SKOR A SPM (Maths) Penerbitan Pelangi Sdn Bhd

2KERTAS MODEL

SIJIL PELAJARAN MALAYSIA 1449/2

MATHEMATICS Dua jam tiga puluh minit

Kertas 2
1

22 jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Kertas soalan ini adalah dalam dwibahasa.
2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam bahasa Melayu.
3. Tunjukkan langkah-langkah penting dalam kerja mengira anda.
4. Anda dibenarkan menggunakan kalkulator saintifik.

Section A
Bahagian A

[52 marks]
[52 markah]

Answer all questions in this section.
Jawab semua soalan dalam bahagian ini.

1. The Venn diagram in the answer space shows set P, set Q and set R such that the universal For
set,  = P  Q  R. On the diagram in the answer space, shade the set Examiner's

Gambar rajah Venn di ruang jawapan menunjukkan set P, set Q dan set R dengan keadaan set semesta, Use
 = P  Q  R. Pada rajah di ruang jawapan, lorekkan set

(a) Pʹ  R
(b) (P  Rʹ)  Q
[3 marks]

[3 markah]

Answer / Jawapan :

(a) Q (b) Q

P P
R R

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–1 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 2. Solution by matrix method is not allowed to answer this question.
Examiner's
Penyelesaian dengan kaedah matriks tidak dibenarkan untuk menjawab soalan ini.
Use
The total price of 2 toothbrushes brand X and a toothbrush brand Y is RM11. The
difference in price of 3 toothbrushes brand X and 2 toothbrushes brand Y is RM6. Find the
price of a toothbrush brand Y.

Jumlah harga bagi 2 batang berus gigi jenama X dan sebatang berus gigi jenama Y ialah RM11. Beza
harga antara 3 batang berus gigi jenama X dan 2 batang berus gigi jenama Y ialah RM6. Cari harga
sebatang berus gigi jenama Y.

[5 marks]

[5 markah]

Answer / Jawapan :

3. IKUT KIRI Diagram 3 shows a rectangular signboard at the side of a
JIKA TIDAK highway. Given the length of the board is (2x – 4) m and
MEMOTONG the width is x m. Calculate the perimeter of the signboard
such that the length of the diagonal is 10 m.
Diagram 3
Rajah 3 Rajah 1 menunjukkan sebuah papan tanda berbentuk segi empat
tepat yang terdapat di tepi lebuh raya. Diberi bahawa panjang
Answer / Jawapan : papan tanda ialah (2x – 4) m dan lebarnya ialah x m. Hitung
perimeter papan tanda itu dengan keadaan panjang pepenjurunya
ialah 10 m.

[4 marks]

[4 markah]

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–2 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2

4. Diagram 4 shows a right prism with a square base KLMN which lies on the horizontal For
plane. Trapezium PQLM is the uniform cross section of the prism. Examiner's

Rajah 4 menunjukkan sebuah prisma tegak dengan tapak segi empat sama KLMN yang terletak pada satah Use
mengufuk. Trapezium PQLM adalah keratan rentas seragam prisma itu.

RS

QP [3 marks]
KN
[3 markah]
8 cm

L 14 cm M

Diagram 4
Rajah 4

(a) Name the angle between the plane SKL and the base KLMN.

Namakan sudut di antara satah SKL dengan tapak KLMN.

(b) Calculate the angle between the plane SKL and the base KLMN.

Hitung sudut di antara satah SKL dengan tapak KLMN.

Answer / Jawapan :
(a)

(b)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–3 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 5. (a) State whether the following sentence is a statement or not a statement.
Examiner's
Nyatakan sama ada ayat berikut adalah suatu pernyataan atau bukan pernyataan.
Use
3x2 + 5x = 10

[1 mark]

[1 markah]

(b) Write a compound statement by combining the two statements given below using the
word ‘or’ or ‘and’ to make a true statement.

Tulis satu pernyataan majmuk dengan menggabungkan dua pernyataan yang diberi di bawah
dengan menggunakan perkataan ‘atau’ atau ‘dan’ untuk menjadikan suatu pernyataan benar.

Statement 1 : Elements of set M = {7, 11, 13} are prime numbers.

Pernyataan 1 : Unsur-unsur set M = { 7, 11, 13} ialah nombor perdana.

Statement 2 : Elements of set N = {16, 25, 37} are perfect squares.

Pernyataan 2 : Unsur-unsur set N = {16, 25, 37} ialah nombor kuasa dua sempurna.

[1 mark]

[1 markah]

(c) Write down premise 2 to complete the following argument:

Tulis premis 2 untuk melengkapkan hujah berikut:

Premise 1 : If a rectangle is a parallelogram, then its diagonals bisect each other.

Premis 1 : Jika sebuah segi empat tepat adalah segi empat selari, maka pepenjuru-pepenjurunya
membahagi dua sama antara satu sama lain.

Premise 2 / Premis 2 :

Conclusion : A kite is not a parallelogram.

Kesimpulan : Lelayang bukan segi empat selari.

[1 mark]

[1 markah]

(d) Write down two implications based on the following statement:

Tulis dua implikasi berdasarkan pernyataan berikut:

∠Q is an obtuse angle if and only if ∠Q is more than 90o and less than 180o.

∠Q adalah sudut cakah jika dan hanya jika ∠Q lebih besar daripada 90o dan kurang daripada 180o.

[2 marks]

[2 markah]

Answer / Jawapan :

(a)

(b)

(c) Premise 2 / Premis 2 :

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–4 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2
(d) Implication 1 / Implikasi 1 :
For
Implication 2 / Implikasi 2 : Examiner's

Use

6. y L(4,3) M(10,r) In diagram 5, point O is the origin. Point K lies on the

x-axis and point N lies on the y-axis. Straight line LM is

parallel to the x-axis and straight line NM is parallel to

the straight line KL. Given the gradient of straight line
1
KO x KL is 2 .

Dalam rajah 5, titik O ialah asalan. Titik K terletak pada paksi-x dan
titik N terletak pada paksi-y. Garis lurus LM adalah selari dengan
N paksi-x dan garis lurus NM adalah selari dengan garis lurus KL.
1
Diagram 5 Diberi kecerunan garis lurus KL ialah 2 .
Rajah 5
Find / Cari

(a) the value of r,

nilai bagi r,

(b) the equation of the straight line KL,

persamaan garis lurus KL,

(c) the coordinates of point N.

koordinat titik N.

[5 marks]

[5 markah]

Answer / Jawapan :
(a)

(b)

(c)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–5 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 7. Diagram 6 shows a solid formed by combining a cylinder and a cone. The volume of a
Examiner's solid is 1 100 cm3.

Use Rajah 6 menunjukkan sebuah pepejal yang terbentuk daripada cantuman sebuah silinder dan sebuah kon.
Isi padu pepejal itu ialah 1 100 cm3.

5 cm

6 cm

Diagram 6
Rajah 6

Using p = 22 , calculate the height, in cm, of the cylinder.
7
Menggunakan p = 22 , hitung tinggi, dalam cm, silinder itu.
7
[4 marks]

[4 markah]

Answer / Jawapan :

1 2 1 2 1 2 8. Given that M =2 –5 1 k5 1 0
3 –6 ,N= p –3 2 and MN = 0 1 .

1 2 1 2 1 2Diberi bahawa M =2 –5 ,N= 1 k 5 1 0 .
3 –6 p –3 2 dan MN = 0 1

(a) Find the value of p and of k.

Cari nilai p dan nilai k.

(b) Hence, by using matrix method, calculate the value of x and of y for the following
equation:

Seterusnya, dengan menggunakan kaedah matriks, hitung nilai x dan nilai y bagi persamaan berikut:

2x – 5y = 8
3x – 6y = 9

[6 marks]

[6 markah]

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–6 SPM Mathematics

1449/2 (b) KERTAS MODEL SPM 2  Kertas 2

Answer / Jawapan : For
(a) Examiner's

Use

9. In Diagram 7, PR and QS are arcs of two circles with common centre O. Given OR = 10 cm
and PQ = 8 cm.

Dalam Rajah 7, PR dan QS ialah lengkok bagi dua buah bulatan dengan pusat sepunya O. Diberi bahawa
OR = 10 cm dan PQ = 8 cm.

Q
S

8 cm

P
R

30°
10 cm 45°

O

Diagram 7
Rajah 7

Using π = 3.142, calculate

Menggunakan π = 3.142, hitung

(a) the perimeter, in cm, of the whole diagram,

perimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm2, of the shaded region. [6 marks]

luas, dalam cm2, kawasan berlorek. [6 markah]

Answer / Jawapan :
(a)

(b)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–7 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 10. Diagram 8 shows five cards labelled with letters. All these cards are put into a box.
Examiner's
Rajah 8 menunjukkan lima keping kad berlabel dengan huruf. Semua kad itu diletakkan di dalam sebuah
Use kotak.

C LUB

Diagram 8
Rajah 8

A card is picked at random from the box. If the card is labelled with a vowel, the card is
removed whereas if the card is labelled with a consonant, the card is put back into the box.
Then, a second card is picked at random from the box.

Sekeping kad dipilih secara rawak daripada kotak itu. Jika kad tersebut berlabel dengan huruf vokal, kad
tersebut dikeluarkan manakala jika kad tersebut berlabel dengan huruf konsonan, kad tersebut diletakkan
semula ke dalam kotak itu. Kemudian, kad kedua dipilih secara rawak daripada kotak itu.

(a) List all sample space. [1 mark]

Senaraikan semua ruang sampel. [1 markah]

(b) List all the possible outcomes of the event and find the probability that

Senaraikan semua kesudahan peristiwa yang mungkin dan cari kebarangkalian bahawa

(i) the second card is labelled with a vowel,

kad kedua berlabel dengan vokal,

(ii) both cards are labelled with the same letter or the second card is labelled with a
letter C.

kedua-dua kad berlabel dengan huruf yang sama atau kad kedua berlabel dengan huruf C.

[4 marks]

[4 markah]

Answer / Jawapan :
(a)

(b) (i)

(ii)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–8 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2

11. Diagram 9 shows the speed-time graphs of a bus and a motorcycle for a period of For
45 seconds. The bus and motorcycle move along the same route. The graphs OM and KLM Examiner's
represent the movement of the bus and the motorcycle respectively.
Use
Rajah 9 menunjukkan graf laju-masa bagi sebuah bas dan sebuah motosikal dalam tempoh 45 saat. Bas
dan motosikal tersebut bergerak di sepanjang jalan raya yang sama. Graf OM dan KLM masing-masing
mewakili pergerakan bas dan motosikal tersebut.

