SKOR A SPM (Maths) Penerbitan Pelangi Sdn Bhd

ANSWERS  

4KERTAS MODEL ∠RSQ = ∠TSU = 180° – 120°
2
Paper 1
= 30°

1. 868 570 = 869 000 (3 significant figures) y° = 120° – 30° – 30°
Answer : D = 60°

y – x = 60 – 23
= 37

2. 3.52 × 10–6 Answer : C
Answer : A
7. Let ∠LKN = y°
3. Let h be the height of the tank.
2x° + y° = 180°
  3.142 × 2 y° = 180° – 2x°
2.8 × 102
× (h – 1.35 × 102)
2
180° – 2x° + x° + 140° + 80° = 360°
= 2.48 × 106
x° = 40

h = 2.48 × 106 + 1.35 × 102 Answer : B
3.142 × (1.4 × 102)2

 = 1.75 × 102 8. ∠PQS = 65°
∠RPQ = 15o
Answer : B
x = 180° – 65° – 15°
4. 31225 = 3 × 53 + 1 × 52 + 2 × 51 + 2 × 50
= 100°
= 41210 y = ∠QPU = 90° – 15°

8 412 = 75°
8 51 4
8 63 x – y = 100° – 75°

06 = 25°

Answer : C

31225 = 6348 9. Scale factor = 4 = 2
Answer : B 2

5.   2 7 Answer : B
  10 111
10. tan w° = y-coordinate
\ 278 = 101112 x-coordinate

p = 1010112 – 101112 = –0.7660
  = 101002 0.6428

0202 = –1.1917

1 0 1 0 1 12 Answer : D
– 1 0 1 1 12
11. y N
1 0 1 0 02
8
Answer : C
6
6. Interior angle = (6 – 2) × 180° P
6 4 M

= 120° 2

Reflex angle, ∠QPU = 360° – 120° O x
2468
= 240°

∠WQP = 180° – 120° xP
180 (5, 5)
= 60°

x° = 360° – 37° – 240° – 60°

= 23 Answer : D

© Penerbitan Pelangi Sdn. Bhd. A – 34 SPM Mathematics

ANSWERS 

12. US = 162 + 122 = 20 cm 17. B

UR = 2 US A 40° xm
5 40° ym

= 2 (20) 18 m 8m U 65°
5 T 10 m C

= 8 cm

QU = 62 + 82 UW = 102 – 82 W
= 10 cm = 6 cm
tan 40° = x tan 65° = y
cos θ° = –   6 = –  3 6
10 5 8
x = 6.71 y = 12.87
Answer : B

13. m° = 45, n° = 225

n – 2m = 225 – 2(45)

= 135

Answer : C BU = 6.71 + 18 = 24.71

14. F Hence, CW = 24.71 – 12.87
E
B = 11.84 m
A
C Answer : A
∠EFA D
Answer : D 18. Latitude of Q = (90° – 60°)S

= 30°S

Longitude of Q = (75° – 25°)W

15. W = 50°W

Position of Q is (30°S, 50°W)

10 – 1.4 Answer : B
= 8.6 m
θ° Q 19. x(x + 3) – 4(x – 2)(x + 2)
V 12 m = x2 + 3x – 4(x2 – 4)
= x2 + 3x – 4x2 + 16
tan θ = 8.6 = –3x2 + 3x + 16
12
Answer : C
= 0.7167
9 – m2 m
θ = 35°38′ 2m2 +m

Answer : A 20. × 3

16. NN = (3 – m)(3 + m)1 × 3 m1 1
2m2 1 +m
3–m
L = 2m
M 42°
Answer : A
110°

21. a = 1 + a
a b
b + a2
K a = ab

∠MKL = 180° – 110° – 42° a2b = b + a2

= 28° a2b – b = a2

Thus, bearing M from K b(a2 – 1) = a2
a2
= 360° – 28° b = a2 – 1

= 332°

Answer : C Answer : C

© Penerbitan Pelangi Sdn. Bhd. A – 35 SPM Mathematics

ANSWERS  

22. 3 – x = x – 3 28. Let x be the number of pairs of shoes produced on
2 Friday.
6 – x = 2(x – 3)
Total pairs of shoes produced
= 2x – 6 = 6 × 50 + 4 × 50 + x
= 500 + x
3x = 12

x = 4 25 × (500 + x) = 200

Answer : B 100

1 500 + x = 800

23. (p4q –2)4 ÷ p3q –5 x = 300

1 Thus, the number of production of pairs of shoes on
Friday is 300.
= (p4)4(q –2)4 ÷ p3q –5
= pq –8 ÷ p3q –5 Answer : B
= p1 – 3q –8 – (–5)
= p–2q –3 29. 3 + 2 + 4 + 10 + x = 26

