Grab A+ Kertas Model SPM Matematik

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1
KERTAS MODEL SPM / SPM MODEL PAPER





Matematik/Mathematics 1449/1
Kertas 1/Paper 1 1 jam 30 minit/1 hour 30 minutes




Kertas soalan ini mengandungi 40 soalan. Jawab semua soalan.
This question paper consists of 40 questions. Answer all questions.

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4
2
1. 0.000867 – 3.2 × 10 = 4. Ungkapkan 3(5 ) + 5 + 19 sebagai suatu nombor dalam
–5
A 8.35 × 10 –4 asas lapan.
4
2
B 8.35 × 10 –2 Express 3(5 ) + 5 + 19 as a number in base eight.
C 5.47 × 10 –4 A 3277 8
D 5.47 × 10 –2 B 3577 8
C 4277 8
D 4577
8
2. Bundarkan 52.10528 betul kepada empat angka bererti.
Round off 52.10528 correct to four significant figures.
A 52.1 5. Diberi 110111 + 101011 = p + 132 , cari nilai p.
2
10
2
8
B 52.11 Given 110111 + 101011 = p + 132 , find the value
10
2
8
2
C 52.105 of p.
D 52.1053 A 5
B 6
C 7
3. Rajah 1 menunjukkan sebuah silinder. D 8
Diagram 1 shows a cylinder.
6. Rajah 2 menunjukkan sebuah pentagon ABCDE. AEG
dan BFG ialah garis lurus.
Diagram 2 shows a pentagon ABCDE. AEG and BFG
are straight lines.

120 cm A


E
B
88° x°

F
0.7 m
Rajah 1 y° D C
Diagram 1 G
Rajah 2
Hitung isi padu, dalam cm , bagi silinder tersebut. Diagram 2
3
Calculate the volume, in cm , of the cylinder.
3
A 0.948 × 10 6 Cari nilai x + y.
B 1.148 × 10 6 Find the value of x + y.
C 1.848 × 10 6 A 68
D 4.62 × 10 5 B 88
C 108
D 540









KM1 K1 – 1 © Pan Asia Publications Sdn. Bhd.

7. Dalam Rajah 3, AF ialah sisi sepunya bagi heksagon 9. Dalam Rajah 5, PQR ialah garis lurus dan PSQ ialah
sekata ABCDEF dan poligon P. segi tiga bersudut tegak.
In Diagram 3, AF is the common side of the regular In Diagram 5, PQR is a straight line and PSQ is a
hexagon ABCDEF and polygon P. right-angled triangle.

R
Q
x°
P
P
B A


F S
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C
Rajah 5
Diagram 5
Diberi PS = 12.5 cm dan PQ = 32.5 cm, cari nilai bagi
D E
kos x°.
Rajah 3 Given PS = 12.5 cm and PQ = 32.5 cm, find the value
Diagram 3
of cos x°.
12
Cari bilangan sisi bagi poligon P. A —–
Find the number of sides of polygon P. 13
5
A 8 B —–
B 10 13
12
C 12 C – —–
D 14 13
5
D – —–
13
8. Dalam Rajah 4, ABC ialah tangen kepada bulatan
dengan pusat O pada titik B. APO, BTQ dan PTSR ialah
garis lurus. 10. Rajah 6 menunjukkan graf bagi fungsi y = sin x dan
In Diagram 4, ABC is a tangent to the circle with y = kos x.
centre O at point B. APO, BTQ and PTSR are straight Diagram 6 shows the graph of the functions y = sin x
lines. and y = cos x.

y
Q
y = kos x
y = cos x
A x° P y = sin x
P
28°
O
T x
0
S
y°
B R Q

Rajah 6
Diagram 6
C
Rajah 4 Tentukan koordinat-x bagi titik P dan titik Q.
Diagram 4 Determine the x-coordinate of points P and Q.
Cari nilai x + y. A 45°, 135°
Find the value of x + y. B 90°, 135°
A 31 C 45°, 225°
B 62 D 90°, 225°
C 93
D 124






© Pan Asia Publications Sdn. Bhd. KM1 K1 – 2

1
KERTAS MODEL SPM / SPM MODEL PAPER





Matematik/Mathematics 1449/2
Kertas 2/Paper 2 2 jam 30 minit/2 hours 30 minutes




Bahagian A
Section A
[40 markah]
[40 marks]
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Jawab semua soalan dalam bahagian ini.
Answer all questions in this section.


