MATHEMATICS
Form 4 •5
Vincent De Selva A/L Santhanasamy (Textbook Author)
Chai Mun
Dr. Nur Hamiza Adenan
Pan Asia Publications Sdn. Bhd.
199101016590 (226902-X)
No. 2-16, Jalan SU 8,
Taman Perindustrian Subang Utama, Seksyen 22,
40300 Shah Alam,
Selangor Darul Ehsan, Malaysia.
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First published 2023
Spotlight A+1 Mathematics Form 4 & 5
ISBN 978-967-466-736-8
Printed by Percetakan Rina Sdn. Bhd. (31964-X)
Extra Features of This Book
CHAPTER Quadratic Functions and concept map
Equations in One Variable
1 The entire content of the
SMART SCOPE Important Learning Standards Page chapter is summarised in the
SMART form of a concept map.
Contains Learning Standards • Identify and describe the characteristics of quadratic expressions in 3
(LS) that need to be achieved 1.1 Quadratic one variable.
in each chapter. Functions and
Equations • Recognise quadratic function as many-to-one relation, hence, 3
describe the characteristics of quadratic functions.
• Investigate and make generalisation about the effect of changing the 4
values of a, b and c on graphs of quadratic functions, f (x) = ax2 + bx + c.
• Form quadratic functions based on situations, and hence relate to 5
the quadratic equations.
• Explain the meaning of roots of a quadratic equation. 5
• Determine the roots of a quadratic equation by factorisation method. 6
• Sketch graphs of quadratic functions. 6
• Solve problems involving quadratic equations. 7
Concept Fo4rm
CHAP.
1
2
Words Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
• Effect of change/ Kesan perubahan One variable, x Power of x is a whole number Highest power of x is 2
• Horizontal line test/ Ujian garis mengufuk
• Many-to-one relation/ Hubungan banyak kepada satu
• Maximum point/ Titik maksimum
• Method of factorisation/ Kaedah pemfaktoran
• Minimum point/ Titik minimum
• Quadratic equation/ Persamaan kuadratik Many-to-one relations Characteristics of quadratic Life situations
Form expressions ax2 + bx + c based on
4 Mathematics Chapter 1 Quadratic Fu••n cRQtauiotaend osrfaa ctnihcda fnEugqneuc/ta iKtoiaondn/a sFruinpnegOrsunibekauVhaaadrniraabtilke identify Quadratic equation ax2 + bx + c = 0
valu•••e sRRVaeoraoilta /rb oPloeu/tn /PS cPeaoDbumlynvebectofaealernchqmytuauoaitbrnadaiesrhaattthiicoenreomqouetastthiooonfdsa are
1CHAP. Example 9 factorisation. quadratic equation Quadratic functions identify Quadratic Functions and Equations form meaning determine
by the method of in One Variable
Determine whether each of the following
of x is the root of the quadratic equation
x2 + x – 6 = 0. describe sketch
(a) x = 1 (b) x = 2 (c) x = –3 1 Write the quadratic equation in the form
ax2 + bx + c = 0.
Solution: 2 Factorise ax2 + bx + c = 0 in the form Characteristics based on Quadratic graphs f (x) = ax2 + bx + c Roots of quadratic equations
(a) When x = 1, x2 + x – 6 = 12 + 1 – 6 (mx + p)(nx + q) = 0. 1
3 State mx + p = 0 or nx + q = 0.
= – 4 ≠ 0 4 Solve the two linear equations in 3 to obtain
∴ x = 1 is not the root of the equation
x2 + x – 6 = 0. x = – —mp– or x = – —nq–. investigate the effect of
Curved- Axis of symmetry of graph change of a, b and c
(b) When x = 2, x2 + x – 6 = 22 + 2 – 6 shape is parallel to the y-axis
=0 Example 10 Maximum or c=0 b=c=0 b=0
∴ x = 2 is the root of the equation minimum point a.0
x2 + x – 6 = 0.
Determine the roots of the following quadratic a.0 a,0 a,0 y a,0 a.0
(c) When x = –3, x2 + x – 6 = (–3)2 + (–3) – 6 equations by the method of factorisation. y y y x y
Ox O
=9–3–6 (a) x2 + 5x = 14 Ox y Ox
=0 (b) (3x + 2)(x – 1) = 3x + 13 Ox Ox
∴ x = –3 is the root of the equation Solution:
x2 + x – 6 = 0.
(a) x2 + 5x = 14 x +7 +7x
Try question 9 in Formative Zone 1.1 x2 + 5x – 14 = 0 x –2 –2x
(x + 7)(x – 2) = 0 x2 –14 +5x
BRILLIANT Tips x + 7 = 0 or x – 2 = 0
x = –7 or x=2
y (b) (3x + 2)(x – 1) = 3x + 13
y = x2 + x – 6 The graph y = x2 + x – 6 3x2 – 3x + 2x – 2 = 3x + 13
3x2 – 4x – 15 = 0
–3 O cuts the x-axis at x = –3 (3x + 5)(x – 3) = 0 3x +5 +5x
x and x = 2. Therefore, the x –3 –9x
2 –15 –4x
roots of the quadratic 3x2
3x + 5 = 0 or x – 3 = 0 Chapter 1 Quadratic Functions and Equations in One Variable Mathematics
equation x2 + x – 6 = 0 are
the x-intercepts of the x = – —53 or x=3 Fo4rm
graph y = x2 + x – 6.
brilliant tips Try question 10 in Formative Zone 1.1
Useful tips to help Sketch graphs of quadratic functions 1.1 Quadratic Functions and (c) The expression 6t2 + pt – 9 contains two CHAP.
students solve Graphs of quadratic functions of the form Equations variables p and t. Therefore, 6t2 + pt – 9 is not
problems in the a quadratic expression in one variable. 1
related subtopics.
y = ax2 + c y = ax2 + bx y = a(x + m)2 Iqdueanydtr=ifayati(acpnxed+xpdmree)s(scqsrxiiob+nens)thine characteristics of (d) The expression y – 7y—21 contains one variable
(a) a . 0 (a) a . 0 one variable y. However, the power of y in the term 7y—21 is
(a) a . 0 (a) a . 0 not a whole number. Therefore, y – 7y—21 is not
y y y 1. Qalgueabdrraaictyiecxperxepsrseiossnioonf thine one variable is an a quadratic expression in one variable.
form ax2 + bx + c, a, b
and c are constants, a ≠ 0 and x is a variable.
Try question 2 in Formative Zone 1.1
2. Cvahraiar–ba–mclp–etO:eristic–s–qno–f qxuadratic expressions in one
c x O – a–b x O –m x • (bE)xpare,ss0ions contain only one variable.
O x (b) a , 0 (b) a , 0 x • The poywer of the variable is a whole number.
(b) a , 0 • The highest power of the variable is 2. Spotlight portal
y y
y Scan the QR code to
O – ab– x O –m BRI–L–mpL–OIANT– –qn– Txips Quadratic equations in one variable browse website or video
c https://bit.ly/3d1ZKeM related to the subtopics
The variable x in quadratic expressions can also be
O represented by other alphabet letters. BRILLIANT Tips
6 1.1.5 1.1.6 1.1.7 If p represents a constant, then 6t2 + pt – 9 is a
quadratic expression in one variable, t.
Example 1
4a2 + b + 3 —r22 – 2r –h2 + 8h – 2 3t2 + —5t Recognise quadratic function as many-
to-one relation, hence, describe the
Identify quadratic expressions in one variable characteristics of quadratic functions
Fo4rm
from the list of expressions above.
Chapter 2 NSuomlubteior nB:ases Mathematics
Example 12 2. nNuummbbee—r2rr2ss –in2r,b–ahse2 +tw8oh –ca2n be converted to 1. Quadratic function is a many-to-one relation. learned.
in base eight and vice versa based on 2. Characteristics of quadratic functions
Convert the tableTryaqbuoevsteio.n 1 in Formative Zone 1.1
(a) 150318 1to10a1n2utombaenruimn bbearseintwbaos.e eight, • The graph has a curved shape.
calculator (b) ExaNmbuampselbetewr2oin • It has a maximum point or a minimum point.
• The axis of symmetry of the graph is parallel
Explains how to use a Solution: CHAP.
scientific calculator in 2Divide theDdeigtietsrmofine whReetphlearce eeaacchhdigoitfof tthhee following to the y-axis.
mathematics calculations. (a) 10111012 = 1 × 26 + 0 × 25 + 1 × 24 + 1 × 23 the numbeerxipnrbeassseions is naumqbueardinrabtaicse eeixgphrtession in one
+ 1 × 22 + 0 × 21 + 1 × 20 BRILLIANT Tips
tAgging ‘Try question ... two into gvroauripasbolef . Give yowuitrhjuthseticfiocrarteisopno.nding
= 64 + 0 + 16 + 8 + 4 + 0 + 1 three digit(saf)ro2mm2 – 9m +t5hree digits in base two. Graphs of quadratic functions y = ax2 + bx + c
in Formative Zone ...’
= 9310 8 93 Remainder tSthhueebrtshiegrqheuteetdno((cbitgt)l)hyit,es65relwtxep23filtt+a–h. c2pext + 1I0gnore any zero in front
= 1358
8 11 … 5 – 9 of the number in base
two.
8 1… 3 tdhigeitcoinrrbeasps(deo)neidygihn–tg.7y—21
(b) 538 = 5 × 81 + 3 80 0… 1
= 40 + 3 2
× 43 Remainder SolutioNnu:mber in a.0 a,0 Fo4rm
ChaptMerax1imQuumadporainttic Functions and Equations in One Variable Mathematics
= 4310 2 21 … 1 (a) The beaxspereeigsshiton 2m2 – 9m + 5 contains one
= 1010112 2 variable m and the highest power of m is
2 10 … 1 CHAP.
Perform com2ep.xupTtrahetesisroeionfonsriienn,vo2onmlev2ivna–gri9aabmdlde+.itio5nis a quadratic
5… 0 and subtr(abc) tiTohneoefxnpuremssbioenrs5ixn3 v–ar2ixou+s 1b0asceosntains one EMxinaimmupmlepo1in1t Example 14 1
1. (Aad) diNtiuonmboe3vefax.rnsrpTuiairhmnebesblbreseeaiorfxsoen.brHtaeiwns,oeo5wosxnei3nev–evvraa2,rrxtiihao+ebusl1heb0i.gaihsseesns:ot tpoawqeuraodfraxtiics
2 2… 1 Axis of Ssykmemtcehtry the graph Afoxirs oef asycmhmeotfry the following Sketch the graph for each of the following
+ 02 12 quadratic functions. quadratic functions. Mark the points where the
2 1… 0 (a) y = x2 – 2 graph cuts the x-axis and the y-axis.
(b) y = –x2 + 4 (a) y = (x – 1)(x – 3)
0… 1 (b) y = –2x2 – 11x – 14
Alternative Method
(a) 10111012 = 001 011 1012 02 02 12 Solution: Solution:
=1 3 58 (a) y y = x2 – 2 (b) y
38
(b) 538 = 5 1.112.1 112.1.2102 4 y = 3–x2 + 4 (a) y y = (x – 1)(x – 3)
3
= 101 0112 (b) Numbers in base three Ox
–2
Calculator + 03 13 23 Ox
(a) Press: MODE MODE 3 BIN 03 03 13 23 Try question 11 in Formative Zone 1.1 O1 x
13 13 23 103 3
2 1 0 1 1 1 0 1 = OCT 23 23 103 113
(b) y = –2x2 – 11x – 14
(b) Press: MODE MODE 3 OCT Example 12 = –(2x2 + 11x + 14)
= –(2x + 7)(x + 2)
2 5 3 = BIN Fo4rm (c) Numbers in base four Sketch the graph for each of the following y
Try question 12 in Formative Zone 2.1
+ 04 14 24 34 quadratic functions. Mark the points where the – 27– –2 O x
Mathemati0c4s C0h4 apte1r42 Nu2m4 ber3B4ases graph cuts the x-axis.
