2jam bhn per. pmr 2010

SEMINAR KECEMERLANGAN
PMR • Tahun 2010

HAJI LAMSAH BIN ARBAI @ BAI
PPT.,PJK
SMK PENGKALAN PERMATANG
KUALA SELANGOR, SELANGOR.

USUCCESS
PLAN FOR
IT.
WILL BE YOURS IF….
DREAM FOR
IT.
WORK FOR
IT.
RACE FOR
IT.

Objectives of the Seminar is to:Objectives of the Seminar is to:
Make sure, you understand theMake sure, you understand the
requirements of the PMRrequirements of the PMR
Mathematics Examination.Mathematics Examination.
Show the simplest steps to solveShow the simplest steps to solve
problems with accurate answer.problems with accurate answer.
Understand the marking scheme.Understand the marking scheme.

GIVE ATTENTION TO:
1. Interpretation of the answer:
(a) (-3)(-2) 6≠
(b)
2
3
2
3
≠
−
−
(d)
0
5
0
≠
(e)
3
1
3
≠(c)
2
3
2
3
≠
−
−

1. Interpretation of answer:
(f)
2
1
8
4
≠
Except in certain cases
(g) 5.3
2
1
3
2
7
== Except in certain cases
(h)
823
≠
(i)
x
yx
x
yx 32
2
64 −
≠
−

THE CORRECT WAY TO WRITE THE FINAL
ANSWER.
24
15
=
8
5
2. 62
4.
3. xx =1
x
x
=
1
5.
yx
xyx
24
186 2
−
=
( )
yx
xyx
24
36 −
y
xy
4
3−
=
1.
= 36
6. 10
=x

THE WRONG WAY TO WRITE THE FINAL
ANSWER
147 −x1. = ( )27 −x
2.
7
15 no need to change to
7
1
2
3.





−
6
4





 −
6
4
( )6,4−≠≠

Things to take note of…
Do NOT use liquid paper;
a) Old words will appear
b) Forget to write the
correct answer
Don’t spend too
much time on
one question.
Before starting, quickly
glance through all the
questions
(Objective: 40, Subjective:
20)
For subjective
questions, show
working
systematically!
Don’t wait till
the last minute
to shade the
answers
(objective)

Questions se-tup
• Objective questions with multiple
choices
(Four choices of answers)
• Choose one correct answer
• Darken/shade the prepared
answer space in objective paper
immediately.

How to get the answer?
• Mental arithmatic
• Do direct calculation
• By using the given answer, work
backwards
•The method of calculationThe method of calculation
need notneed not be shownbe shown..

Objective question answering
skills
SUPERSTEP
TO
SUCCESS!!!

Simplify
1. ( – 4x + 5y ) – (8x – 3y) =
A. 4x – 8y
B. 8y – 12 x
C. 8y + 12x
D. 9x – 5y

SIMPLIFY BY OPENING THE
BRACKET..
( − 4X + 5Y − 8X + 3Y)
−12X + 8Y
REARRANGE
8Y – 12 X
ANSWER B.

What is the answer?
2. ( 4h – 7)2
=
A . 16h – 49
B . 16h2
+49
C . 8h2
−12h + 49
D . 16h2
− 56h + 49

• Simplify the multiplication
• ( 4h – 7)2
=
( 4h – 7)(4h – 7)
16h2
– 28h – 28h + 49
16h2
– 56h + 49
Answer: D
SQUARED THE BRACKET
1
16h2
2
-28h
3
-28h
4
+49

3. Simplify to the most
simplest form.
(2m( m – 5) + 7m)
A. 2m2
+ 2m − 5
B. 2m2
− 12 m
C. 2m2
− 3m
D. 2m2
+ 2m

Simplify the multiplication
2m( m – 5) + 7m
Open the bracket …
2m2
– 10m + 7m
2m2
– 3m
The answer is C.

Simplify this question..
4). ( x – 3y)2
+ 7xy =
A. x2
(x – 2) + xy
B. x2
+7xy – 9y2
C. x2
+ 7y2
+ 8xy
D. x2
+ 9y2
+ xy

Solution …
4) ( x – 3y)2
+ 7xy =
( x – 3y) ( x – 3y) + 7xy
x2
− 3xy −3xy + 9y2
+ 7xy
x2
+ 9y2
− 6xy + 7xy
= x2
+ 9y2
+ xy
answer D.

