Determine domain and range of a function.
1.1.3 Determine the image of a function when the object is given and vice versa.
==========================================================================================================================================
 functions notations
8 In the diagram, set B shows the images of certain elements of set A.
(a) State the type of relation between set A and set B.
(b) Using the function notation, write a relation between set A and set B.
[2 marks] [2006, No.1]
Answer :
(a)
(b)
9 The diagram shows the linear function h.
(a) State the value of m.
(b) Using the function notation, express h in terms of x.
[2 marks] [2007, No.1]
Answer :
(a)
(b)
Set A Set B
4
5
−5
−4
25
16
x h (x)
0
1
m
5
1
2
4
6

7
 the conditions of existence (the conditions such that a relation is a function)*
10 The diagram shows the relation between x and y.
Determine whether the relation is a function. Give reason for your answer. [2 marks]
[Forecast]
Answer :
11 The diagram shows the relation between set P and set Q.
Determine whether the relation is a function. Give reason for your answer. [2 marks]
[Forecast]
Answer :
 given object & image → find the constant
12 The diagram shows the function h : x →
x
x
m −
, x  0, where m is a constant.
Find the value of m. (Ans : 4)
[2 marks] [2006, No.2]
Answer :
O
x
y
• 3
 −3  •
 1  •
 2  •
 3  •
• 2
• 0
Set P Set Q
x
m x
x
−
h
8
1
2
−
function :
→ one to one @ many to one relation
→ each object must mapped with an
image only
to whether a graph is a function :
→ draw a vertical line, and it cut the
graph at only one point

8
13 It is given that the function f (x) = p − 3x, where p is a constant.
Find the value of p such that f (p) = 4. (Ans : −2)
[2 marks] [2013, No.3]
Answer :
14 The diagram shows the function f : x → x − 2m, where m is a constant.
Find the value of m. (Ans : −2)
[2 marks] [2014, No.2]
Answer :
 maps onto inself (unchanged under the mapping) / definition*
15 Given the function f : x → 3x − 2, find
(a) the value of x when f (x) maps onto itself, (Ans : 1)
(b) the value of h such that f (2 − h) = 4h. (Ans : 7
4 )
[4 marks] [2016, No.11]
Answer :
(a) (b)
4 8
x x − 2m
x is maps onto itself under function h

h (x) = x

9
16 A function f is defined by f (x) =
3
2 +
+
x
k
x
, x  h.
(a) Find the value of h.
(b) If x = 2 is unchanged under the mapping of function f, find the value of k.
(Ans : 12)
[3 marks] [Forecast]
Answer :
(a) (b)
 obsolute functions ~ object, image, domain, codomain, range, graph
17 Given the function f : x → │x − 3│, find the values of x such that f (x) = 5.
(Ans : −2, 8 )
[2 marks] [2007, No.2]
Answer :
18 The diagram shows the graph of the function f (x) =  2x − 1 , for the domain 0  x  5.
State
(a) the value of t. (Ans : 2
1 )
(b) the range of f (x) corresponding to the given domain.
[3 marks] [2008, No.1]
Answer :
(a) (b)
O t
y = f (x)
y
1
5
x
x = k
x = −k
5 is maps onto itself under function h

h (5) = 5
f (x) tertakrif
cx + d = 0
f (x) =
f (x) tak tertakrif
cx + d  0

10
19 The diagram shows the graph of the function f : x → | 1 − 2x | for the domain −2  x  4.
State
(a) the object of 7,
(b) the image of 3, (Ans : 5)
(c) the domain of 0  f(x)  5.
[3 marks] [2017, No.9]
Answer :
(a)
(b)
(c)
20 (a) Sketch the graph y =  2x + 6  − 1 on the given axes for the domian −5  x  1. [2 marks]
(b) Hence, find the corresponding range. [1 mark]
[Forecast]
Answer :
(a)
x 0
y 0
(b)
7
O
x
y = f(x)
4
(−2, 5)
x
y
O

11
==========================================================================================================================================
1.2 Composite functions
1.2.1 Describe the outcome of composition of two functions.
1.2.2 Determine the composite functions.
1.2.3 Determine the image of composite functions given the object and vice versa.
1.2.4 Determine a related function given composite function and another function.
1.2.5 Solve problems involving composite functions.
==========================================================================================================================================
 given f(x), g(x) → find fg(x), gf(x), f2
(x), g2
(x)
21 The following information is about the function h and the composite function h2
.
h : x → ax + b, where a and b are constants, and a > 0
h2
: x → 36x + 35
Find the value of a and of b. (Ans : a = 6, b = 5)
[3 marks] [2007, No.3]
Answer :
22 Given the functions f (x) = x − 1 and g (x) = kx + 2, find
(a) f (5),
(b) the value of k such that gf (5) = 14. (Ans : 3)
[3 marks] [2008, No.3]
Answer :
(a) (b)
23 Given the functions g : x → 2x − 3, and h : x → 4x, find
(a) hg (x), (Ans : 8x − 12)
(b) the value of x if hg (x) =
2
1
g (x). (Ans : 2
3
)
[4 marks] [2009, No.2]
Answer :
(a) (b)

12
24 Given the functions g : x → x − 8 and h : x →
2
3 −
x
x
.
(a) State condition for the value of x such that function h is defined. **
(b) Find the value of hg (10). (Ans : 2
1 )
[3 marks] [2010, No.3]
Answer :
(a) (b)
25 It is given that the functions g (x) = 4x − 7 and h (x) = 2x. Find the value of find gh (2). (Ans : 9)
[2 marks] [2011, No.2]
Answer :
26 Given the function f : x → 5x + 6 and g : x → 2x − 1, find gf (x). (Ans : 10x + 11)
[2 marks] [2013, No.2]
Answer :
27 It is given the functions f (x) = 3x and g(x) = h − kx, where h and k are constants. Express h in terms
of k such that gf (1) = 4. (Ans : h = 4 + 3k)
[3 marks] [2015, No.2]
Answer :

13
28 Given the function m : x → px + 1, h : x → 3x − 5 and mh (x) = 3px + q. Express p in terms
of q. (Ans : p = 5
1 q
−
)
[3 marks] [2016, No.12]
Answer :
29 The function m and n are defined by m : x → 3x + 1 and n : x →
1
4
2
−
x
x
.
(a) State the value of x such that n is undefine.
(b) Find the value of k if mn (k) =
5
13
. (Ans : 4)
Answer :
(a) (b)
MIND think :
function g • f exists if and only if range of f  domain of g
function f • g exists if and only if range of g  domain of f
f g
y
X
x
Z
gf
f (x)
Y
gf (x)

14
 given f(x), fg(x) → find g(x)
30 Given the function h (x) =
x
6
, x  0 and the composite function hg (x) = 3x, find
(a) g (x), (Ans : x
2
)
(b) the value of x when gh (x) = 5. (Ans : 15)
[4 marks] [2004, No.3]
Answer :
(a) (b)
31 Given that f (x) = 3x + 4 and fg (x) = 6x + 7, find
(a) fg (4), (Ans : 31)
(b) g (x). (Ans : 2x + 1)
[4 marks] [2012, No.2]
Answer :
(a) (b)
32 Given f (x) = x2
+ 1, find in terms of x, fungsi g(x) such that fg(x) = x2
+ 4x + 5.
(Ans : x + 2)
Answer :

15
 given g(x), fg(x) → find f(x)
33 The diagram shows the relation between set A, set B and set C.
It is given that set A maps to set B by the function
2
1
+
x
and maps to set C by fg : x → x2
+ 2x + 4.
(a) Write the function which maps set A to set B by using the function notation.
(b) Find the function which maps set B to set C.
[ Ans : f (x) = 4x2
+ 3 @ f : x → 4x2
+ 3 ]
[4 marks] [2018, No.22]
Answer :
(a)
(b)
34 The diagram shows the relation of three sets.
It is given that f : x → 2x + 3 and g−1
f : x →
3
x
+ 1, x  0.
(a) If a student writes a = 10, determine the value is correct or wrong. Give your reason.
(Ans : 9)
(b) Find g−1
( x ). (Ans : 3
3
x
x
+
−
, x  3)
[4 marks] [2019, No.8]
Answer :
(a)
(b)
fg : x → x2
+ 2x + 4
A
B
C
f g−1
3 a 2
Set X Set Y Set Z

16
 given f(x) → find f2
(x), f3
(x), f4
(x), . . . , f n
(x) ~ 1
35 It is given that f : x →
1
x
x +
, x  −1.
(a) Find the iterated function f2
(x), f3
(x) and f4
(x).
[ Ans :
2
2 1
( ) x
x
f x +
= , x  1
2
− . . . ]
(b) Hence, determine the general rule fn
(x), where n is a positive integer.
Answer :
(a)
(b)
36 It is given that f : x → ax + b and f3
(x) = 8x + 7.
(a) Find the value of a and of b. (Ans : a = 2, b = 1)
(b) Find the expression of f 4
(x). Hence, determine the general rule f n
(x), where n is a positive
integer. (Ans : 16x + 15)
Answer :
(a)
(b)
f 3
(x) = f [ f 2
(x) ] = f 2
[ f (x) ]
f 4
(x) = f [ f 3
(x) ] = f 2
[ f 2
(x) ]

17
 given f(x) → find f2
(x), f3
(x), f4
(x), . . . , f n
(x) ~ 2
37 It is given that f : x →
3
x
, x  0.
(a) Find the iterated function f2
(x), f3
(x) and f4
(x). [2 marks]
(b) Hence, find the value of
(i) f28
(5), [1 mark]
(ii) f57
(5). [1 mark]
[Forecast]
Answer :
(a)
(b) (i)
(ii)
38 It is given functions f (x) = p, f 2
(x) = q, f 3
(x) = r dan f 4
(x) = s. If f 4
(x) = x,
(a) find in terms of p, q, r, or s for :
(i) f 8n
(x), where n = 1, 2, 3, . . . , [1 mark]
(ii) f 70
(x). [1 mark]
(b) State the value of k, such that f k
(x) = p, where 26  k < 33 and k is an integer.
[1 mark]
[Forecast]
Answer :
(a) (i)
(ii)
(b)
f n
(x) = x  f kn
(x) = x

18
==========================================================================================================================================
1.3 Inverse functions
1.3.1 Describe inverse of a function.
1.3.2 Make and verify conjectures related to properties of inverse functions.
1.3.3 Determine the inverse functions.
==========================================================================================================================================
 describe inverse function
39 In the diagram shows below, the function h maps x to y and the function g maps y to z.
Determine
(a) h −1
(5),
(b) gh (2).
[2 marks] [2005, No.1]
Answer :
(a) (b)
40 The diagram shows the composite function gf that maps x to z.
State
(a) the function that maps x to y.
(b) g−1
( z ) .
[2 marks] [2015,
No.1]
Answer :
(a) (b)
x y z
h g
2
5 8
x y z
gf

19
MIND think :
• to every one to one function f : x → y, there exist an inverse function f −1
: y → x
• if y = f (x) and y = f −1
(x) are inverse to each other → f f −1
(x) = f −1
f (x) = x
 the conditions of existence inverse functions
41 It is given that the relation between set set X = { −4, −3, −1, 1, 3, 4 } and set Y = { 1, 3, 4 } is defined
by the following ordered pairs :
{ (−2k, 4) (−3, 3), (−1, 1), (1, 1), (3, 3), (4, 4)
(a) Find the value of k.
(b) Using the function notation, write the relation between X and set Y.
(c) Determine whether the inverse in (b) is also a function. Give reason for your answer.
Answer :
(a) (b)
(c)
42 (a) State the condition for a function which has an inverse function.
(b) The function g is defined as g (x) = x2
+ 4.
(i) State the type of relation of the function.
(ii) Determine whether g (x) has an inverse function.
Answer :
(a)
(b) (i)
(ii)
x y
f
f −1
X Y ~ f (x) = x + k → f −1
(x) = x − k
~ f (x) = kx → f −1
(x) = x
k
~ f (x) = ax + b → f −1
(x) =
x b
a
−
~ f (x) = x
a
b
+ → f −1
(x) = ( )
a x b
−
~ f (x) = 2
x → f −1
(x) = x
~ f (x) =
ax b
cx d
+
+
→ f −1
(x) =
dx b
cx a
−
− +
(for checking answer)

20
43 The diagram shows five graphs, P, Q, R, S and T.
P Q R S T
(a) State the graph (s) which is /are not a function. Give reason for your answer.
(b) State the graph (s) which has / have inverse function. Give reason for your answer.
Answer :
(a)
(b)
x
y
O
x
y
O
x
y
O
x
y
O
x
y
O

21
 properties of inverse function
44 The diagram in the answer space shows the graph y = f (x) .
(a) On the diagram in the answer space, draw the inverse function of the graph.
(b) From the graph, state the value of x when f (x) = f −1
(x).
Answer :
(a)
(b)
45 The diagram in the answer space shows the graph y = h (x) .
(a) Find h −1
(0).
(b) On the diagram in the answer space, draw the graph y = h −1
(x).
(c) Determine the domain for the graph y = h −1
(x).
Answer :
(a)
(c)
(b)
y
O
4
6
8
2 4 6 8 10 12
x
2
10
12
y = f(x)
y
x
s
t
s t
O
y = h(x)

22
MIND think :
• the graph of y = f (x) and y = f −1
(x) are symmetry about the line y = x
• domain for f −1
(x) = range for f (x)
range for f −1
(x) = domain for f (x)
 find inverse function 1a
46 Given the functions g : x → 3x − 1, find
(a) g (2),
(b) the value of p when g −1
( p ) = 11. (Ans : 32)
[3 marks] [2009, No.3]
Answer :
(a) (b)
47 Given that f : x → x + 5, find
(a) f (3),
(b) the value of k such that 2 f −1
(k) = f (3). (Ans : 9)
[3 marks] [2012, No.3]
Answer :
(a) (b)
y = f (x)
y = f −1
(x)
y
y = x
x

23
 find inverse function 1b
48 Given that g : x → 5x + 1 and h : x → x2
− x + 3. Find
(a) g−1
(3). (Ans : 5
2
)
(b) hg (x). (Ans : 25x2
+ 5x +3)
[4 marks] [2003, No.2]
Answer :
(a) (b)
49 Given the functions g : x → 5x + 2, and h : x → x2
− 4x + 3, , find
(a) g−1
(6), (Ans : 5
4
)
(b) hg (x), (Ans : 25x2
−1)
[4 marks] [2008, No.2]
Answer :
(a) (b)
 find inverse function 1c
50 Given the functions h−1
: x → 4x + m and h : x → 2kx +
8
5
, where m and k are constants, find the
value of m and the value of k. (Ans : k = 8
1
, m = 2
5
− )
[3 marks] [2004, No.2]
Answer :
51 The following information refer to the functions h and g.
h : x → 2x − 3
g : x → 4x − 1
Find g h−1
(x). (Ans : 2x + 5)
[3 marks] [2005, No.3]
Answer :

24
52 Given the functions g : x → 2x + 1 and h : x → 3x +6, find
(a) g −1
(x),
(b) hg −1
(9), (Ans : 18)
[3 marks] [2010, No.2]
Answer :
(a) (b)
53 Given the function g : x → 2x − 8, find
(a) g−1
(x),
(b) the value of p such that 





2
3
2 p
g = 30. (Ans : 9)
[4 marks] [2017, No.10]
Answer :
(a) (b)
 find inverse function 2
54 The function w is defined as
x
x
w
−
=
2
5
)
( , x  2. Find
(a) w−1
(x), (Ans : x
x 5
2 −
)
(b) w−1
(4).
[3 marks] [2005, No.2]
Answer :
(a) (b)

25
55 The inverse function h−1
is defined by h−1
: x →
x
−
3
2
, x  3. Find
(a) h (x). (Ans : x
x 2
3 − )
(b) the value of x such that h (x) = −5. (Ans : 4
1 )
[4 marks] [2011, No.3]
Answer :
(a) (b)
 using concept inverse / given f(x), fg(x) → find g(x)
56 Given the functions f : x → 2x − 1 and f −1
g : x →
6
2
2
−
−
x
x
, x  3. Find the value of gf (3). (Ans : 2
1 )
Answer :
 using f f −1
(x) = x @ f −1
f (x) = x / given g(x), fg(x) → find f(x)
57 Given the function g : x → 3x + 1 and fg : x → 9x2
+ 6x − 4, find
(a) g−1
(x),
(b) f (x). (Ans : x2
− 5)
[3 marks] [2016, No.13]
Answer :
(a)
(b)
f −1
(y) = x
f (x) = y
f [ f −1
(x) ] = x
@
f −1
[ f (x) ] = x

26
58 Given f(x) =
p
x −
3
, x  p and a function g such that 1
( 3) ( 4)
g g p f p
−
− = − . Find the value
of p. (Ans : 4
9 )
Answer :
59 Given f(x) =
2
1
−
x
, x  2 and g(x) =
x
1
+ 2.
(a) Find fg(x). (Ans : x)
(b) State the relation between functions f(x) and g(x).
Answer :
(a)
(b)
 others
60 Given that f : x → x and g : x → x + 4. Express in terms of f or g or both f and g for :
(a) x → 4
+
x ,
(b) x → x2
+ 4,
(c) x → x + 8
Answer :
(a) (b) (c)

27
PAPER 2
 Part A → 6 – 7 marks
61 Given that f (x) → 3x − 2 and g (x) =
5
x
+ 1, find
(a) f −1
(x), [1 mark]
(b) f −1
g (x), (Ans : 15
15
+
x ) [2 marks]
(c) h (x) such that hg (x) = 2x + 6. (Ans : 10x − 4) [3 marks]
[2006, No.2]
Answer :
62 In the diagram, the function f maps set A to set B and the function g maps set B to set C.
Find
(a) in terms of x, function
(i) which maps set B to set A.
(ii) g (x). (Ans : 4x − 3)
[5 marks]
(b) the value of x such that fg (x) = 8x + 1. (Ans : 2) [2 marks]
[2014, No.3]
Answer :
x
A
3x + 2 12x + 5
B C
f g

28
63 It is given that f : x → 2x − 3 dan g : x → 1 − 3x. Find
(a) (i) g (5) ,
(ii) the value of m if f (m +2) =
7
1
g (5) , (Ans : 2
3
− )
(iii) gf (x) . (Ans : −6x + 10)
[5 marks]
(b) Hence, sketch the graph of )
(x
gf
y = for −1  x  3. State the range of y.
(Ans : 0  y  16) [3 marks]
[2018, No.2]
Answer :
FORECAST
64 The diagram shows the mapping from x to y under f : x → px + q and the mapping from z to y under
g : z →
q
z +
3
8
, z 
3
q
− .
Find
(a) the value of p and the value of q, ( Ans : p = 2
1
, q = −1 ) [3 marks]
(b) the function which map y to z, (Ans : y
y
3
8
+
) [2 marks]
(c) the function which map x to z. (Ans : 6
3
14
−
+
x
x
) [2 marks]
Answer :
h
f g
4 1 3
x y z

29
65 In the diagram, the function f (x) = ax + b maps x to y and the function g (y) =
y
a
−
1
, y ≠ 1 maps
y to z.
Find
(a) the values of a and b, (Ans : a = 2, b = 1) [3 marks]
(b) the function which map y to x, (Ans : 2
1
−
y
) [2 marks]
(c) the function that map z to x. (Ans : − z
1 ) [2 marks]
Answer :
66 In the diagram, the function f maps x to y and the function g maps y to z.
Given that h−1
(x) = x + 5 and gh(x) = 3x − 11. Find
(a) in terms of x, function
(i) h (x), [1 mark]
(ii) g (x), (Ans : 3x + 4) [2 marks]
(b) the value of a and of b. (Ans : a = 3, b = −2) [3 marks]
Answer :
1
3
−1
x y z
f g
−2
gh
a b
x y z

30
67 Given that f : x → 3 + 2x and g : x →
1
8
+
x
, x ≠ −1. Express the following in the same form :
(a) f −1
, [1 mark]
(b) g −1
, (Ans : x
x
−
8 ) [2 marks]
(c) g −1
f −1
, (Ans : 3
19
−
−
x
x
) [2 marks]
(d) (fg) −1
, [1 mark]
Answer :
68 Given that f : x →
x
x
−
+
1
1
, x ≠ 1. Express the following in the same form :
(a) f 2
, (Ans : x
1
− ) [2 mark]
(b) f 3
, (Ans : 1
1
x
x
−
+
) [2 marks]
(c) f 4
, (Ans : x) [2 marks]
(d) f 37
, [1 mark]
(e) f 95
. [1 mark]
Answer :
(fg)−1
(x) = g−1
f −1
(x)

31
69 Chelsea runs a physics experiment about the image generated by convex lenses as shown in the following
figure.
It is found that the image of candle that produce is given by f(x) =
10
10
−
x
x
, x  k. Find
(a) the value of k, [1 mark]
(b) the distance of the image of the candle, if the distance of the candle from the lens
is 14 cm, (Ans : 35) [2 marks]
(c) the distance of the candle to the lens if the lens is located in the middle between the candle and
its image. (Ans : 20) [3 marks]
Answer :
f(x)
image of candle
lens
candle
screen
x

32
70 Jackley runs a chemistry experiment for preparation of the Science Practical Examination. The diagram
shows the scale for temperatures in Celsius (°C) and Fahrenheit (°F) on a thermometer. The relation
between temperature in xC and yF is given by y = 1.8x + k, where k is a constant.
(a) Find the value of k. (Ans : 32) [2 marks]
(b) If the temperature of today is 32C, what is its temperature in F (Ans : 89.6) [2 marks]
(c) Form a function that allows us to change the temperature in the unit of Fahrenheit (F) to degrees
Celsius (C). (Ans : x =
32
1.8
y −
) [2 marks]
Answer :
F
100
C
212
50 122
0 32
−20 −4

33
CONTINUOUS EXERCISES
71 The inverse function f −1
(x) is defined by f −1
: x →
2
2
( 10)
5
x − , x  0. Find
(a) f (x), [1 mark]
(b) the domain of f (x). (Answer : x  −4) [1 mark]
Answer :
(a) (b)
72 A function h is defined by h : x → 2
a
x
− , where x  0, and a is a constant. Given
2 1
1
(2) ( 1) 1
2
h h−
+ − = − , find the possible values of a. (Ans : −8, 3)
[4 marks]
Answer :

34
73 The handbags produced by the manufacturer will be sent to retailer before selling them to customers.
During a sales promotion, the manufacturer offer a RM75 rebate from RMx to the retailer. While
retailer offers a 5% discount from RMx to the customers.
(a) Find the price that customer has to pay for a handbag in
term of x. [ Ans : g(x) = 0.95(x + 75) ] [3 marks]
(b) If a customer buys a new handbag at the price of RM499.90.
(i) find the price from the manufacturer, (Ans : 526.21) [2 marks]
(ii) calculate the profit gained by the retailer. (Ans : 48.69) [2 marks]
Answer :
74 It is given that x =
7 ( )
3 ( )
h x
h x
+
−
, where h(x)  0, find h−1
(2). (Ans : 9)
[3 marks]
Answer :
75 The function f and g are defined as f : x → x − 3 and g : x → x2
respectively. Find
(a) gf(x), .
(b) h(x) such that hgf(x) = x2
− 6x + 3. [ Ans : h(x) = x − 6 ]
[3 marks]
Answer :
(a) (b)

35
QUANDRATIC
EQUATIONS AND
QUADRATIC FUNCTIONS
- WORKSHEET
Encik AK Sapri bin Yaakob
Puan Tan Woon Shin

37
WORKSHEET
TOPIC 2 : QUADRATIC FUNCTIONS
==========================================================================================================================================
Revision
==========================================================================================================================================
 roots of a quadratic equation ~ values of x ( variable / unknown ) that satisfy
the equation
1 It is given that −1 is one of the roots of the quadraric equation x2
− 4x − p = 0. Find the value
of p. (Ans : 5)
[2 marks] [2008, No.4]
Answer :
2 It is given −7 is one of the roots of the quadratic equation (x + k)2
= 16, where k is a constant. Find
the values of k. (Ans : 3, 11)
[2 marks] [2015, No.5]
Answer :
3 If
2
3
− is one root of the equation 2x2
− 7x = 18 + px, where p is contants. Find
(a) the value of p, (Ans : 2)
(b) the other root of the equation. (Ans : 6)
Answer :
(a) (b)
MIND think :
• The roots of a quadratic equation :
→ the values of the variable / unknown ( x ) that a quadratic equation.

38
4 If p − 1 and q + 2 are the roots of the quadratic equation x2
+ 5x + 4 = 0. Find the values
of p and q. (Ans : p = −3, q = −3 and p = 0, q = −6)
Answer :
 problems involving quadratic equations
5 The diagram shows the front view of four pieces of wood with the same width. The total front area of
the four pieces of wood is 20 cm2
. The four pieces of wood are used to produce a rectangular photo
frame as shown in Diagram.
Calculate the width, in cm, of the wood. (Ans : 0.5 / 2
1 )
[3 marks] [2016, No.25]
Answer :
MIND think :
• factorization using calculator : CASIO fx-570MS, CANON F-570SG, OLYMPIA ES-
570MS . . .
13 cm
13 cm
8 cm
→ x1 = 3
1
mode
a ? 3 = b ? −5 = c ? −12 =
Example : 3x2
− 5x − 12 = 0 mode mode  2
= → x2 = −1.33 ... shift a b/c − 3
4
( x − 3 ) (3x + 4 ) = 0
NOTE : if appear x only → x1 = x2 (2 equal roots)
( x − root )2
= 0

39
6 The age of Gloria is five times of her son. Four years ago, the product of their ages is 52. Find their
ages now. (Ans : 30, 6)
Answer :
7 Annabella bought a certain number of pens for RM60. If each pen had a discount of 20 cents, she could
have bought 10 more pens for the same amount of money.
Find the number of pens that Annabella bought. (Ans : 50)
Answer :
8 During the training session for the Olympics, Kenney cycled 60 km from Kota Kinabalu to Kota Belut at a
constant speed. If he increases the speed by 10 kmh−1
, he will arrive Kota Belut half an hour earlier.
(a) Form a quadratic equation in terms of x, which shows the movement of
Kenney. (Ans : x2
+ 10x − 1200 = 0) [2 marks]
(b) Calculate the original speed of Kenney. (Ans : 30) [2 marks]
[Forecast]
Answer :

40
9 Boeoon wants to fence his vegetable garden using fencing wire. The area of his vegetable garden is
1787.5 m2
. The length of the garden is 10 m less than twice its width. The bulk price of a roll of 15 m
long fencing wire is RM150, while the retail price of 1 m is RM12.50. Determine the lowest cost, in
RM, needed to buy the fencing wire. (Ans : 1775)
Answer :
==========================================================================================================================================
2.1 Quadratic equations and inequalities
2.1.1 Solve quadratic equations using the method of completing the square and formula.
==========================================================================================================================================
 determine the roots of a quadratic equation ~ formula
10 Solve the quadratic equation 2x (x − 4) = (1 − x) (x + 2). Give your answer correct to four significant
figures. (Ans : −0.2573, 2.591)
[3 marks] [2003, No.3]
Answer :
11 Solve the quadratic equation x (2x − 5) = 2x − 1. Give your answer correct to three decimal
places. (Ans : 0.149, 3.351)
[3 marks] [2005, No.5]
Answer :
12 Eva thrown a ball from a with the height of ball, h m from the ground after t second is given by
h = −4.9t2
+ 18t + 1.5. Determine whether the ball can reach a height of 15 m from the ground. Give
reason for your answer. (Ans : 2.62, 1.05)
Answer :

41
MIND think :
• solve by using formula : x =
2
4
2
b b ac
a
−  −
Example : 3x2 − 5x − 12 = 0 → x =
2
( 5) ( 5) 4(3)( 12)
2(3)
− −  − − −
→ x =
5 169
6

→ x
= 3, 4
3
−
 determine the roots of a quadratic equation ~ completing the square
13 Solve the quadratic equation 2x2
− 6x = x (x + 3) − 4 by using completing the square. Give your answer
correct to three decimal places. (Ans : 8.531, 0.469)
Answer :
14 Solve the quadratic equation 2x (x − 3) = (x + 4)(1 − x) by using completing the square. Give your answer
correct to three significant figures. (Ans : −0.758, 1.76)
Answer :
MIND think :
completing the square
x2
+ bx + c = 0
2 2
2 2
( ) ( )
b b
x c
+ − + = 0
x2
− bx + c = 0
2 2
2 2
( ) ( )
b b
x c
− − − + = 0
ax2
+ bx + c = 0
2
[ ]
b
a
a x x c
+ + = 0
2 2
2 2
[ ( ) ( ) ]
b b
a a
a x c
+ − + = 0
2
2
2 4
( )
b b
a a
a x c
+ − + = 0

42
==========================================================================================================================================
2.1.2 Form quadratic equations from given roots.
==========================================================================================================================================
 form a quadratic equation from given roots ~ 1
15 Form the quadratic equation which has the roots −3 and
2
1
. Give your answer in the form ax2
+ bx +
c = 0, where a, b, and c are constants. (Ans : 2x2
+ 5x − 3 = 0)
[2 marks] [2004, No.4]
Answer :
16 Form the quadratic equation which has the root
2
1
. Give your answer in the form ax2
+ bx + c = 0,
where a, b, and c are constants. (Ans : 4x2
− 4x + 1 = 0)
[2 marks]
[Forecast]
Answer :
17 Given that one of root of a quadratic equation is reciprocal to the other. If 5 is one root of the equation,
form the quadratic equation. Give your answer in the form ax2
+ bx + c = 0, where a, b, and c are
constants. (Ans : 5x2
− 26x + 5 = 0)
[2 marks]
[Forecast]
Answer :
MIND think :
• A quadratic equation, ax2 + bx + c = 0, with roots,  and .
~ (x − ) (x − ) = 0
~ x2 − ( + ) x +  = 0 @ x2 − (SOR) x + POR = 0  SOR = sum of roots, POR =
product of roots
~ SOR =  +  = b
a
− , POR =  = c
a

43
18 It is given that the quadratic equation 3x2
+ 8x + 7 = 0 has roots  and . Form a quadratic equation
with roots 3 and 3. (Ans : x2
+ 8x + 21 = 0)
[3 marks] [2016, No.17]
Answer :
19 It is given that m and n are the roots of the quadratic equation 2x2
− 5x − 3 = 0. Form quadratic equation
with roots m − 1 and n − 1. (Ans : 2x2
− x − 6 = 0)
Answer :
20 It is given that p and q are the roots of the quadratic equation 2x2
= 4 − x. Form quadratic equation
with roots p2
and q2
. [ Note : a2
+ b2
= (a + b)2
− 2ab ] (Ans : 4x2
− 17x + 16 = 0)
Answer :
21 Given  and  are the roots of quadratic equation x2
+ 6x + 3 = 0. Find the value of :
(a)  + ,
(b)


 2
2
+
. (Ans : 10)
Answer :
(a)
(b)

44
22 The roots of the equation x2
+ 3x − 4 = 0 are  and , and the roots of the equation x2
+ 6x + p = 0
are

k
and

k
, find the values of k and p. (Ans : k = −8, p = −16)
[4 marks] [clon 2015, K2, 5(b)]
Answer :
23 It is given that 3 and m + 4 are the roots of the quadratic equation x2
+ (n − 1) x + 6 = 0, where m and
n are constant. Find the value of m and of n. (Ans : m = −2, n = −4)
[3 marks] [2012, No.4]
Answer :
24 Given the quadratic equation 2x2
+ mx − 5 = 0, where m is a constant, find the value of m if
(a) one of the roots of the equation is 2, (Ans : −
2
3 )
(b) the sum of roots of the equation is −4. (Ans : 8)
[4 marks] [2014, No.5]
Answer :
(a) (b)
25 It is given that the quadratic equation hx2
− 3x + k = 0, where h and k are constants, has roots  and
2. Express h in terms of k. (Ans :
k
2 )
[3 marks] [2018, No.21]
Answer :

45
26 One of the root of the equation 9
2
6 2
+
=






− x
x
n
x is one third the other. Find the value
of n. (Ans : 8
45
− )
Answer :
27 Given  and  are the roots of the equation (1− 2x)(x + 5) = k and  = 2. Find the value
of k. (Ans : 14)
Answer :
28 Given
2

and
2

are the roots of the equation kx (x − 1) = 2m − x. If 6
=
+ 
 and 3
=
 , find the
value of k and of m. (Ans : k = 2
1
− , m = 16
3 )
Answer :
==========================================================================================================================================
2.1.3 Solve quadratic inequalities.
==========================================================================================================================================
 quadratic inequalities
29 Find the range of values of x for which x (x − 4)  12. (Ans : −2  x  6)
[3 marks] [2004, No.5]
Answer :

46
30 Find the range of the values of x for (2x − 1)(x + 4) > 4 + x. (Ans : x < −4, x > 1)
[2 marks] [2006, No.5]
Answer :
MIND think :
31 Find the range of the values of x for (x − 3)2
< 5 − x. (Ans : 1 < x < 4 )
[3 marks] [2008, No.6]
Answer :
32 Find the range of value of x for 3x2
− 5x − 16  x (2x + 1). (Ans : −2  x  8)
[3 marks] [2011,
No.6]
Answer :
33 Given that f (x) = −3x2
+ 2x + 13, find the range of values of x for f (x)  5. (Ans : x 
3
4
− , x  2)
[3 marks] [2012, No.6]
y = f (x)
 
x  
x
y = f (x)
a > 0 a < 0
f (x) = ax2
+ bx + c → f (x) = 0 ~ roots = , 
f (x) < 0 →  < x < 
f (x) > 0 → x < , x > 
f (x)  0 →   x  
f (x)  0 → x  , x  

47
Answer :
34 Find the range of values of x such that the quadratic function f (x) = 6 + 5x − x2
is
negative. (Ans : x < −1, x > 6)
[3 marks] [2017, No.12]
Answer :
 problems involving quadratic inequalities
35 Firdaus has a rectangular plywood with dimension 3x metre in length and 2x metre in width. He cut
part of the plywood into square shape with sides of x metre to make a table surface. Find the range of
values of x if the remaining area of the plywood is at least (x2
+ 4) metre2
. (Ans : x  1)
[3 marks] [2018, No.19]
Answer :
==========================================================================================================================================
2.2 Types of roots of quadratic equations

48
2.2.1 Relate types of roots of quadratic equations to the discriminant value.
2.2.2 Solve problems involving types of roots of quadratic equations.
==========================================================================================================================================
 types of roots of the quadratic equation ~ 1
36 It is given that quadratic equation x (x − 5) = 4
(a) Express the equation in the form ax2
+ bx + c = 0.
(b) State the sum of roots of the equation.
(c) Determine the type of roots of the equation.
[3 marks] [2013, No.4]
Answer :
(a) (c)
(b)
37 In the answer space, match the type of roots for each of the given quadratic equation. [3 marks]
[Forecast]
Answer:
MIND think :
x2
3
4
− x +
9
4
= 0 2 different real roots
x (1 − 3x) = 2 2 equal real roots
3x2
= 2x + 7
no real roots /
imaginary roots

49
Types of Roots of Quadratic Equations → by determine the value of discriminant, b2 −
4ac
two different roots
@
intersects at 2 distinct
points

b2 − 4ac > 0
two equal roots @
tangent

b2 − 4ac = 0
no roots /
imaginary roots
@ does not
intersect

b2 − 4ac < 0
real roots

b2
− 4ac  0
38 The quadratic equation x (x + 1) = px − 4 is satisfied by two distinct values of x. Find the range of values
of p. (Ans : p < −3, p > 5)
3 marks] [2003, No.4]
Answer :
39 A quadratic equation x2
+ px + 9 = 2x has two equal roots. Find the possible values of p.
(Ans : −4, 8)
[4 marks] [2006, No.3]
Answer :
40 (a) Solve the following quadratic equation :
3x2
+ 5x − 2 = 0 (Ans : 3
1 , −2)
(b) The quadratic equation hx2
+ kx + 3 = 0, where h and k are constants, has two equal roots.
Express h in terms of k. (Ans : h =
12
2
k
)
[4 marks] [2007, No.4]
Answer :
(a) (b)

50
41 The quadraric equation x2
+ x = 2px − p2
, where p is a constant, has two different roots. Find the range
of values of p. (Ans : p < 4
1 )
[3 marks] [2009, No.4]
Answer :
42 The quadraric equation (1 − p) x2
− 6x + 10 = 0, where p is a constant, has two different roots. Find
the range of values of p. (Ans : p >
10
1 )
[3 marks] [2010, No.5]
Answer :
43 The quadratic equation mx2
+ (1 + 2m) x + m − 1 = 0 has two equal roots. Find the value of m.
(Ans :
8
1
− )
[3 marks] [2011, No.4]
Answer :
44 A quadratic equation x (x − 4) = h − 2k, where h and k are constants, has two equal roots. Express h
in terms of k. (Ans : h = 2k −4)
[3 marks] [2012, No.5]
Answer :
45 Gven the quadratic equation (1 − a) x2
− 2x + 5 = 0 has no roots, find the range of values
of a. (Ans : a <
5
4 )
[2 marks] [2014, No.3]
Answer :

51
46 (a) It is given that one of the roots of the quadratic equation x2
+ (p + 3) x − p2
= 0, where p is a
constant, is negative of the other. Find the value of the product of roots.
(Ans : −9) [2 marks]
(b) It is given that the quadratic equation mx2
− 5nx + 4m = 0, where m and n are constants, has
two equal roots. Find m : n. (Ans : 5 : 4) [2 marks]
[2017, No.13]
Answer :
(a) (b)
47 It is given that quadratic equation (px)2
+ 5qx + 4 = 0 has two equal roots while the quadratic
equation hx2
− x + p = 0 has no roots, where p, q and h are constants. Express the range of q in
terms of h. (Ans : q > 1
5h
, q < − 1
5h
)
[3 marks] [2019, No.9]
Answer :
48 The straight line y = 5x − 1 does not intersect the curve y = 2x2
+ x + p. Find the range of values
of p. (Ans : p > 1)
[4 marks] [2005, No.4]
Answer :
49 It is given that the curve y = (p − 2) x2
− x + 7, where p is a constant, intersects with the straight line
y = 3x + 5 at two points. Find the range of values of p. (Ans : p < 4)
[3 marks] [2018, No.20]
Answer :

52
50 Show that the straight line y = x + k intersects the curve x2
+ y2
− 5x = 4 at two distinct points if
4k2
+ 20k − 57 < 0
Answer :
51 Given that the straight line y = mx + m − 2 is the tangent to the curve y = x2
+ 3x + 1. Find the possible
values of m. (Ans : −1, 3)
Answer :
==========================================================================================================================================
2.3 Quadratic Functions
2.3.1 Analyse and make generalisation about the effects of changes of a, b and c and in
f (x) = ax2
+ bx + c towards the shape and position of the graph.
==========================================================================================================================================
52 The diagram shows the graphs, f (x) = ax2
+ bx + c. Match the following graphs with the possible values
of a.
a = −5 a = 3
a =
1
2
a = −1
x =

53
53 Match the following quadratic functions with the correct graph.
56 The diagram shows the graph for f (x) = −x2
+ 4x + 5, where a = −1, b = 4 and c = 5. Match the graph
of f (x) formed for the change of values in a, b and c.
f (x) = ax2
~ a > 0
f (x) = ax2
~ a < 0
f (x) = ax2
+ c ~ a > 0, c < 0
f (x) = ax2
+ c ~ a < 0, c > 0
f (x) = ax2
~ a < 0
f (x) = ax2
+ bx ~ a > 0, b > 0
f (x) = ax2
+ bx ~ a > 0, b < 0
f (x) = ax2
+ bx ~ a < 0, b > 0
f (x) = ax2
+ bx ~ a < 0, b < 0
f (x) = ax2
+ bx + c ~ a > 0, b > 0
f (x) = ax2
+ b + c ~ a > 0, b < 0
f (x) = ax2
+ bx + c ~ a < 0, b > 0
f (x) = ax2
+ bx + c ~ a < 0, b < 0
a → −4 a → −0.25 b → −4 c decrease by 2
x
5
f (x)
x
5
f (x)
f (x) = −x2
+ 4x + 5
x
5
f (x)
x
5
f (x)
3

54
==========================================================================================================================================
2.3.2 Relate the position of the graph of quadratic functions with type of roots.
==========================================================================================================================================
 relate the position of quadratic function graphs with types of roots for f(x) = 0 ~
1
57 Match the position of the graph of quadratic functions with the correct value of discriminant, b2
− 4ac.
58 The graph of a quadratic function f (x) = px2
− 2x − 3, where p is a constants, does not intersect the x-
axis. Find the range of values of p. (Ans : p <
3
1
− )
[3 marks] [2013, No.5]
Answer :
59 The graph of a quadratic function f (x) = px2
− 8x + q, where p and q are constants, has a maximum
point.
(a) Given p is an integer such that −2 < p < 2, state the value of p.
(b) Using the answer from (a), find the value of q when the graph touches the x-axis at one
point (Ans : −16)
[3 marks] [2015, No.3]
Answer :
(a) (b)
~ the graph interects x-axis
at two different points.
~ two real and different
roots.
~ the graph interects x-axis
at one point only. (x-axis
tangent of the graph)
~ two real and equal roots.
~ the graph does not intersects
the x-axis.
~ no real roots.
(imaginary roots)
x x x
f(x) always positive
f(x) always negative
b2
− 4ac = 0 b2
− 4ac < 0 b2
− 4ac > 0

55
60 Given the quadratic function f (x) = x2
+ 2wx + 3w − 2, where w is a constant, is always positive
(lies completely above the x-axis) when p < w < q. Find the value of p and of q.
(Ans : p = 1, q = 2)
[3 marks] [2016, No.18]
Answer :
61 Show that the graph of a quadratic function f (x) = p2
x2
+ 3px + 2 is always intersect the x-axis for all
values of p.
Answer :
 relate the position of quadratic function graphs with types of roots for f(x) = 0 ~
2
62 The diagram shows the graph of a quadratic function f (x) = n
p
x
+ qx + r such that p, q, r, n and u
are constants.
(a) State the value of n.
(b) If f (x) = 0 and the product of roots is r, state the value of
(i) q,
(ii) p. (Ans : 1)
[3 marks] [2019, No.2]
Answer :
(a) (b) (i)
(ii)
u
O
f (x)
x
−u

56
63 Given function f : x → x2
+ px + q. If f (x) > 0 for x < 2 and x > 3. Find the value of p
and of q. (Ans : p = −5, q = 6)
Answer :
64 Given that f (x) = 2x2
+ px + 16 and that f (x) is only negative when 2 < x < k. Find the values
of p and k. (Ans : p = −12, k = 4)
Answer :
65 The diagram show the graph of a quadratic function f(x) = x2
+ mx − x + 6, where n is a constant.
Find
(a) the roots of the quadratic equation x2
+ mx − x + 6 = 0. (Ans : 2, 3)
(b) the value of m. (Ans : −4)
Answer :
(a) (b)
f(x)
f(x) = x2
+ mx −x + 6
x
k k + 1
O

57
2.3.3 Relate the vertex form of quadratic functions, f (x) = a(x − h)2
+ k with other forms of
quadratic functions.
==========================================================================================================================================
66 Complete the following :
form of
quadratic functions
Vertex Form
f (x) = a (x − h)2
+ k

turning point / vertex
(minimum / maximum point)
~ equation of the axis of symmetry →
~ minimum / maximum value of f (x) →
~ tangent to the graph which parallel to x-axis →
reflected about x-axis
reflected about y-axis
General Form
f (x) = ax2
+ bx + c

axis of symmetry
(symmetrical about)

Intercept Form
f (x) = a (x − p) (x − q)

axis of symmetry
(symmetrical about)

A
C
B
D
A =
B =
C =
D =

max / min value of f(x)

max / min value of f(x)

58
 vertex form ( turning point - minimum @ maximum point ) ~ 1
67 The diagram shows the graph of a quadratic function f (x) = 3 (x + p)2
+ 2, where p is a constant.
The curve y = f (x) has a minimum point (1, q), where q is a constant. State
(a) the value of p,
(b) the value of q,
(c) the equation of the axis of symmetry.
[3 marks] [2005, No.6]
Answer :
(a) (b) (c)
68 The quadratic function f (x) = p (x + q)2
+ r, where p, q and r are constants, has a minimum value of
−4. The equation of the axis of symmetry is x = 3. State
(a) the range of values of p,
(b) the value of q,
(c) the value of r.
[3 marks] [2008, No.5]
Answer :
(a) (b) (c)
x
y
(1, q)
y = f (x)
O