(a) y = x2 + 3x (b) y = –2x2 + 7x
14 ••••123444 24 34 1904× 8 + 3 –14 y = –2x2 – 11x – 14
BRILLIANT Tips1. •• •• •• •• •• •• •• •• •• •• •• •• 24 •• •• •• 34 104 1×141 (b) Place value 25 24 2S3 olu2t2ion:21 20 Try question 14 in Formative Zone 1.1
•• •• •• •• •• •• •• •• •• •• •• •• 34 •• •• •• 104 114 124
•• •• •• •• Digit 1 1 (0a) 0 1 y0 (b) y y = –2x2 + 7x
y = x2 + 3x O x
Table shows the digits for number in base eight The place value of the underlined digit 0 = 23
that correspond to the three digits for number 2–7
in base two. CHAP. ••• (d) Numbers in base five The value of digit 0 = 0 × 23 Solve problems involving quadratic equations
= 0 –3 O x
Number 2+ 05 9 eo1ing5ehstsarae2n5d 35 45
3 (c) Place value 72 71 70 Example 15
05 05 r1esgi1rxo5tyu-pfoeud2r5sa,s 35 45 Digit 4 Try question 12 in Formative Zone 1.1
in base 0 1 2 3 4 5 6 7 15 15 1onee2igs5.hts a3n5d 3 45 105 In the diagram, PQRS is a rectangular plot of land.
eight 10
Number The shaded region that is planted with brinjol has
in base 000 001 010 011 100 101 110 111 25 25 35 45 105 115 The place ovaf lduiegiotf4th=e4un×d7e2rlinEexdadmigpitle4 1=372 an area of 388 m2.
two The value
•• •• •• •• ••••4355 ••••43••••55 45 105 111+1255×1 6×11423855 = 196 Sketch the graph for each of the following
•• •• •• •• •• •• •• •• •• •• •• •• •• 105 115 S 30 m R
•• •• •• •• •• •• •• •• ••
(d) Place value 64 63 qmuina6idm2rautmic6p1fuoninctt.6io0ns. Label its maximum point or xm
+3×1 Digit 2 0 (a) 5y = (x1+ 2)2 1 N 20 m
(b) y = –2x2 + 4x – 2
•• •• •• •• • • • 25 The place ovaf lduiegiotf2th=e2un×d6e4rSlionleudtiodnig:it 2 = 64
2.1.2 2.1.3 75 = 1 × 64 + 1 × 8 + 3 × 1 The value
Number in base eight to represent 75 is 1138. P M xm Q
= 2 592(a) y = (x + 2)2 y
Try question 4 in Formative Zone 2.1 Determine the value of x.
Try question 5 in Formative Zone 2.1 4 Solution:
(–2, 0) O x S 30 m R
xm 20 m
2. The place value for each digit of a number in Example 6 (b) y = –2x2 + 4x – 2 y
base a is a times greater than the place value of N
the digit on its right-hand side. Da inaugmrabmersihnoawcsenrtuaminbbearsbe.asDeetbelormckins==ere––th22pe((rxxen2s–ue–1mn2)t2xbine+gr 1) •(1, 0) x (20 – x) m
For example, the place values for each digit in and represent it in terms of number value. O
–2
the number 320145 are shown in the following (a) y = –2x2 + 4x – 2 P (30 – x) m M x m Q example
table. Area of the shaded region = 388 m2
Try question 13 in Formative Zone 1.1
Place value 54 53 52 51 50
(b)
Digit 32014
3. The value of a digit in the nduigmitbwerithxa is 1.1.7 1.1.8 7
determined by multiplying the the
place value of its corresponding digit. Solution: Examples with complete
solutions to enhance students’
The tagging is located at the end Example 5 (a) Place value 61 60 understanding of the chapters
of the example guides the student learned.
to answer the corresponding Based on the table of place values, determine Digit 14
questions in Formative Zone. the value of the underlined digit in each of the
following numbers. 146 = 1 × 61 + 4 × 60
= 6 + 4
= 1010
(a) 15028 (b) 1100102
(c) 4107 (d) 205116 (b) Place value 52 51 50
Solution: Digit 322
(a) Place value 83 82 81 80 3225 = 3 × 52 + 2 × 51 + 2 × 50
= 75 + 10 + 2
Digit 1502 = 8710
The place value of the underlined digit 5 = 82 Try question 6 in Formative Zone 2.1
The value of digit 5 = 5 × 82
= 320
22 2.1.1
iv
Chapter 1 Quadratic Functions and Equations in One Variable Mathematics Fo4rm
1.1 CHAP. alternative method
1. 9y2 + 16 x2 – 2y2 + 4 6. Diagram shows the graph y = x2. On the 1 Provides alternative solutions to
—v72– – 20 —15 k2 diagram, draw the graphs y = x2 + 4 and certain questions.
y = x2 – 3. Mathematics
Identify quadratic expressions in one variable y
from the list of expressions above. C1
8
2. Determine whether each of the following 6
expressions is a quadratic expression in one
formative zone variable. Give your justification. C2 y = x2 4
2
Questions to test (a) 7c2 + 3 (b) —y12– – 6
students’ understanding at –2 –1 0
the end of each subtopic. (c) a2 + 4b2 + 9 (d) 3p2 – 10p + 5 –2 x
12
3. Given the quadratic function y = –x2 + 4. C2 Make generalisation about the effect of
(a) Explain why the quadratic function is a change for the values of c on the graph
many-to-one relation. y = x2 + c.
(b) Sketch the graph y = –x2 + 4 by describing C4
the characteristics of the quadratic function.
4. On one diagram, draw the graph y = ax2 for 7. Diagram shows the graphs y = x2, y = x2 – 6x Fo4rm
a = —12 and a = – —21 where –3 < x < 3. Hence, and y = x2 + 6x.
determine the relation between the graphs y = x2 + 6x y y = x2 y = x2 – 6x Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
y = —12 x2 and y = – —21 x2. C4 1CHAP. 30 × 20 – —21 (30 – x)(20 – x) – —12 x(20) = 388 Alternative Method
—23 x2 –6 –3 0 x 600 – —12 (600 – 50x + x2) – 10x = 388 y = ax2 + bx + c
36 x = 1, y = 0: a + b + c = 0 ...............
5. Draw the graphs y = and y = 4x2 on the
following diagram. x = –5, y = 0: 25a – 5b + c = 0 ...............
Make a generalisation about the effect of 600 – 300 + 25x – —12 x2 – 10x = 388 x = –2, y = –18: 4a – 2b + c = –18 ...........
change for the values of b on the graph
y 300 + 15x – —12 x2 = 388 – , 24a – 6b = 0
C4 4a – b = 0................................
16 y = x2 + bx. – , 3a – 3b = –18
14 8. In the diagram, PQRS is a trapezium. —12 x2 – 15x + 88 = 0 a – b = –6 ..............................
SR x2 – 30x + 176 = 0
12 – , 3a = 6
a=2
x –8 –8x (x – 8)(x – 22) = 0 From , b=8
10 (x + 5) cm x –22 –22x x = 8 or x = 22 From , 2 + 8 + c = 0
8 x2 +176 –30x c = –10
6 PQ When x = 22, 20 – x = 20 – 22 = –2. Length of NP y = 2x2 + 8x – 10
is a positive quantity. Therefore, x = 8.
4 The length of PQ is 4 cm more than RS. Try question 15 in Formative Zone 1.1 Try question 16 in Formative Zone 1.1
2 y = x2 C3
(a) Form a quadratic function to represent the Example 17
area of the trapezium PQRS.
–2 –1 0 12 x (b) Hence, find a quadratic equation in the
Fo4rm Hence, complete the following generalisation form ax2 + bx + c = 0, given the area of BRILLIANT Tips A quadratic function y = ax2 + bx + c has the
CHAP. about the effect of change for the values of a the trapezium PQRS is 323 cm2. following information.
1. Represent the desired quantity to be found by a
hots1 10. Mathematics on the graph y = ax2. C4 Varia9b.le Determine whether each of the following suitable symbol such as x. • The axis of symmetry is x = 2.
CWhhaepntert1heQuvaadlruaetic oFfunactioinncsreaansdeEs,qutahteiongsrainpOhne values of x is the root of the quadratic equation • The graph cuts the x-axis at x = –1.
11. x2 – 4x – 5 = 0. C2 2. Form a quadratic equation in terms of x from the • The graph has a maximum point.
y = ax2 . 15. In the diagram, KLMN(a)is xa =plo2t of land in the given information.
Determine the roots for each of the following Sketch the graph y = ax2 + bx + c.
quadratic equations by the method of shape of a trapezium(.b) x = –1 3. Solve the quadratic equation by the method of Solution:
spm simulation (c) x = 5 factorisation.
questions factorisation. C3 M y
(a) x2 – 3x – 4 = 0 4. Check the roots of the quadratic equation to
(b) 2x2 + x = 10 determine the value of x that represents the
(c) (4x + 1)(x – 2) = –5 quantity in the problem.
(d) (3x + 4)(x – 2) = 16 – 3x
N
Sketch the graph for each of the following 15 m Q 9 Example 16 x
xm Diagram shows the graph of a quadratic function. 5
quadratic functions. C2 –1O 2
(a) y = x2 + 5 (b) y = 3x2 + 4 y
(c) y = –x2 + 1 (d) y = –2x2 – 3 K xm P L –5 O 1 x
12. Sketch the graph for each of the following Given KL = 10 m and LM = 25 m. The shaded Try question 17 in Formative Zone 1.1
quadratic functions. Mark the points where region is a fish pond with an area of 188 m2.
the graph cuts the x-axis. C2 Calculate the possible values of x. C5 (–2, –18)
16. Diagram shows the graph of the quadratic
Provide a complete solutions with(a) y=x2–2x function y = ax2 + bx + c. Determine the quadratic function in the form BRILLIANT Tips
(b) y = –x2 – 4x y = ax2 + bx + c.
(c) y = 2x2 – x y (2, 48) 1. The maximum point lies on the axis of
(d) y = –4x2 + 8x Solution: symmetry.
the examiner’s comments for the13. Sketch the graph for each of the following y = a(x – 1)(x + 5) 2. The distances of the points of intersection of the
quadratic functions. Label its maximum point When x = –2, y = –18, graph and the x-axis from the axis of symmetry
or minimum point. C2 –18 = a(–2 – 1)(–2 + 5) are equal.
(a) y = 3(x – 2)2 –18 = a(–3)(3)
–2 O x –18 = –9a
SPM simulation HOTS questions.(b) y=–(x+4)2 6 a=2 BRILLIANT Tips
(c) y = –x2 + 6x – 9
(d) y = 4x2 + 24x + 36 Determine the values of a, b and c. C5 y = 2(x – 1)(x + 5)
= 2(x2 + 4x – 5)
14. Sketch the graph for each of the following 17. A quadratic function y = ax2 + bx + c has the = 2x2 + 8x – 10 x = 1 ⇒ x – 1 is a factor
quadratic functions. Mark the points where the following information. C5 x = –5 ⇒ x + 5 is a factor
graph cuts the x-axis and the y-axis. C4
(a) y = –(x – 2)(x + 4) • The minimum point is (–3, –2).
(b) y = (2x – 5)(x – 1) • One x-intercept of the graph
(c) y = 3x2 – 17x + 24
(d) y = –4x2 – x + 18 y = ax2 + bx + c is –2.
(a) Sketch the graph. 8 1.1.8
(b) Determine the values of a, b and c. Chapter 3 Logical Reasoning
Fo4rm
SPM Simulation HOTS Questions EXAMINER’S
COMMENT
1. An aquarium in the shape of a cuboid has length (x + 9) cm, width x cm and height 50 cm. The volume Paper 1
of the aquarium is 31 500 cm3. Calculate the value of x. C3
1. Which of the following sentences is a statement? 6. Which of the following shows a false statement
Examiner’s Comments: 50 cm C1 A What is the value of 100? C2 is changed to a true statement by using the
Volume = 31 500 CHAP. B The square of 8 is 64.
word “no” or “not”?
(x + 9)(x)(50) = 31 500 3 C Multiply both sides of the inequality –y –2 A The symbol π is a rational number.
x2 + 9x = 630 by –1.
D Factorise the quadratic expression x2 + 4x – 5. The symbol π is not a rational number.
x2 + 9x – 630 = 0 B x2 = 8 is a quadratic equation.
(x – 21)(x + 30) = 0 x cm 2. I The lowest common multiple of 3 and 12 summative zone
C2 is 12. x2 = 8 is not a quadratic equation.
x = 21 or x = –30 (x + 9) cm C 20 000 has one significant figure.
Since x . 0, x = 21. II The factors of 14 are 2 and 7.
III Write the formula for the area of 20 000 does not have one significant figure.
D Right-angled triangles have an angle of 90°.
parallelogram ABCD.
IV The sum of two odd numbers is an even Right-angled triangles do not have an angle
of 90°.
number.
10 7. Which of the following compound statements is
C4 true?