1. Rajah 1 menunjukkan sebatang tangga dengan panjang 1.4 m yang disandarkan pada sebuah dinding mencancang.
Diagram 1 shows a ladder with length of 1.4 m resting against a vertical wall.















Rajah 1
Diagram 1
Diberi jarak mengufuk antara bahagian bawah tangga dengan dinding ialah 58 cm.
Given the horizontal distance between the bottom of the ladder with the wall is 58 cm.
(a) Lakarkan satu segi tiga yang terbentuk bagi menunjukkan satah dengan tangga, lantai dan dinding.
Sketch a triangle formed to show the plane with the ladder, floor and wall.
(b) Tandakan sudut di antara satah dengan tangga dan lantai pada segi tiga yang dilakarkan di 1(a).
Seterusnya, hitung sudut tersebut.
Mark the angle between the plane with the ladder and floor on the triangle sketched in 1(a).
Hence, calculate the angle.
(c) Jika bahagian atas tangga dinaikkan sehingga 1.3 m secara mencancang dari lantai, hitung sudut di antara satah
dengan tangga dan lantai yang perlu ditambah atau dikurangkan.
If the top of the ladder is elevated to 1.3 m vertically from the floor, calculate the angle between the plane with the
ladder and floor that need to be added or decreased.
[5 markah]
[5 marks]

Jawapan/ Answer:
(a)



(b)



(c)







KM1 K2 – 1 © Pan Asia Publications Sdn. Bhd.

2. Penyelesaian menggunakan kaedah matriks tidak dibenarkan untuk soalan ini.
Solving using matrix method is not allowed in this question.

Encik Ahmad menjual nasi lemak dan mi goreng di gerai makanannya. Bhala membayar sebanyak RM20 kepada Encik
Ahmad untuk 4 bungkus nasi lemak dan 5 bungkus mi goreng. Nini membeli 8 bungkus nasi lemak dan 15 bungkus mi
goreng daripada Encik Ahmad dengan membayar sebanyak RM54.
Hitung harga, dalam RM, bagi sebungkus nasi lemak dan sebungkus mi goreng yang dijual oleh Encik Ahmad jika
harga nasi lemak adalah lebih rendah daripada harga mi goreng. [4 markah]
Mr Ahmad sells nasi lemak and fried mee at his food stall. Bhala pays RM20 to Mr Ahmad for purchasing 4 packs of
nasi lemak and 5 packs of fried mee. Nini bought 8 packs of nasi lemak and 15 packs of fried mee from Mr Ahmad by
paying RM54.
Calculate the price, in RM, of a pack of nasi lemak and a pack of fried mee sold by Mr Ahmad if the price of nasi lemak
is lower than the price of fried mee. [4 marks]
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Jawapan/ Answer:













3. (a) Nyatakan sama ada setiap pernyataan berikut adalah benar atau palsu.
State whether each of the following statements is true or false.
(i) 2 ialah nombor perdana dan 4 ialah nombor genap.
2 is a prime number and 4 is an even number.
(ii) 7 ialah faktor kepada 40 dan 8 ialah gandaan bagi 2.
7 is a factor of 40 and 8 is a multiple of 2.
[2 markah]
[2 marks]

(b) Lengkapkan hujah berikut:
Complete the following argument:
Premis 1 : Jika x = 3, maka 3x = 15.
Premise 1 : If x = 3, then 3x = 15.
Premis 2 :
Premise 2 :
Kesimpulan : x ≠ 3
Conclusion : x ≠ 3
[1 markah]
[1 mark]

(c) Rajah 3 menunjukkan susunan beberapa keping duit syiling di atas sebuah meja.
Diagram 3 shows the arrangement of several pieces of coins on a table.