Answer : D

512 x = 7
82y
24. 8y = n(S′  T ) = 7 + 4

8y82y = 512 = 11

83y = 83 Answer : B

3y = 3

y = 1

Answer : A 30. Y X Y

X 

25. 3x – 1  2 Z Z

and 15 – 3x  6

x – 3  6 –3x  –9 X
Z
x  9 x  3 Y

x = 4, 5, 6, 7, 8, 9

Answer : D

26. –3  x  4 Answer : C
Answer : B
31. ξ

27. Let x be the number of skirts sold. P K
Hence, number of pants sold = x 35 – 18
32 – 18 18 = 17
= 14

Number of T-shirts sold = 210

140° × Total number of attires = 210 Number of members of the club who do not like travelling
360° or reading

Total number of attires = 540 = 80 – (14 + 18 + 17)
= 31
NNumumbebrerofofblsokuirstsessoslodld = 1
2 Answer : B

Number of blouses sold = 1 x xy
2 32. 4 – 2 = 1

12 x + x + 210 + x = 540 x – 2y = 4

5 x = 330 2y = x – 4
2
x = 132 y = 1 x – 2
2

Thus, number of blouses sold = 1 (132) Gradient = 1
2 2

= 66

Answer : C Answer : B

© Penerbitan Pelangi Sdn. Bhd. A – 36 SPM Mathematics

ANSWERS 

33. P(–2, 2), M(4, 4), J(–2, 0) 37. R ∝ S
 T
Gradient of JM = 4–0
4 – (–2) kS
R =  T
4
= 6   3 = 3
 4
2 k
3
= k = 2

Equation of the line passing through P (–2, 2):   4 = 2 8
 x
y = 2 x + c
3
 8x = 2
2 = 2 (–2) + c
3
x = 4
c = 10 x = 16
3

  3 y = 2 x + 10 Answer : B
3 3

3y = 2x + 10 38. y = –x3 – 4
Answer : C
Answer : C

x 2 1 2 39. (–2   1) 1 –3
34. 30 + 60 + x + 50 = 7 4 2

7x = 2(140 + x) = (–2 × 1 + 1 × 4  –2 × (–3) + 1 × 2)

= 280 + 2x = (2  8)

5x = 280

x = 56 Answer : B

Total number of students = 140 + 56

= 196 1 21 2 1 24 –23 1 32
3 –2 4 4m 8
Answer : A 40. =

35. Let x be the number of Indian workers. 1 2 1 2 4 × 3 + (–2) × 4 = 1 32
3 × 3 + (–2) × 4 4m 8

12 18 + x = 1 1 2 1 24 =132
+ 18 3 1 4m 8

30 + x = 54

x = 24 4 = 32 or 1 = 48m
4m
24 4 16m = 32 4m = 8
P(Indian worker) = 54 = 9
m = 2 m = 2
Answer : D
Answer : A

36. m2n = kp
p
   n = k m2

  n ∝ k p
m2

n varies directly as p and inversely as m2

Answer : C

© Penerbitan Pelangi Sdn. Bhd. A – 37 SPM Mathematics

ANSWERS  
Paper 2

No. Solution and Mark Scheme Marks Total
3
P = Knowledge, K = Method and N = Value
K1
1. y 1. y P2
P1 3
6 y = 3x + 3 6 y = 3x + 3

y = – 3x + 4 4 y = – 3 x + 4 4
4 4

2 2

–4 –2 O x –4 –2 O x
24 24

y=x –2 y=x –2

–4 –4

2. (a) H The straight line y = x is drawn correctly.
EN The region is shaded correctly.