Which of the following are statements? A 5 – 6 = 1 or –8 + 3 = –11 Questions of various
A I, II and III B 0 ÷ 1 = 0 or 1 ÷ 0 = 0
B I, II and IV C 4 × (–2) = –6 or (–3) × (–1) = –3
C I, III and IV D 32 = 6 or 23 = 9
D II, III and IV 8. I 0.0028 = 2.8 × 10–2 or levels of thinking skill
SPM MODEL PAPER 3. Which of the following statements is true? C4 7.03 × 104 = 70 300
C4 A 17 × 10–3 = 0.017
B 38 000 = 3.8 × 103 II 1: 2 =2 : 3 or 40 : 56 = 5 : 8 are provided to evaluate
3
Paper 1 / Kertas 1 C 22 + 32 = 52 III (x – 3)(2x + 1) = 2x2 – 5x – 3
7 3
Instruction: Answer all questions. Time: 1 hour 30 minutes D 10 5 or x2 – 4x + 4 = (x – 2)2
Arahan: Jawab semua soalan. Masa: 1 jam 30 minit the understanding ofIV 30 ÷ 0.01 = 300 or 0.006 × 1 000 = 60
4. Determine the statement that is true.
1. Round off 3.04856 correct to three significant 5. BAUExnpgkr11ae10ps22ska112n0043323473a7sseabnaugami sbaCDetur in base three. C4 A All quadrilaterals have four sides of the same Determine the false compound statements.
figures. nombor dalam asas tiga. A I and III
spm model paper length. B II and III
Bundarkan 3.04856 betul kepada tiga angka bererti. 112210110033 B All quadratic equations have two positive C II and IV each chapter.
A 3.04 C 3.048 D III and IV
B 3.05 D 3.049 roots.
6. BA101111110001101211–11122 001102 = C Some proper fractions are greater than 1.
2. —56—00.—0×0—710—5 = C 11111010111122 D Some cuboids have a square base. Determine the true compound statement.
A 8 × 106 D 9.
B 8 × 107 5. Which of the following is true? C4 A 3 = 0.6 and 0.7 = 7
C 8 × 108 C4 Statement 5 10
SPM format questions D 8 × 109 7. In Diagram 3, QT is a tangent to the circle PQR Truth value B 6 : 12 = 1 : 2 and 3 : 2 = 9 : 4
with centre O at Q. PORT is a straight line. A All obtuse angles lie False 1
Dalam Rajah 3, QT ialah tangen kepada bulatan PQR between 90° and 180°. C 3–1 = 3 and 49 = 7–2
dengan pusat O pada Q. PORT ialah satu garis lurus. True
B All triangles have the D (–1)2 = –1 and –4 = –2
according to the latest SPM3. Diagram 1 shows a cylindrical water pipe with same area. True
radius 2 m. Q
Rajah 1 menunjukkan sebatang paip air yang C Some even numbers are False
berbentuk silinder dengan jejari 2 m. prime numbers.
D Some multiples of 3 are
divisible by 5.
2021 assessment format P 27° x° T FORM 4 ANSWERS
cover all the chapters in O R SPM MODEL PAPER
Forms 4 and 5.
7m
Diagram 1/ Rajah 1 Diagram 3/ Rajah 3
Calculate the volume, in cm3, of the water pipe. Find the value of x. 58
Hitung isi padu, dalam cm3, bagi paip air itu. Cari nilai x.
3 4Use/ Guna π = —272– A 27 C 46
A 1.76 × 107 C 1.76 × 108 B 36 D 63
B 8.8 × 107 D 8.8 × 108 8. In Diagram 4, T is the midpoint of PQ. It is answers
given that cos x° = —53 and sin y° = —78 .
4. In Diagram 2, PQRSTU is a regular hexagon. Dalam Rajah 4, T ialah titik tengah bagi PQ. Diberi
PQH and PRL are straight lines such that
PH = PL. bahawa kos x° = —35 dan sin y° = —78 .
Dalam Rajah 2, PQRSTU ialah sebuah heksagon R
sekata. PQH dan PRL ialah garis lurus dengan
keadaan PH = PL.
TS Complete answers
are provided.
L P y° Scan the QR Code
6 cm x° TQ provided to get
U R 64° K the steps to the
solution.
x° S
Diagram 4/ Rajah 4
PQ H
Diagram 2/ Rajah 2 Find the length, in cm, of PR.
Cari panjang, dalam cm, bagi PR.
Calculate the value of x. A 12 C 16
Hitung nilai x.
A 121 C 133 B 14 D 20
138 ANSWERS
B 129 D Complete answers
http://bit.ly/3vmOen4
429
FORM 4 6. y
y = x2 + 4 8
Chapter 1 Quadratic Functions and Equations in One
Variable
KMODEL Spotlight A+ Mathematics F5.indd 429 05/03/2021 2:50 PM 6
Question Bank 1.1 y = x2 4
2
1. 9y2 + 16, O51nke2
2. (a) Yes; variable, c and the highest power of c x
is 2. –2 –1 O 12
Chapter 1 Quadratic Functions and Equations in One Variable (b) No; The power of y is not a whole number. –2 y = x2 – 3
1. Which of the following characteristics of quadratic function is not correct for f(x) = x2 + 9? (c) No; Two variables, a and b.
A The graph f(x) = x2 + 9 has a minimum point at (0, 9). (d) Yes; One variable, p and the highest power of p
B f is a many to one function. is 2. Graph y = x2 + 4 is the translation of the graph y = x2
C The axis of symmetry for the graph f(x) = x2 + 9 is x = 0. four units upwards, graph y = x2 – 3 is the translation
D The graph f(x) = x2 + 9 cuts the x-axis at x = ±3. 3. (a) Two values x = –1 and x = 1 are mapped to one of the graph y = x2 three units downwards.
value y = 3.
7. The axis of symmetry x = –b.b2 of the graph y = x2 + bx
(b) y changes with the value of
2. Which of the following graphs is correct about the value of a on the graph of the quadratic function y = ax2?
4 8. (a) (x2 + 12x + 35) cm2
Ay Cy y = –x2 + 4 (b) x2 + 12x – 288 = 0
2
y = x2 y = 4x2 –3 –2 –1 O x 9. (a) No (b) Yes (c) Yes
y = 3x2 –2 123
y = x2 10. (a) x = 4 or x = –1 (b) = – 5 or x=2
(d) x = 8 2 = –3
3 x 3
–4 (c) x= 4 or x = 1 or x
question bank Ox Maximum point (0, 4), axis of symmetry x = 0 11. (a) y (b) y
4. y
Drill questions with By Ox
complete answers can be y = x2 4 y = 2–1 x2 5 4
obtained by scanning the Dy 2 Ox Ox
QR Code on the cover of y = x2 x
the book. –3 –2 –1 O 123
–2 (c) y (d) y
–4 y =– 2–1x2
x x 1 Ox
O O y = –5x2 Ox –3
(x + 5) cm
y = – 2–1 x2 Graphs y = 1 x2 and y = – 1 x2 are congruent and
3. The diagram below shows a right-angled triangle. 2 2
reflected on the x-axis.
(x + 10) cm
5. y 12. (a) y (b) y
O
16
14 x –4 O x
2
12
10 y = 4x2
25 cm 8
6
Based on the given information, find the quadratic equation that can be formed. 4 y = 32– x2 (c) y (d) y
2
A x2 + 15x – 250 = 0 C 2x2 + 15x – 500 = 0 y = x2
B x2 + 30x – 500 = 0 D 2x2 + 15x – 750 = 0 x
O x O x
2 2–1 2
4. Which of the following is not a quadratic equation?
A p2 + 8 = 3p C x2 + 3 = 4 –2 –1 O 1
x increases steeper
k= 10 – 3k2 D 13 + 9w – 3w2 = 0 451
B 7
5. Given 4 is a root of the quadratic equation ax2 + 14x – 8 = 0, determine the value of a.
A –4 C3
B –3 D4
1
v
CONTENTS
Formulae viii Revision
date
FORM 4 Relationship and Algebra
Revision Chapter 6 Linear Inequalities in Two 114
date Variables
Relationship and Algebra 1 6.1 Linear Inequalities in Two 116
3 Variables
Chapter 1 Quadratic Functions and 11
Equations in One Variable 6.2 Systems of Linear 120
Inequalities in Two
1.1 Quadratic Functions and Variables
Equations
Summative Zone 126
Summative Zone
Number and Operations 18 Chapter 7 Graphs of Motion 137
Chapter 2 Number Bases 20 7.1 Distance-Time Graphs 139
2.1 Number Bases 32 7.2 Speed-Time Graphs 142
Summative Zone Summative Zone 146
Discrete Mathematics 37 Statistics and Probability 157
Chapter 3 Logical Reasoning 39 Chapter 8 Measures of Dispersion for 159
3.1 Statements 46 Ungrouped Data 161
3.2 Arguments 57 172
Summative Zone 8.1 Dispersion
8.2 Measures of Dispersion
Summative Zone
Chapter 9 Probability of Combined Events 181
Chapter 4 Operations on Sets 70 9.1 Combined Events 183
72
4.1 Intersection of Sets 76 9.2 Dependent Events and 185
4.2 Union of Sets 79 Independent Events
4.3 Combined Operations on
83 9.3 Mutually Exclusive 189
Sets Events and Non-Mutually
Summative Zone Exclusive Events
9.4 Applications of Probability 194
of Combined Events
Chapter 5 Network in Graph Theory 92 Summative Zone 198
5.1 Network 94 Number and Operations 207
Summative Zone 105 209
Chapter 10 Consumer Mathematics: 215
Financial Management
10.1 Financial Planning and
Management
Summative Zone
vi
FORM 5 Revision Chapter 6 Ratios and Graphs of Revision
date Trigonometric Functions date
Relationship and Algebra
Chapter 1 Variation 220 6.1 The Value of Sine, Cosine 346
222 and Tangent for Angle q, 348
1.1 Direct Variation 227
1.2 Inverse Variation 230 0° q 360° 357
1.3 Combined Variation 235 6.2 The Graphs of Sine, Cosine 369
Summative Zone
and Tangent Functions
Summative Zone
Chapter 2 Matrices 241 Statistics and Probability
2.1 Matrices 243 Chapter 7 Measures of Dispersion for 379
Grouped Data
2.2 Basic Operations on Matrices 246 381
7.1 Dispersion 392
Summative Zone 263 7.2 Measures of Dispersion 400
Summative Zone
Number and Operations
Chapter 3 Consumer Mathematics: 269
Insurance
Relationship and Algebra
3.1 Risk and Insurance Coverage 271 414
Chapter 8 Mathematical Modeling 416
Summative Zone 279 8.1 Mathematical Modeling 425
Summative Zone
Chapter 4 Consumer Mathematics: 283
Taxation SPM Model Paper 429
285 Answers 451 – 496
4.1 Taxation 297
Summative Zone
Measurement and Geometry
Chapter 5 Congruency, Enlargement and 301
Combined Transformations
5.1 Congruency 303
5.2 Enlargement 307
5.3 Combined Transformation 315
5.4 Tessellation 323
Summative Zone 329
vii
Formulae
NUMBERS AND OPERATIONS • Area of kite
1
= 2 × product of two diagonals
• am × an = am + n • Area of trapezium
• am ÷ an = am – n = 1 × sum of two parallel sides × height
2
• (am)n = amn • Surface area of cylinder = 2pr2 + 2prh
• m = (am) 1 • Surface area of cone = pr2 + prs
n
an
• Simple interest, I = Prt • Surface area of sphere = 4pr2
( )• Compound interest, MV = P 1+ r nt • Volume of prism = area of cross section × height
n
• Volume of cylinder = pr2h
• Total repayment, A = P + Prt
1
RELATIONSHIP AND ALGEBRA • Volume of cone = 3 pr2h
• Volume of sphere = 4 pr3
3
• Distance = (x2–x1)2 +(y2 –y1)2 1
• Volume of pyramid = 3 × base area × height
( )• Midpoint, (x, y) = + y1 + y2
x1 2 x2 , 2 • Scale factor, k = PA„
PA
• Average speed = Total distance • Area of image = k2 × area of object
Total time
y2 – y1
• m = x2 – x1 STATISTICS AND PROBABILITY
• m = – y-intercept
x-intercept
• Mean, = ∑x
• x N
A–1 = 1 d –b
ad – bc –c a
• Mean, x = ∑fx
f
MEASUREMENT AND GEOMETRY • Variance, s2 = ∑(x – x)2 = ∑x2 – x 2
N N
• Pythagoras Theorem, c2 = a2 + b2 • Variance, s2 = ∑f(x – x)2 = ∑fx2 – x2
∑f ∑f
• Sum of interior angles of a polygon
= (n – 2) × 180° • Standard deviation, s
= ∑(x – x)2 = ∑x2 – x 2
N N
• Circumference of circle = pd = 2pr
• Standard deviation, s = ∑f(x – x)2 = ∑fx2 – x 2
• Area of circle = pr2 ∑f ∑f
n(A)
• Arc length = q • P(A) = n(S)
2pr 360°
• P(Aʹ) = 1 – P(A)
• Area of sector = q
pr2 360°
viii
CHAPTER Network in Graph Theory
5 Important Learning Standards Page
94
SMART • Identify and explain a network as a graph.