59
69 The diagram shows the graph of quadratic function f (x) = − (x + p)2
+ q, where p and q are constants.
State
(a) the value of p,
(b) the equation of the axis of symmetry.
[2 marks] [2009, No.5]
Answer :
(a) (b)
70 The diagram shows the graph of quadratic function f (x) = (x + 3)2
+ 2k − 6, where k is a constant.
(a) State the equation of axis of symmetry of the curve.
(b) Given that the minimum value of the function is 4, find the value of k. (Ans :
5)
[3 marks] [2011, No.5]
Answer :
(a) (b)
x
(−3, 0)
O
y
(0, −9)
y = f (x)
f (x)
x
4
O
f (x) = (x + 3)2
+ 2k − 6

60
71 The diagram shows the graph of a quadratic function f (x) = −(x − 2) 2
+ 3k, where k is a constant.
Given (h, 12) is the maximum point of the graph,
(a) state the value of h and of k, (Ans : h = 2, k = 4)
(b) find the value of p. (Ans : 8)
[3 marks] [2013, No.6]
Answer :
(a) (b)
72 The graph of a quadratic function f (x) = 3 [ 2h − (x − 1)2
], where h is a constant, has maximum point
(1, h − 10).
(a) State the value of h, (Ans : −2)
(b) State the type of roots for f (x) = 0. Justify your answer. (Ans : no roots)
[3 marks] [2019, No.3]
Answer :
(a) (b)
O
p
(h, 12)
x
y

61
73 In the diagram, (3, 10) is the maximum point of a graph with an equation in the forrn of y = a (x + b)2
+
c.
Find
(a) the value of a, (Ans : a = − 3
2 )
(b) the equation of the curve when the graph is reflected about the y-axis .
Answer :
(a) (b)
 vertex form ( turning point - minimum @ maximum point ) ~ 2
74 The diagram shows the graph of the function y = − (x − k)2
− 2, where k is constant.
Find
(a) the value of k,
(b) the equation of the axis of symmetry,
(c) the coordinates of the maximum point.
[3 marks] [2004, No.6]
Answer :
(a) (b) (c)
O
(3, 10)
4
x
y
x
y
(2, −3)
O
−3

62
75 The diagram shows the graph of a quadratic function y = f (x). The straight line y = −4 is a tangent to
the curve y = f (x).
(a) Write the equation of the axis of symmetry of the curve.
(b) Express f (x) in the form of f (x) = (x + b)2
+ c, where b and c are constant.
[3 marks] [2006, No.4]
Answer :
(a) (b)
76 The diagram shows the graph of a quadratic function y = f (x) .
State
(a) the roots of the equation f (x) = 0,
(b) the equation of the axis of symmetry of the curve.
[3 marks] [2010, No.4]
Answer :
(a) (b)
y
x
y = −4
O 1 5
y = f (x)
y
x
5
−1
O
3

63
77 The diagram shows the graph of the quadratic function f (x) = (x − k)2
− 25.
State
(a) the value of k,
(b) the coordinates of the minimum point of the curve,
(c) the range of values of x when f (x) is negative.
[3 marks] [2014, No.4]
Answer :
(a) (b) (c)
78 The diagram shows the graph y = a (x − p)2
+ q, where a, p and q are constants. The straight line y =
−8 is a tangent to the curve at point H.
(a) State the coordinates of H.
(b) Find the value of a. (Ans :
2
1 )
(c) If the graph is reflected about the x-axis, write the equation of the curve. **
[3 marks] [2018, No.18]
Answer :
(a) (b) (c)
O
x
f (x)
8
−2
O
y
x
−1 7
H

64
79 The diagram show the graph of a quadratic function f (x) =  a (x + p)2
+ q , where a > 0.
(a) State the coordinates of point N.
(b) Find the function of the graph. (Ans : a = 3
2
)
Answer :
(a) (b)
 general form → vertex form
80 The quadratic function f (x) = x2
+ 2x − 4 can be expressed in the form f (x) = (x + m )2
− n, where m and
n are constants. Find the value of m and n. (Ans : m = 1, n = 5)
[3 marks] [2007, No.6]
Answer :
81 The quadratic function f ( x ) = −x2
+ 4x + a2
, where a is a constant, has maximum value 8. Find the
values of a. (Ans : 2)
[3 marks] [2009,
No.6]
Answer :
1 N
O
x
(3, 6)
y

65
82 The quadratic function f is defined by f (x) = x2
+ 4x + h, where h is a constant.
(a) Express f (x) in the form (x + m)2
+ n, where m and n are constants.
(b) Given the minimum value of f (x) is 8, find the value of h. (Ans : 12)
[4 marks] [2017, No.11]
Answer :
(a) (b)
83 Prove that y = x2
+ 2x + 7 is always positive for all values of x. Hence state the smallest value
of y. (Ans : 6)
Answer :
84 Find the smallest integer of p such that f (x) = x2
− 4x + p, where p is a constant, is always greater
than 5. (Ans : 10)
Answer :
 vertex form → general form → intercept form
85 The quadratic function f (x) =
2
1 4
3
3 3
x
 
− − +
 
 
can be expressed in the form f (x) = a (x − p) (x − q),
where a, p and q are constants and p > q. Find the value of p and q. (Ans : p = 1, q = 1
3
− )
Answer :

66
==========================================================================================================================================
2.3.4 Analyse and make generalisation about the effects of changes of a, h dan k in quadratic
functions f (x) = a(x − h)2
+ k towards the shape and position of the
graphs.
==========================================================================================================================================
86 The diagram shows the graph for f (x) = −2(x −3)2
+ 5, where a = −2, h = 3 and k = 5. Match the graph
of f (x) formed for the change of values in a, h and k.
87 A quadratic function f (x) = (x + 3)2
+ 2k, where k is a constant, has a minimum value of −6. Find
(a) the value of k,
(b) the equation of the axis of symmetry when the graph moves 2 units to the left,
(c) minimum value of f (x) when the graph moves 10 units upwards.
[4 marks] [Forecat]
Answer :
(a) (b)
(c)
a → −1 h increase by 3 k decrease by 3
a → −7
x
5
f (x)
f (x) = −2(x−3)2
+ 5
2
x
5
f (x)
x
5
f (x)
x
5
f (x)
3 6

67
==========================================================================================================================================
2.3.5 Sketch graphs of quadratic functions.
==========================================================================================================================================
88 The quadratic function f ( x ) = −x2
+ 4x − 3 can be expressed in the form of f (x ) = −(x − 2)2
+ k, where
k is a constant.
(a) Find the value of k. (Ans : 1)
(b) Sketch the graph of the function f (x) on the given axes.
[4 marks] [2010, No.6]
Answer :
(a)
(b)
89 Sketch the graph f (x) = −3x2
+ 12x − 16 for the domain −1  x  5. Hence, state the axis of symmetry
of the graph.
Answer :
y = f(x)
x
O

68
MIND think :
• steps to sketch the graph of a quadratic function
(i) find the maximum / minimum point (turning point) by completing the square
(ii) find the y-intersecpt → the value of c @ substitute x = 0, f (0)
(iii) find the x-intercept → subsitute y = 0
(iv) find the corresponding range, y for the given domain, x
90 Sketch the graph f (x) = 2
1
( 3)
2
x x
 
+ +
 
 
for the domain −4  x  2. Hence, find the quadratik function
that will be formed if the graph f (x) is reflected about the y-axis.
[Ans : f(x) = 2(x− 7
4
)2
− 25
8
]
Answer :
91 Sketch the graph of f (x) = 4 − (x − 1)2
 for the domain −2  x  5. Hence, find the range of f (x)
corresponding to the given domain. (Ans : 0  f(x)  12)
Answer :

69
==========================================================================================================================================
2.3.6 Solve problems involving quadratic functions.
==========================================================================================================================================
92 The diagram shows a bridge supported by parabolic construction.
Given that the equation of the construction is given by y = 2
1000
1
x
− + c metres. The distance between
the two ends of the curve over the bridge is 400 meters, and the height of the bridge from the ground is
6 meters. Find the maximum height of the construction. (Ans : 46)
Answer :
93 The number of food packets prepared by a food stall in a day is given by function f (x) = 3x2
− 24x + 2k −
1, where k is a constant and x is the number of workers in the stall. Find the smallest value of integer
k if the number of food packets provided must exceed 515 packets a day, and the number of workers at
that moment. (Ans : k = 283)
Answer :
94 The function h (t) = −4t2
+ 32t, represents the height of the fireworks, in meters, after t seconds
launched from point P, as shown in the diagram.
Point P is a origin and the fireworks eploded at the highest point. Find the height, in cm, the fireworks
explode. (Ans : 64)
Answer :
P
O


70
95 A football player kicks a ball 5m from the left side of the football middle field. The locus of the
ball is represented by f (x) = kx2
+
18
7
x + 3p, where k and p are constants. The ball reaches a maximum
height of 7 m and touches the surface of the field 60 m from the ball kicked. Find the value
of k and of p. (Ans : k =
900
7
− , p =
108
77 )
Answer :
PAPER 2
 Quadratic Equations ~ Part A → 7 – 8 marks
96 The quadratic equation x2
− 5x + 6 = 0 has roots h and k, where h > k.
(a) Find
(i) the value of h and of k, (Ans : h = 3, k =2)
(ii) the range of x if x2
− 5x + 6 > 0. (Ans : x < 2 , x > 3)
[5 marks]
(b) Using the values of h and of k from (a) (i), form the quadratic equation which has roots h + 2
and 3k − 2. (Ans : x2
− 9x + 20 = 0 ) [2 marks]
[2009, No.2]
Answer :
97 A quadratic equation x2
+ 4(3x + k) = 0, where k is a constant has roots p and 2p, p  0
(a) Find the value of p and of k. (Ans : p = −4 , k = 8 ) [5 marks]
(b) Hence, form the quadratic equation which has the roots p − 1 and p + 6.
(Ans : x2
+ 3x − 10 = 0) [3 marks]
[2012, No.2]
Answer :

71
98 It is given  and  are the roots of the quadratic equation x (x − 3) = 2h − 4, where h is a constant.
(a) Find the range of values of h if   . (Ans : h >
8
7 ) [3 marks]
(b) Given
2

and
2

are the roots of another quadratic equation 2x2
+ kx − 4 = 0, where k is a
constant, find the value of k and of h. (Ans : k = −3, h = 6) [4 marks]
[2015, No.5]
Answer :
99 The diagram shows a cylindrical container with the length of 20 cm placed on the floor against the wall.
Q is a point on the edge of the base of the container. It is given that the distance of point Q is 2 cm
from the wall and 1 cm from the floor.
Mira wants to keep the container in a box with a dimension of 21 cm  7 cm  7 cm. Determine
whether the container can be kept in that box or otherwise. Give a reason for your
answer. (Ans : cannot, diameter 10 > 7)
[6 marks][2017, No.6]
Answer :
Q

72
 Quadratic Functions ~ Part A → 6 – 8 marks
100 The function f (x) = x2
− 4kx + 5k2
+ 1 has a minimum value of r2
+ 2k, where r and k are constants.
(a) By using the method of completing the square, show that r = k − 1. [4 marks]
(b) Hence, or otherwise, find the values of k and r if the graph of the function is symmetrical about
x = r2
− 1. (Ans : k = 0, r = −1 & k = 4, r = 3) [4 marks]
[2003, No.2]
Answer :
101 The diagram shows the curve of a quadratic function f (x) = −x2
+ kx − 5. The curve has a maximum point
at B (2, p) and intersect the f (x) – axis at point A.
(a) State the coordinates of A. [1 mark]
(b) By using the method of completing the square, find the value of k and of p.
(Ans : k = 4, p = −1 ) [4 marks]
(c) Determine the range of values of x, if f (x)  −5. (Ans : 0  x  4 ) [2 marks]
[2008, No.2]
Answer :
x
A
O
f (x)
B (2, p)

73
102 The curve of a quadratic function f (x) = 2(x − h)2
+ 2k intersects the x-axis at points (1, 0) and (5, 0).
The straight line y = −8 touches the minimum point of the curve.
(a) Find the the value of h and of k. (Ans : h = 3, k = −4) [2 marks]
(b) Hence, sketch the graph of f (x) for 0  x  6. [3 marks]
(c) If the graph is reflected about the x-axis, write the equation of the curve. [1 mark]
[2016, No.2]
Answer :
 FORECAST
103 Cherryna bought some calculators for RM704.
(a) Cherryna sold 20 of the calculators at a profit of RM6 each. Write down an expression in terms of
x, for the selling price of each calculator. [1 mark]
(b) Cherryna sold the remaining calculators for RM30 each. Write down in terms of x, the total
amount of money she received for all the calculators. (Ans :
x
14080 + 30x − 480)
[2 marks]
(c) If Cherryna received RM920 altogether, shows that 3x2
− 140x + 1408 = 0. 2 marks]
(d) Find the number of calculators Cherryna bought. (Ans : 32) [2 marks]
Answer :

74
104 The diagram shows two types of tiles bought by Melvin, Riverwood and Riverstone, that are in square
shape.
The length of the Riverstone tile is 2 cm more than the length of the Riverwood tile. Given the total area
of the two tiles is 340 cm2
.
(a) Find the perimeter, in cm, of the Riverstone tile. (Ans : 56) [5 marks]
(b) What is the minimum number of Riverwood tiles needed to fill a square shape living room with a
perimeter of 484 cm ? (Ans : 102) [2 marks]
Answer :
RIVERWOOD
RIVERSTONE

75
105 A page with dimensions 18 cm  25 cm has a border of uniform width x cm surrounding the printed
part of the page, as shown in the diagram.
(a) Write a formula for the area L, in terms of x, of the printed part.
(Ans : L = 450 − 86x + 4x2
) [2 marks]
(b) State the domain dan range of L. (Ans : 0  x < 9, 0 < L  450) [2 marks]
(c) Find the margin (width of the border) that should be used to obtain an area of
348.75 cm2
. (Ans : 1.25) [3 marks]
Answer :
x cm
18 cm
25 cm
x cm
x cm x cm

76
106 The diagram shows the curve of the functions y = −x2
+ 2x + qx − 3 and y = −3 (x − 2)2
+ p that
intersect at two points at the x-axis.
Find
(a) the values of p and q. (Ans : p = 3, q = 2) [4 marks]
(b) the maximum point of each curve. [ Ans : (2, 3), (2, 1) ]
[3 marks]
Answer :
107 The quadratic fuction is defined by f (x) = 24x − 4x2
+ r, where r is a constant.
(a) Express the quadratic function in the form p(x − q)2
+ 16. Hence, find the values of p, q
and r. (Ans : p = −4, q = 3, r = −20) [4 marks]
(b) Find the turning point of the graph f (x). [ Ans : (3, 16) ] [1 mark]
(c) Hence, sketch the graph of the function f (x). [3 marks]
Answer :
y
x
y = −x2
+ 2x + qx −3
y = −3 (x − 2)2
+ p
O

77
108 A bookstore in Telipok town selling Mathematics work books has found out that when books
are sold at a price of RM q per book, the profit, RM U, as a function of the price, q is :
U (q) = 120q – 10q2
(a) Sketch the graph of U (q) = 120q – 10q2
.
[4 marks]
(b) Hence, determine the price of a book that should be established in order to maximise
profit, and state the maximum profit. (Ans
: 6, 360) [2 marks]
Answer :
109 Emily bought a 100 metres wire to fence her rectangular garden with x metres length, as
shown in the diagram.
(a) Show that the area, A = 50x – x2
. [3 marks]
(b) Express the area in the form f (x) = a (x + p)2
+ q. Hence, find the maximum area and the
corresponding value of x. (Ans : 625, 25) [3 marks]
Answer :
x m

78
110 The diagram shows Gregory throws shot put. The thrown form a quadratic function y = f (x)
where y is the height of the shot put and and x is it horizontal distance.
Given that f(x) = a(x − p)2
+ q, based on the information in the given diagram.
(a) Find the values of a, p and q. (Ans : 3, 3, 25
3
− ) [4
marks]
(b) (i) Find the height of the shot put during it released form Gregory's hand.
(Ans : 25
48 ) [2 marks]
(ii) Find the horizontal distance of the shot put when it is at the same height as
(b) (i) again.
(Ans : 6) [1 mark]
Answer :
O
y
x
3
3
8
•

79
111 Given that y = p + qx− x2
= k − (x + h)2
.
(a) Find the values of h and k in term of p and / or q.
(Ans : h = − 2
q
, k = 4
2
q
+ p) [3 marks]
(b) If q = 2, state the axis of symmetry of the curve.
(Ans : x = 1) [2 marks]
(c) Straight line y = 3 touches curve y = p + qx− x2
,
(i) state p in term of q, (Ans : p = 4
12 2
q
−
) [2 marks]
(ii) hence, sketch the graph of the curve. [3 marks]
Answer :

80
112 The roots of the quadratic equations 2
1
( 5)
x k x
k
+ − = − , where k is a constant, are of opposite
/ difference signs. Shows that 0 < k < 5.
Answer :
113 The function f : x → −x2
+ 6x − 5 is defined for x  k, where k is a constant. State the smallest
value of k for which f (x) is a one-to-one relation. (Ans
: 3)
Answer :
114 The graph of the function f (x) = ax2
+ bx + c has equation of axis of symmetry at x = 1 and
passes through the points (0, 1) and (−1, −5). Find the value of a, b and c. (Ans
: a = −2, b = 4, c = 1)
Answer :

81
SYSTEM OF EQUATIONS
- WORKSHEET
Puan Maya Insana Mohd Terang

83
WORKSHEET
TOPIC 3 : SYSTEMS OF EQUATIONS
[ Paper 2, Part A ~ 1 question → 5 – 8 marks ]
==========================================================================================================================================
3.1 Systems of linear equations in three variables
3.1.1 Describe systems of linear equations in three variables.
==========================================================================================================================================
1
Answer :
MIND think :
• The characteristics of systems of liner equation in three variable :
Mark if the equations are system of linear equations in three variable. If not, mark .
 
a ( 5 + 2b) = 1 − c2
2a + c = b
a − b + 5c − 7 = 0
k + 2m = 8
5m = n + 9
n − k + 7 = 0
3(p +
6
q
) = q + 4r
−r + 6q = 2
7q = 2p + 9r − 1
3x + 2(y − z) = 7
xy = 3y + 5z
4x − z + 3y = 5
→ each linear equation has variable.
→ the highest power of each variable is .
variable.

84
==========================================================================================================================================
3.1.2 Solve systems of linear equations in three variables.
==========================================================================================================================================
2 Using the GeoGebra software, determine the type of solution for the following systems of linear
equations in three variable.
Answer :
4x − 7y + 2z = 6
x − 2y = 4
2x − 3y + 2z = −2
x − y − 3z = −6
2x + y + z = 3
−x + 2y + 2z = 1
4x + 8y + 2z = 14
2x − 4y + z = −2
x − 2y + 0.5z = 1.5
one solution
infinite solutions
no solution
The planes intersect
along a straight line
The planes do not
intersect at any point
The planes intersect
at only one point

85
3 (a) Solve the system of linear equations using the elimination method :
x − 3y + z = 2
4x − 4y + z = 7
2x + y − 3z = −4
(Ans : x = 2, y = 1 and z = 3) [6 marks]
[Forecast]
Answer :
MIND think :
• steps in solution : using elimination method
(1) rearrange all the equations in the form, ax + by + cz = d.
(2) do the first elimination. → NOTE : same sign ( − ) , different sign ( + )
(3) do the second elimination. ~ obtained the value of the first variable.
(4) substitute the value of first variable into any equations in (2) to get the value of
second variable.
(5) substitute the values of first and second variable into any equations in (1) to get
the value of third variable.

86
(b) Solve the system of linear equations using the substitution method :
x − 3y + z = 2
4x − 4y + z = 7
2x + y − 3z = −4
(Ans : x = 2, y = 1 and z = 3) [6 marks]
[Forecast]
Answer :
MIND think :
• steps in solution : using substitution method
(1) choose one equation, express one of the variables as subject → example : x in
terms of y and z
(2) substitute it into the others two equations, and express both equations into
same subject. → example : y in terms of z
(3) substitute the equations in (2), to each others ~ obtained the value of the first
variable.
(4) substitute the value of first variable into any equations in (2) to get the value of
second variable.
(5) substitute the values of first and second variable into equation in (1) to get the
value of third variable.

87
(c) Solve the system of linear equations using Gauss elimination :
x − 3y + z = 2
4x − 4y + z = 7
2x + y − 3z = −4
(Ans : x = 2, y = 1 and z = 3) [6 marks]
[Forecast]
Answer :
steps in solution :
using Gauss elimination
ax by cz d
ex fy gz h
px qy rz s
+ + =
+ + =
+ + =

a b c d
e f g h
p q r s
 
 
 
 
 

1 X X X
X X X X
X X X X
 
 
 
 
 

1 X X X
0 X X X
0 X X X
 
 
 
 
 

1 X X X
0 1 X X
0 X X X
 
 
 
 
 

1 X X X
0 1 X X
0 0 X X
 
 
 
 
 
MIND think :

88
2x + 2y − z = 0
4y − z = 1
−x − 2y + z = 2
(Ans : x = 2, y = 5
2
and z = 9) [6 marks]
[Forecast]
Answer :
2x + 2y − z = 0
4y − z = 1
−x − 2y + z = 2
(Ans : x = 2, y = 5
2
[Forecast]
Answer :

89
2x + 2y − z = 0
4y − z = 1
−x − 2y + z = 2
(Ans : x = 2, y = 5
2
[Forecast]
Answer :
2x + y − 3z = 1
3x − y − 4z = 7
5x + 2y − 6z = 5
(Ans : x = 3, y = −2 and z = 1) [6 marks]
[Forecast]
Answer :

90
2x + y − 3z = 1
3x − y − 4z = 7
5x + 2y − 6z = 5
(Ans : x = 3, y = −2 and z = 1) [6 marks]
[Forecast]
Answer :
2x + y − 3z = 1
3x − y − 4z = 7
5x + 2y − 6z = 5
(Ans : x = 3, y = −2 and z = 1) [6 marks]
[Forecast]
Answer :

91
6 Solve the system of linear equations :
x − y + 2z = 3
−3x − 2y + z = −6
4x + z = 11
(Ans : x = 3, y = −2 and z = −1) [6 marks]
[Forecast]
Answer :
7 Solve the system of linear equations :
3x + y − 2z = −7
−x − 3y + 5z = 10
4x − 2y + 3z = 1
(Ans : x = −1, y = 2 and z = 3) [6 marks]
[Forecast]
Answer :

92
==========================================================================================================================================
3.1.3 Solve problems involving systems of linear equations in three variables.
==========================================================================================================================================
8 Three students went to a bookshop to buy rulers, markers and pen. The items bought and
amount spent are shown in the following table.
Students
Number of
rulers
Number of
markers
Number of
pens
Amount paid
(RM)
Lea 2 3 4 11
Melvin 4 3 1 10
Mclarence 1 2 4 8
Find the price, in RM, of a ruler, a marker an a pen.
(Ans : ruler = 0.40, marker = 2.60, pen = 0.60) [7 marks]
[Forecast]
Answer :
9 Product X, Y, Z are assembled from three components A, B, C according to different
proportions. Each product X consists of two components of A, four components of B, and
one component of C; each product of Y consists three components of A, three components
of B, and two components of C; each product of Z consists of four components of A, one
component of B, and four component of C. A total of 750 components of A, 1000
components of B, and 500 components of C are used.
Find the number of products of X, Y, and Z assembled. (Ans : X = 200, Y = 50, Z = 50)
[7 marks]
[Forecast]
Answer :

93
10 Two groups of workers have their drinks at a stall. The first group comprising ten workers
have five cups of tea, two cups of coffee and three glasses of fruit juice at a total cost of
RM11.80. The second group of six workers have three cups of tea, a cup of coffee and two
glasses of fruit juice at a total cost of RM7.10. The cost of a cup of tea and three glasses of
fruit juice is the same as the cost of four cups of coffee.
Find the cost, in RM, of each drink. (Ans : tea = 1, coffee = 1.30, fruit juice = 1.40)
[7 marks]
[Forecast]
Answer :
11 A factory assembles three types of toys Q, R and S. The total time taken to assemble one
unit of R and one unit of S exceeds the time taken to assemble two units of Q by 8
minutes. One unit of Q, two units of R and one unit of S take 31 minutes to assembled.
The time taken to assemble two units of Q, one unit of R and three units of S is 48 minutes.
(a) If x, y and z represent the time, in minutes, taken to assemble each unit of toys Q, R
and S respectively, write a system of linear equations to represent the above
information.
(b) Hence, find the time taken to assemble each type of toy.
(Ans : x = 5, y = 8, z = 10)
[7 marks]
[Forecast]
Answer :

94
12 A fruit stall sells three types of fruits, oranges, apples and pineapples. The monthly cost
being RM6850 for 2150 of fruits. The cost for an orange, an apple and a pineapples are
RM2, RM3 and RM4 respectively. The sale price of an orange, an apple and a pineapples
are RM3, RM4.50 and RM5.50 respectively.
If the fruit stall want makes a monthly profit of RM2975, find the minimum number of fruits
of each type which has to sell. (Ans : orange = 500, apple = 750, pineapple = 900)
[7 marks]
[Forecast]
Answer :
==========================================================================================================================================
3.2 Simultaneous equations involving one linear equation and one non-linear equation
3.2.1 Solve simultaneous equations involving one linear equation and one non-linear equation.
==========================================================================================================================================
13 (a) Solve the following simultaneous equations using the substitution method :
x − 3y + 4 = 0 , x2
+ xy − 40 = 0.
(Ans : x = −6, y = − 3
2 and x = 5, y = 3) [5 marks]
[2008, No.1]
Answer :
NOTE :
• (a + b)2 = a2 + 2ab +
b2
• (a − b)2 = a2 − 2ab +
b2

95
MIND think :
• steps in solving simultaneous equations : using substituition method
(1) From the linear equation, an unknown is expressed in terms of the other
unknown.
(2) Suistituted (1) into the non-linear equation to form a quadratic equation.
(3) Arranged the quadratic equation in general form : ax2 + bx + c = 0
(4) Simplify and solve the quadratic equation by using :
~ factorization / calculator - CASIO fx-570MS, CANON F-570SG, OLYMPIA ES-570MS . . .
→ mode ; mode ; mode ; 1 ;  ; 2 ;
→ key in value a = ; key in value b = ; key in value c = ;
→ x1 , x2 (if x only → x1 = x2 )
(5) Obtain the values of other unknows by substituting the x1 and x2 into (1).
(b) Solve the following simultaneous equations using the elimination method :
x − 3y + 4 = 0 , x2
+ xy − 40 = 0.
(Ans : x = −6, y = − 3
2 and x = 5, y = 3) [5 marks]
[2008, No.1]
Answer :
MIND think :
• steps in solving simultaneous equations : using elimination method
(1) Arranged both equation in general form.
(2) Form quadratic equation :
(i) in terms of x → eliminate y
(ii) in terms of y → eliminate x
~ factorization / calculator - CASIO fx-570MS, CANON F-570SG, OLYMPIA ES-570MS . . .
→ mode ; mode ; mode ; 1 ;  ; 2 ;
→ key in value a = ; key in value b = ; key in value c = ;
→ x1 , x2 (if x only → x1 = x2 )
(4) Obtain the values of other unknows by substituting the x1 and x2 into linear
equation.

96
(c) Solve the following simultaneous equations using the graphical representation method :
x − 3y + 4 = 0 , x2
+ xy − 40 = 0.
(Ans : x = −6, y = − 3
2 and x = 5, y = 3) [5 marks]
[2008, No.1]
Answer :
x −8 −6 −4 −2 0 2 4 6

97
14 (a) Solve the following simultaneous equations using the substitution method :
x − 4y = 9 , 3y2
= 7 −
2
x
.
Give your answers correct to two decimal places.
(Ans : x = 11.56, y = 0.64 and x = 3.76, y = −1.31)
[5 marks][Forecast]
Answer :
MIND think :
• steps in solving simultaneous equations : using substituition method
(1) From the linear equation, an unknown is expressed in terms of the other
unknown.
(2) Suistituted (1) into the non-linear equation to form a quadratic equation.
(3) Arranged the quadratic equation in general form : ax2 + bx + c = 0.
~ formula → x =
2
4
2
b b ac
a
−  −
(5) Obtain the values of other unknows by substituting the x1 and x2 into (1).

98
(b) Solve the following simultaneous equations using the elimination method :
x − 4y = 9 , 3y2
= 7 −
2
x
.
(Ans : x = 11.56, y = 0.64 and x = 3.76, y = −1.31) [5 marks]
[Forecast]
Answer :
MIND think :
• steps in solving simultaneous equations : using elimination method
(1) Arranged both equation in general form.
(2) Form quadratic equation :
(i) in terms of x → eliminate y
(ii) in terms of y → eliminate x
~ formula → x =
2
4
2
b b ac
a
−  −
(4) Obtain the values of other unknows by substituting the x1 and x2 into linear
equation.

99
(c) Solve the following simultaneous equations using the graphical representation method :
x − 4y = 9 , 3y2
= 7 −
2
x
Give your answers correct to ( one / two ) decimal places.
(Ans : x = 11.56, y = 0.64 and x = 3.76, y = −1.31) [5 marks]
[Forecast]
Answer :
y −1.5 −1.2 −0.8 −0.4 0 0.4 0.8 1.5

100
 Part A ~ 1 ( using factorisation / calculator )
15 Solve the simultaneous equations 4x + y = −8 and x2
+ x − y = 2.
(Ans : x = −3, y = 4 and x = −2, y = 0) [5 marks]
[2003, No.1]
Answer :
16 Solve the simultaneous equations x +
2
1
y = 1 and y2
− 10 = 2 x.
(Ans : x = 3, y = −4 and x = − 2
1 , y = 3) [5 marks]
[2005, No.1]
Answer :
17 Solve the following simultaneous equations :
2x − y − 3 = 0 , 2x2
− 10x + y + 9 = 0.
(Ans : x = 1, y = −1 and x = 3, y = 3) [5 marks]
[2007, No.1]
Answer :

101
18 Solve the simultaneous equations :
3x + y = 1 , 5x2
+ y2
+ 4xy − 5 = 0
(Ans : x = −1, y = 4 and x = 2, y = −5) [5 marks]
[2012, No.1]
Answer :
 forecast
19 Solve the simultaneous equations 4x + y = x2
+ x − y = −3.
(Ans : x = −3, y = 9 and x = −2, y = 5) [5 marks]
Answer :
20 Solve the simultaneous equations 0
2
3
4
=
+
−
y
x
and 0
3
4
3
4
=
−
+
y
x
.
(Ans : x = −6, y = 2
3 and x = 4, y = 9) [5 marks]
Answer :

102
21 Given that (2k, 3p) is a solution of the simultaneous equations x − 2y = 8 and
2
1
2
3
2
=
+
y
x
.
Find the value of k and the value of p.
(Ans : k = 1, p = −1 and k = 8, p = 3
4 ) [5 marks]
Answer :
 Part A ~ 2
22 Solve the simultaneous equations p − m = 2 and p2
+ 2m = 8.
Give your answers correct to three decimal places.
(Ans : m = 0.606, p = 2.606 and m = −6.606, p = −4.606) [5 marks]
[2004, No.1]
Answer :

103
23 Solve the simultaneous equations 2x + y = 1 and 2x2
+ y2
+ xy = 5.
Give your answer correct to three decimal places.
(Ans : x = 1.443 , y = −1.886 and x = −0.693 , y = 2.386) [5 marks]
[2006, No.1]
Answer :
24 Solve the simultaneous equations k − 3p = −1 and p + pk − 2k = 0.
(Ans : k = 3.731, p = 1.577 and k = 0.269, p = 0.423). [5 marks]
[2009, No.1]
Answer :
25 Solve the simultaneous equations x − 2y = 7 and xy − x = 9y.
(Ans : x = 4.76, y = −1.12 and x = 13.24 , y = 3.12) [5 marks]
[2010, No.1]
Answer :

104
26 Solve the simultaneous equations y − 2x + 1 = 0 and 4x2
+ 3y2
− 2xy = 7.
(Ans : x = 1.129 , y = 1.258 and x = −0.295, y =−1.590) [5 marks]
[2011, No.1]
Answer :
27 Solve the simultaneous equation x + 2y = 1 and
3 2
5
x y
− = . Give your answer correct to three
decimal places.
(Ans : x = 0.284, y = 0.358 dan x = 2.116, y = −0.558) [5 marks]
[2019, No.1]
Answer :
==========================================================================================================================================
3.1.2 Solve problems involving simultaneous equations ; one linear equation and one non-linear
equation.
==========================================================================================================================================
 problems solving
28 Adam planted vegetables on a piece of land. The shape of the land is a right angled triangle.
Given the longest side of the land is y metre. The other two sides of the land are x metre
and (2x − 1) metre respectively. He fenced the land with 40 metre of barbed wire.
Find the length, in metre, of each side of the land. (Ans : 8, 15, 17) [7 marks]
[2016, No.3]
Answer :

105
29 The diagram shows the plan of a rectangular garden PQRS. The garden consists of a
semicircular pond PTS and grassy area PQRST.
It is given that SR = 6y metre and QR = 7x metre, x  y. The area of the rectangular
garden PQRS is 168 metre2
and the perimeter of the grassy area is 60 metre. The pond with
uniform depth contains 15.4 metre3
of water. By using  =
7
22
, find the depth,
in metre, of water in the pond. (Ans : 0.45). [7 marks]
[2018, No.4]
Answer :
 forecast
30 The sum of two numbers is 9 and the sum of the squares of the numbers is 53. Find the
product of the numbers.
(Ans : 14)
[6 marks]
Answer :
R
S
T
P Q

106
31 Fernandez bought x number of chickens and y number of ducks for RM208. Given that the
total number of chickens and ducks are 20, and the price for each chicken and duck are RMx
and RMy respectively, where y > x. Find the value of x and the value of y.
(Ans : x = 8, y = 12)
[6 marks]
Answer :
32 A piece of wire, 52 cm in length is cut into two different lengths. Each part of the wire is bent
to form a rectangle as shown in the diagram.
If the sum of the area for the both squares is 89 cm 2
, find the values of x and of y.
(Ans : x = 5, y = 8)
[6 marks]
Answer :
y cm
x cm

107
33 Given that the different of the circumferences of the two circles is 4 cm and the sum of their
areas is 52 cm2
. Find the radius of each circle. (Ans : 4, 6)
[6 marks]
Answer :
34 A piece of wire in shape of a circle with radius 14 cm is bent to form a rectangle with sides (2y
+ 20) cm long (x + 10) cm wide. Given that the area of rectangle is 420 cm2
, find the values
of x and the value of y. (Ans : x =4, y = 5 and x = 20, y = −3)
[6 marks]
Answer :
35 The perimeter of a rectangle is 36 cm and the square of its diagonal is 170 cm2
. Find the
length and width of the rectangle.
(Ans : 7, 11)
[6 marks]
Answer :

108
36 The diagram shows a rectangular plank.
A worker wants to cut the plank into two triangular planks. The perimeter of each triangular
plank is 24 cm and the measurement of the longest side of the triangle is (x + y) cm.
Calculate the area, in cm2
, of the plank.
(Ans : 48)
[6 marks]
Answer :
37 The diagram shows a mirror in the shape of a rectangle placed on a table in the shape of a
circle.
If the perimeter of the rectangle and the diameter of the circle are 44 cm and 340 cm
respectively, find the length and width of the rectangle.
(Ans : length = 18, width = 4)
[6 marks]
Answer :
3x cm
y cm

109
38 A piece of wire, 32 cm in length is bent to form a trapezium PQRS as shown in the diagram,
where PQR = SRQ = 90, PQ = y cm, QR = 2x cm, RS = 12 cm and PS =
10 cm .
Find the value of x and of y. (Ans : x = 3, y = 4)
[6 marks]
Answer :
39 Hafizie has a rectangle plot of land. He planted paddy and yam in the area such as the diagram
shown below.
The plantation of yam is in the shape of rectangle. Given that the area of land planted with
paddy is 115 m2
and the perimeter of land planted with yam is 24 m. Find the area of land
planted with yam. (Ans : 35 m2
)
[6 marks]
Answer :
y cm
2x cm
10 cm
12 cm
P
S
R
Q
x m
Yam
Paddy
y m
5 m
15 m

110
40 The diagram shows a prism with a uniform cross section in the shape of right-angled triangle.
Given that the height of the prism is 2x cm. If the total length of the its sides and the total
surface area of the prism are 42 cm and 84 cm2
respectively, find
(a) the values of x and y, where both are not whole numbers,
(Ans : x = 5
4
2 , y = 5
4
4 )
(b) the volume of the prism. (Ans : 125
79
37 )
[7 marks]
Answer :
41 A closed rectangular box has a square base. Given that the total length of its sides is 76 cm and
the total surfaces area of the box is 238 cm2
. Find the length of the base and the height of the
box.
(Ans : x = 3
17 , y = 3
23 and x = 7, y = 5) 6 marks]
Answer :
x cm
y cm
5 cm
2x cm

111
INDICES, SURDS AND
LOGARITHMS
- WORKSHEET
Encik Elbenjoe Wesmin
Puan Nadzrinah binti Ahmad

113
WORKSHEET
TOPIC 4 : INDICES, SURDS AND LOGARITHMS
==========================================================================================================================================
Revision → rewrite ; simplify algebraic expressions using laws of indices ; evaluate
==========================================================================================================================================
 rewrite
1 The following information is regarding a law of indices.
State the value of p and of q.
[2 marks] [2019, No.2]
Answer :
2
Express n in terms of p and q. (Ans : n = 2p + 4q)
Answer :
MIND think :
LAWS OF INDICES
m n m n
a a a +
 =
m
m n m n
n
a
a a a
a
−
 = = ( ) ( )
n m
m m n n
a a a

= =
0
1
a = , where a  0 ( )
n n n
a b ab
 =
n
n
n
a a
b
b
 
=  
 
n n
a b
b a
−
   
=
   
   
1 n
n
a
a
−
= @
1 n
n
a
a−
= n
a = n
a
1
 a = 2
1
a & 3
a = 3
1
a
1 m n
m n
a
a
− +
−
= @ m n
m n
k
ka
a
−
− +
= ( )
1 m n
m n
a
a a
− +
=

@
( )
1 m n
m n
a
a a
− −
=

( )
1
m
n
m m
n n
a a a
= = @ ( )
1
( )
m
m
n
m
n n
a a a
= =
8
( )
q
a = a  a  a  . . .  a , where p and q are constants.
p times
It is given that ( )
( ... ) ( ... )
n
a a a a a a a
      = , where p and q are constants.
(p + q) times q times
... n
a a a a
   =
n times

114
3 (a) Given ( ) 5
3
32
−
=
m n
32
1
, find the value of m and of n. [2 marks]
(b) Given
4
3
5
1 =
n m
5 , find the value of m − n. (Ans : −7) [2 marks]
[Mate SPM, J2010, P1, No.24 / clon Mate SPM, J2016, P1, No.20]
Answer :
(a) (b)
 simplify 1
4 Simplify :
(a) (pa
)b + c
 (pb
)a − c
 (pc
)a + b)
(Ans : 1) [2 marks] [clon UEC, 2013, P1, No.18]
(b) x( a + b) (a − b)
 x(b + c) (b − c)
 x( c − a) (c + a)
(Ans : 1) [2 marks] [UEC, 2004, P1, No.18]
Answer :
(a) (b)
5 (a) Simplify ( )( )
3 3
2 2
3 3 3
x y x xy y
− + + (Ans : x − y) [3 marks] [UEC, 1990, P2, No.4(b)]
(b) Given a2x
= 2, find the value of (ax
+ a−x
)2
. (Ans :
2
9 ) [3 marks] [clon UEC, 2002, P1, No.17]
Answer :
(a) (b)

115
 simplify 2
6 Simplify :
y
x
y
x
5
2
3
4
9
)
6
(
(Ans : 4x3
y5
)
[2 marks] [2014, No.6]
Answer :
7 Simplify :
(a)
( )
3
4
5 2
5
1
2
m m
m

(Ans : 10
17
20
m ) [2 marks] [clon Mate SPM, 2012, No.24]
i.
3
1
10 8 2
4 16 24
(4 )
r s
r s
 
 

 
 
 
(Ans :
18
8
6
r
s
) [2 marks] [clon Mate SPM, 2017, No.24]
Answer :
(a) (b)
 simplify 3
8 Given
k
ah
k
h
k
h x
7
2
4
10
27 
= 3h5
k y
. Calculate the value of a + x + y. (Ans : 16)
Answer :

116
9 Given
2
6
3
2
3
2


y
x
= 18. Find the value of x + y. (Ans : 9)
Answer :
 evaluate
10 Without using a calculator, find the value of 2
3
12  2−1
 27 . (Ans : 4)
Answer :
MIND think :
TABLE FOR NUMBERS POWER OF n
BASE 2 2−3 =
8
1
2−2 =
4
1
2−1 =
2
1
20 = 1 21 = 2 22 = 4 23 = 8 24 = 16 25 =
32
BASE 3 3−3 = 8
1 3−2 =
9
1
3−1 =
3
1
30 = 1 31 = 3 32 = 9 33 = 27 34 = 81 35 =
243
BASE 5 5−3 = 125
1 5−2 =
25
1
5−1 =
5
1
50 = 1 51 = 5 52 = 25 53 =
125
54 =
625
BASE 6 6−3 =
216
1
6−2 =
36
1
6−1 =
6
1
60 = 1 61 = 6 62 = 36 63 =
216
26 =
64
BASE 7 7−3 = 343
1 7−2 = 49
1 7−1 =
7
1
70 = 1 71 = 7 72 = 49 73 =
343
27 =
128
BASE 10 10−3
=
1000
1
10−2 =100
1 10−1 = 10
1
100 =
1
101 =
10
102 =
100
103
= 1000 28 =
256
2 = 2
1
4 3 = 2
1
9 5 = 2
1
25 7 = 2
1
49 2 = 3
1
8 3 = 3
1
27
==========================================================================================================================================

117
4.1 Laws of Indices
4.1.1 Simplify algebraic expressions involving indices using the laws of indices.
==========================================================================================================================================
 simplify 1
11 Simplify : 9n + 2
 [31 + n
 271 − n
] (Ans : 34n
)
Answer :
12 Simplify :
2 6
2 1
16 2
4
n
n
−
+

(Ans : 2−4 − 2n
)
Answer :
13 Simplify :
2 1 6 2
2 5
49 14
56
n n
n
+ −
−

(Ans : 713
 221 − 8n
)
Answer :

118
 simplify 2
14 Given that 3n + 2
− 3n
+ 10(3n − 1
) = h (3n
), where h is a constant. Find the value of h. (Ans : 3
34 )
Answer :
15 Show that 9 is a factor of 4n + 1
+ 4n
− 2 (4n − 1
). [Ans :
2
9 (4n
) ]
Answer :
16 Show that (35
+ 13) 53n
+ 28 ( n + 1)
is a multiple of 8 for all positive integers of n. [Ans : 256(53n
+ 28n
)]
Show that (35
+ 13) 53n
+ 28 ( n + 1)
is divisible by 8 for all positive integers of n.
Answer :

119
==========================================================================================================================================
4.1.2 Solve problems involving indices.
==========================================================================================================================================
 solve equations involving indices 1 ~ with same base 1
17 Solve the equation 32 4x
= 4 8x + 6
. (Ans : 3)
[3 marks] [2004, No.7]
Answer :
18 Solve the equation 82x − 3
=
2
4
1
+
x
. (Ans: 1)
[3 marks] [2006, No.6]
Answer :
19 Given that 9 (3n − 1
) = 27n
, find the value of n. (Ans : 2
1 )
[3 marks] [2007, No.8]
Answer :
20 Solve the equation :
16 2x − 3
− 84x
= 0 (Ans : −3)
[3 marks] [clon 2008, No.7]
Answer :

120
21 Given 3n − 3
 27n
= 243, find the value of n. (Ans : 2 )
[3 marks] [2009, No.7]
Answer :
27 (32x + 4
) = 1 (Ans :
2
7
− )
[3 marks] [2012, No.7]
Answer :
23 Given
1
3
125
25
−
+
p
h
= 1, express p in term of h. (Ans : p = 3
9
2 +
h
)
[3 marks] [2017, No.5]
Answer :
24 Solve the equation 2
16 +
x
= 1
1
4 2
x x+

. (Ans : −1)
Answer :

121
25 Given 2x − 3
=
5
3
(0.125)
y
, express y in term of x. (Ans :
3
5
x
y
−
= )
[3 marks] [clon UEC, 1984, P2, No.2(b)]
Answer :
26 Solve the equation 2
(3 ) 27 2187
x x x
 = . (Ans : −7, 1)
Answer :
27 Solve the equation x
9 . x
2
2 = 216 (Ans :
2
3 )
Answer :
2 . x
5 = 0.01 ( )
4
1
10x +
( Ans : −
3
2 )
Answer :