5 . The difference between median
and mode in the number row
2,4,5,4,3,2,2,5,7 is
A. 1
B. 2
C. 3
D. 4

Solution.
Determine the median for
2, 2, 2, 3, 4 , 4, 5, 5, 7
The median is 4.
And mode is 2
the difference between median
and mode is 2
answer : B

1. 923 758 becomes 924 000
after it is rounded off to
A one
B ten
C hundred
D thousand

923 758
3 000 3 758 4 000
Answer D

2. Diagram 2 shows some of the factors of
270. The possible value of y is
A 4
B 6
C 8
D 12
2
3y
9
10
270
DIAGRAM 2
15
270
6
= 45

3. The water level in a container is 2.0 m on
Sunday. The water level drops by 25 % on
Monday. It rises by 40% on Tuesday as
compared to Monday’s level. What is the
height, in m, of the water level on
Tuesday?
A 0.9
B 1.2
C 1.9
D 2.1

Sunday = 2.0m
Monday = 2.0 – (25% × 2.0)
= 2.0 – 0.5
= 1.5
Tuesday = 1.5 + (40% × 1.5)
= 1.5 + 0.6
= 2.1

3
2
2
1
4. A group of students consists of 42 boys and 70 girls.
of the boys and
Calculate the percentage of students who attended
the youth camp.
A 43.75
B 56.25
C 60.00
D 62.50
of the girls attended a youth camp.

Number of students attended the
youth camp
= ( x 42 boys) + ( x 70 girls)
3
2
2
1
Therefore % of students attended
the youth camp
=
63
112
x 100 = 56.25 %
= 28 boys + 35 girls = 63 students
Total number of the student = 112

5. Diagram 5 shows four rectangles drawn on a
square grid.
Among A, B , C and D , choose the
rectangle with the smallest fraction shaded.
DIAGRAM 5
A
B C
D

A. 4
12
=
1
3
B. 4
16 =
1
4
C. 4
20
= 1
5
D. 8
24
=
1
3

6. Diagram 6 shows the change in the reading of a
weighing machine when two cakes of equal mass
are removed.
Find the mass, in g, of each cake that is removed
A 42
B 85
C 425
D 850
0.35 kg 0.18 kg
2 cakes are removed
DIAGRAM 6
0.35 kg 0.18 kg
0.35kg – 0.18kg = 0.17kg
0.17kg ÷ 2 = 0.085kg
0.085kg × 1000 = 85 g

7. In Diagram 7, PQST and RSUV are rectangles. R
and U are the midpoints of QS and ST
respectively.
R
P
Q
V U
S
T
11 cm
8 cm
DIAGRAM 7
Find the area of the shaded region,
A 44 cm2
B 55 cm2
C 66 cm2
D 88 cm2
V

Area of shaded region
= area of PQT + RSUV
Area of PQT
= ½ × 11cm × 8cm = 44cm2
Area of RSUV = 4cm × 5.5cm = 22cm
Area of shaded region = 44cm2
+ 22cm2
= 66cm2

8. In Diagram 8, PQUV is a square and RSTU is a
rectangle. T
S
R
U
QP
V
7 cm
5 cm
DIAGRAM 8
Given that PR = RS, perimeter of the whole diagram is
A 37 cm
B 47 cm
C 57 cm
D 67 cm

T
S
R
U
QP
V`
7 cm
5 cm
10cm
10cm
5cm 5cm
5cm
Perimeter = 5cm + 5cm + 5cm + 5cm + 10
+ 7 + 10 = 47cm

9. Given that the mean of 5, 7, 6, 3,
p, 8 and 7 is 6, find the value of p is
A. 5
B. 6
C. 7
D. 8

5 + 7 + 6 + 3 + p + 8 + 7
7
= 6
36 + p = 6 x 7
36 + p = 42
p = 42 – 36
p = 6

53o
32o
yo
O
P
Q
R
DIAGRAM 10
10. Diagram 10 is a circle with center O. PQR is a straight line.
The value of y is
A 53o
B 69o
C 74o
D 85o
90o
– 32o
= 58o
yo
= 180o
– (58o
+ 53o
)
= 69o

DIAGRAM 11
11 Diagram 11 shows a set of numbers.
7, 3, 6, 9, 8,
4, 2, 3
The difference between mode and median of the
above numbers is
A 0.25
B 0.50
C 1.00
D 2.00
mode = 3
median =
2 , 3 , 3 , 4 , 6 , 7 , 8 , 9
4 + 6
2
= 5
median – mode = 5 – 3
= 2

If you are smart enough,If you are smart enough,
nothing can be a problem.nothing can be a problem.