Rajah 3
Diagram 3
(i) Buat satu kesimpulan secara induksi tentang bilangan duit syiling pada baris ke-n.
Make one conclusion by induction for the number of coins in the nth row.
(ii) Cari bilangan baris jika baris terakhir bagi susunan tersebut terdiri daripada 258 keping duit syiling.
Find the number of row if the last row of the arrangement is made up of 258 pieces of coins.
[2 markah]
[2 marks]
© Pan Asia Publications Sdn. Bhd. KM1 K2 – 2

Jawapan Lengkap (Kertas1)
JAWAPAN / ANSWERS Complete Answers (Paper 1)
https://bit.ly/3gz6lAU







Kertas Model 1/ Model Paper 1 3 × 1 – 2:
12x + 15y – (8x + 15y) = 60 – 54
Kertas 1/ Paper 1 12x + 15y – 8x – 15y = 6
4x = 6
1. A 2. B 3. D 4. B 5. D 6
6. A 7. C 8. B 9. C 10. C x = —
4
11. D 12. B 13. A 14. A 15. D x = 1.50
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16. B 17. A 18. A 19. C 20. A Gantikan x = 1.50 ke dalam 1,
21. D 22. B 23. C 24. B 25. D Substitute x = 1.50 into 1,
26. C 27. A 28. D 29. C 30. A 4(1.50) + 5y = 20
31. D 32. D 33. B 34. C 35. D 6 + 5y = 20
36. D 37. B 38. A 39. B 40. D 5y = 20 – 6
14
y = —–
Kertas 2/ Paper 2 5
y = 2.80
Bahagian A/ Section A ∴ Harga sebungkus nasi lemak ialah RM1.50 dan harga
1. (a)
sebungkus mi goreng ialah RM2.80.
The price of a pack of nasi lemak is RM1.50 and the price of
a pack of fried mee is RM2.80.
3. (a) (i) Benar/ True
(ii) Palsu/ False
(b) 3x ≠ 15
(c) (i) Daripada rajah,
From the diagram,
3 = 2 + 2 1 – 1
4 = 2 + 2 2 – 1
θ 6 = 2 + 2 3 – 1
10 = 2 + 2 4 – 1
58 cm ∴ 2 + 2 n – 1 , n = 1, 2, 3, 4, ...
58
(b) kos θ/ cos θ = —–– (ii) 2 + 2 n – 1 = 258
140
58
1
θ = kos / cos —–– 2 2 n – 1 = 258 – 2
–1
–1
140
= 256
2
n – 1
θ = 65° 32ʹ 2 n – 1 = 2
8
Sudut di antara satah dengan tangga dan lantai ialah n – 1 = 8
65° 32ʹ. n = 9
The angle between the plane with the ladder and the floor ∴ Terdapat 9 baris susunan duit syiling.
is 65° 32ʹ. There are 9 rows of coin arrangement.
130
(c) sin θ = —–– 4. (a)
140
–1 130
1
θ = sin —–– 2 Jarak (km)
Distance (km)
140
θ = 68° 13ʹ 50
Sudut di antara satah dengan tangga dan lantai yang perlu 40
ditambah 30
The angle between the plane with the ladder and the floor 20
that need to be added 10
= 68° 13ʹ – 65° 32ʹ Masa (minit)
O 10 20 30 40 50 60 70 80 90 100 110120 Time (minute)
= 2° 41ʹ
2. Katakan/ Let (b) (i) Laju/ Speed
20 – 0
x = harga sebungkus nasi lemak = —–———
25
the price of a pack of nasi lemak —– – 0
y = harga sebungkus mi goreng 60
–1
the price of a pack of fried mee = 48 km j / km h –1
4x + 5y = 20 ..................1 (ii) Tempoh masa Armel berhenti di Taman Beverly
Duration of time Armel stopped at Taman Beverly

8x + 15y = 54 ................2
= 45 – 25
= 20 minit/ minutes
© Pan Asia Publications Sdn. Bhd.
J – 1