122
29 Solve the equation 2x + 4
− 2x + 3
= 1. (Ans : −3)
[3 marks] [2005, No.7]
Answer :
30 Solve the equation 3x +2
− 3x
=
9
8
. (Ans : −2)
[3 marks] [2010, No.7]
Answer :
31 Solve the equation : 23x
= 8 + 23x −1
(Ans :
3
4 )
[4 marks] [2011, No.7]
Answer :
32 Given p
2 + p
2 = k
2 , express p in term of k. (Ans : p = k −1)
[2 marks] [2018, No.16]
Answer :

123
 solve equations involving indices 4
33 Given 32x
= k, 3y
= h and 3y + 2x
= 7 + 9x
. Express k in term of h. (Ans : k = 1
7
−
h
)
[3 marks] [2015, No.7]
Answer :
34 Given 52x
= m, 5y
= n and 2
5(25 ) 1 5
x x y
−
− = . Express m in term of n. (Ans : m = 5 1
n
n −
)
[3 marks] [clon 2015, No.7]
Answer :
 solve equations involving indices 5 ~ solve by “elimination” or “substituition”
35 Solve the simultaneous equations : 3x
× 9y − 1
= 243 and y
x
4
23
= 32 (Ans : x = 3, y = 2)
Answer :
36 Solve the simultaneous equations : 3x
× 92y
= 1 and 55x
× 25y
=
25
1
(Ans : x = − 9
4 , y = 9
1 )
Answer :

124
 solve equations involving indices 6
37 Given 3(4p
) = 5(2q
) and 9(8p
) = 10(4q
), show that 2p + 1
= 5.
Answer :
38 Given 3p
= 5q
= 15r
, express r in terms of p and q. (Ans : q
p
pq
r +
= )
[3 marks] [2016, No.15]
Answer :
39 Given 2a
= 5b
= 10c
, , express a in terms of b and c. (Ans : bc
b c
a −
= )
Answer :

125
 solve equations involving indices 7 ~ let ax = y, form a quadratic equation
40 Solve the equation : 3 × 9x
= 2 × 3x
+ 1. (Jwp : 0)
Answer :
41 Solve the equation : 23 + 2x
+ 4 = 33 × 2x
. (Ans : −3, 2)
Answer :
 solve daily problems
42 The number of chairs produced by a furniture factory in January 1990 was 5000 units. It is estimated
that the production of chairs will increases with a rate of 20% each year. After t year, the number of
chairs that produced is given by P(1 + k)t
, where P is the number of chairs produced on January 1990
and k is the yearly rate of incensement in production of chairs.
(a) State the value of P and of k. [1 mark]
(b) Find the number of chairs that produced in January 2002. State your answer in the nearest
integer. (Ans : 44581) [2 marks] [Forecast]
Answer :
(a)
(b)

126
43 The population of a country increases according to the function, P = 2400000 e0.03t
where t is
the number of years after 1990 and e = 2.718.
(a) What is the population of the country in 1990 ?
(b) What is the population of the country in 2010 ? (Ans : 4372813)
[2 marks]
[Forecast]
Answer :
(a)
(b)
44 The temperature of a liquid decreases from 100°C to x°C is according to the equation x =
( )
100 0.98
t
when the liquid is cooled for t seconds.
(a) State the percentage of change in temperature for each second.
[1 mark]
(b) Find the difference in liquid temperature between fifth and sixth second, in nearest
Celsius degree. (Ans : 2) [2 marks]
Answer :
(a)
(b)
45 The half life of a radioactive sample is the time taken for its radioactivity to become half its initial
value. The half life of the sample is 50 minutes. Its radioactivity, R, counts per minute,
is given by R = 1024
1
2
1
+






n
, where n is the number of half life experienced
by the sample. The radioactivity of the sample reduces to 4n
counts per minute at t minutes.
Find the value of t. (Ans : 150)
[4 marks]
[Forecast]
Answer :

127
==========================================================================================================================================
4.2 Laws of surds
4.2.1 Compare rational numbers and irrational numbers, and hence relate surds to irrational
numbers.
==========================================================================================================================================
 conversion of recurring decimal to fractional form
46 Express the recurring decimal 0.969696 ………. as a fraction in its simplest form.
(Ans : 33
32 )
[4 marks] [2004, No.12]
Answer :
47 Given is a recurring decimal where h and k are postive integers. Find the
value of h + k.
(Ans : 7)
Answer :
48 Given p = is a recurring decimal. Express p as a fraction in its simplest form.
(Ans :
33300
35117 )
Answer :
k
h
= 0.16 . . .
1.05456

128
 relate surds to irrational numbers ~ a non-recurring decimal
49 Mark (  ) if the term is surd, mark (  ) if the term is not a surd.
(a) 8 (b) 3
120 (c) 4
4096 (d) 6 64
512
(e) 7 97
798
MIND think :
• surds → numbers with radicals ( , 3
, . . . , n
) , have infinite decimal places and are non-recurring
(irrational numbers ~ cannot converted to fractions)
==========================================================================================================================================
4.2.2 Make and verify conjectures on
(i) a b
 (ii) a b

and hence make generalisation.
==========================================================================================================================================
50 State the following value of surd in five decimal places :
3  5 3 5

51 State the following value of surd in five decimal places :
14  2 14 2

 Generalisation ~ for a > 0, b > 0 → a b
 =
 Generalisation ~ for a > 0, b > 0 → a b
 =

129
==========================================================================================================================================
4.2.3 Simplify expressions involving surds.
==========================================================================================================================================
 surd in the form of a b , where a and b are integers, and a is a largest value
52 Solve the following without using a scientific calculator :
(a) If 2 = 1.41, find the value of 32 . (Ans : 5.64) [2 marks] [UEC, 1989, P1, No.12]
(b) If 60 = 7.7, find the value of 540 . (Ans : 23.1) [2 marks] [UEC, 1997, P1, No.12]
Answer :
(a)
(b)
 addition and subtraction
53 Simplify the following expressions :
(a) 12 3
− (Ans : 3 ) [2 marks] [UEC, 1982, P1, No.1]
(b) 32 50
− (Ans : 2
− ) [2 marks] [UEC, 1992, P1, No.4]
Answer :
(a) (b)
(a) 5 24 4 54 6
+ + (Ans : 23 6 ) [2 marks] [UEC, 1990, P1, No.11]
(b) 10 90 40
− − (Ans : 4 10
− ) [2 marks] [UEC, 1994, P1, No.15]
Answer :
(a) (b)

130
(a) 3 12 2 3 48
+ − (Ans : 4 3 ) [2 marks] [UEC, 1988, P1, No.24]
(b) 45 125 2 20
+ − (Ans : 4 5 ) [2 marks] [UEC, 2007, P1, No.39]
Answer :
(a) (b)
56 Simplify : 2 18 3 8 3 32 50
+ + − (Ans : 19 2 ) [2 marks] [UEC, 1996, P2, No.3(a)]
Answer :
(a)
54 6 24
3 2 3
+ − (Ans :
5 6
6 ) [2 marks] [UEC, 1999, P2, No.4(a)]
(b)
3
48 27 243
2
− + (Ans :
17 3
2 ) [2 marks] [UEC, 2008, P1, No.10]
Answer :
(a) (b)

131
58 (a) Simplify :
40
2 50 90
2
+ − (Ans : 10 2 2 10
+ ) [2 marks]
(b) Given that 2 = 1.414, 3 = 1.732, find 48 6 3
−  . (Ans : 2.686) [2 marks]
[UEC, 2009, P2, No.5(b) / UEC, 2014, P1, No.11]
Answer :
(a) (b)
 multiplication
59 Without using a scientific calculator, find
(a) ( )
2
2 3 2
− . (Ans : 14 4 6
− ) [2 marks] [UEC, 1986, P1, No.15]
(b)
2
1
1
2
 
−
 
 
. (Ans : 3
2
2
− ) [2 marks] [UEC, 2010, P1, No.12]
Answer :
(a) (b)
MIND think :
2 2 2
( ) 2
a b a b ab
+ = + + 2 2 2
( ) 2
a b a b ab
− = + − 2 2
( )( )
a b a b a b
+ − = − a a a
 =
60 Without using a scientific calculator, find
(a) ( )( )
5 2 3 5 2 3
+ − . (Ans : 13) [2 marks] [UEC, 1982, P1, No.2]
(b) ( ) ( )
2 2
2 2 3 2 2 3
− − + . (Ans : 24 2
− ) [2 marks] [UEC, 2010, P2, No.4(a)]
Answer :
(a) (b)

132
61 Given that a = 3 2
+ , b = 3 2
− , find the value of a2
b − ab2
. (Ans : 14 2 )
[3 marks] [UEC, 2014, P1, No.18]
Answer :
==========================================================================================================================================
4.2.4 Simplify expressions involving surds by rationalising the denominators.
==========================================================================================================================================
 division 1
62 Without using a scientific calculator, simplify each of the following :
(a)
28
7
(Ans : 2) [2 marks] [UEC, 1987, P1, No.19]
(b)
20 10 2
10
+
(Ans : 2 10 2 5
+ ) [2 marks] [UEC, 1984, P1, No.9]
Answer :
(a) (b)
63 Without using a scientific calculator, simplify :
1
1 2 3
2
 . (Ans : 3 2 )
[2 marks] [UEC, 2002, P1, No.7]
Answer :

133
 division 2
64 Simplify each of the following :
(a)
1
5 2
−
(Ans : ( )
1
3
5 2
+ ) [2 marks] [clon UEC, 1985, P1, No.5]
(b)
1
3 2
+
(Ans : 2 3
− ) [2 marks] [UEC, 1995, P1, No.15]
Answer :
(a) (b)
(a)
2
6 2
+
(Ans : 6 2
− ) [2 marks] [UEC, 2006, P1, No.3(a)]
(b)
8
5 3
−
(Ans : 10 6
+ ) [2 marks] [UEC, 2011, P1, No.10]
Answer :
(a) (b)
66 Given that 2 = 1.414, 3 = 1.732, find the value of
4
3 2
−
. (Ans : 12.584)
[3 markah] [UEC, 2010, P1, No.8]
Answer :
~ p a − q b p a + q b
conjugate surd

134
 division 3
(a)
2 1
2 1
+
−
(Ans : 3 2 2
+ ) [2 marks] [UEC, 1976, P1, No.10]
(b)
2 1
2 1
−
+
(Ans : 3 2 2
− ) [2 marks] [UEC, 1988, P1, No.25]
Answer :
(a) (b)
68 Simplify :
3 2
3 2
+
−
(Ans : 5 2 6
+ )
[2 marks] [UEC, 1998, P1, No.16]
Answer :
 division 4
(a)
2
8
1 2
+
−
(Ans : −2) [3 marks] [UEC, 2001, P2, No.3(b)]
(b)
6 3
8
6 3
+
−
−
(Ans : 3) [3 marks] [UEC, 2013, P2, No.2(a)]
Answer :
(a) (b)

135
70 (a) Simplify : ( )
2 1
1 2
2 1
− −
−
(Ans : 2 3 2
− ) [3 marks] [UEC, 1995, P2, No.4(a)]
(b) If 3 = 1.732, find the value of
3 2
1
3 2
+
+
−
. (Ans : −12.928) [4 marks] [UEC, 2003, P2, No.5(a)]
Answer :
(a) (b)
 division 5
(a)
1 1
3 2 3
−
−
(Ans :
3 2 2 3
3
+
) [3 marks] [UEC, 2000, P2, No.5(a)]
(b)
1 1
6 2 6 2
−
− +
(Ans : 2) [3 marks] [UEC, 2008, P2, No.4(a)]
Answer :
(a) (b)
(a)
1 1
3 2 3 2
−
− +
(Ans : 2 2 ) [3 marks] [UEC, 1993, P1, No.18]
(b)
1 1
5 3 5 3
−
+ −
(Ans : 3
− ) [3 marks] [UEC, 2012, P1, No.9]
Answer :
(a) (b)

136
 division 6
73 (a) Simplify :
5 3 5 3
5 3 5 3
− +
+
+ −
(Ans : 8) [3 marks]
(b) Express
1 2 1 2
5 3 5 3
+ −
+
+ −
in terms of 5 6
a b
+ . (Ans : 5 6
− ) [3 marks]
[UEC, 2002, P2, No.5(b) / UEC, 1976, P2, No.3(b)]
Answer :
(a) (b)
74 Simplify :
3 2 4 3
5 2 7 3
− +
+
+ −
(Ans :
65 16 2 11 3
46
− +
)
Answer :
75 Given that 3 = 1.732, find the value of
3 1 1
3 1 3 1
+
−
− +
correct to two decimal
place. (Ans : 3.37)
[4 marks] [UEC, 2009, P2, No.5(a)]
Answer :

137
 division 7
76 Simplify each of the following
(a)
6 1 1
3
3
12 3 2
+ −
+
(Ans : 2 3
− ) [3 marks] [UEC, 2004, P2, No.5(a)]
(b)
1 2 3
2 1 5 3 5 2
+ −
− + −
(Ans : 1 3
− ) [3 marks] [UEC, 2007, P2, No.4(a)]
Answer :
(a) (b)
77 Simplify :
1 1 2
2 3 1 2 1 3
+ −
+ + +
(Ans : 0)
[3 marks] [UEC, 2015, P2, No.4(a)]
Answer :
==========================================================================================================================================
4.2.5 Solve problems involving surds.
==========================================================================================================================================
 solve problems 1 ~ daily problems
78 The length and width of a rectangle are ( )
5 2
+ cm and ( ( )
2 5 2
− ) cm respectively. Find
the area, in cm2
, of the rectangle in the form of 10
a b
+ . (Ans : 8 10
+ )
Answer :

138
79 The diagram shows a right-angled triangle PQR.
Given that the area of the triangle PQR is
7
2
cm2
. Find
(a) the height, in cm, of the triangle in the form of 2
a b
+ , (Ans : 5 3 2
− )
(b) the length, in cm, of PR. (Ans : 86 )
Answer :
(a)
(b)
 solve problems 2 ~ solve equation 1
80 Solve the equation 3 2 4 5 1
y
+ = + , express your answer in the form of 5
a b
+ , where a
and b are rational numbers. (Ans : 21 4 5
− )
Answer :
( 5 3 2
+ ) cm
P Q
R

139
81 If 5 8
x x
= + , find the value of x in the form of
a
b
. (Ans :
10 2
2
+
)
Answer :
82 The diagram shows three circles.
Cirlce with centre P has a radius of 2 cm, and circle with centre Q has a radius of 1 cm. MN is a
common tangent and all circles touch one another. Find the radius, in cm, of the smaller circle in the
form of 2
a b
+ . (Ans : 6 4 2
− )
Answer :
P
Q
M N

140
 solve problems 3 ~ solve equation 2
83 Solve the equation : 10 3
x x
= − (Answer : 4)
Answer :
84 Solve the equation : 2 2 7 0
x x
+ − = (Answer : −2)
Answer :
85 Solve the equation : 2
3 5 19 2
x x
+ − = (Answer : 2)
Answer :
86 Solve the equation : 4 3 1 4 2
x x
+ + − = (Answer : 3
4
− , 1
4
)
Answer :

141
87 Solve the equation : 3 1 3 2 1
x x
− − − = (Answer : 3
4
)
Answer :
88 Solve the equation : 3 5 2 8
x x x
− + + = + (Answer : 4)
Answer :
==========================================================================================================================================
4.3 Laws of logarithms
4.3.1 Relate equations in the form of indices and logarithms, and hence determine the logarithm of
a number.
==========================================================================================================================================
y = x
O
x
y
f(x) = ax
y = ax
= x
a > a 
&
ao
= 1 
a1
= a 
log x
a a =
10
log a =
y >
log x
a
a =
HINT
y = ax

142
90 Convert the following to logarithmic form and vice-versa :
Base 2 Base 3 Base 5
Index Logarithmic Index Logarithmic Index Logarithmic
2−4
=
1
16
−4 = 3
1
log
81
5−4
=
1
625
−3 = 2
1
log
8
3−3
=
1
27
−3 =
5
1
log
125
2−2
=
1
4
−2 = 3
1
log
9
5−2
=
1
25
−1 = 2
1
log
2
3−1
=
1
3
−1 = 5
1
log
5
20
= 1 0 = 3
log 1 50
= 1
1 = 2
log 2 31
= 3 1 = 5
log 5
22
= 4 2 = 3
log 9 52
= 25
3 = 2
log 8 33
= 27 3 = 5
log 125
24
= 16 4 = 3
log 81 54
= 625
5 = 2
log 32 35
= 243 5 =
5
log 3125
26
= 64 6 = 3
log 729
7 = 2
log 128 37
= 2187 3 = 3
logx x
28
= 256 8 =
3
log 6561
2 = 2
logx x
9 = 2
log 512 1 = logx x
210
= 1024 −2 =
2
logx x−
−1 =
1
logx x−
0 = 0
logx x
91 Evaluate each of the following without using a scientific calculator :
(a) 2 5
1 1
log log
8 125
 (Ans : 9) [2 marks] [UEC, 2015, P1, No.19]
(b) log5125 + 2
1
log
4
− 3
log 3 (Ans : 1
2
) [2 marks] [UEC, 1998, P1, No.18]
Answer :
(a) (b)

143
(a) ( ) 2
log 10 10 log 0.25
− (Ans : 7
2
) [2 marks] [UEC, 1992, P1, No.5 / UEC, 2001, P1,
No.15]
(b)
3
27 2
1
2log 3 log
2
 
−  
 
(Ans : 11
3
) [2 marks] [clon UEC, 2011, P1, No.19]
Answer :
(a) (b)
(a) 25 5 25
1
log 5 log 25 log
5
+ − (Ans : 3) [2 marks] [UEC, 1993, P1, No.21]
(b) 81 16 9
1 1
log 9 3log 2 log
2 9
+ + (Ans : 0) [2 marks] [UEC, 2014, P1, No.19]
Answer :
(a) (b)
94 Solve each of the following :
(a) log 343 x =
1
3
(7)
(b) lg x = −2
(0.01)
(c) log x 7776 = 5
(6)
(d) log x
81
1 = −4
(3)

144
(a) 3
log k = 2 (Jwp : 81) [2 marks]
(b) 2
log 16
n
= 2 (Jwp : 2) [2 marks]
[Forecast]
Answer :
(a) (b)
(b) 2
27
log 3y =
3
1
(Ans : 1) [3 marks]
(a) ( )
5 5
log log x = 5
log 4 (Ans : 625) [2 marks]
[Forecast]
Answer :
(a) (b)
(a)
log
10
3 27
x
= (Ans : 1000) [3 marks] [UEC, 1987, P1, No.18]
(b)
10
1
log
5 t
+ = 10. (Ans : 4) [3 marks] [Forecast]
Answer :
(a) (b)

145
lg =
3
lg
10 . (Ans :
1000)
Answer :
99 (a) Given 10
log a = −2b, express 100b
in term of a.
(Ans : 1
a
) [3 marks] [UEC, 2012, P1, No.11]
(b) Given that x = 4
log5 . Find the value of ( )
5
x
. (Ans : 2) [3 marks] [Forecast]
Answer :
(a) (b)
100 Given that 2
log x = 0.66 and 2
log y = 1.68, find the value of x2
y. (Ans : 8)
[3 marks] [UEC, 2007, P1, No.29]
Answer :
101 Given  
3 2 2
log log (2 1) log 4
x − = , find the value of x. (Ans : 256 2
1
)
Answer :
102 Given 3
log ( 3)
m n
+ = and 2
log ( 4) 1
m n
− = − . Show that m2
− m − 12 =
2
1
× 6n
.
Answer :

146
103 By using a scientific calculator, complete each of the following :
10
log
y x
=
x 300 3 0.003
y 3.477 1.477 −0.523 −1.523
==========================================================================================================================================
4.3.2 Prove laws of logarithms.
==========================================================================================================================================
 using law ~ type 1
105 (a) Without using a scientific calculator, find the value of
2log 4
5
5 . (Ans : 16) [2 marks]
(b) Solve the equation 27 = 8 x
2
log
. (Ans : 3) [3 marks]
[Forecast]
Answer :
(a) (b)
x = ap
~ logarithmic form
y = aq
xy = ( ) ( )
xy = a
( Product law )
x
y
=
a
x
y
=
( Division law )
xn
= ( ) n
( Power law )
xn
=
if a, x, y are positive, and a  1

147
(a) 3 1
log log
x
x
 (Ans : −3) [2 marks] [UEC, 2004, P1, No.17]
(b)
log5 log125
2
+
(Ans : 2log5 ) [2 marks] [UEC, 1988, P1, No.20]
Answer :
(a) (b)
107 Given log 2 = 0.301 and log 7 = 0.845, find the values of the following logarithms :
(a) log 2000 (Ans : 3.301) [2 marks]
(b) log 0.07 (Ans : −1.155) [2 marks]
[Forecast]
Answer :
(a) (b)
108 Given log5 2 = 0.43 and log5 7 = 1.21, find the values of the following logarithms :
(a) log5 70 (Ans : 2.64) [3 marks]
(b) log5 1.6 (Ans : 0.29) [3 marks]
[Forecast]
Answer :
(a) (b)
(a) log 15 (Ans : 1.1761) [3 marks]
(b) log 0.4444 . . . (Ans : −0.3522) [4 marks]
[UEC, 2015, P2, No.3(a) / Forecast]
Answer :
(a) (b)

148
(a) log 12 − log 1.5 (Ans : 0.903) [3 marks] [UEC, 2003, P2, No.5(b)(i)]
(b) log
2
3
+
1
3
log 64 (Ans : 0.426) [4 marks] [UEC, 2012, P2, No.3(b)]
Answer :
(a) (b)
111 Simplify :
0
5
log400
(400 25)
log 25
−  (Ans : log 2)
[3 marks] [UEC, 2008, P1, No.15]
Answer :
(a)
log9 log25
log15
+
(Ans : 2) [3 marks] [UEC, 2000, P2 No.5(b)]
(b) 3
log (2 7 1)
+ + 3
log (2 7 1)
− . (Ans : 3) [3 marks] [UEC, 1996, P1 No.19]
Answer :
(a) (b)
(a)
2 2
2 2
2
log log
log
x y
xy
+
(Ans : 2) [3 marks] [UEC, 1997, P1 No.23]
(b)
9
log
2
log
3
log
6
8
log
m
m
m
m
−
−
. (Ans : 3) [3 marks] [Forecast]
Answer :
(a) (b)

149
(a) 1.5
3 3 3
3 1
log 81 log 36 log 64
4 2
− + (Ans : 0) [4 marks] [UEC, 2008, P2, No.4(c)]
(b) 3 3 3
1 4
log 324 3log 2 log
2 81
− + (Ans : 2) [3 marks] [UEC, 2010, P2, No.3(b)]
Answer :
(a) (b)
 using law ~ HOTS
115 Find the value of ( )2
log5 (log2)(log50)
+ . (Ans : 1)
[3 marks] [UEC, 2006, P2, No.1(c)]
Answer :
116 If a2
+ b2
= 6ab, prove that ( )
1
log log log
2 2
a b
a b
−
 
= +
 
 
.
[4 marks] [UEC, 1979, P2, No.6(c)]
Answer :
117 (a) Given 23x
= 9(32x
). Prove that
8
log log 9
9
p p
x = . [4 marks]
(b) Given 3m
= 12n
. Show that 2
log 3 =
n
m
n
−
2
. [4 marks]
[Forecast]
Answer :
(a) (b)

150
118 (a) Given 5
log 3 k
= . If 52λ − 1
= 15, exprress λ in term of k. (Ans : λ =
2
2
+
k ) [3 marks]
(b) Given 3
log 5 = 1.465. With using a scientific calculator, solve the equation
5 (3x − 2
) = 5 . (Ans : 1.2675) [3 marks]
[Forecast]
Answer :
(a) (b)
=========================================================================================================================================
4.3.3 Simplify algebraic expressions using the laws of logarithms.
==========================================================================================================================================
 simplify 1
(a) 10
3log 2
p − (Ans :
3
10 100
log
p
 
 
 
) [3 marks]
(b) 2
2log x − 3
2
9log x + log2 (2x + 1) (Ans : log2 ( )
x
x 1
2 + ) [3
marks]
[Forecast]
Answer :
(a) (b)

151
(a) 9
log x − 9
2log y +
2
1
(Ans : 3
9 2
log x
y
 
 
 
) [3 marks]
(b) 3 + log 2 x +
1
2
log 2 y − 2 log 2 (x − y) (Ans : log2 (
8
2
( )
x y
x y
−
) ] [3 marks]
[Forecast]
Answer :
(a) (b)
 simplify 2
121 (a) If 10
log 2 = a, express 10
log 25 in term of a. (Ans : 2(1 − a)] [3 marks] [UEC, 1987, P1 No.20]
(b) Given 10
log 25 = p, express 10
log 50 in term of p. (Ans : 1
2
1
+
p ) [3 marks]
[Forecast]
Answer :
(a) (b)
122 Given 5
log 2 m
= and 5
log 7 p
= , express 5
log 4.9 in terms of m and p. (Ans : 2p − m − 1)
[4 marks] [2004, No.8]
Answer :

152
123 Given that log 2
m p
= and log 3
m r
= , express 





4
27
log
m
m in terms of p and r.
(Ans : 3r − 2p + 1) [4 marks] [2005, No.9]
Answer :
124 Given that 2
log x h
= and 2
log y k
= , express
3
2
log
x
y
in terms of h and k.
(Ans : 3h − k )[3 marks] [2011, No.8]
Answer :
125 Given that log 2 a
= and log3 b
= , express 10
1
log 2
3
 
−
 
 
in terms of a and b.
(Ans : 1 − a − b) [3 marks] [UEC, 1999, P1 No.19]
Answer :
126 Given that log 2 a
= and log3 b
= , express 10
3
log
2
+ 10
5
log
4
+ 10
9
log
8
in terms of a
and b. (Ans : 1 − 7a + 3b)
[4 marks] [UEC, 1997, P2 No.?]
Answer :

153
==========================================================================================================================================
4.3.4 Prove
log
log
log
c
a
c
b
b
a
= and use the relationship to determine the logarithm of a number.
==========================================================================================================================================
128 Convert each of the following to “ bases 10 ” and “ natural logarithm ”. Hence, determine the values.
logarithm 9
log 7 2.1
log 4000
2
1
log
12
16
3
log 36 20
3
4
log
3
base 10
natural
logarithm
value
NOTE : natural logarithm → loge @ In
 changing base 1
129 Given a =
3
1
x
, find
(a) log x a, (Ans : −3)
(b) 2log a x. (Ans : 3
2
− )
[3 marks] [2013, No.7]
Answer :
(a) (b)
~ index form
loga
x b
=
if a, b, c are positive, and a  1, c  1
~ take logc on both sides
x
a b
=
log log
x
c c
a b
=

if c = b
HINT
loga b = k → logb a =
log n
a
b = =

154
130 Given log k 9 = 2, find the value of
(a) k, (Ans : 3)
(b) 9
1
log
k
 
 
 
. (Ans : − 2
1 )
[3 marks] [2014, No.7]
Answer :
(a) (b)
131 Given h
8
log = k, express in terms of k :
(a) h
2
log , (Ans : 3k ) [2 marks]
(b) 512
logh . (Ans : k
3 ) [2 marks]
[Forecast]
Answer :
(a) (b)
132 (a) Given that mt
= 8. Express 2
log m in terms or t. (Ans : t
3 ) [3 marks]
(b) Given p = 5
log x , express log 25
x in term of p. (Ans : p
2 ) [3 marks]
[Forecast]
Answer :
(a) (b)

155
133 (a) If logb x = k, express 1
log
b
x in term of k. (Ans : −k ) [3 marks]
(b) If log 7
b
= v, express 7
log b in term of v. (Ans : v
2
) [3 marks]
[Forecast]
Answer :
(a) (b)
134 Given that 9
log 15 = m, express 5
log 9 in term of m. (Ans : 1
2
2
−
m
) [3 marks]
[Forecast]
Answer :
 changing base 2
(a) 9
log 3 − 1.4
49
log
25
(Ans : 4
7
− ) [3 marks]
[Forecast]
(b) 2
log 8 + 4
log 2 (Ans : 1
4
6 ) [3 marks] [UEC, 2007, P2 No.3(b)]
Answer :
(a) (b)

156
(a)
a
a
a
a
125
5
log
log
4
log
16
log
+ (Ans : 5) [3 marks]
(b) 4
log 128 + log m + 1
3
1
1
+
m
(Ans : 6
19 ) [3 marks]
[Forecast]
Answer :
(a) (b)
137 (a) Simplify : 4
log a  25
log 8  3
log 125
a
(Ans : 27
4
) [3 marks]
(b) Given log 3 q  log p 81  log q p3q = 9, find the value of q.
(Ans :
4
3 ) [3 marks]
[Forecast]
Answer :
(a) (b)
 changing base 3
138 Given that 2
log b x
= and 2
log c y
= , express 4
8
log
b
c
 
 
 
in terms of x and y.
[ Ans : 2
1
(3 + x − y) ] [4 marks] [2007, No.7]
Answer :

157
139 Given that 2
log 3 a
= and 2
log 5 b
= , express 8
log 45 in terms of a and b.
[Ans : 3
1 (2a + b ) [3 marks] [2010, No.8]
Answer :
140 Given logm C x
= , express in term of x
(a)
1
logm
C
 
 
 
,
(b) 3
log m
Cm . (Ans : 2x + 6)
[4 marks] [2019, No.10]
Answer :
(a) (b)
141 (a) Given r = 2m
and t = 2n
, express 8 4
32
log
rt
 
 
 
in terms of m and n.
[ Ans : 3
1 (5 − m − 4n)] [4 marks]
(b) Given 3p
= 5 and 9q
= 2, express 3
log 50 in terms of p and q.
(Ans : 2p + 2q) [4 marks]
[Forecast]
Answer :
(a) (b)

158
142 Given 2
log 3 p
= and 3
log 7 q
= . Express 3
log 31.5 in terms of p and q.
(Ans : 2 + q − p
1 ) [4 marks] [Forecast]
Answer :
143 Given that 9
log p x
= and 3
log q y
= , express 3
81
log
p
q
 
 
 
in terms of x and y.
(Ans : 4 + 2x − y) [4 marks] [Forecast]
Answer :
 changing base 4
144 Given log 2
p x
= and log 5
p y
= , express 2
5
log 8p in terms of x and y.
(Ans : y
x 2
3 +
) [3 marks] [2015, No.6]
Answer :
145 Given log 7
a r
= , express in term of r :
(a) log 49
a , (Ans : 2r)
(b) 2
7
log 343a . (Ans : 3 + r
2 @ r
r 2
3 + )
[4 marks] [2016, No.14]
Answer :
(a) (b)

159
146 Given that log 5
a p
= and log 7
a q
= , express 3
35
log a in terms of p and q.
(Ans : q
p+
3
) [3 marks] [Forecast]
Answer :
147 Given log 3
c h
= and log 2
c k
= , express 2
log 72 in terms of h and k.
(Ans :
2 3
h k
k
+
Answer :
148 Given log 5
x m
= and log 7
x n
= . Express 2
7
log 25x in terms of m and / or n.
(Ans : n
m 4
4 +
Answer :
149 Given 2
log 3 p
= and 3
log 7 q
= . Express 42
log 63 in terms of p and q.
(Ans : pq
p
q
p
+
+
+
1
)
2
(
Answer :

160
==========================================================================================================================================
4.3.5 Solve problems involving the laws of logarithms.
==========================================================================================================================================
 solve index equation ~ involving logarithm
(a) 3
2
9
m
= 20 (Ans : 2.045) [3 marks]
(b) 23x − 1
− 7 = 0 (Ans : 1.269) [3 marks]
[Forecast]
Answer :
(a) (b)
152 Solve the equation 42x − 1
= 7x
. (Ans: 1.677)
[4 marks] [2003, No.6]
Answer :
(a)
x
−






1
3
4
= 5x
. (Ans : 0.152) [3 marks]
(b) 42x − 1
= 3x + 2
(Jwp : 2.141) [3 marks]
[Forecast]
Answer :
(a) (b)
y = x
O
x
y
f(x) = ex
log10 =
In e =
log
10 x
=
HINT
y = ex
In x
e =
In x
e =
=
NOTE
log x
a
a = x In = loge
natural logarithms

161
(a) 32x
. 4x
= 7x + 1
. (Ans : 1.188) [3 marks]
(b) 33x
. 2x − 2
= 5 3 − 2x
. (Ans : 0.862) [3 marks]
[Forecast]
Answer :
(a) (b)
155 Solve the equation 42x
+ 42x − 1
= 4. (Ans : 0.420)
Answer :
156 Solve the equation : 2x + 1
+ 3(21 − x
) = 7. (Ans : 0.585, 1)
Answer :
 solve natural logarithmic equation
(a) ( )
5
In 3 5
2
x − = . (Ans : 5.727) [2 marks]
(b) ( )2
In 3 5
x + = (Ans : 9.182, −15.182) [2 marks]
[Forecast]
Answer :
(a) (b)

162
(a)
2
x
e +
= 10 (Ans : 0.303) [2 marks]
(b)
2
3 9.6
x
e = (Ans : 0.582) [2 marks]
[Forecast]
Answer :
(a) (b)
MIND think :
In x y
=  y
x e
=
2 7 3
x x
e e
= − (Ans : 1.099, −0.693)
Answer :
(a) 1 3
2 0
x x
e e
+
− = (Ans : 0.153) [4 marks]
(b) 2 1 1
6 0
x x
e e
+ −
− = (Ans : −0.2.0) [4 marks]
[Forecast]
Answer :
(a) (b)

163
 solve logarithmic equation ~ use antilog
(a)
log2
6 x
= 5 (Ans : 1.864) [3 marks]
(b) 2
log 3 log 5
n = (Ans : 1.605) [3 marks]
[Forecast]
Answer :
(a) (b)
 solve logarithmic equation ~ same base
162 Solve the equation log3 4x − log3 (2x − 1) = 1. (Ans : 2
3
)
[3 marks] [2005, No.8]
Answer :
163 Given that log2 xy = 2 + 3 log2 x − log2 y, express y in terms of x. (Ans : y = 2x)
[4 marks] [2006, No.7]
Answer :
164 Solve the equation 2 + log3 (x − 1) = log3 x . (Ans : 8
9 )
[3 marks] [2006, No.8]
Answer :

164
1 + log 2 (x − 2) = log 2 x (Ans : 4)
[3 marks] [2012, No.8]
Answer :
log 3 2 + log 3 (x − 4) = 1 (Ans : 2
11 )
[3 marks] [2013, No.8]
Answer :
 solve logarithm equation ~ change base 1
167 Given that 2
log T − 4
log V = 3, express T in terms of V. (Ans : T = 8 v )
[4 marks] [2003, No.5]
Answer :
168 Given that 4
log x = 2
log 3, find the value of x. (Ans : 9)
[3 marks] [2008, No.8]
Answer :

165
169 Given that 8
log p − 2
log q = 0, express p in terms of q. (Ans : p = q3
)
[3 marks] [2009, No.8]
Answer :
324
logm − m
m
2
log = 2
(Ans : 3)
[4 marks] [2017, No.6]
Answer :
 solve logarithm equation ~ change base 2
171 (a) Given P = Q
a
log , state the conditions of a. [1 mark]
(b) Given y
3
log =
3
log
2
xy
, express y in terms of x. (Ans : y = 2
1
x
) [3 marks]
[2018, No.17]
Answer :
(a) (b)
(a) 4
log (4 5)
x + =
2
1
+
4
log
1
7x
. (Ans : 2
1
) [4 marks]
(b) 3
log x = 5 − 6log 3
x (Ans : 9, 27) [4 marks]
[Forecast]
Answer :
(a) (b)

166
log x + 2 = log 27
x (Ans :
27
1 , 3)
[4 marks]
[Forecast]
Answer :
=========================================================================================================================================
4.4 Applications of indices, surds and logarithms
4.4.1 Solve problems involving indices, surds and logarithms.
==========================================================================================================================================
174 In a culture experiment, a scientist took some bacteria. The function that shows the number of bacteria
after t hour is given by P(t) = 50 (10)0.3t
.
(a) How many bacteria are taken at the beginning of the experiment ?
(b) When will the number of bacteria reach one million ? (Ans : 14.34)
[4 marks]
[Forecast]
Answer :
(a) (b)
175 The population in Pekan Telipok is expected to increase at an annual rate of 3%. The increase in
population follows the equation P = P0 (R)t
, where P0 is the original population and P is the new
population after t years.
(a) State the value of R.
(b) Find the value of t such that the population doubles the original population. (Ans : 23.45)
[4 marks]
[Forecast]
Answer :
(a) (b)

167
176 A liquid cools from its original temperature of 100°C to x°C in t seconds. Given that x =
( )
100 0.98
t
, find
(a) the temperature of the liquid after cooled for 8.5 seconds,
(Ans : 84.22)
(b) the time, in second, for the temperatute of the liquid dropped by 60°C.
(Ans : 45.35)
Answer :
(a) (b)
177 (a) Mclarance invested RM50000 in a bank and did not withdraw any money from his
account. After n years, his savings becomes
6
50000
5
n
 
 
 
. Calculate the minimum number
of years for his saving exceeded one million ringgit for the first time.
(Ans : 17) [3 marks]
(b) The population of a country can be estimated with the growth model, P = 2400000 e0.03t
where t is the number of years after 1990. When the population of the country will
exceed 4.42 millions for the first time ?
(Ans : 2011) [3 marks]
[Forecast]
Answer :
(a) (b)

168
178 (a) The price of a computer can be determine with the equation,
n log 10 





−
m
2
1 = log 10 x − log 10 y.
In this equation, the computer with m years of usage and the price RM y will drop to RM x after
being used for n years. A computer is bought at RM10000 has 5 years of usage. If the price of the
computer drops to RM2000, find the years of usage for that computer. Ans : 3.15) [3 marks]
(b) In astronomy, the diameter of a planet, d km, is calculated by using the formula
log 0.2
10
10 5001 10
d g
= − , where g is the absolute magnitude of the planet. Find the diameter
of the planet when given the absolute magnitude is 18.48 km. (Ans : 35.08) [3 marks]
[Forecast]
Answer :
(a) (b)
PAPER 2
179 (a) Simplify : ( )
2
log 2 1
x + − 2
4
5log x + 2
4log x (Ans : log2 ( )
x
x 1
2 + ) [4 marks]
(b) Hence, solve the equation : ( )
2
log 2 1
x + − 2
4
5log x + 2
4log x = 3
( Ans : 6
1 ) [2 marks]
[2011, No.2]
Answer :
(a) (b)

169
180 It is given that p = 2x
and q = 2y
.
(a) Express
x
y
x
4
8 +
in terms of p and q. (Ans : pq3
) [3 marks]
(b) Find
2
4
4
log
p
q
in terms of x and y. [ Ans :
2
1 (2 + 2x − y) [5 marks]
[2014, No.4]
Answer :
(a) (b)
181 Express 2
2n +
− 1
2n +
+ 1
2n −
in the form 1
(2 )
n
p −
, where p is a constant.
Hence, solve the equation 8 ( 2
2n +
− 1
2n +
+ 1
2n −
) =
2
5(2 )
n
. (Ans : p = 5 ; n = −1, 2)
[6 marks]
[2019, No.2]
Answer :

170
 FORECAST
182 Two experiments are carried out by a scientist and obtain the following two equations:
log x (y + 7) = 2 − log x 3
9x + 1 = y
3
81
Find the value of x and of y that satisfy both experiments.
(Ans : x = 3, y = −4)
[6 marks]
Answer :
183 Given log 40
x p
= and log 50
x q
= .
Express log 2
x and log 5
x in terms of p and q. (Ans : log 2
x =
2
5
p q
−
, log 5
x =
3
5
q p
−
)
[6 marks]
Answer :
184 Given 4
log x a
= and 2
log y b
= .
If xy = 128 and
y
x
= 4. Find the value of a and of b. (Ans : a = 2 4
1
, b = 2 2
1
)
[6 marks]
Answer :

171
185 Given 3
logb xy m
= and 3 2
logb x y n
= .
Find logb xy in terms of m and n. (Ans :
2
14
m n
+
)
[6 marks]
Answer :
186 Find the value of x which satisfy the equation
2 2
8 5
log log
x y
x
xy xy
+ + = . (Ans : 2)
[4 marks] [clon SBP 2020, No.15]
Answer :
187 Solve the equation 2 2
log 64 2 log
x
x
= + . (Ans : 2, 1
8
)
[4 marks] [clon Johor 2020, No.17]
Answer :
188 Given 5
log 4 x
= and 8
log 80 y
= . Express x in terms of y. (Ans : 2
3 4
y
x −
= )
[4 marks] [clon YIK 2020, No.10]
Answer :

172
PROGRESSION
- WORKSHEET
Encik Adry Colodius

174
WORKSHEET
TOPIC 5 : PROGRESSIONS
==========================================================================================================================================
5.1 Arithmetic progressions
5.1.1 Identify a sequence as an arithmetic progression and provide justification.
==========================================================================================================================================
 conditions → has common difference, d & d  0
1 (a) Determine whether the following sequence is an arithmetic progression.
2 , 8 , 18 , 32 , . . .
(b) Given a reason for the answer in (a).
Answer :
(a) (b)
2 Given x, y and z are three successive terms in an arithmetic progression. Express y in terms of x and
z.
Answer :
3 Complete the following network diagram, where the relationship of the network is a consecutive term in
an arithmetic progression.
Answer :
(a) (b) (c)
13 21
4
lg p
9lg p
5
3lg p 44k
14k
26k

175
MIND think :
==========================================================================================================================================
5.1.2 Derive the formula of the nth
term, Tn, of arithmetic progressions, and hence use the formula
in various situations.
5.1.3 Derive the formula of sum of the first n terms, Sn, of arithmetic progressions, and hence use
the formula in various situations.
==========================================================================================================================================
4 Complete the following table, to derive the formula of the nth
term, Tn, of arithmetic progressions :
An arithmetic progression with first term, a and common difference, d
T1 a
T2 a + d
T3
T4
T5
Tn
5 Complete the following table, to derive the formula of sum of the first n terms, Sn, of arithmetic
progressions :
Sum
T1 Tn = a + a + (n − 1)d = 2a + (n − 1)d
T2 Tn − 1 = =
T3 Tn − 2 = =
T4 Tn − 3 = =
T5 Tn − 4 = =
Tn − 1 T2 = =
Tn T1 = =
Sum Sn =
• Arithmetic Progression
→ a sequence of number where each term is obtained by a constant to the term before it.
→ has common , d = Tn − Tn−1, where d  ; n > @ n  .
• Arithmetic Mean → 1 1
2
n n
T T
− +
+
= , where n is a positive integer and n  .