Table shows the time allocation forTable shows the time allocation for
a test.a test.
TestTest Time AllocationTime Allocation
Paper 1Paper 1 11 ¼ hours¼ hours
BreakBreak 20 minutes20 minutes
Paper 2Paper 2 1 ½ hours1 ½ hours
All candidates must be in theAll candidates must be in the
examination hall 10 minutes beforeexamination hall 10 minutes before
Paper 1 starts. Paper 2 ends at 1.05Paper 1 starts. Paper 2 ends at 1.05
p.m. At what time must thep.m. At what time must the
candidates be in the hall beforecandidates be in the hall before
paper 1 starts?paper 1 starts?
A 9.35 a.m.A 9.35 a.m. C 10.00 a.m.C 10.00 a.m.
B 9.50 a.m.B 9.50 a.m. D 10.10 a.m.D 10.10 a.m.
CCT Question

SOLUTIONSOLUTION
Convert 1.05 p.m.Convert 1.05 p.m.
in 24-hour system.in 24-hour system.
Find the durationFind the duration
of the papersof the papers
including break.including break.
Subtract it.Subtract it.
Write the answer.Write the answer.
1.05 + 1200 =1.05 + 1200 =
0105 + 1200 = 13050105 + 1200 = 1305
1¼ hours + 20 mints +
1½ hours + 10 mints =
3hrs 15mins = 0315
1305 – 0315 = 09501305 – 0315 = 0950
That is,That is,
9.50 a.m.9.50 a.m.
To solve, work backwards.To solve, work backwards.
Answer:Answer: BB

In diagram 16, PQRS is aIn diagram 16, PQRS is a
circle with centre O and PSTcircle with centre O and PST
is a straight line. Find theis a straight line. Find the
value ofvalue of xx..
AA 1515oo
CC 4545oo
BB 3535oo
DD 5050oo
85 °
x °
100°
O
T
P
Q
S
R

Knowing that angles subtended at the circumferenceKnowing that angles subtended at the circumference
by an arc are half the one at centre,by an arc are half the one at centre, ∠∠ PPQSQS == ½½∠∠ POSPOS..
ThereforeTherefore ∠∠ PQSPQS == 5050oo
..
SinceSince PQRSPQRS is cyclic quadrilateral,is cyclic quadrilateral,
∠∠PQRQR == ∠∠RSTRST = 85= 85oo
Therefore,Therefore, ∠∠SQRSQR == ∠∠PQRPQR −− ∠∠PQSPQS
= 85= 85oo
−− 5050oo
== 3535oo
SOLUTIONSOLUTION
Answer:Answer: BB
85 °
x °
100°
O
TP
Q
S
R

I swear!I swear!
I didn'tI didn't
use theuse the
calculator.calculator.

QUESTION FORMAT
• Answers must be written in the
space provided in question
paper.
• Marks given based on steps
made during calculation and
accuracy of answer.

If a question needs candidates
to show the proper solving steps
..,
BUT
Only the final answer written in
the space provided...
NO
Marks will be given, because
it is assumed candidates did
not Do calculation.

• A. Calculate the value of
6(57 ÷ 3 − 4) + 51
Solution:
6(19 − 4) + 51
6(15) + 51 K1
90 + 51
141 N1

Solve each of the following linear
equation
a) 2q = −5q −14
b) 3(4m − 1) = 2m + 5
Solution:
a) 2q + 5q = −14
7q = −14
q = −2 P1

• Factorise completely
• b ) p(5 – x) + 2q(x – 5)
5p – px + 2qx – 10q
5p – 10q – px + 2qx
5( p – 2q)− x( p −2q)
(p – 2q)(5 – x)
K1
K1

Given p – 2 = q2
+ 3,
express q in the term of p
• Rearrange the expression
q2
+ 3 = p – 2
q2
= p – 2 – 3
q = 32 −−p
K1
Not
complete yet
5−pq = N1