2. (a) ∠GNF or ∠FNG

20 cm D G
A F

M C
B 9 cm

∠GNF or ∠FNG

(b) sin ∠GNF = 9 (b) sin ∠GNF = 9 K1
15 15 N1
  ∠GNF = sin–1 9
15 36.87° or 36°52ʹ

= 36.87° or 36°52ʹ

3. Volume of the solid 3. 1 × 22 × 72 × h K1 4
= volume of the cone + volume of the hemisphere 37 5

3 542 = 1 × 22 × 72 × h + 1 × 4 × 22 × 73 1 × 4 × 22 × 73 K1
3 37 23 7 23 7

154h = 3 542 – 2 156 3 542 = 1 × 22 × 72 × h + 1 × 4 × 22 × 73 K1
3 33 3 37 23 7 N1

= 462 9 cm
h = 9 cm

4. x + y = 6 …….…a 4. x + y = 6 K1
5x + 3y = 22 …..b 5x + 3y = 22 K1
2y = 8 K1
a × 5: 5x + 5y = 30 .......c y=4 N1
x=2 N1
c – b: 2y = 8

y = 4

Substitute y = 4 into a: x + 4 = 6

x = 2

© Penerbitan Pelangi Sdn. Bhd. A – 38 SPM Mathematics

No. Solution and Mark Scheme ANSWERS 
5. (a) Some
5. (a) Some Marks Total
P1 5

(b) If the length of every side KLMN is the same, (b) If the length of every side KLMN is the P1
then KLMN is a square. (False) same, then KLMN is a square. (False)

(c) Number of sides, n = Number of vertices, v (c) n = v K1
n = v, such that n = 5, 6, 7, … n = 5, 6, 7, … N1

(d) Premise 1: (d) Premise 1: If n(P  Q) = n(P) + n(Q), P1
If n(P  Q) = n(P) + n(Q), then intersection of then intersection of set P and set Q is

set P and set Q is an empty set. an empty set.

6. Volume of the solid = 300 6. (x + 2) × 10 × x or 1 × (x + 2) × 2 × 10 K1 4
2
Volume of the cuboid + volume of the right prism

= 300 (x + 2) × 10 × x + 1 × (x + 2) × 2 × 10 = 300 K1
2
(x + 2) × 10 × x + 1 × (x + 2) × 2 × 10 = 300
102x2 + 20x + 10x + 20 = 300
(x + 7)(x – 4) = 0 K1

10x2 + 30x – 280 = 0

x2 + 3x – 28 = 0 x=4 N1

(x + 7)(x – 4) = 0

x + 7 = 0 or x – 4 = 0
x = 4
x = –7

(Not acceptable)

7. (a) ad – bc = 0 7. (a) 1 (6) – 5m = 0 P1 6
2 N1
1 (6) – 5m = 0
2 m= 3
5m = 3 5

m = 3
5

1 2 1 1 2(b) p = 1 6 3 10 1 2 1 1 2(b) p= 1 6 3 10 K2
q –5 1 –8 q –5 1 –8 N1
1 (6) – (5)(–3) 2 1 (6) – (5)(–3) 2 N1
2 2

1 1   = 1
18
(6)(10) + (3)(–8) p=2

(–5)(10) + 1 (–8) q = –3
2

1 2 = 1 36 Note:
18 –54

1 2 = 2 1 2 1 2 1. Ifp=2 as final answer,
–3 q –3

p=2 award N1.
q = –3

© Penerbitan Pelangi Sdn. Bhd. A – 39 SPM Mathematics

ANSWERS  

8. (a) S(–6, 0) No. Solution and Mark Scheme Marks Total
8. (a) S(–6, 0) P1 5
K1
(b) Gradient of ST = 6–0 (b) 0 = 1 (–6) + c N1
6 – (–6) 2
K1
= 6 y = 1 x + 3 N1
12 2
P1 5
= 1 K1
2 N1
K1
Using (–6, 0), N1

y = mx + c K1 6
N1
0 = 1 (–6) + c
2

 c = 3

Equation of ST :

y = 1 x + 3
2

(c) Let the coordinates of point U is (x, 0) (c) 0 – (–4) = 1 or equivalent
x – (–6) 2
1
G radient of RU = 2 x-intercept = 2

0 – (– 4) = 1
x – (–6) 2

x 4 6 = 1
+ 2

x + 6 = 8

x = 2

x-intercept = 2

9. (a) S = {(G, G), (G, 5), (G, 7), (B, G), (B, 5), 9. (a) S = {(G, G), (G, 5), (G, 7), (B, G),
  (B, 7), (R, G), (R, 5), (R, 7), (3, G),   (B, 5), (B, 7), (R, G), (R, 5),
(3, 5), (3, 7) } (R, 7), (3, G), (3, 5), (3, 7)}

(b) (i) {(G, G)} (b) (i) {(G, G)}
112
P(both balls are labelled with letter G)

= 1
12

(ii) {(G, G), (G, 5), (G, 7), (B, 7), (R, 7), (3, 7)} (ii) {(G, G), (G, 5), (G, 7), (B, 7),
(R, 7), (3, 7)}
P(the first ball is labelled with a letter G or
21
the second ball is labelled with number 7)