176
Sum of first n terms, Sn

where
MIND think :
 given three @ more consecutive terms → concept d, Tn, Sn / arithmetic mean 1
6 The first three terms of an arithmetic progression are k − 3, k + 3, 2k + 2. Find
(a) the value of k,
(Ans : 7)
(b) the sum of the first 9 terms of the progression.
(Ans : 252)
[3 marks]
[2003, No.7]
Answer :
(a) (b)
7 Three consecutive terms of an arithmetic progression are 5 − x, 8, 2x. Find the common difference of the
progression. (Ans : 14)
[3 marks] [2007, No.10]
Answer :
n is a integer,
n > @ n 
Sn =
Sn =

T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 + T13 + T14 + T15
T1 = S1 = a
T2 = S2 − S1
T3 = S3 − S2

Tn =

177
8 The first three terms of an arithmetic progression are 3h, k, h + 2.
(a) Express k in terms of h. (Ans : k = 2h + 1)
(b) Find the 10th
term of the progression in terms of h. (Ans : 9 − 6h)
[4 marks] [2010, No.11]
Answer :
(a) (b)
9 It is given that x, 5, 8, …, 41, … is an arithmetic progression.
(a) State the value of x.
(b) Write the three consecutive terms after 41.
[3 marks] [2011, No.9]
Answer :
(a) (b)
10 It is given that 11, y + 4 and 3y − x are three consecutive terms of an arithmetic progression.
(a) Express y in terms of x. (Ans : y = x −3)
(b) Find the common difference if x = 8. (Ans : −2)
[4 marks] [2012, No.10]
Answer :
(a) (b)

178
11 (a) State the conditions for a sequence is an arithmetic progression. **
(b) The first three terms of a arithmetic progression are h, 8 and k. Find the value of h + k.
(Ans : 16) [2 marks] [2013, No.9]
Answer :
(a) (b)
 arithmetic mean 2
12 If 25 terms are added into between 4 and 82 to form an arithmetic progression, find the sum
of the 25 terms that are added. (Ans : 1075)
Answer :
13 Given an arithmetic progression, p,
5
2
6 , . . . ,
5
3
13 , q. The sum of all the terms is 270.
Find the number of terms in the progression. (Ans : 27)
Answer :
 concept, formula 1
14 The volume of water in a tank is 450 litres on the first day. Subsequently, 10 litres of water is
added to the tank everyday. Calculate the volume, in litres, of water in the tank at the end of the
7th
day.
(Ans : 510) [2 marks] [2004, No.11]
Answer :

179
15 The first three terms of an arithmetic progression are 5, 9, 13. Find
(a) the common difference of the progression,
(Ans : 4)
(b) the sum of the first 20 terms after the 3 rd term.
(Ans : 1100)
[4 marks] [2005, No.11]
Answer :
(a) (b)
16 The diagram shows three square cards
The perimeters of the cards form an arithmetic progression. The terms of the progression are in ascending
order.
(a) Write down the first three terms of the progression,
(b) Find the common difference of the progression.
[3 marks] [2009, No.10]
Answer :
(a) (b)
 concept, formula 2 [ given 3 values of consecutive terms → solve equation ]
17 Given an arithmetic progression −7, −3, 1, …………., state three consecutive terms in this progression
with sum up to 75. (Ans : 21, 25, 29)
[3 marks] [2004, No.10]
Answer :
3 cm
7 cm
5 cm

180
18 The first three terms of an arithmetic progression are 46, 43, 40. The nth term of this progression is
negative. Find the least value of n. (Ans : 17 )
[3 marks] [2008, No.10]
Answer :
 concept, formula 3 [ solve an equation ]
19 In an arithmetic progression, the common difference is −5. Given the sum of the first 10 terms of the
progression is 45, find
(a) the first term of the progression, (Ans : 27)
(b) the tenth term of the progression. (Ans : −18)
[4 marks] [2013, No.10]
Answer :
(a) (b)
20 The third term of an arithmetic progression is 4 and the fourth term is 7.
(a) State the common difference of the progression.
(b) Find the sum of the first 25 terms of the progression. (Ans : 850)
[4 marks] [2019, No.6]
Answer :
(a) (b)

181
 concept, formula 4 [ solve 2 equations ]
21 The 9th
term of an arithmetic progression is 4 + 5p, and the sum of the first four term of the progression
is 7p − 10, where p is a constant. Given that the common difference of the progression is 5, find the
value of p. (Ans : 8)
[3 marks] [2006, No.9]
Answer :
22 The second term of an arithmetic progression is −3 and the sixth term is 13. Find the first term and the
common difference of the progression. (Ans : a = −7, d = 4)
[3 marks] [2011,
No.10]
Answer :
23 In an arithmetic progression, the sum of the first four terms is 14 and the sixth term is −7. Find the first
term and the common difference of the progression. (Ans : a = 8, d = −3)
[3 marks] [2015, No.9]
Answer :
24 The seventh term of an arithmetic progression is 12. The tenth term of the progression is greater than the
second term by 8. Find the first term and the common difference of the progression. (Ans : a = 6, d = 1)
Answer :

182
25 The sum of the first and seventh terms of an arithmetic progression is 6, and the ninth term of the series
is double the sixth term. Find the first term and the common difference of the progression.
(Ans : a = −6, d = 3) [4 marks] [Forecast]
Answer :
26 An arithmetic progression has 14 terms. The sum of the odd terms is 140, and the sum of the even terms
is 161. Find the first term and the common difference of the progression. (Ans : a = 2, d = 3)
[4marks] [Forecast]
Answer :
 given formula Tn
27 The n-th term of an arithmetic progression is given by Tn = 11 − 3n. Find
(a) the common difference, (Ans : −3)
(b) the sum of the second five terms. (Ans : −65)
Answer :
(a) (b)
-

183
28 The nth
term, Tn, of a progression is given by 4n − 7.
(a) Show that the progression is an arithmetic progression.
(b) Express Tn − 1 in term of n.
(Ans : 4n − 11)
Answer :
(a) (b)
 given formula Sn
29 The sum of the first n terms of an arithmetic progression is given by Sn = )
1
3
(
2
+
n
n
. Find
(a) the sum of the first 5 terms, (Ans : 40)
(b) the 5th
term. (Ans : 14)
[4 marks] [2010, No.9]
Answer :
(a) (b)
30 It is given that the sum of the first n terms of an arithmetic progression is Sn =
2
n
[ 13 − 3n ]. Find the
n term. (Ans : 8 − 3n)
[3 marks] [2017, No.8]
Answer :

184
31 It is given that the sum of the first m terms of an aritmetic progression is Sm =
1
2
k +
(a + 7), such that k
is a constant, a is the first term and 7 is the last term.
(a) Express k in terms of m. (Ans : k = m − 1)
(b) State the range of values of k. (Ans : k > 0 @ k  1)
[2 marks] [2019, No.5]
Answer :
(a) (b)
==========================================================================================================================================
5.1.4 Solve problems involving arithmetic progressions.
==========================================================================================================================================
 daily problems
32 A stall selling ‘teh tarik’ gives choice to customers of using either condensed milk or evaporated milk in
their drink. On a particular day the stall has 70 cans of condensed milk dan 48 cans of evaporated milk.
The stall used 5 cans of condensed milk and 3 cans of evaporated milk in a day. After how many days,
the remainder cans of both milk are the same ? (Ans : 11)
[3 marks] [2016, No.21]
Answer :
33 A student has a wire with the length of 13.16 m. The student divided the wire into several pieces. Each
piece is to form a square. The diagram shows the first three squares formed by the student.
How many squares can be formed by the student ?
(Ans : 14)
[3 marks] [2018, No.15]
Answer :
4 cm
4 cm
7 cm
7 cm
10 cm
10 cm

185
34 Find the total number of integers between 100 and 200 that are multiples of 9. (Ans : 11)
Answer :
35 Given that 52
. 54
. 56
. 58
. . . . . 52n
= (0.04)−28
. Find the value of n. (Ans : 7)
Answer :
36 In a game, Julita is given 130 cuboid to form a pyramid. She needs to arrange a cuboid in the first row,
three cuboids in the second row, five cuboids in the third row, and so on. Find the number of cuboids
there were left. (Ans : 9)
Answer :
37 A piece of wire is used to form 20 circles as shown in the diagram.
The radius of the circles form an arithmetic progression. Given that the smallest circle has a radius of 10
cm and the biggest circle has a circumference of 96 m. Find the radius, in m, for second circle.
(Ans : 12) [4 marks] [Forecast]
Answer :
. . .

186
38 A wire of length p cm is cut into 30 pieces. The length of these pieces form an arithmetic
progression. Given that the length of the longest piece is 99 cm, and the sum of the
three shortest pieces is 45 cm. Find the length, in cm, of the shortest piece.
(Ans : 12)
Answer :
39 Two particles are moves simultaneously from the both end of a straight tube with a length of 14.3 m. One
of the particle moves 51 cm in the 1st
second, 49 cm in the 2nd
second, 47 cm in the 3rd
second, and so
on. The other particle moves 30cm in the 1st
second, 31 cm in the 2nd
second, 32 cm in the 3rd
second,
and so on. Find the time, in second, that it would take for the two particles to meet ?
(Ans : 20)
Answer :
==========================================================================================================================================
5.2 Geometric progressions
5.2.1 Identify a sequence as a geometric progression and provide justification.
==========================================================================================================================================
 conditions → has common ratio, r & r  1
40 (a) Determine whether the following sequence is an arithmetic progression or a geometric progression.
16x, 8x, 4x, ……….
(b) Give a reason for the answer in (a).
[2 marks] [2007, No.9]
Answer :
(a) (b)

187
41 Given x, y and z are three successive terms in a geometric progression. Express y in terms of x and
z.
Answer :
42 Complete the following network diagram, where the relationship of the network is a consecutive term in
a geometric progression.
Answer :
MIND think :
1
3
1
12
3
4
1 1
4
• Geometric Progression
→ a sequence of number where each term is obtained by a constant with the previous term.
→ has common , r =
1
−
n
T
n
T
, where r  ; n > @ n  .
• Geometric Mean → ( )( )
1 1
n n
T T
− +
 = , where n is a positive integer and n  .

188
==========================================================================================================================================
5.2.2 Derive the formula of the nth
term, Tn, of geometric progressions, and hence use the formula
in various situations.
5.2.3 Derive the formula of sum of the first n terms, Sn, of geometric progressions, and hence use
the formula in various situations.
==========================================================================================================================================
43 Complete the following table, to derive the formula of the nth
term, Tn, of geometric progressions :
A geometric progression with first term, a and common ratio, r
T1 a
T2 ar
T3
T4
T5
Tn
44 Complete the following table, to derive the formula of sum of the first n terms, Sn, of geometric
progressions :
nth
term r  nth
term
T1 = a rT1 = ar
T2 = ar rT2 =
T3 = rT3 =
T4 =
rTn−2 =
Tn−1 = rTn−1 =
Tn = rTn =
Sum Sn (1) (2)
Sum of first n terms, Sn

where
(1) − (2)
commonly used when
| r | < 1 → −1 < r < 1
(2) − (1)
commonly used when
| r | > 1 → r < −1, r > 1
n is a integer,
n > @ n 

189
MIND think :
 given three @ more consecutive terms→ concept r, Tn , Sn / geometric mean
45 Given a geometric progression y, 2,
y
4 , p, ……., express p in term of y. (Ans : p = 2
8
y
)
[2 marks] [2004, No.9]
Answer :
46 The first three terms of a sequence are 2, x, 8. Find the positive value of x so that the sequence is
(a) an arithmetic progression,
(Ans : 5)
(b) a geometric progression.
(Ans : 4)
[2 marks] [2005, No.10]
Answer :
(a) (b)
47 It is given that the first four terms of a geometric are 3, −6, 12 and x. Find the value of x.
(Ans : −24) [2 marks] [2008, No.9]
Answer :
T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 + T13 + T14 + T15
T1 = S1 = a
T2 = S2 − S1
T3 = S3 − S2

Tn =

190
48 The first three terms of a geometric progression are x, 6, 12. Find
(a) the value of x, (Ans : 3)
(b) the sum from the fourth term to the ninth term. (Ans : 15)
[4 marks] [2009, No.11]
Answer :
(a) (b)
49 The first three positive terms of a geometric progression are 2, p and 18. Find the value of p and the
common ratio of the progression. (Ans : p = 6, r = 3)
[3 marks] [2012, No.9]
Answer :
50 It is given that (x + 1), (2x − 7) and 




 +
4
1
x
are three consecutive terms of a geometric progression
with a common ratio of
2
1
. Find
(b) the first term if (x + 1) is the 12th
term of the progression. (Ans : 12288)
[4 marks] [2016, No.22]
Answer :
(a) (b)
51 The ratio of three numbers f, g, h is 1 : 3 : 7. If each of the numbers is added by 2, its form a geometric
progression. Find the value of f + g + h. (Ans : 22)
Answer :

191
 concept, formula 1 [ given 3 values of consecutive terms → solve equation ]
52 The sum of the first n term of the progression 8, 24, 72, …………….. is 8744. Find
(a) the common ratio of the progression,
(Ans : 3)
(b) the value of n.
(Ans : 7)
[4 marks] [2005, No.12]
Answer :
(a) (b)
 concept, formula 2 [ solve an equation ]
53 In a geometric progression, the first term is a and the common ratio is r. Given that the third term of
the progression exceeds the second term by 12a, find the values of r. (Ans : −3, 4)
[3 marks] [2012, No.11]
Answer :
54 A geometric progression has a positive terms. The sixth term is four times the fourth term, and the third
term is 1, find
(a) the common ratio, (Ans : 2)
(b) the first term. (Ans : 1
4
)
Answer :
(a) (b)

192
55 The common ratio of a geometric progression is
2
1
. The sum of the first four terms after the third term is
15. Find the first term of the progression. (Ans : 64)
Answer :
56 A geometric progression and arithmetic progression have the same first term of 3. The common ratio and
the common difference of both progressions are also the same. The fifth term of the geometric progression
is 48, and the sum of the first n terms of the arithmetic progression is equal to the fourth term of the
geometric progression. Find
(a) the common ratio, (Ans : 2)
(b) the value of n. (Ans : 4)
Answer :
(a) (b)
57 The ratio of the sum of the first two terms of a geometric progression to the sum of its first and third term
terms is 2 : 5. Find the possible value of the common ratio. (Ans : 2
1
− , 3)
Answer :
58 The sum of the first six terms of a geometric progression is 28 times the sum of the first three terms. Find
the common ratio of the progression. (Ans : 3)
Answer :

193
 concept, formula 3 [ solve 2 equations ~ 1 ]
59 In a geometric progression, the third term is −3 and the sixth term is 24. Find the sum of the first eleven
terms. (Ans : −512 4
1
)
Answer :
60 The sum of the first and third terms of a geometric progression is 20, whereas the sum of the fourth and
sixth terms is 540. Find the first term and the common ratio of the progression. (Ans : a = 2, r = 3)
Answer :
61 The third term of a geometric progression exceeds the second term by 4, and the fourth term
of the progression exceeds the third term by 3. Find the first term and the common ratio of the
progression. (Ans : a = 64
3
− , r = 3
4
)
Answer :
62 The sum of the first three term of a geometric progression is 21, and the sum of the next three terms is 168.
Find the first term and the common ratio of the progression. (Ans : a = 3, r = 2)
Answer :

194
 concept, formula 4 [ solve 2 equations ~ 2 ]
63 Three consecutive terms of a geometric progression are 32, p, q. It is given that the sum of these three
terms is 26. Find the possible values of p and of q. (Ans : p = −8, q = 2 ; p = −24, q = 18)
[3 marks] [2019, No.7]
Answer :
64 Given that p, 20, q, where p < q, are three consecutive numbers of an arithmetic progression,
whereas p, 12, q are three consecutive numbers of a geometric progression. Find the values of
p and q. (Ans : p = 4, q = 36)
Answer :
==========================================================================================================================================
5.2.4 Determine the sum to infinity of geometric progressions, and hence use the formula in various
situations, S and hence use the formula in various situations.
==========================================================================================================================================
65 (a) Consider a geometric progression with | r | < 1, state the formula of Sn, that commonly used.
(b) If the value of n increases and get closer to infinity (n → ) :
(i) state the value of rn
,
(ii) hence, determine the sum to infinity of geometric progressions.
Answer :
(a) (b) (i)
(ii)

195
 sum to infinity 1
66 The first three terms of a geometric progression are 27, 18, 12. Find the sum to infinity of the geometric
progression. (Ans : 81)
[3 marks] [2007, No.11]
Answer :
67 Given the geometric progression −5,
3
10
,
9
20
− , … , find the sum to infinity of the progression.
(Ans : −3) [3 marks] [2009, No.9]
Answer :
68 Find the sum to infinity of the geometric series 1 −
4
1
+
16
1
−
64
1
+ . . . . (Ans : 5
4
)
Answer :
69 Given k = a + 4 +
5
4
+
25
4
+ . . . is a infinite geometric series.
Find the values of a and k. (Ans : a = 20, k = 25)
Answer :

196
 sum to infinity 2
70 In a geometric progression, the first term is 4 and the common ratio is r.
(a) State the condition for r such that the sum to infinify of the progression is exist. **
(b) Given that the sum to infinity of this progression is 16, find the value of r. (Ans : 0.75)
[2 marks] [2008, No.11]
Answer :
(a) (b)
71 It is given that 1, x2
, x4
, x6
, … is a geometric progression and it sum to infinity is 3. Find
(a) the common ratio in terms of x,
(b) the positive value of x. (Ans : 3
2 )
[3 marks] [2010, No.10]
Answer :
(a) (b)
72 It is given that x2
, x4
, x6
, x8
, … is a geometric progression such that 0 < x < 1. The sum to infinity of
this progression is
3
1
. Find
(a) the common ratio of this progression in terms of x,
(b) the value of x. (Ans : 2
1 )
[3 marks] [2011, No.11]
Answer :
(a) (b)

197
 sum to infinity 3 → problems solving
73 In a geometric progression, the first term is 64 and the fourth term is 27. Calculate
(a) the common ratio,
(Ans : 4
3
)
(b) the sum to infinity of the geometric progression.
(Ans : 256)
[4 marks] [2003, No.8]
Answer :
(a) (b)
74 The third term of a geometric progression is 16. The sum of the third term and the fourth term is 8. Find
(a) the first term and the common ratio of the progression,
(Ans : a = 64, r = 2
1
− )
(b) the sum to infinity of the progression.
(Ans : 42 3
2
)
[4 marks] [2006, No.10]
Answer :
(a) (b)
75 It is given that p, 2, q are the first three terms of a geometric progression. Express in terms of q
(a) the first term and the common ratio of the progression, (Ans : a = q
4 , r = 2
q
)
(b) the sum to infinity of the progression. [ Ans : )
2
(
8
q
q −
]
[4 marks] [2018, No.14]
Answer :
(a) (b)

198
76 A geometric progression has a positive terms. The first term of the progression exceeds the second term
by 16, and the sum to infinity of the progression is 36. Find the first term and the common ratio of the
progression. (Ans : a = 24, r = 1
3
)
Answer :
 sum to infinity 4 → recurring decimal
77 Express the recurring decimal 0.969696 ………. as a fraction in its simplest form. (Ans : 33
32 )
[4 marks] [2004, No.12]
Answer :
78 Given 0.16
h
k
•
= is a recurring decimal where h and k are positive integers. Find the value of h + k.
(Ans : 7)
Answer :

199
79 Given 1.05456
p
• • •
= is a recurring decimal. Express p as a fraction in its simplest form. (Ans : 33300
35117
)
Answer :
 given formula Tn
80 It is given that the n-term of a geometric progression is Tn =
2
3 1
−
n
r
, r  k. State
(a) the value of k,
(b) the first term of the progression.
[2 marks] [2017, No.7]
Answer :
(a) (b)
81 The nth
term of a geometric progression is given by Tn = b (22n − 1
). If the third term of the progression is
96, find
(a) the value of b, (Ans : 3)
(b) the sixth term of the progression. (Ans : 6144)
Answer :
(a) (b)

200
 given formula Sn
82 It is given the sum of the first n terms of the geometric progression is Sn =
2
5
(3n
− 1). Find
(a) the first term of the progression, (Ans : 5)
(b) the common ratio of the progression. (Ans : 3)
[3 marks] [2014, No.8]
Answer :
(a) (b)
==========================================================================================================================================
5.2.5 Solve problems involving geometric progressions.
==========================================================================================================================================
 daily problems
83 Adam has just completed his diploma in engineering field. He was offered a job from two different
companies. Syarikat Satria offered him an initial salary of RM 36000 per annum with 5% yearly
increment from the basic salary. Syarikat Perdana offered an initial salary of RM 30000 per annum with
9% yearly increment from the basic salary. Adam decided to choose the company which offered higher
income and save 20% of his salary for further study after working for 10 years.
Which company should Adam choose and how much his total saving after working for 10 years. [Round
off your answer to the nearest RM ] (Ans : Perdana, 91158)
[4 marks] [2014, No.10]
Answer :

201
84 Mohan took 4 minutes to complete the first kilometer of a 15 km run. He could not sustain his stamina
thus for each subsequent kilometer, he took
8
1
more time compared to the time he took for the previous
kilometer. The participants who finished the run more than two hours are not qualified for the state level
run. Did Mohan qualified ? Show calculation to support your answer. (Ans : 155.27 > 2 hours)
[3 marks] [2016, No.23]
Answer :
85 The diagram shows a chess board.
Bervelly puts RM1, RM2, RM4, RM8, . . . consecutively in the first square, second
square, third square, . . . , and so on. If Bervelly has RM2500, find the
number of squares on the chess board that would fill the money.
(Ans : 11)
Answer :

202
86 The diagram shows part of the squares which is drawn consecutively. The length of the sides of the squares
form a geometric progression.
Given that the first square has a side of x cm, and the ratio of the length of the sides
of the fourth square to the length of side of the first square is 8 : 27. If the sum of the
areas of the first three square is
81
3325
cm2, find the value of x.
(Ans : 5)
Answer :
87 The diagram a square ABCD with side length of 14 cm.
The second square is formed by connecting the midpoints of the sides of the given
square. The third square is formed by connecting the midpoints of the sides of the
second square and so on. Find the sum to infinity of the areas, in cm2, of the squares.
(Ans : 392)
Answer :
D
F
E G
H
A B
C
P Q
R
S

203
88 The increasement of the number of tourists to a resort form a geometric progression. In the second month,
the number of tourist increased by 120, the third month increased by 240. The incresement of the
subsequent month is 480, and so on. The number of tourists in the fifth month is 2040. If the number of
tourists is exceeds 10000, a special promotion will be given. At which month, the special promotion will
be started ? (Ans : 9)
Answer :
PAPER 2
Arithmetic Progression
89 The diagram shows part of an arrangement of bricks of equal size.
The number of bricks in the lowest row is 100. For each of the other rows, the number
of bricks is 2 less than in the row below. The height of each brick is 6 cm. Ali builds a
wall by arranging bricks in this way. The number of bricks in the highest row is 4.
Calculate
(a) the height, in cm, of the wall, (Ans : 294) [3 marks]
(b) the total price of the bricks used if the price of one brick is 40 sen.
(Ans : 1019.20) [3 marks]
[2005, No.3]
Answer :
6 cm

204
90 The diagram shows the side elevation of part of stairs built of cement blocks.
The thickness of each block is 15 cm. The length of the first block is 985 cm. The
length of each subsequent block is 30 cm less than the preceding block as shown in the
diagram.
(a) If the height of the stairs to be built is 3 m, calculate
(i) the length of the top most block,
(Ans : 415 cm)
(ii) the total length of the block.
(Ans : 14000 cm)
[5 marks]
(b) Calculate the maximum height of the stairs. (Ans : 495 cm) [3 marks]
[2007, No.6]
Answer :
15 cm
925 cm
955 cm
985 cm

205
91 The diagram shows the arrangement of cylinders having the same radius, 3 cm. The height of the first
cylinder is 4 cm and the height of each subsequent cylinder increase by 2. [ volume of cylinder = r2
h
]
(a) Calculate the volume, in cm3
, of the 17th
cylinder, in term of . (Ans : 324) [3 marks]
(b) Given the total volume of the first n cylinders is 1620 cm3 , find the value of n.
[2010, No.3]
Answer :
92 Two companies, Delta and Omega, start to sell cars at the same time.
(a) Delta sells k cars in the first month and its sales increase constantly by m cars
every subsequent month. It sells 240 car in the 8th month and the total sales for
the first 10 month are 1900 cars. Find the value of k and m.
(Ans : k = 100, m = 20) [5 marks]
(b) Omega sells 80 cars in the first month and its sales increase constantly by 22 cars
every subsequent month. If both companies sell the same number
of car in the nth, find the value of n.
[2006, No.3]
Answer :
4 cm 6 cm 8 cm

206
93 At a certain day, a breeder has 3000 ducks in his farm to supply to a wholesaler. He starts selling 250
ducks on the next day and subsequently for the following days. The breeder feeds the ducks before selling.
If the cost to breed a duck is RM0.50 per day, calculate the total cost until his remaining ducks
are 500. (Ans : 9625) [6 marks]
[2015, No.4]
Answer :
94 Height of the wall is 2 m. The side length of the first coloured rectangle is 5 cm and the side length of
each subsequent coloured rectangle increases by 3 cm.
It is given that the total number of the coloured rectangles is 54.
(a) Find
(i) the side length, in cm, of the last coloured rectangle, (Ans : 164)
(ii) the total length, in cm, of the painted wall. (Ans : 4563)
[4 marks]
(b) Which coloured rectangle has an area of 28000 cm2
. Hence, state the colour of that particular
rectangle. (Ans : 46, red)
[3 marks] [2017, No.4]
Answer :
R B Y R B Y R B Y

207
95 The sum of the first n terms of an arithmetic progression, Sn is given by Sn =
2
)
33
(
3 −
n
n
. Find
(a) the sum of the first 10 terms, (Ans : −345) [1 mark]
(b) the first term and the common difference, (Ans : a = −48, d = 3) [3 marks]
(c) the value of q, given that the qth
term is the first positive term of the
progression. (Ans : 18) [2 marks]
[2018, No.1]
Answer :
Geometric Progression
96 The diagram shows the arrangement of the first three of an infinite series of similar triangles. The first
triangle has a base of x cm and a height of y cm. The measurements of the base and height of each
subsequent triangle are half of the measurements of its previous one
(a) Show that the areas of the triangles form a geometric progression and state the
common ratio. (Ans : 4
1
) [3 marks]
(b) Given that x = 80 cm and y = 40 cm,
(i) determine which triangle has an area of
4
1
6 cm2
, (Ans : 5)
(ii) find the sum to infinity of the areas, in cm2
, of the triangles. (Ans : 2133 3
1
)
[5 marks][2004, No.6]
Answer :
y cm
x cm

208
97 Muthu started working for a company on 1 January 2002 with an initial annual salary of RM 18000.
Every January, company increased his salary by 5 % of the previous year’s salary. Calculate
(a) his annual salary, to the nearest RM, for the year 2007, (Ans : 22973) [3 marks]
(b) the minimum value of n such that his annual salary in the nth
year will
exceed RM 36000, (Ans : 16) [2 marks]
(c) the total salary, to the nearest RM, paid to him by the company, for the years 2002
to 2007. (Ans : 122434 ) [2 marks]
[2008, No.3]
Answer :
98 Amir drops a ball from a height of H cm above the floor. After the first bounce, the ball reaches a height
of H1 cm = 0.8H. After the second bounce, the ball reaches a height of H2 , where H2 cm = 0.8H1. The
ball continuous bouncing in this way until it stops. Given that H = 200, find
(a) the number of bounces when the maximum height of the ball from the floor is less than 50 cm for
the first time, (Ans : 7) [4 marks]
(b) the total distance, in cm, travelled by the ball until it stops. (Ans : 1800) [2 marks]
[2009, No.6]
Answer :

209
99 It is given that … , 567 , y , 5103 , … is part of geometric progression and the sum of the first five terms
of the progression is 847. Find
(a) the common ratio, (Ans : 3) [2 marks]
(b) the first term, (Ans : 7) [2 marks]
(c) the smallest value of n such that the n-th term exceeds 10000.
(Ans : 8) [2 marks]
[2011, No.3]
Answer :
100 A wire is cut into n parts. The length of each part increase and form a geometri progression. It is given
that the length of the fifth part of the wire is 4 times the length of the third part of the wire.
(a) Calculate the common ratio. (Ans : 2) [2 marks]
(b) If the total length of the wire is 1533 cm and the length of the first part of the wire is 3 cm, calculate
(i) the value of n, (Ans : 9)
(ii) the length, in cm, of the last part of the wire. (Ans : 768)
[4 marks]
[2013, No.2]
Answer :

210
FORECAST
Arithmetic Progression
101 A wire of length 108 cm was cut to form eight circles as shown in the diagram.
The difference of the diameter between the subsequent circle is 1 cm. Find
(a) the diameter of the largest circle, Ans : 17) [4
marks]
(b) the maximum number of circles that it would form is the length of the wire
is 300 cm. (Ans : 16) [3 marks]
Answer :
102 The diagram shows the position of A and B on a straight line such that the distance of AB is 60 m.
Particel P moves from A to B with a initial speed of 10ms−1
, and the subsequent speed increase
constantly by 2 ms−1
each second. The particle Q moves with a constant speed of 8 ms−1
from B to A.
Given particles P and Q moves at the same time. Find
(a) the value of t, if particle P and Q meets after t seconds, (Ans : 3) [5 marks]
(b) the distance travelled by particle P from the beginning until it meets particle Q.
Answer :
B
60 m
A
P → 10 ms−1
8 ms−1
 Q

211
103 A wire of length 262.5 cm is cut into 30 pieces. The length of these pieces form an arithmetic progression.
If the difference between the length of the longest piece and the shortest piece is 14.5 cm, find
(a) the length of the shortest piece, (Ans : 1.5) [4 marks]
(b) the difference between the lengths of the fifth piece and the tenth piece. (Ans : 2.5) [3 marks]
Answer :
104 A wire of length y cm is bent to form the arcs of semi-circles as shown in the diagram
Given that the radius of the smallest semi circle is 2 cm, and the radius of the subsequent semi circle
increase constantly by 3 cm over the preceding semi circle, until the biggest semi circle has a radius of 59
cm
(a) Find the length of the wire that used to form the first eight semi circle. (Ans : 100 ) [4 marks]
(b) Determine whether exist the possible value of y is 500  cm. (Ans : no) [3 marks]
Answer :

212
105 Find the sum of integers between 10 and 150 that cannot be exactly divided by 3 or 4. (Ans : 5599)
Answer :
106 The digram shows the arrangement of a few chairs. The height of each chair is 68 cm. When the chairs
are arranged, there is a 5 cm gap in between two chairs. The arranged chairs will kept in the store.
(a) Find the maximum number of chairs that can be arranged if the height of the store
is 3 m. (Ans : 47) [3 marks]
(b) 15 stack of chairs were kept in the store with the condition that the first stack will have the maximum
number of chair and the arrangement of chairs for subsequent stacks decrease by 2. Do you agree
that there are 500 chairs in the store ? Show your calculations. (Ans : 495)
[3 marks]
[ clon text book form 4 ]
Answer :
Geometric Progression
107 Tin is extracted from tin ores at a mine in Pahang. In its first year of operation, the mine was able to
produce 8000 kg of tin a year. With increasing difficulty in mining, tin production in each subsequent year
decreased by 10% from the previous year. Assume this mining situation lasts for an indefinite period of
time.
(a) Calculate the maximum quantity of tin that can be extracted. (Ans : 80000) [3 marks]
(b) Due to economic factors, tin mining will be discontinued if its annual production is less than 1000
kg. Calculate the number of years that the mine will operate. (Ans : 20) [4 marks]
Answer :

213
108 Bacteria Y reproduce by binary fission process as shown in the diagram. Under the optimum condition,
the process will take 20 minutes.
A study on reproduction is carried out in 10 samples of bacteria Y. The first sample contains 2 bacteria
Y. The number of bacteria in the next sample is 3 times the number of its previous sample.
(a) What is the total number of bacteria Y in the entire sample at the beginning of the
study ? (Ans : 59048) [3 marks]
(b) After a certain period, it is found that the number of bacteria in the 6th
sample is 124416. Determine
the period, in minutes. (Ans : 160) [4 marks]
Answer :
109 The table dan diagram show the knock-off system for eight teams which participating in a competition
Round Number of match
1 4
2 2
3 1
To determine the winning team, three rounds are required with the total matches of 4 + 2 + 1 = 7. In a
competition, there are 256 teams participating. Find
(a) the number of rounds, (Ans : 8) [3 marks]
(b) the total matches. (Ans : 255) [3 marks]
Answer :
Initially
After 20 minutes
Team 1
Team 2
Team 3
Team 4
Team 5
Team 6
Team 7
Team 8
Champion

214
Arithmetic Progression & Geometric Progression
110 The first three terms of a geometric progression are also the first, ninth and eleventh terms respectively of
an arithmetic progression.
(a) Find the common ratio of the geometric progressions. (Ans : 4
1 ) [4 marks]
(b) If the sum to infinity of the geometric progression is 8, find
(i) the first term, (Ans : 6)
(ii) the common difference of the arithmetic progression. (Ans : 16
9
− )
[4 marks]
Answer :
111 The diagrams show the numbers in base two which are arranged in a few rows.
(a) Find the total number in the first n row, in term of n. [ Ans : Sn = ( )
1
2
+
n
n
] [3 marks]
(b) Hence, find the sum of the numbers of the first six row. (Ans : 2097151) [4 marks]
Answer :
1
2
23
24
25
22
26
27
28
29

215
112 Diagram shows a series of shapes produced by a combination of rectangles with sides of 2 cm.
If y represents the perimeter of the shape produced by x equal rectangular pieces.
(a) Form an equation relating x dan y. (Anz: y = 4x +4)
(b) Hence, find the perimeter of the shape that has 40 equal squares. (Ans: 164)
Jawapan :
(a) (b)

216
LINEAR LAW
- WORKSHEET
Encik Mohd Zulkarnain bin Zulkifli

217

218
WORKSHEET
TOPIC 6 : LINEAR LAW
==========================================================================================================================================
6.1 Linear and non-linear relations
6.1.1 Differentiate between linear and non-linear relations based on tables of data and graphs.
6.1.2 Draw lines of best fit for graph of linear relations with and without the use of digital
technology.
[ lines of best fit need not necessarily pass through any of the points ]
6.1.3 Form equations of lines of best fit.
6.1.4 Interpret information based on lines of best fit.
6.2 Linear law and non-linear relations
6.2.1 Apply linear law to non-linear relations
6.3 Application of linear law
6.3.1 Solve problems involving linear law.
[ problem-based learning may be involved ]
==========================================================================================================================================
MIND think :
Match the following :
Equation of Straight Line, y = mx + c ; where m = gradient, c = y-intercept
m =
2
1
2
1
x
x
y
y
−
−
=
1
2
1
2
x
x
y
y
−
−
m =
intercept
intercept
y
x
 
−
− 
−
 
Laws of Logarithms
log = 1
a a log 1 = 0
a
log = log
n
a a
x n x
log ( ) = log + log
a a a
xy x y log = log log
a a a
x
x y
y
 
−
 
 
The graph which forms a straight line.
The graph which does not forms a straight line.
A linear relation.
A non-linear relation.
m = vertical distance
horizontal distance
horizontal distance
vertical
distance
vertical
distance
horizontal distance
m = ( )
vertical distance
horizontal distance
−

219
Line of Best Fit
passes 3 points
SPM 2010
passes 2 points
SPM 2016
passes 4 points
SPM 2017
passes 4 points
SPM 2018
passes 2 points
SPM 2019
==========================================================================================================================================
6.2 Linear law and non-linear relations
6.2.1 Apply linear law to non-linear relations
==========================================================================================================================================
1 Reduce the following no-linear relations to the linear form, Y = mX + c
 type 1 ~ USING MULTIPLICATION @ DIVISION
(1) y = ax2
+ bx → Y =
x
y
, X = x (2)
x
q
px
x
y
+
= → Y = y, X = x2
(3) y =
x
p
+ x
q → Y = x
y , X = x (4) y − h =
x
hk
→ Y = xy, X = x
(5) 1
3 2
=
+
+
b
y
a
x
→ Y = y2
, X = x + 3 (6) 1
2
−
=
x
q
y
p
→ Y =
y
1
, X =
2
1
x
(wrongly recoreded)
extreme point

220
(7) y =
b
x
a
−
→ Y =
y
1
, X = x (8) nx = py + xy → Y =
y
1
, X =
x
1
(9) T = 2 
g
L
→ Y = T 2
, X = L (10) 4kx = (y − h)2
→ Y = y , X = x
 type 2 ~ USING LAW OF LOGARITHMS
(1) y = abx
→ Y = y
10
log , X = x (2)
2
10 bx
a
y +
= → Y = 10
log y , X = x2
(3) y = hk2x
→ Y = y
10
log , X = x (4) y =
x
h
k
→ Y = y
10
log , X = x

221
(5) y = pqx − 2
→ Y = y
10
log , X = x − 2 (6) y =
1
+
x
k
p
→ Y = y
10
log , X = x + 1
(7) y = (1 + k) 2
h
x → Y = y
10
log , X = x
10
log (8) y = b
a
x
→ Y = y
10
log , X = x
10
log
 apply 1
2 Diagram (a) shows the curve y = −3x2
+ 5. Diagram (b) shows the straight line graph obtained when
y = −3x2
+ 5 is expressed in the linear form Y = 5X + c.
Diagram (a) Diagram (b)
Express X and Y in terms of x and / or y. (Ans : Y = 2
x
y
, X = 2
1
x
)
[3 marks] [2006, No.11]
Answer :
X
y = −3x2
+ 5
O
y Y
x O
−3

222
3 The variables x and y are related by the equation
y
x
= x
x
p
5
+ , where p is a constant. When the
equation
y
x
= x
x
p
5
+ is express in linear form, the straight line obtained is Y = pX +
2
q
.
(a) Express X and Y in term of x and / or y. (Ans : Y = y
1
, X = 2
1
x
)
(b) Find the value of q. (Ans : 10)
Answer :
(a) (b)
 apply 2
4 The diagram shows a straight line graph of
x
y
against x.
Given that y = 6x − x2, calculate the values of k and h. (Ans : h = 3, k = 4)
[4 marks] [2004, No.13]
Answer :
(2, k)
x
O
y
x
(h, 3)

223
5 The variables x and y are related by the equation y2
= 2x (10 − x). A straight line graph is obtained by
plotting
x
y2
against x, as shows in diagram.
Find the value of p and of q. (Ans : p = 10, q = 14)
[3 marks] [2007, No.12]
Answer :
 apply 3 ~ 1
6 The variables x and y are related by the equation hy = kx2
+ hk. A straight line graph is obtained by
plotting y against x2
as shown in the diagram.
Given the gradient of the straight line is 3, find the value of h and of k. (Ans : h = 2, k = 6)
[3 marks] [2010,
No.12]
Answer :
2
y
x
x
(3, q)
(p, 0)
O
y
O
x
(0, 6)

224
x
x
p
y
12
)
1
(
3 +
−
= , where p is a constant. The
diagram shows the straight line QR obtained by plotting xy against x2
.
(a) Express the equation
x
x
p
y
12
)
1
(
3 +
−
= in its linear form, which is used ot obtain the straight
line graph shown in diagram. [Ans : xy = ( ) 4
2
3
1
+
−
x
p
]
(b) Given that the gradient of QR is −2, find the value of p and of t. (Ans : p = −5, t = 3
2 )
[4 marks] [2011, No.12]
Answer :
(a) (b)
8 Given the variable x and y are related by the equation x − py = qxy, where p and q are constants.
If the vertical-axis is represented by
y
1
. Explain, how the p and q can be obtained.
Answer :
O Q
xy
x2
R (0, 6t)

225
 apply 4
y
x
= px2
− qx where p and q are constants. Diagram
(i) and Diagram (ii) show the straight line graphs obtained by plotting the relations from the equation.
(i) (ii)
Express p in terms of q. (Ans : p =
18
6
q +
)
[3 marks] [2019, No.11]
Answer :
 apply 5
10 x and y are related by the equation y = px2
+ qx, where p and q are constants. A straight line is obtained
by plotting
x
y
against x, as shown in diagram below.
Find the values of p and q. (Ans : p = −2, q = 13)
[4 marks] [2003, No.10]
Answer :
3
y
x
X
O k + 3
O
Y
x
−6k
x
(2, 9)
(6, 1)
y
x
O

226
2
1
x
q
y
p
−
= . The diagram shows shows the straight
line graph obtained by plotting
y
1
against
2
1
x
.
Find the value of
(a) p, (Ans : 2
1 )
(b) q. (Ans : 5
2
− )
[4 marks] [2012, No.12]
Answer :
(a) (b)
12 The variable x and y are related by the equation y = 2x2
−
x
q
, where q is a constant. A straight line is
obtained by plotting xy aginst x3
, as shown in Diagram.
Find the value of h and of q. (Ans : h = 5, q = −3)
[3 marks] [2016, No.16]
Answer :
y
1
O
(5, 6)
2
1
x
2
xy
x3
3
(h, 13)
O

227
13 The variables x and y are related by the equation y = x +
2
x
r
, where r is a constant. The diagram
shows a straight line graph obtained by plotting (y − x) against
2
1
x
.
Express h in terms of p and r. (Ans : h = r
p
10
)
[3 marks] [2017, No.19]
Answer :
14 The variables x and y are related by the equation y = (x + 2) 





+ 3
x
m
, where m is a constant. When
the graph of
2
+
x
y
is plotted against
x
1
, a straight line passing through the points A and B is obtained.
Find
(a) the coordinates of point B,
(b) the value of m. (Ans : 2)
[4 marks]
[Forecast]
Answer :
(a) (b)
O
(y − x)
2
1
x
, 5
2
h
p
 
 
 
B
2
y
x +
1
x
O
A (4, 11)

228
15 x and y are related by the equation
x
b
ax
x
y
+
=
2
, where a and b are constants. A straight line passes
through point (4, 6) with a gradient of
2
1
is obtained by plotting
x
y
against x2
. Calculate the values of
a and b. (Ans : a = 2
1
, b = 4)
Answer :
 apply 3 ~ 2
16 The variables x and y are related by the equation y = x
k
5
, where k is a constant. Diagram below shows
the straight line graph obtained by plotting log10 y against x.
(a) Express the equation y = x
k
5
in its linear form used to obtain the straight line
graph shown in diagram.
(b) Find the value of k. ( Ans : 100
1
)
[4 marks] [2008, No.12]
Answer :
(a) (b)
x
O
(0, −2)
10
log y

229
17 The variables x and y are related by y =
2
1 px
, where p is a constant. A straight line with a gradient of
3 is obtained by plotting log2y against x. Find
(a) the value of p, (Ans : 8)
(b) the y-intercept of the straight line. (Ans : −1)
Answer :
(a) (b)
 apply 5 ~ 2
18 The variables x and y are related by the equation y = kx4
, where k is a constant.
(a) Convert the equation y = kx4
to linear form.
(b) The diagram shows the straight line obtained by plotting log10 y against log10 x.
Find the value of
(i) log10 k, (Ans : 3)
(ii) h. (Ans : 11)
[4 marks] [2005, No.13]
Answer :
(a) (b) (i)
(ii)
10
log y
O
10
log x
(0, 3)
(2, h)

230
19 The variables x and y are related by the equation y = 1000px
, where p is a constant. The diagram shows
the straight line grapah obtained by plotting log10y against x.
(a) Express the equation y = 1000px
in linear form used to obtain the straight line graph shown
above. [ Ans : y = (log 10 p)(x) + 3 ]
(b) Find the value of h and of p. (Ans : h = 3, p = 100
1 )
[4 marks] [2013, No.12]
Answer :
(a) (b)
20 x and y are related by the equation y = axn
, where a and n are constants. When the graph of 2
log y
is plotted against 2
log x , a straight passing through the points (1, 5) and (3, 11) is obtained. Find the
value of a and of n. (Ans : a = 4, n = 3)
Answer :
(1, 1)
(0, h)
x
log10 y
O

231
 apply 6
21 Diagram (a) shows the graph of a non linear equation. Diagram (b) shows the straight line graph obtained
when the non linear equation is expressed in linear form.
Calculate the value of c, p and q. (Ans : p = −5, q = 5
1
, c = 1)
Answer :
22 The diagram (a) shows part of the curve y = ax2
+ bx, where a and b are positive constants
The graph of the curve y = ax2 + bx is converted to its linear form , a straight line is
obtainted as shown in diagram (b). Find
(a) the values of p and q, (Ans : p = 3, q = 9)
(b) the values of a and b. (Ans : a = 2, b = −3)
Answer :
(a) (b)
O
x
c
y
x
y
(1, −4)
(2, −18)
y
x
= px + c
x
q
O
O
y
x
(6, 54)
(3, 9)
(p, 3)
(6, q)
x
x
y
O

232
23 Diagram (a) shows the curve y = abx
, where a and b are constants. Diagram (b) shows the straight line
graph obtained when y = abx
is expressed in the linear form.
Find
(a) the values of p and q, (Ans : p = 0.7782, q = 1.7324)
(b) the values of a and b. (Ans : a = 2, b = 3)
Answer :
(a) (b)
 apply 7
24 The diagram shows part of the graph 5
3
+
=
x
y which passes points P (1, 8) and Q (3, 6).
Sketch the graph xy against x that shows the coordinates of point P and
point Q. [Ans : P (1, 8), Q (3, 18)]
Answer :
O
O
x
y
x
lg y
(3, 54)
(1, 6) (1, p)
(3, q)
O
3
5
y
x
= +
x
y
Q (3, 6)
P (1, 8)
O

233
25 The diagram shows a curve. The variables x and y are related by the linear equation y2
= Ax + B, where
A and B are constants.
(a) Calculate the value of A and B. (Ans : A = 2, B = −6)
(b) Sketch the graph linear of y2
against x.
Answer :
(a) (b)
26 The diagram shows a curve. The variables x and y are related by the equation
ab
x
bx
y
+
= , where a
and b are constants.
(a) Sketch the graph linear of
y
1
against
x
1
.
(b) Calculate the values of a and b. (Ans : a = 2, b = 3)
Answer :
(a) (b)
O
y
x
P (5, 2)
Q (11, 4)
O
O
x
y
(3, 1)
3
6,
2
 
 
 
O

234
==========================================================================================================================================
6.1.3 Form equations of lines of best fit.
==========================================================================================================================================
 type 1
27 The diagram shows the straight line graph obtained by plotting (y − x) against x2
.
Express y in terms of x. (Ans : y = 3x2
+ x + 6)
[3 marks] [2015, No.10]
Answer :
28 The diagram shows the graph of a straight line
y
x2
against
x
1
.
Based on the diagram, express y in terms of x. (Ans : y = x
x
10
3
3
2
−
)
[3 marks] [2018, No.13]
Answer :
O
x2
( y − x )
6
−2
y
x2
x
1
O
−5
(6, 4)