Paper 2 topics with ‘BIG’ marks !!!!
TOPIC MARKS
1. Algebraic Expressions
(Form1,2 & 3)
i. Expansion
ii. Factorization
iii. Algebraic Fractions
iv. Algebraic formulae
2 marks
3 marks
3 marks
3 marks
2. Loci in Two Dimensions
(Form 2)
5 marks
3. Transformations
1 or 2 questions
4 to 6 marks

4. Statistics (Form 2 & 3)
~ pictograph (2004), bar chart
(2004), pie chart (2005), line
graph (2006),bar chart (2007)
4 marks
5. Geometrical Constructions
(Form 2) 6 marks
6. Graph of Functions
(Form 3)
4 marks
TOTAL (approximately) : 35 marks

FRACTIONSFRACTIONS
4 basic operations
Simplifying
Involving LCM
Transferring the answer

1. Calculate the value of






+÷
5
2
3
1
1
5
3
2







+÷
5
2
3
1
1
5
3
2








+÷=
5
2
3
4
5
13






÷=
15
26
5
13
N1
K1
2
1
1=2/3
130
195
26
15
5
13
=





×=







−×−
4
3
5
1
3
2
1
and express the answer as a
fraction in its lowest term.
[2 marks]







−×−
20
11
3
5






12
11
K1
N1

Calculate the value ofCalculate the value of
5
3
2
5
2
3
1
2 ÷





−
and express the answerand express the answer
as a fraction in itsas a fraction in its
lowest term.lowest term.

13
5
15
6
15
35
×





−
5
3
2
5
2
3
1
2 ÷





−
5
13
5
2
3
7
÷





−
STEPS TO BE SHOWNSTEPS TO BE SHOWN
13
5
15
29
×
39
29
=
=
=
=
1
3
K1
N1

4
1
1
16
4
5
15 −÷
1.Calculate 15 ÷ + ( −16)
5
4
15 × − 16
12 − 16
− 4
K1
N1

DECIMALSDECIMALS
 Place value
 4 basic operations
 Change decimal to fractions and
vice versa
 Bracket operations

)
4
3
1(2.004.6 −−÷−1. Calculate
75.1
2.0
04.6
+
−
75.1
2
4.60
+−
75.12.30 +−
45.28−
K1
N1

)
5
2
1(3.043.5 −−÷3. Calculate
4.1)3.043.5( +÷
4.11.18 +
19.5
K1
N1

Calculate the value and express the answer
correct to two decimal places
0.456
6
1
12 ×−− 







)0.076-(-12=
0.07612+=
N1
K1
12.076=
12.08=

Calculate the value and express the answer
correct to two decimal places
( ) 0.33.12
5
2
2 ÷−−
0.33.122.4 ÷+=
10.42.4 +=
N1
K1
12.80=

3
4
(6 − 0.24) ÷2. Calculate
Correct to two decimal places
5.76 ×
4
3 K1
1.44 × 3
= 4.32 N1

SQUARES, SQUARE ROOTS
• CUBES AND CUBES ROOTS

1. a. Find the value
9
4
5
9
49
3
7
or
3
1
2 P1

1. b. Calculate the value 3
064.0 - (-1)2
14.0 − K1
6.0− N1

2. a Find the value of
36.0
6.0= P1
b. Calculate
23
)275( −+
2
)]3(5[( −+
4
K1
N1

3a. Find the value of 3
0.3
0.30.30.3 ××=
0.027= P1

64)2( 3
×−3.a. find the value of
88×−
64− P1
b. Calculate
9
7
14 −
3
4
4 −
3
8
K1
N1

1. a. Simplify
2435
)()( qq ÷−−
815−
q
7
q= P1
b.
2739 2
=× −x
322
333 =× −x
322 =−+ x
3=x
K1
N1

2. a. Simplify
542
)( mm ÷
58−
m3
m P1
b. 2
2216 =× m
24
222 =× m
24 =+ m
2−=m
K1
N1