Speed (m s–1)
Laju (m s–1)

30 L M
20

8K 15 t Time (s)
O 45 Masa (s)

Diagram 9
Rajah 9

(a) State the uniform speed, in m s–1, of the motorcycle. [1 mark]

Nyatakan laju seragam, dalam m s–1, motosikal itu. [1 markah]

(b) Calculate the value of t. [2 marks]

Hitung nilai t. [2 markah]

(c) Find the different of the total distance travelled between the bus and motorcycle.

[3 marks]

Cari beza bagi jumlah jarak yang dilalui antara bas dengan motosikal itu. [3 markah]

Answer / Jawapan :
(a)

(b)

(c)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–9 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

Section B
Bahagian B

[48 marks]
[48 markah]

Answer any four questions from this section.
Jawab mana-mana empat soalan daripada bahagian ini.

For 12. (a) xvCLea=onlmgu–k2epa.s5lpeokdtafaennyTJxaaw=dbhul2eea. nl11xdini=rut–ha2ne.g5ajnaawsnwadpeaxrns=,pba2acg. ei p, efrosramthaeaneqyu=a–ti  o8xndyen=ga–n  m8xenbuyliswnriilatii-nngila[[d22ioymmwaapnararkbktahishlae]]
Examiner's

Use

(b) Use graph paper for this part of the questions. You may use a flexible ruler.

Gunakan kertas graf untuk ceraian soalan ini. Anda boleh menggunakan pembaris fleksibel.

By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 4 units on the y-axis, draw
tDheenggarnapmhenogfgyun=ak–a n 8xskfaolar
8 –4 < x < 4 and –16 < y < 16. cm kepada 4 unit [4 marks]
x
2 cm kepada 1 unit pada paksi-x dan 2 pada paksi-y,

lukis graf y = –   untuk –4 < x<4 dan –16 < y < 16. [4 markah]

(c) From the graph in 12(b), find

Daripada graf di 12(b), cari

(i) the value of y when x = 1.3,

nilai y apabila x = 1.3,

(ii) the value of x when y = 4.4.

nilai x apabila y = 4.4.

[2 marks]

[2 markah]

(d) Draw a suitable straight line graph on the graph in 12(b) to find the values of x which
satisfy the equation x2 – x – 4 = 0 for –4  x  4 and –16  y  16. State these

values of x. [4 marks]
Lukis satu garis lurus yang sesuai pada graf di 12(b) untuk mencari nilai-nilai x yang memuaskan
persamaan x2 – x – 4 = 0 untuk –4  x  4 dan –16  y  16. Nyatakan nilai-nilai x ini. [4 markah]

Answer / Jawapan :

(a) x – 4 –2.5 –1 –0.5 0.5 1 2 3.2 4
y –2.5 –2
2 8 16 –16 –8

Table 1
Jadual 1

(c) (i) y =
(ii) x =

(d) x =

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–10 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2

13. (a) Diagram 10.1 shows point R drawn on a Cartesian plane. For
Examiner's
Rajah 10.1 menunjukkan titik R dilukis pada suatu satah Cartes.
Use
y

4
R(2,3)

2

–4 –2 O x
–2 24

Diagram 10.1
Rajah 10.1

Transformation G is a reflection in the line y = 1.
1 2 Transformation H is a translation
–3 .
2
Transformation T is an anticlockwise rotation of 90°at centre (0, 2).

State the coordinates of the image of point R under the following transformations:

Penjelmaan G ialah satu pantulan pada garis y = 1.

1 2 Penjelmaan H ialah satu translasi–3 .
2
Penjelmaan T ialah satu putaran 90o arah lawan jam pada pusat (0, 2).
Nyatakan koordinat imej bagi titik R di bawah penjelmaan berikut:

(i) HG (ii) TH
[4 marks]

[4 markah]

(b) In Diagram 10.2, ABCDE, PQRST and NQKLM are three pentagons drawn on a
Cartesian plane.

Dalam Rajah 10.2, ABCDE, PQRST dan NQKLM ialah tiga buah pentagon yang dilukis pada suatu
satah Cartes.

y

L
10

D M8 K
S R

6
CT

E4

A B N2 P

Q

–8 –6 –4 –2 O x

246

Diagram 10.2
Rajah 10.2

(i) NQKLM is the image of ABCDE under the combined transformation WV.
Describe in full, the transformation:

NQKLM ialah imej bagi ABCDE di bawah gabungan penjelmaan WV. Huraikan selengkapnya
penjelmaan:

(a) V, (b) W.

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–11 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For (ii) Given that the area of the shaded region is 108 m2. Calculate the area, in m2, of
Examiner's
PQRST.
Use
Diberi bahawa luas kawasan berlorek ialah 108 m2. Hitung luas, dalam m2, PQRST.

[8 marks]

[8 markah]

Answer / Jawapan :

(a) (i)

(ii)

(b) (i) (a) (b)

(ii)

14. The frequency polygon in Diagram 11 shows the marks of 36 students in a Mathematics
test.

Poligon kekerapan dalam Rajah 11 menunjukkan markah bagi 36 orang murid dalam satu ujian Matematik .

Number of students
Bilangan murid

10

8

6

4

2

0 Marks
33 38 43 48 53 58 63 68 Markah

Diagram 11
Rajah 11

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–12 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2

(a) Based on the frequency polygon in Diagram 11, complete Table 2 in the answer For
Examiner's
space. [4 marks]
Berdasarkan poligon kekerapan dalam Rajah 11, lengkapkan Jadual 2 di ruang jawapan. [4 markah] Use

(b) Calculate the estimated mean mark of a student. [3 marks]

Hitung min anggaran markah bagi seorang murid. [3 markah]

(c) Use graph paper for this part of questions.

Gunakan kertas graf untuk ceraian soalan ini.

By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 4 students on

the vertical axis, draw an ogive for the data. [4 marks]
Dengan menggunakan skala 2 cm kepada 5 markah pada paksi mengufuk dan 2 cm kepada 4 orang
murid pada paksi mencancang, lukis ogif bagi data tersebut. [4 markah]

(d) Based on the ogive drawn in 14(b), state the third quartile. [1 mark]

Berdasarkan ogif yang dilukis di 14(b), nyatakan kuartil ketiga. [1 markah]

Answer / Jawapan :

(a) Marks Midpoint Upper boundary Cumulative frequency

Markah Titik tengah Sempadan atas Kekerapan longgokan

31 – 35 33 35.5 0

Table 2
Jadual 2

(b)
(d)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–13 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 15. You are not allowed to use graph paper to answer this question.
Examiner's
Anda tidak dibenarkan menggunakan kertas graf untuk menjawab soalan ini.
Use
(a) Diagram 12.1 shows a solid right prism with a rectangular base EFGH on a horizontal
plane. The surface ADJKHE is the uniform cross section of the prism. Edges AE, KH
and DJ are vertical. Rectangle JKLM is a horizontal plane and rectangle ABCD is an
inclined plane.

Rajah 12.1 menunjukkan sebuah pepejal berbentuk prisma tegak dengan tapak segi empat tepat
EFGH terletak di atas satah mengufuk. Permukaan ADJKHE ialah keratan rentas seragam prisma itu.
Tepi AE, KH dan DJ adalah tegak. Segi empat tepat JKLM adalah satah mengufuk dan segi empat tepat
ABCD ialah satah condong.

B

A C
7 cm
2 cm
E
D M 5 cm
J L
F
K 3 cm
G

7 cm 6 cm
H

Diagram 12.1
Rajah 12.1

Draw to full scale, the plan of the solid. [3 marks]

Lukis dengan skala penuh, pelan pepejal itu. [3 markah]

Answer / Jawapan :
(a)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–14 SPM Mathematics

1449/2 KERTAS MODEL SPM 2  Kertas 2

(b) B Another solid prism with right angle triangle For
PQR as its uniform cross section is joined to Examiner's
A Q the prism in Diagram 12.1 at the horizontal
C plane LMS. Given that PQ = 5 cm and Use
7 cm QM = 3 cm. The combined solid is as shown
5 cm M 5 cm in Diagram 12.2.
E P R
X DT Sebuah pepejal lain berbentuk prisma tegak dengan
3 cm segi tiga bersudut tegak PQR sebagai keratan rentas
S F L seragamnya dicantumkan kepada prisma dalam
J K Rajah 12.1 pada satah mengufuk LMS. Diberi
3 cm bahawa PQ = 5 cm dan QM = 3 cm. Gabungan
G pepejal adalah seperti yang ditunjukkan dalam
Rajah 12.2.
7 cm 6 cm

H Y

Diagram 12.2
Rajah 12.2

Draw to full scale,

Lukis dengan skala penuh,

(i) the elevation of the combined solid on a vertical plane parallel to EH as viewed

from X. [4 marks]
dongakan gabungan pepejal itu pada satah mencancang yang selari dengan EH sebagaimana
dilihat dari X. [4 markah]

(ii) the elevation of the combined solid on a vertical plane parallel to GH as viewed

from Y. [5 marks]
dongakan gabungan pepejal itu pada satah mencancang yang selari dengan GH sebagaimana
dilihat dari Y. [5 markah]

Answer / Jawapan :
(b) (i)

(ii)

© Penerbitan Pelangi Sdn. Bhd.   KM2–K2–15 SPM Mathematics

KERTAS MODEL SPM 2  Kertas 2 1449/2

For 16. D(42oS, 105oW), E(42oS, 38oW), F and G are four points on the surface of the earth. DF
Examiner's is the diameter of the parallel of latitude of 42oS.

Use D(42oS, 105oB), E(42oS, 38oB), F dan G adalah empat titik pada permukaan bumi. DF ialah diameter selarian
latitud 42oS.