= 6
12

= 1
2

10. (a) 36θ0° × 2 × 22 × 40 = 20 10. (a) θ ×2 × 22 × 40 = 20
7 360° 7
θ
360° = 7
88 28.64°

θ = 28.64°

© Penerbitan Pelangi Sdn. Bhd. A – 40 SPM Mathematics

ANSWERS 

No. Solution and Mark Scheme Marks Total
K1
(b) Area of the painted region (b) 7 × 22 × 402 or 7 × 22 × 102
88 7 88 7 K1
  = 4 7 × 22 × 402 – 7 × 22 × 102
88 7 88 7 K1
7 22 7 22 N1
= 402 – 102 88 × 7 × 402 – 88 × 7 × 102

= 1 500 cm2

  4 7 × 22 × 402 – 7 × 22 × 102
88 7 88 7

1 500

11. (a) 1.1 – 0.6 = 0.5 11. (a) 30 min N1 6
0.5 × 60 = 30 min

(b) Speed = 180 (b) 180 K1
1.5 1.5 N1

= 120 km h–1 120 km h–1

(c) (i) Distance from town X (c) (i) 99 km N1
= 180 – 81

= 99 km

(ii) 1.1 × 60 = 1 hr 6 min (ii) 10.36 – 1.06 K1
9.30 a.m. N1
Started time = 10.36 – 1.06

= 9.30 a.m.

12. (a) When x = –3, y = 12+ 5(–3) – 2(–3)2 = –21 12. (a) –21 N1 12
When x = 3, y = 12 + 5(3) – 2(3)2 = 9 9 N1

x –3 3 (b) Axes are drawn in correct directions P1
y –21 9 with uniform scale for –3  x  4.

(b) y All points are plotted correctly or K2
curve passes through all the points for
20 –3  x  4 and –21  y  15.

15 All smooth and continuous curve N1
without any straight line passes through
10 all 8 points using the given scale for
8 –3  x  4 and –21  y  15.

5

–2.6 –1.7 O 2.1 x
–3 –2 –1 1234

–5

–10

–15

–20

© Penerbitan Pelangi Sdn. Bhd. A – 41 SPM Mathematics

ANSWERS  No. Solution and Mark Scheme 
(c) (i) x = –1.7 ± 0.1
(c) From the graph, Marks Total
(i) x = –1.7 (ii) y = 8 ± 0.5 P1
(ii) y = 8
P1

(d) y = 12 + 5x – 2x2…..a (d) Equation of the straight line is K1
0 = 11 – x – 2x2……b y = 6x + 1
a – b : y = 1 + 6x
Equation of the straight line is y = 6x + 1 The straight line y = 6x + 1 is drawn K1
From the graph, x = –2.6, 2.1 correctly.

13. (a) (i) (–3, 3) ⎯R→ (3, 3) Values of x: –2.6, 2.1 N2
(ii) (–3, 3)  ⎯ T→ (1, 5)  ⎯ T→ (5, 7)
(iii) (–3, 3) ⎯R→ (3, 3)  ⎯ T→ (7, 5) 13. (a) (i) (3, 3) P1 12
P2
(ii) (5, 7) P2

(iii) (7, 5)


(b) (i) P is a reflection in the line CD. (b) (i) Reflection in the line CD. P2

(ii) Rotation of 90° anticlockwise about centre (ii) Rotation of 90o anticlockwise P3
(0, 6). about centre (0, 6).

(iii) Area of shaded region (iii) 22(Area of ABCD) – Area of ABCD K1
= 22(Area of ABCD) – Area of ABCD
118.8 N1
= 3(Area of ABCD)
= 3(39.6)
= 118.8 m2

14. (a) Class Upper 14. (a) 12
interval boundary
Midpoint Frequency Class Upper
interval boundary
Midpoint Frequency

35 – 39 37 39.5 2 35 – 39 37 39.5 2
40 – 44 42 44.5 9 40 – 44 42 44.5 9
45 – 49 47 49.5 10 45 – 49 47 49.5 10
50 – 54 52 54.5 6 50 – 54 52 54.5 6
55 – 59 57 59.5 7 55 – 59 57 59.5 7
60 – 64 62 64.5 4 60 – 64 62 64.5 4
65 – 69 67 69.5 2 65 – 69 67 69.5 2