3. a .simplify
10
4433
4
)()(16
xy
yx
10
169
4
16
xy
yx
=
68
4 yx P1

b.
x
x
5
125
5 2
=−
x
x
5
5
5
3
2
=−
xx −−
= 32
55
xx −=− 32
52 =x
2
5=x
K1
N1

ALGEBRAIC EXPRESSIONS
• EXPANSION
• FACTORIZATION

8. Simplify
(a) 2(n + 5) − 3
= 2n + 10 − 3
= 2n + 7 N1
(b) 3(4m – 3k) – (5k – m)
12m – 9k – 5k + m
13m – 14k
K1
N1

yy 255 2
+−
prpstqrqst 8383 +−+−
1. Factorise completely a.
b.
)5(5)5(5 +−−− yyoryy P1
prqrpstqst 8833 ++−−
))(83( pqrst ++−
K1
N1
)(8)(3 pqrpqst +++−

qrpq 64 +
)5(4 ++ xx
2, Factorise completely a.
b.
)32(2 rpq + P1
xx 54 2
++
452
++ xx
)4)(1( ++ xx
K1
N1

)2(45 −−− kk
)1(4)2( 2
−+− yy
3. Simplify to its simplest form a.
b.
845 +−− kk
k98 − P1
44442
−++− yyy
2
y
K1
N1

77 Factorise completelyFactorise completely
(a) 2y + 6(a) 2y + 6
(b) 12 – 3x(b) 12 – 3x22

7 (b) 12 – 3x7 (b) 12 – 3x22
=3(4 – x=3(4 – x22
))
=3(2 + x)(2 – x)=3(2 + x)(2 – x)
7 (a) 2y + 67 (a) 2y + 6
= 2(y + 3)= 2(y + 3)P1
K1
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN
N1

88 ExpandExpand
(a) q(2 + p)(a) q(2 + p)
(b) (3m – n)(b) (3m – n)22

8 (b) (3m – n)8 (b) (3m – n)22
= (3m)= (3m)22
– 2(3m)(n)+(n)– 2(3m)(n)+(n)22
= 9m= 9m22
– 6mn + n– 6mn + n22
K1
N1
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN
8 (a) q(2 + p)8 (a) q(2 + p)
= 2q + pq= 2q + pq P1

Factorize completelyFactorize completely
55hh((kk – 3) – 2(3 –– 3) – 2(3 – kk))

55hh((kk – 3) – 2(3 –– 3) – 2(3 – kk))
55hh((kk – 3) – 6 + 2– 3) – 6 + 2kk
55hh((kk – 3) + 2– 3) + 2kk – 6– 6
55hh((kk – 3– 3) + 2() + 2(kk – 3– 3))
(5(5hh + 2)(+ 2)(kk – 3)– 3)
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN
K1
N1

qq 4
1
3
4
−1.
qq 12
3
12
16
−
q12
13
K1
N1

2.
)2(3
4
2
1
qpqp +
+
+
)2(3
43
qp +
+
)2(3
7
qp +
K1
N2

ExpressExpress
as a single fractionas a single fraction
in its simplest form.in its simplest form.
n
w
n 6
32
3
5 −
+

( )
( ) n
w
n 6
32
32
52 −
+=
n
w
n 6
32
6
10 −
+=
n
w
n 6
32
3
5 −
+
n
w
6
3210 −+
=
n
w
6
312 −
=
n
)w(
6
43 −
=
n
w
2
4 −
=



SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN
1
2
K1
K1
N1

17. Express as a single fraction in its
simplest form
6b
2b
3b
2 −
−
6b
b-6
=
K1
K1
N1
b6
2b
b3x2
4 −
−=
6b
2b-4 +
=

17. Express as a single fraction in its simplest form.
9ab
b2
3a
2 −
−
9ab
)b-2(-3b2×
=
9ab
b2-6b +
=
9ab
2-b7
=
N1
K1
K1

LINEAR EQUATION
• SOLVE THE VALUE

pp =−− 6)3(4
5
73
2
k
k
−
=
1. a.
b.
pp =−− 6124
183 =p
6=p
kk 7310 −=
317 =k
17
3
=k
K1
N1
K1
N1

432 −= nn
3
4
23
−=
−
n
n
2. a
2.b.
4−=− n
4=n P1
12423 −=− nn
21243 +−=− nn
10=n
K1
N1

6435 +=− xx
31)46(
2
3
−=−+ ff
3 a.
3.b.
3645 +=− xx
9=x
6212182 −=−+ ff
186210 −−=− f
8010 =f
8=f
P1
K1
N1

kk −−= 14
9)1(3 =+− pp
4. a.
b.
142 −=k
7−=k P1
933 =−− pp
122 =− p
6−=p
K1
N1

LOCUS IN TWO DIMENSION
• DRAW THE LOCUS
• MARK THE INTERSECTION

6. Loci in Two Dimensions.
i) A locus must be drawn in a full line.
Do not draw dotted line.
ii) If the locus is a straight line, use a
ruler to draw.
iii) If the locus is a circle or part of a
circle, use a pair of compasses to
draw.