(a) State the location of point F. [2 marks]

Nyatakan kedudukan titik F. [2 markah]

(b) Calculate the shortest distance, in nautical mile, from D to F measured along the

surface of the earth. [2 marks]
Hitung jarak terpendek, dalam batu nautika, dari D ke F yang diukur sepanjang permukaan bumi.
[2 markah]

(c) Calculate the distance, in nautical mile, from D to the east of E measured along the

common parallel of latitude. [3 marks]
Hitung jarak, dalam batu nautika, dari D ke arah timur ke E diukur sepanjang selarian latitud
sepunya. [3 markah]

(d) An aeroplane took off from E and flew due north to G. The average speed of the
aeroplane was 600 knots and the time taken was 5 hours 45 minutes. Calculate

Sebuah kapal terbang berlepas dari E dan terbang ke utara ke G. Purata laju kapal terbang itu ialah
600 knot dan masa yang diambil ialah 5 jam 45 minit. Hitung

(i) the distance, in nautical mile, from E to G measured along the meridian,

jarak, dalam batu nautika, dari E ke G yang diukur sepanjang meridian,

(ii) latitude of G. [5 marks]

latitud G. [5 markah]



Answer / Jawapan :
(a)

(b)

(c)

(d) (i) (i)

© Penerbitan Pelangi Sdn. Bhd. END OF QUESTION PAPER SPM Mathematics
KERTAS SOALAN TAMAT

  KM2–K2–16

ANSWERS

1KERTAS MODEL 8. ∠RQP = ∠QPR = 57°
∠QRP = ∠SRU = 180°– 2(57°)
Paper 1
= 66°

m° = 360° – 66° – 93° – 122°

1. 0.0 0 0 0 0 4 3 1 = 4.31 × 10–6 = 79
Answer : D
Answer : C

9. ∠SRP = 160° ÷ 2

2. 1 5 2 6 = 80°

∠PRQ = 180°– 80° = 100°

2  5 ∠RPQ = 180° – 100° – 30°
= 1 500 (correct to two significant figures)
Answer : C = 50°

∠OPQ = 90°

x° = 90° – 50°

= 40

3. Mass of 68 atoms Answer : C
3.44 × 10–25 × 68
= 2.3392 × 10–23 × 103 g 10.
= 2.34 × 10–20 g
Q P
Answer : D N

4. 250 × Height = 35 000

Height = 140 cm M

= 140 × 10 mm

= 1 400 mm Centre Direction Angle
M Anticlockwise 90°
= 1.4 × 103 mm

Answer : B Answer : B

5. 18 – P2 = 58 11. P
18 – 58 = P2
R

100102 – 1012 = 11012 C

Answer : B S
Q

6. 1010002 – 11002 = 111002 Answer : C
Answer : C
12. y
(6 – 2) × 180°
7. Interior angle = 6 = 120° 1

∠BAC = (180° – 120°) ÷ 2 x
0 180° 360°
= 30° –1

∠CAF = ∠DCA = 120° – 30° Answer : B

= 90°

y° = 360° – 30° – 180° 13. QR = 122 + 52

= 150° = 13 PR
x° = 360° – 120° – 120° – 90° QR
cos x° = –   (Quadrant II)
= 30°
x + y = 30 + 150 = 180 = –   12
13
Answer : C
Answer : A

© Penerbitan Pelangi Sdn. Bhd. A – 1 SPM Mathematics

ANSWERS  

14. MQ R = (3 – 2p) × 1 SPM Mathematics
2q 2
P SV
K = 3 – 2p
4q
W

UN Answer : A

∠MNK T  20. x
+
x y = 3w

Answer : B x = 9w2
+
1.02 x y
15. tan 18° = RT x = 9w2(x + y)
x = 9w2x + 9w2y
RT = 3.14

P x – 9w2x = 9w2y

x(1 – 9w2) = 9w2y

xm Q 34° x = 9w2y
1 – 9w2
S
18° 1.02 m Answer : D

R T 21. r + s = 4r + 3
s
tan 34° = x s(r + s) = 4r + 3
3.14
rs + s2 = 4r + 3
x = 2.12
rs – 4r = 3 – s2
PQ = 2.12 – 1.02
r(s – 4) = 3 – s2
= 1.1 3 – s2
r = s–4
Answer : B
Answer : C
380
16. sin θ = 1 600 22. 3a + 2b = 64

θ = 13.74° or 13°44′ 3a = 64 – 2b

Answer : C a = 64 – 2b
3

17. A 10 cm B Answer : B
60°
30°

90° – 60° N 23. y –5 × 2y 3
= 30° C
= 2y (–5) + 3
= 2y –2

Bearing of A from C = 2
= 360° – 30° y2
= 330°
Answer : D Answer : A

1
24. m2n2(2m7m3n3)3
18. Latitude of Q = 0°
Longitude of Q = (50° – 30°)W = m2n2 × (3mn)
m
= 20°W
= 3m2n3
Location of Q = (0°, 20°W)

Answer : C Answer : A

19. 9 – 4p2 ÷ 6 + 4p 25. x  2x – 1
14pq 7p 4
1
= (3 – 2p)(3 + 2p) × 7p –x  – 4
14pq 6 + 4p
x  1
= (3 – 2p)(3 + 2p) 1 7p 1 4
7p(2q) 2(3 + 2p) 1
1 ×

Answer : A

© Penerbitan Pelangi Sdn. Bhd. A – 2

ANSWERS 

26. 1 x  3 7 – 2x  3 35. P(letter A is not chosen) = 1– 2
3 = 6 14
x  9 –2x  –4 7
x  2

2  x  9 Answer : D

x = 2, 3, 4, 5, 6, 7, 8

Answer : D 36. Number of girls = 60 – 20 = 40

P(girls) = 40 – 15
60 – 15
27. Mode = 5
Median = 4 = 5
Different = 5 – 4 = 1 9

Answer : B Answer : C

28. Total number of pineapples sold 37. P ∝ 1
= (4 + 7 + 6 + 5) × 20  M
= 440
P = k
Answer : D  M

29. y 1

k = PM or PM 2

Answer : A

Ox 38. j ∝  3k
l

Answer : D j = k  3k
 l

30. Modal score = 2 4 = k  38
Answer : B 6

k = 12

31. P′ = {c, d, e, f } ∴ j =  12  364
Answer : B 3

j = 16

32. n(P  Q) = n(Q′) Answer : A

3 + 4 + 1 + x + 2 + 7 = 3 + 4 + 2x + 5

17 + x = 12 + 2x 1 2 39. (5  4
1) ×  3 = (5 × 4 + 1 × 3)
x = 5
= (23)
n() = 3 + 4 + 1 + 5 + 2 + 7 + 5 + 2(5)
Answer : A
= 37

Answer : A 1 2 1 2 40. 3Q +  2 4
6 –3 = 8 10
6 12
33. 5x + y = 7
1 2 1 23Q =  10
y = –5x + 7 8 12  – 24
6 6 –3
Gradient = –5
1 2 =
Answer : B 66
0 15

34. m = 10 – 5 = 1 1 2Q =  1 6 6
10 – 0 2 3 0 15

Let (x, 0) be the intersection point with x-axis. 1 2 = 22
05
5 – 0 1
0 – x = 2 Answer : A

x = –10

x-intercept = –10

Answer : A

© Penerbitan Pelangi Sdn. Bhd. A – 3 SPM Mathematics

ANSWERS  
Paper 2

No. Solution and Mark Scheme Marks Total
P1 3
P = Knowledge, K = Method and N = Value

1. (a) 1. (a)

P Q P Q
R R

(b) (b) P2

P Q P Q
R R

2. (100 + p)(150 + p) = 2(100)(150) 2. (100 + p)(150 + p) = 2(100)(150) K1 4
p2 + 250p – 15 000 = 0 K1
p2 + 250p – 15 000 = 0 (p – 50)(p + 300) = 0 K1
p = 50 N1
(p – 50)(p + 300) = 0 K1 5
3. 2x + y = 19 K1
p – 50 = 0 or p + 300 = 0 3x – 2y = –3 K1
7x = 35 or equivalent N1
p = 50 p = –300 x=5 K1
y=9
(Not accepted) P1 3
4. (a) ∠JKG or ∠GKJ
3. 2x + y + 15 = 34 K1
2x + y = 19...........................a N1

2y – 3 = 3x

3x – 2y = –3...........................b

a × 2:  4x + 2y = 38..........c

b + c: 7x = 35

x = 5

Substitute x = 5 into a,
2(5) + y = 19

y = 9

4. (a) K M
5 cm G
E
12 cm F

H 8 cm J

∠JKG or ∠GKJ

(b) GK = 122 + 52  (b) 8
= 13 cm ∠JKG = tan–1 13

 ∠JKG = 8
tan–1 13 31.6° or 31°36ʹ

= 31.6° or 31°36ʹ

© Penerbitan Pelangi Sdn. Bhd. A – 4 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total
K1 4
5. Let t be the height of the pyramid. 5. (3 × 5 × 4) K1
K1
Volume of composite solid  1 ×t N1
= volume of cuboid + volume of pyramid
3
×5× 4

  100 = (3 × 5 × 4) + 1 × 5 ×4 × t  100 = (3 × 5 × 4) + 1 ×5 × 4× t
3 3

100 = 60 + 20 t t=6
3
20
40 = 3 t

t = 6

6. (a) mSR = mPQ = –3 6. (a) mSR = mPQ = –3 P1 5
5 = –3(2) + c K1
Using y = mx + c and (2, 5) y = –3x + 11 N1
5 = –3(2) + c
c = 11 K1
∴ y = –3x + 11 N1

(b) y = –3x + 11 (b)

When y = 0, 0 = –3x + 11

0 = –3x + 11 11
3
x = 11 x =
3

x-intercept of straight line PQ is 11 .
3

7. (a) Length of time the bus stopped 7. (a) 5 or 0.42 hour N1 5
12
= (35 – 10) ÷ 60

= 5 or 0.42 hour
12

(b) Speed = 20 – 0 (b) 20 – 0 K1
(40 ÷ 60) (40 ÷ 60) N1
K1
= 30 km h–1 30 km h–1 N1

(c) Average speed = 20 + 20 (c) 20 + 20
(90 ÷ 60) (90 ÷ 60)

= 80 km h–1 or 26.7 km h–1 80
3 3 km h–1 or 26.7 km h–1

8. (a) True 8. (a) True P1 5

(b) Implication 1: If 2p – 3 = 7, then p = 5 (b) Implication 1: If 2p – 3 = 7, then p = 5 P1
Impication 2: If p = 5, then 2p – 3 = 7 Impication 2: If p = 5, then 2p – 3 = 7 P1

(c) (n2 × 3) – 2, n = 1, 2, 3, 4, … (c) (n2 × 3) – 2 K1
n = 1, 2, 3, 4, … K1

9. (a) Length of the shaded region    9. 1 ×2 × 22
(a) 2(7) or 2 4 7 × 3.5 K1
= PR + RQ + arc PS + arc SQ

  1 ×2 22 or equivalent
= 2(7) +2 4 × 7 × 3.5


= 14 + 11   1 22
2(7) + 2 4 ×2× 7 × 3.5 or equivalent K1
= 25 cm

25 cm N1

© Penerbitan Pelangi Sdn. Bhd. A – 5 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total

(b) Area of the shaded region  (b) 90° 22 K1
2 360° × 7 × 3.52 or 3.5 × 3.5
= sector of PS + sector of SQ

+ area of square RTSU or equivalent

  90° × 22
= 2 360° 7 × 3.52 + 3.5 × 3.5   90° × 22
360° 7 × 3.52 K1
2 + 3.5 × 3.5
= 19.25 + 12.25

= 31.5 cm2 31.5 cm2 N1

10. (a) Inverse matrix 1 210. (a) 1 5 –9 N1 6
2 –2 4
1 5 2–9
= 1 –2
4×5–9×2 4

1 2 =1 5 –9
2 –2 4

(b) Let x be the price of a coconut and y be the 1 21 2 1 2(b) 49 x = 74 P1
25 y 40
price of a papaya. K2
N1
1 21 2 1 2 4 9 x 74 1 2 1 21 2 x=1 5 –9 74 N1
2 5 y = 40 y 2 –2 4 40

1 2 1 21 2x = 1 5 –9 74 x=5
y 2 –2 4 40 y=6

1 2 = 1
2
5 × 74 + (– 9) × 40 Note:
(–2) × 74 + 4 × 40

1 2 = 1 10 1 2 1 2 1. Ifx = 5 as final answer,
2 12 y 6

1 2 =  5 award N1.
6
2. Do not accept solution solved not

Thus, price of a coconut is RM5 and price of a using matrix method.
papaya is RM6.

