This document provides details of an academic excellence workshop for SPM Mathematics held on 20 September 2016 at SMK Mukah. It includes the following: 1. An analysis of SPM Mathematics question answers from 2005 to 2016, showing the distribution of scores among categories A, B, C and D. 2. Examples of past year SPM questions on topics like linear equations, linear inequalities, and solid geometry. The questions are explained step-by-step with working. 3. Practice questions similar to those found in SPM exams, with solutions shown. The workshop aimed to help students understand common SPM Mathematics questions and improve their exam skills through worked examples and practice.

1. BENGKEL
KECEMERLANGAN AKADEMIK
MATEMATIK (1449) SPM 2016
SMK MUKAH
20 SEPTEMBER 2016
0900 – 1030
Disediakan oleh: Panitia Matematik
1 Mathematics SPM

2. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 1
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
2
0
2
1
2
2
2
3
2
4
2
5
2
6
Then,
KNOWLEDGE = 11 + 14 + 15 + 23 + 12 + 5 + 4 + 7 + 5 = 96%
HARDWORK = 8 + 1 + 18 + 4 + 23 + 15 + 18 + 11 = 98%
Both are important, but fall just short of 100%
But,
ATTITUDE = 1 + 20 + 20 + 9 + 20 + 21 + 4 + 5 = 100%
2 Mathematics SPM

3. ANALISIS JAWAPAN SPM
Tahun A B C D
2005 8 11 11 10
2006 10 10 10 10
2007 6 15 12 11
2008 10 11 9 10
2009 9 10 11 10
2010 10 9 10 11

4. ANALISIS JAWAPAN SPM
Tahun A B C D
2011 11 10 10 9
2012 11 8 10 11
2013 9 10 11 10
2014 10 10 11 9
2015 9 9 11 11
2016 8 – 11

5. Mathematics SPM5
Question 1
KETAKSAMAAN LINEAR
[ 3 marks]

6. Mathematics SPM6
Symbol Definition Line
> Lebih besar daripada Garis putus-putus
------------------
< Kurang daripada
 Lebih besar dan sama
dengan
Garis solid
 Kurang daripada dan
sama dengan

7. Mathematics SPM7
Example 5 : (SPM Nov 2005)
On the graph in the answer space, shade the
region which satisfies the three inequalities
y  2x + 10 , x < 5 and y  10 .
Answer :
x < 5
√ K1
√ K2

8. Mathematics SPM8
Example 6 :
State the three inequalities that satisfied the shaded
region in the graph below.
Answer :
(i) 3y  x + 12 ; (ii) y  2x + 4 ; (iii) x < 2
√ N1 √ N1 √ N1

9. Mathematics SPM9
LATIHAN: (SPM Jun 2016)
Pada graf tersebut, lorek rantau yang memuaskan
ketiga-tiga ketaksamaan
y  x + 8 , y  x and y  6 . (3 markah)
Answer :

10. Mathematics SPM10
JAWAPAN: (SPM Nov 2016)

11. Mathematics SPM11
Question 2
Persamaan Linear
[ 4 marks]

12. Example 7 : (SPM Nov 2015)
Calculate the value of x and of y that satisfy the following
simultaneous linear equations:
82
445


yx
yx
12 Mathematics SPM

13.  
6
366
441040
44285
)1(int)3(
)3(28:)2(
)2(82
)1(445







y
y
yy
yy
osubstitute
yxfrom
yx
yx



Method 1 : Substitution method
√ K1
√ N1
 
4
628
:)3(int6



x
oysubstitute
√ N1√ K1
13 Mathematics SPM

14. Method 2 : Elimination method
4
123:)1()3(
)1(445
)3(1642
2)2(:)2(
)2(82
)1(445







x
x
yx
yx
from
yx
yx




√ K1
√ N1
6
122
824
)2(int4




y
y
y
oxsubstitute
√ K1 √ N1
14 Mathematics SPM

15.      
     