235
29 The diagram shows a straight line graph of
x
y
against x .
Express y in term of x. (Ans :
3
2
2
y x x
= + )
Answer :
 type 2
30 A straight line is obtained by plotting log10 y against log10 x, as shown in diagram below.
Find the relation between y and x. (Ans : 4
100
3
x
y = )
Answer :
O
2
y
x
(4, 6)
x
O
10
log y
10
log x
A
1 1
,
4 3
 
 
 
B (2, −2)

236
31 The diagram shows a straight line graph lg y against x2
.
Given the gradient of the straight line PQ is 2, and P lies on lg y -axis.
(a) Find the coordinates of point P.
(b) Express y in term of x. (Ans : y = 1
2 2
10 +
= x
y )
Answer :
(a) (b)
32 The diagram shows the graf lg y against x. Given the length of AB is 4 5 unit and point A lies on x-
axis
Express y in term of x.
(Ans : y = 4
2
10 −
x
)
Answer :
O
P
T
x2
lg y
Q (4, 9)
O A
B (6, 8)
x
lg y

237
33 The diagram shows a straight line graph y against log10 x.
Given the straight line passes thtough points (−3, 0) and (0, 2).
Find the value of y when x = 1000. (Ans : 4)
Answer :
PAPER 2
 Part B → 10 marks
==========================================================================================================================================
6.1.2 Draw lines of best fit for graph of linear relations with and without the use of digital
technology.
6.1.4 Interpret information based on lines of best fit.
6.3 Application of linear law
6.3.1 Solve problems involving linear law.
==========================================================================================================================================
 type 1a ~ apply linear law to non-linear relations
34 Use graph paper to answer this question.
The table shows the values of two variables, x and y, obtained from an experiment. Variables x and y
are related by the equation y = px +
px
r
, where p and r are contants
x 1.0 2.0 3.0 4.0 5.0 5.5
y 5.5 4.7 5.0 6.5 7.7 8.4
(a) Plot xy against x2
, by using a scale of 2 cm to 5 unit on both axes.
Hence, draw the line of best fit. [5 marks]
(b) Use the graph from (a) to find the value of
(i) p, (Ans : 1.373)
(ii) r. (Ans : 5.488)
[5 marks][2005, No.7]
Answer : REFER GRAPH
O
y
(0, 2)
10
log x
(−3, 0)

238
are related by the equation y = 2kx2
+ x
k
p
, where p and k are contants.
x 2 3 4 5 6 7
y 8 13.2 20 27.5 36.6 45.5
(a) Plot
x
y
against x, using a scale of 2 cm to 1 unit on both axes.
(i) p, (Ans : 0.75)
(ii) k, (Ans : 0.25)
(iii) y when x = 1.2. (Ans : 4.32)
[6 marks]
[2007, No.7]
are related by the equation
y
k
=
x
p
+ 1, where k and p are contants.
x 1.5 2.0 3.0 4.0 5.0 6.0
y 2.502 0.770 0.465 0.385 0.351 0.328
(a) Based on the table, construct a table for the values of
x
1
and
y
1
. [2 marks]
(b) Plot
y
1
against
x
1
, using a scale of 2 cm to 0.1 unit on the
x
1
-axis and 2 cm to 0.5 unit on
the
y
1
-axis. Hence, draw the line of best fit. [3 marks]
(c) Use the graph from (b) to find the value of
(i) k, (Ans : 0.2564)
(ii) p, (Ans : −1.333)
[5 marks]
[2009, No.8]

239
are related by the equation 1
+
= px
y
n
, where n and p are contants.
x 0.1 0.2 0.3 0.4 0.5 0.6
y 0.303 0.364 0.465 0.588 0.909 1.818
y
1
. [1 mark]
(b) Plot
y
1
against x, using a scale of 2 cm to 0.1 unit on the x-axis and 2 cm to 0.5 unit on the
y
1
(i) y when x = 0.38, (Ans : 0.5714)
(ii) n, (Ans : 0.2597)
(iii) p, (Ans : −1.428)
[6 marks]
[2011, No.7]
2
1
kx
kx
h
y +
= , where h and k are contants.
x 1 2 3 4 5 6
y 2.601 0.551 0.194 0.089 0.040 0.017
(a) Based on the above table, construct a table for the values of x2
y. [ 1 mark]
(b) Plot x2
y against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.5 unit on the
x2
y -axis.
(i) y when x = 2.5, (Ans : 0.32)
(ii) k, (Ans : 3
1 )
(iii) h, (Ans : −0.1333)
[6 marks]
[2012, No.7]

240
39 Use the graph paper to answer this question.
are related by the equation nx = py + xy, where n and p are constants.
x 1.5 2.0 3.0 4.0 5.0 6.0
y 2.020 0.770 0.465 0.385 0.351 0.339
(a) Based on the table above, construct a table for the values of
y
1
and
x
1
. [2 marks]
(b) Plot
y
1
against
x
1
, using a scale of 2 cm to 0.1 unit on
x
1
y
1
(c) Using the graph in (b), find the value of
(i) n, (Ans : 0.2632)
(ii) p, (Ans : −1.316)
[5 marks]
[2015, No.11]
40 Use the graph paper provided to answer this question.
The table shows the values of two variables, x and y, obtained by an experiment. The variables x and
y are related by the equation y − h =
x
hk
, where h and k are constants.
x 1.5 2.0 3.5 4.5 5.0 6.0
y 4.5 5.25 5.5 6.3 6.34 6.5
(a) Plot xy against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the
xy-axis.
(b) Using the graph in (a), find
(i) the value of h and of k, (Ans : h = 50 64
49
/ 50.766, k = 3249
256
− / −0.079)
(ii) the correct value of y if one of the values of y has been wrongly recoreded during the
experiment
(Ans : 6)
[6 marks]
[2017, No.9]

241
 type 1b ~ form equations of lines of best fit
The table shows the values of two variables, x and y, obtained from an experiment. A straight line will
be obtained when a graph of
x
y2
against
x
1
is plotted.
x 1.25 1.43 2.00 2.50 4.00 5.00
y 4.47 4.38 4.18 3.87 2.83 2.24
x
1
and
x
y2
. [2 marks]
(b) Plot graph
x
y2
against
x
1
, using a scale of 2 cm to 0.1 unit on the
x
1
- axis and 2 cm to 2
units on the
x
y2
- axis.
(c) Using the graph in (b),
(i) find the value of y when x =2.7, (Ans : 3.747)
(ii) express y in terms of x. (Ans : y = x
4
25− )
[5 marks]
2018, No.11]
The table shows the values of two variables, x and y, obtained from an experiment.
x 1 1.2 1.4 1.6 1.8
y 79 56 36.5 18.4 0.9
(a) Plot xy aginst x3
, by using a scale of 2 cm to 1 unit on the x3
-axis and 2 cm to 10 units on the
xy-axis.
(b) Use the graph from (a),
(i) determine the linear equation connecting x and y, (Ans : xy = −15.89x3
+ 94.5)
(ii) find the value of x, when y =
x
50
. (Ans : 1.409)
[6 marks]
[forecast]

242
 type 2 ~ USING LAW OF LOGARITHMS
The table shows the values of two variables, x and y, obtained from an experiment. It is knows that x
and y are related by the equation y =
2
x
pk , where p and k are contants.
x 1.5 2.0 2.5 3.0 3.5 4.0
y 1.59 1.86 2.40 3.17 4.36 6.76
(a) Plot log y against x2
, by using a scale of 2 cm to 2 units on the x2
the log10 y-axis.
(b) Use the graph in (a) to find the value of
(i) p, (Ans : 1.259)
(ii) k. (Ans : 1.109)
[5 marks]
[2003, No.7]
are related by the equation y = pkx
x 2 4 6 8 10 12
y 3.16 5.50 9.12 16.22 28.84 46.77
(a) Plot log10 y against x, by using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.2 unit on
the log10 y-axis.
(i) p, (Ans : 1.820)
(ii) k. (Ans : 1.309)
[6 marks]
[2004, No.7]

243
are related by the equation y = pkx + 1
x 1 2 3 4 5 6
y 4.0 5.7 8.7 13.2 20.0 28.8
(a) Plot log y against (x + 1), using a scale of 2 cm to 1 unit on the (x + 1)-axis and 2 cm to 0.2 unit
on the log10 y-axis.
(i) p, (Ans : 1.778)
(ii) k. (Ans : 1.483)
[5 marks]
2006, No.7]
are related by the equation y = hk2x
, where h and k are contants.
x 1.5 3.0 4.5 6.0 7.5 9.0
y 2.51 3.24 4.37 5.75 7.76 10.00
(a) Based on the table, construct a table for the values of log10 y. [1 mark]
(b) Plot log10 y against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.1 unit on the
log10 y-axis.
(i) x when y = 4.8, (Ans : 5)
(ii) h, (Ans : 1.905)
(iii) k, (Ans : 1.096)
[5 marks]
[2008, No.8]

244
are related by the equation y =
k
hx
, where h and k are contants.
x 3 4 5 6 7 8
y 2.57 3.31 4.07 4.90 6.31 7.94
(a) Plot log10 y against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.1 unit on
the log10 y -axis.
(i) h, (Ans : 1.252)
(ii) k, (Ans : 0.7586)
(iii) y when x = 2.7. (Ans : 2.40)
[6 marks]
2010, No.7]
x
k
h
y = , where h and k are constants.
x 4 6 8 10 12 14
y 2.82 2.05 1.58 1.23 0.89 0.66
(a) Based on the table above, construct a table for the values of log10 y. [1 mark]
(b) Plot log10 y against x, using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.1 unit on
the log10 y-axis.
(i) y when x = 2, (Ans : 3.758)
(ii) h, (Ans : 5.012)
(iii) k, (Ans : 1.155)
[6 marks]
2014, No.9]

245
The table shows the values of two variables, x and y, obtained from an experiment. The variables x
and y are related by the equation y =
b
a
x
, where a and b are constants.
x 0.34 0.43 0.55 0.85 1.08 1.42
y 47.68 25.12 12.58 4.17 2.51 1.38
(a) Based on the table, construct a table for the values of 10
log x and 10
log y . [2 marks]
(b) Plot 10
log y against 10
log x , using a scale of 2 cm to 0.1 unit on the X-axis and
2 cm to 0.2 unit on the Y-axis.
Hence, Draw the line of best fit. [3 marks]
(i) a, (Ans : −3.020)
(ii) b. (Ans : 0.400)
[5 marks]
[2019, No.11]
FORECAST
x
b
x
a
y +
= , where a and b are constants.
x 1 2 3 4 5
y 0.64 4.79 9.67 14.82 19.89
(a) Plot xy against x, by using a scale of 2 cm to 1 unit on the both axes.
(b) Use the graph in (a), find
(i) the values of a and b, (Ans : a = 2.3, b = −1.5)
(ii) the value of y, when x = 3.6. (Ans : 12.66)
[6 marks]

246
The table shows the values of two variables, x and y, obtained from an experiment. Variables x and y are
related by the equation 4k2
x = (y − c)2
, where k and c are constants.
x 0 100 400 900 1600 2500
y 20 30 40 50 60 70
(a) Plot y against x , by using a scale of 2 cm to 10 unit on the both axes.
(i) the value of k and c, (Ans : k = 2
1
, c = 20)
(ii) the value of x, when y = 55, (Ans : 1225)
(iii) the value of y, when x = 500. (Ans : 42)
[6 marks]
related by the linear equation
a
x 3
+
+
b
y2
= 1, where a and b are constants.
x 1 2 3 4 5
y 1.31 1.39 1.47 1.55 1.62
(a) Plot a linear graph to represent the data. [5 marks]
[ HINT : 1
x y
a b
+ = ~ equation of straight line in intercept form ]
(b) Use the graph from (a), find
(i) the value of a and b, (Ans : a = −3.5, b = 0.8)
(ii) the corresponding value of x, when y = 1.5. (Ans : 3.4)
[5 marks]

247
related by the equation y2
= m (x2
+ 2x) + n, where m and n are constants.
x 1 2 3 4 5
y 2.18 2.65 3.24 3.84 4.53
.
(a) Plot y2 against (x + 1)2, using a scale of 2 cm to 4 units on the (x + 1)-axis and 2
cm to 2 units on the y2-axis.
(b) Use the graph from (a), to find the value of
(i) m, (Ans : 0.4792 ~ 0.4971)
(ii) n. (Ans : 3.1971 ~ 3.2792)
[5 marks]
The table shows the values of displacement, s meter and time, t seconds of a moving particle, obtained from
an experiment. Given s and t are related by the equation s = ut + 2
2
1
at , where u and a are the initial
velocity and acceleration of the particle respectively. A pair of data is misrecorded because of a mistake.
Time, t seconds 20 50 80 110 140 180
Displacement, s meter 6 22.5 48 x 119 198
(a) Plot
t
s
against t, using a scale of 2 cm to 40 units on the t-axis and 2 cm to 0.1 unit on the
t
s
-
axis.
(b) Marks  on the graph to represent the exactly point of the misrecorded data. [1 mark]
(c) Use the graph from (a) to find the value of
(i) initial velocity, (Ans : 0.2)
(ii) acceleration, (Ans : 0.01)
(iii) x. (Ans : 82.5)
[5 marks]

248
The swing time of a pendulum with a length of L, is known to satisfy the non linear relation T = 2
L
g
,
where g is a constant. The table shows the corresponding values of L and T obtained from
an experiment.
L 20 40 60 80 100
T 0.95 1.28 1.58 1.81 2.04
(a) Plot T2
against L, by using a scale of 2 cm to 20 units on the L-axis and 2 cm to 1 unit on the
T2
-axis.
(i) the value of g, (Ans : 947.7 ~ 995.7)
(ii) the value of T, when L = 50cm, (Ans : 1.449)
(iii) the value of L, when T = 1.79s. (Ans : 77 ~ 78)
[6 marks]
Some load with a mass of m kg is hung at the end of a spring and swung vertically. The swing rate, f
swings per second, for each load is determined. The table shows the results of the experiment.
Mass of load, m kg 0.01 0.02 0.04 0.06 0.08
Swing rate, f swing per second 20 14 10 8 7
It is known that the rate of swing, f swings per second, and mass of load, m kg, is related by the
equation 2
1
f km = , whete k is a contant.
(a) Plot f 2
against
m
1
, by using a scale of 2 cm to 20 units on the
m
1
-axis and 2 cm to 50 units on
the f 2
(i) the swing rate for a load with a mass of 0.05kg, (Ans : 8.944)
(ii) the mass of the load that makes 15 swings per second, (Ans : 18)
(Give your answer correct to nearest gram)
(iii) the value of k. (Ans : 0.2558)
[5 marks]
(c) When the spring is replaced by another spring, it is found that the relation between f and m
becomes
2 1
2
f
m
= . Draw the graph that is formed on the same axis. [1 mark]

249
related by the equation y = ab−x
, where a and b are constants.
x 1 2 3 4 5 6
y 41.7 34.7 28.9 27.5 20.1 16.7
(a) Plot log10 y against x by using a scale of 2 cm to 1 unit on the x-axis and 2 cm to
0.2 unit on the log10 y -axis. Hence, draw the line of best fit. [4 marks]
(b) Use your graph from (a) to find
(i) and mark  on the graph to represent the exactly point of the misrecorded
data, then estimate a more accurate value for the data,
(Ans : 23.99)
(ii) the value of a and of b, (Ans : a = 50.12, b = 1.202)
(iii) the value of y when x = 3.5. (Ans : 26.3)
[6 marks]
related by the equation y = pqx − 1
, where p and q are constants.
x 3 4 5 6 7 8
y 12.1 6.46 3.47 1.89 0.95 0.53
(a) Plot log10 y against (x − 1), by using a scale of 2 cm to 1 unit on the (x − 1)-axis
and 2 cm to 0.2 unit on the log10 y -axis.
(b) From the graph paper in (a), find the value of
(i) p and q, (Ans : p = 42.66, q = 0.5346)
(ii) x when y = 5.0. (Ans : 4.4)
[6 marks]

250
related by the equation y = (1 + k) 2
c
x , where k and c are constants.
x 3.2 6.3 10 30 50 80 100
y 17.8 25 31.6 54.7 70.1 89.4 100
(a) Express the non linear equation to linear form. [2 marks]
(b) Plot a linear graph to represent the data. [4 marks]
(c) Use the graph from (a), to find the value of
(i) k, (Ans : k = 9)
(ii) c. (Ans : 1)
[4 marks]
At time t = 0, a bacteria colony has 1000 bakteria. The bacteria population of the colony, y at any time
t hours is given by the formula y = y0 ent
. The bacteria population of the colony at certain times t are
recorded in the table.
t 05 1 1.5 2 2.5
y 2718 7389 20086 54598
14841
0
(a) Show that y0 = 1000. [1 mark]
(b) Plot 10
log y against t by using a scale of 2 cm to 0.5 unit on the both axes.
(c) Use the graph from (b), find [ Use 10
log e = 0.4343 ]
(i) the value of n, (Ans : 1.996)
(ii) the minimum time taken for the bacteria population to exceed 100000.
(Ans : 2.3)
[5 marks]

251
The table shows the data obtained by Lea and Rowena in a Chemical experiment. The data show the
relationship between the reaction rate of a chemical of X mol s−1
and the temperature of T C.
Reaction rate of chemical, X mol s–1
05 22 40 59 78
Temperature, T C – 75 – 44 321 214 682
It is known that the temperature T and the reaction X are related by the equation 10 X
T Ab
+ = , where A
and b are constants.
(a) Write T + 10 = AbX
in the form of linear equation.
[ Ans : 10
log ( 10)
T + = 10
log A + ( )
10
log b X ] [1 mark]
(b) Plot 10
log ( 10)
T + against X by using a scale of 2 cm to 1 unit on the X-axis and 2 cm to 0.2 unit
on the 10
log ( 10)
T + -axis.
(c) Use the graph from (b), find
(i) the value of A and of b, (Ans : A = 2.0895, b = 1585)
(ii) the value of X when T = 0 C. (Ans : 3.45)
[5 marks]
The table shows the values of two variables, x and y, obtained from an experiment. The variables x and
y are related by the equation 2 2 2
2
y a x abx b
= + + , where a and b are constants.
x 2 4 6 8 10
y 40.23 25.81 14.58 6.54 1.68
(a) Express 2 2 2
2
y a x abx b
= + + in linear form, Y = mX + c. (Ans : y ax b
= + ) [2
marks]
[ HINT : (a + b)2
= a2
+ 2ab + b2
]
(b) Using a scale of 2 cm to 2 units on the x-axis and 2 cm to 1 unit on the y -axis, plot y against
x and draw the line of best fit. [4 marks]
(c) Use the graph from (b),
(i) estimate the value of a and b. (Ans : a = −0.6286, b = 7.6)
(ii) find the value of x when y = 17.64. (Ans : 5.4)
[4 marks]
[Kedah2020, No.11]

252
The table shows the values of two variables, x and y, obtained from an experiment. The variables x and
y are related by the equation
m
y
x n
=
+
, where m and n are constants.
x 0.1 1.5 2.5 3.5 4.5 6.5
y 3.9 2.4 1.8 1.5 1.3 1.0
(a) Based on the table, construct a table for the values of xy. [1 mark]
(b) plot xy against y, using a scale of 2 cm to 0.5 unit on the horizontal axis and 2 cm to 1 unit on
the vertical axis.
(i) m, (Ans : 2.085)
(ii) n, (Ans : 8.5)
(iii) the gradient of the straight line obtained if
1
y
is plotted against x. (Ans : 0.1176)
[6 marks]
[YIK2020, No.11]

253
COORDINATE
GEOMETRY
- WORKSHEET
Encik Suhairul bin Hadlee

255
WORKSHEET
TOPIC 7 : COORDINATE GEOMETRY
==========================================================================================================================================
Revision [ Distance, Midpoint, Gradient, Equation of straight line, Intersection point of
two straight lines ]
==========================================================================================================================================
 distance = 2 2
2 1 2 1
( ) ( )
x x y y
− + −
1 Given the points A (3, 3), B (5, −2), and C (−2, 5). Show that the triangle ABC is isosceles.
Answer :
2 Given P (1, 7), Q and R (6, 8) are three points on on a Cartesian plane. If the area of triangle PQR is
6 unit2
, find the perpendicular distance from Q to PR. (Ans : 2.353)
Answer :
3 The distance between points C (−k, 5k) and D (k, 4k) is 80 , find the values of k. (Ans : −4, 4)
Answer :
4 Given P (2, 4), Q (7, 3), and R (t, 6) are three points on on a Cartesian plane. If the length of PQ is half
the length of PR, find the possible values of t. (Ans : −8, 12)
Answer :

256
 midpoint, 1 2 1 2
( , ) ,
2 2
x x y y
x y
+ +
 
=  
 
5 Show that the line joining A (−2, 2) and B (6, 10) and the line joining C (0, 8) and D (4, 4) bisect one
another. [ Hint : bisect one another  have same midpoint ]
Answer :
6 Given A (2, 3), B (5, 4), C (6, 7), and D (h, k) are vertices of a parallelogram. Find the values of h
and k. (Ans : h = 3, k = 6)
Answer :
7 The coordinates of points A, B and C are (−3, 4), (−1, −2) and (k, 4) respectively. Find the possible
values of k if the distance between the midpoint of AB and C is 3 2 . (Ans : −5, 1)
Answer :
 gradient, 2 1 1 2
2 1 1 2
y y y y
m
x x x x
− −
= =
− −
8 Point (h, 3) lies on the straight line that joins the points (3, 9) and (−2, 6). Find the value
of h. (Ans : −7)
Answer :

257
9 Given points P(−5, m), Q (3, n), and R (11, 1) are collinear. Show that m − 2n + 1 = 0.
Answer :
MIND think :
10 The straight line 2y = 3x + h + 4 intersect the y-axis at 5k, where h and k are constants. Express h in
terms of k. (Ans : h = 10k − 4)
[2 marks] [2016, No.8]
Answer :
11 Given that the line 2x − 3y − 12 = 0 meets the y-axis at P. Express the equation of the line in gradient from
. Hence, state the coordinates of P. (Ans : y = 3
2
x − 4)
Answer :
m = vertical distance
horizontal distance
horizontal distance
vertical
distance
vertical
distance
horizontal distance
m = ( )
vertical distance
horizontal distance
−
 the equation of a straight line ~ gradient form, y = mx + c
m = gradient
c = y-intercept

258
12 The diagram shows a straight line PQ with the equation 1
2
10
=
+
k
y
x
.
(a) h, (Ans : 2)
(b) k. (Ans : −4)
[3 marks] [2012, No.14]
Answer :
(a) (b)
13 A straight line which passes through point (0, 8) has a gradient of 6, find the x-intersept
of the straight line. Hence, state the equation of the straight line in
intercept form. (Ans : 3
4
− ) [3 marks] [Forecast]
Answer :
14 Express the equation 6x − 7y − 5 = 0 in the intercept form. Hence, state the gradient of
the straight line. (Ans : 1
7
5
6
5 =
−
y
x
, 7
6
)
Answer :
 the equation of a straight line ~ intercept form, 1
x y
a b
+ =
a = x-intercept
b = y-intercept
-intercept
-intercept
y
m
x
 
= − 
 
y
x
P (5h, 0)
Q (0, −8)
O

259
 intersection point of two straight lines
15 The straight line that has a gradient of 2 and and passes through the point (4, −1)
intersect the straight line x + y + 4 = 0 at point Q. Find the coordinates
of point Q. [ Ans : ( 3
5
, − 3
17
) ]
Answer :
16 The straight line AB passes through the points (−6, −2) and (2, 6). The straight line CD has
a gradient of 3 and passes through the point (2, 8). Find the intersection point of the
straight line AB and CD. [ Ans : (1, 5) ]
Answer :
==========================================================================================================================================
7.1 Divisor of a line segment
7.1.1 Relate the position of a point that divides a line segment with the related ratio.
==========================================================================================================================================
17 Sketch the following situasi on the given line sements. Hence, state the ratio of the following case.
(a) B divides AC internally in the ratio 2 : 3. (b) Point B lies on AC such that AB : AC = 3 :
4.
(c) The straight line AB is extended to point C
such that its distance from point B is twice
the distance of AB.
(d) Point R divides the line segment PQ
such that 2PQ = 3RQ.
AB : AC =
BC : AC =
AB : BC = RQ : PR =

260
(e) A line segment such that
3
5
=
ML
KM
. (f) A line segment such that
3
8
PM PQ
= . .
==========================================================================================================================================
7.1.2 Derive the formula for divisor of a line segment on a Cartesian plane, and hence use the
formula in various situations.
==========================================================================================================================================
18 In the diagram, point P (x, y) is a point which divides line segment AB in the ratio m : n.
Show that 1 2 1 2
( , ) ,
nx mx ny my
P x y
m n m n
+ +
 
=  
+ +
 
.
Answer :
MIND think :
ML : KL = PM : MQ =
n
m
O
B (x2, y2)
x
A (x1, y1)
P (x, y)
y
• If m = n, P will become the of line segment AB.

261
19 The points A (2h, h), B (p, t) and C (2p, 3t) are on a straight line. B divides AC internally in the ratio 2
: 3. Express p in term of t. (Ans : p = −2t)
[3 marks] [2003, No.9]
Answer :
20 The diagram shows a straight line AC.
The point B lies on AC such that AB : BC = 3 : 1. Find the coordinates of B. [ Ans : ( 2
5
, 4
3
) ]
[3 marks] [2009, No.15]
Answer :
21 A straight line passes through A (−2, −5) and B (6, 7)
(a) Given C (h, 10) lies on the straight line AB, find the value of h. [HINT : collinear] (Ans : 8)
(b) Point D divides the line segment AB in the ratio 1 : 3. Find the coordinates of D.
[ Ans : (0, −2) ]
[4 marks] [2010, No.13]
Answer :
(a) (b)
O
y
x
C (4, 0)
B (h, k)
A (−2, 3)

262
22 A straight line passes through P (3, 1) and Q (12, 7). The point R divides the line segment PQ such that
2PQ = 3RQ. Find the coordinates of R. [ Ans : (6, 3) ]
[3 marks] [2017, No.18]
Answer :
23 The diagram shows the straight line PQ with equation
5
x
+
7
y
= 1 intersects the straight line AB at
point P.
(a) State the y-intercept of PQ.
(b) Find the coordinates of B if BP = 2PA. [ Ans : (9, 8) ]
[3 marks] [2014, No.12]
Answer :
(a)
(b)
24 It is given that A (1, 3) and B (4, 7) lie on a Cartesian plane.
(a) State the distance AB. (Ans : 5)
(b) The straight line AB is extended to point C such that its distance from point B is twice the
distance of AB. Find the coordinates of C. [ Ans : (10, 15) ]
[3 marks] [2019, No.13]
Answer :
(a)
(b)
A (3, −4)
O
x
y
Q
P (5, 0)
B

263
25 Point R (−3, 5) internally divides the line segment joining the points P (−6, 7) and Q (a, b) such that
1
4
PR PQ
= . Find the values of a and b. (Ans : a = 6, b = −1)
Answer :
26 The points A (−9, −2), B (h, 0), and C (3, 6) are on a straight line. Find
(a)
BC
AB
, (Ans : 3
1
)
(b) the value of h. (Ans : h = −6)
Answer :
(a) (b)
27 C (p, −1) is the point dividing the line joining A (−6, −5) and B (1, 9) internally in the ratio h : k. Find
(a) the values of h and k,
(Ans : h = 2, k = 5)
(b) the value of p.
(Ans : −4)
Answer :
(a) (b)

264
==========================================================================================================================================
7.1.3 Solve problems involving divisor of a line segment.
==========================================================================================================================================
28 The diagram shows the position of two bees, P and Q.
The coordinates of bee P and bee Q are (−4, 4) and (8, −12) respectively. Both bees fly towards each
other on a straight line with different velocity. The velocity of bee P is three times more than velocity of
bee Q. Find the distance of bee P from its initial point when it meets with bee Q. (Ans : 15)
[3 marks] [2015, No.12]
Answer :
29 The digram show three traffic lights A, B and C along a straight line on a Cartesian plane.
It is given that the distance from B to C is
5
8
times the distance from A to C. Find the coordinates of
traffic light C. [ Ans : (11,−3) ]
Answer :
O
x
Bee P
y
Bee Q
y
O
x
A (−5, 5)
C
B (1, 2)

265
30 The diagram shows the plan of a straight highway between two towns, P and Q on the Cartesian plane.
An engineer wants to build two rest houses between the two towns such that the two rest houses divide
the road into three equal parts of equal distance.
Find the coordinates of the two rest houses. [Ans : (1, 2.5), (6, 4) ]
[3 marks] [Clon textbook form 4]
Answer :
==========================================================================================================================================
7.2 Parallel lines and perpendicular lines
7.2.1 Make and verify conjectures about gradient of :
(i) parallel lines,
(ii) perpendicular lines
and hence, make generalisations.
==========================================================================================================================================
31 The diagram shows a line segment on a Cartesian plane.
Show that the gradient of a straight line, m = tanθ .
Answer :
y
O
x
P (−4, 1)
Q (11, 5.5)
O
(x2, y2)
x
(x1, y1)
y


266
32 Match the following straight lines with the correct value of tanθ .
Answer :
MIND think :
33 The diagram shows two straight lines, L1 and L2 on a Cartesian plane.
Given L1 is parallel to L2, show that m1 = m2, where m1 is the gradient of L1 and m2 is the gradient
of L2.
Answer :
O
x
O

0 <  < 90
 = 0  = 90 90 <  < 180

x
x
x
y y y
y
O
O
tan  = 0
tan  < 0 tan  > 0 tan  =  (undefined)
• m = , with  being the angle formed between the straight line and the postive ,
and     180.
O
x
y
1 2
L1 L2

267
34 The diagram shows two straight lines, L1 and L2 on a Cartesian plane.
Given L1 is perpendicular to L2, show that m1  m2 = −1, where m1 is the gradient of L1 and m2 is the
gradient of L2.
Answer :
MIND think :
 parallel lines
35 The equation of two straight lines are
3
x
+
4
y
= 2 and 3y = 1 − 4x. Determine whether the lines are parallel
to each other. (Ans : parallel)
Answer :
O
x
y
L1
L2
1
2
1
tanα =
tanθ
tan(θ 90 ) = tanα
+  −


HINT
• Two straight lines, L1 and L2 are parallel to each other 
• Two straight lines, L1 and L2 are perpendicular to each other 

268
36 The straight line
6
x
+
h
y
= 1 has a y-intercept of 2 and is parallel to the straight line y + kx = 0.
Determine the value of h and of k. (Ans : h = 2, k =
3
1
[3 marks] [2007, No.13]
Answer :
37 The straight line y = −3x + 8 is parallel to the straight line y = (k + 2) x + 7, where k is a constant.
Determine the value of k. (Ans : −5)
[2 marks] [2014, No.13]
Answer :
38 Given the points P (2, −3), Q (5, −1), R (−8, m), and S (−5, 4). Find the value of m if the straight lines
PQ and RS are parallel. (Ans : 2)
Answer :

269
39 The diagram shows three straight lines, such that k, p, q and r are constants.
Express
(a) k in terms of q, (Ans : k = 6
q
− )
(b) r in terms of k and p. (Ans : r =
3p
k
− )
[3 marks] [2019, No.14]
Answer :
(a) (b)
40 The straight line which joining the points (2k, −k) and (2, −4) is parallel to the straight line y = 1 − 3x.
Find the value of k. (Ans : 5
2
)
Answer :
41 The table shows three equations of straight line, KL, MN and PQ.
Straight Line Equation
KL 3x − 4y = 7
MN y =
4
3
x − 18
PQ 3x − 2y − 17 = 0
Which straight line did not intersect to each other ?
Give reason for your answer.
Answer :
y
x
ky = 3x + 6
1
x y
p q
− =
O
r

270
 perpendicular lines
42 The equation of two straight lines are 1
3
5
=
+
x
y
and 5y = 3x + 24. Determine whether the lines are
perpendicular to each other. (Ans : perpendicular)
[3 marks] [2003, No.11]
Answer :
43 The following information refers to the equations of two straight lines, JK and RT, which are perpendicular
to each other.
JK : y = px + k
RT : y = (k −2) x + p
where p and k are
constants.
Express p in terms of k. (Ans : p = k
−
2
1
)
[2 marks] [2005, No.14]
Answer :
44 The following information refers to the equation of two straight lines, AB and CD.
AB : y − 2kx − 3 = 0
CD :
h
x
3
+
4
y
= 1
where h and k are constants.
Given the straight lines AB and CD are perpendicular to each other, express h in terms
of k. (Ans : h = 3
8
k)
[3 marks] [2018, No.10]
Answer :

271
45 The straight line y = 2x − 4 is perpendicular to the line segment which joins the points P and Q.
Find the value of h (Ans : 3
1
5 )
Answer :
46 If the straight line y = 8x − 6 is perpendicular to the straight line which joins points (2, 3) and (4, p), find
the value of p. (Ans : 4
11
)
Answer :
47 The coordinate of three points P, Q, and R are (1, 1), (2, m), and (5, −1) respectively. If PQR = 90,
find the possible values of m. (Ans : 2)
Answer :
48 Two points have coordinates E (2, 4) and F (8, 6). If C (2, t) lies on the perpendicular bisector of EF, find
the value of t. (Ans : 14)
Answer :
P
8
0,
3
 
 
 
y = 2x − 4
O
x
y
Q (h, 0)

272
49 The coordinates of the points A, B, and C are (−4, 2), B (−1, 4), and C (3, −2) respectively. Show that
ABC is right-angled triangle.
Answer :
50 The line y = px + q is perpendicular to the line y − 2x = 5 and passes through the point (1, −5). Find the
values of p and q. (Ans : p = 2
1
− , q = 2
1
4
− )
Answer :
==========================================================================================================================================
7.2.2 Solve problems involving equations of parallel and perpendicular lines.
[ the use of dynamic software is encouraged ]
==========================================================================================================================================
 the equation of a straight line that passes through a fixed point and parallel to a
given line
51 Find the equation of the line that passing through the point (−4, 8) and is parallel to the line which joins
points (3, 5) and (4, −1). (Ans : y = −6x − 16)
Answer :
52 Find the equation of the straight line which passes through the point P (−3, 6) and is parallel to the straight
line 4x − 2y + 1 = 0. (Ans : y = 2x + 12)
Answer :

273
53 Find the equation of the line that is parallel to x + 3y = 2 and passing through the x-intercept of
1
5
4
=
−
y
x
. (Ans : y = 3
4
3
1
+
− x )
Answer :
54 Find the equation of the straight line that is parallel to 5x + 2y = 8 and bisect the line segment joining the
points (3, 1) and (−1, −5). (Ans : y = x
2
5
− + 2
1
)
Answer :
 the equation of a straight line that passes through a fixed point and
perpendicular to a given line.
55 The diagram shows a straight line PQ with the equation
2
x
+
3
y
= 1. The point P lies on the x-axis and
the point Q lies on the y-axis .
Find the equation of the straight line that is perpendicular to PQ and passing through
the point Q. (Ans : y = 3
2
x + 3)
[3 marks] [2004, No.14]
Answer :
P
Q
O
x
y

274
56 Find the equation of the straight line that passes through point (−1, 9) and perpendicular to the line
3
x
−
6
y
= 1. (Ans : y = 2
17
2
1
+
− x )
Answer :
57 Given the points A (3, 3), B (5, −2), and C (−2, 5). Find the equation of the straight line
that passes through point A and perpendicular to the straight line BC.
(Ans : y = x)
Answer :
58 The coordinates of point A and point B are (−2, 6) and (4, 2) respectively. Point P divides AB internally
in the ratio 3 : 1. Find the equation of the straight line which passes through point P and perpendicular
to the straight line AB. (Ans : y = x
2
3
− 4
3
)
Answer :

275
 the equation of the perpendicular bisector of a straight line
59 A straight line 1
6
2
=
+
y
x
cuts the x-axis at P and y-axis at Q. Find
(a) the gradient of the straight line, (Ans : −3)
(b) the equation of the perpendicular bisector of the straight line. (Ans : 3
8
3
1 +
= x
y )
[3 marks] [2011, No.13]
Answer :
(a) (b)
60 The diagram shows a straight line AB.
Find
(a) the midpoint of AB, [ Ans : (7, 5) ]
(b) the equation of the perpendicular bisector of AB. (Ans : y = −4x +33)
[4 marks] [2012, No.13]
Answer :
(a) (b)
61 Given ABCD is a rhombus with vertices A (−1, 1) and C (5, 7). Find the equation of the straight
line BD. (Ans : y = −x + 6)
Answer :
O
x
y
A (−1, 3)
B (15, 7)

276
 the intersection point of two straight lines which perpendicular to each others
62 The diagram shows the straight line AB which is perpendicular to the straight line CB at the point B.
The equation of the straight line CB is y = 2x − 1. Find the coordinates of B.
[ Ans : (2, 3) ]
[3 marks] [2006, No.12]
Answer :
63 The diagram shows two straight lines on a Cartesian plane.
(a) State the value of q. (Ans : − 3
1 )
(b) Find the coordinates of F. [ Ans : (−3, −5) ]
[3 marks] [2016, No.9]
Answer :
(a) (b)
B
O
x
y
A (0, 4)
C
y = 3x + 4
O
y
x
y = qx − 6
F

277
64 Find the equation of the straight line passing through point (4, 5) and perpendicular to
the line 1
2
4
=
+
y
x
. These two lines intersect at P. Find the coordinates of P.
[ Ans : y = 2x − 3, (2, 1) ]
Answer :
 problems involving equations of parallel and perpendicular lines.
65 The diagram shows the locations of the school, hall and park drawn on a Cartesian plane.
A clock tower will be built such as its distance from school and hall are the same, and
the closest to the park. Find the coordinates of the clock tower.
[ Ans : (200, −300) ]
Answer :
O
School (−400, 150)
Park (300, −250)
x
y
Hall (200, 450)

278
==========================================================================================================================================
7.3 Areas of polygons
7.3.1 Derive the formula of area of triangles when the coordinates of each vertex are known.
[ the use of digital technology is encouraged ]
7.3.4 Make generalisation about the formula of area of polygons when the coordinates of each vertex
are known, and hence use the formula to determine the area of polygons.
==========================================================================================================================================
66 The diagram shows two triangles, ABC and ACD, drawn on a Cartesian plane.
(a) Show that :
(b). Based on (a), make a conclusion by induction for the area of a polygon with n sides.
Answer :
MIND think :
Shoelace Algorithm :
O
x
A (x1, y1)
B (x2, y2)
C (x3, y3)
y
D (x4, y4)
• area of  =
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
= 1 2 2 3 3 1 1 2 2 3 3 1
1
( ) ( )
2
x y x y x y y x y x y x
+ + − + +
(+) value ~ coordinates of vertices are arranged anticlockwise
(−) value ~ coordinates of vertices are arranged clockwise
~ If three points A, B and C are collinear → area of ABC =
• area of quadrilateral =
3
1 2 4 1
3
1 2 4 1
1
2
x
x x x x
y
y y y y
= 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 1
1
( ) ( )
2
x y x y x y x y y x y x y x y x
+ + + − + + +
• area of polygon with n-sides =
3
1 2 1
3
1 2 1
. . .
1
. . .
2
n
n
x x
x x x
y y
y y y
= 1 2 2 3 3 4 1 1 2 2 3 3 4 1
1
( . . . + ) ( . . . + )
2
n n
+ + + − + + +
(i) area of  ABC = 1 2 2 3 3 1 1 2 2 3 3 1
1
2
x y x y x y y x y x y x
+ + − − − 
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
(ii) hence, area of quadrilateral ABCD = 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 1
1
2
+ + + − − − −

279
==========================================================================================================================================
7.3.2 Determine the area of triangles by using the formula.
7.3.3 Determine the area of quadrilaterals by using the formula.
==========================================================================================================================================
 area 1
67 By considering the area of triangle PQR, show that points P (2, 5), Q (−1, −1) and R (−4, −7) are
collinear.
Answer :
68 By considering the area of quadrilateral PQRS, determine whether P (−5, 5), Q (−3, −2), R (−1, 1), and
S (3, 7) are collinear.
Answer :
69 The diagram shows a parallelogram OABC with A (0, 5) and B (4, 7).
Find the area of parallelogram OABC. (Ans : 20)
Answer :
A
B
C
O
x
y

280
70 The diagram shows a parallelogram PQRS.
Find the area of parallelogram PQRS. (Ans : 24)
Answer :
71 Find the area of a pentagon PQRST with vertices P (−2, 1), Q (1, 5), R (4, 4), S (6, −1)
and T (2, −3). (Ans : 37.5)
Answer :
 area 2
72 The vertices of a triangle are A (5, 2), B (4, 6) and C (p, −2). Given that the area of the triangle is 30 units2
,
find the values of p. (Ans : −9, 21)
[3 marks] [2007, No.14]
Answer :
73 The points (0, 3), (2, t) and (−2, −1) are the vertices of a triangle. Given that the area of the triangle is
4 units2
, find the values of t. (Ans : 3, 11 )
[3 marks] [2008, No.14]
Answer :
P
O
Q (4, 6)
R (7, 5)
S (4, −2)
x
y

281
74 The points P (k, 3), Q (9, k), R (2k, 7), and S (−3, 3k + 2) are vertices of a quadrilateral. Find the value
of k if the area of PQRS is 30 unit2
, where k > 0, and k is an integer. (Ans : 2)
Answer :
75 The area of a parallelogram with vertices A (−1, −2), B (1, −1), C (3, 4), and D (p, q) is 23 unit2
. Show
that 2q − 3p = 18.
Answer :
==========================================================================================================================================
7.3.5 Solve problems involving areas of polygons.
==========================================================================================================================================
76 The diagram shows the position of three campsites A, B and C at a park of a riverbank drawn on a Cartesian
plane, such that A and B lie on the same straight riverbank.
Shah wants to cross the river from campsite C to the opposite riverbank where the campsites A and B
located. Find the shortest distance, in m, that he can take to cross the river. Give your answer correct
to four decimal places. (Ans : 7.1587)
[4 marks] [2018, No.23]
Answer :
O
C (4, −3)
B (7, 5)
A (−2, 3)
x (m)
y (m)

282
77 Given that the straight line 2y = x + 6 intersects x-axis and y-axis at point A and point B respectively.
If C is (−4, 5), find the area of triangle ABC. (Ans : 12)
Answer :
78 Given that the straight line 2x + y − 8 = 0 intersects the straight line y = k, x-axis and y-axis at points A, B
and C respectively. If the area of  OAC is 12 unit2
, find the value of k. (Ans : 2)
[3 marks] [clon textbook form 4]
Answer :
79 The digram shows a triangle ABC with an area of 18 unit2
. The equation of straight line BC is y − x + 1 =
0.
Find the coordinates of point B. [ Ans : (3, 2)]
Answer :
B
x
y
A (−2, 3)
C (−3, −4)
O

283
==========================================================================================================================================
7.4 Equations of loci
7.2.1 Represent graphically, the locus that satisfies these conditions :
(i) the distance of a moving point from a fixed point is constant,
(ii) the ratio of a moving point from two fixed points is constant,
and hence determine the equation of the locus.
==========================================================================================================================================
 graphical of a locus
80 In the answer space, choose P, Q or R, which represents each of the following locus.
P Q R
Answer :
MIND think :
Equation of circle with centre (a, b) and radius r units → 2 2 2
( ) ( )
x a y b r
− + − =
(x, y)
r
(x1, y1)
r m
(x, y)
(x1, y1) (x2, y2)
n
m
n
(x, y) (x2, y2)
(x1, y1)
(a) Locus of a moving point which the distance from a fixed point is always constant.
(b) Locus of a moving point which the distance from two fixed points is always constant
in the ratio m : n.
(c) Locus of a moving point which the distance from two fixed points is always the same.