11. (a) 11. (a) P2 6

(4, S) (7, S) (8, S) (9, S) (13, S) (19, S) (4, S) (7, S) (8, S) (9, S) (13, S) (19, S)
(4, R) (7, R) (8, R) (9, R) (13, R) (19, R) (4, R) (7, R) (8, R) (9, R) (13, R) (19, R)
(4, M) (7, M) (8, M) (9, M) (13, M) (19, M) (4, M) (7, M) (8, M) (9, M) (13, M) (19, M)

(b) (i) {(4, M), (8, M)} (b) (i) {(4, M), (8, M)} P1
N1
P(an even number and letter M) 1
9
= 2 = 1
18 9

(ii) {(7, S), (13, S), (19, S), (4, R), (7, R),
(8, R), (9, R), (13, R), (19, R), (7, M),
(13, M), (19, M)} (ii) {(7, S), (13, S), (19, S), (4, R), P1
(7, R), (8, R), (9, R), (13, R), N1
P(a prime number or letter R) (19, R), (7, M), (13, M), (19, M)}

= 12 = 2 2
18 3 3

© Penerbitan Pelangi Sdn. Bhd. A – 6 SPM Mathematics

ANSWERS 

12. (a) When x = –1, y = (–1)3 – 10(–1) + 15 = 24 No. Solution and Mark Scheme Marks Total
When x = 3, y = 33 – 10(3) + 15 = 12 12. (a) 24 N1 12

x –1 3 12 N1
y 24 12

(b) y (b) Axes are drawn in correct directions P1
40 with uniform scale for –3  x  4 and
35 3  y  39.
30
25 All the points are plotted correctly or K2
20
curve passes through all the points for
15 –3  x  4 and 3  y  39.
10
All smooth and continuous curve without

any straight line passes through all 9 N1

correct points using the given scale for
–3  x  4 and 3  y  39.

5

–3 –2 –1 O 0.7 1 2 3.9
x
(c) From the graph,
(i) when x = –0.5, y = 20 3 3.5 4
(ii) when y = 35, x = 3.9
(c) (i) y = 20 ± 0.5 P1
(ii) x = 3.9 ± 0.1 P1

(d) y = x3 – 10x + 15................a (d) Identify equation y = 5x + 5 K1
0 = x3 – 15x + 10................b K1
Straight line is y = 5x + 5 drawn
a – b : y = 5x + 5 correctly. N1
N1
The suitable straight line is y = 5x + 5 Values of x:
0.7 ± 0.1
The values of x which satisfy the equation 3.5 ± 0.1
x3 − 15x + 10 = 0 are 0.7 and 3.5.

13. ( a) (i) R(4, 3) ⎯T→ (6, 6) 13. (a) (i) (6, 6) P1 12
(ii) (a) Q(–2, –4) ⎯T→ (0, –1)
(b) Q(–2, –4) ⎯P→ (4, –4) ⎯T→ (6, –1) (ii) (a) (0, –1) P1
(b) (6, –1) P2

(b) (i) (a) M is a rotation of 90° clockwise about (b) (i) (a) Rotation of 90° clockwise P3
centre (1, 3). about centre (1, 3) P3
(b) Enlargement of scale factor
(b) N is an enlargement of scale factor 2 with centre (2, 0)
2 with centre (2, 0).

(ii) Area of EFGH (ii) 22 × 30 K1
= k2 × area of ABCD
= 22 × 30 120 unit2 N1

= 120 unit2

© Penerbitan Pelangi Sdn. Bhd. A – 7 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total

14. (a) Class Interval Midpoint Frequency 14. (a) Class Interval Midpoint Frequency 12
16 – 20 18 0
21 – 25 23 4 16 – 20 18 0
26 – 30 28 7
31 – 35 33 12 21 – 25 23 4
36 – 40 38 8
41 – 45 43 4 26 – 30 28 7
46 – 50 48 3
51 – 55 53 2 31 – 35 33 12
56 – 60 58 0
36 – 40 38 8

41 – 45 43 4

46 – 50 48 3

51 – 55 53 2

56 – 60 58 0

All class intervals correct P1
All midpoints correct P1
All frequencies correct P2

(b) Estimated mean (4 × 23) + (7 × 28) + (12 × 33) +

(4 × 23) + (7 × 28) + (12 × 33) + (8 × 38) (8 × 38) + (4 × 43) + (3 × 48) +

= + (4 × 43) + (3 × 48) + (2 × 53) (b) (2 × 53) K2
4 + 7 + 12 + 8 + 4 + 3 + 2 4 + 7 + 12 + 8 + 4 + 3 + 2

= 1 410
40

= 35.25 or 35 1 35.25 or 35 1 N1
4 4

(c) 12 (c) Axes are drawn in correct directions P1
11 with uniform scale for 18  x  58 and
10 0  y  12. Horizontal axis is labelled

9 with midpoint. Vertical axis is labelled
8
7 with frequency.
6
Frequency 5 All points plotted correctly or straight K2
4
3 line passes through all points for
2 18  x  58 and 0  y  12.
1
0 Frequency polygon with correct scale. N1

18 23 28 33 38 43 48 53 58
Reward point

(d) Number of customers with reward points of (d) 9 N1
more than 40

= 4 + 3 + 2
= 9

© Penerbitan Pelangi Sdn. Bhd. A – 8 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total

15. (a) E/D K J/C 15. (a) 12

E/D K J/C

7 cm

7 cm

F/A 3 cm G 6 cm H/B

F/A 3 cm G 6 cm H/B

Correct shape with rectangles EKGF K1
and KJHG. All solid lines.

FG  GH  HJ K1

Measurements correct to ±0.2 cm N1
(one way) and right angles at vertices
= 90° ± 1.

(b) (i) (b) (i)

F/E F/E

7 cm G/K 6 cm H/J 7 cm G/K 6 cm H/J
T/U 4 cm T/U 4 cm

A/D 4 cm M/L S/R B/C A/D 4 cm M/L S/R B/C
1 cm
1 cm

Correct shape of polygon FGHBSTMA. K1

All solid lines.

SB  AM = BH  GH  AD K1

Measurements correct to ±0.2 cm N2
(one way) and right angles at vertices
= 90° ± 1.

(ii) F E (ii)

3 cm F E
J/K
H/G 7 cm 2 cm H/G 3 cm
T U T
2 cm J/K
B/S/M/A C/R/L/D B/S/M/A 2 cm
U
7 cm 2 cm

C/R/L/D

© Penerbitan Pelangi Sdn. Bhd. A – 9 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total

Correct shape with rectangles FEJH K1
and HJCB. All solid lines.
Dotted line TU. K1
K1
CU = UJ  JE  BC

Measurements correct to ±0.2 cm N2
(one way) and right angles at vertices
= 90° ± 1.

16. (a) Latitude of D = 40°N 16. (a) 40°N P1 12
Longitude of D
= (180° – 10°)E (180° – 10°)E K1
= 170°E
(40°N, 170°E) N1
Location of D = (40°N, 170°E) K1

(b) ∠VOT = 3 000 = 50° (b) 3 000 = 50° K1
60 60 N1
K2
Longitude of T N1
(170° – 50°)E K1
= (170° – 50°)E 120°E
K1
= 120°E N1

(c) Distance from C to D (c) 50° × 60 × cos 40°
= 50° × 60 × cos 40°
= 2 298.13 nautical miles 2 298.13
(d) 40° × 60
(d) Distance from C to T
= 40° × 60 2 298.13 + 2 400
600
= 2 400 nautical miles

Total time taken

= 2 298.13 + 2 400 7.83 hours
600

= 7.83 hours

2KERTAS MODEL 4. Volume of 30 spheres

Paper 1 = 4 × 22 × 503 × 30
3 7

= 15 714 285.71 cm3

1. 0.01356 = 0.014 (2 significant figures) Volume of 20 cylinders

= 15 714 285.71
20
5=5
Answer : B = 785 714.29 cm3

= 7.86 × 105 cm3

2. 6 8 4 0 . 3 = 6.8403 × 103 kg m–3 Answer : A
Answer : D
5. 24038
3. Thickness = 392 The value of digit 4
5 600 = 4 × 82
= 256
= 0.07
Answer : C
= 7 × 10–2 mm

Answer : C

© Penerbitan Pelangi Sdn. Bhd. A – 10 SPM Mathematics

ANSWERS 

6. 1 0 1 1 12 11. tan θ = 3
+ 1 1 02 4
BD
1 1 1 0 12 AB = 3
4
Answer : B
BD = 3 × 8 =6 cm
4
BD
7. ∠SRQ = 180o – 40o DE = 3
2
= 140o

Interior angle, DE = 2 × 6 =4 cm
3
(n – 2) × 180o
n = 140o BE = 6 + 4 = 10 cm

180n – 360o = 140n t an α = 2
3
40n = 360o BE
BC
n = 9 = 2
3
Answer : D
BC = 3 × 10 = 15 cm
2
8. ∠ABE = 180° – 116° – 28°
Thus, AC = 8 + 15 = 23 cm
=36°

∠EBC = 36° = 18° Answer : B
2
3
x° = ∠BDC = 180° – 78° – 18° 12. sin ∠SWR = 5

= 84° SR 3
WR 5
Answer : D =

9. Angle in the alternate segment, SR = 3 × 15
∠KML = 68o 5

= 9 cm

∠KOL = 68° × 2 WS = 152 – 92 = 12 cm
VS = 162 + 122 = 20 cm
= 136°

∠KLO = 180° – 136°
2
cos x° = –  2102 = –  35
= 22° Answer : A

Angle in the alternate segment, 13. OQ
x° = ∠KLM R
= 22° + 13°
= 35°

Answer : B

10. Scale factor, k = Length of PQʹ
Length of PQ

= 16 The correct rotation is
8
Shape Direction of rotation Image
=2 Q Clockwise R