96
5.1 Network • Compare and contrast
(i) directed graphs and undirected graphs. 97
(ii) weighted graphs and unweighted graphs. 100
100
• Identify and draw subgraphs and trees.
• Represent information in the form of networks.
• Solve problems involving networks.
Words
• Network/ Rangkaian
• Graph/ Graf
• Point/ Bintik
• Vertex/ Bucu
• Edge/ Tepi
• Degree/ Darjah
• Simple graph/ Graf mudah
• Directed graph/ Graf terarah
• Undirected graph/ Graf tak terarah
• Loop/ Gelung
• Multiple edges/ Berbilang tepi
• Weighted graph/ Graf berpemberat
• Unweighted graph/ Graf tak berpemberat
• Subgraph/ Subgraf
• Graph theory/ Teori graf
92
Concept
Road Simple Multigraphs Directed Undirected
networks graphs graphs graphs
represent are identify
compare
Networks identify Network in Graph Theory identify Graphs Chapter 5 Network in Graph Theory Mathematics
identify draw
compare
represent represent
Transportation Social Subgraphs Weighted Unweighted
networks networks identify draw graphs graphs
Trees
93
CHAP. Fo4rm
5
Chapter 9 Probability of Combined Events Mathematics Fo4rm
9.1 Combined Events Example 1
Describe combined events and list down the In an experiment, Fahmi chooses a number at
possible combined events random from {1, 3, 4, 6} and Ganan chooses a
number at random from {3, 6, 9}. Given A is the
1. Combined event is the event that is produced event Fahmi chooses an even number and B is the
from the union or intersection of two or more event Ganan chooses an odd number.
events. Describe, in words and listing, the outcomes of
each of the combined events.
Combined Event A or Event A and (a) A and B (b) A or B
event event B event B Solution:
A>B A = Event Fahmi chooses an even number
Notation A<B = {(4, 3), (4, 6), (4, 9), (6, 3), (6, 6), (6, 9)}
Both A and B = Event Ganan chooses an odd number
A occurs or B B occur = {(1, 3), (1, 9), (3, 3), (3, 9), (4, 3), (4, 9), (6, 3), (6, 9)}
Meaning occurs or both A (a) The combined event A and B is the event
and B occur Fahmi chooses an even number and Ganan
chooses an odd number.
Venn ξA B ξA B A > B = {(4, 3), (4, 9), (6, 3), (6, 9)}
diagram (b) The combined event A or B is the event Fahmi
chooses an even number or Ganan chooses
an odd number.
A < B = {(1, 3), (1, 9), (3, 3), (3, 9), (4, 3), (4, 6),
(4, 9), (6, 3), (6, 6), (6, 9)}
Try question 1 in Formative Zone 9.1
BRILLIANT Tips Example 2
Combined event for three events A, B and C. A number is chosen at random from {10, 12,
(a) A < B < C 13, 15, 18, 20, 23, 24}. Given A is the event of
choosing a prime number, B is the event of
ξ B choosing a multiple of 3 and C is the event of CHAP.
A choosing a number with the sum of digits greater
than or equal to 5. 9
List down all the outcomes for each of the
following combined events.
(a) A and C (b) B and C
(c) A or C (d) B or C
C Solution:
B
(b) A > B > C A = Event of choosing a prime number
= {13, 23}
ξ B = Event of choosing a multiple of 3
A = {12, 15, 18, 24}
C = Event of choosing a number with the sum of
the digits greater than or equal to 5
= {15, 18, 23, 24}
(a) A and C = A > C
= {23}
(b) B and C = B > C
= {15, 18, 24}
C (c) A or C = A < C
= {13, 15, 18, 23, 24}
(d) B or C = B < C
= {12, 15, 18, 23, 24}
Try question 2 in Formative Zone 9.1
9.1.1 183
Fo4rm Mathematics Chapter 7 Graphs of Motion
3. Solve problems involving speed-time graphs
Speed Speed Speed
Example 12
O Time O Time O Time The diagram below shows the speed-time graph
for the movement of two particles P and Q in a
Interpretation of speed-time graph period of 30 seconds.
Speed (m s–1)
Positive gradient Zero gradient Negative gradient 20 B
represents represents zero represents D
acceleration acceleration deceleration 16 C
(Speed keeps on (Uniform (Speed keeps on u
increasing) speed) decreasing)
A F E Time (s)
O 18 30
Example 11 Graph AB represents the movement of particle P
whereas graph CDE represents the movement of
The speed-time graph below shows Maria’s run in particle Q.
a sports practice. (a) Determine the value of u.
(b) By using the value of u obtained in (a), find
Speed (m s–1) B
10 A the acceleration of particle Q in the first
CHAP. 18 seconds.
(c) Calculate the difference between the distance
7 travelled by particles P and Q in the period of
30 seconds.
C
O5 15 18 Time (s) Solution:
(a) Find the acceleration, in m s–2, for the (a) Gradient AD = gradient AB
(i) first 5 s, u 20
(ii) duration of time from 5 s until 15 s, 18 = 30
(iii) last 3 s.
u = 20 × 18
(b) Hence, describe Maria’s run in the period of 30
18 s.
= 12
(b) Acceleration = gradient CD
Solution: = – 161–812
= – 148
(a) (i) Acceleration = gradient OA
10
= 5
= 2 m s–2 = – 29 m s–2
(ii) Acceleration = gradient AB
= 0 m s–2 (c) Distance travelled by particle P
= area of triangle ABE
(iii) Acceleration = gradient BC 1
= – 181–015 = 2 × 30 × 20
= – 130 m s–2 = 300 m
Distance travelled by particle Q
BRILLIANT Tips = area of trapezium ACDF + area of triangle
DEF
Negative acceleration is known as deceleration. = 1 × (16 + 12) × 18 + 1 × 12 × 12
2 2
(b) For the first 5 s, Maria ran from 0 m s–1 until = 252 + 72
10 m s–1 with acceleration 2 m s–2. After = 324 m
that, she ran with speed 10 m s–1 for 10 s Difference between the distance travelled by
and decreased her speed with deceleration particles P and Q = 324 – 300
3 13 m s–2 until finally stopped. = 24 m
Try question 5 in Formative Zone 7.2 Try question 6 in Formative Zone 7.2
144 7.2.3 7.2.4
Fo5rm Mathematics Chapter 6 Ratios and Graphs of Trigonometric Functions
6.1 The Value of Sine, Cosine and (b) y 140° lies in
Tangent for Angle θ, x quadrant II.
0° < θ < 360° 140°
Make and verify conjecture about the O
value of sine, cosine and tangent for
angles in quadrants II, III and IV with the (c) y
corresponding reference angle
1. A Cartesian plane is divided into four parts 230° 230° lies in
(called quadrant) by the x-axis and the y-axis. O x quadrant III.
The quadrants are named as quadrant I,
quadrant II, quadrant III and quadrant IV in the (d) y 310° lies in
anticlockwise direction. x quadrant IV.
310°
y O
90°
180° Quadrant Quadrant 0° x
II I 360°
O Try question 1 in Formative Zone 6.1
Quadrant Quadrant
III IV
270° 3. y
2. y
P(x, y)
P 1 y
θ
O xQ x
θ
O x
CHAP. An angle θ is measured by rotating the line OP sin θ = —PO–QP— = —1y– = y
∴ sin θ = y-coordinate
6 in the anticlockwise direction from the positive cos θ = —OO–QP— = —1x– = x
x-axis at the origin. ∴ cos θ = x-coordinate
tan θ = —OP–QQ— = —xy–
Example 1 ∴ tan θ = —xy--–cc—oooo—rrdd—iinn—aa—ttee
Sketch on separate diagrams to represent the BRILLIANT Tips
angles (a) 50°, (b) 140°, (c) 230° and (d) 310°.
Hence, determine the quadrant for each of the C • sin θ = —ABCC––
angles lies. Hypotenuse • cos θ = —AAC–B–
Solution: • tan θ = —ABC–B–
(a) y
50° 50° lies in Opposite side
O x quadrant I.
θ
AB
Adjacent side
348 6.1.1
Chapter 1 Quadratic Functions and Equations in One Variable Mathematics Fo4rm
1.1 Quadratic Functions and (c) The expression 6t2 + pt – 9 contains two CHAP.
Equations variables p and t. Therefore, 6t2 + pt – 9 is not
a quadratic expression in one variable. 1
Identify and describe the characteristics of
quadratic expressions in one variable (d) The expression y – 7y—12 contains one variable
1. Quadratic expression in one variable is an y. However, the power of y in the term 7y—12 is
not a whole number. Therefore, y – 7y—12 is not
algebraic expression of the form ax2 + bx + c, a, b a quadratic expression in one variable.
and c are constants, a ≠ 0 and x is a variable.
2. Characteristics of quadratic expressions in one Try question 2 in Formative Zone 1.1
variable:
• Expressions contain only one variable. Quadratic equations in one variable
• The power of the variable is a whole number. https://bit.ly/3d1ZKeM
• The highest power of the variable is 2.
BRILLIANT Tips
The variable x in quadratic expressions can also be BRILLIANT Tips
represented by other alphabet letters.
Example 1 If p represents a constant, then 6t2 + pt – 9 is a
quadratic expression in one variable, t.
4a2 + b + 3 —r22 – 2r –h2 + 8h – 2 3t2 + —5t
Recognise quadratic function as many-
Identify quadratic expressions in one variable to-one relation, hence, describe the
from the list of expressions above. characteristics of quadratic functions
1. Quadratic function is a many-to-one relation.
Solution: 2. Characteristics of quadratic functions
—r22 – 2r, –h2 + 8h – 2 • The graph has a curved shape.
Try question 1 in Formative Zone 1.1 • It has a maximum point or a minimum point.
• The axis of symmetry of the graph is parallel
Example 2
to the y-axis.
Determine whether each of the following
expressions is a quadratic expression in one BRILLIANT Tips
variable. Give your justification.
(a) 2m2 – 9m + 5 Graphs of quadratic functions y = ax2 + bx + c
(b) 5x3 – 2x + 10
(c) 6t2 + pt – 9 a.0 a,0
(d) y – 7y—21 Minimum point Maximum point
Solution:
Axis of symmetry Axis of symmetry
(a) The expression 2m2 – 9m + 5 contains one
variable m and the highest power of m is
2. Therefore, 2m2 – 9m + 5 is a quadratic
expression in one variable.
(b) The expression 5x3 – 2x + 10 contains one
variable x. However, the highest power of x is
3. Therefore, 5x3 – 2x + 10 is not a quadratic
expression in one variable.
1.1.1 1.1.2 3
Fo4rm Mathematics Chapter 5 Network in Graph Theory
Example 11 Example 13
The diagram below shows a graph G.
The diagram on the right shows DC
graph G. AB D
Determine whether each of the C
following is a subgraph of G.
(a) (b) C
D B BA E
A B D Graph H has the following vertices and edges.
(c) (d) A Determine whether H is a subgraph of G. Give
C B your justification.
D (a) Vertices of H = {A, B, C, D}
Edges of H = {(A, B), (A, C), (B, C), (C, D)}
CHAP. (b) Vertices of H = {A, B, C, E}
Edges of H = {(A, B), (A, C), (C, E)}
5 A B D
Solution:
Solution: (a) H is a subgraph of G because the vertices and
(a) A subgraph of G.
(b) A subgraph of G although its shape is different edges of H are the same as G.
(b) H is not a subgraph of G because {(C, E)} is
from G but it has the same vertices {B, C, D}
and edges {(B, C), (B, D), (C, D)}. not an edge of G.
(c) Not a subgraph of G because G does not
contain the edge {(A, D)}. Try question 13 in Formative Zone 5.1
(d) A subgraph of G although its shape is different
from G but it has the same vertices {A, B, D} Example 14
and edge {(B, D)}.
Try question 11 in Formative Zone 5.1 Draw four possible subgraphs for each of the
Example 12 following graphs G.
(a) (b)
The diagram below shows a directed graph G.
DE
D
E D
C FA C
B
B AB E
AC Solution:
State whether each of the following directed (a)
graphs is a subgraph of G. Give your justification. DD ED ED
(a) A (b)
DB A E C FC FC F
AB B
E (b)
C D DD D
Solution: A CB C C
E E B
(a) Not a subgraph of G because the edge
{(C, D)} is not in its graph. E E
(b) A subgraph of G because the vertices {A, B, Try question 14 in Formative Zone 5.1
D, E} and edges {(A, B), (A, E), (D, E), (E, D)}
are the same as graph G.