     
6,4,
6
4
36
24
6
1
8541
8442
6
1
8
4
51
42
1425
1
8
4
21
45








































































yxso
y
x
y
x
Method 3 : Kaedah Matriks
√ K2
√ N1 √ N1
15 Mathematics SPM

16. Check answer : Calculator Scientific
16 Mathematics SPM
MODE  5: EQN  1 :
anX+bnY=cn

17. Example 8 :
Diberi jumlah 50 tiket telah dijual dengan harga RM 2080
dalam satu konsert. Jika harga satu tiket ialah RM35
ataupun RM50, cari bilangan tiket bagi harga RM35 dan
bilangan tiket bagi harga RM50 yang telah dijual.
Penyelesaian :
Katakan x = bilangan tiket bagi harga RM35
y = bilangan tiket bagi harga RM50
20805035
50


yx
yx
17 Mathematics SPM

18.      
22,28,
22
28
330
420
15
1
2080
50
135
150
135501
1
2080
50
5035
11






















































yxmaka
y
x
y
x
Method 3 : Kaedah Matriks
√ K2
√ N1 √ N1
18 Mathematics SPM

19. Mathematics SPM19
Question 4
Lines and Planes in 3-Dimensions
[ 3 marks]

20. WON Technique
Mathematics SPM20
Example 12 :
Given that V and W are midpoints of UT and PS.
Name the angle between line VQ and the base PQRS.

21. WON Technique
Mathematics SPM21
Step 1 : Arrange the line and the plane in two rows.
Then draw 3 boxes.
VQ
PQRSW
Step 2 : Find out the same alphabet. Slash the same
alphabet.
VQ
PQRSW

22. WON Technique
Mathematics SPM22
Step 3 : Write V in the first box.
VQ
PQRSW
Step 4 : Look at V in the diagram. Choose the slash
alphabet
nearest to V. Write in the centre box.
VQ
PQRSW
V
V Q

23. WON Technique
Mathematics SPM23
Step 5 : Look at V in the diagram.
Choose the non-slash alphabet nearest to V.
Write in the last box.
VQ
PQRSW
 VQW
V Q W

24. Mathematics SPM24
LATIHAN : (SPM Jun 2016)
Diagram 4 shows a right pyramid with height 7cm.
M is the midpoint of AB. Given AB = 4cm and BC
= 6cm.
(a) On the diagram, draw the orthogonal projection
of the line ME and the base ABCD.
(b) Calculate the angle between the line ME and
the
plane ABCD. (3

25. Mathematics SPM25
JAWAPAN : (SPM Jun 2016)

26. Mathematics SPM26
Question 7
Gradient and Area under a Graph
[ 5 ~ 6 marks]

27. (I) Distance-time Graph
Mathematics SPM27
(a) Find distance or period of time, when object
stationary
(b) Gradient = Speed = Rate of change of distance
(c) Average speed =

28. (II) Speed-time Graph
Mathematics SPM28
(a) Find speed or period of time, when object at ‘uniform speed’
(b) Gradient = Acceleration = Rate of change of speed
(c) Average speed =
* (d) Total distance travelled = Area under the graph

29. Mathematics SPM29
LATIHAN : (SPM Jun 2016)
Diagram 9 shows a speed time-graph for the movement of
two particles, A and B, for a period of 30 seconds. The graph
PR represents the movement of particle A and the graph
PQR represents the movement of particle B. Both particles
start at the same point and move along the same route.
Diagram 9

30. Mathematics SPM30
Example 15 : (SPM Jun 2016)
(a) State the uniform speed, in ms-1, of particle B.
(b) Calculate the rate of change of speed, in ms-2, of particle
B.
12 ms-
1
2
2
1
24
12
306
012







ms
Particle A
Particle B
(6,
12)
(30,
0)
√ N1
√ K1
√ N1

31. Mathematics SPM31
Example 15 : (SPM Jun 2016)
(c) Find the difference between the distance, in m, travelled
by
particle A and particle B for the period of 30 seconds.
Distance travelled by particle A = Area under the graph
PR
=
12
30
m180
1230
2
1