Area of ∆PQR = Area ∆PQʹRʹ
k2
64 Answer : B
= 4

= 16 14. 50°

Area of shaded region = 64 – 16

= 48 cm2 5m

Answer : A Answer : B

© Penerbitan Pelangi Sdn. Bhd. A – 11 SPM Mathematics

ANSWERS  

15. y 20. mn + 2n ÷ 8mn
4 – n2 2–n
1
0 x = (2 n(m + 2) n) × (2 – n)
–1 – n)(2 + 8mn
90° 180° 270°

= (m + 2)
8m(2 + n)
Answer : C
Answer : D

16. S 2p p–4
q+ 4q
45° 21. 4 =
65°

8pq = (q + 4)(p – 4)

xm 8pq = pq – 4q + 4p –16

7pq + 4q = 4p – 16

q(7p + 4) = 4(p – 4)
4(p – 4)
45° 65° q = 7p + 4
P 10 m Q y m R

tan 45° = 1 Answer : A

10 x y = 1 22. 5w + 2 – w– 3 = 1
+ 3 2

x = 10 + y …...…..a 2(5w + 2) – 3(w – 3) = 6

x 10w + 4 – 3w + 9 = 6
yx
tan 65° = 2.145 7w = –7

y = …........b w = –1

Substitute b into a: Answer : C

x = 10 + x 3 33
2.145
23. (p8q–4)4 × p–3q2 = (p8)4(q–4)4 × p–3q2

2.145x = 21.45 + x = p6q–3 × p–3q2

1.145x = 21.45 = p q6 +(–3) (–3) + 2

x = 18.73 = p3q–1

Answer : B = p3
q

17. C Answer : B

MD E 2
N F
B 24. w3 = 3w2
AG
Answer : C

H 25. 3 – 2x  –5 and 3 – 2x  –1
–2x  –4
∠HNA –2x  –8 x  2
Answer : C
x  4

18. ∠POQ = 4 200 = 70° 2 4
60 Answer : B

Longitude of Q = (70° – 50°)S 26. Let x be the number of cars sold in March

= 20oS

Answer : B 13000 (30 + x + 70 + 40) = x

19. 5x(x – 3y) – (x – y)2 30 (140 + x) = x
= 5x2 – 15xy – (x2 – 2xy + y2) 100
= 5x2 – 15xy – x2 + 2xy – y2
= 4x2 – y2 – 13xy 4 200 + 30x = 100x

Answer : A 70x = 4 200

x = 60

Answer : C

© Penerbitan Pelangi Sdn. Bhd. A – 12 SPM Mathematics

ANSWERS 

27. k  6 32. Let OP is x units.
Thus, k = 7
Answer : B Hence, OQ = 2x

mPQ = –   x
2x

28. –(x + 3)2 = 0 = –   1
2

x + 3 = 0 Using y = mx + c and (–8, 2),
2 = –  1 (–8) + c
x = –3
2
y c = –2

–3 O x Thus, y = –  1 x – 2
2

Answer : B
Answer : C
33. Using y = mx + c and (2, 5),
29. G HG H 5 = –5(2) + c
J c = 15
J
 Hence, B(0, 15)

(H  J)ʹ G OC = 172 – 152 B
= 8 17 units 15 units
G H Thus, C(–8, 0)

Answer : B CO

A 34. x x = 5
+ 150 8
(H  J)ʹ  G
8x = 5(x + 150)

Answer : A = 5x + 750

3x = 750

x = 250

30. ξ Answer : D

K L 35. Let x be the number of boys and y be the number of girls
2 1 M2
need to be added.
1 x = 3

2 60 5
x = 36
n[(K  M)] = 3
Answer : A Number of girls = 60 – 36 = 24

24 + y = 1
60 + y 2
31.
48 + 2y = 60 + y
Swimming
Gymnasium 23 y = 12

40 – 15 – 12 – 10 15 12 48 – 12 – 23 – 3 Answer : B
=3 10 = 10
36. m ∝ p
Cycling
m = kp

k = m
 p
n(cycling only)
= 100 – (48 + 15 + 10) 10
= 27 =  4
= 5
Answer : C
Answer : A

© Penerbitan Pelangi Sdn. Bhd. A – 13 SPM Mathematics

ANSWERS  

37. H ∝ G2 1 2 1 2 4 12 8
 F 39. w (3  2) =  3w 6

1 2 1 2
Answer : C 12 8 = 12 8
3w 2w 3w 6

38. r ∝ 1 2w = 6
pq
k w = 3
pq
r = Answer : B

k = rpq 40. 3(40) + 2(25) = x
= 3 × 2 × 4 4(40) + 5(25) = y

= 24

r = 24 Equations in matrix form:
2×6 x
1 2 1 2 1 2 34 2 40 = y
= 2 5 25

Answer : D Answer : C

Paper 2

No. Solution and Mark Scheme Marks Total
P1 3
P = Knowledge, K = Method and N = Value

1. (a) Q 1. (a) Q

P P
R R

(b) Q (b) Q P2

P P
R R

2. Let x = price of toothbrush brand X 2. 2x + y = 11 K1 5
y = price of toothbrush brand Y 3x – 2y = 6 K1
7x = 28 or equivalent K1
 2x + y = 11 ...…............…a x=4 N1
3x – 2y = 6 ……............…. y=3 N1

a × 2 : 4x + 2y = 22 …… SPM Mathematics

 +  : 7x = 28

x = 4

Substitute x = 4 into a: 2(4) + y = 11

y = 3

Hence, the price of a toothbrush brand Y is RM3.

© Penerbitan Pelangi Sdn. Bhd. A – 14

ANSWERS 

No. Solution and Mark Scheme Marks Total
K1 4
3. (2x – 4)2 + x2 = 102 3. (2x – 4)2 + x2 = 102 K1
N1
5x2 – 16x – 84 = 0 N1

(x – 6)(5x + 14) = 0 (x – 6)( 5x + 14) = 0 P1 3

x – 6 = 0 or 5x + 14 = 0

x = 6 x = 14 x=6
5 28 cm

(Not accepted)

Perimeter = 2[2(6) – 4] + 2(6)

= 28 cm

4. (a) RS 4. (a) ∠SKN or ∠NKS

QP
KN

8 cm

L 14 cm M (b) tan ∠SKN = 8 K1
14 N1
∠SKN or ∠NKS
29.74o or 29o44
(b) tan ∠SKN = 8
14

  ∠SKN = tan–1 8
14
= 29.74o or 29o44

5. (a) Not a statement. 5. (a) Not a statement. P1 5
5
(b) Elements of set M = {7, 11, 13} are prime (b) Elements of set M = {7, 11, 13} are P1
numbers or elements of set N = {16, 25, 37}
prime numbers or elements of set
are perfect squares. N = {16, 25, 37} are perfect squares.

(c) The diagonals of a kite do not bisect each other. (c) The diagonals of a kite do not bisect P1
each other.

(d) Implication 1: (d) Implication 1: P1
If ∠Q is an obtuse angle then ∠Q is more than If ∠Q is an obtuse angle then ∠Q is P1

90o and less than 180o. more than 90o and less than 180o.

Implication 2: Implication 2:
If ∠Q is more than 90o and less than 180o then If ∠Q is more than 90o and less than

∠Q is an obtuse angle. 180o then ∠Q is an obtuse angle.

6. (a) r = 3 6. (a) r = 3 P1

(b) Using m = 1 and (4, 3), (b) 3 = 1 (4) + c K1
2 2 N1

y = mx + c y = 1x + 1
2
3 = 1 (4) + c
2

 c = 1
Equation of KL: y = 1 x + 1

2

© Penerbitan Pelangi Sdn. Bhd. A – 15 SPM Mathematics

ANSWERS  

(c) 3 = 1 (10) + c No. Solution and Mark Scheme Marks Total
2 (c) 3 = 1 (10) + c K1
2
 c = –2 N(0, –2) N1
Thus, the coordinates of point N is (0, –2).

7. Volume of cylinder + volume of cone = 1 100 7. 22 × 52 × h K1 4
7
22 × 52 × h + 1 × 22 × 52 × 6 = 1 100 1 × 22 × 52 × 6 K1
7 37 37 K1
22 × 52 × h + 1 × 22 × 52 × 6 = 1 100 N1
550 h + 1 100 = 1 100 7 37 P2 6
77 h = 12 cm
h = 12 cm

8. (a) N = M–1 1 28. –6 5
k (a) 1 –3 2
1 2 1 1p –3 5 = 1 –6 5 2 (2)(–6) – (–5)(3)
–3 2
2 (2)(–6) – (–5)(3)

1 2 = 1 –6 5 Note:
3–3 2
1 21 or  –6 5 , award P1.
Thus, p = 3, k = –6
(2)(–6) – (–5)(3) –3 2

1 21 2 1 2(b) 2 x 8 1 21 2 1 2(b) x 8 P1
3 –5 y = 9 2 –5 y = 9 K1
–6 3 –6
N1
1 2 1 21 2 x = 1 –6 5 8 1 2 1 21 2 x –6 5 8 N1
y 3 –3 2 9 y = 1 –3 2 9
(2)(–6) – (–5)(3)
1  = 1
3
(–6)(8) + (5)(9) 1 2 1 21 2 orx
(–3)(8) + (2)(9) y =1 –6 5 8
3 –3 2 9
1 2 = 1 –3
3 –6 x = –1

1 2 = –1 y = –2
–2

x = –1, y = –2 Note:

1 2 1 2 1. x = –1 as a final answer,
y –2

award N1

2. Do not accept any solutions solved

not using matrix method.