Try question 12 in Formative Zone 5.1
98 5.1.3
Chapter 2 Number Bases Mathematics Fo4rm
Convert numbers from one base to another Solution: CHAP.
using various methods
(a) 5710 == 32 + 16 + 8 + 1 2
1. The conversion of numbers from one base 25 + 24 + 23 + 20
to base ten can be carried out by using place = 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 20
values. + 0 × 21 + 1 x 20 1
= 1110012
Example 7 Place value 25 24 23 22 21
Convert each of the following numbers to a Digit 11100
number in base ten. (b) 2 57 Remainder
2
((ca)) 4101301105 2 ((bd)) 16253284 2 28 … 1
2
Solution: 2 14 … 0
2
7… 0
(a) Place value 24 23 22 21 20 3… 1
Digit 11010 1… 1
110102 0… 1
= 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20
= 16 + 8 + 0 + 2 + 0 5710 = 1110012
= 2610
(b) Place value 42 41 40 Calculator
123 Press: MODE MODE 3 DEC
2 5 7 = BIN
Digit
1234 = 1 × 42 + 2 × 41 + 3 × 40 Try question 8 in Formative Zone 2.1
= 16 + 8 + 3
= 2710 Example 9
(c) Place value 53 52 51 50 By using place val(ube) sb, acosenveeigrth6t,5 10 to a number in
(a) base five, (c) base three.
4031
Digit Solution:
40315 = 4 × 53 + 0 × 52 + 3 × 51 + 1 × 50 (a) 6510 == 50 + 15
= 500 + 0 + 15 + 1 2 × 25 +
= 51610 3 × 5
= 2 × 52 + 3 × 51 + 0 × 50
= 2305
(d) Place value 82 81 80
Place value 52 51 50
Digit 652
Digit 230
6528 == 6 × 82 + 5 × 81 + 2 × 80 (b) 6510 == 64 + 1
384 + 40 + 2 82 + 1
= 42610
= 1 × 82 + 0 × 81 + 1 × 80
= 1018
Try question 7 in Formative Zone 2.1
Place value 82 81 80
2. The conversion of numbers in base ten to a Digit 101
number in another base can be carried out by
using place values and division. (c) 6510 == 54 + 9 + 2 + 2
2 × 27 + 9
= 2 × 33 + 1 × 32 + 0 × 31 + 2 × 30
= 21023
Example 8
Convert 5710 to a number in base two by using Place value 33 32 31 30
(a) place values, (b) division. Digit 2102
Try question 9 in Formative Zone 2.1
2.1.2 23
Fo4rm Mathematics Chapter 4 Operations on Sets
(b) Solve problems involving the union of sets
ξ ●19 ●18
●6
A Example 14
●7 B ●17
●5 ●10
A residential area has 105 families. A total of 40
●8 ●20 ●16 families own motorcycles. The number of families
●15 that own cars only is twice the number of families
that own motorcycles only. The number of families
CHAP. ●9 ●11 ●12 ●13 ●14 that own motorcycles and cars are the same as the
Try question 4 in Formative Zone 4.2 number of families that do not own motorcycles
4 or cars.
(a) Draw a Venn diagram to show the information
Operations on sets
https://bit.ly/2IRwsBs above.
(b) Hence, determine the number of families that
Example 13
own motorcycles or cars.
Solution:
(a) j = {families in the residential area}
A = {families that own motorcycles}
B = {families that own cars}
Given the universal set = {x : 4 < x , 16, ξ B
x is an integer}, P = {x : x is a perfect square}, A
Q = {x : x is a multiple of 3} and R = {x : x is a prime
number}. x y 2x
(a) Determine n[(P Q R)]. y
(b) Shade the region that represents (P Q R)
on a Venn diagram.
n(A) = 40
Solution: x + y = 40........ a
(a) = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} n(j) = 105
P = {4, 9} x + y + 2x + y = 105
Q = {6, 9, 12, 15} 3x + 2y = 105...... b
R = {5, 7, 11, 13} a × 2, 2x + 2y = 80........ c
P Q R = {4, 5, 6, 7, 9, 11, 12, 13, 15}
(P Q R) = {8, 10, 14} b – c, x = 25
n[(P Q R)] = 3 From a, 25 + y = 40
y = 15
Alternative Method ξ B
A
n(j) = 12 25 15 50
n(P Q R) = 9 15
n[(P Q R)] = 12 – 9
=3 (b) n(A B) = 25 + 15 + 50
(b) = 90
ξ Q The number of families that own motorcycles
P ●6 or cars is 90.
●4 ●9 ●12 Try question 6 in Formative Zone 4.2
●15
●8 ●14
●5 ●7
●10 ●11
●13 R
Try question 5 in Formative Zone 4.2
78 4.2.2 4.2.3
Chapter 4 Operations on Sets Mathematics Fo4rm
4.3
1. The Venn diagram below shows the elements of 3. The Venn diagram below shows three sets A, B,
sets A, B, C and the universal set j. C and the universal set j.
ξ C B ξA
●c
A ●a BC
●b ●d ●g
●f ●h
●e
List down the elements of C4 On a separate Venn diagram, shade the region CHAP.
that represents C4
(a) A > B C, (b) (A C) > B, (a) (A C > B), (b) [A (B > C)]. 4
(c) A B > C, (d) (A > C) B. 4. In the Venn diagram below, j is the universal
set. P, Q and R are three sets. The number of
2. In the Venn diagram below, the number of elements in each set are shown.
elements of sets A, B, C and the universal set j
are shown. ξQ
P
ξ B x 2 y 26 R
A
x
25 6 7 10 13
4 C
Given n(j) = 109 and n(Q) = 71. C5
5 (a) Find the values of x and y.
(b) Hence, determine
Determine C4
(a) n[(A > C B)], (b) n{[A (B > C)]}. (i) n(P > Q R),
(ii) n[(P > R) Q].
SPM Simulation HOTS Questions EXAMINER’S
COMMENT
1. The diagram below shows a Venn diagram with Answer: C
the universal set j, sets P, Q and R. Examiner’s Tips:
Candidates should not guess the answer for this
ξ Q question. Actually, candidates are required to
P R shade the region that represents each set in the
options.
ξ Q ξ Q
P R P R
Which of the following represents the shaded
region? C4
A (P Q) > R C (Q > R) > P
B (Q R) > P D (P > Q) R (P < Q) > R (Q < R) > P
Examiner’s Comments: ξ
P
ξ Q ξ Q Q
P R P R R
Q > R (Q > R) > P (P > Q) < R
81
Chapter 4 Operations on Sets Mathematics Fo4rm
Paper 1
1. Given set A = {1, 3, 4, 5, 7, 9} and set 6. The diagram below shows a Venn diagram with CHAP.
C2 B = {2, 3, 4, 5, 6, 8, 10}. Which of the following C4 the universal set j = P Q R.
4
describes the intersection of sets A and B? P QR
A Set of integers from 3 to 5. ●3
B Set of integers from 3 to 6. ●1 ●5 ●7 ●2
C Set of integers between 3 and 5. ●6
D Set of integers between 3 and 6. ●4
2. Given set A = {5, 7, 11, 13} and set List down all the elements of the set (P > Q > R).
C2 B = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Determine A {1, 3, 6}
B {1, 2, 4, 7}
A > B in the set builder notation. C {2, 3, 4, 6, 7}
A {x : 4 < x , 12, x is an odd number} D {1, 2, 3, 4, 6, 7}
B {x : 4 < x < 13, x is a whole number}
C {x : 5 < x < 12, x is an even number} 7. A group of 49 customers bought at least one
D {x : 5 < x < 13, x is a prime number} C3 type of fruits from mangosteen, cempedak and
3. Given set F = {letters of the word KOMPAS} and rambutan. The Venn diagram below shows the
C2 set G = {letters of the word PEMBARIS}. Which number of customers who bought mangosteen
and cempedak only.
of the following Venn diagrams describes F > G? PR Q
A C
FG F ●B G 10 18
●K ●E
●K ●A ●B ●A
●M
●P ●E ●I
●M ●O ●P ●I
●S ●O ●R ●S ●R Given P = {customers who bought mangosteen},
B D Q = {customers who bought cempedak} and
R = {customers who bought rambutan}. The
F GF G number of customers who bought mangosteen
●B
●K ●A ●E ●K ●A ●B and cempedak is three times the number of
●M
●O ●I ●S
●P ●O ●I ●P customers who bought rambutan. Determine the
●S ●R ●M ●R ●E
number of customers who bought all the three
4. Given set A = {2, 4, 6, 8, 10, 12}, set fruits.
C4 B = {1, 3, 4, 6, 8, 9} and set C = {3, 4, 5, 7, 8, 10, 11}. A 5 C 19
B 7 D 21
Find A > B > C.
A {4, 8} 8. A school has 33 teachers. An interview was
B {6, 8} C3 carried out to find out the number of teachers
C {4, 6, 8}
D {4, 8, 10} who have visited Hokkaido in Japan and Jeju
Island in Korea. In the interview, it was found
5. Given the universal set j = {x : 4 < x < 20, x is that each teacher has visited at least one of the
C4 an even number}, set M = {multiples of 4} and tourist places. A total of 18 teachers have visited
Hokkaido and 26 teachers have visited Jeju
set N = {even numbers between 4 and 16}. Island. Find the number of teachers who have
Determine (M > N). visited Hokkaido and Jeju Island.
A {4, 6, 10, 14, 16, 18} A 8
B {4, 6, 10, 14, 16, 20} B 9
C {4, 6, 10, 14, 16, 18, 20} C 11
D {4, 6, 8, 10, 12, 14, 16, 20} D 13
83
Fo4rm Mathematics Chapter 4 Operations on Sets
25. The universal set j = P Q R. Which of the 26. The Venn diagram below shows the relation
C4 following shaded regions in the Venn diagram C5 among three sets A, B and C such that the
represents [(P Q) > R]?
A Q R universal set j = A B C.
P AB
C
B Q R
4 x 5 11
P QR
15
CHAP.
Given that n(A) = n(B > C), determine
4 C n[(A C) > B].
P A 12
B 15
D QR C 23
D 27
P
Paper 2
1. Given set A = {6, 8, 10, 12, 16, 24} and set B = {7, 8, 9, 11, 13, 16, 24}. [3 marks]
C2 (a) Describe A ∩ B by using
(i) description,
(ii) listing,
(iii) set builder notation.
(b) Complete the following Venn diagram. B
A
[2 marks]
2. The Venn diagram below shows the relation among sets A, B and C. [1 mark]
C2 [1 mark]
A [1 mark]
●1 ●3 C [1 mark]
●2 B ●4
●5 ●9 ●6
●7 ●12
●16
●10 ●8
●11
Find
(a) B > C,
(b) A ∪ B,
(c) A ∩ (B ∪ C),
(d) (A ∩ B) ∪ C.
86
SPM MODEL PAPER
Paper 1 / Kertas 1
Instruction: Answer all questions. Time: 1 hour 30 minutes
Arahan: Jawab semua soalan. Masa: 1 jam 30 minit
1. Round off 3.04856 correct to three significant 5. UExnpgkraepsska2n423473a7sseabnaugami sbaetur in base three. asas tiga.
figures. nombor dalam
A 111022110033 C 112210110033
Bundarkan 3.04856 betul kepada tiga angka bererti. B D
A 3.04 C 3.048
B 3.05 D 3.049
6. BA10 1111110001101211–11122 001102 =
2. —56—00.—0×0—710—5 = C 11111010111122
A 8 × 106 D
B 8 × 107
C 8 × 108 7. In Diagram 3, QT is a tangent to the circle PQR
D 8 × 109 with centre O at Q. PORT is a straight line.
Dalam Rajah 3, QT ialah tangen kepada bulatan PQR
dengan pusat O pada Q. PORT ialah satu garis lurus.
3. Diagram 1 shows a cylindrical water pipe with Q
radius 2 m.
Rajah 1 menunjukkan sebatang paip air yang P 27° x T SPM MODEL PAPER
berbentuk silinder dengan jejari 2 m. O R
7m Diagram 3/ Rajah 3
Diagram 1/ Rajah 1
Calculate the volume, in cm3, of the water pipe. Find the value of x.
Hitung isi padu, dalam cm3, bagi paip air itu. Cari nilai x.
3 4 Use/ Guna π = —272– A 27°
A 1.76 × 107 C 1.76 × 108 B 36° C 46°
D 63°
B 8.8 × 107 D 8.8 × 108 8. In Diagram 4, T is the midpoint of PQ. It is
given that cos x = —53 and sin y = —87 .
4. In Diagram 2, PQRSTU is a regular hexagon. Dalam Rajah 4, T ialah titik tengah bagi PQ. Diberi
PQH and PRL are straight lines such that bahawa kos x = —35 dan sin y = —78 .