6. Loci in Two Dimensions.
iv) Draw the locus fully on the given
diagram.
v) Label the locus correctly.
vi) Mark the intersection(s) of the 2
loci correctly.

LOCI IN TWO DIMENSIONS .
A B
i). Always equidistant from two points (AB)
ii). Always equidistant from PR
P Q
R
S
1) e.g

Always equidistant from a point
e.g
i) The locus of X such that XJ = JM
J K
LM
X
2)

ii) The locus of Y such that QY = 4cm
4cm
6 cm
P Q
RS
Y

Always equidistant from a line or two
lines.
e.g.
i) Locus K is 3 cm from AB
3.
A
B
3 cm 3 cm
K

ii) Always equidistant 2 lines.
eg. Locus W is always equidistant from
PQ and RS or PW = WS
P Q
RS
W

2. Diagram 5 in the answer space shows a rectangle PQRS and
SP = 2cm. W and Y are two points moving in the diagram.
On the diagram, draw
(a) the locus of point W such that WS = WP,
(b) the locus of point Y such that QY = 1.5 cm,
(c) mark with all the possible points of intersection between
locus W and locus Y.
[5 marks]
⊗
S R
QP
DIAGRAM 5
(2)
(2)
(1)

2. Diagram 5 in the answer space shows a rectangle PQRS and SP = 2cm.
W and Y are two points moving in the diagram.
On the diagram
(a) construct locus W such that WS = WP,
(b) construct locus Y such that QY = 1.5 cm,
[5 marks]
⊗
S R
QP
DIAGRAM 5
W
Y
⊗
(2)
(2)
(1)
3cm

Diagram 8 in the answer
space shows four isosceles triangles
PMQ, QMR, RMS and PMS.
W, X and Y are three moving points
in the diagram.

15.a. W is a moving point such that it
is equidistant from points S and Q.
b. X is a moving point such that
it is 2 cm. from point M
c. Y is a moving point that is
equidistant from line PQ and SR.
hence, mark and label all the
intersection points of locus X and
locus Y

15. a. State the locus of W
by using the letters in the diagram
b. Draw the locus of X and locus Y
P Q
R
M
S
Straight line PMR/PR
locus X
Locus Y
K1
P2
N1
K1

TRANSFORMATIONS
• TRANSLATION
• ROTATION
• REFLECTION
• ENLARGEMENT

TRANSFORMATIONSTRANSFORMATIONS
Describe the transformations
Writing technique
Spelling

12
10
8
6
4
2
0 2 4 6 8 10 12
P’
P
P’P’ is the image ofis the image of PP under transformationunder transformation MM..
Describe in full transformation M.Describe in full transformation M.

12
10
8
6
4
2
0 2 4 6 8 10 12
PP
P’
TranslationTranslation 




−
5
4
 
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN

K1
N1
5. Draw and label of 90o
anticlock wise
rotation from origin
A’
5 B
4
3
2
1
54321-1-2-3-4-5
-1
-2
-3
-4
D’
C’B’
C
D
A

2
5
R
Q
P
T
DIAGRAM 2
Diagram 2 in the answer space shows triangle PQR drawn on
a grid of equal squares. Draw the image of triangle
PQR under a clockwise rotation of 90o
with T as the centre of
rotation.
[2 marks]

In Diagram 2, the kites ABCD
in the answer space was drawn
on a grid of equal squares.
On the diagram, draw and label
A’B’C’D’ the image of ABCD
under a reflection at line MN

Diagram 4 in the answer space
shows triangle PQR
drawn on a grid of equal squares.