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

© Penerbitan Pelangi Sdn. Bhd. A – 42 SPM Mathematics

ANSWERS 
Marks Total
No. Solution and Mark Scheme
K2
(b) Estimated mean 37(2) + 42(9) + 47(10) + 52(6) +
N1
37(2) + 42(9) + 47(10) + 52(6) + 57(7) + (b) 57(7) + 62(4) + 67(2)
40
= 62(4) + 67(2)
40

= 2 015 50.38
40

= 50.38

(c) (c) Axes are drawn in correct directions P1
with uniform scale for 34.5  x  69.5
Frequency and 0  y  10.
10
Vertical axis is labelled with frequency. K2
9 Horizontal axis is labeled with upper
8
7 boundary.
6
5 7 rectangles are drawn using upper N1
4 boundary. Correct histogram is drawn.
3
2
1

O 34.5 39.5 44.5 49.5 54.5 59.5 64.5 69.5
Marks

(d) 4 + 2 = 6 students (d) 6 N1
15. (a) 12
15. (a) K/M K/M

E/D 5 cm H/C E/D 5 cm H/C
J/L 3 cm J/L 3 cm
2 cm 2 cm
F/A G/B F/A G/B

Correct shape of rectangle EFGH and K1
semicircle JHK.

The lines are all solid. K1
EH = FG  JE  HG = EF

Correct measurement ±0.2 cm and all N1
angles at the vertices = 90° ± 1°.

© Penerbitan Pelangi Sdn. Bhd. A – 43 SPM Mathematics

ANSWERS  

No. Solution and Mark Scheme Marks Total

(b) (i) G/F H/E/J K (b) (i)
7 cm
R G/F H/E/J K
1.5 cm
2 cm U R 2 cm U
S W T W T
V 1.5 cm V
3.5 cm S 7 cm

3.5 cm

P/Q 4 cm B/A 2 cm C/D/L 4 cm M

P/Q 4 cm B/A 2 cm C/D/L 4 cm M

Correct shape of rectangles HKMC, K1
GHCB, RWVS and SVBP.

All lines are solid. Dotted lines WU and K1
VT.

KM  MC = BP  PS  BC = GW  K1
RS

Correct measurement ±0.2 cm and all N1
angles at the vertices = 90° ± 1°.

(ii) J F/E G/H (ii)
3.5 cm 7 cm J
2 cm S/V/T F/E G/H
R/W/U 3.5 cm 2 cm 3.5 cm 7 cm
R/W/U S/V/T
5 cm 3.5 cm
5 cm

Q/L 3 cm P/A/D 5 cm B/C

Q/L 3 cm P/A/D 5 cm B/C

Correct shape of rectangle FGBP, K1

polygons RQPS and JFSR.

All lines are solid lines. K1

BG  PB = FG = RQ  FS = SP  QP K1
= JF  JR

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

16. (a) (i) L(32°S, 120°E) 16 (a) (i) L(32°S, 120°E) P1 12
(ii) M(32°N, 120°E) (ii) M(32°N, 120°E) P1

(b) Shortest distance K to M (b) [180 – 2(32)] × 60 K2
= [180 – 2(32)] × 60 N1
= 6 960 nm 6 960

© Penerbitan Pelangi Sdn. Bhd. A – 44 SPM Mathematics

ANSWERS 

(c) Difference of longitude = 60° No. Solution and Mark Scheme Marks Total
(c) (120° – 60°) × 60 × cos 32° K2
Distance P to M
= (120° – 60°) × 60 × cos 32° 3 052.97 or 3 053 N1
= 60° × 60 × cos 32°
= 3 052.97 nm

(d) (i) Distance K to M (d) (i) 180° × 60 × cos 32° + 2(32) × 60 K1
= 180° × 60 × cos 32° 12 998.92 or 12 999
= 9 158.92 nm
N1
Distance M to L
= 2(32) × 60
= 3 840 nm

Total distance = 9 158.92 + 3 840

= 12 998.92 nm

(ii) Speed = distance travelled (ii) 12 998.92 K1
time taken 900 N1

900 = T1i2me99t8a.k9e2n 14.44


T ime taken = 12 998.92
900

= 14.44 hours

© Penerbitan Pelangi Sdn. Bhd. A – 45 SPM Mathematics



© Penerbitan Pelangi Sdn. Bhd. A – 46 SPM Mathematics

© Penerbitan Pelangi Sdn. Bhd. A – 47 SPM Mathematics



© Penerbitan Pelangi Sdn. Bhd. A – 48 SPM Mathematics