1. JABATAN PELAJARAN PERAK
ANSWER TO SCORE
MATHEMATICS SPM
(SECONDARY SCHOOL)
TOPICAL EXERCISES

2. 1
PART A
SETS
1. The Venn diagram in the answer space shows sets X, Y and Z such that the universal set,
.X Y Z   
On the diagrams in the answer space, shade
(a) the set Y Z 
(b) the set ( ) .X Y Z 
[3 marks]
Answer:
(a)
(b)
2. The Venn diagram in the answer space shows sets P, Q and R with  = PQR.
On the diagram provided in answer space, shade
a) P  Q'
b) Q  ( P  R )
[3 marks]
Answer :
a) P Q
R

3. 2
b)
P Q
R
3. On the diagrams provided in the answer space, shade the region which
represent the following operation of set.
(a) Set A  B'
(b) (b) A  (B  C)'
[3 marks]
Answer:
( a )
(b)
4 The Venn Diagram in the answer space shows sets A, B and C.
On the diagram provided in the answer space, shade
a the set A  B'
b the set A  B  C '
Answer:
(a) (b)
CBA
C
B
A
A
B C
A
B
C

4. 3
5  The Venn diagram in the answer space shows sets P,Q and R such that the universal set,
 = P  Q  R.
 On the diagram in the answer space, shade
 a  the set P/
 Q
 b  the set ( P  Q /
)  R.
  Answer:
  (a)

 (b)

P
Q
R
P
Q
R

5. 4
LINEAR INEQUALITIES
1. On the graph in the answer space, shade the region which satisfies the three
inequalities 3 12,y x  2 4y x   and 2x  .
[3 marks]
Answer :
2. On the graph provided, shade the region which satisfies the three inequalities
y  2x + 8, y  x and y < 8.
[3 marks]
Answer :
y
=
x
y=2x+8
x
y
O
2 4y x  
3 12y x 
O x
y

6. 5
3. On the graph in the answer space, shade the region which satisfies the three inequalities
y  – 2x + 10, x < 5 and y  10.
[3 marks]
Answer :
x
y
O
y = 10
y=-2x+10
5
4. On the graph provided, shade the region which satisfies the three inequalities
y ≥ -2x + 5, y ≥ x and y < 5.
[3 marks]
Answer :
y = -2x + 5
y = x
y
x
O

7. 6
5. On the graph provided, shade the region which satisfies the three inequalities
y ≥ x + 3, x > -3 and 3x + y ≤ 2.
[3 marks]
Answer:
y = x + 3
3x + y = 2
y
x
- 3 O
I

8. 7
SIMULTANEOUS LINEAR EQUATIONS
1. Calculate the values of m and n that satisfy the following simultaneous
linear equations.
54
1332


nm
nm [ 4 marks]
2. Calculate the values of p and q that satisfy the following simultaneous
linear equations.
243
132
2
1


qp
qp [ 4 marks]
3. Calculate the values of r and s that satisfy the following simultaneous
linear equations.
7
2
3
62


sr
sr
[ 4 marks]

9. 8
4. Calculate the values of g and h that satisfy the following simultaneous
linear equations.
1834
12


hg
hg
[ 4 marks]
5. Calculate the values of x and y that satisfy the following simultaneous
linear equations.
164
3
2
3


yx
yx [ 4 marks]
6. Calculate the values of a and b that satisfy the following simultaneous
linear equations.
93
5
3
1
2


ba
ba [ 4 marks]

10. 9
7. Calculate the values of d and e that satisfy the following simultaneous
linear equations.
732
24


ed
ed
[ 4 marks]
8. Calculate the values of h and k that satisfy the following simultaneous
linear equations.
4
2
1
434


kh
kh
[ 4 marks]
9. Calculate the values of m and n that satisfy the simultaneous linear
equations.
52
72


nm
nm
[ 4 marks]

11. 10
10. Calculate the values of u and v that satisfy the simultaneous linear equations
46
734


vu
vu
[ 4 marks]
.

12. 11
QUADRATIC EXPRESSIONS AND EQUATIONS
1. Solve the equation 2
1
52 2



x
xx
[ 4 marks]
2. Solve the quadratic equation y
y
3
3
52 2


[ 4 marks]
3. Solve the quadratic equation 6
2
)1(3


p
pp
[ 4 marks]

13. 12
4. Using factorization, solve the following quadratic equation.
qq 17154 2
 [ 4 marks]
5. Solve the quadratic equation
r
r
r
2
36
1

 [ 4 marks]
6. Solve the quadratic equation
2
562


w
w
[ 4 marks]

14. 13
7. Solve the quadratic equation 7)2)(12(  aa [ 4 marks]
8. Solve the quadratic equation 1
2
32

p
p [ 4 marks]
9. Solve the following quadratic equation
)3(5)3)(1(  eee [ 4 marks]

15. 14
10. Solve the equation 52)4( 2
 yy [ 4 marks]

16. 15
SOLID GEOMETRY
1.
DIAGRAM 1
Diagram 1 shows a composite solid formed by the combination of a cuboid and a half-
cylinder. Using π =
7
22
calculate the volume, in cm3
, of the solid. [4 marks]
21 cm
8 cm
P Q
R
S
T
U
V
W
7 cm

17. 16
2.
DIAGRAM 2
Diagram 2 shows a solid cylinder of height 15 cm and diameter 14 cm. A cone with height 6 cm is
taken out of the solid. Calculate the volume, in cm3
, of the remaining solid. (Use π =
7
22
).
` [4 marks]
15 cm

18. 17
3.
DIAGRAM 3
Diagram 3 shows a solid cuboid ABCDEFGH. A right pyramid with its base EFGH is
taken out of the solid. The volume of the remaining solid is 480 cm3
. Calculate the height
of the solid, h.
(Use π =
7
22
). [4 marks]

19. 18
4. The diagram 4 shows a composite solid, formed by the combination of a cylinder to a right
prism. Trapezium ABEF is the uniform cross-section of the prism. AB = BC= 10 cm.
The diameter of the cylinder is 7 cm and the volume of the composite solid is 1075.5 cm3
.
Using
7
22
 , calculate
(a) the volume, in cm3
, of the right prism
(b) the length h in cm, of the cylinder.
[5 marks]
DIAGRAM 4

20. 19
5. The diagram 5 below shows a combined solid of a right prism and a right pyramid which are
joined at the plane PQRS. T is vertically above the base PQRS. Trapezium PQKJ is the uniform
cross section of the prism.
The height of the pyramid is 6 cm and PQ = 18 cm.
(a) Calculate the volume, in cm3
, of the right pyramid.
(b) It is given that the volume of the combined solid is 448 cm3
. Calculate the length, in cm, of
PJ.
[5 marks]
DIAGRAM 5

21. 20
MATHEMATICAL REASONING
1. a) State whether the following statement is true or false
b) Write down two implications based on the following sentence
c) State the converse of the following statement and hence determine whether its
converse is true or false
if 2x > 8, then x > 4
Answer :
(a)
(b) Implication 1 :
Implication 2 :
(c)
2. (a) Complete each statement in the answer space with the quantifier “all” or
“some” so that it will become a true statement.
(b) State the converse of the following statement and hence, determine
whether its converse is true or false.
(c) Write down Premise 2 to complete the following argument :
Premise 1 : If m is less than zero, then m is a negative number.
Premise 2 : _________________________________________________
Conclusion : - 4 is a negative number. [5 marks]
Answer :
(a) (i) of the multiples of 5 are even numbers.
(ii) hexagons have six sides.
(b)
(c) Premise 2 :
If p > 7, then p > 4
9 > 8 or 4² = 8
x³ = 64 if and only if x = 4

22. 21
3. (a) State whether the sentence below a statement or non-statement ?
(b) Statement 1 : a0
= 1 for any values of a except a = 0
Statement 2 : All prime numbers are odd numbers
Combine the above statement to form a new statement which is
(i) true
(ii) false
(c) Make a general conclusion by induction for the following number pattern:
5 = 3(2 ) – 1
10 = 3(22
) – 2
21 = 3(23
) – 3
60 = 3(24
) – 4
……………… [5 marks]
Answer :
(a)
(b) Conclusion :
(c) Conclusion :
4. (a) State whether each of the following statement is true or false.
(i) 23
= 6 or = 3.5
(ii) ( -4 ) x ( -5 ) = 20 and -4 > -2
(b) Complete the premise in the following argument:
Premise 1 : If the determinant of a matrix = 0, then the matrix does not
have an inverse.
Premise 2 : _______________________________________________
Conclusion :Matrix A does not have an inverse.
(c) Write down two implications based on the following sentence.
A  B if and only if A B = A'
[5 marks]
Answer :
(a) (i)
(ii)
p + q = 2
2
7

23. 22
(b) Premise 2 :
(c) Implication 1 :
Implication 2 :
5. (a) State whether the following statement is true or false.
(b) Write down two implications based on the following statement:
(c) State the converse of the following statement and hence determine whether
its converse is true or false
if x > 9, then x > 5
[5 marks ]
Answer :
(a)
(b) Implication 1 :
Implication 2 :
(c)
6. (a) State whether each of the following statement is true or false.
(i) 42
= 8 or = -2
(ii) a  { a, b, c } and -3 > -7
(b) Write down premise 1 to complete the following argument.
Premise 1 :_________________________________________________
Premise 2 : 6 × p  42
Conclusion : p  7
(c) Form a general conclusion by induction for the number sequence
11, 23, 43, 71, … which follow the pattern
11 = 4(1)2
+ 7
23 = 4(2)2
+ 7
43 = 4(3)2
+ 7
71 = 4(4)2
+ 7
……………… [5 marks ]
-2 ( 3 ) = 6 or -4 > -5
p3
= 8 if and only if p = 2
3
8

24. 23
Answer :
(a) (i)
(ii)
(b) Premise 1 :
(c) Conclusion:
7. (a) Determine whether each of the following sentence is a statement.
(i) 9 is a prime number.
(ii) 2x(x – 2) = 0
(b) Fill in the suitable quantifier to make the following statement become true.
(c) Write down two implications based on the following sentence
3m > 15 if and only if m > 5 [5 marks]
Answer :
(a) (i)
(ii)
(b)
(c)
8. (a) State whether the following statement is true or false.
(i) 8 + 2 = 10 and 2 < -3
(ii) All perfect square numbers are even numbers
.
(b) Complete the following argument.
Premise 1 : If 3t = 0, then t = 0.
Premise 2 : ________________________________________________
Conclusion :3t  0
(c) State the converse of the following statement and hence, determine
whether its converse is true or false.
[5 marks]
…………. prime number are odd numbers
If x3
= 27 then x = 3

25. 24
Answer :
(a) (i)
(ii)
(b) Premise 2 :
(c)
9. (a) Is the sentence below a statement or non-statement?
(b) Write down two implications based on the following sentence:
(c) Make a general conclusion by induction for the following number pattern :
2 = (0)2
+ 2
3 = (1)2
+ 2
6 = (2)2
+ 2
11 = (3)2
+ 2
…. = ……….. [5 marks]
Answer :
(a)
(b) Implication 1 :
Implication 2 :
(c)
10. (a) State whether each of the following statement is true or false
(i) (-2)2
= 4 or ( -3 )3
= -9
(ii)
1 1
2 2
1
16 8 and 25
5

 
(b) Complete the following argument.
Premise 1 :_________________________________________________
Premise 2 : xn
+ x is not a quadratic expressions.
Conclusion : n  2.
(c) Write down two implications based on the following sentence.
[5 marks]
3 and 4 are factors of 8
m is an even number If and only if m can be divided by 2.
1 – q > 2 if and only if q < -1

26. 25
Answer :
(a) (i)
(ii)
(b) Premise 1 :
(b) (c) Implication 1 :
Implication 2 :

27. 26
THE STRAIGHT LINE
1. In Diagram 1, O is the origin, point R lies on the x-axis and point P lies on the y-axis. Straight
line PU is parallel to x-axis and the straight line PR is parallel to straight line ST. The equation
of straight line PR is x + 2y = 14.
R xO
T(2,-5)
S
U
x + 2y = 14
y
••
•
•
•
(a) State the equation of the straight line PU. [2 marks]
(b) Find the equation of the straight line ST and hence,
state its x-intercept. [3 marks]
Answer:
P
DIAGRAM 1

28. 27
2. The diagram below shows that the straight line EF and GH are parallel.
DIAGRAM 2
Find
(a) the equation of EF. [3 marks]
(b) the y - intercept and x - intercept of EF [2 marks]
Answer:

29. 28
K O
N
y
x
P
M
L (4, 7)
DIAGRAM 3
3. In the following diagram, O is the origin, point K and point P lies on the x-axis and point N lies
on the y-axis. Straight line KL is parallel to straight line NP and straight line MN is parallel to the
x-axis. The equation of straight line NP is x – 2y – 18 = 0.
(a) State the equation of the straight line MN.
(b) Find the equation of the straight KL and hence, state the coordinate of the point K.
[5 marks]
Answer:

30. 29
4. In the diagram 4, the gradient of the straight line KLM is
2
1
 . Find
(a) the value of p.
(b) the x-intercept of the straight line MN.
Answer:

K(0, p)
0
y
x
N
M
L(2, 4)
DIAGRAM 4

31. 30
5. The diagram 5 below shows STUV is a trapezium
i) Given that gradient of TU is -3, find
a) the coordinates of point T. [2 marks]
b) the equation of straight line TU. [1 marks]
ii) The equation of straight line TV [2 marks]
Answer:
DIAGRAM 5