(iii) Purata laju/ Average speed ∴ Potongan bulanan PCB adalah mencukupi untuk
20 + 30 + 50
= —––————– membayar cukai pendapatan.
120/60 Monthly deduction of MTD is enough to pay the income
= 50 km j / km h –1
–1
tax.
5. (a) Luas kawasan berlorek 9. (a) 2x – 10 = 0
Area of the shaded region 2x = 10
42° 22 28° 22 10
= —––– × —– × 21 – —––– × —– × 7 2 x = —–
2
360° 7 360° 7 2
13 x = 5
= 149—– cm 2
18 ∴ Jarak di antara rumah Lisa dan rumah Melissa ialah
(b) Perimeter kawasan berlorek 5 km.
Perimeter of the shaded region The distance between Lisa’s house and Melissa’s house
= OA + AB + BP + PR + RQ + QO is 5 km.
42° 22 28° 22
= 21 + —––– × 2 × —– × 21 + 7 + 7 + —––– × 2 × —– × 7 (b) y = 2x – 10 .........................1
360° 7 360° 7
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+ 7 y + 2x = 20 .........................2
Gantikan 1 ke dalam 2,
37
= 60—– cm
45 Substitute 1 into 2,
6. (a) Katakan/ Let A = Mukah, B = Sibu, C = Sarikei, D = Kapit, 2x – 10 + 2x = 20
E = Bintulu, F = Miri dan/ and G = Limbang 4x = 20 + 10
30
x = —–
G 4
15
x = —–
E 2
F 15
A Gantikan x = —– ke dalam 1,
2
15
Substitute x = —– into 1,
B 2
15
1 2
y = 2 —– – 10
2
C D y = 5
15
1
2
(b) V = {A, B, C, D, E, F, G} Koordinat bagi perpustakaan ialah —–, 5 .
E = { (A, B), (A, C), (A, E), (B, C), (B, D), (B, E), (C, D), 2
15
1
2
(D, E), (D, F), (E, F), (F, G)} The coordinates of the library is —–, 5 .
2
n(V) = 7 Persamaan garis lurus bagi laluan di antara rumah Lisa dan
n (E) = 11
sekolah ialah y = 2x.
7. Menggunakan teorem Pythagoras, The equation of the straight line of the path between Lisa’s
Using Pythagoras theorem, house and school is y = 2x.
x 2 + (3x – 3) = (2x + 3) 2 y = 2x .................................3
2
x 2 + 9x – 18x + 9 = 4x + 12x + 9 Gantikan 3 ke dalam 2,
2
2
10x – 4x – 18x – 12x = 0 Substitute 3 into 2,
2
2
6x – 30x = 0 2x + 2x = 20
2
20
6x(x – 5) = 0 x = —–
x = 0 atau/ or x = 5 4
Perimeter bagi segi tiga PQR x = 5
Perimeter of triangle PQR Gantikan x = 5 ke dalam 3,
= 5 + [3(5) – 3] + [2(5) + 3] Substitute x = 5 into 3,
= 5 + 12 + 13 y = 2(5)
= 30 cm y = 10
8. (a) Pendapatan bercukai/ Chargeable income Koordinat bagi sekolah ialah (5, 10).
= RM93 600 – (RM9 000 + RM7 000 + RM2 400 The coordinates of the school is (5, 10).
+ RM2 400 + RM2 200) Jarak di antara sekolah dan perpustakaan
= RM70 600 Distance between the school and the library
15
2
1
2
Cukai pendapatan/ Income tax = ABBBBBBBBBBBBB 2
—– – 5 + (5 – 10)
= RM4 600 + (RM70 600 – RM70 000) × 21% 2
= RM4 600 + RM126 = 5.590 km 3
x
= RM4 726 10. (a) —–—— = —
7
16 + x
(b) Jumlah PCB yang dibayar oleh Puan Wendy 7x = 3(16 + x)
Total MTD paid by Madam Wendy 7x = 48 + 3x
= RM450 × 12 7x – 3x = 48
= RM5 400 4x = 48
Lebihan cukai pendapatan yang dibayar x = —–
48
Excess income tax paid 4
= RM5 400 – RM4 726 x = 12
= RM674
© Pan Asia Publications Sdn. Bhd. J – 2