284
 locus of a moving point which the distance from a fixed point is always constant
81 Point P moves such that its distance is always 5 unit from Q (−3, 4).
(a) Describe fully the locus of P. **
(b) Find the equation of the locus of P. (Ans : x2
+ y2
+ 6x − 8y = 0)
[3 marks] [2010, No.14]
Answer :
(a)
(b)
82 Find the equation of locus of the moving point P such that its distance from the midpoint of A (2, 0) and
B (4, −2) is 3 unit. (Ans : x2
+ y2
− 6x + 2y + 1 = 0)
Answer :
83 A point P moves along the circumference of a circle with centre A (2, 3). The circumference passes through
point Q (−2, 0). Find the equation of the locus of P. (Ans : x2
+ y2
− 4x − 6y − 12 = 0)
Answer :
84 The points P (2, 6), Q (−4, −2) and R lie on the circumference of the circle with diamerer PQ. Find the
equation of the locus of the moving point R. (Ans : x2
+ y2
+ 2x − 4y − 20 = 0)
Answer :

285
85 Given the points A (4, 6) and B (2, 4). Find the equation of the locus of the moving point P such that the
triangle APB always has a right angle at P. (Ans : x2
+ y2
− 6x − 10y + 32 = 0)
Answer :
86 The coordinate of points A and B are (−2, 1) and (5, −6) respectively. Find the equation of the locus of
the moving point R such that ARB is always 90. (Ans : x2
+ y2
− 3x + 5y − 16 = 0)
Answer :
87 Given the locus of P with centre A (h, k) and diameter 13 units is x2
+ y2
− 5x − 12y = 0. Find the
coordinates of point A. [ Ans : ( 2
5
, 6) ]
Answer :
88 The diagram show a locus P (x, y) which moves such that its distance from point M (0, h) is always equal.
Given the equation of the locus of P is x2
+ y2
− 10y = 0. Find the value of h. (Ans : 5)
Answer :
P (x, y)
M
x
y
O

286
 locus of a moving point which the distance from two fixed points is always
constant in the ratio m : n.
89 The point A is (−1, 3) and the point B is (4, 6). The point P moves such that PA : PB = 2 : 3. Find the
equation of the locus of P. (Ans : 5x2
+ 5y2
+ 50x − 6y − 118 = 0)
[3 marks] [2004, No.15]
Answer :
90 A point R moves such that its distance from the points A (−2, 3) and B (5, −1) are in the ratio 2 : 1. Find
the equation of the locus of R. (Ans : 3x2
+ 3y2
− 44x + 14y + 91 = 0)
Answer :
91 The coordinates of point A and B are (−3, −5) and B (1, −2) respectively. A point P moves such that
2AP = 3PB. Find the equation of the locus of P.
(Ans : 5x2
+ 5y2
− 42x − 4y − 91 = 0)
Answer :
92 Find the locus of the moving point M such that its distance from point A (1, −1) is three time its distance
from point B (−2, 3). (Ans : 8x2
+ 8y2
+ 38x − 56y + 115 = 0)
Answer :

287
93 A point R moves from two fixed points P (1, 0) and Q (−2, 3) such that RP =
2
1
RQ. Find the equation
of the locus of R. (x2
+y2
−4x + 2y − 3 = 0)
Answer :
94 Given the points P (1, −3) and Q (3, −1). Find the equation of the locus of Z if PZ = 2RZ such that PQ
= 2QR and PQR is a straight line. (Ans : x2
+ y2
− 10x − 2y + 18 = 0)
Answer :
 locus of a moving point which the distance from two fixed points is always the
same.
95 The diagram shows a straight line passing through S (3, 0) and T (0, 4)
(a) Write down the equation of the straight line ST in the form
a
x
+
b
y
= 1.
(b) A point P (x, y) moves such that PS = PT.
Find the equation of the locus of P. (Ans : 6x − 8y + 7 = 0)
[4 marks] [2008, No.13]
Answer :
(a)
(b)
x
y
S (3, 0)
O
T (0, 4)

288
96 The diagram shows the straight line 2y = 3x + 4 on a Cartesian plane.
A point T moves such that its distance from point P and point Q are equal.
(a) Describe fully the locus of P.
(b) Find the equation of the locus of T. (Ans : 6x + 9y − 5 = 0)
Answer :
(a)
(b)
97 A (6, 1) and B (2, 4) are two fixed points. A point C moves such that CAB = CBA. Find the equation
of the locus of C. (Ans : 8x − 6y − 17 = 0)
Answer :
Q
2y = 3x + 4
P
O
x
y

289
==========================================================================================================================================
7.2.2 Solve problems involving equations of loci.
[ usage of dynamic geometry software need to be involved ]
==========================================================================================================================================
98 A point R moves such that its distance from the point M (2, 0) is equal to its distance from the line x =
−2. Find the equation of the locus of R. (Ans : y2
= 8x)
Answer :
99 A point P moves such that its distance from a point A (2, 1) and its distance to the x-axis is always
equal. Find the equation of the locus P. (x2
− 4x − 2y + 5 = 0)
Answer :
100 A point M moves such that its distance from the point R (
2
3
− , 0) is twice its distance from the y-axis. Find
the equation of the locus of M. (Ans : 12x2
− 4y2
− 12x − 9 = 0)
Answer :
101 A point P moves such that the sum of distance OP and PN is always 2 units. O is the origin, N lies on
x-axis and PN is parallel to y-axis. Show that the equation of the locus of P is y = 1 −
2
4
1
x .
Answer :

290
102 A money is tied to one vertex of the cage which measures 1 m  1 m with a rope. The length of the rope
is
1
2
2
m. On the diagram in the answer space, sketch the locus if the monkey moves clockwise around
the cage with the rope taut.
Answer :
103 A point T moves such that its distance from point (3, 1) is 2
2 units. Show that the straight line y − x
= 2 is the tangent of locus of T. (Ans : x2
+ y2
− 6x − 2y + 2 = 0)
Answer :
104 Given points A (1, 3) and B (4, 1). A point P moves such that its distance from B is twice its distance
from A. Show that the locus of P does not intersect x-axis. (Ans : 3x2
+ 3y2
− 22y + 23 = 0)
Answer :
R
Q
S
1 m
P
1 m
1
2
2
m

291
105 The diagram shows a field in the shape of a rectangle, PQRS which measures 7 m  6 m.
Rachel wants to move from one point on the PS so the distance from points A and B
is always the same.
(a) Find the distance from P where she should start. (Ans : 5
4
)
(b) State the pattern of her journey.
Answer :
(a)
(b)
106 The diagram shows the x-axis and y-axis which represent th floor and wall respectively.
A ladder, AB of length 10 m leaning against the wall touches the floor and wall at
points A (a, 0) dan B (0, b) respectively.
(a) Write the equation which relates a and b.
(b) Given R (x, y) is a point on the ladder AB such that the ratio AR : RB = 1 : 3.
Show that the locus of point R when both ends of the ladder slide along the x-axis
and y-axis is 100
16
9
16 2
2
=
+ y
x .
Answer :
(a)
(b)
R
Q
P
S
2 m
B
A
2 m
6 m
7 m
4 m
3 m
O
x
B (0, b)
y
R (x, y)
A (a, 0)

292
PAPER 2
 Part A ~ 6 – 8 marks
107 The diagram shows a straight line CD which meets a straight line AB at point D. The point C lies on the
y-axis.
(a) Write down the equation of AB in the form of intercepts. (Ans : 1
6
9
=
−
y
x
) [1 mark]
(b) Given that 2AD = DB, find the coordinates of D. [Ans : (3, −4) ] [2 marks]
(c) Given that CD is perpendicular to AB, find the y-intercept of CD. (Ans : 2
1
) [3 marks]
[2004, No.2]
Answer :
108 Solutions by scale drawing will not be accepted.
In the diagram, the straight line AB has an equation y + 2x + 8 = 0. AB intersects the x-axis at point A
and intersects the y-axis at point B.
Point P lies on AB such that AP : PB = 1 : 3. Find
(a) the coordinates of P, [ Ans : (−3, −2) ] [3 marks]
(b) the equation of the straight line that passes through P and perpendicular to AB.
(Ans : y = 2
1
2
1
−
x ) [3 marks][2007, No.2]
Answer :
O
D
C
x
B (9, 0)
y
A (0, −6)
B
P
A
O
x
y
y + 2x + 8 = 0

293
The diagram shows the straight line AC which intersects the y–axis at the point B.
The equation of AC is 3y = 2x − 15. Find
(a) the equation of the straight line which passes through point A and perpendicular
to AC. (Ans : y = x
2
3
− − 2
23
) [4 marks]
(b) (i) the coordinates of B, [ Ans : (0, −5) ]
(ii) the coordinates of C, given AB : BC = 2 : 7. [ Ans : ( 2
21 , 2) ]
[3 marks][2010, No.5]
Answer :
The diagram shows a straight line AB.
(a) Calculate the area of triangle AOB. (Ans : 9) [2 marks]
(b) Point C divides the straight line AB internally in the ration AC : CB = 3 : 2. Find coordinates
of C. [ Ans : ( 5
12 , 5
2 ) ] [2 marks]
(c) Point P moves such that its distance from A is always twice its distance from B. Find the equation
of the locus of P. (Ans : x2
+ y2
− 18x + 8y + 45 = 0 ) [3 marks]
[2011, No.5]
Answer :
O
C
B
x
y
A (−3, −7)
O
x
y
A (−3, 4)
B (6, −2)

294
111 Solutions by scale drawing is not accepted.
The diagram shows a campsite OPQR that has been cleared by a group of scouts. Points A, B and C
are the centre of the tents A, B and C. A, B and C are colliner.
Given the distance of point C is 3 times from point B and 4 times from point A.
(a) Find
(i) the area, in m2, of the campsite OPQR,
(Ans : 94)
(ii) the coordinates of A.
[ Ans : (−1, 2) ]
[4 marks]
(b) A scout spread sulphur powder around tent C such that the distance of the
sulphur powder track from the centre of tent C is always 3 m.
Find the equation of the track of the sulphur powder.
(Ans : x2 + y2 −22x − 20y + 212 = 0) [3 marks]
[2015, No.3]
Answer :
x (m)
Q (14, 8)
R (15, 1)
C (11, 10)
B (2, 4)
P (−5, 3) A
O
y (m)

295
The diagram shows the locations of town A and town B draw on a Cartesian plane.
PQ is a straight road such that the distance from town A and town B to any point on the road is always
equal.
(a) Find the equation of PQ.
(Ans : 3x + y + 3 = 0 / y = −3x − 3) [3 marks]
(b) Another straight road, ST with an equation y = 2x + 7 is to be built.
(i) A traffic light is to be installed at the crossroads of the two roads. Find the
coordinates of the traffic light. [ Ans : (−2, 3) ]
(ii) Which of the two roads passes through town C 





− 1
,
3
4
?
[4 marks]
[2017, No.5]
Answer :
x
Town B
(2, 1)
Town A
(−4, −1)
O
y
P
Q

296
113 Solution by scale drawing is not accepted
The diagram shows a triangle OAB.
(a) Given the area of triangle OAB is 30 unit2, find the value of k.
(Ans : 6) [2 marks]
(b) Find the perpendicular distance from O to AB. ** (Ans : 4.472) [2 marks]
(c) Point Q (2, 4) lies on the straight line AB.
(i) Find AQ : QB. (Ans : 2 : 1)
(ii) Point P moves such that PB = 2PQ.
Find the equation of the locus P. (Ans : 3x2 + 3y2 − 4x − 28y + 40 = 0)
[5 marks]
[2018, No.3]
Answer :
x
A (−6, 8)
B (k, 2)
O

297
 Part B ~ 10 marks
114 Solutions to this question by scale drawing will not be accepted.
A point P moves along the arc of a circle with with centre A (2, 3). The arc passes through Q
(−2, 0) and R (5, k).
(a) Find
(i) the equation of locus of the point P,
(Ans : x2
+ y2
− 4x − 6y − 12 = 0)
(ii) the values of k.
(Ans : −1, 7)
[6 marks]
(b) The tangent of the circle at point Q intersects the y-axis at point T. Find the area of triangle
OQT.
(Ans : 3
2
2 ) [4 marks]
[2003, No.11]
Answer :

298
115 Solution to this question by scale drawing will not be accepted
(a) Find
(i) the equation of the straight line AB,
(Ans : y = 2x + 17) [2 marks]
(ii) the coordinates of point B.
[ Ans : (−8, 1) ] [3 marks]
(b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find the
coordinates of D.
[ Ans : (−14, −11) ] [2 marks]
(c) A point P moves such that its distance from point A is always 5 units. Find the equation
of the locus of P.
(Ans : x2
+ y2
+ 8x − 18y + 72 = 0) [3 marks]
[2005, No.9]
Answer :
B
2y + x + 6 = 0
x
y
O
A (−4, 9)

299
The diagram shows the triangle AOB where O is the origin. Point C lies on the straight line
AB.
(a) Calculate the area, in unit2
, of triangle AOB. (Ans : 9) [2 marks]
(b) Find the shortest distance from O to AB. ** (Ans : 1.664) [2 marks]
(c) Given that AC : CB = 3 : 2, find the coordinates of C. [ Ans : ( 5
12
, 5
2
) ] [2 marks]
(d) A point P moves such that its distance from point A is always twice its distance from
point B.
(i) Find the equation of the locus of P. (Ans: x2
+ y2
− 18x + 8y + 45 = 0)
(ii) Hence, determine whether or not this locus intercepts the y-axis. (Ans : no)
[6 marks]
[2006, No.9]
Answer :
C
O
y
x
B (6, −2)
A (−3, 4)

300
The diagram shows a triangle OPQ. Point S lies on the line PQ.
(a) A point W moves such that its distance from point S is always 2
2
1
units. Find the
equation of the locus of W.
(Ans : 4x2
+ 4y2
−24x − 8y + 15 = 0) [3 marks]
(b) It is given that point P and point Q lies on the locus of W. Calculate
(i) the value of k,
(Ans : 3)
(ii) the coordinates of Q.
[ Ans : ( 2
1
4 , −1) ]
[5 marks]
(c) Hence, find the area, in unit2
, of triangle OPQ. (Ans : 2
15
)
[2 marks]
[2008, No.10]
Answer :
Q
x
P 





k
,
2
3
S (3, 1)
y
O

301
The diagram shows a trapezium OABC. The line OA is perpendicular to the line AB, which
intersect the y–axis at the point Q. It is given that the equation of OA is y = x
2
3
− and the
equation of AB is 6y = kx + 26.
(a) Find
(i) the value of k,
(Ans : 4)
(ii) the coordinates of A.
[ Ans : (−2, 3) ]
[4 marks]
(b) Given AQ : QB is 1 : 2, find
(i) the coordinates of B, [Ans : (4, 7)]
(ii) the equation of the straight line BC. (Ans : y = x
2
3
− + 13)
[4 marks]
(c) A point P (x, y) moves such that 2PA = PB. Find the equation of the
locus of P. (Ans : 3x2
+ 3y2
+ 24x − 10y − 13 = 0) [2 marks]
[2009, No.9]
Answer :
C
Q
B
3
2
y x
= −
A
O
6y = kx + 26
x
y

302
The diagram shows a quadrilateral PQRS. The straight line PQ is perpendicular to the straight
line PS. Point T lies on the straight line PS.
Find
(a) the equation of the straight line PS, (Ans : y = −2x + 8) [3 marks]
(b) the coordinates of S, [ Ans : (5, −2) ] [2 marks]
(c) the coordinates of T if PT : TS = 1 : 3, [ Ans : (2, 4) ] [2 marks]
(d) the coordinate of R if the area of quadrilateral PQRS is 30 unit2
.
[ Ans : (7, 5) ] [3 marks]
[2012, No.10]
Answer :
R
7x − 2y = 39
S
T
O
Q (5, 8)
P (1, 6)
y
x

303
The diagram shows a rectangle ABCD. The equation of the straight line AB is y = 2x + 3.
Find
(a) the equation of the straight line DC, (Ans : y = 2x − 7) [2 marks]
(b) the equation of the straight line AD, (Ans : y = x
2
1
− + 2
1 ) [3 marks]
(c) the coordinates of D, [ Ans : (3, −1) ] [2 marks]
(d) the area, in unit2
, of rectangle ABCD. (Ans : 30) [3 marks]
[2013, No.9]
Answer :
y = 2x + 3
x
y
B
A (−1, 1)
O
D
C (6, 5)

304
121 The diagram shows a quadrilateral PQRS. Point R lies on the y-axis
The equation of a straight line PS is 2y = 5x − 21.
(a) Find
(i) the equation of straight line PQ, (Ans : y = − 5
2 x − 5
9
)
(ii) the coordinates of P. [ Ans : (3, −3) ]
[6 marks]
(b) A point T moves such that its distance from point S is always 5 units. Find the equation
of the locus of T. (Ans : x2
+ y2
−10x − 4y + 4 = 0) [4 marks]
[2014, No.10]
Answer :
Q (−2, −1)
S
O
P
R
y
x
y = 2

305
122 Solution by scale drawing is not accepted.
The diagram shows the path of a moving point P (x, y). P always moves at a constant distance
from point A.
B (−1, −2) and R (−5, q) lie on path of point P. The straight line BC is a tangent to the path
and intersects the x-axis at point C. Find
(a) the equation of the path of point P, (Ans : x2
+ y2
+ 4x − 2y − 5 = 0)
[3 marks]
(b) the possible values of q, (Ans : 0, 2) [2 marks]
(c) the area of ABC. (Ans : 10) [5 marks]
[2019, No.9]
Answer :
x
y
A (−2, 1)
P (x, y)
O
B

306
FORECAST
123 The diagram shows the surface of a rectangular wall, PQRS which measures 4 m  3m.
Syafiqah want to mark a point A so that the lamp can be installed at that position exactly. She
has made the measurement using two ropes QT and PU which intersect at points A that
required. Find
(a) the distance ST and TU so that the two ropes intersect at A, (Ans : ST = 2
5
, TU = 4
5
)
[4 marks]
(b) the ratio of TA : TQ, (Ans : 1 : 6) [2 marks]
(c) the area of quadrilateral AQRU. (Ans : 5.2) [2 marks]
Answer :
P
R
A
T U
Q
S
1 m
0.5 m
4 m
3 m

307
The diagram shows the points A (−2, 4), B (1, −1), C (6, 2) and D on a Cartesian plane.
It is given that lines AD and BC are parallel and angle ACD = 90. Find
(a) the equation of the line AD and CD, (Ans : AD : 5
26
5
3
+
= x
y , CD : y =4x −22)
[5 marks]
(b) the coordinates of point D, [ Ans : (8, 10) ] [2 marks]
Answer :
O
x
y
A (−2, 4)
D
B (1, −1)
C (6, 2)

308
The diagram shows a rectangle PQRS.
Given that the equation of the straight line PR is y = 2 + x. Point T lies on the straight line PR
such that PT : TR = 2 : 1. Find
(a) the equation of the straight line SR, (Ans : 3
16
3
1
+
= x
y ) [3 marks]
(b) the coordinates of point T, [ Ans : ( 3
10
, 3
16
) ] [3 marks]
(c) the area of triangle PST. (Ans : 3
20
) [2 marks]
Answer :
O
R
x
y
Q (6, 4)
S (−1, 5)
P
T

309
In the diagram, APB and CPD are straight lines.
Given that P is the midpoint of AB, and CD : PD = 4 : 3. Find
(a) the coordinates of point P, [ Ans : (3, 4) ] [1 mark]
(b) the coordinates of point D, [ Ans : ( 2
9
, 1) ] [2 marks]
(c) the coordinates of the intersection point of the straight line AC and DB that are
produced. [ Ans : ( 2
7
, 13) ] [5 marks]
Answer :
B (4, 7)
A (2, 1)
x
y
D
5
, 5
2
C
 
 
 
O
P

310
 Part B ~ 10 marks
The diagram shows a rhombus on a Cartesian plane. Diagonal AC and BD intersect at point
(3, 6).
(a) Find
(i) the coordinates of point B, [ Ans : (−3, 8) ] [1 mark]
(ii) the equation of the line AC, (Ans : y = 3x − 3) [2 marks]
(iii) the equation of the line CD. (Ans : y = −2x + 22) [3 marks]
(b) A point P moves such that its distance from point A is always 4 units. Find the equation
of the locus of point P. (Ans : x2
+ y2
− 2x − 15 = 0 ) [2 marks]
Answer :
x
D (9, 4)
C
B
A
O
y

311
The diagram shows the vertices of a kite KLMN on a Cartesian plane. LM is parallel to the y-
axis.
Find
(a) the coordinates of the point M, [ Ans : (−1, −2) ] [4 marks]
(b) the area of the kite KLMN, (Ans : 15) [3 marks]
(c) the equation of the locus of point P such that KP = 2KL.
(Ans : x2
+ y2
− 4x − 8y − 20 = 0) [3 marks]
Answer :
K (2, 4)
N (3, 1)
O
M
L (−1, 3)
x
y

312
In the diagram, ABC is a triangle such that ABC = 90.
Find
(a) the value of t, (Ans: 8) [2 marks]
(b) the area of triangle ABC, (Ans : 6) [2 marks]
(c) the perpendicular distance from B to AC, (Ans: 2.353) [3 marks]
(d) the equation of the straight line that passes through point B and perpendicular to the
straight line AC. (Ans: y = −5x + 20) [3 marks]
Answer :
O
x
y
A (1, 7) C (6, t)
B (3, 5)

313
The diagram shows a triangle ABC that has the vertices A (8, 5), B (4, −3), and C (−3, 4). AD
is perpendicular to BC, and CE is perpendicular to AB. AD and CE intersect at point H.
Find
(a) the coordinates of point D, [ Ans : (2, −1) ] [3 marks]
(b) the coordinates of point H, [ Ans : ( 3
11
, 3
2
) ] [3 marks]
(c) the ratio AD : HD, (Ans : 18 : 5) [2 marks]
(d) the area of triangle AHC. (Ans : 3
2
21 ) [2 marks]
Answer :
B (4, −3)
C (−3, 4)
A (8, 5)
O
x
y
H
D
E

314
In the diagram, P (−4, 10), Q (0, 6) and R (−2, 2) are the midpoints of the straight lines AB, BC
and AC respectively such that ARQP forms a parallelogram.
(a) Find
(i) the equation of the straight line AB, (Ans : y = 2x + 18) [2 marks]
(ii) the equation of the perpendicular bisector of the straight line AC.
(Ans : y = x + 4) [2 marks]
(b) The straight line AB that is produced intersects the perpendicular bisector of the straight
line AC at point S. Find the coordinates of point S.
[ Ans : (−14, −10) ] [3 marks]
(c) Calculate the area of triangle PQR. (Ans : 12)
Hence, find the area of triangle ABC. (Ans : 48)
[3 marks]
Answer :
x
y
O C
Q
B
P
A
R

315
VECTORS
- WORKSHEET
Encik Hartono bin Josed

317
WORKSHEET
TOPIC 8 : VECTORS
==========================================================================================================================================
8.1 Vectors
8.1.1 Compare and contrast between vectors and scalars, and hence identify whether a quantity is
a vector or a scalar by providing justifications.
==========================================================================================================================================
1 Identify whether each of the following quantities is a scalar of a vector by marking ( ✓ ).
Quantity Scalar Vector Quantity Scalar Vector
force speed
distance resistance
volume displacement
weight mass
work area
velocity time
length momentum
impulse temperature
power energy
pressure acceleration
2 Identify whether each of the following quantities is a scalar of a vector by marking ( ✓ ).
Quantity Scalar Vector
(a) A bus is moving at a speed of 80 kmh−1
due east.
(b) Batrisyia walks 500 m from house to school.
(c) A beg weight 250 N is moved as high as 2 m from the floor.
(d) The body temperature of Maven is 36.5C.
(e) The density of a solid Y is 2.3g cm3
.
MIND think :
Match the following :
scalar quantity a quantity that has magnitude and direction
tensor at level zero
vector quantity a quantity that has magnitude but no direction
tensor at level one

318
==========================================================================================================================================
8.1.2 Represent vectors by using directed line segments and vector notations, and hence determine
the magnitude and direction of vectors.
==========================================================================================================================================
3 Beginning from the point provided, draw and label the following vectors :
(a) AB
→
represents a displacement of 60 m to the right.
(b)
~
p represents a velocity of 40 km/h due south.
(c) u represents an acceleration 5 2 ms−2
due northwest.
Answer :
(a)
1 unit represents 10 m
(b)
1 unit represents 8 km/h
(c)
1 unit represents 1 ms−2
4 Determine the magnitude and direction of each of the following vectors :
(a) (b)
Answer : (Ans : N 36.87 E @ 036.87)
(a)
(b)
r
L
K

319
5 Two bus, P and Q are moving away from town O. Bus P moves due south while bus Q moves due west.
Given that | |
OP
→
= 40 km and | |
OQ
→
= 96 km, after both buses travelled for one hours. Find the
distance between the two buses. (Ans : 104)
Answer :
6 A car moves due north from A to B with a distance of 6 km, then moves due east from B to C with
a distance of 8 km and finally turn back to A. Find the magnitude and displacement direction from
C to A. (Ans : 10 km, S 53.13 W @ 233.13)
[3 marks]
Answer :
MIND think :
• the bearing of a point B from a point A is the angle measured in a clockwise direction
from the north-line of A to the line joining A and B.
• written in a three-digit form, from 000 to 360
A
N

B
 = bearing of B from A
North
South
East
West
North-east
South-east
North-west
South-west

320
 same vector / negative vector
7 The diagram shows a parallelogram, PQRS. The points A, B, C and D are the midpoints of PQ, QR, RS
and ST respectively.
Given that
~
SD u
→
= ,
~
SO v
→
= and
~
SC w
→
= . State the vectors for the following in terms of
~
u ,
~
v or
~
w .
(a) RB
→
(b) QO
→
(c) OB
→
(d)
AO
→
[clon textbook form 4]
Answer :
8 The diagram shows a regular hexagon with centre O.
Given that
~
AB p
→
= ,
~
OB q
→
= and
~
BC r
→
= . State the vectors for the following in terms of
~
u ,
~
v
or
~
w .
(a) ED
→
(b) CD
→
(c) EF
→
(d) OE
→
Answer :
O
~
u
P A Q
R
S
B
C
D
~
v
~
w
A B
C
D
E
F O
~
p
~
q
~
r

321
MIND think :
9 Given that (2x2
− 7x + 3)
~
b =
~
0 , find the value of x. (Ans : 2
1
, 3)
Answer :
10 If
~
a and
~
b are non-parallel vectors and (m + 3)
~
a + (2n − m + 1)
~
b is a zero vector, find the values of
m and n. (Ans : m = −3, n = −2)
Answer :
• A Zero vectors, , has magnitude, and its cannot be determined.
• Two vector are the same if and only if both the vectors have the same and
.
• A vector is negative if the vector consists of magnitude, but in the
and .
direction, that is AB
→
= .

322
==========================================================================================================================================
8.1.3Make and verify conjectures about the properties of scalar multiplication on
vectors.
==========================================================================================================================================
11 The diagram shows four vectors.
State the following vectors in terms of a .
(a) AB
→
(b) CD
→
(c) EF
→
Answer :
12 The diagram shows the vector SR
→
=
~
z .
State the vector QP
→
in term of
~
z .
Answer :
B
A
a
C
D E
F
R
S
P
Q

323
13 The diagram shows the vector
→
AB =
~
2u .
State the vector
→
CB in term of
~
u .
Answer :
MIND think :
B
A
C
• Multiplication of scalar k with vector
~
a produces vector
~
k a , with the conditions :
→
~
| |
k a =
→ if k 0, then
~
k a is in the same direction with
~
a .
→ if k 0, then
~
k a is in the opposite direction with
~
a .

324
14 The diagram in the answer space shows the vector
→
AB .
On the diagram, construct the vector
→
PQ such that
→
PQ =
3
4 →
BA .
Answer :
15 The diagram in the answer space shows the vector
→
AB .
On the diagram, construct the vector
→
PQ such that
→
PQ =
2
3
−
→
BA .
Answer :
==========================================================================================================================================
8.1.4 Make and verify conjectures about parallel vectors.
==========================================================================================================================================
 parallel vectors
16 Given =
~
2 x , =
~
2 y and =
~
4 x . Which pairs of vectors are parallel ?
Hence, determine the relation between the two parallel vectors.
Answer :
P
B
A
P
A
B
→
AB
→
CD
→
EF

325
MIND think :
17 Given
→
PQ =
~
3u ,
→
RS =
~
2u
− and
→
TU =
~
3v . Which pairs of vectors are parallel ?
Hence, determine the relation between the two parallel vectors.
Answer :
18 PQRS is a trapezium such that PQ parallel to SR, PQ = 4 cm, and SR = 10 cm. Express
→
RS in terms of
→
PQ . (Ans :
→
RS = 5
2
−
→
PQ )
Answer :
19 In the diagram, ABC is a triangle,
→
BC and
→
DE are two parallel vetors.
Given 
→
BC  = 4 cm and 
→
DE  = 6 cm. Express
(b)
→
AB in terms of
→
AD , (Ans : AB
→
= 2
3
AD
→
)
(c) CE
→
in terms of AE
→
. (Ans : CE
→
= 1
3
AE
→
)
Answer :
(a) (b)
•
~
a is parallel to
~
b  , where  is a constant.
• A, B, C are collinear 
•
~
a and
~
b are not parallel and non-zero, and
~
~
b
k
a
h =  h = k = .
D
B
E
C
A

326
MIND think :
area
area
ABC
ADE


= ( )2
DE
BC = ( )2
AD
AB = ( )2
AE
AC area
area
ABE
CDE


= ( )
2
AB
CD = ( )
2
AE
ED =
( )
2
BE
EC
area
area
ABC
ACD


=
CD
BC / luas
luas
ABC
ABD


=
CD
BC
20 In the diagram, PQ and RS are parallel.
Given that RS : PQ = 1 : 3,  PQ
→
 = 12 cm, PS
→
= 8p and QT
→
= 9q . Find
(a)  SR
→
 , (Ans : 4)
(b) PT
→
in terms of p, (Ans : 6 p )
(c) RT
→
in terms of q , (Ans : 3q
− )
Answer :
 collinear
21 Given
→
ST =
~
12 a and
→
TU =
~
8 a , show that S, T and U are collinear.
Answer :
BC AC
AB
DE AD AE
= =
A
B C
D E
A B
C D
E AB AE BE
CD ED EC
= =
A
B C D
T
S
P Q
R

327
22 Given points K, M and N are collinear with MN
→
=
3
5
KN
→
. Express
→
KM in terms
of
→
MN . (Ans : KM
→
= 2
3
MN
→
)
Answer :
23 Given points A, B and C are collinear with AB
→
=
~
ha and BC
→
=
~
(3 2)
k a
− , where k is a constant. If
5 AB
→
= 2 AC
→
, express k in terms of h. (Ans : = 3 4
6
h
k +
= )
Answer :
 not parallel and non-zero
24 The vector a and b are non-zero and non-parallel. It is given that (h + 3) a = (k − 5) b , where h and
k are constants. Find the value of
(a) h, (Ans : −3)
(b) k. (Ans : 5)
[2 marks]
[2008, No.15]
Answer :
(a) (b)

328
25 The diagram shows two vectors,
→
OA = x and
→
OB = y.
Find the value of h and k such that (h − 2) x = (3h + k) y.
(Ans : h = 2, k = −6)
[3 marks]
Answer :
26 If
~
a and
~
b are non-zero and non-parallel vectors such that (2m + n − 1)
~
a − (m − n + 7)
~
b = 0 find the
values of m and n. (Ans : m = −2, n = 5)
Answer :
==========================================================================================================================================
8.2 Addition and subtraction of vectors
8.2.1Perform addition and substraction involving two or more vectors to obtain a
resultant vector.
==========================================================================================================================================
 addition and subtraction of parallel vector
(a) 2
~
u +
3
1
~
u +
3
2
~
u = (b) 8
~
c − 5
~
c −
~
c =
A
O
B

329
 addition and subtraction of non-parallel vector
(a) 4
~
x +
2
1
2
~
y +
2
1
~
x +
~
y = (b) 6
~
c − 5
~
d − 4
~
c + 3
~
d =
 addition and subtraction of non-parallel vector ~ triangle law
29 The diagram shows a triangle PQR.
The point T lies on QR such that QT : TR = 3 : 1. Express in terms of a and b :
(a)
→
QR ,
(b)
→
PT . (Ans : 3 a + 2
3 b )
[3 marks] [2008, No.16]
Answer :
(a) (b)
R
T
4 a
P
6 b
Q

330
Given that
→
PQ = 3 a ,
→
PR = 6 b and point S lies on QR such that QS : SR = 2 : 1, express in terms of
a and b :
(a)
→
QR ,
(b)
→
SP . (Ans : − a − 4b )
[4 marks] [2009, No.14]
Answer :
(a) (b)
31 The diagram shows a triangle PQR and M is a point on AB.
Given that
→
OA = 5 a ,
→
OB = 4 b and 2AM = 3MB, find
(a)
→
AB ,
(b)
→
OM . (Ans : 2 a + 5
12 b )
[4 marks] [2010, No.16]
Answer :
(a) (b)
P Q
R
S
A
M
B
O

331
32 The diagram shows a trapezium PQRS with QR = 2PS.
Express in terms of a and / or b
(a)
→
SR −
→
PR ,
(b)
→
QP . (Ans : 8 a −7b )
[3 marks] [2014, No.15]
Answer :
(a) (b)
 addition and subtraction of non-parallel vector ~ triangle law @ parallelogram
law 1
33 The diagram shows a parallelogram ABCD with BED as a straight line.
Given that
→
AB = 6p,
→
AD = 4q, and DE = 2EB, express, in terms of p and q :
(a)
→
BD . (Ans : −6p + 4q)
(b)
→
EC . (Ans : 2p + 3
8 q)
[4 marks] [2003, No.14]
Answer :
(a) (b)
S
R
Q
P
4 a
7 b
A B
C
D
E

332
34 The diagram shows a rectangle OABC and the point D lies on the straight line OB.
It is given that OD = 3DB. Express
→
OD in terms of x and y . [ Ans :
4
3
( 5 y + 9 x ) ]
[3 marks] [2007, No.15]
Answer :
35 The diagram shows a parallelogram ABCD.
Point E lies on AB such that AE : EB = 2 : 1. It is given that EB
→
= 4u and AD
→
= 3v .
Express in terms of u and v
(a) AE
→
,
(b) ET
→
.
(Ans : 3
2
v 2u
− )
[3 marks] [2019, No.15]
Answer :
(a) (b)
O A
B
D
C
9 x
5 y
D C
T
E B
A

333
 addition and subtraction of non-parallel vector ~ polygon law
36 The diagram shows a pentagon JKLMN.
Given that KL
→
=
1
3
JN
→
, 2 NM
→
= JK
→
, JN
→
=
~
x and JK
→
=
~
y , express LM
→
in terms of
~
x
and
~
y . (Ans : 2
3 ~
x − 1
2
~
y )
Answer :
37 The diagram shows a quadrilateral such that
→
PS = (m − 1) x ,
→
SR = n y and
→
QR = n x . m and n
are constants.
If
→
PQ = 3 x + 




 +
5
1
m
y , find the values of m and n. (Ans : m = 4
21 , n = 4
5 )
Answer :
J K
L
M
N
~
y
~
x
R
Q
S
T

334
law 2(a)
→
OP = and
→
OQ =
~
y .
Express in terms of and
~
y :
(a)
→
OR ,
(b)
→
RT . (Ans :
~
y + 2 )
Answer :
(a)
(b)
~
x
Q
R
O
T
P
~
x
~
x

335
→
OP = and
→
OQ =
~
y .
Express in terms of and
~
y :
(a)
→
OR ,
(b)
→
PT . (Ans : + 2
3
~
y )
Answer :
(a)
(b)
40 The diagram shows the vectors
→
OA ,
→
OB and
→
OP drawn on a grid of equal squares with sides of 1
unit.
Determine
(a) 
→
OP  , (Ans : 2
3 )
(b)
→
OP in terms of a and b .
[2 marks] [2012, No.15]
Answer :
(a)
(b)
~
x
R
T
Q
P
O
~
x
~
x
P
B
A
O
a
b

336
41 The diagram shows vectors AB
→
, AC
→
and AD
→
drawn on a square grid with sides of 1 unit.
(a) Find BA
→
− .
(b) Given AB
→
= b and AC
→
= c , express in terms of b and c ,
(i) BC
→
,
(ii) AD
→
. (Ans : 2 c − b )
[3 marks] [2017, No.3]
Answer :
(a)
(b) (i)
(ii)
A D
B
C

337
law 2(b)
42 The diagram shows two vectors, OA
→
=
~
a and OB
→
=
~
b .
(a) State vector
→
OQ in term of
~
a and
~
b .
(b) Given that
→
OP =
~
~
2 b
a +
− , mark and label the point P on the above diagram.
Answer :
(a) (b) refer the diagram
43 The diagram shows the vectors OP
→
, OQ
→
dan OM
→
drawn on a square grid.
(a) Express OM
→
in the form p
h + q
k , where h and k are constants. (Ans : p + q
2 )
(b) On the diagram, mark and label the point N such that MN
→
+ OQ
→
= 2 OP
→
.
[3 markah] [2018, No.8]
Answer :
(a) (b)
O
B
A
Q
O
M
Q
P
p
q

338
44 The diagram shows three vectors drawn on square grids.
(a) Draw the vector 2
~
a −
~
b +
~
2
1
c on the square grids in the answer space, start
from the given point.
(b) If 
~
c  = 4, find  2
~
a −
~
b +
~
2
1
c .
Answer :
(a)
(b)
~
a
~
b
~
c

339
==========================================================================================================================================
8.2.2 Solve problems involving vectors.
==========================================================================================================================================
 solve problems 1
45 Janelle rows her boat from point K across the river due north with a velocity of 7 kmh−1
. The river stream
flows due west with a with velocity of 12 kmh−1
.
(a) Sketch the digram which shows the movement of the boat and the river stream.
(b) After affected by the river stream, calculate for the boat :
(i) new velocity, (Ans : 13.89)
(ii) new direction. (Ans : 300.26)
Answer :
(a) (b) (i)
(ii)
46 An aeroplane is flying to the south from airpoint A to airport B for 1000 km in 2 hours. The wind blows
from east with a velocity of 250 kmh−1
.
(a) Sketch the digram which shows the movement of the aeroplane and the wind blow.
(b) Hence, find
(i) the velocity of the plane without the influence of the wind, ( Ans : 433.01)
(ii) the original direction of the aeroplane. (Ans : 150)
Answer :
(a) (b) (i)
(ii)

340
47 The diagram shows the positions of L, M and N along a river.
The width of the river is 50 m, M is due north of L and the velocity of the downstream river flow is
2.5 m/s. Nathaniel wanted to row his boat from L across the river to M, but the boat was swept by the
current flow and stopped at N in 15 seconds. Calculate the speed, in m/s, of Nathaniel’s boat. (Ans :
6.08)
Answer :
48 Given the position vector for three toy cars are OP
→
= x + 3y, OQ
→
= 2x + 5y and OR
→
= kx + 4y, where
k is a constant. These toy cars are placed in a straight line, find the value of k . (Ans : 2
3
)
Answer :
49 Given that O, P, Q, and R are four points such that
→
OP = p,
→
OQ = q and
→
OR = 4 p . M is the
midpoint of PQ, and the line OM is extended to a point S such that
→
→
= OM
OS
5
8
.
(a) Express in terms of p and q :
(i)
→
OS . [ Ans : 5
4
( p +q ) ]
(ii)
→
QR . (Ans : 4 p −q )
[4 marks]
(b) Hence, show that point S lies on QR and state the ratio of QS : SR.
(Ans : 1 : 4) [3 marks] [Forecast]
Answer :
L
M N
85 m
current flow

341
50 In the diagram, P is the midpoint of OA, and Q is a point on AB such that
→
→
= QB
AQ 3 .
(a) Given that
~
5a
OA =
→
and
~
10b
OB =
→
. Express in terms of
~
a and
~
b :
(i)
→
BP . (Ans : 2
5
~
a − 10
~
b )
(ii)
→
OQ , (Ans : 4
5
~
a + 2
15
~
b )
[4 marks]
(b) Given that
→
→
= BP
BG  and
→
→
= OQ
OG  , find
(i) the values of  and , (Ans :  = 5
2
, .= 5
4
)
(ii) the ratio of area of triangle OGP : area of triangle QGB.
(Ans : 6 : 1)
[6 marks]
[Forecast]
Answer :
O
Q
P
B
G
A

342
51 The diagram shows a triangle, ACE. It is given that
~
x
k
AE =
→
,
~
3 x
BD =
→
and
~
)
1
( y
h
ED −
=
→
,
where k and h are constants.
If
~
~
6
2 y
x
AB +
=
→
, find
(a) the values of h and k, (Ans : h = 7, k = 5) [4 marks]
(b) the area of triangle BCE, if the area of triangle ABE is 18 unit2
.
(c) the area of triangle BCD, if the area of triangle ABE is 15 unit2
.
(Ans : 13.5) [2 marks]
(d) the area of triangle BCD, if the area of triangle ACE is 30 unit2
.
(Ans : 10.8) [2 marks]
[Forecast]
Answer :
D
B
C
A E

343
==========================================================================================================================================
8.3 Vectors in a cartesian plane
8.3.1 Represent vectors and determine the magnitude of the vectors in the Cartesian
plane.
==========================================================================================================================================
 vectors in the Cartesian plane ~ 1
52 The diagram shows five points, P, Q, R, S and T on a grid.
Express
(a) PR
→
, QR
→
, TR
→
, SR
→
, PQ
→
and PT
→
in the form 







y
x
,
(b) RP
→
, RQ
→
, RT
→
, RS
→
, QP
→
and TP
→
in the form x i + y j.
Answer :
(a) PR
→
= QR
→
= TR
→
=
SR
→
= PQ
→
= PT
→
=
(b) RP
→
= RQ
→
= RT
→
=
RS
→
= QP
→
= TP
→
=
P
R
Q
S
T

344
 vectors in the Cartesian plane ~ 2
→
OP and
→
QO .
Express
(a)
→
OP in the form 







y
x
,
(b)
→
OQ in the form x i + y j.
[2 marks] [2003, No.12]
Answer :
(a) (b)
54 The diagram shows a parallelogram OABC, drawn on a Cartesian plane.
Express
(a)
→
CB in the form 







y
x
,
(b)
→
BA in the form x i + y j.
[2 marks]
[Forecast]
Answer :
(a) (b)
O
y
x
Q (−8, 4) P (5, 3)
B
x
y
O
C (−5, 3)
A (2, 1)

345
MIND think :
 magnitude of a vector
55 Given that OAB is a right-angled triangle with AOB = 90,
~ ~
4 2
OA i j
→
= + and
~ ~
3 6
OB i j
→
= − + . Find
the area of triangle AOB. (Ans : 15)
Answer :
MIND think :
56 Given that a = −2 i + h j. Find the values of h such that  a = 20 . (Ans : 4)
Answer :
57 Given that
~
~
3 j
k
i
OP +
=
→
and
~
4 j
OQ =
→
. If OP and OQ are the two sides of a rhombus, find the value
of k. (Ans :  7 )
Answer :
→
OA = 
→
AO =
A (−3, 2)
in the form of compenent
~
i and
~
j
in the form column vector
→
OA = 
→
AO =
 
 
 
 
 
 
~
~
~
j
y
i
x
r +
= =
x
y
 
 
 
 magnitude
~
r ,
~
r =

346
==========================================================================================================================================
8.3.2Describe and determine the unit vector in the direction of a vector.
==========================================================================================================================================
 unit vector in the direction of a vector
58 The diagram shows vector
→
OA drawn on a Cartesian plane.
(a) Express
→
OA in the form 







y
x
.
(b) Find the unit vector in the direction of
→
OA .
[2 marks]
[2005, No.15]
Answer :
(a) (b)
59 The diagram shows the vector
→
OR .
Express in the form x i + y j :
(a)
→
OR ,
(b) the unit vector in the direction of
→
OR .
[3 marks] [2010, No.15]
Answer :
(a) (b)
y
2
x
A
4
6
2 4 6 8 10 12
O
R (3, 4)
x
y
O

347
60 The length of vector
~
u is 13 units and the direction is opposite with vector −3i + 2j.
Find the vector of
~
u . [ Ans : 3 13i 2 13 j
− ]
Answer :
61 Given that
~
~
~
3 j
k
i
p +
= and
~
~
~
)
3
(
5
1
j
k
i
p +
=

, find the possible values of k. (Ans :  4)
[2 marks] [forecast]
Answer :
MIND think :
62 Given that
5
74
i k j
+
is a unit vector, find the possible values of k. (Ans : 7)
Answer :
63 Given that
^
~
v =
3
(1 )
5
h i j
− − , find the possible values of h. (Ans : 1
5
, 9
5
)
Answer :
 unit vector in the direction
~
r ,

~
r =
•
~
~
~
j
y
i
x
r +
= =
x
y
 
 
 
Note :

~
r =

~
2r =

~
3r . . .
• if
~
r is a unit vector  
~
r  =

348
64 Given that 







=
k
h
x and 






−
=
h
k
y . If the unit vector in the direction of 2 x is
3
2








k
h
.
Find the value of  3 y . (Ans :
2
9 )
Answer :
==========================================================================================================================================
8.3.3 Perform arithmetic operations onto two or more vectors.
==========================================================================================================================================
 arithmetic operations / vectors in the Cartesian plane 1
65 The diagram shows a parallelogram, OPQR, drawn on a Cartesian plane.
It is given that
→
OP = 6i + 4j and
→
PQ = −4i + 5j. Find
→
PR .
(Ans : −10i + j)
[3 marks]
[2005, No.16]
Answer :
y
x
O
P
R
Q

349
66 The diagram shows a parallelogram ODEF drawn on a Cartesian plane.
It is given that
→
OD =
−
−
+ j
i 2
3 and
→
DE =
−
−
+
− j
i 3
5 . Find
→
DF . (Ans :
−
−
+
− j
i
8 )
[3 marks] [2011, No.16]
Answer :
 arithmetic operations / vectors in the Cartesian plane 2
→
OA and
→
AB .
Express
(a)
→
OA in the form 







y
x
, (b)
→
AB in the form x i + y j.
[2 marks] [2006, No.13]
Answer :
(a) (b)
O
D
y
E
x
F
B
−5
A (4, 3)
x
y
O

350
→
PQ and
→
RS .
Express
(a)
→
PQ in the form 







y
x
, (b)
→
SR in the form x i + y j.
Answer :
(a) (b)
69 The diagram shows two vectors, and .
If point N lies on PQ such that
→
PN =
→
NQ
2
1
. Find the
→
ON in the form x i + y j.
[ Ans : : −i + 3j ]
Answer :
O
x
y
S (−4, 4)
P (4, 2)
Q (1, 2)
R (−2, 1)
→
OP
→
OQ
O
y
x
Q (3, 5)
P (−3, 2)

351
 arithmetic operations / magnitude of a vector
70 Given that
~
~
3
2 j
i
AB +
−
=
→
,
~
~
4
3 j
i
BC −
=
→
, and B (−1, 5). Find
(a) the coordinates of point A, [ Ans : (1, 2) ]
(b) the length of AC. (Ans : 2 )
Answer :
(a) (b)
71 Given that a = 13 i + j and b = 7 i − k j, find
(a) a − b , in the form x i + y j, [ Ans : 6 i + ( 1 + k ) j ]
(b) the values of k if  a − b  = 10. (Ans : −9, 7)
[4 marks] [2009, No.13]
Answer :
(a) (b)
72 It is given that vector r = 







− 2
8
and vector s = 







7
h
, where h is a constant.
(a) Express the vector r + s , in terms of h.
(b) Given that  r + s  = 13 units, find the positive value of h. (Ans : 4)
[4 marks] [2011, No.17]
Answer :
(a) (b)

352
 arithmetic operations / unit vector in the direction of a vector 1
73 Given that O (0, 0), A (−3, 4), and B (2, 16), find in terms of the unit vectors, i and j,
(a)
→
AB ,
(b) the unit vector in the direction of
→
AB . [ Ans : 13
1
(5 i + 12j)
]
[4 marks] [2004, No.16]
Answer :
(a) (b)
74 The following information refers to the vectors a and b .