32. 31
PERIMETER AND AREA
1.
18 cm
Diagram 1
15 15
E
A
B
O F
C D
In Diagram 1, OAB, OCD and OFE are three sectors in a circle centre O. AOF, BCO and EDO are
straight lines. C and D are midpoints of OB and OE respectively.
Using
22
7
  , calculate
a) the perimeter , in cm, of the whole diagram,
b) the area, in cm2
, of the shaded region. [6 marks]

33. 32
2.
Diagram 2
S
O
R
P Q120
Diagram 2 shows two sector ORS and OPQ with the same centre O. P and Q are midpoints of OR
and OS respectively and OS = 28 cm. Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]

34. 33
3.
In Diagram 3, OFIE is a quadrant of a circle with centre O and GH is an arc of another circle with
centre O. OFG and OIH are straight lines.
OF = FG = 14 cm, and 35GOH  
.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Diagram 3
35
H
E
G
F
I
O

35. 34
4.
14 cm
Diagram 4
O
P
Q
R
T
U
S
Diagram 4 shows two semicircles RSO and RTU with centre Q and O respectively. OPQ is a
quadrant of a circle with centre O. RQ = QO. ROU is a straight line.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]

36. 35
5.
H
G
F
E
I
Diagram 5
J
K
In diagram 5, two arcs of two equal circles, GFE and GIJ, with centres H and K respectively. It is
given that GH = 14 cm.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]

37. 36
6.
.
Diagram 6
O S
Q
RP
150
Diagram 6 shows two sectors OPQ and OQR with centre O. OS = SR = 7 cm. POSR is a straight
line.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]

38. 37
MATRICES
1a) The inverse matrix of is
4
5 3
p
q
 
 
 
. Find the value of p and of q.
b) Using matrices, calculate the values of x and y that satisfy the following simultaneous
equations :
3x – 2y = 6 [6 marks]
5x – 4y = 4
Answer :
1a) 1b)
2.a) Given that F = and the inverse matrix of F is find the value
of m and of n.
b) Hence, using matrices, calculate the value of p and of q that satisfies the following
equation
2
5
p
F
q
   
   
   
[6 marks]
Answer :
2a) 2b)








45
23






n
m
2
3
,
2
34
14
1








m

39. 38
3.a) Given that find matrix A.
b) Hence, using the matrix method, find the value of d and e which satisfy the
simultaneous equations below.
-d + 2e = 11
-3d + 5e = 15 [6 marks]
Answer :
3a) 3b)
4. Given matrix M = and matrix MN =
a) Find the matrix N.
b) Hence, calculate by using the matrix method, the values of r and s that satisfy the
following simultaneous equations :
4r + 5s = 1
6r + 8s = 1 [6 marks]
Answer :
4a) 4b)
,
10
01
53
21














A






86
54






10
01

40. 39
5. Given the matrix K is ,
a) Find the matrix L so that KL =
b) Hence, calculate the values of h and k, which satisfy the matrix equation:
[6 marks]
Answer :
5a) 5b)
6. Given matrix M =
1 2
2 6
 
 
 
, find
a) the inverse matrix of M.
b) hence, using matrix, find the values of x and y that satisfy the following
simultaneous equations :
x  2y = 3
–2x + 6y = 4 [6 marks]
Answer :
6a) 6b)






















11
7
58
34
k
h






10
01








58
34

41. 40
7. It is given that matrix A =
1
3 4
a 
 
 
does not have an inverse matrix.
a) Find the value of a.
b) If a = 2, find the inverse matrix of A and hence, using matrices, find the values of
x and y that satisfy the following simultaneous linear equations.
x  2y = 6
3x + 4y = 2 [6 marks]
Answer :
7a) 7b)
8a) Find matrix G such that
3 1 3 1
4 2 4 2
G
   
   
   
b) Using matrices, calculate the values of p and q that satisfy the following matrix
equation.
2p + 4q = 4
p + 3q = 8 [5 marks]
Answer :
8 a) 8b)

42. 41
9 a) Find the inverse of matrix
4 5
1 2
 
 
 
b) Hence, using matrices, calculate the values of m and n that satisfy the following
simultaneous equations :
– 4m + 5n = 1
–m + 2n = 2 [6 marks]
Answer :
9a) 9b)

43. 42
10a) Given matrix S =
6 3
4 m
 
 
 
, find the value of m if matrix S has no inverse.
b) Given the matrix equations
and
i) Find the value of h and k
ii) Hence, find the value of x and y. [6 marks]
Answer :
10a) 10bi)
10bii)
11. It is given matrix U =
1 2
3 2
  
 
 
and matrix V =
2 2
3
v
u
 
 
 
such that UV =
1 0
0 1
 
 
 
a) find the values of u and v.
b) using the matrix method, calculate the values of x and y that satisfy the following
simultaneous equations
x 2 y = 3
3x + 2y = 5 [6 marks]
Answer:
11a)
11b)




















1
4
85
67
y
x 8 41
5 7 1
x k
y h
    
    
    

44. 43
GRADIENT AND AREA UNDER GRAPH
1. Diagram 1 shows a distance-time graph for the journey of a car and lorry from town Q.
Diagram 1
Graph ABCD represents the journey of the car and graph ACE represents the journey of the
lorry. Both vehicles depart from town Q at the same time along the same road.
(a) State the length of time, in hours, during which the car is stationary.
(b) Calculate the average speed in km/h for the total distance of the car.
(c) At the certain time during the journey, both vehicles meet at the same location.
(i) Find the distance, in km, between that location and town Q.
(ii) State the time taken by the lorry to reach that location from town Q.
[ 5 marks ]
Answer:
(a)
(b)
(c) (i)
(ii)
70
142
195
Time (hour)
Distance (km)
0 3.21.41.0
A
E
D
B C

45. 44
2. The diagram shows a speed-time graph of the movement of a particle for a period of 35 s.
(a) State the uniform speed, in ms-1
, of the particle.
(b) Given the distance travelled by the particle at a uniform speed is 250m,
calculate
(i) the value of t,
(ii) the average of speed, in ms-1
, of the particle for the period of 35 seconds.
[ 6 marks ]
Answer:
(a)
(b) (i)
(ii)
Time (s)
Speed (m s-1
)
20
12
10 t 35

46. 45
3. Diagram 2 shows a distance-time graph for the journey of a bus and van.
Diagram 2
Graph AFB represents the journey of the van from town X to town Y. The graph CDEFG
represents the journey of the bus from town Y to town X. The van leaves town X and the
bus leaves town Y at 9.45 p.m. and they travel along the same road.
(a) (i) At what time do the vehicles meet ?
(ii) Find the distance, in km, from town Y when the vehicles meet.
(b) State the length of time, in minutes, during which the bus is stationary.
(c) Calculate the average speed, in km h-1
, of the bus for the whole journey.
[ 5 marks ]
Answer:
(a) (i)
(ii)
(b)
(c)
Time
(minutes)0
10 50 70 150
B
Distance (km)
65
100
A
C
G
DTown Y E
F
Town X

47. 46
4. Diagram 3 shows a speed-time graph for a particle, of 30 seconds.
Speed (m s-1
)
(a) The distance travelled during the last seconds is 106 m, find the value of v.
(b) Calculate the acceleration, in m s-2
, of the particle in the last 5 seconds.
(c) The total distance travelled before 10 seconds is 230 m. Calculate the average
speed for the whole journey.
[ 6 marks ]
Answer:
(a) (i)
(b)
(c)
10 25 30 Time (s)
21
0
Diagram 3
v
25

48. 47
5. The diagram 4 shows the speed-time graphs of two particles M and N over a period of 100
seconds. ABCD is the speed-time graph of particle M and EFG is the speed-time graph of
particle N.
(a) State the length of time in seconds, during which the particle N is constant speed.
(b) Calculate the rate of change of speed, in ms-2
, of the particle M in the first 20s,
(c) Given that the distance travelled by both particles over the 100 s period is the
same and the distance traveled by particles N is 8500 m. Calculate the value of v.
[ 6 marks ]
Answer:
(a)
(b) (i)
(ii)
Speed (ms-1
)
20 40 60 10080
Time (s)
0
v
80
145
G
D
F
E
B
C
A
Diagram 4

49. 48
PROBABILITY
1.
Students
Probability of choosing:
Biology Physics Chemistry
Ahmad
4
3
8
1
Ling
5
1
5
2
The table 1 shows the probability of how Ahmad and Ling might choose their subjects in
Form 6. The table is incomplete.
Calculate the probability that
(a) Ahmad will choose Chemistry and Ling will choose Biology,
(b) Ahmad and Ling will choose the same subjects. [5 marks]
Answer:
(a)
(b)
2. In bag A, there are 4 blue balls and 3 yellow balls while in bag B, there are 5 blue balls and
6 yellow balls. A ball is drawn at random from bag A. After its colour is recorded, it is put
into beg B. Then a ball is drawn at random from beg B. Find the probability that
(a) the colour of the two balls are blue,
(b) two different colour balls are drawn. [5 marks]
Answer:
(a)
(b)
Table 1

50. 49
3. A company van carries a group of workers consists of 6 males and 8 females. They are
dropped off at random at various houses along the route as shown in the diagram below to
serve customers.
(a) If two workers are dropped off at house A, calculate the probability that both are females,
(b) Two male workers are dropped off at house A. If another two workers are then dropped
off at house B, calculate the probability that at least one of them is female.
[5 marks]
Answer:
(a) (b)
4 In a lucky draw, there are three categories of gifts consisting of 4 hotel vouchers, 6 shopping
vouchers and 10 restaurant vouchers. All vouchers are placed inside similar envelopes and
put inside a box. The honored guests of the lucky draw are requested to draw at random two
envelopes from the box.
Calculate the probability that the first honored guest draw
(a) the first envelope with a hotel voucher and the second with a shopping voucher,
(b) two envelopes with vouchers of the same categories. [5 marks]
Answer:
(a)
(b)
Other houses●
Company House A
House B
●
●
●

51. 50
5. The table below shows the number of participants in a quiz contest according to school and
gender.
School
Number of Participants
Male Female
SMK Ria 4 4
SMK ACS 5 2
SMK Puteri 3 6
(a) If two participants are chosen at random from SMK ACS, calculate the probability that
both of them are males.
(b) If two participants are chosen from the male participants, calculate the probability that
both of them are from the same school. [5 marks]
Answer :
(a)
(b)
6. Bottle A contains three red marbles and two yellow marbles. Bottle B contains four red marbles
and three yellow marbles. A marble is chosen at random from each bottle.
Find the probability that
(a) both marbles are in yellow,
(b) both marbles are in different color. [5 marks]
Answer:
a)
b)
Table 2

52. 51
LINES AND PLANES IN 3-DIMENSIONS
1. Diagram 1 shows a pyramid with a horizontal rectangular base ABCD. M is the
mid-point of CD and apex V is 9 cm vertically above the point M
Identify and calculate the angle between the line VB and the base ABCD.
24 cm
14 cm
B
A
D
M
C
V
(4 marks)
Answer :
2. Diagram 2 shows a right prism. Right angled triangle PQR is the uniform cross-
section of the prism. Calculate the angle between the plane RTU and the plane
PQTU. (3 marks)
Answer :
DIAGRAM 1
T
P
5 cm
Q
DIAGRAM 2
R
S
U
18 cm
12 cm

53. 52
3. Diagram 3 shows a prism with a horizontal square base HJKL. Trapezium EFLK is
the uniform cross-section of the prism. The rectangular surface DEKJ is vertical while
the rectangular surface GFLH is inclined.
Calculate the angle between the line DL and the base HJKL (4 marks)
Answer :
4. Diagram 4 show s a right prism . Plan ABFE, ADHE and BCGF are rectangular
planes . Planes ABCD and EFGH are parallel trapezium. It also given  BAD and
ADC = 90

, FG = 5 cm, AB = 2 cm and CG = 12 cm. Identify and calculate the angle
between the plane ADG and AEHD.
5 cm
4 cm
H
G
FE
D C
BA
[4 marks]
Answer :
8 cm
G
L
H
K
J
D
F
E L
5 cm
6 cm DIAGRAM 3
DIAGRAM 4

54. 53
5. Diagram 5 shows a right prism with horizontal rectangle base. Right triangle
RSW are the uniform cross section of the prism.
Calculate the angle between plane SRV and plane RSTU (3 marks)
Answer :
6. Diagram 6 shows a right prism. The base PQRS is a horizontal rectangle. Right angled
triangle QRU is the uniform cross section of the prism. V is the midpoint of PS.
Identify and calculate the angle between the line UV and the plane PQRS.
(4 marks)
Answer :
DIAGRAM 5
5 cm
10 cm
12 cm
T
U
W
S
R
VV
T
U
W
S
R
DIAGRAM 6
S
Q
U
T
P
10 cm
R
12 cm
5 cmV

55. 54
7. Diagram 7 shows a right prism. Right angled triangle PQR ia the uniform cross-
section of the prism. Calculate the angle between the line TR and the plane
PQTU. (4 marks)
Answer :
18 cm
12 cm
5 cm S
T
U
Q
P
R
DIAGRAM 7