JAWAPAN / ANSWERS









Kertas Model 1/ Model Paper 1 y° = 180° – 108° – 56°
= 16°
Kertas 1/ Paper 1 x + y = 52 + 16
= 68
1. A 0.000867 – 3.2 × 10 –5
= 8.67 × 10 – 0.32 × 10 –4
–4
= (8.67 – 0.32) × 10 –4 7. C Sudut pedalaman bagi heksagon sekata
©PAN ASIA PUBLICATIONS
= 8.35 × 10 –4 Interior angle of regular hexagon
(6 – 2) × 180°
= —–—————–
2. B 52.10528 720° 6
= —–––
6
Tambah 1 kepada digit 0 kerana digit seterusnya, = 120°
5, adalah sama dengan 5. Sudut pedalaman bagi poligon P
Add 1 to the digit 0 since the next digit, 5, is Interior angle of polygon P
equal to 5. = 360° – 120° – 90°

= 52.11 (4 angka bererti/ significant figures) = 150°
Bilangan sisi bagi poligon P
Number of sides for polygon P
3. D Jejari/ Radius = 0.7 m 360°
= 70 cm = —–————–
180° – 150°
Isi padu silinder/ Volume of cylinder = 12
= πr h
2
22
2
= —– × 35 × 120 8. B ∠POB = 180° – 28° – 90°
7
= 462 000 = 62°
= 4.62 × 10 5 x + y = 62
2
2
2
4. B 3(5 ) + 5 + 19 9. C QS = 32.5 – 12.5 2
4
= 3(625) + 25 + 19 = 900
900
= 1919 QS = ABBB
10
= 30 cm
8 1919 Baki/ Remainder kos x°/ cos x° = – —–––
30
8 239 ... 7 32.5
12
8 29 ... 7 = – —–
13
8 3 ... 5
0 ... 3 10. C sin 45° = kos 45° / cos 45°

sin 225° = kos 225° / cos 225
∴ 3(5 ) + 5 + 19 = 3577
2
4
8 ∴ x = 45°, 225°
5. D 110111 + 101011 = p + 132
3
a b
3 –2
3 –9
–2 4
2 2 10 8 11. D ABBBB × 3a b ÷ 6a b
p = 110111 + 101011 – 132 8 — 1
10
2
2
–2 4
3 –9 3
3 –2
p = (1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 + 1 × 2 ) = (a b ) × 3a b ÷ 6a b
0
3
4
2
1
5
10
3 –2
–2 4
–3
+ (1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 ) = ab × 3a b ÷ 6a b
2
3
1
4
5
0
3
– (1 × 8 + 3 × 8 + 2 × 8 ) = —(a 1 + (–2) – 3 )(b –3 + 4 – (–2) )
0
2
1
p = 8 10 6
1
10
–4
3
∴ p = 8 = —(a )(b )
2
b
3
= ——
6. A Sudut pedalaman bagi pentagon 2a 4
Interior angle of pentagon 256
x
(5 – 2) × 180° 12. B 32 = ——
= —–—————– 8 x
5 32 × 8 = 256
x
x
= 108° 256 = 256
x
∠ABF = 360° – (2 × 108° + 88°) x = 1
= 56°
x° = 108° – 56° 5p 2p – 4
3
= 52° 13. A —– + 3 , ——––
5
© Pan Asia Publications Sdn. Bhd.
J – 1