5. Draw and label the image
M
N
K1
D
C
B
N1
A
A’
B’ C’
D’

6. Draw and label the image under
reflection on the line MN
K1
M
N
P
P’
N1

Transformations
(iii) Reflection
Eg : Diagram 3 in te answer space shows two
quadrilaterals, ABCD and A’B’C’D’, drawn on a grid of
equal squares. A’B’C’D’ is the image of a reflection. On the
diagram in the answer space, draw the axis of reflection.
Answer :
A
B
C D
A’
B’C’
D’
DIAGRAM 3
[2 marks]

(iii) Reflection
Answer :
A
B
C D
A’
B’C’
D’
DIAGRAM 3
[2 marks]

Diagram 2 in the answer space
shows two quadrilaterals. JKLM and
J’K’L’M’, drawn on the grid of equal
squares. J’K’L’M’is the image of
JKLM under an enlargement. Mark
the P as the center of enlargement

Question 6
K’
J’
M’ L’
P2
J
K
L
M
K’
L’
M’
J’

Question 6
K’
J’
M’ L’
P
P2

y
x
1
2
3
4
5
6
0–1 1 2 3 4 5 6
A’
A
B’
B
D
C’
C
• K
E
•K’

On the grid, draw the
image of triangle PQR
Under an enlargement with
scale factor 2 at centre M

K1
N1
7. Draw an enlargement image
P’
Q
R’
Q’
R
M
P

21. Diagram 2, in the answer space shows a
triangle L drawn on a grid of equal
squares.
On the diagram in the answer space,
draw the image of triangle L under
an enlargement with centre Q and scale
factor 3.
[2 marks]

Q
DIAGRAM 2
L
•
L’
•
• •

CONSTRUCTIONS
• CONSTRUCT THE DIAGRAM
• FOLLOW THE INSTRUCTIONS

Set square and protractor cannot be used to answer thisSet square and protractor cannot be used to answer this
questionquestion
Diagram 5 in the answer space shows a straight lineDiagram 5 in the answer space shows a straight line PQPQ..
a)a) Using a pair of compasses and ruler, constructUsing a pair of compasses and ruler, construct
ι.ι. ∆∆PQRPQR withwith PQPQ = 8 cm,= 8 cm, PRPR = 5 cm and= 5 cm and ∠∠RPQRPQ = 60= 60oo
ii.ii. bisector ofbisector of ∠∠ PQRPQR
b)b) Based on the diagram constructed, measure the distanceBased on the diagram constructed, measure the distance
between the pointbetween the point QQ and the straight lineand the straight line PRPR..
QP

QP
R
S
d) QS = 7.05 cmd) QS = 7.05 cm
6060oo
Set square and protractor cannot be used to answer this questionSet square and protractor cannot be used to answer this question
Diagram 5 in the answer space shows a straight lineDiagram 5 in the answer space shows a straight line PQPQ..
a)a) Using a pair of compasses and ruler, constructUsing a pair of compasses and ruler, construct
ι.ι. ∆∆PQRPQR withwith PQPQ = 8 cm,= 8 cm, PRPR = 5 cm and= 5 cm and ∠∠RPQRPQ = 60= 60oo
ii.ii. bisector ofbisector of ∠∠ PQRPQR
b)b) Based on the diagram constructed, measure the distanceBased on the diagram constructed, measure the distance
between the pointbetween the point QQ and the straight lineand the straight line PRPR..
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN

19. a (i). Construct triangle ABC
(ii). Construct AD line
b (i). Measure the angle of ABC
(ii). measure AD
1
1
1
1
1
1
6 cm 8 cm
AD = 5.9±0.1 cm
A
B C
D
ABC = 740
±1

STATISTICS
• CONSTRUCT PIE CHART
• CONSTRUCT BAR CHART
• CONSTRUCT LINE GRAPH
• CONSTRUCT PICTOGRAM

Construct a pie chart for the
above data.
The following data shows the
number of stamps each
student have.
Abu Bakar Chua David Ellan Fadil
30 50 20 40 60 40
STATISTICSSTATISTICS

First, find the value of angles that
represents the number of stamps
according to the table.
Abu
0
360
240
30
×⇒
Bakar
Chua
David
Ellan
0
360
240
50
×⇒
0
360
240
20
×⇒
0
360
240
40
×⇒
0
360
240
60
×⇒
Fadil
0
360
240
40
×⇒
= 45o
= 75o
= 30o
= 60o
= 90o
= 60o

Using a protractor, draw the angles andUsing a protractor, draw the angles and
label the sectors on the pie chart.label the sectors on the pie chart.
45o
90o
60o
75o
60o
30o
Abu
Bakar
Ellan
Fadil
David
Chua
Title : Number of stamps each student have.

Grade A B C D E
Number of
Stud 25 38 12 15 10
2.