56. 55
PART B
GRAPH OF FUNCTIONS
1 (a) Complete Table 1 in the answer space for the equation y =
x
6
by writing down
the values of y when x = -2 and x = 0.5 [2 marks]
(b) (For this part of the questions, use a graph paper)
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 1 unit on the
y-axis, draw the graph of y =
x
6
for -4 ≤ x ≤ 4. [4 marks]
(c) From your graph, find
i. the value of y when x = -2.7,
ii. the value of x when y = 3.6. [2 marks]
(d) Draw a suitable straight line on the graph in (b) to find the values of x which
satisfy the equation 023
6
 x
x
for -4 ≤ x ≤ 4. [4 marks]
Answer:
(a)
X -4 -3 -2 -1 -0.5 0.5 0.8 1.5 2.5 4
Y -1.5 -2 -3 -6 -12 12 4 1.5
Table 1
(c) i) y =
ii) x =
(d) x =
2. (a) Complete Table 2 in the answer space for the equation y =
x
72
by writing
down the values of y when x = 2.4, x = 6 and x = 12.5 [3 marks]
(b) (For this part, use a graph paper)
By using a scale 1 cm to 1 unit on the x-axis and 1 cm to 2 units on the y-axis, draw
the graph of y =
x
72
for 2 ≤ x ≤1 4. [4 marks]
(c) From your graph, find
a. the value of y when x = 4.3,
b. the value of x when y = 22. [2 marks]
(d) Draw a suitable straight line on your graph to find all the values of x which
satisfy the equation 015
2
36

x
x
State these values of x. [3 marks]

57. 56
Answer:
(a)
X 2 2.4 3 4 6 8 10 11.6 12.5 14
Y 36 24 18 9 7.2 6.2 5.14
Table 2
(c) (i) y =
(ii) x =
(d) x =
3. a) Complete Table 3 in the answer space for the equation y = 2x2
– 5x – 3.
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw
the graph of y = 2x2
– 5x – 3 for -3 ≤ x ≤ 5. [4 marks]
c) From your graph, find
i) the value of y when x = -2.4,
ii) the value of x when 2x2
– 5x – 3 = 0. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which satisfy
the equation 2x2
– 8x = 7 for -3 ≤ x ≤ 5.
State these values of x. [4 marks]
Answer:
a)
c) i) y =
ii) x =
d) x =
x -3 -2 -1 0 0.5 1 2 3 4 5
y 30 4 -3 -6 -5 0 9 22
Table 3

58. 57
4. a) Complete Table 4 in the answer space for the equation y = -2x² + 5x + 8 by writing
down the values of y when x = -2 and x = 3 [2 marks]
b) For this part of question , use the graph paper provided . You may use a flexible
curve ruler.
By using a scale 2 cm to 1 unit on the x axis and 2cm to 5 unit on they axis , draw
the graph of y = -2x² + 5x +8 for -3.5 ≤ x ≤ 4 [4 marks]
c) From the graph, find
(i) the value of y when x = 3.3
(ii) the value of x when y = -16
[2 marks]
d) Find and draw a suitable straight line on your graph to determine the values of x
which satisfy the equation 2x – 2x ² = -10 for -3.5 ≤ x ≤ 4
State the values of x [4 marks]
Answer :
a)
x -3.5 -3 -2 -1 0 1 2 3 4
y -34 -25 1 8 11 10 -4
c) i) y =
ii) x =
d) x =
5. a) Complete Table 5 in the answer space for the equation y = x3
– 13x + 18 .
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the
graph of y = x3
– 13x + 18 for -4 ≤ x ≤ 4. [4 marks]
c) From your graph, find
i. the value of y when x = -1.5,
ii. the value of x when y = 25. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which satisfy the
equation x3
– 11x – 2 = 0 for -4 ≤ x ≤ 4. State these values of x. [4 marks]
Table 4

59. 58
Answer:
(a)
X -4 -3 -2 -1 0 1 2 3 4
Y 6 36 30 18 6 6 30
c) i) y =
ii) x =
d) x =
6. a) Complete Table 6 in the answer space for the equation y = x3
+ x2
– 12x – 5.
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw
the graph of y = x3
+ x2
– 12x – 5 for -4 ≤ x ≤ 4. [4 marks]
c) From your graph, find
i. the value of y when x = 0.5,
ii. the value of x when y =11.9. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which
satisfy the equation x3
+ x2
– 10x = 0 for -4 ≤ x ≤ 4.
State these values of x. [4 marks]
Answer:
(a)
X -4 -3 -2 -1 0 1 2 3 4
Y -5 13 7 -5 -15 -17 27
c) i) y =
ii) x =
d ) x =
Table 5

60. 59
TRANSFORMATION
1. Diagram 3 shows three pentagons EFGHJ, EKLMN and KPQRS on a Cartesian
plane.
Diagram 3
(a)
Transformation T is the translation 





2
3
Transformation R is a reflection about the line KMN.
State the coordinates of the image of point J under the following
transformations:
(i) TR
(ii) RT
[4 marks]
(b) EKLMN is the image of EFGHJ under transformation V and EKLMN is the
image of KPQRS under transformation W.
Describe in full the transformation
(i) V
(ii) W
[5 marks]

61. 60
(c) Given that the area of EFGHJ is 26.8cm2
, calculate the area of shaded region
[3 marks]
Answer:
(a)
(i)
(ii)
(b)
(i) V =
(ii) W=
(c)
2 Diagram 5 shows quadrilateral ABCD, PQRS and KLRM drawn on a Cartesian
plane.
Diagram 5
(a)
Transformation T is the translation 





 2
4
.
Transformation V is a reflection in the line y = 1 .
State the coordinates of the image of point A under the following
transformations:
(i) Translation T,
(ii) Combined transformation VT
[3 marks]

62. 61
(b) (i) KLRM is the image of ABCD under the combined transformations WU.
Describe in full, the transformation U and the transformation W.
(ii) Given that the shaded region KLQPSM represents a region of area 120 m2
.
Calculate the area, in m2
, of the region represented by PQRS.
[9 marks]
Answer:
(a)
(i)
(ii)
(b)
(i) U =
V =
(ii)
3. (a) Diagram 3 shows the point F on a Cartesian plane.
DIAGRAM 3
x
y
2 4 6 8 10 12 14 16
2
4
6
8
10
0
F

63. 62
Transformation S is a translation 





 2
5
.
Transformation T is a reflection in the x = 9.
(i) State the coordinates of the image of point F under transformation S.
(ii) State the coordinates of image of point F under transformation TS. [3 marks]
Answer:
(a) (i)
(ii)
(b) Diagram 4 shows three triangle PQR, ACG and EFG on a Cartesian
plane.
DIAGRAM 4
Triangle ACG is the image of triangle PQR under transformation V.
Triangle EFG is the image of triangle ACG under transformation W.
(i) Describe in full transformation :
(a) V
(b) W [6 marks]
(ii) Given that the area of triangle EFG represents a region of area 72 unit2
.
x
y
E
2 4 6 8 10 12 14 16
2
4
6
8
10
O
F
C
A
G
P
RQ

64. 63
Calculate the area, in unit2
, of the region represented by triangle PQR.
[3 marks]
Answer:
(b) (i) (a)
(b)
(ii)
4. (a) Transformation R is a 90° clockwise rotation about the centre (2, 2).
Transformation T is a translation 





 3
4
.
State the coordinates of the image for coordinate (6 , 4) under the following
transformations:
(i) R2
.
(ii) TR. [4 marks]
Answer:
(a) (i)
(ii)
(b) Diagram 5 shows quadrilateral , ABCD, PQRS and EFGH, drawn on a Cartesian
plane.
DIAGRAM 5
-12 -10 -8 -6 -4 -2 2 4 6 8
10 12
F
E
GH
S
QP
DC
B
A
y
xO
R
6
4
2
- 2
- 4

65. 64
PQRS is the image of ABCD under transformation S and EFGH is the image of PQRS
under transformation Q.
(i) Describe in full transformation :
(a)Transformation S
(b)Transformation Q [5 marks]
(ii) Given the area of ABCD is 64 unit2
, calculate the area of shaded region.
[3 marks]
Answer:

66. 65
STATISTICS
1. The data in Diagram 1 shows the monthly pocket money, in RM, received by 40 students.
56 44 60 64 52 53 55 35
36 47 54 59 34 54 52 48
49 51 62 58 38 63 49 43
45 38 48 57 44 49 46 40
32 41 46 56 42 48 51 39
Diagram 1
(a) Based on the data in Diagram 1 and using a class interval of RM5, complete
Table 1 in the answer space. [4 marks]
(b) From the table in a),
(i) state the modal class,
(ii) calculate the mean monthly pocket money of the students. [4 marks]
(c) By using a scale of 2 cm to RM5 on the x-axis and 2 cm to 1 student on the y-axis,
draw a histogram based on the data.
[4 marks]
Answer:
(a)
Pocket money(RM) Frequency Midpoint
31-35
36-40
Table 1
(b) (i)
(ii)
(c)

67. 66
2. The data in Diagram 2 shows the marks obtained by 42 students in a Mathematics final
exam.
51 20 45 31 26 40 30
48 32 37 41 25 36 38
46 38 28 37 39 23 39
33 35 42 29 38 31 49
42 34 26 35 43 42 22
26 47 40 48 44 34 54
Diagram 2
(a) Based on the data in Diagram 2 and using a class interval 5 marks, complete
Table 2 in the answer space. [4 marks]
(b) From the table in a),
(i) state the modal class,
(ii) calculate the mean mark for the Mathematics final exam and give your
answer correct to 2 decimal places. [4 marks]
(c) By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on
the vertical axis, draw a histogram based on the data. [4 marks]
Answer:
(a)
Marks Frequency Midpoint
20-24
25-29
Table 2
(b) (i)
(ii)
(c)

68. 67
3. The data below shows the height, in cm for a group of 40 students in a class.
132 141 146 156 142 148 151 139
136 147 154 159 134 154 152 148
140 153 162 155 138 163 149 143
156 144 160 164 152 151 158 135
145 138 148 157 144 149 146 149
a) Based on the data, complete the Table 3 in the answer space. [3 marks]
b) From the table in a),
i) State the modal class,
ii) Calculate the mean height of the students. [4 marks]
c) By using a scale of 2 cm to 5cm on the x-axis and 2cm to 1 student on
the y-axis, draw a frequency polygon based on the data. [5 marks]
Answer:
a)
Height (cm) Midpoint Frequency
131-135
136-140
Table 3
b i)
ii)
c)

69. 68
4. The data shows the time taken by 42 participants to complete a road race.
51 20 45 31 26 40 30
25 32 37 41 21 36 38
46 38 28 37 39 23 39
33 35 42 29 38 31 23
42 34 26 35 43 28 22
25 47 31 48 44 34 54
a) Using the data above and a class interval of 5 minutes, complete Table 4 in the answer
space. [3marks]
Answer:
Minutes Midpoint Frequency
20-24
25 -29
Table 4
b) Based on your table in (a),
Calculate the mean for the time taken and give your answer correct to two decimal places.
c) By using a scale of 2 cm to 5 minutes on the horizontal axis and 2 cm to 1 participant on the
vertical axis, draw a frequency polygon based on the data. [5 marks]
d) Based on the polygon in (c), state one piece of information. [1 mark]

70. 69
5. The data shows the length, in cm, of 50 pieces of wood.
18 42 23 13 29 33 30 38 30 27
7 25 35 11 27 35 27 34 16 26
16 21 41 31 20 33 27 28 17 32
26 20 30 31 26 32 23 19 22 37
21 24 25 28 35 12 29 32 33 31
a) Using the data and a class interval of 5 cm, complete Table 5 in the answer space.
[5 marks]
b) For this part of the question, use the graph paper.
By using a scale of 2 cm to 5 cm on the x-axis and 2 cm to 5 pieces of wood on the y-axis, draw
an ogive based on the data. [5 marks]
c) (i) From your ogive in (b), find the first quartile,
(ii) Hence, explain briefly the meaning of the first quartile. [2 marks]
Answer:
5a)
Length(cm) Upper
boundary
Frequency Cumulative Frequency
0 4
5  9
6. A donation drive for the School Building Fund accumulated a substantial amount of money from
400 people as shown in Table 6.
Collection
(RM)
1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90
Frequency 10 20 40 60 85 90 50 35 10
Table 6
a) State the modal class for the above data. [1 mark]
b) Construct a cumulative frequency table. [2 marks]
c) By using a scale of 2 cm to RM10 on the x-axis and 2 cm to 40 people on the y-axis, plot an
ogive for the data. [5 marks]
d) From the ogive, find the
(i) median, [1 mark]
(ii) percentage of donors who donated less than RM30, [1 mark]
(iii) number of donors who donated at least RM65. [2 marks]
Table 5

71. 70
PLAN AND ELEVATION
1 (a) Diagram 1(i) shows a solid right prism. The base BCKJ is on horizontal plane.
EFGM and CDLK are vertical planes whereas EDLM is a horizontal plane.
The plane AFGH is inclined. Hexagon ABCDEF is the uniform cross section of the prism.
The sides AB, FE and DC are vertical.
Draw in full scale, the plan of the solid . [3 marks]
(b) A half-cylinder is joined to in Diagram 1(i) at the vertical plane BCQP to form a combined solid as
shown in Diagram 1(ii)
The height of the half-cylinder is 2 cm
b ii)
Draw in full scale ,
i. the elevation of the combined solid on a vertical plane parallel to BC as viewed from X [4 marks]
ii. the elevation of the combined solid on a vertical plane parallel to CK as viewed from Y. [5 marks]
C
3 cm
K
L
M
E
D
G
H
F
A
B
J
8 cm
3 cm
4 cm
6
6
DIAGRAM 1(i)
X C
K
L
M
E
D
G
H
F
A
B
J
8 cm
3 cm
4 cm
DIAGRAM 1(ii)
P
Q
Y
3 cm