PH = PL.
Dalam Rajah 2, PQRSTU ialah sebuah heksagon R
sekata. PQH dan PRL ialah garis lurus dengan
keadaan PH = PL.
TS
L
U R 64° K P y Q
6 cm x T
x
PQ H S
Diagram 2/ Rajah 2 Diagram 4/ Rajah 4
Calculate the value of x. Find the length, in cm, of PR.
Cari panjang, dalam cm, bagi PR.
Hitung nilai x. A 12 C 16
A 121° C 133°
B 129° D 138° B 14 D 20
429
Mathematics SPM Model Paper
Paper 2 / Kertas 2
Time: 2 hours 30 minutes / Masa: 2 jam 30 minit
Section A/ Bahagian A
[40 marks/ 40 markah]
Instruction: Answer all questions in this section.
Arahan: Jawab semua soalan dalam bahagian ini.
1. (a) The Venn diagram in Diagram 1.1 in the answer space shows sets P and Q such that the universal set
ξ = P Q. Shade the set Pʹ.
Gambar rajah Venn dalam Rajah 1.1 pada ruang jawapan menunjukkan set P dan set Q dengan keadaan set semesta
ξ = P Q. Lorek set Pʹ.
(b) The Venn diagram in Diagram 1.2 in the answer space shows sets H, J and K such that the universal set
ξ = H J K. Shade the set (H J) K.
Gambar rajah Venn dalam Rajah 1.2 pada ruang jawapan menunjukkan set H, set J dan set K dengan keadaan set
semesta ξ = H J K. Lorek set (H J) K.
[4 marks/ 4 markah]
Answer/ Jawapan: (b) H J
(a) P
Q K
SPM MODEL PAPER Diagram 1.1/ Rajah 1.1 Diagram 1.2/ Rajah 1.2
2. Diagram 2 shows a composite solid consisting of a right prism and a cuboid.
Rajah 2 menunjukkan sebuah gabungan pepejal yang terdiri daripada sebuah prisma tegak dan sebuah kuboid.
6 cm
3 cm
4 cm 8 cm
5 cm
Diagram 2/ Rajah 2 [4 marks/ 4 markah]
Calculate the volume, in cm3, of the composite solid.
Hitung isi padu, dalam cm3, bagi gabungan pepejal itu.
Answer/ Jawapan:
436
ANSWERS Complete Answers
https://bit.ly/3RWKRxJ
FORM 4 6. y = x2 + 4 8 y
Chapter 1 Quadratic Functions and Equations in One
Variable
6
1.1 y = x2 4
2
1. 9y2 + 16, O15nke2 variable, c and the highest power of c x
2. (a) Yes; –2 –1 O 12
is 2. –2 y = x2 – 3
(b) No; The power of y is not a whole number.
(c) No; Two variables, a and b.
(d) Yes; One variable, p and the highest power of p
is 2. Graph y = x2 + 4 is the translation of the graph y = x2
four units upwards, graph y = x2 – 3 is the translation
3. (a) Two values x = –1 and x = 1 are mapped to one of the graph y = x2 three units downwards.
value y = 3.
7. The axis of symmetry x = –b.b2 of the graph y = x2 + bx
(b) y changes with the value of
4 8. (a) (x2 + 12x + 35) cm2
y = –x2 + 4 (b) x2 + 12x – 288 = 0
2
–3 –2 –1 O x 9. (a) No (b) Yes (c) Yes FORM 4 ANSWERS
–2 123 5
10. (a) x = 4 or x = –1 (b) = – 2 or x=2
(d) x = 8 = –3
–4 (c) x= 3 or x = 1 x 3 or x
4
Maximum point (0, 4), axis of symmetry x = 0 11. (a) y (b) y
4.
y
4 y = –21 x2 5 4
2 Ox Ox
x
–3 –2 –1 O 123 (c) y (d) y
–2
–4 y = – 1–2x2 1 Ox
Ox –3
Graphs y = 1 x2 and y = – 1 x2 are congruent and
2 2
reflected on the x-axis.
5. y 12. (a) y (b) y
16 O
14 x –4 O x
2
12
10 y = 4x2
8 y = 2–3 x2 (c) y (d) y
6
4 y = x2 O x O x
2 x –21 2
–2 –1 O 1 2
increases steeper
451
Question Bank
Chapter 1 Quadratic Functions and Equations in One Variable
1. Which of the following characteristics of quadratic function is not correct for f(x) = x2 + 9?
A The graph f(x) = x2 + 9 has a minimum point at (0, 9).
B f is a many to one function.
C The axis of symmetry for the graph f(x) = x2 + 9 is x = 0.
D The graph f(x) = x2 + 9 cuts the x-axis at x = ±3.
2. Which of the following graphs is correct about the value of a on the graph of the quadratic function y = ax2?
C
A y y
y = x2 y = 4x2
y = 3x2
y = x2
Ox Ox
B y D y
y = x2 y = x2
x O x
O (x + 5) cm y = –5x2
y = – 12– x2
3. The diagram below shows a right-angled triangle.
(x + 10) cm
25 cm
Based on the given information, find the quadratic equation that can be formed.
A x2 + 15x – 250 = 0 C 2x2 + 15x – 500 = 0
B x2 + 30x – 500 = 0 D 2x2 + 15x – 750 = 0
4. Which of the following is not a quadratic equation?
A p2 + 8 = 3p C x2 + 3 = 4
x
k= 10 – 3k2 D 13 + 9w – 3w2 = 0
B 7
5. Given 4 is a root of the quadratic equation ax2 + 14x – 8 = 0, determine the value of a.
A –4 C 3
B –3 D 4
1
Answers
1. y 9. a = –1 0
Shape
(0, 9) f(x) = 0, 3x – x2 = 0
x(3 – x) = 0
x = 0 or x = 3
y
Ox
Answer: D
2. Answer: C O x
3
3. (x + 10)2 + (x + 5)2 = 252
x2 + 20x + 100 + x2 + 10x + 25 = 625
2x2 + 30x – 500 = 0 Answer: A
10. y = a(x + 4)(x – 1)
x2 + 15x – 250 = 0 = ax2 + 3ax – 4a
Answer: A a = 1, y = x2 + 3x – 4
a = 2, y = 2x2 + 6x – 8
4. x2 + 3 = 4 a = 3, y = 3x2 + 9x – 12
x Answer: D
The power of x is –1.
11. y
Answer: C
y = (x + 3)2
5. x = 4, a(4)2 + 14(4) – 8 = 0
16a + 56 – 8 = 0
16a = –48
a = –3 y = (x – 1)2
Answer: B
6. 2x2 – 13x + 15 = 0 1
(2x – 3)(x – 5) = 0 –3 O 1 x
–1
2x – 3 = 0 or x – 5 = 0
3
x = 2 or x = 5 –2 y = (x – 1)2 – 2
Answer: A
7. (y + 3)2 = 4 12. (a) y
y2 + 6y + 9 = 4 16
y = –x2 + 16
y2 + 6y + 5 = 0
(y + 1)(y + 5) = 0
y = –1 or y = –5
Answer: B
8. x2 + 8x + 16 = 0 x
(x + 4)2 = 0 4
x = –4 or x = –4 (same sign) –4 O
(b) y
x2 + 7x + 6 = 0
(x + 1)(x + 6) = 0 y = x2 + 5x
x = –1 or x = –6 (same sign)
x2 – 2x – 8 = 0
(x + 2)(x – 4) = 0
x = –2 or x = 4 (different sign)
x2 – 11x + 24 = 0
(x – 3)(x – 8) = 0 –5 O x
x = 3 or x = 8 (same sign)
Answer: C
4
New
Topics
CHAP. Fo4rm
10
208
Concept
Fixed Emergency Fixed Variable M A R Mathematics Chapter 10 Consumer Mathematics: Financial Management
saving fund Credit card Measurable Attainable Realistic
Active Passive Loan
Incomes Savings Debts Expenditures S SMART T
Specific concept Time-bound
involve construct Financial Management describe guided by
Personal financial plan Effective process of
financial management
based on
Financial goals Setting financial goals
. 5 years , 1 year Evaluating financial status
Long-term Short-term Creating financial plan
evaluate
Feasibility of financial plans Carrying out financial plan
influenced by
Inflation, health, sources of Review and revising the progress
incomes and expenditures
Chapter 10 Consumer Mathematics: Financial Management Mathematics Fo4rm
10.1 Financial Planning and Example 1
Management Match the following.
Pay car instalment for every
Describe effective financial management month.
process
1. Financial planning is a process of evaluating Give six months salary as Needs
the current and future financial status of an yearly bonus. Wants
individual or organisation by using the sources Provide lunch to all workers.
of incomes and expenditures to determine the Needs
cash flow, asset value and financial plan of the Buy personal Takaful
future. insurance.
2. Financial management is a process involving Solution:
the use of sources of incomes and assets on
the expenditures, savings, investments and Pay car instalment for every
protection to fulfil the financial goals of an month.
individual or organisation.
Give six months salary as
3. The process of financial management involves yearly bonus.
five main steps.
Setting financial Provide lunch to all workers. Wants
1 goals
Rerevvieipswrinoaggnrtdehses5
Buy personal Takaful
fEivnalaunactiialng insurance.
status
Financial Try question 1 in Formative Zone 10.1
management
2 process (b) Evaluating financial status.
• The current financial status can be
finanCcriaelaptilnagn 3 Cafirnryanincgiaol uptla4n determined based on cash flow statement CHAP.
and net asset value statement.
• The cash flow statement gives information 10
on how the sources of incomes are
acquired and spent.
(a) Setting financial goals. BRILLIANT Tips
• The financial goal is usually set at the
beginning of a year. Positive cash flow occurs when the total incomes is
• When setting a financial goal, all exceeding the total expenditures whereas negative
expenditures have to be given priority to cash flow occurs when the total incomes is less than
needs than wants. the total expenditures.
• Financial goals are categorised as short-
term (less than one year), medium-term Total incomes Total incomes
(one year up to five years) and long-term . total expenditures , total expenditures
(more than five years).
Cash flow
Positive cash flow Negative cash flow
10.1.1 209
Fo4rm Mathematics Chapter 10 Consumer Mathematics: Financial Management
• The asset value statement gives 2. • Expenditures are comprised of fixed
information associated with current expenditures (house rental, car instalment,
assets (properties) and liabilities (debts). insurance payment) and variable expenditures
(utility bills, medical cost, car fuel).
• Assets are cash and investments such as
savings, shares, unit trusts and properties. • Variable expenditure is the amount of money
spent on an item or service that varies
• Liabilities are loans, debts of credit cards, according to the needs and wants of a person.
rental and utility bills.
• Fixed expenditure is the amount of money
• The financial status needs to be examined spent on an item or a service that is fixed over a
at least six months once to check the period of time.
achievement of financial goals.
Example 2 • Financial plan is usually prepared
monthly for a term of one year.
The information below shows the total incomes
and total expenditures of Alwi and Basri in a Example 3
month.
The information below shows the incomes and
Alwi : Total incomes = RM2 340 expenditures of Nasrul in May.
Total expenditures = RM2 070
Basri : Total incomes = RM2 580 Active incomes: RM3 470
Total expenditures = RM2 750 Passive incomes: RM650
Determine the monthly cash flow of Alwi and Basri. Fixed expenditures: RM2 860
Hence, state the individual who has the better Variable expenditures: RM1 280
monthly financial status. Give your justification.
Solution: (a) Find
Monthly cash flow of Alwi = 2 340 – 2 070 (i) the total incomes of Nasrul in May,
= RM270 (ii) the total expenditures of Nasrul in May.
Monthly cash flow of Basri = 2 580 – 2 750
= –RM170 (b) Hence, state whether Nasrul is a person who
The monthly cash flow of Alwi is positive. is wise in the financial management in May.
Therefore, the monthly financial status of Alwi is Give your justification.
better.
CHAP. Try question 2 in Formative Zone 10.1 Solution:
10 (c) Creating financial plan. (a) (i) Total incomes = 3 470 + 650
• Financial plan is an important component = RM4 120
in driving the achievement of financial (ii) Total expenditures = 2 860 + 1 280
goal based on the expected financial = RM4 140
value and the actual financial value. (b) The total expenditures exceeded the
• The financial value is determined total incomes in May. Therefore, Nasrul is
based on incomes, savings, debts and not a person who is wise in the financial
expenditures. management in May.
BRILLIANT Tips Try question 3 in Formative Zone 10.1
1. • Incomes are comprised of active incomes (d) Carrying out financial plan.