9. (a) Perimeter  9. (a) 45° × 2 × 3.142 × 10 or K1 6
360° K1
= OR + arc RP + PQ + arc QS + OS  30° × 2 × 3.142 × 18 N1
 = 10 + 45° × 2 × 3.142 × 10 360°
360°   10 + 45° × 2 × 3.142 × 10
360°
  + 8 + 30° × 2 × 3.142 × 18 + 18   + 8 + 30° × 2 × 3.142 × 18 + 18
360° 360°

= 10 + 7.855 + 8 + 9.426 + 18 53.28 cm

= 53.28 cm

© Penerbitan Pelangi Sdn. Bhd. A – 16 SPM Mathematics

ANSWERS 

S No. Solution and Mark Scheme Marks Total
K1
(b) Let h = height of ∆OPS P (b) 30° × 3.142 × 182 or 1 × 18 × 5
sin 30° = h h 360° 2 K1

10 10 cm 30° or 45° × 3.142 × 102 N1
h = 5 cm 360°
O
30° × 3.142 × 182 – 1 × 18 × 5
Area of the shaded region 360° 2
= Area of sector QOS – Area of ∆OPS
+ 45° × 3.142 × 102
 + Area of sector ORP 360°

= 30° × 3.142 × 182 – 1 × 18 × 5 79.11
360° 2

 + 45° × 3.142 × 102
360°

= 84.834 – 45 + 39.275

= 79.11 cm2

10. (a) S = {(C, C), (C, L), (C, U), (C, B), (L, C), 10. (a) S = {(C, C), (C, L), (C, U), (C, B), P1 5
 (L, L), (L, U), (L, B), (U, C), (U, L), (L, C), (L, L), (L, U), (L, B),
 (U, B), (B, C), (B, L), (B, U), (B, B)} (U,C), (U, L), (U, B), (B, C),
(B, L), ( B, U), (B, B)}

(b) (i) {(C, U), (L, U), (B, U)} (b) {(C, U), (L, U), (B, U)} K1
N1
P(second card is labelled with a vowel) 1 K1
= 3 = 1 N1
5
15 5 P1 6
{(C, C), (L, L), (B, B), (L, C), (U, C),
(B, C)}
(ii) {(C, C), (L, L), (B, B), (L, C), (U, C),
(B, C)} 2
5
P(both cards are labelled with the same
letter or the second card is labelled with
a letter C)

= 6 = 2
15 5

11. (a) 30 m s–1 11. (a) 30 m s–1

(b) 3t0––1250 = 20 – 8 (b) 30 – 20 = 20 – 8 K1
15 – 0 t – 15 15 – 0 N1

t – 15 = 10 × 15 27.5
12

t = 27.5

(c) Total distance of the motorcycle (c) 1 × (8 + 30) × 27.5 or (45 – 27.5) K1
= 1 × (8 + 30) × 27.5 + (45 – 27.5) × 30 2 K1
× 30 or 1 × 45 × 30 N1
2
= 522.5 + 525 2

= 1 047.5 m  1 × (8 + 30) × 27.5 + (45 – 27.5)
2
Total distance of the bus   × 30 – 1 × 45 × 30 or equivalent
= 1 × 45 × 30 2

2 372.5
= 675 m

Different = 1 047.5 – 675 = 372.5 m

© Penerbitan Pelangi Sdn. Bhd. A – 17 SPM Mathematics

ANSWERS  

12. (a) When x = – 2.5, y = –   8 = 3.2 No. Solution and Mark Scheme Marks Total
–2.5 12. (a) 3.2 K1 12
When x = 2, y = –  8 = – 4
2 –4 K1

x –2.5 2

y 3.2 – 4

(b) y (b) Axes are drawn in correct directions P1
with uniform scale for –4 < x < 4 and
16 –16 < y < 16.

12 All points are plotted correctly or K2

8 curve passes through all the points for
–4 < x < 4 and –16 < y < 16.

4 All smooth and continuous curve without N1

–1.8 –1.6 2.5 x any straight line passes through all
–4 –3 –2 –1 O 123 4
9 points using the given scale for
–4 –4 < x < 4 and –16 < y < 16.
–6
–8

–12

–16

(c) From the graph (c) (i) –6 ± 0.4 P1
(i) when x = 1.3, y = –6 (ii) –1.8 ± 0.1 P1
(ii) when y = 4.4, x = –1.8

(d) x2 – x – 4 = 0 (d) Identify equation y = –2x + 2 K1

x2 – x = 4 …….a

a ÷ x: x – 1 = 4 …… Straight line y = –2x + 2 is drawn K1
x correctly.

 × –2: –2x + 2 = – 8x  …. Values of x:
2.5 ± 0.1
–2x + 2 = y –1.6 ± 0.1 N1
N1
The suitable straight line is y = –2x + 2.
The values of x which satisfy the equation are

2.5 and –1.6.

13. (a) (i) (2 , 3) ⎯G→ (2, –1) ⎯H→ (–1, 1) 13. (a) (i) (–1, 1) P2 12
(ii) (2 , 3) ⎯H→ (–1, 5) ⎯T→ (–3, 1) P2
(ii) (–3, 1)

(b) (i) (a) V is a rotation of 90o clockwise about (b) (i) (a) Rotation of 90o clockwise P3
centre (–2, 0). about centre (–2, 0). P3

(b) W is an enlargement of scale factor (b) Enlargement of scale factor 2
2 with centre (3, 2). with centre (3, 2).

(ii) Area of NQKLM – Area of PQRST = 108 (ii) 108 K1
22 × Area of PQRST – Area of PQRST 3 N1

= 108 36 m2
Area of PQRST = 108

3
Area of PQRST = 36 m2

© Penerbitan Pelangi Sdn. Bhd. A – 18 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total

14. Mid- Upper Cumulative 14. (a) 12
Marks point boundary frequency
Marks Mid- Upper Cumulative
31 – 35 33 35.5 0 point boundary frequency
36 – 40 38 40.5 2
41 – 45 43 45.5 7 31 – 35 33 35.5 0
46 – 50 48 50.5 16
51 – 55 53 55.5 26 36 – 40 38 40.5 2
56 – 60 58 60.5 32
61 – 65 63 65.5 36 41 – 45 43 45.5 7

46 – 50 48 50.5 16

51 – 55 53 55.5 26

56 – 60 58 60.5 32

61 – 65 63 65.5 36

All class intervals correct. P1
All midpoints correct. P1
All upper boundaries correct. P1
All cumulative frequencies correct. P1

(b) Estimated mean 38(2) + 43(5) + 48(9) + 53(10) +

38(2) + 43(5) + 48(9) + 53(10) + (b) 58(6) + 63(4) K2
36 N1
= 58(6) + 63(4)
36 51.47
= 1 853
36

= 51.47

(c) Cumulative frequency (c) Axes are drawn in correct directions P1
with uniform scale of 35.5 < x < 65.5
36 and 0 < y < 36.

32 Horizontal axis is labelled with upper
boundary.
28
Vertical axis is labelled with cumulative
24 frequency.

20 All points are plotted correctly using K2
upper boundary.
16

12 All points joined correctly to construct N1
an ogive.

8

4

0 40.5 56 65.5
35.5
45.5 50.5 55.5 60.5
Marks

(d) 3 × 36 = 27 (d) 56 ± 0.5 N1
4

From the graph, third quartile = 56

© Penerbitan Pelangi Sdn. Bhd. A – 19 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total
12
15. (a) B/F C/M L/G 15. (a) B/F C/M L/G

6 cm 6 cm

A/E 2 cm D/J 5 cm K/H A/E 2 cm D/J 5 cm K/H

Correct shape with rectangles BADC K1
and CDKL. All lines are solid.

LK = CD = BA  DK = CL  AD = BC K1

Measurements correct to ±0.2 cm N1
(one way) and right angles at vertices
90° ± 1°.

(b) (i) A/B (b) (i) A/B

P/Q 5 cm R P/Q 5 cm R
1 cm 3 cm 1 cm 3 cm
K/L K/L
D/C 3 cm D/C 3 cm

7 cm H/G 7 cm

J/S/M J/S/M

E/F 7 cm E/F 7 cm H/G

Correct shape of rectangle PRKJ K1
and polygon ADJKHE. All lines are
solid. K1
N2
AE = EH  PR = JK  RK
= KH  DJ  PD

Measurements correct to ±0.2 cm
(one way) and right angles at
vertices = 90° ± 1° except vertices
D/C.

(ii) A B (ii) A B
P 1 cm P 1 cm
5 cm R/Q 5 cm R/Q
1 cm 1 cm
DT C DT C
2 cm
K/J S L/M K/J S 2 cm

3 cm L/M

G/F 3 cm

H/E 6 cm H/E 6 cm G/F

Correct shape of rectangle ABGH. K1
All lines are solid lines.

© Penerbitan Pelangi Sdn. Bhd. A – 20 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total

Dotted line CT. K1
K1
HG = AB  PR = SL  KH = LG
 DK = CL = AD  BR = RC =
DT = KS

Measurements correct to ±0.2 cm N2
cm (one way) and right angles at
vertices = 90° ± 1°.

16. (a) Longitude of F 16. (a) (180° – 105°)E P1 12
= (180° – 105°)E F(42°S, 75°E) N1
= 75°E

Location of F = (42°S, 75°E)

(b) ∠DOF = 180o – 2(42o) (b) [180o – 2(42o)] K1
5 760 N1
= 96o

Shortest distance from D to F
= DSF
= 96o × 60

= 5 760 nm

(c) Distance from D to the east of E (c) (105o – 38o) × 60o × cos 42o K2
= (105o – 38o) × 60 × cos 42o N1
= 2 987.4 nm 2 987.4 K1
(d) (i) 600 × 5.75 N1
(d) (i) Distance from E to G K1
= 600 × 5.75 3 450 K1
= 3 450 nm N1
(ii) 3 450
(ii) 3 450 = 57.5° 60
60
(57.5o – 42o)N
Latitude of G
= (57.5o – 42o)N 15.5° or 15o30N
= 15.5° or 15o30N

3KERTAS MODEL 2. 20 197, P, 20 205, 20 209, 20 213
   +4 +4 +4 +4
Paper 1 P = 20 201 = 2.0201 × 104
1. 0  5, therefore 1 is maintained.
0.007105 = 0.0071 (two significant figures) Answer : B
Answer : B
3. 1.6 × 105 + 66 000 = 160 000 + 66 000
© Penerbitan Pelangi Sdn. Bhd.
= 226 000

= 2.26 × 105

Answer : A

A – 21 SPM Mathematics

ANSWERS  

4. Given volume of block = 10 m3, 10. y
mass of cube = 2.5 g = 0.0025 kg, 4
density = 2 800 kg m–3 J

2 800 = Mass of block 2
10


Mass of block = 2 800 × 10 O 2K 4 6 x
–2 8

= 28 000 kg

Number of cubes = 28 000
0.0025

= 11 200 000 The coordinates of the centre of rotation are (1, 2).
Answer : A
= 1.12 × 107
11. Polygon B is not an image of polygon P under a rotation.
Answer : B Answer : B

5. 42308 12. cos x = –0.7314 for 180°  x  360° (quadrants III and
IV are considered)
Digit 2 in base 10 = 2 × 82
Since the value of cos x is negative, x is in quadrant III.
= 128 ∝ = cos–1 0.7314 = 43°
x = 180° + 43° = 223°
Answer : C Answer : A

6. 12 12 12 12
+
1 0 1 1 12
1 1 0 12

1 0 0 1 0 02

Answer : A

7. ∠QRS = (6 – 2) × 180° = 120°
6
∠QRU = 60° 13. y
1
∠PRU = 30°

x = 180° – 30° – 100° = 50° y = sin x
180º
Answer : B 0 x

360º

–1

8. ∠STP = (5 – 2) × 180° = 108° B represents the graph of y = sin x for 0°  x  180°.
5 Answer : B

x = 108° ÷ 3 = 36°

∠TPQ = 108°

y = 108°  y and ∠TPQ are corresponding angles.