72. 71
2. Diagram 2(i) shows a solid right prism with a rectangular base ABCD. ABMKGF is the
uniform cross- section of the prism. Rectangle GHJK is an inclined plane. EFGH and
JKMN is a horizontal plane.
The edges AF, BM, CN and DE are verticals. BM = CN = 2cm, AF = DE = 4cm
AB =DC = 7cm, KM = JN = 5cm and FG = EH = 3cm.
S 2 cm
5 cm
6 cm
7 cm
3 cm
4 cm
J N
MK
H
G
E
F
D
C
B
A
a) Draw full scale, the elevation of the solid on a vertical plane parallel to AB as
viewed from X. [3 marks]
b) A prism is joined to the solid in Diagram 2(i) at the vertical plane CDEHJN.The
combined solid is as shown in Diagram 2(ii). Right angle triangle PQR and DSC is the
uniform cross-section of the prism.
 QPR =  SDC = 900
, DP = CR = SQ = 6cm and PQ = 3 cm.
S
3 cm
R
Q
P
2 cm
5 cm
6 cm
7 cm
3 cm
4 cm
J N
MK
H
G
E
F
D
C
B
A
Draw to full scale,
(i) The elevation of the combined solid on a
vertical plane parallel to BC as viewed
from Y. [4 marks]
(ii) The plan of the combined solid. [5 marks]
DIAGRAM 2(i)
DIAGRAM 2ii)
Y
X

73. 72
3. (a) Diagram 3(i) shows a solid prism.
Hexagon ABCDEF is the uniform cross section of the prism. The base ALGF is on the
horizontal plane. The sides BA, CD and EF are vertical whereas the sides BC and DE are
horizontal.
Draw in full scale, the plan of the solid prism. [3 marks]
(b) A solid prism with triangle AFM as its uniform cross section is joined at the vertical plane
ABCDEF to form a combined solid as shown in Diagram 3(ii).
Draw in full scale,
i. the elevation of the combined solid on a vertical plane parallel to GF as viewed from X.
[4 marks]
ii. the elevation of the combined solid on a vertical plane parallel to AF as viewed from Y.
[5 marks]
P
F
G
H
E
DI
J
K B
C
A
L
2 cm
4 cm
4 cm
4 cm
8 cm
5 cm
DIAGRAM 3(i)
DIAGRAM 2(ii)
6 cm
Y
M
FG
H
E
D
I
J
K B
C
A
L
2 cm
4 cm
4 cm
X
Q N
4 cm

74. 73
4. Diagram shows a solid right prism with rectangular base ABCD on a horizontal plane. The
surface BFGJKC is its uniform cross-section. The rectangle LKJI is an inclined plane and the
square EFGH is a horizontal plane. The edges GJ and HI are vertical and HI = GJ = 2 cm.
(a) Draw full scale, the elevation of the combined solid on a vertical plane parallel to JI as
viewed from X. [3 marks]
(b) A half-cylinder of diameter 4 cm is joined to the prism in diagram 4(i) at the vertical plane
LDPM. The combined solid is as shown in Diagram 4 (ii)
Draw full scale,
(i) the plan of the combined solid. [4 marks]
(ii) the elevation of the combined solid on a vertical plane parallel to AB as
viewed from Y.
[5marks]
C
B
E
A
DF
G
L
H
X
I J
K
4
8
8
6
DIAGRAM 4(i)
M
P
Q
N
C
B
E
A
DF
G
L
H
Y
I J
K
4 cm
8 cm
8 cm
6 cm
DIAGRAM 4(ii)

75. 74
5. Diagram 5(i) shows a solid right prism with square base ABCD on a horizontal plane.
ABKIGF is the uniform cross-section of the prism. AF, LC and BK are vertical edges.
Rectangle IJLK is a horizontal plane and rectangle EFGH is an inclined plane.
Answer :
(a) Draw full- scale, the plan of the solid. [3 marks]
(b) A prism with isosceles triangle as its uniform cross-section such that Q is 2 cm above the
midpoint of JL is joint to the prism at the plane MPLJ. The length of PL is 2 cm. The
combined solid is shown in the diagram 5(ii)
cm
Draw full-scale,
i. the elevation of the combined solid on a vertical plane parallel to AB as viewed
from R [4 marks]
ii. the elevation of the combined solid on a vertical plane parallel to BC as viewed
from S. [5 marks]
S
R
P
N
M
Q
2 cm
4 cm
5 cm
L
K
J
I
H
G
F
E
D
C
B
A
DIAGRAM 5 (ii)
2 cm
2 cm
2 cm
4 cm
5 cm
L
K
J
I
H
G
F
E
D
C
B
A
A
DIAGRAM 5 (i)

76. 75
EARTH AS A SPHERE
1. Table 1 shows the latitudes and longitudes of four points A, B, C and D, on the surface of
the earth.
Point Latitude Longitude
A
B
C
D
200
S
x0
N
200
S
300
N
250
W
250
W
y0
E
y0
E
(a) Q is a point on the surface of the earth such that AQ is the diameter of the earth.
State the position of Q. [2 marks]
(b) Calculate
(i) the value of x, if the distance from A to B measured along the meridian is
4800 nautical mile,
(ii) the value of y, if the distance from A due east to C measured along the
common parallel of latitude is 3664.8 nautical mile. [7 marks]
(c) An aero plane took off from A and flew due east to C along the common parallel of
latitude and then due north to D.
If the average speed for the whole flight is 500 knots, calculate the time taken for the
whole flight. [3 marks]
2. P(600
N, 600
W), Q, R and S are four points on the surface of the earth. PQ is the diameter
of the parallel of latitude 600
N. S lies 4800 nautical mile due to south of P and R lies 450
due east of P.
(a) State the longitude of Q. [2 marks]
(b) Find the position of R. [3 marks]
(c) Calculate the distance, in nautical mile, from P to Q measured along the parallel
latitude. [3 marks]
(d) An aero plane took off from Q and flew towards P using the shortest distance, as
measured along the surface of the earth, and then flew due south to S, with an average
speed of 600 knots.
Calculate the time, in hours, taken for the flight. [4 marks]
Table 1

77. JABATAN PELAJARAN NEGERI PERAK
KERTAS MODEL PEPERIKSAAN SEBENAR SPM
MATEMATIK
Kertas 2
Dua jam tiga puluh minit
Kod Pemeriksa
Bahagian Soalan
Markah
Penuh
Markah
Diperoleh
A
1 3
2 4
JANGAN BUKA KERTAS SOALAN INI
SEHINGGA DIBERITAHU
1. Kertas soalan ini mengandungi dua bahagian :
Bahagian A dan Bahagian B. Jawab semua
soalan daripada Bahagian A dan empat soalan
dalam Bahagian B.
2. Jawapan hendaklah ditulis dengan jelas dalam
ruang yang disediakan dalam kertas soalan.
Tunjukkan langkah-langkah penting. Ini boleh
membantu anda untuk mendapatkan markah.
3. Rajah yang mengiringi soalan tidak dilukis
mengikut skala kecuali dinyatakan.
4. Satu senarai rumus disediakan di halaman 2 & 3
5. Anda dibenarkan menggunakan kalkulator
saintifik yang tidak boleh diprogram.
3 4
4 4
5 4
6 5
7 5
8 6
9 6
10 5
11 6
B
12 12
13 12
14 12
15 12
16 12
Jumlah
Kertas soalan ini mengandungi 15 halaman bercetak.
SULIT
1449/2
Matematik
Kertas 2
2009
2
1
2 jam
1449/2
1449/2  2009 Hak Cipta JPN Perak [Lihat sebelah
SULIT
NAMA :
TINGKATAN :
76

78. SULIT
1449/2
77
Untuk
Kegunaan
Pemeriksa
MATHEMATICAL FORMULAE
The following formulae may be helpful in answering the questions. The symbols given are the
ones commonly used.
RELATIONS
1. nmnm
aaa 
 .
2. nmnm
aaa 

3.   mnnm
aa 
4. 









ac
bd
bcad
A
11
5.
)(
)(
)(
Sn
An
AP 
6. )(1)'( APAP 
7.
2 2
1 2 1 2
tan ( ) ( )dis ce x x y y   
8. Midpoint
  




 

2
,
2
, 2121 yyxx
yx
9. Average speed =
distance travelled
time taken
10. Mean =
sum of data
number of data
12. Pythagoras Theorem
222
bac 
13 2 1
2 1
y y
m
x x



14.
intercept
intercept



x
y
m
11. Mean =
sfrequencieofsum
frequency)rkof(classmasum 

79. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa
SHAPE AND SPACE
1. Area of trapezium =
2
1
 sum of parallel sides  height
2. Circumference of circle= d = 2r
3. Area of circle = r2
4. Curved surface area of cylinder = 2rh
5. Surface area of sphere = 4r2
6. Volume of right prism = cross sectional area  length
7. Volume of cylinder = r2
h
8. Volume of cone =
3
1
r2
h
9. Volume of sphere =
3
4
r3
10. Volume of right pyramid = 
3
1
base area  height
11. Sum of interior angles of a polygon = ( n  2 )  180 0
12. 0
arc of length angle subtended at centre
circumference of circle 360

13. 0
area of sector angle subtended at centre
area of circle 360

14. Scale factor, k =
PA
PA '
15. Area of image= k 2
 area of object
78

80. SULIT
1449/2
79
Untuk
Kegunaan
Pemeriksa
SECTION A
[ 52 marks]
Answer all question in this section.
1. On the graph provided, shade the region which satisfies the three inequalities
y ≤ x + 5, x + y ≤ 5 and y > 1. [3 marks]
Answer:
2. Calculate the values of e and f that satisfy the simultaneous linear equations.
104
1625


fe
fe
[4 marks]
Answer:
x
y
y = x + 5
x + y = 5
O
1 -

81. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa 3. Diagram 1 below shows a right prism with PST is the uniform cross section of the prism.
Diagram 1
A solid cylinder of diameter 6 cm is taken out from the solid prism. Using
7
22
 ,
calculate the volume in cm3
of the remaining solid.
[4 marks]
Answer:
4. Solve the quadratic equation 6
2
)52(

xx
Answer:
[4 marks]

82. SULIT
1449/2
81
Untuk
Kegunaan
Pemeriksa
5. Diagram 2 shows a right prism with an isosceles triangle base, STU.
The isosceles triangle STU is the uniform cross-section of the prism.
ST = SU and W is the midpoint of TU.
Calculate the angle between the line PW and the base STU. [4 marks]
Answer:
6. (a) State whether the following sentence is a statement or a non-statement.
Give a reason for your answer.
(b) State whether each of the following statements is true or false.
(i) { 0 } is an empty set or  is an empty set.
(ii){ } is and empty set and  is also an empty set.
(c ) Complete the following argument.
Premise 1 : If a > 3, then 5 a > 15.
Premise 2 : 5 a < 15.
Conclusion :________________________________________________
[5 marks]
Answer:
(a) _____________________________________
(b) i.___________________
ii. _________________
(c) Conclusion: ____________________________________
13
P
W
10
12
Q
R
S T
U
DIAGRAM 2
2 + 7 = 1 + 6

83. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa
7. Bottle A contains three red marbles and two yellow marbles. Bottle B contains four red
marbles and three yellow marbles. A marble is chosen at random from each bottle.
Find the probability that
(a) both marbles are in yellow,
(b) both marbles are in different color. [5 marks]
Answer:
8.
.
T
Diagram 6
O S
Q
R
P
150
Diagram 3 shows two sectors OPQ and ORS with centre O. OP = 8 cm, QR = 6 cm and
OS = 2 OT. POTS is a straight line.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Answer :
3

84. SULIT
1449/2
83
Untuk
Kegunaan
Pemeriksa
9. The following diagram shows the distance-time graph of the journeys of two particles S
and T.
Graph ABCD represents the journey of the particle S from station X to Y. The graph EF
represents the journey of the particle T from station Y to X.
Both particles leave station X and station Y respectively at the same time along the same
road.
(a) State the length of time, in minute, during which the particle S is
stationary.
(b) If the journey starts at 1500, at what time do the particles meet?
(c) Given that the distance travelled by particle S over the 60 minutes period is 186km.
Calculate the value of v.
(d) Find the distance, in kilometres, from station Y when the particles meet.
(e) Calculate the average speed in m/s of particle S for the whole journey.
[ 6 marks ]
Answer:
A
D
F
CB
600
v
145 E
Speed (ms-1
)
Time (minute)
20 40 100
80
Station
X
Station
Y

85. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa
10. In Diagram 4, KL, LN and MN are straight lines. L is on y-axis. LN is parallel to
x-axis and KL is parallel to MN.
Diagram 4
The equation of straight line KL is 2x + y + 4 = 0, find
(a) The equation of straight line LN,
(b) The equation of straight line MN hence state its x-intercept.
[ 5 marks ]
Answer :
11. It is given matrix U =
1 2
3 2
  
 
 
and matrix V =
2 2
3
v
u
 
 
 
such that UV =
1 0
0 1
 
 
 
a) find the values of u and v.
b) using the matrix method, calculate the values of x and y that satisfy the following
simultaneous equations
32  yx
523  yx [6 marks]
Answer :
NL
M(1,6)
K
O
x
y

86. SULIT
1449/2
85
Untuk
Kegunaan
Pemeriksa
SECTION B
[ 48 marks]
Answer any four questions from this section.
12. (a) Table 1 shows values of x and y for equation 823
 xxy .
Find the values of v and w.
[ 2 marks ]
(b) For this part of the question, use graph paper. You may use a flexible curve
ruler.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to10 unit on the y-axis,
Draw the graph of 823
 xxy for 43  x .
[ 4 marks ]
(c) From your graph, find
(i) the value of y when x = 15,
(ii) the value of x when y = 20.
[ 2 marks]
(a) Draw a suitable straight line on your graph to find the values of x that satisfy the
equation 0493
 xx for 43  x .
State the values of x.
[ 4 marks]
Answer :
x 3 2 1 0 1 2 3 4
y 29 v 7 8 9 4 w 48
TABLE 1

87. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa 13. Diagram 5 shows four quadrilateral ABCD, ADEF, GHJK and LPNM drawn on
Cartesian plane.
Diagram 5
(a) Transformation M is a reflection about the line x = 3.
Transformation N is the translation 





 3
5
State the coordinates of the image of point (2,4) under the following transformations:
i) MN
ii) NM
(b) ADEF is the image of ABCD under transformation V and GHJK is the image of ADEF
under transformation W.
Describe in full
i) Transformation V
ii) A single transformation which is equivalent to transformation WV.
(c) LPNM is the image of ABCD under an enlargement.
i) State the coordinates of the centre of the enlargement.
ii) Given that the area LPNM is 72.8 unit2
, calculate the area of ABCD.
[12 marks]

88. SULIT
1449/2
87
Untuk
Kegunaan
Pemeriksa
Answer:
a. i) ____________________________________________________
ii) ____________________________________________________
b. i) _____________________________________________________
ii) ____________________________________________________
c. i) _____________________________________________________
ii) ________________________________

89. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa 14. The data in Diagram 6 shows the mass in kg of 50 students in a class.
40 52 47 64 52 35 54 43 39 54
43 39 54 57 53 39 55 38 46 42
45 53 60 38 54 62 47 45 52 53
38 55 62 54 45 54 36 58 44 42
52 48 49 46 49 48 45 54 41 46
Diagram 6
(a) Based on the data in Diagram 7 and using a class interval of 5 kg, complete
Table 2 in the answer space. [4 marks]
(b) From the table in a),
(i) state the modal class,
(ii) calculate the mean mass of the students and give your answer correct to 2
significant figures. [4 marks]
(c) By using a scale of 2 cm to 5 kg on the x-axis and 2 cm to 1 student on the y-axis,
draw a histogram based on the data.
[4 marks]
Answer:
(a)
Mass Frequency Midpoint
35-39
40-44
Table 2
(b)
(c)

90. SULIT
1449/2
89
Untuk
Kegunaan
Pemeriksa
15. Diagram 8 shows a right prism with a horizontal rectangular base ABCD. BCHGF is the
uniform cross -section of the prism. GHIJ is a horizontal plane. ABFE and DCHI are
vertical planes.
(a) Draw to full scale, the elevation of the solid on a vertical plane parallel to BC as seen
from X [3 marks]
Answer:
D
B
G
E
A
J
1 cm
C
I
F
H
X
4 cm
3 cm
3 cm
2 cm
DIAGRAM
8(i)

91. SULIT
1449/2
Untuk
Kegunaan
Pemeriksa (b) A solid right prism with uniform cross-section KLM is removed from the solid. The
remaining solid is shown in diagram 8(ii).
Draw at full scale
(i) the plan of the remaining solid . [4 marks]
(ii) the elevation of the remaining solid on a vertical plane parallel to AB as
seen from Y. [5 marks]
Answer:
N
L
K M
1 cm
Y
D
B
G
E
A
J
1 cm
C
I
H
4 cm
3 cm
3 cm
2 cm
DIAGRAM 8(ii)

92. SULIT
1449/2
91
Untuk
Kegunaan
Pemeriksa
16. P(650
S, 400
E), Q(650
S, 600
W), R and S are four points on the surface of the earth. PR
is the diameter of the parallel of latitude 650
S.
(a) (i) State the longitude of R.
(ii) Calculate the shortest distance, in nautical mile, from P to R measured along the
surface of the earth. [4 marks]
(b) S lies north of Q and the distance of SQ measured along the surface of the earth is
5100 nautical mile. Calculate the latitude of S. [3 marks]
(c) An aero plane took off from P and flew due west to Q and then flew due north to S.
The average speed for the whole flight was 560 knots.
Calculate
(i) the distance, in nautical mile, taken by the aero plane from P to Q measured
along the common parallel of latitude,
(ii) the total time, in hours, taken for the whole flight. [5 marks]
Answer:
END OF QUESTION PAPER

93. JABATAN PELAJARAN PERAK
ANSWER TO SCORE
MATHEMATICS SPM
(SECONDARY SCHOOL)
TOPICAL EXERCISES
ANSWERS

94. 92
PART A
SETS
1. The Venn diagram in the answer space shows sets X, Y and Z such that the universal set,
.X Y Z   
On the diagrams in the answer space, shade
(a) the set Y Z 
(b) the set ( ) .X Y Z 
[3 marks]
Answer:
(a)
[ 1 ]
(b)
[ 2]

95. 93
2. The Venn diagram in the answer space shows sets P, Q and R with  = PQR.
On the diagrams provided in answer space, shade
a) P  Q´
b) Q  ( P  R )
[3 marks]
Answer :
a) P Q
[ 1 ]
R
b) P Q
[ 2 ]
R
3. On the diagrams provided in the answer space, shade the region which
represent the following operation of set.
(a) Set A  B’
(b) (b) A  (B  C)’
[3 marks]
Answer / Jawapan:
( a ) (b)
[ 1 ] [ 2 ]

96. 94
4 The Venn Diagram in the answer space shows sets A, B and C.
On the diagrams provided in the answer space, shade
(a) The set A  B´
(b) The set A  B  C ´ [ 3 marks ]
Answer:
(a) (b)
5 The Venn diagram in the answer space shows sets P,Q and R such that the universal set,
 = P  Q  R.
On the diagrams in the answer space, shade
(a) the set P´ Q
(b) the set ( P  Q´)  R. [ 3 marks ]
Answer:
(a) (b)
R
Q
P
A B C A
B
C
P
Q
R
[ 1 ] [ 2 ]
[ 1 ] [ 2 ]

97. 95
LINEAR INEQUALITIES
1. On the graph in the answer space, shade the region which satisfies the three
inequalities 3 12,y x  2 4y x   and 2x  . [3 marks]
Answer :
Straight line 2x drawn correctly. [ 1 ]
Region shaded correctly. [ 2 ]
2. On the graph provided, shade the region which satisfies the three inequalities
y  2x + 8, y  x and y < 8. [3 marks]
Answer :
y = 8
y = 2x + 8
y = x
y
x
Straight line y = 8 drawn correctly. [ 1 ]
Region shaded correctly. [ 2 ]
42  xy
R
123  xy
2x
y
x

98. 96
3. On the graph in the answer space, shade the region which satisfies the three inequalities
y  – 2x + 10, x < 5 and y  10. [3 marks]
Answer :
x = 5
y = - 2x + 10
y = 10
y
x
Straight line x=5 drawn correctly. [ 1 ]
Region shaded correctly. [ 2 ]
4. On the graph provided, shade the region which satisfies the three inequalities
y ≥ -2x + 5, y ≥ x and y < 5. [3
marks]
Answer :
Straight line y=5 drawn correctly. [ 1 ]
Region shaded correctly. [ 2 ]
y = 5
y = x
y = -2x + 5

99. 97
5. On the graph provided, shade the region which satisfies the three inequalities
y ≥ x + 3, x > -3 and 3x + y ≤ 2. [3 marks]
Straight line x = -3 drawn correctly. [ 1 ]
Region shaded correctly. [ 2 ]
x = -3
y = x + 3
3x + y = 2

100. 98
SIMULTANEOUS LINEAR EQUATIONS
1 . Calculate the values of m and n that satisfy the following simultaneous linear
equations.
54
1332


nm
nm [ 4 marks]
Answer:
2
2814
13)45(32
45




m
m
mm
mn
3
)2(45


n
n
2. Calculate the values of p and q that satisfy the following simultaneous linear
equations.
243
132
2
1


qp
qp [ 4 marks]
Answer:
6
244
243
264




p
p
qp
qp
5
204
264



q
q
qp
3. Calculate the values of r and s that satisfy the following simultaneous linear
equations.
7
2
3
62


sr
sr
[ 4 marks]
Answer:
2
84
1423
62




r
r
sr
sr
4
622


s
s
4. Calculate the values of g and h that satisfy the following simultaneous linear
equations.
1834
12


hg
hg
[ 4 marks]
Answer:
2
2211
183)21(4
21




h
h
hh
hg
3
)2(21


g
g
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]

101. 99
5. Calculate the values of x and y that satisfy the following simultaneous linear
equations.
164
3
2
3


yx
yx [ 4 marks]
Answer:
4
287
164
1264




y
y
yx
yx
3
124
16)4(4



x
x
x
6. Calculate the values of a and b that satisfy the following simultaneous linear
equations.
93
5
3
1
2


ba
ba [ 4 marks]
Answer:
2
63
93
156




a
a
ba
ba
3
9)2(3


b
b
7. Calculate the values of d and e that satisfy the following simultaneous linear
equations.
732
24


ed
ed
[ 4 marks]
Answer:
1
1111
73)42(2
42




e
e
ee
ed
2
)1(42


d
d
8. Calculate the values of h and k that satisfy the following simultaneous linear
equations.
4
2
1
434


kh
kh
[ 4 marks]
Answer:
4
205
1624
434




k
k
kh
kh
2
42
4)4(
2
1



h
h
h
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]

102. 100
9. Calculate the values of m and n that satisfy the simultaneous linear equations.
52
72


nm
nm
[ 4 marks]
Answer:
1
33
1042
72




n
n
nm
nm
3
62
7)1(2



m
m
m
10. Calculate the values of u and v that satisfy the simultaneous linear equations.
46
734


vu
vu
[ 4 marks]
Answer:
3
1
927
73)64(4
64




v
v
vv
vu
2
)
3
1
(64


u
u
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]

103. 101
QUADRATIC EXPRESSIONS AND EQUATIONS
1. Solve the equation 2
1
52 2



x
xx
[ 4 marks]
Answer:
2,
2
1
0)2)(12(
0232
2252
2
2




x
xx
xx
xxx
2. Solve the quadratic equation y
y
3
3
52 2


[ 4 marks]
Answer:
5,
2
1
0)5)(12(
0592
952
2
2




y
yy
yy
yy
3. Solve the quadratic equation 6
2
)1(3


p
pp
[ 4 marks]
Answer:
3,
3
4
0)3)(43(
01253
12233
2
2




p
pp
pp
ppp
4. Using factorization, solve the following quadratic equation.
qq 17154 2

[ 4 marks]
Answer:
5,
4
3
0)5)(34(
015174 2



q
qq
qq
5. Solve the quadratic equation
r
r
r
2
36
1

 [ 4 marks]
Answer:
2,
2
3
0)2)(32(
062 2



r
rr
rr
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]

104. 102
6. Solve the quadratic equation
2
562


w
w
[ 4 marks]
Aswer:
4,
2
3
0)4)(32(
01252 2



w
ww
ww
7. Solve the quadratic equation 7)2)(12(  aa
[ 4 marks]
Answer:
3,
2
3
0)3)(32(
0932 2



a
aa
aa
8. Solve the quadratic equation 1
2
32

p
p
[ 4 marks]
Answer:
2,
2
1
0)2)(12(
0232 2



p
pp
pp
9. Solve the following quadratic equation
)3(5)3)(1(  eee
[ 4 marks]
Answer:
3,6
0)3)(6(
01832



e
ee
ee
10. Solve the equation 52)4( 2
 yy
[ 4 marks]
Answer:
7,3
0)3)(7(
021102



y
yy
yy
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]
[ 1 ]
[ 1 ]
[ 1 ] [ 1 ]

105. 103
SOLID GEOMETRY
1.
DIAGRAM 1
Diagram 1 shows a composite solid formed by the combination of a cuboid and
a half-cylinder.Using π =
7
22
,
calculate the volume, in cm3
, of the solid. [4 marks]
Solution:
= 7821  [ 1 ]
2
1
x
7
22
x 42
x 21 [ 1 ]
= 7821  +
2
1
x
7
22
x 42
x 21 [ 1 ]
= 1704 [ 1 ]
2.
DIAGRAM 2
Diagram 2 shows a solid cylinder of height 15 cm and diameter 14 cm. A cone with
height 6 cm is taken out of the solid. Calculate the volume in cm3
of the remaining solid.
(Use π =
7
22
). `[4 marks]
Solution : 157
7
22 2
 [ 1 ]
67
7
22
3
1 2
 [ 1 ]
[ 1 ]
= 2002 [ 1 ]
P Q21 cm
8 cm
R
S
T
U
V
W
7 cm
15 cm
157
7
22 2
 - 67
7
22
3
1 2


106. 104
3.
Diagram 3 shows a solid cuboid ABCDEFGH. A right pyramid with its base EFGH is
taken out of the solid. The volume of the remaining solid is 480 cm3
. Calculate the height
of the solid, h.
(Use π =
7
22
). `[4 marks]
Solution : 89h [ 1 ]
h 89
3
1
[ 1 ]
hh  89
3
1
89 = 480 [ 1 ]
h = 10 [ 1 ]
4. The diagram 4 shows a composite solid, formed by the combination of a cylinder to a right
prism. Trapezium
ABEF is the uniform cross-section of the prism. AB = BC= 10 cm.
The diameter of the cylinder is 7 cm and the volume of the composite solid is 1075.5 cm3
.
Using
7
22
 , calculate
(a) the volume, in cm3
, of the right prism
(b) the length, h, in cm, of the cylinder. [5 marks]
Diagram 3
Diagram 4