(salary, allowance, commission) and passive • When carrying out a financial plan,
incomes (rental, interest, dividend). payment for fixed expenditures has
to be given priority to avoid charge or
• Active income is the income received by a additional interest that needs to pay on
person for performing a job that involves time late payment.
and energy. • Expenditures that are planned are
required to make comparison to the
• Passive income is the earnings received from actual expenditures for a particular
investments in which the person is not actively month to control wastage.
involved. • If the monthly net income is surplus,
the money can be used for saving and if
deficit, the expenditures of the month are
required to examine for the subsequent
month.
210 10.1.1
Chapter 3 Consumer Mathematics: Insurance Mathematics Fo5rm
3.1 Risk and Insurance Coverage 10. Compensation that is paid by the insurance
company is based on the principle of indemnity.
Explain the meaning of risk and the According to the principle of indemnity, the
importance of insurance coverage, and insurance company will only recover the
hence determine the types of life insurance policyholder as in the original state, that is,
and general insurance for protecting a the condition before the loss incurred.
variety of risks 11. Compensation can be paid in one of the
1. When an individual experiences an unexpected following ways:
• Payment
event or disaster, such a situation is known as • Recovery or replacement
risk.
2. Risk is the possibility of the occurrence of an 12. Types of insurance
unavoidable disaster.
CHAP.
3. Examples of risks are road accidents, home theft,
home fires and so on. 3
4. Taking out insurance is a measure to protect risk. Life insurance General insurance
5. Insurance is an economic mechanism that
A Life insurance
transfers risk from one party, that is, the 1. Life insurance is an insurance policy that
insurance policyholder to another party, that is,
the insurance company. provides protection to the policyholder in
the event of the occurrence of something
transfer risk undesirable such as death, critical illness, loss of
ability and hospitalisation.
Insurance Pays a premium Insurance
policyholder Pays compensation company 2. Every individual is encouraged to take out a life
insurance for the purpose of:
6. Advantages of taking out insurance to • Financial protection to family members after
policyholders: the policyholder dies.
• Receive compensation in the event of the • The policyholder may receive compensation
occurrence of an insurable risk. if he/she suffers from a total and permanent
• No need to withdraw money from own funds disability.
to cover the risk.
B General insurance
7. The amount of compensation paid by the 1. General insurance is an insurance policy that
insurance company depends on the total amount
of insurance taken by the policyholder. protects an individual from any losses and
damages incurred such as the loss of property
8. The risk that is insured must occur during the due to theft or fire, death due to accident or
period when the insurance is in force. injury and liability arising from third parties
caused by such individual.
9. The compensation that is paid by the insurance 2. Characteristics of general insurance:
company to the policyholder depends on the • Needs to be renewed annually
amount of the actual loss and does not exceed • Using the principle of indemnity
the amount of the loss.
3.1.1 271
Fo5rm Mathematics Chapter 3 Consumer Mathematics: Insurance
3. Types of general insurance: 4. Four types of motor insurance policies:
(a) Act policy
Motor – Protects liability of the third party
insurance for death or bodily injury (excluding
passengers)
Medical Fire (b) Third party policy
and health insurance – Additional protection on the act policy
insurance which provides additional protection to
the third party
Types of – This policy covers losses on property that
general is suffered by the third party
insurance (c) Third party, fire and theft policy
Personal – Additional protection on the third
Travel accident party policy which provides additional
insurance protection to policyholder
3CHAP. insurance – This policy covers losses to own vehicle
caused by accidental fire or theft
Robbery (d) Comprehensive policy
insurance – Additional protection on the third party,
fire and theft policy which provides
(a) Fire insurance additional protection to policyholder
– Provides protection against fire risk – This policy covers losses and damages to
own vehicle due to accident
(b) Personal accident insurance
– Provides coverage to policyholder who 5. Group insurance:
suffers bodily injury, disability or death • Provides coverage to a group of individuals,
resulting directly from accident. This usually employees of a company such as bank
insurance is different from life insurance, employees or pupils in schools and students
and medical and health insurance. in institutes of higher education
• Employees who are covered under this policy
(c) Robbery insurance receive financial protection in the events of
– Provides protection due to robbery death, disability, hospitalisation and surgery
in accordance to the claim limits set out in the
(d) Travel insurance policy
– Protects policyholder against losses
incurred during travel such as death and Investigate, interpret and perform
permanent disability, loss of luggage, calculations involving insurance rates and
passport and money, medical expenses premiums
and others.
1. Premium is the amount of money payable by
(e) Medical and health insurance the policyholder to the insurance company to
– Pays compensation to policyholder who cover the risk.
requires health care due to health and
illness treatments 2. Actuary will take into accounts the past statistical
records that are related to a risk in the calculation
(f) Motor insurance of premiums.
– Provides coverage against any loss or
damage related to the use of motor
vehicle
272 3.1.1 3.1.2
Chapter 3 Consumer Mathematics: Insurance Mathematics Fo5rm
(a) Comprehensive policy for Peninsular BRILLIANT Tips CHAP.
Malaysia:
3
Basic premium The premium rate for third party, fire and theft
= Rate for the first RM1 000 + RM26 for policy is 75% of the basic premium of the
comprehensive policy.
each RM1 000 or part thereof on value
exceeding RM1 000
(b) Comprehensive policy for Sabah and Gross premium for third party policy: RM
Sarawak: 151.20
Basic premium Basic premium
= R ate for the first RM1 000 + RM20.30 Minus NCD 25% 37.80
for each RM1 000 or part thereof on Gross premium
value exceeding RM1 000 Try question 3 in Formative Zone 3.1 113.40
Example 3 BRILLIANT Tips
Mr Hafiz has a car of model X to use in Peninsular 1. No Claim Discount (NCD) clause will be issued
Malaysia. The information of the car is as follows. if no claim is made against the motor insurance
during the coverage period before the policy
Sum insured : RM94 000 renewal is made.
Age of vehicle : 5 years 2. NCD will reduce the total premium payment for
the insurance policyholder.
Engine capacity : 1 799 cc
3. The benefit of NCD will be lost if the policyholder
NCD : 25% has made a claim for a personal or a third party
damage.
Calculate the gross premium of Mr Hafiz's car
under the comprehensive policy, the third party, Solve problems involving insurance
fire and theft policy, and the third party policy. including deductible and co-insurance
A Deductible
Solution: 1. Deductible is a provision in an insurance
Gross premium for comprehensive policy:
contract that requires the policyholder to borne
The first RM1 000 RM a small portion of the losses incurred.
RM26 × 93 (each RM1 000 339.10 2. Under this provision, a certain amount that has
balance) 2 418.00 been set will be deducted from the amount of
compensation that should have been paid by the
Basic premium 2 757.10 insurance company.
Minus NCD 25% 689.28 3. Deductible is a provision that exists in the
medical and health insurance as well as the
Gross premium 2 067.82 property insurance.
4. Deductible is not used in the life insurance.
BRILLIANT Tips
—RM—9—4—R0M0—01—–0—R0M0—1—0—00– = RM93
Gross premium for third party, fire and theft policy:
Basic premium RM
Minus NCD 25%
2 067.83
516.96
Gross premium 1 550.87
3.1.2 3.1.3 275
Chapter 4 Consumer Mathematics: Taxation Mathematics Fo5rm
4.1 Taxation purchase of certain goods or use of services.
For example, cigarette tax.
Explain the purpose of taxation • Government policy implementation tool.
• Taxes are used as a tool to control inflation
1. Tax is a method used by the government to and economic recession.
collect revenue from individuals, companies 5. As responsible citizens, we must pay taxes that
and other entities, to channel in the country’s are levied on time.
development projects besides providing various 6. Paying of taxes is not something that is a burden,
public facilities for the well-being of citizens. in fact it helps to improve the development of the
nation.
2. In Malaysia, taxes are usually collected by
the Inland Revenue Board (IRB), the Royal Describe various taxes, hence the
Malaysian Customs Department (RMCD) and consequences of tax evasion from legal and
several other government departments. financial aspects
3. Taxes are usually collected in the form of money
based on certain conditions.
4. The purpose and the importance of taxes that are 1. In Malaysia, tax collection is managed by two CHAP.
levied by the government: main agencies under the Ministry of Finance,
• To smoothen the administration of the namely the Inland Revenue Board (IRB), the 4
country and to help citizens through the Royal Malaysian Customs Department (RMCD)
development of projects. and the Royal Malaysian Excise.
Tax
2. Types of taxes in Malaysia:
Government Citizens Service tax Income Road tax
– Salary to public – Construction of Sales tax tax Property
assessment
sector employees hospitals Types of
– Other expenses – Construction of taxes tax
for managing schools Quit rent
the government – Free textbooks
administration
• Entrepreneurs
pay taxes to
Government
Tax is channeled to underprivileged group A Income tax
through various types of assistance 1. Income tax is a tax paid by individuals,
• Taxes paid are used as economic stimulants or
movers. companies and other entities that earn
• Taxes can be levied on certain goods or more than a certain amount after deducting
services so that the citizens can reduce the tax exemptions and tax relief allowed by the
government for the period of the concern year
of assessment.
4.1.1 4.1.2 285
Fo5rm Mathematics Chapter 4 Consumer Mathematics: Taxation
2. Failure to submit D Quit rent
Make omission annual report if 1. Besides property assessment tax, property
on any income eligible for tax
in Income Tax payment owners also need to pay quit rent.
Return Form 2. Quit rent is a tax that is levied on private land.
(ITRF) 3. Quit rent is paid by the landowner to the state
authority through the State Land Office.
Income tax 4. Consequences faced by the landowner for failing
to pay quit rent:
Fine of RM1 000 Fine of RM200 up • Arrears will be charged in addition to quit
up to RM10 000 or to RM20 000 or rent.
imprisonment or both imprisonment of not • Notice of claim will be issued. If quit rent is
and penalty of 200% exceeding 6 months not settled within 3 months from the notice of
of the underpaid or both (Income Tax instruction, the land may be seized (National
taxes (Income Tax Act Act 1967 (Act 53) Land Code 1965 Section 100).
1967 (Act 53) Section Section 112(1))
113(1)(a)) E Sales tax
CHAP. 1. Sales tax is levied on final goods that are sold
4 B Road tax by domestic firms and imported goods from
abroad.
1. Motor vehicle owner must pay road tax before 2. Sales tax is a level of indirect tax that is
the vehicle is legally allowed to drive on the road. administered by the Royal Malaysian Customs
Department through the provision of the Sales
2. Vehicle owners Tax Act 2018.
Motor vehicle can be fined of not 3. Sales Tax Registration Threshold when the firm’s
exceeding RM2 000 sales value exceeds RM500 000 per year.
owners who fail to in accordance with 4. Sales tax is set at various rates, i.e. 5% or 10%.
renew the expired the Road Transport 5. However, not all goods that are manufactured in
Act 1987 (Act 333) Malaysia are subject to sales tax.
road tax (no Section 23(1)
insurance)
C Property assessment tax 6. Consequences faced by the business firms/
companies if evading from paying sales tax:
1. Property assessment tax is levied on properties First offence of tax evasion
such as industrials and vacant lands. Minimum fine of 10 times up to 20 times
the amount of sales tax or imprisonment of
2. Property assessment tax is also known as not exceeding 5 years or both (Sales Tax Act
assessment rate. 2018 (Act 806) Section 86(1) and 86(2)).
3. Property will
Property owners who be seized or
do not pay property auctioned
assessment tax
286 4.1.2
Chapter 4 Consumer Mathematics: Taxation Mathematics Fo5rm
Tax payable for the assessment year of 2019 9 231
Minus: Monthly tax deduction (PCB) 1 050
Tax balance for the assessment year of 2019
8 181
Try question 3 in Formative Zone 4.1
BRILLIANT Tips Rebate
• Tax rebate deducted from the tax levied
A separate tax assessment is more economical than
a joint tax. A joint assessment will only result in the to determine the tax payable for the
reduction of the total tax payable if either husband assessment year.
or wife has a low amount of income. • Divided into two, namely zakat or
fitrah tax rebate and individual tax
8. Differences between tax relief and rebate: rebate of RM400 for the taxpayer who
Tax relief has a chargeable income not exceeding
RM35 000.
• Expenditures that are allowed to be
deducted from the total annual income to 9. Calculation of road tax: CHAP.
reduce the amount of income to be taxed. • Road tax is levied based on the engine capacity
of the vehicle used by the owner. 4
• The higher is the engine capacity of vehicle,
the higher is the road tax rate being levied.
Example 4
The table below shows the road tax rates for private cars in Peninsular Malaysia.