x + y = 36° + 108° = 144°

Answer : C 14. T

9. QR = QP, so ∠PRQ = ∠RPQ. US R

∠SRP = 100° ÷ 2 = 50° The P between Q
∠SOP = 360° – 100° = 260° angle
∠RPO = 360° – 260° – 30° – 50° = 20° the plane UPR and
∠RPQ = 90° – 20° = 70° the plane UPQ
∠PRQ = 70°
Answer : C
x = 180° – 70° – 70° = 40°

Answer : D

© Penerbitan Pelangi Sdn. Bhd. A – 22 SPM Mathematics

ANSWERS 

15. G 20. 2p – 8 = 4(3 – p)
65°
2p – 8 = 12 – 4p
20 m
65° 2p + 4p = 12 + 8

EF 6p = 20

p = 20
6

tan 65° = 20 = 10
EF 3

EF = 20 Answer  : C
tan 65°
21. 22 × 33
= 9.33 m
1

Answer : C (27 × 4–6)3

= 22 × 33

1

16. R (33 × 2–12)3
40º
= 22 × 33
P 40º T 31 × 2–4

= 22 – (–4) × 33 – 1

= 26 × 32

= 576

Q 12 m S Answer  : D

RT 22. Price of curry puff = RMg
tan 40° = 12 Price of nasi lemak = RMh
RT = 12 tan 40°
4g + h = 4.1
= 10.07 m
4g = 4.1 – h
1 4.1 – h
Since PQ = 2 × RS, so TS = RT = 10.07 m g = 4

RS = 10.07 + 10.07 Answer  : A

= 20.14 m

Answer : C 23. Volume of water in 3 cups and 1 bottle
= 2 000 ml – 500 ml
17. North/ Utara = 1 500 ml

120° Let x = volume of water in a cup
y = volume of water in a bottle
100°
3x + y = 1 500 ….…… a
x 3x = y ………………… b

Substitute b into a:

Bearing of farm from house, x = 180° + 100° 3x + 3x = 1 500  Substitute to find the value of x.
6x = 1 500
= 280° x = 250

Answer  : D

18. Longitude of R = 20°W Answer  : A
Longitude of P = 60°E
Difference in longitude = 60° + 20° = 80°  2 –3 m–3
n2
Answer  : C 24. 2 m3n–1 ×

= 2m–2n3 × m–3
n2
= 2m–2 + (–3) n3 – 2
19. 2(p – 1)(p + 5) = 2(p2 + 5p – p – 5)
= 2m–5n
= 2(p2 + 4p – 5) 2n
= m5
= 2p2 + 8p – 10

Answer  : B Answer  : A

© Penerbitan Pelangi Sdn. Bhd. A – 23 SPM Mathematics

ANSWERS  

25. h – 1  h + 5 30. y = 8 – x3
4 The function is a cubic function and a  0.

h – 4  4h + 20 So, the shape is . y-intercept is 8.
Answer : A
–4 – 20  4h – h

–24  3h

3h  –24

h  –8 31.  = {40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}
A = {41, 43, 45, 47, 49, 50}
Answer  : C A = {40, 42, 44, 46, 48}

26. x  4 ; 11 – 3x  5 Answer : A

–3x  5 – 11

–3x  –6 32. Since  = X  Y, set X  Y is equivalent to φ.
Answer : B
x  2

Hence, x = 2, 3, 4

Answer  : B of MN 1
3
33. Gradient =

27. Percentage of reference books M(x, 0) and N(6, 3)
= 100% – 25% – 10%
36 – 0 = 1
= 65% – x 3

x = 65 × 360° 6 – x = 9
100
x = 6 – 9

= 234° x = –3

Answer  : C Hence, coordinates of M = (–3, 0)

28. If x = 3, median = 14th data = 2 Answer : C
If x = 4, median = average of 14th and 15th data
34. P(0, y), Q(5, 10), R(5, 0)

=2 Since SR = 3 units, coordinates of S = (2, 0)

If x = 5, median = 15th data = 2 PS = 25

If x = 6, median = average of 15th and 16th data (2 – 0)2 + (0 – y)2 = 25

= 2.5 4 + y2 = 20

Hence, the maximum value of x is 5. y2 = 16

Answer  : C y = 4 (since P is located above x-axis)

29. \ P(0, 4)

Midpoint Frequency Gradient of PQ = 10 – 4
8 5 5–0
13 9
18 12 = 6
23 15 5
28 9
= 1 1
5

Answer : B

Mean = Sum of (midpoint × frequency) 35. P(green marble) = 1 – 1 – 1
Sum of frequencies 82
3
= (8 × 5) + (13 × 9) + (18 × 12) + (23 × 15) + (28 × 9) = 8
5 + 9 + 12 + 15 + 9
n(green marble) = 15

= 40 + 117 + 216 + 345 + 252 Number of marbles in the box = n(S)
50 n(green marble)
P(green marble) = n(S)
= 970
50 3 × n(S) = 15
8
= 19.4 n(S) = 40

Answer : C Answer : A

© Penerbitan Pelangi Sdn. Bhd. A – 24 SPM Mathematics

ANSWERS 

36. n(S) = 40 1 21 2 1 2 39. 2x 1 = 6
14 2 y
Let A = {multiple of 5}

= {15, 20, 25, 30, 35, 40, 45, 50} 1 2 1 2 2× 1 + x × 2 = 6
1 × 1 + 4 × 2 y
n(A) = 8
n(A) = 40 – 8 1 2 1 22 + 2x = 6
1+8 y
= 32

P(not a multiple of 5) = 32 2 + 2x = 6
40 2x = 4
x = 2
=4
Answer : D 5 1 + 8 = y
y = 9

37. P a r Answer : A
P = kr
1 = k(6) 1 2 40. Let A = ab
cd
2
k = 1 1 2 1 2 1 2 2ab + 4 2 = 82
c d 3 1 39
12
2a + 4 2b + 2
\ P = 1 r 2c + 3 2d + 1
12 1 2 1 2 = 82
39

Answer : D 2a + 4 = 8

m 2a = 4
r
38. p a 6 = 3(16) a = 2
r
km 2b + 2 = 2
p = r r = 48 b = 0
6
9 = k(6) 2c + 3 = 3
2 r = 8 c = 0

18 = 6k Hence, x = 8 2d + 1 = 9
2d = 8
k = 3 d = 4

∴ p = 3m 1 2 ∴ A = 2 0
r 04

Answer: C Answer : D

Paper 2

1. y No. Solution and Mark Scheme Marks Total
y = 2x + 1 P = Knowledge, K = Method and N = Value 3

1. y

y = 2x + 1

O y = x O y = x
–1 –1

x+y=3 x+y=3

The straight line y = –1 is drawn correctly. K1
The region is shaded correctly. P2

© Penerbitan Pelangi Sdn. Bhd. A – 25 SPM Mathematics

ANSWERS  No. Solution and Mark Scheme 
2. (a) ∠VBC or ∠CBV
2. (a) Angle between the planes VAB and ABCD Marks Total
= ∠VBC or ∠CBV P1 3

(b) t an ∠VBC = 7 (b) tan ∠VBC = 7 or equivalent K1
6 6 N1

∠VBC = tan–1   7 49.40° or 49°24ʹ
6

= 49.40° or 49°24ʹ

3. 5x + 2 = x2 3. 3x2 – 5x – 2 = 0 K1 4
3 K1
5x + 2 = 3x2 (x – 2)(3x + 1) = 0
or equivalent N1
N1
3x2 – 5x – 2 = 0 x = 2,
x = –  1
(x – 2)(3x + 1) = 0
3
x = 2, x = –  1
3

4. Let x = number of shirt A 4. 2x + 2y = 4 008 or 1.5x + 1.5y = 3 006 or K1 4
y = number of shirt B equivalent

x + y = 2 004 ……................…a 0.5x = 500 or 0.5y = 502 or equivalent K1
1.5x + 2y = 3 508 …….........…b

a × 2: 2x + 2y = 4 008 ………c x = 1 000 N1
y = 1 004 N1
c – b: 0.5x = 500

x = 1 000

Substitute x = 1 000 into a: Note:
Do not accept solution solved using matrix
1 000 + y = 2 004 method.

y = 1 004

∴ The number of shirt A = 1 000
The number of shirt B = 1 004

5. πr2h + 2  πr3 = 339 3 5. 22 × 32 × h or 2 × 22 × 33 K1 3
37 7 37
K1
    22 × 32 × h + 2 × 22 × 33 = 339 3    22 × 32 × h + 2 × 22 × 33 = 339 3 N1
7 37 7

198 h + 396 = 339 3 7 37 7

77 7 10 m

198 h = 1 980
77
h = 10

∴ The height of the cylinder is 10 m.

© Penerbitan Pelangi Sdn. Bhd. A – 26 SPM Mathematics

ANSWERS 
Marks Total
No. Solution and Mark Scheme
K1 6
6. (a) Perimeter of the whole diagram  6. (a) 30° × 2 × 22 × 7 or 142 + 142
= OC + CD + DB + BA + OA 360° 7 K1

 = 7 + 30° × 2 × 22 × 7 + 7 + 142 + 142  + 14   7 + 30° × 2 × 22 × 7 + 7 N1
360° 7
360° 7
= 7 + 3.67 + 7 + 19.80 + 14
+ 142 + 142  + 14

= 51.47 cm

51.47 cm

(b) The area of the shaded region    (b) 1 × 14 × 14 or 90° × 22 × 72 K1
= Area of triangle OAB – Area of sector ODE 2 360° 7 K1
N1
    = 1 × 14 × 14 – 90° × 22 × 72     1 × 14 × 14 – 90° × 22 × 72
2 360° 7 2 360° 7
= 98 – 77 59.5 cm2 or 59 1 cm2 or 119 cm2

2 22
= 59.5 cm2 or 59 1 cm2 or 119 cm2

22

7. (a) not a statement 7. (a) not a statement P1 6
(b) If 15x  45, then x  3. P1
(b) Implication 1 : If x  3, then 15x  45. P1
If 15x  45, then x  3.
Implication 2: (c) ABCD is a trapezium. P1
If x  3, then 15x  45. (d) 3n2 + 1 K1
n = 1, 2, 3, 4, ... N1
(c) ABCD is a trapezium.

(d) 4 = 3(1) + 1 = 3(12) + 1
13 = 3(4) + 1 = 3(22) + 1
28 = 3(9) + 1 = 3(32) + 1
49 = 3(16) + 1 = 3(42) + 1

Conclusion:
3n2 + 1, n = 1, 2, 3, 4, ...