107. 105
Solution :
(a)   1081014
2
1
 [ 1 ]
= 960 [ 1 ]
(b) h






2
2
7
7
22
[ 1 ]
  1081014
2
1
 + h






2
2
7
7
22
= 1075.5 [ 1 ]
h = 3 [ 1 ]
5. The diagram 5 below shows a combined solid consists of a right prism and a right pyramid
which are joined at the plane PQRS. T is vertically above the base PQRS. Trapezium
PQKJ is the uniform cross section of the prism.
The height of the pyramid is 6 cm and PQ = 18 cm.
(a) Calculate the volume, in cm3
, of the right pyramid.
(b) It is given that the volume of the combined solid is 448 cm3
. Calculate the length, in cm, of
PJ.
[5 marks]
Solution :
(a) 6818
3
1
 [ 1 ]
= 288 [ 1 ]
(b)  PJ1814
2
1
 [ 1 ]
6818
3
1
 +  PJ1814
2
1
 = 448 [ 1 ]
PJ = 10 [ 1 ]
Diagram 5

108. 106
MATHEMATICAL REASONING
1. a) State whether the following statement is true or false
b) Write down two implications based on the following sentence
c) State the converse of the following statement and hence, determine whether its
converse is true or false.
“ If 2x > 8, then x > 4 “
[5 marks]
Answer :
(a) True [ 1 ]
(b) Implication 1 : If x³ = 64 then x = 4 [ 1 ]
Implication 2 : If x = 4 then x³ = 64 [ 1 ]
(c) If x > 4 , then 2x > 8.
True. [ 2 ]
2. (a) Complete each statement in the answer space with the quantifier “all” or
“some” so that it will become a true statement.
(b) State the converse of the following statement and hence, determine
whether its converse is true or false.
(c) Write down Premise 2 to complete the following argument :
Premise 1 : If m is less than zero, then m is a negative number.
Premise 2 : _________________________________________________
Conclusion : - 4 is a negative number. [5 marks]
Answer :
(a) (i) Some of the multiples of 5 are even numbers. [ 1 ]
(ii) All hexagons have six sides. [ 1 ]
(b) If p > 4 then p > 7 [ 1 ]
False [ 1 ]
(c) Premise 2 : - 4 is less than zero [ 1 ]
If p > 7, then p > 4
9 > 8 or 4² = 8
x³ = 64 if and only if x = 4

109. 107
3. (a) State whether the sentence below a statement or non-statement ?
(b) Statement 1 : a0
= 1 for any values of a except a = 0.
Statement 2 : All prime numbers are odd numbers.
Combine the above statements to form a new statement which is
i) true
ii) false
(c) Make a general conclusion by induction for the following number pattern:
5 = 3(2 ) – 1
10 = 3(22
) – 2
21 = 3(23
) – 3
60 = 3(24
) – 4
……………… [5 marks]
Answer :
(a) Non statement [ 1 ]
(b) i) a0
= 1 for any values of a except a = 0 or all prime numbers are odd numbers.
ii) a0
= 1 for any values of a except a = 0 and all prime numbers are odd
numbers.
[ 2 ]
(c) Conclusion : 3(2n
) – n , n = 1,2,3,4… [ 2 ]
4. (a) State whether each of the following statement is true or false.
(i) 23
= 6 or = 3.5
(ii) ( -4 ) x ( -5 ) = 20 and -4 > -2
(b) Complete the premise in the following argument:
Premise 1 : If the determinant of a matrix = 0, then the matrix does not
have an inverse.
Premise 2 : _______________________________________________
Conclusion :Matrix A does not have an inverse.
(c) Write down two implications based on the following sentence.
A  B if and only if A B = A' [5 marks]
Answer :
(a) (i) True [ 1 ]
(ii) False [ 1 ]
(b) Premise 2 : The determinant of matrix A = 0 [ 1 ]
(c) Implication 1 : If A  B then A  B = A' [ 1 ]
Implication 2 : If A  B = A' then A  B [ 1 ]
p + q = 2
2
7

110. 108
5. (a) State whether the following statement is true or false.
(b) Write down two implications based on the following statement:
c) State the converse of the following statement and hence, determine whether its
converse is true.
“ If x > 9, then x > 5 “
Answer :
(a) True [ 1 ]
(b) Implication 1 : If p3
= 8 then p = 2 [ 1 ]
Implication 2 : If p = 2 then p3
= 8 [ 1 ]
(c) If x > 5, then x > 9 [ 1 ]
False [ 1 ]
6 (a) State whether each of the following statement is true or false.
(i) 42
= 8 or = -2
(ii) a  { a, b, c } and -3 > -7
(b) Write down premise 1 to complete the following argument.
Premise 1 :_________________________________________________
Premise 2 : 6 × p  42
Conclusion : p  7
(c) Form a general conclusion by induction for the number sequence
11, 23, 43, 71 , … which follow the pattern
11 = 4(12
) + 7
23 = 4(22
) + 7
43 = 4(32
) + 7
71 = 4(42
) + 7
……………… [5 marks ]
Answer :
(a) (i) True [ 1 ]
(ii) False [ 1 ]
(b) Premise 1 : If p = 7, then 6 × p = 42 [ 1 ]
(c) Conclusion: 4n2
+ 7, n = 1,2,3,4… [ 2 ]
-2 ( 3 ) = 6 or -4 > -5
p3
= 8 if and only if p = 2
3
8

111. 109
7. (a) Determine whether each of the following sentence is a statement.
(i) 9 is a prime number.
(ii) 2x(x – 2) = 0
(b) Fill in the suitable quantifier to make the following statement become true.
(c) Write down two implications based on the following sentence:
“ 3m > 15 if and only if m > 5 “
[6 marks]
Answer :
(a) (i) Statement [ 1 ]
(ii) Non Statement [ 1 ]
(b) Some [ 1 ]
(c) Implication 1 : If 3m > 15 then m > 5 [ 1 ]
Implication 2 : If m > 5 then 3m > 15 [ 1 ]
8. (a) State whether the following statement is true or false.
(i) 8 + 2 = 10 and 2 < -3
(ii) All perfect square numbers are even numbers
.
(b) Complete the following argument.
Premise 1 : If 3t = 0, then t = 0.
Premise 2 : ________________________________________________
Conclusion : 3t  0
(c) State the converse of the following statement and hence, determine
whether its converse is true or false.
[5 marks]
Answer :
(a) (i) False [ 1 ]
(ii) False [ 1 ]
(b) Premise 2 : t  0 [ 1 ]
(c) If x = 3, then x3
= 27 [ 1 ]
True [ 1 ]
…………. prime numbers are odd numbers
If x3
= 27 then x = 3

112. 110
9. (a) Is the sentence below a statement or non-statement?
(b) Write down two implications based on the following sentence:
(c) Make a general conclusion by induction for the following number pattern :
2 = (0)2
+ 2
3 = (1)2
+ 2
6 = (2)2
+ 2
11 = (3)2
+ 2 [5 marks]
Answer :
(a) Statement [ 1 ]
(b) Implication 1 : If m is an even number then m can be divided by 2. [ 1 ]
Implication 2 : If m can be divided by 2 then m is an even number. [ 1 ]
(c) n 2
+ 2 , n = 0, 1, 2,3 … [ 2 ]
10. (a) State whether each of the following statement is true or false
(i) (-2)2
= 4 or ( -3 )3
= -9
(ii) and
(b) Complete the following argument.
Premise 1 :_________________________________________________
Premise 2 : xn
+ x is not a quadratic expression.
Conclusion : n  2.
(c) Write down two implications based on the following sentence.
[5 marks]
Answer :
(a) (i) True [ 1 ]
(ii) False [ 1 ]
(b) Premise 1 : If n = 2, then x n
+ x is a quadratic expression [ 1 ]
(b) (c) Implication 1 : If 1 – q > 2 then q < -1 [ 1 ]
Implication 2 : If q < -1 then 1 – q > 2 [ 1 ]
3 and 4 are factors of 8
m is an even number If and only if m can be divided by 2.
8162
1

5
1
25 2
1


1 – q > 2 if and only if q < -1

113. 111
THE STRAIGHT LINES
1. In Diagram 1, O is the origin, point R lies on the x-axis and point P lies on the y-axis.Straight
line PU is parallel to x-axis and the straight line PR is parallel to straight line ST. The
equation of straight line PR is x + 2y = 14.
R xO
T(2,-5)
S
U
x + 2y = 14
y
••
•
•
•
(a) State the equation of the straight line PU. [2 marks]
(b) Find the equation of the straight line ST and hence,
state its x-intercept. [3 marks]
Answer:
a) Equation PR, y = 7
2
1
 x [ 1 ] (b) mST =
2
1

Point P = (0, 7) y – (-5) =
2
1
 (x – 2) [ 1 ]
Equation PU, y = 7 [ 1 ] y =
2
1
 x – 4 [ 1 ]
x- intercept = -8 [ 1 ]
2. The diagram below shows that the straight line EF and GH are parallel.
DIAGRAM 2
Find
(a) the equation of EF. [3 marks]
(b) the y - intercept and x - intercept of EF [2 marks]
P
DIAGRAM 1

114. 112
K O
N
y
x
P
M
L (4, 7)
Answer:
(a) mGH =
)1(4
)5(2


=
5
7
[ 1 ] (b) y – intercept = 5 [ 1 ]
-2 = c )5(
5
7
[ 1 ] x –intercept = -
5/7
5
c = 5
y = 5
5
7
x [ 1 ] =
7
25
[ 1 ]
3. In the following diagram, O is the origin, point K and point P lies on the x-axis and point N lies
on the y-axis. Straight line KL is parallel to straight line NP and straight line MN is parallel to the
x-axis. The equation of straight line NP is x – 2y – 18 = 0.
(a) State the equation of the straight line MN.
(b) Find the equation of the straight KL and hence, state the coordinate of the point K.
[5 marks]
Answer:
(a) Equation NP, y = 9
2
1
x [ 1 ] (b) mKL =
2
1
Point N = (0, -9) y – 7 =
2
1
(x – 4) [ 1 ]
Equation MN, y = - 9 [ 1 ] y =
2
1
x + 5
0 =
2
1
x + 5
K = ( -10 , 0 ) [ 1 ]
Diagram 3

115. 113
4. In the diagram 4, the gradient of the straight line KLM is
2
1
 . Find
(a) the value of p.
(b) the x-intercept of the straight line MN.
[5 marks]
Answer:
(a) mKL =
2
1
20
4


p
[ 1 ] (b) mLM =
2
1
2
40



x
[ 1 ]
p = 5 [ 1 ] x = 10
x-intercept = 10 [ 1 ]
5. The diagram 5 below shows STUV is a trapezium
Given that gradient of TU is -3, find
(a) the coordinates of point T. [2 marks]
(b) the equation of straight line TU. [1 marks]
(c) the equation of straight line TV [2 marks]
Answer:

K(0, p)
0
y
x
N
M
L(2, 4)
Diagram 4
Diagram 5

116. 114
a) The gradient =
06
2

 p
-3 =
6
2 p
[ 1 ]
-18 = 2 – p
p= 20
Coordinates of point T = (0 , 20) [ 1 ]
b) m = -3, c = 20
y = -3x + 20 [ 1 ]
c)
3
10
60
020



[ 1 ]
20
3
10
 xy [ 1 ]

117. 115
PERIMETER AND AREA
1.
18 cm
Diagram 1
15 15
E
A
B
O F
C D
In Diagram 1, OAB, OCD and OFE are three sectors in a circle centred O. AOF, BCO and EDO
are straight lines. C and D are midpoints of OB and OE respectively.
Using
22
7
  , calculate
a) the perimeter , in cm, of the whole diagram,
b) the area, in cm2
, of the shaded region. [6 marks]
Answer:
a)
15 22 150 22
2 18 2 9
360 7 360 7
or
   
        
   
[ 1 ]
=
15 22 150 22
2(18) 2(9) 2 2 18 2 9
360 7 360 7
   
           
   
[ 1 ]
= 87 cm [ 1 ]
b)
15 22 150 22
18 18 9 9
360 7 360 7
or
   
        
   
[ 1 ]
=
15 22 150 22
18 18 9 9
360 7 360 7
   
         
   
[ 1 ]
=
1 297
148
2 2
or or 148.5 cm2
[ 1 ]

118. 116
2.
Diagram 2
S
O
R
P Q120
Diagram 2 shows a sector two sectors ORS and OPQ with the same centre O. P and Q are
midpoints of OR and OS respectively and OS = 28 cm.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Answer:
a)
120 22 240 22
2 28 2 14
360 7 360 7
or
   
        
   
[ 1 ]
=
120 22 240 22
2(14) 2 28 2 14
360 7 360 7
   
          
   
[ 1 ]
=
1 436
145
3 3
or or 145.33 cm [ 1 ]
b)
120 22 120 22
28 28 14 14
360 7 360 7
or
   
        
   
or
240 22
14 14
360 7
 
   
 
[ 1 ]
=
120 22 120 22 240 22
28 28 14 14 14 14
360 7 360 7 360 7
      
                
      
[ 1 ]
=
2 3080
1026
3 3
or or 1026.67 cm2
[ 1 ]

119. 117
3.
In Diagram 3, OFIE is a quadrant of a circle with centre O and GH is an arc of another circle with
centre O. OFG and OIH are straight lines.
OF = FG = 14 cm, and  35GOH
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Answer :
a) 











 28
7
22
2
360
35
14
7
22
2
360
55
or [ 1 ]
= 











 28
7
22
2
360
35
14
7
22
2
360
55
)14(4 [ 1 ]
= 56.86
9
779
9
5
86 oror [ 1 ]
b) 