=
8
2
a , 






−
=
4
1
b
Find
(a) the vector 2 a − b , [ Ans : 







12
5
]
(b) the unit vector in the direction of 2 a − b . [ Ans :
13
1








12
5
]
[4 marks] [2007, No.16]
Answer :
(a) (b)
75 Given STUV is a parallelogram,
~
~
3
2 j
i
TV +
=
→
and
~
~
2
2 j
i
UV −
−
=
→
. Find the unit vector in the direction
of TU
→
in terms of
~
i and
~
j . [ Ans :
41
1 (4i + 5j) ]
Answer :

353
76 Given that a = 2i − j and b = 3i + j. Find the vector in the same direction and parallel 2b − 4a and has a
magnitude of 5 10 . [ Ans : 5 (− i + 3j) ]
Answer :
77 The diagram shows a regular hexagon with centre O.
(a) Express AC
→
+ CE
→
+ CB
→
as a single vector. (Ans : AF
→
)
(b) Given OA
→
= a , OB
→
= b , and the length of each side of the hexagon is 3 units, find the unit
vector in the direction of AB
→
in terms of a and b . (Ans : 3
b
a +
−
)
[3 marks] [2016, No.10]
Answer :
(a) (b)
A B
C
D
E
F O

354
78 Given that
→
AB = 







−1
q
q
and
→
OA = 







2
2
. If
→
OB is a unit vector, find the possible values
of q. (Ans : −2, −1)
Answer :
 arithmetic operations
79
r = 3a + 4b,
s = 4a − 2b,
t = pa + (p + q)b, where p and q are constants.
Use the above information to find the values of p and q when t = 2r − 3s.
(Ans : p = −6, q = 20)
[3 marks] [2003, No.13]
Answer :
80 Given that A (−2, 6), B (4, 2), and C (m, p), find the value of m and of p such that
→
AB +
→
BC
2 =
~
~
12
10 j
i − . (Ans : m = 6, p = −2)
[4 marks] [2004, No.17]
Answer :

355
81 A (2, 3) and B (−2, 5) lie on a Cartesian plane. It is given that 3 OA
→
= 2 OB
→
+ OC
→
. Find
(a) the coordinates of C, [ Ans : (10, −1) ]
(b) | AC
→
| . (Ans : 4 5 )
[4 marks] [2018, No.9]
Answer :
(a) (b)
82 It is given that P (2, m), Q (h, 6), v = 2i − j , w = 9i + 3j and PQ
→
= 2v + kw , such that m, h
and k are constants. Express h in terms of m. (Ans : h = 30 −3m)
[3 markah] [2019, No.16]
Answer :
 parallel vectors / collinear ~ 1
83 The following information refers to the vector a and b .
a = 





− 4
6
m
, b = 





5
2
It is given that a = kb , where a is parallel to b and k is a constant. Find the value of
(a) k, (Ans : 3)
(b) m. (Ans : 19)
[3 marks] [2012, No.16]
Answer :
(a) (b)
84 Given u = 





4
3
and v = 





−1
6
k
, find

356
(a) the unit vector in the direction of u , [ Ans :
5
1






4
3
]
(b) the value of k such that u and v are parallel. (Ans : 9)
[4 marks] [2013, No.15]
Answer :
(a) (b)
85 The diagram shows a trapezium ABCD.
Given p = 







4
3
and q = 






 −
2
1
k
, where k is a constant, find the value of k.
(Ans : 2
5 )
[3 marks] [2017, No.4]
Answer :
86 Given that 







=
2
2
~
p , 






−
=
6
9
~
q and 







=
4
~
m
r . If
~
~
2 q
p + is parallel to
~
r , find the value
of m. (Ans : −2)
Answer :
87 Vector 







b
a
has a magnitude of 5 unit, and parallel to 






−
2
4
. If a > b, find the value of a and
of b. (Ans : a = 2, b = −1)
D C
A
B
q
p

357
Answer :
88 The points P, Q, and R are collinear. It is given that
~
~
2
4 b
a
PQ −
=
→
and
~
~
)
1
(
3 b
k
a
QR +
+
=
→
, where
k is a constant. Find
(a) the value of k. (Ans : −2.5)
(b) the ratio of PQ : QR. (Ans : 4 : 3)
[4 marks] [2006, No.14]
Answer :
(a) (b)
89 Given that
~ ~
(2 1) 3
AB k p q
→
= − + . If AB
→
is extended to point C such that
~ ~
6
BC k p hq
→
= + , express k
in terms of h. (Ans : k = 2
4 1
h
h−
)
Answer :
90 It is given
→
OP = 







4
k
,
→
OQ = 







3
1
and
→
OR = 







− 2
h
, where h and k are constants. Express h in
terms of k, if points P, Q and R lie on a straight line. (Ans : h = 6 − 5k)
[3 marks] [2015, No.15]
Answer :
91 Given
~
~
j
i
OA +
=
→
,
~
~
3
5 j
i
OC +
=
→
and
~
~
3 j
i
OD 
+
=
→
. If point D lies on AC, find the value
of . (Ans : 2)
Answer :

358
92 Given that
~
~
10 j
i
OP +
−
=
→
,
~
~
2
3 j
i
OQ +
=
→
, and
~
4 i
OR =
→
. Show that P, Q, and R are collinear.
Answer :
93 Given that
~
2
OS j
→
= ,
~ ~
10
3
OK i j
→
= + , and
~
6
OR i
→
= . Determine whether S, K, and R are in a straight
line. Prove your answer mathematically. (Answer : no)
Answer :
94 Given p = 




−
3
4
and q = 





k
2
, find
(a) p , (Ans : 5)
(b) the value of k such that p + q is parallel to the x-axis. (Ans : −3)
[3 marks] [2014, No.16]
Answer :
(a) (b)
95 Given that 







−
=
1
3
~
a , and 






−
=
5
2
~
b . If 2p
~
a + 5
~
b is parallel to y-axis, find the value
of p. (Ans : 3
5 )
Answer :

359
96 Given that 






 −
=
1
1
~
m
p and 







=
8
9
~
q , where m is a constant. Find the value of m if
~
p is
perpendicular / orthogonal to
~
q . (Ans : 9
1
)
Answer :
MIND think :
~
a perpendicular to
~
b

(
~
a
m ) (
~
b
m ) = −1 @
~ ~
. 0
a b =
•
~
a is parallel to x-axis  the constant of j =
•
~
a is parallel to y-axis  the constant of i =

360
==========================================================================================================================================
8.3.4 Solve problems involving vectors.
==========================================================================================================================================
97 A motoboat cross a river with the enjine that can move with a speed of 7
~
j . The speed for the current
of the river and the wind blows against the motorboat are 2
~
i − 3
~
j and −4
~
i + 6
~
j respectively. Find
the resultant vector cause to the motoboat. (Ans : −2
~
i + 10
~
j )
Answer :
98 A ball is throw horizontally with an acceleration of 2 ms−2
from the top of a tower. The ball will drop
freely with an acceleration due to the gravity force, g ms−2
. By using
~
i as the unit vector in the direction
of horizontal acceleration, and
~
j as the unit vector in the drop direction due to the gravity force.
(a) Find the resultant vector of the ball in terms of
~
i and
~
j . [1 mark]
(b) Hence, by using g = 10, calculate the magnitude of the resultant vector of
the ball. (Ans : 10.20) [2 marks]
Answer :
[Forecast]
(a) (b)
99 A particle moves with the velocity vector,
~
v = ( 2
~
i − 3
~
j ) ms−1
. If it started from the position
~
i + 4
~
j . Find
(a) the speed, in ms−1
, of the particle, (Ans : 13 )
(b) the position of the particle after 3 seconds, (Ans : 7
~
i − 5
~
j )
(c) the duration, in second, for the particle to reach the position 11(
~
i −
~
j ). (Ans : 5)
Answer :
(a) (c)
(b)

361
MIND think :
100 Car P left town P (0, 0) with velocity of
~ ~
(6 8 )
P
v i j
= + kmh−1
. At the same time, car Q left
town Q (100, 40) with velocity of
~ ~
( 4 4 )
Q
v i j
= − + kmh−1
. Find the time when the car P will cross
car Q. (Ans : 10)
Answer :
101 A particle is moving from the point P (7, 15) with the velocity vector of (3
~
i − 2
~
j ) ms−1
. After t
seconds leaving P, the paticle is on point S.
(a) Find
(i) the speed of the particle, (Ans : 13 )
(ii) the position of the particle from O after 4 seconds. [ Ans : (19, 7) ]
(b) When will the particle reside on the right side of the origin ? (Ans : 7.5)
Answer :
(a) (i) (b)
(ii)
102 The diagram shows the location of Pay’house, the school and the public library at point O, point A and
point B respectively on the Cartesian plane. The shortest distance between Pay’house and the school is
7.5 km while the shortest distance between the school and the public library is 19.5 km.
Given the vector from Vivi’s house to the school is 3 i + 4 j , express the vector from Pay’s house to
the public library in the form of x i + y j . (Ans : 8 i + 16 j )
Answer :
speed = of velocity vector
y
16
x
O
B
A

362
PAPER 2
 Part A ~ parallel 1 → 6 – 8 marks
103 Given that 







=
7
5
AB , 







=
3
2
OB and 







=
5
k
CD , find
(a) the coordinates of A, [ Ans : (–3, –4) ] [2 marks]
(b) the unit vector in the direction OA , (Ans : j
i 5
4
5
3
−
− ) [2 marks]
(c) the value of k, if CD is parallel to AB . (Ans : 7
25 ) [2 marks]
[2003, No.6]
Answer :
104 It is given that
→
AB = j
i 2
3 +
− and
→
AC = j
i 5
7 +
−
(a) Find
(i)
→
BC , (Ans : i
4
− + j
3 )
(ii) the unit vector in the direction
→
BC . (Ans : 5
3
4 j
i+
−
)
[4 marks]
(b) Given
→
AD = j
i
p 15
− , where p is a constant and
→
AD is parallel to
→
BC , find the
value of p. (Ans : 20) [3 marks]
[2012, No.5]
Answer :

363
 Part A ~ parallel 2
105 The diagram shows the position and the direction of boats A, B and C in a solar boat competition.
Both boat A and boat B move in the direction of the water current. The velocity of
the water current is given by w = 





+ j
i
2
1
ms−1. Given the velocity of boat A is a =
( )
j
i +
2 ms−1 and the velocity of boat B is b = ( )
j
i 3
6 + ms−1.
(a) Determine how many times the resultant velocity of boat B compare to the
resultant velocity of boat A. (Ans : 3
7 ) [4 marks]
(b) On the way to the finishing line, boat C is facing a technical problem and off track.
The velocity of boat C is c = 





− j
i
2
3
2 ms−1. Find
(i) the resultant velocity of boat C, (Ans : 3i − j )
(ii) the unit vector in the direction of boat C, [ Ans :
10
1 (3 i − j ) ]
[3 marks]
[2016, No.5]
Answer :
C
B
A
Starting line

364
106 In the diagram, ABCD is a quadrilateral. AED and EFC are straight lines.
It is given that
→
AB = 20
~
x ,
→
AE = 8
~
y ,
→
DC = 25
~
x − 24
~
y , AE =
4
1
AD and EF =
5
3
EC.
(a) Express in term of
~
x and
~
y :
(i)
→
BD , (Ans : −20
~
x + 32
~
y )
(ii)
→
EC , (Ans : 25
~
x )
[3 marks]
(b) Show that the points B, F and D are collinear.
[3 marks]
(c) If 
~
x  = 2, and 
~
y  = 3, find 
→
BD . (Ans : 104) [2 marks]
[2005, No.6]
Answer :
D
F
E
B
A
C

365
107 The diagram shows a trapezium ABCD.
It is given that
→
AB = 2
~
y ,
→
AD = 6
~
x ,
→
AE =
3
2 →
AD and
→
BC =
6
5 →
AD
(a) Express
→
AC in term of
~
x and
~
y . (Ans : 5
~
x + 2
~
y ) [2 marks]
(b) Point F lies inside the trapezium ABCD such that 2
→
EF = m
→
AB , m is a constant.
(i) Express
→
AF in term of m,
~
x and
~
y . (Ans : 4
~
x + m
~
y )
(ii) Hence, if the point A, F and C are collinear, find the value of m.
(Ans : 5
8 )
[5 marks]
[2006, No.5]
Answer :
F
E
B C
D
A

366
108 The diagram shows a quadrilateral PQRS. The straight line PR intersects the straight line QS at point
T.
It is given that QT : TS = 2 : 3,
→
PQ = u
10 ,
→
PS = v
25 and
→
QR = v
u 15
+
−
(a) Express in terms of u and v ,
(i)
→
QS , (Ans : u
10
− + v
25 )
(ii)
→
PT , (Ans : u
6 + 10v )
[3 marks]
(b) Find the ratio PT : TR. (Ans : 2 : 1)
[5 marks]
[2013, No.3]
Answer :
S
P Q
R
T

367
109 The diagram shows a trapezium OPQR and point T lies on PR.
It is given that
→
OR = b
18 ,
→
OP = a
6 and
→
OR = 2
→
PQ .
(a) Express in terms of a and b ,
(i)
→
PR , (Ans : a
6
− + b
18 )
(ii)
→
OQ , (Ans : a
6 + b
9 )
[3 marks]
(b) It is given that
→
PT = k
→
PR , where k is a constant. Find the value of k if the point
O, T and Q are collinear. (Ans : 3
1 ) [5 marks]
(c) If the area of triangle QTR = 45 unit2, and the perpendicular distance from P to
OR is 4 units, find b . (Ans : 3.75) [2 marks]
[2014, No.5]
Answer :
P Q
R
T
O

368
The diagram shows the positions of jetty O and kelongs, K, L, R, S and T in the sea.
Kelong L is situated 400 m from jetty O and kelong R is situated 600 m from jetty O
in the direction of OL. Kelong S is situated 300 m from jetty O and kelong T is situated
600 m from kelong S in the direction of OS. Kelongs L, K and T are situated on a
straight line such that the distance of kelong K from kelong T is 5 times its distance
from kelong L.
(a) By using p to reperesent 100 m in the direction of OR and q to represent 150
m in the direction of OT, express in terms of p and q
(i) OK
→
(Ans : 10
3
p + q )
(ii) RK
→
(Ans : 8
3
− p + q )
[3 marks]
(b) If Joe uses a binocular to observe kelong R from kelong S, determine whether
kelong R can be seen without being blocked by kelong K or otherwise.
Prove your answer mathematically.
(Ans : KS
→
= 10
3
− p + q ; can be seen) [5 marks]
[2019, No.6]
Answer :
Land
Sea
T
S
K
L R

369
 Part A ~ triangle law 1
111 In the diagram, ABCD is a quadrilateral. The diagonals BD and AC intersect at point R. Point P lies
on AD.
It is given that AP =
3
1
AD, BR =
3
1
BD,
→
AB = x and
→
AP = y .
(a) Express in terms of x and y :
(i)
→
DB , (Ans : x − 3 y )
(ii)
→
AR . (Ans : 3
2 x + y )
[3 marks]
(b) Given that
→
DC = k x − y and
→
AR = h
→
AC , where h and k are constants, find
the value of h and of k. (Ans : h = 2
1 , k = 3
4 ) [4 marks]
[2008, No.6]
Answer :
R
P
A B
C
D

370
112 The diagram shows triangle ABC. The straight line AQ intersect the straight line BR at P.
It is given that AR = 3RC, BQ =
3
2
BC,
→
AB = x
3 and
→
AC = y
4 .
(i)
→
BC , (Ans : x
3
− + y
4 )
(ii)
→
AQ . (Ans : x + y
3
8 )
[3 marks]
→
AP =
→
AQ
h and
→
AP =
→
AR +
→
RB
k , where h and k are constants,
find the value of h and of k. (Ans : h = 11
9 , k = 11
3 ) [5 marks]
[2009, No.5]
Answer :
C
A B
P
Q
R

371
 Part B ~~ parallel 3 → 10 marks
113 The diagram shows a parallelogram ABCD. Point P lies on the straight line AB and point Q lies on the
straight line DC. The straight line AQ is extended to the point R such that AQ = 2QR.
It is given that AP : PB = 3 : 1, DQ : QC = 3 : 1,
→
AP = u
6 and
→
AD = v .
(a) Express, in terms of u and v :
(i)
→
AQ , (Ans : v + u
6 )
(ii)
→
PC . (Ans : u
2 + v )
Hence, show that the points, P, C and R are collinea [6 marks]
(b) It is given that u = i
3 and v = i
2 + j
5 .
(i) Express
→
PC in terms of i and j , ( Ans : i
8 + j
5 )
(ii) Find the unit vector in the direction of
→
PC . (Ans :
89
5
8
−
−
+ j
i
)
[4 marks]
[2011, No.10]
Answer :
A P B
Q
R
D
C

372
 Part B ~~ triangle law 1
114 The diagram shows triangle OAB. The straight line AP intersects the straight line OQ at R. It is given
that OP =
3
1
OB, AQ =
4
1
AB, =
~
6 x , and
→
OA =
~
2 y .
(a) Express in terms of
~
x and / or
~
y
(i)
→
AP , (Ans : −2
~
y + 6
~
x )
(ii)
→
OQ . (Ans : 2
3
~
y + 2
9
~
x )
[4 marks]
(b) (i) Given that
→
AR = h
→
AP , state
→
AR in terms of h,
~
x and
~
y .
[ Ans : h (−2
~
y + 6
~
x ) ]
(ii) Given that
→
RQ = k
→
OQ , state
→
RQ in terms of k,
~
x and
~
y .
[ Ans : k ( 2
3
~
y + 2
9
~
x ) ]
[2 marks]
(c) Using
→
AR and
→
RQ from (b), find the value of h and of k.
(Ans : h = 2
1 , k = 3
1 ) [4 marks]
[2004, No.8]
Answer :
→
OP
O
R
Q
P B
A

373
115 The diagram shows triangle AOB. The point P lies on OA and the point Q lies on AB. The straight line
BP intersects the straight line OQ at the point S.
It is given that OA : OP = 4 : 1, AB : AQ = 2 : 1,
→
OA = 8 x and
→
OB = 6 y .
(a) Express in terms of x and / or y .
(i)
→
BP , (Ans : 2 x − 6 y )
(ii)
→
OQ . (Ans : 4 x + 3 y )
[3 marks]
(b) Using
→
OS = h
→
OQ and
→
BS = k
→
BP , where h and k are constants, find the value
of h and of k. (Ans : h =
5
2
, k =
5
4
)
[5 marks]
(c) Given that │ x │ = 2 units, │ y │ = 3 units and AOB = 90, find │
→
AB │.
(Ans : 580 ) [2 marks]
[2007, No.8]
Answer :
A
Q
S
P
B
O

374
116 The diagram shows a triangle ABC.
It is given AP : PB = 1 : 2, BR : RC = 2 : 1,
→
AP = x
2 , and
→
AC = y
3 .
(a) Express in terms of x and y ,
(i)
→
CP , (Ans : y
3
− + x
2 )
(ii)
→
CR , (Ans : y
− + x
2 )
[3 marks]
(b) Given x = i
2 and y = i
− + , find
→
CR . (Ans : 41 ) [2 marks]
(c) Given
→
CQ = m
→
CP and
→
QR = n
→
AR , where m and n are constants, find the
value of m and of n. (Ans : m = 5
3 , n = 5
2 ) [5 marks]
[2015, No.9]
Answer :
C
R
B
P
Q
A
j
4

375
 Part B ~~ triangle law 1 / parallel
117 The diagram shows a triangle PQR. The straight line PT intersects with the straight line QR at point S.
Point V lies on the straight line PT.
It is given that QS
→
=
3
1
QR
→
, PR
→
= x
6 and PQ
→
= y
9 .
(a) Express in terms of x and / or y :
(i) QR
→
(ii) PS
→
. (Ans : y
6 + x
2 )
[3 marks]
(b) It is given that PV
→
= m PS
→
and QV
→
= n ( x − y
9 ), where m and n are constants.
Find the value of m and of n. (Ans : m = 8
3 , n = 4
3 ) [5 marks]
(c) Given PT
→
= x
h + y
9 , where h is a constant, find the value of h.
(Ans : 3) [2 marks]
[2017, No.8]
Answer :
Q
S
P
T
V
R

376
 Part B ~~ triangle law 2 → 10 marks
118 The diagram shows triangle OAB. The point C lies on OA and the point D lies on AB. The straight line
OD intersects the straight line BC at the point E.
It is given that
→
OA = x ,
→
OB = y ,
→
OC =
3
2 →
OA and
→
AB = 2
→
AD .
(i)
→
BC , (Ans : y
− + x
3
2 )
(ii)
→
OD . (Ans : y
2
1 + x
2
1 )
[4 marks]
→
OE =
→
OD
h and
→
BE =
→
BC
k , where h and k are constants. Express
→
OE
(i) in terms of h, x and y , [ Ans : h ( y
2
1 + x
2
1 ) ]
(ii) in terms of k, x and y . (Ans : y y
k
− + x
k
3
2 )
[3 marks]
(c) Hence, find the value of h and of k. (Ans : h = 5
4 , k = 5
3 ) [3 marks]
[2010, No.9]
Answer :
E
A
C
O
D
B

377
119 The diagram shows triangles OAQ and OPB where point P lies on OA and point Q lies on OB. The
straight lines AQ and PB intersect at point R.
It is given that OA
→
= x
18 , OB
→
= y
16 , OP : PA = 1 : 2, OQ : QB = 3 : 1, PR
→
= m PB
→
, QR
→
= nQA
→
, where m and n are constants.
(a) Express OR
→
in terms of :
(i) m, x dan y , (Ans : x
6 − x
m
6 + y
m
16 )
(ii) n, x dan y . (Ans : y
12 − y
n
12 + x
n
18 )
[4 marks]
(b) Hence, find the value of m and of n. (Ans : m = 3
2 , n = 9
1 ) [4 marks]
(c) Given | x | = 2 unit, | y | = 1 unit and OA is perpendicular to OB,
calculate | PR
→
| . (Ans : 3
40 ) [2 marks]
[2018, No.8]
Answer :
B
A
R
Q
O
P

378
FORECAST
120 A boat is set on a course of N 30 E with a speed of 12 knots. However, the water current is
flowing at 5 knots towards the east.
(a) Sketch the digram which shows the movement of the boat and the water current.
[1 mark]
(b) Find :
(i) the magnitude, (Ans : 15.13) [2 marks]
(ii) the direction, (Ans : 046.63) [2 marks]
of the resultant velocity of the boat.
(c) If the boat intends to move northward, show that the direction at which must be steered
in order to do so is N 24.62 W. [2 marks]
Answer :

379
121 The diagram shows a right-angled triangle ABC. M is the midpoint of AB. MN is parallel to BC and P
is a point on CN such that CP = 2PN. Given
~
a
BM =
→
and
~
2b
BC =
→
.
(a) Express in terms of a and / or b :
(i)
→
MN ,
(ii)
→
CA, (Ans : −2b + 2a )
(iii)
→
BP . (Ans : 3
4 b + 3
2 a )
[4 marks]
(b) (i) If │ a │ = 12 units, │b │ = 9 units, find the area of triangle ABC. (Ans : 216 )
(ii) Find the shortest distance for P to BC. (Ans : 12)
[3 marks]
Answer :
A
B
P
N
M
C

380
122 The diagram shows a triangle OAB and OAC. The straight lines OB and AC intersect at point K such
that AK : AC = 1 : 3. Given
→
OA = 3 a and
→
OC = h c , where h is a constant.
Find
(a)
→
AK in terms of h, a and c , [ Ans : 3
1 (−3a + h c ) ] [2 marks]
(b)
→
OK in terms of h, a and c , (Ans : 2a + 3
h c ) [1 mark]
Hence, if
→
KB = 10a + 5 c , find the value of h.` (Ans : 3) [4 marks]
Answer :
O
B
C
K
A

381
123 The diagram shows a map of part of Telipok District, with the condition that all roads are straight. The
mosque is equidistance from the school and the Community Hall, while the fountain is equidistance from
the bus station and the mosque.
The Telipok District Council has decided to build a straight road from the school to the library through the
fountain. The distance from the school to the library is k times the distance from the school to the
fountain. The distance from the Community Hall to the library is twice the distance from the library to the
bus station. Given that the displacement of bus stations and mosques from the school is
~
u and
~
v
respectively.
(a) Find the value of k. (Ans :
3
4 ) [5 marks]
(b) Given that the cost of road construction from school to fountain is RM600000. Find the cost, in
RM, of the construction of the same type of road from the fountain to the library. (Ans : 200000)
[2 marks]
Answer :
School Mosque Community Hall
Bus Station
Library
Fountain

382
124 The diagram shows a triangle OPR and QSP is a straight line.
It is given that
→
OP = p ,
→
OQ = q and
→
OR = 4
→
OQ .
(a) Express in terms of p and q .
(i)
→
PQ , (Ans : − p + q ) [1 mark]
(ii)
→
PR . (Ans : − p + 4q ) [1 mark]
→
PS = m
→
PQ and
→
PT = k
→
PR .
By using
→
OT = 2
→
OS , find the value of m and of k.
(Ans : k = 3
1 , m = 3
2 ) [5 marks]
Answer :
R
Q
P
S
T
O

383
 Part A ~~ triangle law 2
125 The diagram shows a parallelogram PQRS, where L is the midpoint of RS. QR is produced to N such
that QR = RN and QL is produced to meets SN at M.
It is given that
→
PQ = 3 a and
→
QR = 2b .
(a) Express in terms of a and / or b :
(i)
→
QL , (Ans : 2b − 2
3 a )
(ii)
→
SN , (Ans : 3a + 2b )
[2 marks]
(b) Given that
→
QM =
→
QL
h and
→
NM =
→
NS
k , find
→
QM in terms of
(i) h, a and b , ( Ans : 2hb − 2
3 h a )
(ii) k, a and b . (Ans : 4b −3ka −2k b )
Hence, find the value of h and of k. (Ans : h = 3
4 , k = 3
2 ) [5 marks]
Answer :
M
N
S
R
Q
P
L

384
126 The diagram shows a trapezium PQRS. U is the midpoint of PQ and
→
PU =
→
SV
2 . PV and TU are two
straight lines intersecting at W where TW : WU = 1 : 3 and PW = WV.
It is given that
~
12a
PQ =
→
,
~
18b
PS =
→
and
~
18b
QR =
→
−
~
5a
(a) Express in terms of
~
a and / or
~
b
(i)
→
SR , (Ans : 7
~
a )
(ii)
→
PV , (Ans : 3
~
a + 18
~
b )
(iii)
→
PW . (Ans : 2
3
~
a + 9
~
b )
[4 marks]
(b) Using PT : TS = h : 1, where h is a constant, express
→
PW in terms of h,
~
a
and / or
~
b . [ Ans : 2
3
~
a + )
1
(
2
27
+
h
h
~
b ]
Hence, find the value of h. (Ans : 2)
[4 marks]
Answer :
R
Q
P
S
T
U
V
W

385
127 The diagram shows a rectangle OABC and the point D lies on the straight line OB.
It is given that OD = kDB. Express
→
OD in terms of k, x and y . [ Ans : 1
k
k+
( 5 y + 9 x ) ]
[3 marks] [clon 2007, No.15]
Answer :
128 The diagram shows the point A on the Cartesian plane.
(a) State OA
→
in the form of
x
y
 
 
 
.
(b) Point A is reflected about the y-axis to point A. It is given OB
→
= i m j
+ and unit vector of
'
A B
→
is
10
24
n
 
 
 
, where m and n are constants. Find the value of m and n.
(Ans : m = 15, n = 1
26
)
[4 marks] [2020, No.14]
Answer :
(a)
(b)
O A
B
D
C
9 x
5 y
y
x
O
A (4, 3)

386
SOLUTION OF
TRIANGLES
- WORKSHEET
Encik Mohd Salleh Ambo

388
WORKSHEET
TOPIC 9 : SOLUTION OF TRIANGLES
[ Part C → 10 marks ]
==========================================================================================================================================
9.1 Sine Rule
9.1.1 Make and verify conjectures on the relationship between the ratio of length of sides of a triangle
with the sine of the opposite angles, and hence define the sine rule.
[ the use of digital technology is encouraged ]
9.1.2 Solve triangles involving sine rule.
9.1.3 Determine the existence of ambiguous case of a triangle, and hence identify the conditions for such
cases.
9.1.4 Solve triangles involving ambiguous cases.
9.1.5 Solve problems related to triangles using the sine rule.
9.2 Cosine rule
9.2.1 Verify the cosine rule.
9.2.2 Solve triangles involving the cosine rule.
9.2.3 Solve problems involving the cosine rule.
9.3 Area of a triangle
9.3.1 Derive the formula for area of triangles, and hence determine the area of a triangle.
9.3.2 Determine the area of a triangle using the Heron’s formula.
9.3.3 Solve problems involving areas of triangles .
9.4 Application of sine rule, cosine rule and area of a triangle
9.4.1 Solve problems involving triangles.
==========================================================================================================================================

389
==========================================================================================================================================
9.1 Sine Rule
9.1.1 Make and verify conjectures on the relationship between the ratio of length of sides of a
triangle with the sine of the opposite angles, and hence define the sine rule.
==========================================================================================================================================
Show that
sin sin
a b
A B
= .
[2 marks]
Answer :
Show that
sin sin
A B
a b
= .
[2 marks]
Answer :
MIND think :
SINE RULE
~ For any triangle ABC,
sin
a
A
= OR
sin A
a
=
C
b a
c
h
D
A B
A B D
h
C
a
b
c

390
==========================================================================================================================================
9.1.3 Determine the existence of ambiguous case of a triangle, and hence identify the conditions for
such cases.
==========================================================================================================================================
3 The diagram shows an incomplete triangle ABC.
If A, length of a and length of b are fixed.
For each of the following cases, state the number of possible triangle/s that can be formed.
Answer :
a < h a = h a  b h < a < b
4 In the answer space, mark ( ✓ ) for the triangle which exist ambiguous case. If ( ✓ ), sketch the another
difference triangle on the same diagram.
Answer :
(a) (b) (c)
(d) (e) (f)
MIND think :
AMBIGUOUS exists if :
b
A
h B
C
a
B
c
65
5 cm
8 cm
35
5 cm
8 cm
5 cm
8 cm
115
35
5 cm
8 cm
35
8 cm
5 cm
35
8 cm
5 cm
~ given sides and non-included angle.
~ A is an angle, and a b.
A B
C
a
b
A B
C
b
a

391
5 PQR is a triangle where PQ = 12 cm, QR = 6.8 cm, and RPQ = 33. Sketch the two possible triangles
PQR. Hence, find the two possible values of PQR. (Ans : 40.97, 73.03)
[5 marks]
Answer :
6 In a triangle DEF, DE = 8 cm, DF = 10 cm, and DFE = 50. Sketch the two possible
triangles DEF. Hence, find the probable values of the length of EF.
(Ans : 4.122, 8.734)
[5 marks]
Answer :
==========================================================================================================================================
9.2 Cosine rule
9.2.1 Verify the cosine rule.
==========================================================================================================================================
7 The diagram show a triangle ABC.
Show that 2 2 2
2 cos
a b c bc A
= + − .
[3 marks]
Answer :
C
b a
c
h
D
A B

392
Show that
2 2 2
cos
2
b c a
A
bc
+ −
= .
[3 marks]
Answer :
MIND think :
COSINE RULE
b2
= OR cos B =
c2
= OR cos C =
==========================================================================================================================================
9.3 Area of a triangle
9.3.1 Derive the formula for area of triangles, and hence determine the area of a triangle.
9.3.2 Determine the area of a triangle using the Heron’s formula.
==========================================================================================================================================
Show that the area of triangle ABC =
1
sin
2
ab C =
1
sin
2
bc A .
[2 marks]
Answer :
A B
D
h
C
a
b
c
B
c a
b
h
D
A C

393
Show that the area of triangle ABC =
1
sin
2
ab C .
[2 marks]
Answer :
MIND think :
HERON’S FORMULA
• semi perimeter, s =
• area of triangle ABC =
A B D
h
C
a
b
c
A C
B
a
c
b

394
PAPER 2
 2 dimensional surfaces
11 The diagram shows a quadrilateral ABCD such that ABC is acute.
(a) Calculate
(i) ABC, (Ans : 57.23)
(ii) ADC, (Ans : 106.07)
(iii) the area, in cm2, of quadrilateral ABCD. (Ans : 82.37)
[8 marks]
(b) A triangle ABC has the same measurements as those given for triangle ABC,
that is, AC = 12.3 cm, CB = 9.5 cm, and BAC = 40.5, but which is different
in shape to triangle ABC.
(i) Sketch the triangle ABC,
(ii) State the size of ABC. (Ans : 122.77)
[2 marks]
[2004, No.13]
Answer :
A
B
9.8 cm
D
C
40.5
5.2 cm
12.3 cm
9.5 cm

395
(a) Calculate the length, in cm, of AC. (Ans : 19.27) [2 marks]
(b) A quadrilateral ABCD is now formed so that AC is a diagonal, ACD = 40 and
AD = 16 cm. Calculate the two possible values of ADC.
(Ans : 50.73, 129.27) [2 marks]
(c) By using the acute angle ADC from (b), calculate
(i) the length, in cm, of CD, (Ans : 24.89)
(ii) the area, in cm2, of the quadrilateral ABCD. (Ans : 290.1)
[6 marks]
[2005, No.12]
Answer :
A
B
C
15 cm
20 cm
65

396
13 The diagram shows a quadrilateral ABCD.
The area of triangle BCD is 13 cm2
and BCD is acute. Calculate
(a) BCD, (Ans : 60.074) [2 marks]
(b) the length, in cm, of BD, (Ans : 5.5738) [2 marks]
(c) ABD, (Ans : 116.54) [3 marks]
(d) the area, in cm2
, of quadrilateral ABCD. (Ans : 35.439) [3 marks]
[2006, No.13]
Answer :
D
C
B
A 9 cm
5 cm
6 cm
40

397
14 The diagram shows a quadrilateral ABCD.
(a) Calculate
(i) the length, in cm, of AC, (Ans : 13.359)
(ii) ACB. (Ans : 23.89)
[4 marks]
(b) Point A lies on AC such that AB = AB.
(i) Sketch ABC.
(ii) Calculate the area, in cm2
, of ABC. (Ans : 13.785)
[6 marks]
[2007, No.15]
Answer :
D
C
B
5.6 cm
16.4 cm
50
105
A
6 cm

398
15 In the diagram, ABC is a triangle. ADFB, AEC and BGC are straight lines. The straight line FG is
perpendicular to BC.
It is given that BD = 19 cm, DA = 16 cm, AE = 14 cm, DAE = 80 and FBG = 45
(a) Calculate the length, in cm, of
(i) DE, (Ans : 19.34)
(ii) EC. (Ans : 16.21)
[5 marks]
(b) The area of triangle DAE is twice the area of triangle FBG.
Calculate the length, in cm, of BG. (Ans : 10.502) [4 marks]
(c) Sketch triangle ABC which has a different shape from triangle ABC such that AB = AB,
AC = AC and ABC = ABC. [1 mark]
[2008, No.14]
Answer :
E
A
B C
G
D
80
45
F

399
16 The diagram shows a trapezium KLMN. KN is parallel to LM and LMN is obtuse.
Find
(a) the length, in cm, of LN, (Ans : 23.23) [2 marks]
(b) the length, in cm, of MN, (Ans : 21.76) [3 marks]
(c) LMN, (Ans : 98.20) [3 marks]
(d) the area, in cm2
, of triangle LMN. (Ans : 60.31) [2 marks]
[2009, No.12]
Answer :
N
M
L
K
12.5 cm
5.6 cm
32
80

400
The diagram shows triangle ABC and triangle CDE where BCE and ACD are straight lines.
(i) BC, (Ans : 2.207)
(ii) DE, (Ans : 7.072)
[5 marks]
(b) Point C lies on BE such that AC = AC.
(i) Sketch triangle ACB.
(ii) Find ACB. (Ans : 75)
(iii) Calculate the area, in cm2
, of triangle ACB. (Ans : 8.264)
[5 marks]
[2010, No.13]
Answer :
A
B
C
D
E
4 cm
2.5 cm
50
25
6 cm

401
The diagram shows PQR and TQR.
It is given that PQR = 87.95, PQ = 10 cm, PR = 12 cm and TQ = QR = 7 cm.
(a) Find
(i) PRQ, (Ans : 56.39)
(ii) the length, in cm, of TR, (Ans : 7.750)
(iii) the area, in cm2
, of PQT. (Ans : 12.39)
[7 marks]
(b) In the diagram, SQR is the image of TQR under the reflection in the QR.
Find the length, in cm, of PS. (Ans : 16.62) [3 marks]
[2011, No.14]
Answer :
R
Q
P T
12 cm
87.95
10 cm
7 cm
S
P T R
12 cm
87.95
Q
10 cm
7 cm

402
The diagram shows triangle ABC such that ABC = 34 and AB = 9 cm.
It is given that the area of triangle ABC is 28 cm2
.
(a) Calculate
(i) the length, in cm, of BC, (Ans : 11.13) [2 marks]
(ii) the length, in cm, of AC, (Ans : 6.228) [2 marks]
(iii) ACB. (Ans : 53.91) [2 marks]
(b) Point C lies on BC such that AC = AC.
(i) Sketch the triangle ABC.
, of the triangle ABC. (Ans : 9.544)
[4 marks]
[2012, No.14]
Answer :
A
C
B
9 cm
34

403
20 The diagram shows trapezium PQRS.
(a) Calculate
(i) QPR, (Ans : 44.42)
(ii) the length, in cm, of PS. (Ans : 5.214)
[5 marks]
(b) The straight line PQ is extended to Q such that QR = QR.
(i) Sketch the trapezium PQRS,
, of  QQR. (Ans : 4.900)
[5 marks]
[2013, No.13]
Answer :
Q
S R
5 cm
7 cm
4 cm
P
110

404
21 The diagram shows two triangles ABC and BDE.
It is given that BE = 8.5 cm, DE = 4.6 cm and AC = 5.8 cm.
(a) Calculate
(i) the length, in cm, of BC, (Ans : 10.18)
(ii) the length, incm, of CD, (Ans : 2.2)
, of ABC. (Ans : 20.88)
[8 marks]
(b) (i) Sketch a ABC which has a different shape from ABC such that AB = AB, AC = AC
and ABC = ABC.
(ii) Hence, state the size of BAC. (Ans : 11)
[2 marks]
[2014, No.13]
Answer :
140
101
34
A
B
C
D
E

405
22 The diagram shows a quadrilateral PQRS.
(a) Find
(i) the length, in cm, of QS, (Ans : 14.19)
(ii) QRS, (Ans : 141.36)
, of the quadrilateral PQRS. (Ans : 72.38)
[8 marks]
(b) (i) Sketch a triangle SQR which has a different shape from triangle SQR such that SR =
SR, SQ = SQ and SQR = SQR.
(ii) Hence, state SRQ. (Ans : 38.64)
[2 marks]
[2015, No.14]
Answer :
P
8 cm
78
32 Q
9 cm
6 cm
R
S

406
23 The diagram shows a cyclic quadrilateral ABCD.
(a) Calculate
(i) the length, in cm, of AC, (Ans : 9.672)
(ii) ACD, (Ans : 62.21)
[6 marks]
(b) Find
(i) the area, in cm2
, of ABC. (Ans : 27.57)
(ii) the shortest distance, in cm, from point B to AC. (Ans : 5.701)
[4 marks]
[2016, No.15]
Answer :
A
D
C
B
8 cm
7 cm
3 cm
80

407
The diagram shows a quadrilateral ABCD such that AC and BD are straight lines.
It is given that the area of ABC = 6 cm2
and ABC is obtuse.
(a) Find
(i) ABC, (Ans : 121)
(ii) the length, in cm, of AC, (Ans : 6.532)
(iii) BAC, (Ans : 27.34)
[7 marks]
(b) Given BD = 7.3 cm and BCD = 90, calculate the area, in cm2
,
of ACD. (Ans : 17.808) [3 marks]
[2019, No.13]
Answer :
A
D
3.5 cm
4 cm
B
C

408
 Part C ~ 2 dimensional and 3 dimensional surfaces
25 The diagram shows a tent VABC in the shape of a pyramid with triangle ABC as the horizontal base. V
is the vertex of the tent and the angle between the inclined plane VBC and the base is 50.
Given that VB = VC = 2.2 m and AB = AC = 2.6 m. Calculate
(a) the length of BC if the area of the base is 3 m2,
(Ans : 2.700) [3 marks]
(b) the length of AV if the angle between AV and the base is 25,
(Ans : 3.149) [3 marks]
(c) the area of triangle VAB. (Ans : 2.829) [4 marks]
[2003, No.15]
Answer :
A
V
B
C

409
The diagram shows a quadrilateral ABCD on a horizontal plane.
VBDA is a pyramid such that AB = 12 m and V is 5 m vertically above A. Find
(a)  BDC, (Ans : 56.88) [2 marks]
(b) the length, in cm, of BD. (Ans : 21.01) [3 marks]
(c) the area, in m2
, of inclined plane BVD. (Ans : 62.64) [5 marks]
[2017, No.15]
Answer :
B
V
C
A 10 m
D
64
22 m
20.5 m

410
The diagram shows a transparent prism with a rectangular base ABCD. The inclined surface ABFE is a
square with sides 12 cm and the inclined surface CDEF is a rectangle. AED is a uniform cross section
of the prism. BDE is a shaded plane in the prism.
It is given that ADE = 37 and EAD = 45. Find
(a) the length, in cm, of DE, (Ans : 14.099) [2 marks]
(b) the area, in cm2
, of the shaded plane, (Ans : 119.06) [6 marks]
(c) the shortest length, in cm, from point E to the straight line BD. (Ans : 10.204) [2 marks]
[2018, No.14]
Answer :
E
A
F
B
C
D