5p 9 2p – 4
—– + — , ——––
3 3 5 20. A Bulan Januari Mac Ogos
5(5p + 9) , 3(2p – 4) Month January March August
25p + 45 , 6p – 12
25p – 6p , –12 – 45 Kos
19p , –57 perubatan 12 000 8 500 22 000
(RM)
57
p , – —– Medical
19
p , –3 cost (RM)
Boleh
6hk + 2h 2 9hk + 3h k
2
3
2
14. A ————– ÷ ———–—– menuntut
3k 6kh Ya Tidak Ya
2h(3k + h) 6kh 2 pampasan Yes No Yes
2
= ——–——– × ———–—– Can make
3k
3hk(3k + h)
2
4h a claim
= —–
3k Pampasan
©PAN ASIA PUBLICATIONS
yang
15. D 6p + 4q + 3 = 166q + 3 diperoleh 12 000 – 10 000 22 000 – 10 000
3
6p = 166q + 3 – 4q – 3 (RM) = 2 000 – = 12 000
3
6p = 162q Claim
3
162q received
p = —–—
3
6 (RM)
p = 27q
3
3
p = ABBB Jumlah pampasan yang diperoleh oleh Sharon
27q
q
p = 3AB Amount of claim obtained by Sharon
3
= RM2 000 + RM12 000
= RM14 000
16. B (3x – 2y) + x(2x + y)
2
= 9x – 12xy + 4y + 2x + xy
2
2
2
= 11x – 11xy + 4y 2 21. D Pintasan-y/ y-intercept = 16
2
16 – 2x = 0
3
16 = 2x 3
17. A F 16
x = —–
3
2
x = 8
3
x = 2
Pintasan-x/ x-intercept = 2
Graf berbentuk/ The shape of graph is .
Q P Jadi/ Hence, a = –2 , 0
∠FPQ Maka/ Thus, y = 16 – 2x 3
∴ ∠FPQ

22. B Pintasan-y/ y-intercept = 3
1
18. A —x + 3 = 0
R 2 1
1 S T —x = –3
2
2 3 4 x = –6
A B C D
Pintasan-x/ x-intercept = –6
1
m = —, maka graf berbentuk
2
1
m = —, thus the shape of graph is
2
R  S ʹ  T ʹ
= {1, 2, 3}  {1, 3, 4} {1, 2} 23. C Persamaan PQ ialah/ Equation of PQ is
= {1} 2
Maka, rantau A. y = —x + 4 ......... 1
3
Therefore, region A. Gantikan y = 0 ke dalam 1,
Substitue y = 0 into 1,
2
19. C ξ = {28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41} —x + 4 = 0
P = {28, 31, 33, 35, 37, 39, 40} 3 2
P ʹ = {29, 30, 32, 34, 36, 38, 41} —x = –4
3
n(Pʹ) = 7 x = –6
Pintasan-x ialah –6.
The x-intercept is –6.
© Pan Asia Publications Sdn. Bhd. J – 2

JSU MATEMATIK K1 / MATHEMATICS P1 SPM Set 1

Catatan: Penyediaan JSU ini adalah berdasarkan tanggapan penulis terhadap sampel
soalan yang terpapar dalam Format Baharu 2021. Jadi, kami meminta maaf sekiranya
terdapat ketidaktepatan dalam maklumat JSU.

No. Konstruk Aras Topik Item
Construct Level Topic Item
1 Mengingat dan R Bentuk Piawai (T3)
Memahami Standard Form (F3) Pilih A
Recall and
Understanding
2 Mengingat dan R Bentuk Piawai (T3)
Memahami Standard Form (F3) Pilih B
Recall and
Understanding
3 Mengingat dan R Bentuk Piawai (T3)
Memahami Standard Form (F3) Pilih D
Recall and
Understanding
4 Mengingat dan R Nombor asas (T4)
Memahami Number Bases (F4) Pilih B
Recall and
Understanding
5 Mengaplikasi S Nombor Asas (T4) Pilih D
Application Number Bases (F4)
R
Mengingat dan
Poligon (T2)
6 ©PAN ASIA PUBLICATIONS
Memahami Polygon (F2) Pilih A
Recall and
Understanding
7 Mengaplikasi S Poligon (T2) Pilih C
Application Polygon (F2)
8 Mengaplikasi S Bulatan (T2) Pilih B
Application Circles (F2)
9 Mengingat dan R Nisbah Trigonometri (T3)
Memahami Trigonometric Ratio (F3) Pilih C
Recall and
Understanding
10 Menganalisis T Nisbah dan Graf Fungsi Trigonometri
Analysing (T5) Pilih C
Ratio and Graphs of Trigonometric
Functions (F5)
11 Mengingat dan R Indeks (T3)
Memahami Indices (F3) Pilih D
Recall and
Understanding
12 Mengaplikasi S Indeks (T3) Pilih B
Application Indices (F3)

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