 2828
7
22
360
35
1414
7
22
360
55
or [ 1 ]
= 























 1414
7
22
360
35
2828
7
22
360
35
1414
7
22
360
55
[ 1 ]
= 78.273
9
2464
9
7
273 oror cm2
[ 1 ]
Diagram 3
35 
H
E
G
F
I
O

120. 118
4.
14 cm
Diagram 4
O
P
Q
R
T
U
S
Diagram 4 shows two semicircles RSO and RTU with centre Q and O respectively. OPQ is a
quadrant of a circle with centre O. RQ = QO. ROU is a straight line.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Answer :
45 22
2 7
360 7
 
   
 
or
180 22
2 14
360 7
 
   
 
[ 1 ]
a) =
45 22 180 22
2(7) 14 2 7 2 14
360 7 360 7
   
           
   
[ 1 ]
=
1 155
77
2 2
or or 77.5 cm [ 1 ]
45 22
7 7
360 7
 
   
 
or
180 22
14 14
360 7
 
   
 
or
180 22
7 7
360 7
 
   
 
[ 1 ]
b) =
45 22 180 22 180 22
7 7 14 14 7 7
360 7 360 7 360 7
      
                
      
[ 1 ]
=
1 1001
250
4 4
or or 250.25 cm2
[ 1 ]

121. 119
5.
H
G
F
E
I
Diagram 5
J
K
In diagram 5, two arcs of two equal circles, GFE and GIJ, with centres H and K respectively. It is
given that GH = 14 cm.
Using
22
7
  , calculate
a) the perimeter, in cm, of the whole diagram
b) the area, in cm2
, of the shaded region. [6 marks]
Answer :
a)
120 22
2 14
360 7
 
   
 
or
240 22
2 14
360 7
 
   
 
[ 1 ]
=
120 22 240 22
2(14) 2 14 2 14
360 7 360 7
   
          
   
[ 1 ]
= 116 cm [ 1 ]
b)
120 22
14 14
360 7
 
   
 
or
240 22
14 14
360 7
 
   
 
[ 1 ]
=
120 22 240 22
14 14 14 14
360 7 360 7
   
         
   
[ 1 ]
= 616 cm2
[ 1 ]

122. 120
MATRICES
1a) The inverse matrix of is
4
5 3
p
q
 
 
 
. Find the value of p and of q.
b) Using matrices, calculate the values of x and y that satisfy the following simultaneous
equations :
3x – 2y = 6 [6 marks]
5x – 4y = 4
1a)
4 21
5 312 ( 10)
4 21
5 32
 
  
    
 
  
  
p = 2 [ 1 ]
q =
1
2
 [ 1 ]
1b)
3 2
5 4
 
 
 
=
6
4
 
 
 
[ 1 ]
4 2 61
5 3 42
x
y
    
     
     
[ 1 ]
16
21
x
y
   
   
   
x = 16 [ 1 ]
y = 21 [ 1 ]
2.a) Given that F =
3
2
m
n
 
 
 
and the inverse matrix of F is
4 31
214 m
 
 
 
, find the value of m and
of n.
b) Hence, using matrices, calculate the value of p and of q that satisfies the following
equation
2
5
p
F
q
   
   
   
[6 marks]
Answer :
2a) F 1 31
26
n
mmn
 
  
  
mn 6 = 14
n = 4 [ 1 ]
4m  6 = 14
m = 5 [ 1 ]
2b)
5 3
2 4
p
q
  
  
  
=
2
5
 
 
 
[ 1 ]
4 3 21
2 5 514
p
q
    
    
    
[ 1 ]
p =
1
2
 [ 1 ]
q =
3
2
[ 1 ]
3.a) Given that find matrix A.
b) Hence, using the matrix method, find the value of d and e which satisfy the
simultaneous equations below.
-d + 2e = 11
-3d + 5e = 15 [6 marks]
Answer :
3a) A =
5 21
3 15 6
 
 
   
[ 1 ] 3b)
1 2
3 5
 
 
 
11
15
d
e
   
   
   
[ 1 ]
=
5 2
3 1
 
 
 
[ 1 ]
5 2 11
3 1 15
d
e
    
    
    
=
25
18
 
 
 
[ 1 ]
d = 25 [ 1 ]
e = 18 [ 1 ]








45
23






y
x
,
10
01
53
21














A

123. 121
4. Given matrix M = and matrix MN =
a) Find the matrix N.
b) Hence, calculate by using the matrix method, the values of r and s that satisfy the
following simultaneous equations :
4r + 5s = 1
6r + 8s = 1 [6 marks]
4a) N =
8 51
6 432 30
 
 
  
(1)
= (1)
4b)
4 5 1
6 8 1
r
s
    
    
    
(1)
8 5 11
6 4 12
r
s
    
    
     
(1)
=
13
2
5
 
 
   
r =
13
2
(1)
s = 5 (1)
5. Given the matrix K is ,
a) Find the matrix L so that KL =
b) Hence, calculate the values of h and k, which satisfy the matrix equation:
[6 marks]
Answer :
5a) L =
5 31
8 420 24
 
 
   
=
5 31
8 44
 
 
 
(1)+(1)
5b)
4 3 7
8 5 11
h
k
     
    
     
(1)
5 3 71
8 4 114
h
k
     
    
     
(1)
=
21
124
 
 
 
h =
1
2
(1)
k = 3 (1)
6. Given matrix M =
1 2
2 6
 
 
 
, find
a) the inverse matrix of M.
b) hence, using matrices,find the values of x and y that satisfy the following
simultaneous equations :
x  2y = 3
–2x + 6y = 4 (6 marks)






86
54






10
01








46
58
2
1








58
34






















11
7
58
34
k
h






10
01

124. 122
Answer :
6a) M 1
=
6 21
2 16 4
 
 
  
=
6 21
2 12
 
 
 
[ 1 ] + [ 1 ]
6b)
1 2 3
2 6 4
x
y
    
    
     
[ 1 ]
6 2 31
2 1 42
x
y
    
    
    
[ 1 ]
x = 5 [ 1 ]
y = 1 [ 1 ]
7. It is given that matrix A =
1
3 4
a 
 
 
does not have an inverse matrix.
a) Find the value of a.
b) If a = 2, find the inverse matrix of A and hence, using matrices, find the values of
x and y that satisfy the following simultaneous linear equations.
x  2y = 6
3x + 4y = 2 [6 marks]
Answer :
7a) 4 – 3a = 0
3a = 4
a = 
4
3
[ 1 ]
7b) A 1
=
4 21
3 14 6
 
 
    
=
4 21
3 12
 
 
  
[ 1 ]
1 2 6
3 4 2
x
y
     
    
    
[ 1 ]
4 2 61
3 1 22
x
y
    
    
      
[ 1 ]
=
201
162
 
 
 
x = 10 [ 1 ]
y = 8 [ 1 ]
8a) i) Find matrix G such that
3 1 3 1
4 2 4 2
G
   
   
   
ii) Find the determinant of matrix C 







22
14
b) Using matrices, calculate the values of p and q that satisfy the following matrix
equation.
2p + 4q = 4
p + 3q = 8 (6 marks)
Answer :
8 a) i) G = [ 1 ]
ii) = 6)1)(2()2)(4(  [ 1 ]
8b
2 4 4
1 3 8
p
q
    
    
    
[ 1 ]
3 4 41
1 2 86 4
p
q
    
    
    
[ 1 ]
=
201
122
 
 
 
p =  10 [ 1 ]
q = 6 [ 1 ]






10
01

125. 123
9.a) Find the inverse of matrix
4 5
1 2
 
 
 
b) Hence, using matrices, calculate the values of m and n that satisfy the following
simultaneous equations :
– 4m + 5n = 1
–m + 2n = 2 (6 marks)
Answer :
9a)
2 51
1 48 5
 
 
   
[ 1
]
= 








41
52
3
1
[ 1
]
9b)
4 5 1
1 2 2
m
n
    
    
     
[ 1 ]




















2
1
41
52
3
1
m
m
[ 1 ]
m = 4 [ 1 ]
n = 3 [ 1 ]
10.a) Given matrix S = 





 m4
36
, find the value of m if matrix S has no inverse.
b) Given the matrix equations
and
8 41
5 7 1
x k
y h
    
    
    
i) Find the value of h and k.
ii) Hence, find the value of x and y. [6 marks]
Answer :
10a) 6m + 12 = 0
m = 2 [ 1 ]
10b i) h = 5630
= 26 [ 1 ]
k = 6 [ 1 ]
ii) =
8 6 41
5 7 126
  
  
  
[ 1 ]
=
1
1
2
 
 
  
 
x = 1 [ 1 ]
y = [ 1 ]




















1
4
85
67
y
x






y
x
2
1


126. 124
GRADIENT AND AREA UNDER THE GRAPH
1. Diagram 1 shows a distance-time graph for the journey of a car and lorry from town Q.
Diagram 1
Graph ABCD represents the journey of the car and graph ACE represents the journey of
the lorry. Both vehicles depart from town Q at the same time along the same road.
(a) State the length of time, in hours, during which the car is stationary.
(b) Calculate the average speed in km/h for the total distance of the car.
(c) At the certain time during the journey, both vehicles meet at the same location.
(i) Find the distance, in km, between that location and town Q.
(ii) State the time taken by the lorry to reach that location from town Q.
[ 5 marks ]
Answer:
(a) 0.14.1  BC tt
= 0.4 hours [ 1 ]
(b) Average speed =
takentimetotal
traveleddistancetotal
=
2.3
142
[ 1 ]
= 44.38 km/h [ 1 ]
(c) (i) 70km [ 1 ]
(ii) 1.4 hours [ 1 ]
2. The diagram shows a speed-time graph of the movement of a particle for a period of 35 s.
0
A
Distance (km)
Time (hour)
1.0 1.4 3.2
E
D142
70
195
B C
20
12
10 t Time (s)
Speed (m s-1
)
35

127. 125
(a) State the uniform speed, in ms-1
, of the particle.
(b) Given the distance travelled by the particle at a uniform speed is 250m,
calculate
(i) the value of t,
(ii) the average of speed, in ms-1
, of the particle for the period of 35 seconds.
[ 6 marks ]
Answer:
(a) 20 m s-1
[ 1 ]
(b) (i) At uniform speed,
Distance = 250 m
20 )10(  t = 250 [ 1 ]
10t = 12.5
t = 22.5 s [ 1 ]
(ii)
=
35
25)5.1225(
2
1
10)2012(
2
1

[ 1 ]
= 15.29 m/s [ 1 ]
3. Diagram 2 shows a distance-time graph for the journey of a bus and van.
Diagram 2
Graph AFB represents the journey of the van from town X to town Y. The graph CDEFG
represents the journey of the bus from town Y to town X. The van leaves town X and the
bus leaves town Y at 9.45 p.m. and they travel along the same road.
(a) (i) At what time do the vehicles meet ?
(ii) Find the distance, in km, from town Y when the vehicles meet.
(b) State the length of time, in minutes, during which the bus is stationary.
(c) Calculate the average speed, in km h-1
, of the bus for the whole journey.
[ 5 marks ]
Answer:
(a) (i) 9.45 pm + 70 minutes = 10.55 pm [ 1 ]
(ii) 100km km65 = 35km [ 1 ]
(b) 50 4010  minutes [ 1 ]
(c) =
60150
100

[ 1 ]
= 40 km/h [ 1 ]
150
C
Time
(minutes)
Town X
A
G
100
0
50
B
F
ED
10
Town Y
70
65
Distance (km)

128. 126
4. Diagram 3 shows a speed-time graph for a particle, of 30 seconds.
Speed (m s-1
)
(a) The distance traveled during the 5 last second is 106m. Find the value of v.
(b) Calculate the acceleration, in m s-2
, of the particle in the last 5 seconds.
(c) The total distance traveled before 10 seconds is 230m. Calculate the average
speed for the whole journey.
[ 6 marks ]
Answer:
(a) 1065
2
1
 v [ 1 ]
v = 42.4 m s-1
[ 1 ]
(b)
5
4.420 
[ 1 ]
= - 8.48 m s-2
[ 1 ]
(c)
30
106]15)214.42(
2
1
[230 
[ 1 ]
= 27.05 m/s [ 1 ]
10 25 30 Time (s)
Diagram 3
v
21
0
25

129. 127
5. The diagram 4 shows the speed-time graphs of two particles M and N over a period of 100
seconds. ABCD is the speed-time graph of particle M and EFG is the speed-time graph of
particle N.
(a) State the length of time, in seconds, during which the particle N is constant speed.
(b) Calculate the rate of change of speed, in ms-2
, of the particle M in the first 20s.
(c) Given that the distance travelled by both particles over the 100 s period is the same
and the distance traveled by particle N is 8500m. Calculate the value of v.
[ 6 marks ]
Answer:
(a) 100 – 60 = 40 s [ 1 ]
(b)
020
080


[ 1 ]
= 4 ms-2
[ 1 ]
(c)   85006040145
2
1
 vv [ 2 ]
v = 70 ms-1
[ 1 ]
E
B
G
D
0
v
80
145
40 60 10020 80
F
Speed (ms-1
)
Time (s)
C
A
Diagram 4

130. 128
PROBABILITY
1.
Students
Probability of choosing:
Biology Physics Chemistry
Ahmad
4
3
8
1
Ling
5
1
5
2
The table 1 shows the probability of how Ahmad and Ling might choose their subjects in Form
6. The table is incomplete.
Calculate the probability that
(a) Ahmad will choose Chemistry and Ling will choose Biology,
(b) Ahmad and Ling will choose the same subjects. [5 marks]
Answer:
(a)
8
1
8
1
4
3
1  and
5
2
5
2
5
1
1 
20
1
40
2
5
2
8
1