411
FORECAST
(a) Calculate the length, in cm, of AC. (Ans : 9.587) [2 marks]
(b) If the length of AB is extanded to D such that angle ADC = 48, find the length,
in cm, of BD. (Ans : 8.673) [3 marks]
(c) Find the area, in cm2, of triangle ADC. (Ans : 44.16) [3 marks]
(d) Find the shortest distance, in, from C to AD. (Ans : 6.860) [2 marks]
Answer :
110
B
4.2 cm
7.3 cm
A
C

412
(a) Calculate obtuse angle ABC. (Ans : 126.54) [3 marks]
(b) Sketch and label another triangle which is different from triangle ABC, such that
the lengths of AB and AC, and the ACB are maintained.
Hence, find the area, in cm2, of ABC. (Ans : 32.34) [4 marks]
(c) If the length of AB is reduced while the length of AC and ACB are
maintained, so that only one triangle ABC can be formed.
(i) Sketch the triangle ABC. [1 mark]
(ii) Calculate the length, in cm, of AB. (Ans : 5.785) [2 marks]
Answer :
9 cm
40
B
7.2 cm
A
C

413
(a) Calculate PRQ. (Ans : 56.75) [3 marks]
(b) Sketch and label another triangle that is different from triangle PQR in the
diagram, such that the lengths of PR and RQ, and the angle RPQ remain the
same.
Hence, using the cosine rule, find the length of PQ. (Ans : 2.061) [4 marks]
(c) If the length of RQ is shortened while the length of PR and RPQ are
maintained, such that only one triangle PQR can be formed
(i) Sketch the triangle PQR. [1 mark]
(ii) Find the area, in cm2, of the new triangle formed. (Ans : 6.155) [2 marks]
Answer :
P
Q
R
5 cm
4 cm
50

414
31 The diagram shows a quadrilateral PQRS.
(i) PR, (Ans : 17.16)
(ii) PS, (Ans : 8.775)
(iii) RPS, (Ans : 72.09)
[7 marks]
(b) Point P lies on PR such that PS = PS.
Calculate the area, in cm2, of triangle PPS. (Ans : 22.53) [3 marks]
Answer :
R
P
S
Q
25
40
30
8 cm
16.7 cm

415
32 The diagram shows two triangles ABC and BCD, where ACB and BCD are obtuse.
If the area of triangle ABC is 20 cm2, calculate
(a) BAC, (Ans : 24.62) [2marks]
(b) the length, in cm, of BC, (Ans : 5.784) [2 marks]
(c) the length, in cm, of AD. (Ans : 12.14) [6 marks]
Answer :
C B
A
D
36
8 cm
12 cm
9.2 cm

416
33 (a) In the diagram, CFE is a straight line.
Calculate the length, in cm, of CFE. (Ans : 22.79)
[4 marks]
(b) In the diagram, P, Q, R, and S are four points on a horizontal groud. A
surveyor want to measure the distance PR and PS. The surveyor knows that
PQ = 3.2 km.
(i) The distance PS cannot be measure directly because there is a lake
between P and S. By measure, the surveyor found that QPS = 7330
and PQS = 4430.
Find the distance, in km, PS. (Ans : 2.540) [3 marks]
(ii) The distance PR also cannot be measure directly because there is a forest
between R and P. By measure, the surveyor found that RQ = 4.4 km and
PQR = 6220.
Find the distance, in km, PR. (Ans : 4.065) [3 marks]
Answer :
F E
D
29.6
9.9 cm
10 cm
15 cm
C
R
Q
S
P
Forest
Lake

417
34 (a) The diagram shows a triangle PQR. PR is horizontal.
Calculate
(i) the bearing of Q from R, (Ans : 020.13)
(ii) the area of the new triangle if PR is extended, while the lengths of PQ
and QR, and QPR are maintained. (Ans : 24.76)
[5 markah]
(b) In the diagram, KLMN is a cyclic quadrilateral of a circle with centre O.
Calculate
(i) the length, in cm, of LN, correct to two decimal places, (Ans : 9.64)
(ii) KNL. (Ans : 56.06)
[5 marks]
Answer :
R
Q
P
28
5 cm
10 cm
O
K
L
M
N
60
10 cm
7 cm
4 cm

418
35 (a) In the diagram, PTR and QTS are straight lines.
Given that RS = 6 cm, RT = 4 cm, TS = 3 cm, and PQ = 14 cm, calculate
(i) RTS, (Ans : 117.28)
(ii) the length, in cm, of QT. (Ans : 12.07)
[4 marks]
(b) The diagram shows a circle with centre O and a radius of 6 cm.
Given that PRQ = 50 and PR = 5 cm, calculate
(i) the length, in cm, of PQ, (Ans : 9.193)
(ii) the area, in cm2, of triangle PSQ. (Ans : 12.52)
[6 marks]
Answer :
T
P
50
R
Q
S
Q
O
S
P
R
50

419
36 (a) In the diagram, ABCD is a semi circle with centre O and radius 10 cm.
Given that BC = 4 cm and CBD = 32. Calculate
(i) the length, in cm, of CD, (Ans : 10.596)
(ii) the length, in cm, of AD. (Ans : 14.501)
[6 marks]
(b) The diagram shows a pyramid with a horizontal triangular base ABC.
Given AB = 8 cm, BC = 10 cm and ABC = 90. Vertes D is 7 cm vertically
above B. Calculate the area, in cm2, of the slanting surface ADC. (Ans : 60.07)
[4 marks]
Answer :
O
A
C
B
D
32
C
A
D
B
10 cm
7 cm
8 cm

420
37 (a) In the diagram, sin PQR =
13
12
, where PQR is obtuse.
Calculate
(i) the length, in cm, of PR, (Ans : 8.39)
(ii) PSR. (Ans : 41.70)
[5 marks]
(b) The diagram shows a cuboid with a square base of 4 cm, and a height of 6 cm.
Calculate
(i) EGB, (Ans : 66.91)
(ii) the area, in cm2, of triangle EGB. (Ans : 18.76)
[5 marks]
Answer :
S
4 cm
6 cm 12 cm
P
Q
R
72
B
F
H G
E
A
C
D

421
38 (a) In the diagram, cos QPR =
5
3
− .
Calculate
(i) the length, in cm, of RQ, (Ans : 14.422)
(ii) PSR . (Ans : 71.66)
[5 marks]
(b) The diagram shows a pyramid with a horizontal quadrilateral base.
The apex V is 12 cm vertically above P and RPS is 50. Calculate
(i) the length, in cm, of RS, (Ans : 6.130)
(ii) the area, in cm2, of triangle VRS. (Ans : 39.84)
[5 marks]
Answer :
R
P
Q
80
10 cm
5 cm
S
6 cm
P
Q
V
R
S
50
8 cm
3 cm
4 cm

422
39 (a) In the diagram, sin PQR =
5
3
, where SPR is obtuse.
.
Calculate
(i) the length of PR, correct to three significant figures, (Ans : 7.66)
(ii) SPR. (Ans : 102.45)
[5 marks]
(b) The diagram shows a cuboid, where PQ = 8 cm, QR = 6 cm and MP = 4 cm. L is
the midpoint of QR and PK : KQ = 1 : 3.
Calculate
(i) the length, in cm, of NL, (Ans : 9.434)
(ii) the area, in cm2, of triangle KLN. (Ans : 24.92)
[5 marks]
Answer :
R
Q
S
P
60
12 cm
7 cm
22 cm
N T
U
M
K
P Q
L
R
S

423
40 (a) The diagram shows a right prism with a horizontal rectangular base ABCD. Trapezium BCRQ is
the uniform cross section. T is the midpoint of RS.
If RQ =
3
1
BC, calculate the angle between the line CT and the base ABCD. (Ans : 27.92)
[2 marks]
(b) The diagram shows a solid formed by a right prism and a cuboid. The rectangular surface, QRUT,
is inclined. The angle between the line TR and the plane PQRS is 30.
Calculate
(i) the lengths, in cm, of PQ, TR, and AR, (Ans : 6.664, 12, 14.42) [4 marks]
(ii) the area, in cm2
, of triangle TAR. (Ans : 83.13) [4 marks]
Answer :
Q
T
A
P
D
R
S
B
C 12 cm
10 cm
5 cm
R
Q
P
S
T
U
A B
42
10 cm
C
D
6 cm

424
41 (a) The diagram shows a baseball diamond playing field in the shape of a square of side 18 m. The
pitching rubber is located 13.8 m from the home plate on a line joining home plate and second
base.
Calculate the distance, in m, from the pitching rubber to
(i) second base, (Ans : 11.66) [2 marks]
(ii) third base. (Ans : 12.77) [2 markah]
(b) In a triangle DEF, DE = 8 cm, DF = 10 cm, and DFE = 50. Sketch the two possible triangles
DEF. Hence, find the probable values of the length of EF. (Ans : 4.122, 8.734)
[6 markah]
Answer :
Pitching rubber
pembaling
1st
base
Home plate
3rd
base
2nd
base
18 m

425
42 Coast guard station A is located 184 km due west of station B. A cruiser at sea sends an emergency
call that is received by each station. The call to station A indicates that the location of the cruiser is 40
east of north and the call to the station B indicates that the location of the cruiser is 30 west of north.
(a) How far is each station from the cruiser? (Ans : 150, 169.58) [3 marks]
(b) A speedboat capable of speeding 100 km per hour is despatched from the station B to the cruiser.
After travelling for
2
1
an hour, the speedboat encounters heavy crosswinds and strong currents
from the west. The crew finds that the speedboat of off course by 20.
(i) How far is the speedboat from the cruiser ? (Ans : 104.43) [2 marks]
(ii) Through what angle should the speedboat turn to correct its course? (Ans : 29.44)
[2 marks]
(iii) How much time has been added to the trip because of this problem? (Ans : 2.658)
[1 mark]
(c) Since the cruiser needs assitance badly, a helicopter which is capable flying 200 km per hour is
deispatched from station A. After 15 minutes journey, the pilot discovered that he was 10 off
course.
What is the average speed, in kmh−1
, should the pilot maintain, so that the total time to reach the
cruiser is not more than 30 minutes. (Ans : 241.3) [2 marks]
Answer :
A B
184 km
40 30
B
20

A
10


426
43 Given that the lengths of the two sides of a parallelogram are 68.2 cm and 83.3 cm. One of its diagonal
has a length of 42.5 cm. Calculate
(a) the angles of the parallelogram, (Ans : 30.56, 149.44)
(b) the length, in cm, of another diagonal, (Ans : 146.20)
(c) the area, in cm2
, of the parallelogram. (Ans : 2888.48)
[7 marks]
Answer :
44 In a parallelogram, the adjacent angles of a diagonal of length 76.33 cm are 52.2 and 41.45
respectively. Find
(a) the length, in cm, of the sides of the parallelogram, (Ans : 50.63, 60.44)
, of the parallelogram. (Ans : 3053.62)
[7 marks]
Answer :

427
45 The angles of a triangle are in the ratio 5 : 10 : 21, and the shortest side is 35.64 cm. Find
(a) the length, in cm, of the longest side, (Ans : 81.46)
, of the triangle. (Ans : 1112.00)
[5 marks]
Answer :
46 The perimeter of PQR is 40 cm and P : Q : R = 1 : 2 : 6. Find
(a) the length, in cm, of the sides p, q and r. (Ans : 7.392, 13.890, 18.717)
, of the PQR. (Ans : 44.459)
[10 marks]
Answer :
47 The length of the sides of a triangle are x −2, 2x + 7 and 2x + 8. Given the perimeter of the triangle is
63 cm.
, of the triangle. (Ans : 107.99)
[5 marks]
Answer :

428
48 Darren want to prepare a greeting card in the shape of a triangle. The lengths of two sides of the triangle
are 8 cm and 11 cm respectively, and the area of the is 30 cm2
.
(a) Sketch the two possible triangles.
(b) Hence, find the possible lengths, in cm, of the third side. (Ans : 7.501, 17.713)
[7 marks]
Answer :
49 A regular pentagon has sides of 5 cm. Find
(a) the length, in cm, of its diagonal. (Ans : 8.090)
, of the regular pentagon. (Ans : 43.009)
[7 marks]
Answer :
50 If the three sides of triangle are a, b and 2 2
a b ab
+ + .
Find the greater angle of this triangle. (Ans : 120)
[3 marks]
Answer :

429
51 Given that the ratio of the sides of a triangle is 7 : 4 3 : 13, find its smallest angle. (Ans : 20.92)
[3 marks]
Answer :
52 Given PQR in which sin P : sin Q : sin R = 3 : 5 : 7. Find its largest angle. (Ans : 120)
[3 marks]
Answer :
53 Given ABC in which (b + c) : (c + a) : (a + b) = 7 : 8 : 9. Find sin A : sin B : sin C.
(Ans : 5 : 4 : 3)
[3 marks]
Answer :
54 Given ABC in which (b + c) : (c + a) : (a + b) = 4 : 5 : 6. Find A. (Ans : 120)
[3 marks]
Answer :

430
INDEX NUMBERS
- WORKSHEET
Encik Rayner Doukim
Encik Patrick Tan

432
WORKSHEET
TOPIC 10 : INDEX NUMBERS
[ Part C → 10 marks ]
==========================================================================================================================================
10.1 Index numbers
10.1.1 Define index numbers and describe the use of it.
10.1.2 Determine and interpret index numbers.
10.1.3 Solve problems involving index numbers.
10.2 Composite index
10.2.1 Determine and interpret composite index with and without the weightage.
10.2.2 Solve problems involving index numbers and composite index.
==========================================================================================================================================

433
1 The diagram is a bar chart indicating the weekly cost of the items P, Q, R, S, and T for the year 1990.
Table below show the prices and the price indices for the item.
Items Price in 1990 Price in 1995 Price index in 1995 based on 1990
P x RM 0.70 175
Q RM 2.00 RM 2.50 125
R RM 4.00 RM 5.50 y
S RM 6.00 RM 9.90 150
T RM 2.50 z 120
(a) Find the value of x, y, and z. (Ans : x = 0.40, y = 137.5, z = 3.00) [3 marks]
(b) Calculate the composite index for the items in the year 1995 based on the
year 1990. (Ans : 140.92) [2 marks]
(c) The total monthly cost of the items in the year 1990 is RM456. Calculate the corresponding total
monthly cost for the year 1995. (Ans : RM 642.60) [2 marks]
(d) The cost of the items increase by 20% from the year 1995 to the year 2000. Find the composite
index for the year 2000 based on the year 1990. (Ans : 169.104) [3 marks]
[2003, No.13]
Answer :
T
12
R
Q
P S
Weekly Cost (RM)
15
24
5
30
33
Items
0

434
2 The table shows the price indices and percentage of usage of four items, P, Q, R, and S, which are the
main ingredients in the production of a type of biscuit.
Item
Price index for the year 1995
based on the year 1993
Percentage of usage
(%)
P 135 40
Q x 30
R 105 10
S 130 20
(a) Calculate
(i) the price of S in the year 1993, if its price in the year 1995 is RM37.70. (Ans : 29)
(ii) the price index of P in the year 1995 based on the year 1991 if its price index in the year 1993
based on the year 1991 is 120. (Ans : 162)
[5 marks]
(b) The composite index number of the cost of biscuit production for the year 1995 based on the year
1993 is 128, calculate
(i) the value of x, (Ans : 125)
(ii) the price of a box of biscuit in the year 1993, if the corresponding price in the year 1995
is RM32. (Ans : 25)
[5 marks]
[2004, No.12]
Answer :

435
3 The table shows the price and price indices for the four ingredients, P, Q, R, and S, used in making
biscuits of a particular kind. The pie chart represents the relative amount of the ingredients P, Q, R, and
S, used in making these biscuits.
Ingredients
Price per kg (RM) Price index for the
year 2004 based on
the year 2001
Year 2001 Year 2004
P 0.80 1.00 x
Q 2.00 y 140
R 0.40 0.60 150
S z 0.40 80
(a) Find the value of x, y and z, (Ans : x = 125, y = 2.80, z = 0.50) [3 marks]
(b) (i) Calculate the composite index for the cost of making these biscuits in the year 2004 based on
the year 2001. (Ans : 129.44)
(ii) Hence, calculate the corresponding cost of making these biscuits in the year 2001, if the cost
in the year 2004 was RM2985. (Ans : 2306.09)
[5 marks]
(c) The cost of making these biscuits is expected to increase by 50% from the year 2004 to the year
2007. Find the expected composite index for the year 2007 based on the year 2001. (Ans : 194.16)
[2 marks]
[2005, No.13]
Answer :
P
Q
R
S
100
60
120

436
4 A particular kind of cake is made by using four ingredients, P, Q, R and S. The table shows the prices
of the ingredients.
Ingredient
Price per kilogram (RM)
Year 2004 Year 2005
P 5.00 w
Q 2.50 4.00
R x y
S 4.00 4.40
(a) The index number of ingredient P in the year 2005 based on the year 2004 is 120. Calculate the
value of w. (Ans : 6) [2 marks]
(b) The index number of ingredient R in the year 2005 based on the year 2004 is 125. The price per
kilogram of ingredient R in the year 2005 is RM 2.00 more than its corresponding price in the year
2004. Calculate the value of x and y. (Ans : x = 8, y = 10) [3 marks]
(c) The composite index for the cost of making the cake in the year 2005 based on the year 2004 is
127.5. Calculate
(i) the price of a cake in the year 2004, if its corresponding price in the year 2005 is RM30.60,
(Ans : 24)
(ii) the value of m if the quantities of ingredients P, Q, R and S used are in the ratio of 7 : 3
: m : 2. (Ans : 4)
[5 marks]
[2006, No.15]
Answer :

437
5 The table shows the prices and the price indices of five components, P, Q, R, S and T, used to produce
a kind of toy. The diagram shows a pie chart which represents the relative quantity of components used.
Component
Price (RM) for the year Price index for the year 2006
2004 2006
P 1.20 1.50 125
Q x 2.20 110
R 4.00 6.00 150
S 3.00 2.70 y
T 2.00 2.80 140
(a) Find the value of x and of y. (Ans : x = 2.00, y = 90) [3 marks]
(b) Calculate the composite index for the production cost of the toys in the year 2006 based on the year
2004. (Ans : 123.5) [3 marks]
(c) The price of each component increase by 20% from the year 2006 to the year 2008. Given that
the production cost of one toy in the year 2004 is RM55, calculate the corresponding cost in the
year 2008. (Ans : 81.51) [4 marks]
[2007, No.13]
Answer :
R
Q
P
S
T
72
36
144

438
6 The table shows the prices and the price indices of four ingredients, fish, flour salt, and sugar, used to
make a type of fish cracker. The diagram shows a pie chart which represents the relative quantity of the
ingredient used.
Ingredients
Price (RM) per kg for the year
2004 2005
Fish 3.00 4.50 150
Flour 1.50 1.80 h
Salt k 0.90 112.5
Sugar 1.40 1.47 105
(a) Find the value of h and of k. (Ans : h = 120, k = 0.80) [3 marks]
(b) Calculate the composite index for the cost of making these crackers in the 2005 based on the
year 2004. (Ans : 126.375) [3 marks]
(c) The composite index for the cost of making these crackers increases by 50% from the year 2005 to
the year 2009. Calculate
(i) the composite index for the cost of making these crackers in the year 2009 based on the year
2004, (Ans : 189.5625)
(ii) the price of a box of these crackers in the year 2009 if its corresponding price in the year
2004 is RM 25. (Ans : 47.39)
[4 marks]
[2008, No.13]
Answer :
Sugar
10 %
Fish
30%
Flour
45%
Salt
15%

439
7 The table shows the prices, the price index and weightages for four types of stationery P, Q, R and S.
Stationary
Price (RM) per unit
Weightage
Year 2007 Year 2008
P 2.80 2.10 x 4
Q 4.00 4.80 120 2
R 2.00 y 130 3
S z 5.80 116 m
(a) Find the value of
(i) x, (Ans : 75)
(ii) y, (Ans : 2.6)
(iii) z, (Ans : 5)
[3 marks]
(b) The composite index for the price of the stationery in the year 2008 based on the year 2007 is 108.4.
Calculate the value of m. (Ans : 6) [3 marks]
(c) The total expenditure for the stationery in the year 2007 is RM 525, Calculate the corresponding
total expenditure in the year 2008, (Ans : 569.1) [2 marks]
(d) The price index for Q in the year 2009 based on the year 2007 is 132. Calculate the price index
for Q in the year 2009 based on the year 2008. (Ans : 110) [2 marks]
[2009, No.13]
Answer :

440
8 The table shows the price indices for three items, P, Q and R used in the production of a type of bag.
Item
Price index in the year 2006
P 125 150
Q 116 x
R y 120
(a) Find the price index of item P in the year 2008 based on the year 2006.
(Ans : 120) [2 marks]
(b) The price of item Q in the year 2004 is RM 7.50 and its price in the year 2008 is RM 10.50. Find
(i) the value of x. (Ans : 140)
(ii) the price of item Q in the year 2006. (Ans : 8.70)
[3 marks]
(c) The composite index for the production cost of the bag in the year 2006 based on the year 2004 is
118.5. The cost of the items, P, Q and R used are in the ratio of 2 : 1 : 3. Find the value of y.
(Ans : 115) [3 marks]
(d) Given the price of the bag in the year 2006 is RM 47.40, find the corresponding price of the bag
in the year 2004. (Ans : 40) [2 marks]
[2010, No.15]
Answer :

441
9 The table shows the price, price indices and percentage expenditure of four ingredients, P, Q, R and S,
used in making of a kind of food.
Ingredient
Price (RM) per kg
Percentage
expenditure (%)
2005 2007
P 4.00 5.00 x 16
Q 3.00 y 150 12
R 8.00 10.00 125 48
S z 3.00 120 24
(a) Find the values of x, y and z. (Ans : x = 125, y = 4.50, z = 2.50) [4 marks]
(b) Calculate the composite index for the cost of making the food in the year 2007 based on the
year 2005. (Ans : 126.8) [2 marks]
(c) The cost of making a packet of the food in the year 2005 was RM 50.00. Calculate the corresponding
cost in the year 2007, (Ans : 63.40) [2 marks]
(d) The cost of all the ingredients increased by 15 % from the year 2007 to the year 2009. Find the
composite index for the year 2009 based on the year 2005. (Ans : 145.82) [2 marks]
[2011, No.13]
Answer :

442
10 The table shows the price indices of three types of fuel for the year 2008 based on the year 2006. The
diagram shows a pie chart which represents the proportion of the fuel used in a factory.
Fuel Price index for the year 2008
Diesel 150
Petrol 120
Gas 110
(a) If the factory spends RM 9000 per week for diesel in the year 2008, find the corresponding
expenditure for diesel in the year 2006. (Ans : 6000) [2 marks]
(b) Calculate the composite index for the fuel expenditure of the factory in the year 2008 based on the
year 2006. (Ans : 133) [3 marks]
(c) The fuel expenditure used by the factory is RM 30000 per week in the year 2006. Calculate it
corresponding fuel expenditure in the year 2008. (Ans : 39900) [2 marks]
(d) The price of diesel increases by 30%, the price of petrol increases by 20% while the price of gas
remains unchanged from the year 2008 to the year 2010. Calculate the composite index for the fuel
expenditure of the factory in the year 2010 based on the year 2006. (Ans : 162.7) [3 marks]
[2012, No.13]
Answer :
Petrol
72
180
Diesel
Gas

443
11 The table shows the price indices, changes in price indices and weightages of four item A, B, C and D,
which are the main items used to make a tin of biscuit.
Item
Changes in price index from the
year 2012 to the year 2014
Weightage
A 112 No change 1
B 140 10% decrease 4
C x No change 2
D 130 5% increase 3
(a) Calculate
(i) the price of item B in the year 2010 if its price in the year 2012 is RM8.40, (Ans : 6.00)
(ii) the price of item D in the year 2012 if its price in the year 2010 is RM4.50. (Ans : 5.85)
[3 marks]
(b) The composite index for the cost of making a tin of biscuit in the year 2012 based on the year 2010
is 132. Calculate the value of x. (Ans : 129) [2 marks]
(c) Hence, calculate the composite index for the cost of making a tin od biscuit in the year 2014 based
on the year 2010. (Ans : 128.35) [3 marks]
(d) Calculate the cost of making a tin of biscuit in the year 2014 if the corresponding cost in the year
2010 is RM20. (Ans : 25.67) [2 marks]
[2013, No.14]
Answer :

444
12 The table shows the price indices and the weightages of four ingredients, P, Q, R and S, used in the
making of a cake. The composite index for the cost of making the cake in the year 2014 based on the year
2013 is 106.
Ingredient
Weightage
P 115 3
Q 95 1
R 100 4
S m 2
(a) Calculate the price of ingredient Q in the year 2014 if its price in the year 2013 is
RM 20. (Ans : 19) [2 marks]
(b) Find the percentage of price change from the year 2013 to the year 2014 for
ingredient S. (Ans : 10) [4 marks]
(c) The composite index for the cost of making the cake increased at the same rate from the year 2014
to the year 2015, calculate **
(i) the composite index for the expenses in the year 2015 based on the year
2013, (Ans : 112.36) [2 marks]
(ii) the price of the cake in the year 2015 if its corresponding price in the year 2013 is
RM75 (Ans : 84.27) [2 marks]
[2014, No.15]
Answer :

445
13 The table shows the price indices for the year 2013 and 2015 based on the year 2011 of the three
materials A, B, and C used in making a type of shoe.
Material
A 106 120
B 105 125
C 110 m
(a) The price of material C in the year 2011 is RM12.00 and its price in the year 2015 is RM15.60.
Find
(i) the value of m, (Ans : 130)
(ii) the price of the material C in the year 2013. (Ans : 13.20)
[3 marks]
(b) The composite index for the production cost of the shoe in the year 2013 based on the year 2011
is 106.7. The ratio of the materials A, B and C used are 2 : h : 3. Find
(i) the value of h, (Ans : 5)
(ii) the corresponding price of the shoe in the year 2011 if the price of the shoe in the year 2013
is RM58.20. (Ans : 54.55)
[5 marks]
(c) Find the price index of material B in the year 2015 based on the
year 2013. (Ans : 119.05) [2 marks]
[2015, No.15]
Answer :

446
14 The table shows the price indices and change in price indices of four raw materials A, B, C and D, used
to produce a type of biscuits in a factory.
Raw material
Price index in 2011
based on 2008
Change in price index from
2011 to 2015
A 140 15 % increase
B 120 5 % increase
C 160 Unchange
D 150 10 % decrease
The diagram is a bar chart which represents the mass of the raw materials used to make the biscuits in
2008.
(a) The price of raw material A in 2011 is RM70. Find the corresponding price
in 2008. (Ans : 50) [2 marks]
(b) Find the price indices of all the four raw materials in 2015 based on 2008. [3 marks]
(c) (i) Calculate the composite index for the cost of producing the biscuits in 2015 based
on 2008. (Ans : 149)
(ii) Hence, find the cost of producing the biscuits in 2008 if the corresponding cost in 2015
is RM268.20. (Ans : 180)
[5 marks]
[2016, No.12]
Answer :
0
5
10
15
20
25
30
A B C D
Raw material
Mass (kg)

447
15 The table shows the prices and the price indices of three types of ingredients A, B and C, used in the
production of a type of fish ball.
Ingredinet
Price (RM) per kg for the
year Price index for the year 2016
Weightage
2014 2016
A 5.00 6.64 132.8 50
B y 3.00 x 20
C 0.50 0.95 190 1
(a) The price of ingredinet B is increased by 20% from the year 2014 to the year 2016.
(i) State the value of x.
(ii) Find the value of y. (Ans : 2.50)
[3 marks]
(b) Calculate the composite index for the cost of making the fish balls for the year 2016 based on the
year 2014. (Ans : 130) [2 marks]
(c) It is given that the composite index for the cost of making the fish balls increased by 40% from the
year 2012 to the year 2016.
(i) Calculate the composite index for the cost of making the fish balls in the year 2014 baded on
the year 2012. (Ans : 107 13
9 )
(ii) The cost of making a fish ball is 10 sen in the year 2012. Find the maximum number of
fish balls that can be produced using an allocation of RM80 in the year 2016. (Ans : 571)
[5 marks]
[2017, No.13]
Answer :

448
16 The table shows the information related to four ingredients, P, Q, R and S, used in the production of a
type of noodle.
Ingredient Change in price from the
year 2013 to the year 2017
Percentage
of usage (%)
P 40% increase 10
Q 20% increase 10
R 60% increase
S 10% decrease 50
The production cost for this noodle is RM47600 in the year 2017
(a) If the price of ingredient Q in the year 2013 is RM4.20, find its price in the year
2017. (Ans : 5.04) [2 marks]
(b) Percentage of usage for several ingredients were given in the table. Calculate the corresponding
production cost in the year 2013. (Ans : 40000) [5 marks]
(c) The production cost is expected to increase by 50% from the year 2017 to the year 2019.
Calculate the percentage of changes in production cost from the year 2013 to the
year 2019. (Ans : 78.5) [3 marks]
[2018, No.13]
Answer :

449
17 The table shows information related to five cake ingredients, J, K, L, M and N used by a baker in his
business.
Ingredient
Price Index for the
year 2018 based on
the year 2016
Change in the price
index from the year
2018 to the year
2020
Price index for the
year 2020 based on
the year 2016
Weight
J 124 No change 124 5
K 115 40% increase x 6
L 130 No change 130 p
M 140 10% decrease y 4
N 120 No change 120 2
The composite index for the cost of making the cakes in the year 2020 based on the year 2016 is 136.
(a) (i) Find lthe value of x and of y.
(ii) Calculate the price for ingredient M in the year 2016 if the price in the year 2020
is RM6.30. (Ans : 5)
[4 marks]
(b) Calculate the p. (Ans : 3) [3 marks]
(c) The cost of baking a cake in the year 2016 is RM25. Find the selling price of a cake in the year
2020, if the baker intends to make a profit of 80%. (Ans : 61.20) [3 marks]
[2019, No.15]
Answer :

450
FORECAST
18 The table shows the prices and the price indices of three types of ingredients A, B and C, used in the
production of a type of fish ball.
Ingredinet Price index for the year 2016 based on the year 2014
A 132.6
B x
C 125
D y
(a) The price of ingredinet B is decreased by 5% from the year 2014 to the year 2016 and the price
of ingredinet D is unchnaged from the year 2014 to the year 2016. State the value of :
(i) x [1 mark]
(ii) y [1 mark]
(b) Calculate the composite index for the cost of making the fish balls for the year 2016 based on the
year 2014. (Ans : 113.15) [2 marks]
(c) It is given that the composite index for the cost of making the fish balls increased by 35% from the
year 2012 to the year 2016.
(i) Calculate the composite index for the cost of making the fish balls in the year 2014 baded on
the year 2012. (Ans : 119.31)
(ii) The cost of making a fish ball is 20 sen in the year 2012. Find the maximum number of
fish balls that can be produced using an allocation of RM150 in the year 2016. (Ans : 555)
[6 marks]
Answer :

451
19 (a) The price indices of an item for the year 2005 based on the year 2000 and the year 1995 are 120
and 135 respectively. Given that the price of the item is RM 45 in 2000, find the price of the item
in 1995. (Ans : 40) [2 marks]
(b) The table shows the prices of three items, A, B, and C, in the year 1996 and 1998, together with
their weightages.
Item Price (RM) in 1996 Price (RM) in 1998 Weightaget (%)
A 70.00 105.00 y
B 80.00 100.00 x
C 60.00 67.50 2x
(i) Using the year 1996 as the base year, calculate the price index of items A, B,
and C. (Ans : 150, 125, 112.5) [3 marks]
(ii) Given the composite price index of these items in the year 1998 based on the year 1996 is
140, find the values of x and y. (Ans : x = 10, y = 70) [5 marks]
Answer :

452
20 The table shows the price indices and weightages of five items, A, B, C, D, and E in the year 2002
based on the year 2000.
Item Price index Weightage
A 132 23
B 130 13
C p q
D 136 9
E 106 5
The total weightages is 63 and the composite index in the year 2002 based on the year 2000 is 126.
(a) Find
(i) the values of q, [1 mark]
(ii) the percentage of price change from the year 2000 to the year 2002 for
item C. (Ans : 12.15) [3 marks]
(b) Find the value of price index of item D in the year 2000, using the year 2002 as the
base year. (Ans : 73.529) [2 marks]
(c) From the year 2002 to the year 2003, the cost of item B increases by 10%, item D decreases 5%,
item E increases 15%, and the remaining items remain unchanged. Calculate the composite index
of the items in the year 2003 based on the year 2000. (Ans : 128.97) [4 marks]
Answer :

453
21 (a) The table shows the prices and price indices of an item from the year 1999 to 2002.
Year 1999 2000 2001 2002
Price (RM) 160 180 a b
Price index (1999 = 100) 100 c 90 d
Price index (2000 = 100) e 100 f 120
Find the value of a, b, c, d, e, and f.
(Ans : a = 144 b = 216, c = 112.5, d = 135, e = 88.89, f = 80) [6 marks]
(b) The table shows the prices and mass of four items, using in making a type of cake, in the year
1999 and 2001.
Item Price in 1999 Price in 2001 Mass (kg)
A RM2.00 RM2.50 10
B RM12.00 RM14.40 5
C RM1.00 RM1.30 20
D RM5.00 RM6.00 15
Calculate the composite index for the cost of making these cake in the year 2001 based on the
year 1999. (Ans : 125) [4 marks]
Answer :

454
22 The table shows the cost of maintaining three type of machines by a factory in the year 2002 and 2003,
price indices in the year 2003 based on the year 2002, with their respective weightages.
Machine 2002 2003 Price index Weightage
P RM 12,000 RM 15,000 125 2
Q RM 7,000 y 150 3
R RM 5,000 RM 5,500 z 10
(a) Calculate
(i) the value of y and z, (Ans : y = 10500, z = 110)
(ii) the composite index of the cost of maintaining the machines in the year 2003 based on the
year 2002. (Ans : 120)
[5 marks]
(b) The composite indes continues to increase at the same rate from the year 2003 to the year 2004.
(i) Calculate the composite indices of the cost of maintaining the machines in the year 2004,
using the year 2002 as the base year. (Ans : 144)
(ii) If the cost of maintaining the machines in the year 2002 is RM 450,000, calculate the mean
of the cost of maintaining the machines form the year 2002 to 2004. (Ans : 546000)
[5 marks]
Answer :

455
23 (a) The table shows the information related to four types of transportations fares between two towns.
Transport
Change in price from the
year 2013 to the year 2017 Weightage
Aeroplane 40% increase 2
Taxi 30% increase 3
Bus 10% increase 4
Train 22% increase 1
Calculate
(i) the taxis fare in the year 2013, if the taxis fare in the year 2017 is RM80. (Ans : 61.54)
(ii) the price index of transportations fares in the year 2017 based on the year 2013.
(Ans : 123.2)
(iii) the price index of aeroplanes fare in the year 2017 based on the year 2015, if the price
index in the year 2015 based on the year 2013 is 120. (Ans : 116.67)
[7 marks]
(b) The price index of a certain item for the year 2000 based on the year 1997 is 130. Based on the
year 2000, the price index of the item in the year 2002 is 108. Calculate the price index of the
item in the year 2002 with the year 1997 taken as the base year.
(Ans : 140.4) [3 marks]
Answer :

456
24 The table shows the prices for four types of books in a bookstore for three consecutive years.
Book
Price for the year (RM) Price Index Price Index
Weightage
2000 2001 2002
2001
(2000 = 100)
2002
(2000 = 100)
P w 20 30 150 225 6
Q 50 x 65 115 130 5
R 40 50 56 125 140 3
S 80 z 150 y y 2
(a) Find the values of w, x, y and z. (Ans : w = 13.3, x = 57.5, y = 187.5, z = 150) [4 marks]
(b) State the price index for the year 2002 based on the year 2001 for book R. [Ans : 112] [1 mark]
(c) Calculate the composite index for the price of the book for the year 2002 based on the
year 2001. (Ans : 125.075) [3 marks]
(d) A school spent RM4865 buying the books for school library in 2002. Find the estimated allocation
for the books in 2003 if the composite index for the year 2003 based on the year 2002 is equal to
the composite index for the year 2002 based on the year 2001. (Ans : 6084.90) [2 marks]
Answer :

457
25 The table shows the price index of the cost of few items needed to run a company, with their respectively
weightages.
Item
Price index
Weightage
2004
(2002 = 100)
2005
(2002 = 100)
A 101 102 3
B 118 121 6
C 105 107 k
D 103 110 4
E 105 120 5
Given the composite index in the year 2004 based on the year 2002 is 107.9.
(a) If the expenses of item D in the year 2004 is RM2678, find its expenses in the
year 2005. (Ans : 2860) [2 marks]
(b) If the expenses of item B in the year 2005 is RM3993, find the increase in its expenses compare to
the year 2004. (Ans : 99) [2 marks]
(c) Calculate the value of k. (Ans : 2) [3 marks]
(d) Based on the year 2004, calculate the composite index for the year 2005.
(Ans : 106.033) [3 marks]
Answer :

458
26 The table shows the price for an item in 2000 and 2015.
Year Price
2000 RM12
2015 RM15
(a) If the rate of price increase from 2015 to 2020 is twice the rate of price increase from 2000 to
2015, determine the price of that item in 2020. (Ans : 22.50)
(b) Calculate the price index in the year 2020 based on the year 2000. (Ans : 187.5)
[5 marks]
Answer :
27 The table shows the number of visitors who visit Mount Kinabalu National Park in 2010 and 2015.
Year Number of visitors
2010 2.54 million
2015 3.86 million
(a) Determine the number of visitors in 2020 if the increase for the number of visitors in 2015 to
2020 is twice the increase from 2010 to 2015. (Ans : 6.5)
(b) Calculate the index for the number of visitors in the year 2020 based on the year 2015. State the
interpretation based on the index number obtained. (Ans : 168 76
193
)
[5 marks]
Answer :

459
28 The table shows the price indices and weightage of four items P, Q, R and S.
Item
based on the year 2011 Weightage
P 105 10% increase 5
Q 120 No change k
R 160 No change 5
S 130 5% decrease 1
Calculate :
(a) (i) The price of item P in the year 2013 if its price in the year 2011 is
RM10.20. (Ans : 10.71)
(ii) The price of item S in the year 2011 if its price in the year 2013 is
RM8.60. (Ans : 6.62)
[3 marks]
(b) The composite index for the four items in the year 2013 based on the year 2011 is 129. Calculate
the value of k. (Ans : 4) [3 marks]
(c) If the price of the four items P, Q, R and S is RM20.80 in the year 2011, calculate the price of
the four items in the year 2015. (Ans : 21.42) [4 marks]
Answer :

460
29 The table shows the price indices and weightages of four items, A, B, C and D used in a production of
a type of food.
Ingredient
based on the year 2018 Weightage
A 100 4
B 120 3
C m 2
D 106 1
The composite index for the cost of making the food for the year 2020 based on the year 2018 is 105.6
(a) Find the value of m and hence, give your comment on the change in price of
item C. (Ans : 95) [4 mark]
(b) Calculate the price of item B in the year 2020 if its price in the year 2018 was
RM7.80. (Ans : 9.36) [2 marks]
(c) The composite index for the cost of making food is expected to increase by 15% form the year
2020 to the year 2022. Calculate
(i) the composite index for the year 2022 based on the year 2018, (Ans : 121.44)
(ii) the price of 1kg of the food in the year 2022 if its price in the year 2020 is
RM85.00. (Ans : 97.75)
[4 marks]
[YIK2020, No.13]
Answer :

461
30 The table shows the prices and the price indices of four types of material R, S, T and U used in the
production of a type of shoes.
Material
Price (RM) per unit
Percentage of
usage (%)
Year 2015 Year 2020
R 4.00 6.00 150 20
S q 2.23 p 30
T 0.80 1.00 125 10
U 0.50 0.67 134 m
(a) The price of ingredient S is increased by 20% from the year 2015 to the year 2020.
(i) State the value of m,
(ii) State the value of p,
(iii) Find the value of q, (Ans : 1.83)
[4 marks]
(b) Calculate the composite index for the production of the shoes in the year 2020 based on the
year 2015. (Ans : 132.1) [2 marks]
(c) It is given that the composite index for the cost of the production of shoes increased by 35% form
the year 2013 to the year 2020.
(i) Calculate the composite index for the cost of the production of shoes in the year 2015 based
on the year 2013. (Ans : 120.20)
(ii) The cost of the production was RM7.50 in the year 2013. Find the maximum number of
shoes that can be produced using allocation of RM488.00 in the year 2020. (Ans : 48)
[4 marks]
[Pulau Pinang2020, No.12]
Answer :

462
31 A hawker sells “Yong Tau Fu” with the selective ingredients of brown squids, chicken balls, fish balls,
tofu and water spinach.
Ingredient
Price (RM) per unit
Percentage of
usage (%)
Year 2017 Year 2020
Brown Squids 25.00 28.50 x 10
Chickne ball 7.20 9.00 125 w
Fish ball 8.00 y 140 25
Tofu 12.00 13.50 112.5 15
Water spinach z 4.80 120 20
(a) Find the value of w, x, y and z. (Ans : x = 114, y = 11.20, z = 4) [4 marks]
(b) Calculate the composite index for cost of preparing the “Yong Tau Fu” in the year 2020 based on
the year 2017. (Ans : 124.775) [2 marks]
(c) If the hawker spends RM250 daily to buy the ingredient in the year 2017, find the total cost to buy
the ingredients in March 2020. (Ans : 9670.14) [2 marks]
(d) The cost of all the ingredient increases by 14% from the year 2020 to the year 2021. Find the
composite index for the year 2021 based on the year 2017. (Ans : 142.24) [2 marks]
[Terengganu2020, No.12]
Answer :

463
32 The diagram is a bar chart which represents the ratio of production of clothes for the factory in the year
2020. The table shows the price indices for four types of clothes A, B, C and D produced by a factory.
Material Price index in 2018 based on 2016 Price index in 2020 based on 2018
A 105 107
B 110 x
C 95 110
D 115 125
(a) (i) The price of clothes B in the year 2016 is RM35, and its price in the year 2018 is RM42.
Find the value of x. (Ans : 109.09)
(ii) Find the price of clothes in the year 2018. (Ans : 38.50)
[4 marks]
(b) Find the price index of clothes A in the year 2020 based on the year 2016.
(Ans : 112.35) [2 marks]
(c) (i) Calculate the composite index for the production cost of the clothes in the year 2020 based
on 2018. (Ans : 115.127)
(ii) Hence, given the total price of the four types is RM150 in the year 2018, find the
corresponding price of the clothes in the year 2020. (Ans : 172.69)
[4 marks]
[Kedah2020, No.15]
Answer :
0
2
4
6
8
A B C D
Cothes
Ratio of production

464
1) Can be viewed or downloaded using the following link.
bit.ly/MTWORKSHEETANSWER
or
2) Scan this QR code.
WORKSHEET ANSWERS

465
COORDINATORS
1) Dg Rukayah Ag Mahmun Jabatan Pendidikan Negeri Sabah
2) Nor Khamisah @ Julie Atin Jabatan Pendidikan Negeri Sabah
PANELS
1) Lee Chiong Tee
(Penyelaras/Ketua Zon 2)
SM St Peter, Telipok
2) Dr. Ma Chi Nan
(Ketua Zon 1)
SM St Michael, Penampang
3) Peter Wong Yung Ming
(Ketua Zon 3)
Sekolah Tinggi Kota Kinabalu
4) Norfadzilah Lee
(Ketua Zon 4)
SMJK Tiong Hua, Sandakan
5) Ong Choon Keat
(Ketua Zon 5)
SM St John, Tuaran
6) Hafizi Fazli Bakar
(Ketua Zon 6)
SMK Kemabong, Tenom
7) Rayner Doukim SMK Tandek, Kota Marudu
8) Patrick Tan SMK Tandek, Kota Marudu
9) AK Sapri Yaakub SMK Menumbok, Kuala Penyu
10) Tan Woon Shin SMK St Paul, Beaufort
11) Maya Insana Mohd Terang SMK Madai, Kunak
11) Mohd Salleh Ambo SMK Terusan, Lahad Datu
12) Mohd Zulkarnain bin Zulkifli SMK Pamol, Beluran
13) Adry Colodius SMK Nabawan, Pensiangan
14) Elbenjoe Wesmin SMK Pinggan-Pinggan, Pitas
15) Nadzrinah binti Ahmad SMK Pitas 2, Pitas
16) Suhairul bin Hadlee SMKA Mohamad Ali, Ranau
17) Lai Hoi Yong SMK Elopura, Sandakan
18) Sandra Chong Ket San SMK Sandakan
19) Hartono bin Josed SMK Pengiran Omar, Sipitang
20) Jamaliah Mohd. Elmi SMK Tamparuli, Tuaran
21) Dennis Chua Ah Thin SMK Pekan Telipok, Tuaran
“Success is not final; failure is not fatal:
It is the courage to continue that counts”
ACKNOWLEDGEMENT

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