Engine capacity Base rate Road tax rate
RM20.00 Progressive rate
1 000 cc and below RM55.00 -
1 001 cc – 1 200 cc RM70.00 -
1 201 cc – 1 400 cc RM90.00 -
1 401 cc – 1 600 cc RM200.00 -
1 601 cc – 1 800 cc RM280.00
1 801 cc – 2 000 cc + RM0.40 for each cc exceeding 1 600 cc
+ RM0.50 for each cc exceeding 1 800 cc
Mr Yusuf and his wife bought two cars with engine capacity of 850 cc and 1 850 cc respectively for private use
in Pahang. Calculate the road tax that is levied on both the cars.
Solution:
Car road tax (850 cc) = RM20.00
Car road tax (1 850 cc) = Base rate + Progressive rate
= RM280 + (1 850 – 1 800) × RM0.50
= RM280.00 + RM25
= RM305.00
Try question 4 in Formative Zone 4.1
4.1.3 293
Fo4rm Mathematics Chapter 5 Network in Graph Theory
BRILLIANT Tips 1. Difference between directed graph and
undirected graph.
In a graph that has loops and multigraph, the degree Graph
of a vertex is determined in the same way as simple
graphs. Directed graph Undirected graph
C C
Example 7 A BA B
A, B, C and D are four vertices of a graph. Without D D
using graph, determine the degree for each of the
vertices with the following edges. Each edge is marked Each edge is not
(a) {(A, B), (A, D), (B, D), (C, D), (C, D), (C, D)} with an arrow. marked with an arrow.
(b) {(A, D), (A, D), (B, B), (B, C), (B, C), (B, D), (C, C)} The edge A • • B The edge A • • B
is written in the is written in the form
5CHAP. Solution: ordered pair (B, A). (A, B).
(a) Vertex A : 2 edges {(A, B), (A, D)}
º Degree (A) = 2 (A, B) ≠ (B, A) (A, B) = (B, A)
Vertex B : 2 edges {(A, B), (B, D)} The edges of the graph The edges of the graph
º Degree (B) = 2 are {(A, C), (A, D), are {(A, B), (A, C),
Vertex C : 3 edges {(C, D), (C, D), (C, D)} (B, A), (B, C), (D, A)} (A, D), (A, D), (B, C)}
º Degree (C) = 3
Vertex D : 5 edges {(A, D), (B, D), (C, D), (C, D), BRILLIANT Tips
(C, D)}
º Degree (D) = 5 1. For the directed graph, (A, A) represents a directed
(b) Vertex A : 2 edges {(A, D), (A, D)} loop.
º Degree (A) = 2
Vertex B : 4 edges {(B, B), (B, C), (B, C), (B, D)} or
º Degree (B) = 4
Vertex C : 3 edges {(B, C), (B, C), (C, C)} AA
º Degree (C) = 3
Vertex D : 3 edges {(A, D), (A, D), (B, D)} 2. For the undirected graph, (A, A) represents an
º Degree (D) = 3 undirected loop.
Try question 7 in Formative Zone 5.1
A
BRILLIANT Tips Example 8
Match the following.
The degree of a vertex is found by determining the (a)
number of edges that involve the vertex.
Compare and contrast Directed graph
(i) directed graphs and undirected graphs (b) Undirected
(ii) weighted graphs and unweighted graphs graph
(c)
96 5.1.1 5.1.2
Chapter 5 Network in Graph Theory Mathematics Fo4rm
Solution: BRILLIANT Tips
(a)
The weight on the edge of a weighted graph may
represents distance, time, cost, price and so on.
Directed graph Example 10
(b)
State whether each of the following graphs
represents a weighted graph or an unweighted
graph.
(a) (b)
3
Undirected 5
(c) graph
10 2
7 CHAP.
(c) 2 11
5
Try question 8 in Formative Zone 5.1 65 4
2
Example 9
3
Solution:
A graph has vertices {A, B, C, D} and edges {(A, A), (a) A weighted graph
(A, B), (A, D), (B, C), (C, A), (C, B)}. (b) An unweighted graph
(a) Mark 3 to show the correct graph. (c) A weighted graph
Directed graph Undirected graph Try question 10 in Formative Zone 5.1
Give your justification. Identify and draw subgraphs and trees
(b) Hence, draw a diagram to represent the
1. Subgraph of a graph is the graph consisting of
graph. the particular vertex and edge of the original
Solution: graph.
(a) Directed graph 3 BRILLIANT Tips
{(B, C)} and {(C, B)} are directed edges that 1. If H is the subgraph of graph G, then
connect vertex B and vertex C. (a) the vertices of H are also the vertices of G,
(b) the edges with its two vertices of H are also
(b) the edges with its two vertices of G.
C 2. A subgraph may have shapes that are different
A from the original graph but its vertices and edges
must be the same.
DB
Try question 9 in Formative Zone 5.1 3. Each graph is a subgraph of itself.
2. Difference between weighted graph and B CB B
unweighted graph. AC C
Graph BA
C
Weighted graph Unweighted graph Subgraphs AB
A of graph
5 B B C
2
C C A
79 AB B
A
A C
Each edge is associated Each edge is not CB B A
with a number called associated with a A C
weight. number. C
5.1.2 5.1.3 A
97
Fo4rm Mathematics Chapter 5 Network in Graph Theory
Represent information in the form of Solution:
networks
Langkawi Kota Kinabalu
Example 18
The diagram below shows some roads in an area. Kota Bharu Sibu
Pulau Tawau
Pinang Kuching
Kuala Johor Bahru
Lumpur
Try question 19 in Formative Zone 5.1
By using the road junction as vertex and the road Example 20
Five friends Asiah, Khairul, Natifah, Roslan and
5CHAP. as edge, draw an undirected graph to represent Zeti made conversation by video calls in the
the network of the roads. following situation.
Solution: Asiah called Khairul and Roslan but not Natifah
or Zeti. Zeti only called Roslan and Natifah. Zeti
Try question 18 in Formative Zone 5.1 had never called Khairul although Khairul always
called Zeti. Often Natifah and Roslan will call
one another as Khairul and Natifah.
Draw a directed graph to represent the
conversation by video calls described in the
situation.
Solution:
Natifah
Khairul
Example 19 Zeti
The diagram below shows some locations of Roslan Asiah
airports in Malaysia.
Try question 20 in Formative Zone 5.1
Langkawi Kota Kinabalu
Kota Bharu Sibu Solve problems involving networks
Pulau Tawau
Pinang Kuching
Kuala Example 21
Lumpur
Johor Bahru The diagram below shows a network of roads
connecting five places A, B, C, D and E.
A company Air X scheduled 10 flights on a day
as follows. D
200
• Pulau Pinang to Johor Bahru
• Kuching to Tawau B 340 100
• Tawau to Kuching C 270
• Kuching to Kuala Lumpur
• Kota Kinabalu to Kuala Lumpur 400 120 E
• Kuala Lumpur to Langkawi A 350
• Langkawi to Pulau Pinang
• Johor Bahru to Sibu The weights stated on each edge represent the
• Pulau Pinang to Kota Bharu distances, in m, between two places. A path is
• Kuala Lumpur to Kuching a route along the edges of the graph from one
Draw a directed graph to represent the information vertex to the other vertices without repeating any
of the flights. one vertex.
By listing down all the possible paths, determine
the shortest path from A to C and state its length.
100 5.1.4 5.1.5
Concept
Mathematical Modeling
involving
given Linear model given Quadratic model Exponential model
y = mx + c y = ax2 + bx + c y = a(b)x
if a . 0 and b . 1 if a . 0 and 0 , b , 1
The gradient m and paAracbooolradainnadteth(ex1v,eyr1t)eoxn(hth, ke): Exponential Exponential Compound Chapter 8 Mathematical Modeling Mathematics
the y-intercept, c: y = a(x − ℎ)2 + k function shows function shows interest
model
y = mx + c pAarcaoboorladiannadteth(xe1,xy-1in) toenrctehpet: growth decay
The gradient m and a y = a(x − p)(x − q)
such that such that A(t) = P 11 + —nr 2nt,
cyo−oryd1i=namte((xx−1, yx11)): STohlvreeethceoothradrneinedae(tqxeu3s,a(ytx3i)1o,:nys1)b, (yxu2,syin2)g Exponential Exponential decay
Twoyco=anomrddx(ixn+2a,tcye2as)n:(dx1, y1) y = ax2 + bx + c growth model is where A(t) is the
y = a(1 + r)t, model is amount of savings
y − y1 = m(x − x1) where a is the y = a(1 − r)t, after t years, P is
initial value, r is where a is the
the growth rate initial value, r is the principal of
and t is the time the decay rate and the savings, r is the
t is the time yearly interest rate,
n is the number
of interests being
compound per
year and t is the
number of years
CHAP. Fo5rm
8
415
Fo5rm Mathematics Chapter 8 Mathematical Modeling
This situation involves the linear function because B Quadratic modeling
the first differences are a constant, that is, RM39. 1. Quadratic modeling is related to a quadratic
The first differences is the gradient, m, that is, the
rate of change of y. function. The basic equation for quadratic
Hence, y = 39x + c. function is y = ax2 + bx + c, where a is the leading
coefficient, b is the center coefficient and c is the
When (0, 200), y-intercept.
200 = 39(0) + c 2. The value of the leading coefficient a can
c = 200 determine the shape of the curve of a parabola.
y = 39x + 200 (a) When a , 0,
When x = 60, • the parabola opens downward.
y = 39(60) + 200 • the function has a maximum value known
y = 2 540
as vertex (h, k).
Thus, the membership fee to be paid for 5 years
is RM2 540. y
Try questions 3 & 4 in Formative Zone 8.1 10
(0, 6)
5
Example 6 –10 –5 0 5 10 x
The graph below shows the journey distance,
y km, of a helicopter for x seconds. –5
y (km)
8 (b) When a . 0,
(2, 7) • the parabola opens upward.
• the function has a minimum value known
6 as vertex (h, k).
4 y
2 24 x (s) 15
10
–2 0
–2
(a) Based on the graph, write the equation 5 (0, 1) x
involved. –10 –5 0 5 10
(b) The total journey for the helicopter to reach 3. Strategies that can be proposed in the
the designated destination is 18 200 km. How construction of a quadratic model to solve a
much time is required by the helicopter to problem:
reach the destination? (a) Identify the variables involved along with
their units.
Solution: (b) Identify the important information of
(a) m = —yx22—–––yx—11 the situation such as the maximum point,
m = —72—–– —00 minimum point and related points.
m = 3.5 (c) Identify the solution. Typically, the solution
Thus, y = 3.5x will involve the use of a table of values to
obtain the formula for the function of model
(b) y = 3.5x to be solved.
18 200 = 3.5x (d) Construct a formula for the function
x = 5 200 involved.
CHAP. Thus, the time required by the helicopter to (e) Solve the function using the constructed
formula.
8 reach the destination is 5 200 seconds.
420 8.1.2
Chapter 8 Mathematical Modeling Mathematics Fo5rm
(a) Exponential growth model Use the constructed model to solve the problem.
The exponential function shows growth if y = 19 000(1 + 0.15)9
= 66 839.6495
a . 0 and b . 1. y ≈ 66 840
The exponential growth model: Thus, the population of the city in the 9th year is
y = a(1 + r)t 66 840 people.
where a = the initial value
r = the growth rate Try questions 7 & 8 in Formative Zone 8.1
t = the time
y
3 Example 12
The value of a car worth RM350 000 will be
2 depreciated every year by 11%.
(a) If y represents the value of the car, in RM, for
1
t years, model the situation.
–2 –1 0 1 t (b) What is the value of the car after 9 years?
Solution:
(b) Exponential decay model (a) The initial value is 350 000, that is,
The exponential function shows decay if
a = 350 000.
a . 0 and 0 , b , 1. The decay rate is 11%, that is, r = 0.11.
The exponential decay model: Thus, the exponential decay model is
y = a(1 – r)t y = 350 000(1 – 0.11)t
where a = the initial value y = 350 000(0.89)t
r = the decay rate (b) y = 350 000(0.89)9
t = the time = 122 624.7413
y ≈ 122 624.74
y Thus, the value of the car after 9 years is
3 RM122 624.74.
Try question 9 in Formative Zone 8.1
2
8. The compound interest model is the interest
1 calculated with the reference to the original
principal as well as the accumulated interest
–1 0 12 t from the previous retention period.
9. The compound interest model:
Example 11 1 2A(t) = P 1 + —nr nt
The population of a city is expected to increase by where A(t) = the amount of savings after t years
15% annually. The population of today is 19 000 P = the principal of the savings
people. State the model involved and hence find r = the yearly interest rate
the population of the city in the 9th year. n = the number of interests being
Solution: compound per year
t = the number of years
The initial value is 19 000, that is, a = 19 000.
The growth rate is 15%, that is, r = 0.15. CHAP.
Hence, the exponential growth model is
y = 19 000(1 + 0.15)t 8
8.1.2 423
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