8. (a) y-coordinate of M = 2 × 3 = 6 8. (a) (0, 6) N1 5
Coordinates of M = (0, 6)
(b) 6 – 4 K1
(b) Given M(0, 6) and N(5, 4), 0–5 N1
–   2
mMN = 6–4 = –  2
0–5 5 5

© Penerbitan Pelangi Sdn. Bhd. A – 27 SPM Mathematics

ANSWERS  

(c) PQ and MN are parallel, No. Solution and Mark Scheme Marks Total
(c) 3 = –  2 (0) + c or y – 3 = –  2 (x – 0) K1
mPQ = mMN = –  2 55
5 or equivalent

Substitute (0, 3) into y = mx + c y = –  2 x + 3 or equivalent P1
3 = –  2 (0) + c 5

5
c=3

or
y – 3 = –  2 (x – 0)

5
y = –   2 x + 3

5

∴ y = –  2 x + 3 or equivalent
5

9. (a) S = {(1, A), (1, B), (1, C), (2, A), (2, B), (2, C), 9. (a) S = {(1, A), (1, B), (1, C), (2, A), (2, B), P2 6

(3, A), (3, B), (3, C), (4, A), (4, B), (4, C), (2, C), (3, A), (3, B), (3, C), (4, A),

(5, A), (5, B), (5, C), (6, A), (6, B), (6, C)} (4, B), (4, C), (5, A), (5, B), (5, C),

(6, A), (6, B), (6, C)}

Accept 1 or 2 mistake award P1

(b) (i) {(4, A), (4, B), (4, C)} (b) (i) {(4, A), (4, B), (4, C)} K1
N1
P(begins with 4) = 3 3 1 K1
18 18 6
or N1
=1
6

(ii) {(2, B), (2, C), (3, B), (3, C), (5, B), (5, C)} (ii) {(2, B), (2, C), (3, B), (3, C),
(5, B), (5, C)}

P(consists of prime number and consonant) 6 or 1
18 3
= 6
18

= 1
3

10. (a) Duration of time = 8 – 5 10. (a) 3 s N1 6

=3s K1
N1
(b) Speed in first 5 seconds (b) 50 – 0 K1
5–0 K1
= 50 – 0 N1
5–0

= 10 m s–1 10 m s–1

(c) Total distance travelled = 50 + 50 (c) 100 m

= 100 m

Average speed = Total distance travelled 10 = 100
Total time k

10 = 100 k = 10
k
10k = 100


k = 10

© Penerbitan Pelangi Sdn. Bhd. A – 28 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total
K1 6
1 2 1 2 11. (a) R =a b 11. (a) 4m – (–6)(2) = 0 N1
c d = 4 –6 m = –3
2m P1
K1
R has no inverse, N1
N1
ad – bc = 0

4m – (–6)(2) = 0

4m + 12 = 0

4m = –12

m = –3

1 21 2 1 2(b) 2–1 x –7 1 21 2 1 2(b) 2–1 x = –7
–3 4 y = 13 –3 4 y 13

1 2 1 21 2x 4 1 –7 1 2 1 21 2x
y 3 2 13 y
= 1 = 2(4) 1 4 1 –7
2(4) – (–1)(–3) – (–1)(–3) 3 2 13

1 2 = 1
5
4(–7) + 1(13)
3(–7) + 2(13) x = –3

1 2 = 1 –15 y = 1
5 5

1 2 = –3 Note: x
1 1 2 1 2 1. Ify=
–3
x = –3, y = 1 1  as final answer,

award N1.

2. Do not accept solution solved not

using matrix method.

12. (a) When x = –3, 12. (a) –10 K1 12
2 K1
y = (–3)3 – 4(–3) + 5
= –10

When x = 1,
y = (1)3 – 4(1) + 5

= 2

x –3 1
y –10 2

(b) (b) Axes are drawn in correct directions P1
with uniform scale for –4  x  3.5 and
y 2 cm – 43  y  33.9
30 y = 8x + 12 2 cm
20 All the points are plotted correctly or K2
x
10 3 curve passes through all the points for
– 4  x  3.5 and – 43  y  33.9

–4 –3 –2 –1 0 12 All smooth and continuous curve without N1
–10
any straight line passes through all 9

correct points using the given scale for
– 4  x  3.5 and – 43  y  33.9

–20

y = x3 – 4x + 5 –30 Note:
–40 1. 7 or 8 points are plotted correctly,

award K1
2. Ignore the curve out of range

© Penerbitan Pelangi Sdn. Bhd. A – 29 SPM Mathematics

ANSWERS  No. Solution and Mark Scheme 
(c) (i) 10  y  11
(c) (i) When x = 2.5, y = 10.5 Marks Total
(ii) When y = –30, x = –3.65 (ii) –3.7  x  –3.6 P1
P1
(d) y = x3 – 4x + 5 ...........a
0 = x3 – 12x – 7 ........  (d) Identify equation y = 8x + 12 K1
a –  : y = 8x + 12
Equation of the straight line: Straight line y = 8x + 12 is drawn K1
y = 8x + 12 correctly.
From the graph, x = –3.1, –0.6
Values of x: N1
–3.15  x  –3.05 N1
–0.65  x  –0.55

Note:
Values of x obtained by calculation,

award N0

13. (a) Mass (kg) Frequency Midpoint 13. (a) Mass (kg) Frequency Midpoint 12
35 – 39 0 37
40 – 44 2 42 35 – 39 0 37 P1
45 – 49 3 47 P1
50 – 54 8 52 40 – 44 2 42 P1
55 – 59 12 57
60 – 64 9 62 45 – 49 3 47
65 – 69 4 67
70 – 74 2 72 50 – 54 8 52
75 – 79 0 77
55 – 59 12 57

60 – 64 9 62

65 – 69 4 67

70 – 74 2 72

75 – 79 0 77

All class intervals are correct.
All frequencies are correct.
All midpoints are correct.

(b) (i) Modal class = 55 – 59 (b) (i) 55 – 59 P1

(ii) Estimated mean mass (2 × 42) + (3 × 47) + (8 × 52)

(2 × 42) + (3 × 47) + (8 × 52) + (12 × 57) + (9 × 62) + (4 × 67)

+ (12 × 57) + (9 × 62) + (4 × 67) (ii) + (2 × 72) K2
2 + 3 + 8 + 12 + 9 + 4 + 2 N1
+ (2 × 72)
= 2 + 3 + 8 + 12 + 9 + 4 + 2
57.375
= 2 295
40

= 57.375

© Penerbitan Pelangi Sdn. Bhd. A – 30 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total

(c) (c) Axes are drawn in correct directions P1
with uniform scale for 37  x  77 and
Frequency 2 cm 0  y  12.
2 cm
12 Horizontal axis is labelled with midpoint.
10
Vertical axis is labelled with frequency.
8
6 All the points are plotted correctly using K2
4 midpoint.
2
All the points joined correctly to N1
construct a frequency polygon.

0
37 42 47 52 57 62 67 72 77
Mass (kg)

(d) Percentage of students who have a mass of (d) 67.5% N1

exceeding 54 kg

= 12 + 9 + 4 + 2 × 100%

2 + 3 + 8 + 12 + 9 + 4 + 2

= 27 × 100%
40

= 67.5%

14. (a) (i) K(5, 6) ⎯P→ (5, 4) 14. (a) (i) (5, 4) P1 12
(ii) K(5, 6) ⎯ T→ (2, 8) ⎯P→ (2, 2) (ii) (2, 2) P2

(b) (i) (a) N = Rotation 90° anticlockwise at (b) (i) (a) Rotation 90° anticlockwise at P3
centre (2, 5) P3
centre (2, 5)
(b) M = Enlargement with scale factor 2
at centre Q (b) Enlargement with scale factor
2 at centre Q

(ii) Area of HJLQ (ii) 22 × 16 K1
= Area of DEFG 64 – 16 K1
= 16 m2 48 N1

Area of ABCQ
= 22 × 16
= 64 m2

Area of shaded region
= Area of ABCQ – Area of HJLQ
= 64 – 16
=48 m2

© Penerbitan Pelangi Sdn. Bhd. A – 31 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total
15. (a) 12
15. (a) J/T 4 cm K/S J/T 4 cm K/S

5 cm 2 cm 2 cm
L/R L/R

5 cm

N/Q 2 cm M/P N/Q 2 cm M/P

Correct shape with polygon JKLMN. All K1
lines are solid lines.

JN  JK  KL = NM K1

Correct measurement ±0.2 cm (one N1
way) and ∠J = ∠K = ∠N = 90° ± 1°.

(b) (i) C/D 7 cm F/E (b) (i)

C/D 7 cm F/E

7 cm K/J S/T K/J S/T
2 cm L 2 cm
7 cm R
LR

3 cm 3 cm
M/N 4 cm P/Q/G
B/A

B/A M/N 4 cm P/Q/G

Correct shape with square CFPB, K1
rectangle KSRL and rectangle
LRPM. All lines are solid lines.

CF = CB  KS = LR = MP  RP K1
 SR

Correct measurement ±0.2 cm N2
(one way) and all angles at the
vertices = 90° ± 1°.

(ii) E/G F/T/Q P S/R (ii) E/G F/T/Q P S/R

7 cm 4 cm 7 cm 4 cm

2 cm 2 cm
J/N M 2 cm K/L J/N M 2 cm K/L

D/A 2 cm C/B D/A 2 cm C/B

© Penerbitan Pelangi Sdn. Bhd. A – 32 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total

Correct shape with rectangle K1
EFCD and square FSKJ. All lines

are solid lines.

Dotted line PM K1

ED  SK = PM  DC = JM = MK K1

Correct measurement ±0.2 cm N2
(one way) and all angles at the
vertices = 90° ± 1°.

16. (a) Latitude of R = 50°N 16. (a) (180° – 45°)E P1 12
(50°N, 135°E) N1
Longitude of R = (180° – 45°)E

= 135°E

Location of R = (50°N, 135°E)

(b) The shortest distance from Q to R (b) (180 – 50 – 50) × 60 K2
= (180 – 50 – 50) × 60 4 800 N1
= 80 × 60
= 4 800 nautical miles (c) (80 – 45) × 60 × cos 50° K2
1 349.85 N1
(c) The distance from P due east to Q
= (80 – 45) × 60 × cos 50°
= 35 × 60 × cos 50°
= 1 349.85 nautical miles

(d) (i) ∠HOQ = 5 820  (d) (i) 5 820 – 50° S K1
60 60 N1
47°S K1
= 97°
N1
Latitude of H = (97° – 50°)S

= 47°S 5 820
Total time taken
Total distance (ii) 580 =
Total time taken
(ii) Average speed = or equivalent

580 = 5 820 10.03
Total time taken

Total time taken = 5 820
580

= 10.03 hours

© Penerbitan Pelangi Sdn. Bhd. A – 33 SPM Mathematics