(b)
8
3
40
15
5
2
8
1
5
1
8
1
5
2
4
3



2 In bag A, there are 4 blue balls and 3 yellow balls while in bag B, there are 5 blue balls and
6 yellow balls. A ball is drawn at random from bag A. After its colour is recorded, it is put
into beg B. Then a ball is drawn at random from beg B. Find the probability that
(a) the colour of the two balls are blue,
(b) two different colour balls are drawn. [5 marks]
Answer:
(a)
7
2
12
6
7
4


[ 1 ]
[ 1 ]
[ 2 ]
[ 1 ]
[ 1 ]
[ 1 ]
Table 1

131. 129
(b)
84
39
12
5
7
3
12
6
7
4


3 A company van carries a group of workers consists of 6 males and 8 females. They are
dropped off at random at various houses along the route as shown in the diagram below
to serve customers.
(a) If two workers are dropped off at house A, calculate the probability that both are females,
(b) Two male workers are dropped off at house A. If another two workers are then dropped
off at house B, calculate the probability that at least one of them is female.
[5 marks]
Answer:
(a) (b)
13
4
182
56
13
7
14
8



11
10
132
120
11
4
12
8
11
7
12
8
11
8
12
4



4 In a lucky draw, there are three categories of gifts consisting of 4 hotel vouchers, 6 shopping
vouchers and 10 restaurant vouchers. All vouchers are placed inside similar envelopes and
put inside a box.
The honored guests of the lucky draw are requested to draw at random two envelopes from
the box.
Calculate the probability that the first honored guest draw
(a) the first envelope with a hotel voucher and the second with a shopping voucher,
(b) two envelopes with vouchers of the same categories. [5 marks]
Answer:
(a)
95
6
19
6
20
4


Other houses●
Company House A
House B
●
●
●
[ 1 ]
[ 1 ]
[ 2 ]
[ 1 ]
[ 2 ]
[ 1 ]
[ 1 ]
[ 1 ]

132. 130
(b)
95
33
380
32
19
9
20
10
19
5
20
6
19
3
20
4



5. The table below shows the number of participants in a quiz contest according to school and
gender.
School
Number of Participants
Male Female
SMK Ria 4 4
SMK ACS 5 2
SMK Puteri 3 6
(a) If two participants are chosen at random from SMK ACS, calculate the probability that
both of them are males.
(b) If two participants are chosen from the male participants, calculate the probability that
both of them are from the same school. [5 marks]
Answer:
(a)
42
20
42
20
6
4
7
5



(b)
72
19
144
38
144
62012
12
2
12
3
12
4
12
5
12
3
12
4





[ 2 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 2 ]
[ 1 ]

133. 131
LINES AND PLANES IN 3-DIMENSIONS
1. Diagram 1 shows a pyramid with a horizontal rectangular base ABCD. M is the
mid-point of CD and apex V is 9 cm vertically above the point M
Identify and calculate the angle between the line VB and the base ABCD. [ 4 marks ]
24 cm
14 cm
B
A
D
M
C
V Answer :
Identify VBM [ 1 ]
Length of BM = 25 [ 1 ]
tan VBM =
25
9
[ 1 ]
VST 19.80 or 19 48’ [ 1 ]
2. Diagram 2 shows a right prism. Right angled triangle PQR is the uniform cross-
section of the prism. Calculate the angle between the plane RTU and the plane
PQTU. [ 3 marks ]
Answer :
Identify RTQ [ 1 ]
tan RTQ =
18
12
[ 1 ]
RTQ 33.69 or 33 41’ [ 1 ]
DIAGRAM 1
T
P
5 cm
Q
DIAGRAM 2
R
S
U
18 cm
12 cm

134. 132
3. Diagram 3 shows a prism with a horizontal square base HJKL. Trapezium EFLK is
the uniform cross-section of the prism. The rectangular surface DEKJ is vertical while
the rectangular surface GFLH is inclined.
Calculate the angle between the line DL and the base HJKL [ 4 marks ]
Answer :
Identify DLJ [ 1 ]
Length of LJ = 10 cm [ 1 ]
tan DLJ =
10
5
[ 1 ]
DLJ 26.57 or 2634’ [ 1 ]
4. Diagram 4 shows a right prism . Plane ABFE, ADHE and BCGF are rectangular
planes .
Planes ABCD and EFGH are parallel trapezium. It also given  BAD and
ADC = 90

, FG = 5 cm, AB = 2cm and CG = 12 cm.
Identify and calculate the angle between the plane ADG and AEHD.
[ 3 marks ]
5 cm
4 cm
H
G
FE
D C
BA
Answer :
Identify GDH [ 1 ]
tan GDH =
12
5
[ 1 ]
GDH 22.62 or 22 37’ [ 1 ]
5. Diagram 5 shows a right prism with horizontal rectangle base URST. Right
triangle RSW are the uniform cross section of the prism.
Calculate the angle between plane SRV and plane RSTU [ 3 marks]
Answer :
Identify VST [ 1 ]
tan VST =
12
5
[ 1 ]
VST 22.62 or 2237’ [ 1 ]
8 cm
G
L
H
K
J
D
F
E L
5 cm
6 cm DIAGRAM 3
DIAGRAM 4
DIAGRAM 5
5 cm
10 cm
12 cm
T
U
W
S
R
VV
T
U
W
S
R

135. 133
7. Diagram 6 shows a right prism. The base PQRS is a horizontal rectangle. Right angled
triangle QRU is the uniform cross section of the prism. V is the midpoint of PS.
Identify and calculate the angle between the line UV and the plane PQRS.
[ 4 MARKS]
Answer :
Identify UVR [ 1 ]
Length of VR = 13 [ 1 ]
tan UVR =
13
5
[ 1 ]
UVR 21.04 or 212’ [ 1 ]
8. Diagram 7 shows a right prism. Right angled triangle PQR ia the uniform cross-
section of the prism. Calculate the angle between the line TR and the plane PQTU.
[ 3 marks]
Answer :
Identify RTQ [ 1 ]
tan RTQ =
18
12
[ 1 ]
UVR 33.69 or 3341’ [ 1 ]
DIAGRAM 6
S
Q
U
T
P
10 cm
R
12 cm
5 cmV
18 cm
12 cm
5 cm S
T
U
Q
P
R
DIAGRAM 7

136. 134
12
10
8
6
4
2
-2
-4
-6
-8
-10
-12
y
-4 -2 2 4
y =3x-2
PART B
GRAPH OF FUNCTIONS
1 (a) Complete Table 1 in the answer space for the equation y =
x
6
by writing down
the values of y when x = -2 and x = 0.5 [2 marks]
(b) For this part of the question, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 1 unit on the
y-axis, draw the graph of y =
x
6
for -4 ≤ x ≤ 4. [4 marks]
(c) From your graph, find
a. the value of y when x = -2.7,
b. the value of x when y = 3.6. [2 marks]
(d) Draw a suitable straight line on the graph in (b) to find the values of x which
satisfy the equation 023
6
 x
x
for -4 ≤ x ≤ 4. [4 marks]
Answer:
(a)
X -4 -3 -2 -1 -0.5 0.5 0.8 1.5 2.5 4
Y -1.5 -2 -3 -6 -12 12 7.5 4 2.4 1.5
Table 1 [2]
(b)
x
y
6

[4]
y=3x-2

137. 135
(c) i) y = -2.2 [1]
ii) x = 1.67 [1]
(d) Identify equation of y = 3x – 2 [1]
Straight line y = 3x – 2 correctly drawn [1]
x = -1.2, 1.7 [2]
2. (a) Complete Table 2 in the answer space for the equation y =
x
72
by writing
down the values of y when x = 2.4, x = 6 and x = 12.5 [3 marks]
(b) For this part, use a graph paper.
By using a scale 1 cm to 1 unit on the x-axis and 1 cm to 2 units on the y-axis, draw
the graph of y =
x
72
for 2 ≤ x ≤1 4. [4 marks]
(c) From your graph, find
a. the value of y when x = 4.3,
b. the value of x when y = 22. [2 marks]
(d) Draw a suitable straight line on your graph to find all the values of x which
satisfy the equation 015
2
36

x
x
State these values of x. [3 marks]
Answer:
(a)
X 2 2.4 3 4 6 8 10 11.6 12.5 14
Y 36 30 24 18 12 9 7.2 6.2 5.76 5.14
Table 2
[3]

138. 136
(b)
35
30
25
20
15
10
5
2 4 6 8 10 12 14
[4]
(c) (i) y = 16.7 [1]
(ii) x = 3.3 [1]
(d) Identify equation of y = - x + 30 [1]
Straight line y = - x + 30 correctly drawn [1]
x = 2.7 [1]
3. a) Complete Table 3 in the answer space for the equation y = 2x2
– 5x – 3.
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw
the graph of y = 2x2
– 5x – 3 for -3 ≤ x ≤ 5. [4 marks]
c) From your graph, find
i) the value of y when x = -2.4,
ii) the value of x when 2x2
– 5x – 3 = 0. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which satisfy
the equation 2x2
– 8x = 7 for -3 ≤ x ≤ 5.
State these values of x. [4 marks]
x
y
72

y = - x + 30

139. 137
Answer:
a)
[2]
b) graph
[4 ]
c) i) 20.4 ≤ y ≤ 20.6 [1]
ii) -0.6 ≤ x ≤ -0.4 [1]
2.9 ≤ x ≤ 3.1
d) Identify equation y = 3x + 4 [1]
Straight line y = 3x + 4 correctly drawn [1]
- 0.80 ≤ x ≤ -0.70 [1]
- 4.80 ≤ x ≤ - 4.70 [1]
x -3 -2 -1 0 0.5 1 2 3 4 5
y 30 15 4 -3 -5 -6 -5 0 9 22
40
35
30
25
20
15
10
5
-5
-10
-10 -5 5 10
-0.75 4.75
20.5
-2.4
y x  = 3x+4
y x  = 2x2-5x-3
Table 3

140. 138
4. a) Complete Table 4 in the answer space for the equation y = -2x² + 5x + 8 by writing
down the values of y when x = -2 and x = 3 [2 marks]
b) For this part of question , use the graph paper provided . You may use a flexible
curve ruler.
By using a scale 2 cm to 1 unit on the x axis and 2cm to 5 unit on they axis , draw
the graph of y = -2x² + 5x +8 for -3.5 ≤ x ≤ 4 [4 marks]
c) From the graph, find
(i) the value of y when x = 3.3
(ii) the value of x when y = -16
[ 2 marks]
d) Find and draw a suitable straight line on your graph to determine the values of x
which satisfy the equation 2x – 2x ² = -10 for -3.5 ≤ x ≤ 4
State the values of x [4 marks]
Answer :
a) y = -2x² + 5x + 8
x -3.5 -3 -2 -1 0 1 2 3 4
y -34 -25 -10 1 8 11 10 5 -4
[2]
b) Graph
15
10
5
-5
-10
-15
-20
-5 -4 -3 -2 -1 1 2 3 4 5
y= 3x-2
y=-2x2+5x+8
[4]
c) i) 2.5 ≤ y ≤ 2.7 [1]
ii) -2.45 ≤ x ≤ -2.40 [1]
Table 4

141. 139
d) Identify equation y = 3x - 2 [1]
Straight line y = 3x -2 correctly drawn [1]
-1.8 ≤ x ≤ - 1.7 [1]
2.7 ≤ x ≤ 2.8 [1]
5. a) Complete Table 5 in the answer space for the equation y = x3
– 13x + 18 .
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the
graph of y = x3
– 13x + 18 for -4 ≤ x ≤ 4. [4 marks]
c) From your graph, find
i. the value of y when x = -1.5,
ii. the value of x when y = 25. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which satisfy the
equation x3
– 11x – 2 = 0 for -4 ≤ x ≤ 4. State these values of x. [4 marks]
e)
Answer:
(a)
X -4 -3 -2 -1 0 1 2 3 4
Y 6 30 36 30 18 6 0 6 30
[2]
c) i) y=34, 33 ≤ y ≤ 35 [1]
ii) x=3.85 , 3.80 ≤ x ≤ 3.90 [1]
45
40
35
30
25
20
15
10
5
-5
-6 -4 -2 2 4 6
y=x^3-13x+18
y=-2x+20
3.85
34
3.35
-0.25
-3.2
[4]
Table 5

142. 140
d) y=x3
-13x+18
0=x3
-11x-2 (-)
--------------------------
y = -2x+20 [1]
3.3 ≤ x ≤ 3.5 [1]
-0.3 ≤ x ≤ -0.1 [1]
-3.3 ≤ x ≤ -3.1 [1]
6. a) Complete Table 6 in the answer space for the equation y = x3
+ x2
– 12x – 5.
[2 marks]
b) For this part, use a graph paper.
By using a scale 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw
the graph of y = x3
+ x2
– 12x – 5 for -4 ≤ x ≤ 4. [4 marks]
c) From your graph, find
i. the value of y when x = 0.5,
ii. the value of x when y =11.9. [2 marks]
d) Draw a suitable straight line on your graph to find all the values of x which
satisfy the equation x3
+ x2
– 10x = 0 for -4 ≤ x ≤ 4.
State these values of x. [3 marks]
Answer:
(a)
X -4 -3 -2 -1 0 1 2 3 4
Y -5 13 15 7 -5 -15 -17 -5 27
[2]
20
15
10
5
-5
-10
-15
-20
-25
-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6
h x  = -2x-5
g x  = x3+x2 -12x-5
A
[4]