√ K1
Lakarkan bentuk geometri

32. Mathematics SPM32
Example 15 : (SPM Jun 2016)
(c) Find the difference between the distance, in m, travelled
by
particle A and particle B for the period of 30 seconds.
Distance travelled by particle B = Area under the graph
PQR
=
12
30
m216
12)306(
2
1


6
The difference = 216 – 180 = 36 m
√ K1
√ N1
Lakarkan bentuk geometri

33. Mathematics SPM33
Question 8
Solid Geometry ( Volume )
[ 4 marks]

34. Mathematics SPM34
Example 16 : (SPM Nov 2007)
Diagram 6 shows a solid, formed by joining a cylinder to a
right prism. Trapezium AFGB is the uniform cross-section of
the prism.
AB = BC = 9 cm. The height of the cylinder is 6 cm and its
diameter is 7 cm.
Diagram 6
Calculate the volume, in cm3, of the solid.
[Use ]7
22

9 cm
9 cm
6 cm
7 cm

35. Mathematics SPM35
Example 16 : (SPM Nov 2007)
(i) Volume of
cylinder
231
6
2
7
7
22
2
2








 hr
(ii) Volume of prism
 
756
98129
2
1








 Ah
(iii) Volume of the solid = 231 +
756
= 987 cm3
√ K1 √ K1
√ N1
√ K1

36. Mathematics SPM36
LATIHAN : (SPM Jun 2016)
A container contains 1386 cm3 of water. Then of the water
is poured into the right prism container as shown in Diagram
3. Trapezium ABCD is the uniform cross section of the prism.
Find the depth, in cm, of the water. (3 marks)
[Use ]
7
22

4
3

37. Mathematics SPM37
JAWAPAN : (SPM Jun 2016)

38. Mathematics SPM38
Question 10
Matrices
[ 6 marks]

39. Mathematics SPM39
Inverse matrix
Let , then
.







dc
ba
A
No Inverse matrix
If a matrix has no inverse, then ad – bc = 0
.
ad – bc =
0










ac
bd
bcad
A
11

40. Mathematics SPM40
LATIHAN: (SPM Jun 2016)
(a) It is given that is the inverse matrix of .
Find the value of p and of q.
;






 q
p
5
3






35
47
p =  4 q = 7
√ N1 √ N1

41. Mathematics SPM41
Example 18 : (SPM Jun 2016)
(b) Write the following simultaneous linear equations as a
matrix
form :
7x + 4y = 5
5x + 3y = 3
Hence, using matrix method, calculate the value of x and
of y.
Answer :


































































4
3
3755
3453
3
5
75
43
5437
1
3
5
35
47
y
x
y
x
y
x
y
x
 x = 3 , y =  4
√ K1
√ K1
√ N1√ N1

42. Mathematics SPM42
Question 11
Probability II
[ 5 ~ 6 marks]

43. Mathematics SPM43
LATIHAN: (SPM Jun 2016)
Diagram 11 shows two boxes, P and Q. Box P contains five
tokens labelled with letter and box Q contains three tokens
labelled with number. Two tokens are picked at random. The
first token is picked from box P and the second token is
picked from box Q.
Diagram 11

44. Mathematics SPM44
JAWAPAN: (SPM Jun 2016)
(a) List all the elements in the sample space.
6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
n(S) = 15 √ P2

45. Mathematics SPM45
JAWAPAN: (SPM Jun 2016)
(b) By listing down all the possible outcomes of the event, find
the
probability that
(i) a token labelled with letter A and a token labelled with
an
even number are picked,
X = { (A, 6), (A, 8) }
P(X) =
15
2
√ K1
√ N1
6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
√ √

46. Mathematics SPM46
JAWAPAN: (SPM Jun 2016)
(b) By listing down all the possible outcomes of the event, find
the
probability that
(ii) a token labelled with a vowel or a token labelled with
number 9 is picked.6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
X = { (K, 9), (E, 6), (E, 8), (E, 9), (L, 9), (A, 6), (A, 8), (A,
9), (H, 9)}
P(X) =
15
9
√ K1
√ N1
√
√
√
√
√
√
√
√
√