GRADE
Number of
Students
5
10
15
20
25
30
35
40
A B C D E

GRAPH OF FUNCTION
• DRAW THE AXIS
• MARK THE LOCATIONS
• JOIN THE POINTS

1. Use the real graph paper not a square grid1. Use the real graph paper not a square grid
paper.paper.
2.2. Always remember that the big squareAlways remember that the big square
of the graph paper measures 2 cm.of the graph paper measures 2 cm.
3.3. Identify the lowest and the highest value ofIdentify the lowest and the highest value of
the x-axis and y-axis.the x-axis and y-axis.
4.4. Label the x-axis and the y-axis correctly andLabel the x-axis and the y-axis correctly and
uniformly.uniformly.
5.5. Relate the number on the value table withRelate the number on the value table with
coordinates. [eg. ( - 3,15), (-2,5) ….]coordinates. [eg. ( - 3,15), (-2,5) ….]
6.6. Then mark all the points correctly using smallThen mark all the points correctly using small
cross or dots.cross or dots.
7.7. Draw smooth curve that passes through all theDraw smooth curve that passes through all the
GRAPH OF FUNCTIONS.

Use graph paper provided to answer thisUse graph paper provided to answer this
question.question.
Table 2 shows the values of twoTable 2 shows the values of two
variablesvariables xx andand yy, of a function., of a function.
x -3 -2 -1 0 1 2 3
y
-
32
-
11
-2 1 4 13 34
Draw the graph of the functionDraw the graph of the function
using a scale of 2 cm to 1 unit atusing a scale of 2 cm to 1 unit at
thethe xx-axis and 2 cm to 10 units-axis and 2 cm to 10 units
at theat the yy –axis.–axis.

X
y
10
-10
20
30
40
-20
-30
-40
- 3 -2 -1 0 1 2 3
⊗
⊗
⊗
⊗
⊗
⊗
⊗
1.

-3 -2 -1 0 1 32 x
y
×
×
×
×
×
×
× -10
-20
10
-30
20
30
40
-40
SS
TT
EE
PP
SS
TT
OO
BB
EE
SS
HH
OO
WW
NN

x -3 -2 -1 0 1 2.5 4 5
y 15 5 -1 -3 -1 9.5 29 47
2.

-3 -2 -1 0 1 32 4 5 x
×
×
×
×
×
×
y
35
10
5
15
-5
20
25
40
-10
30
45
50
×
×

X
Y
-4
-27
-2.5
-7.5
-1
3
0 1 2 3 4 5
5 3 -3 -13 -27 -45

y
-10
-35
-40
-30
-45
-25
-20
-5
-50
-15
5
10
-3 -2 -1 0 1 32 4 5 x-4-5
×
×
×
×
×
×
×
×
×

-3 - 2 -1 0 1 2 3 4
y
4
8
12
×
× ×
×
× ×
×
×

SOLID GEOMETRYSOLID GEOMETRY
 Net and layout
 Surface area
 Volume of solid geometry and
combinations
 Basic characteristics

6 . Diagram 3 shows a prism
with a rectangle base.
Draw a full scale net of the prism
on the grid in the answer space.
The grid has equal squares with sides of 1 unit.
[3 marks]
DIAGRAM 3
3 units
4 units
8 units

7. Draw the net of the prism
8 unit
3 unit 5 unit4 unit
K2
N1
3 unit
3 unit

7. Solid Geometry
Eg : Diagram 4 shows a cuboid.
DIAGRAM 4
Draw a full scale the net of the pyramid on the grid in the answer
space. The grid has equal squares with sides of 1 unit.
1 unit
2 units
3 units

WHAT YOU THINK IS WHAT YOU
ARE
Notes:
All the guides given is not a guarantee that you will
achieve an excellent result in mathematics. What
is important is that you have the determination to
succeed, prepared to work hard and consistently
practise past year’s questions and forever asking
when confused.
This is the real key to success.
You determine your own success.

daripadadaripada
Lamsah bin Arbai @ BaiLamsah bin Arbai @ Bai PPT. PJKPPT. PJK
 603 - 32893611
 013 - 3464160
e- lamsaharbai@yahoo.com

semalam,semalam, hari inihari ini dan esokdan esok

semalam,semalam,
Semalam …
Telahku campakkan benih kekecewaan
Ku lontarkan segala kedukaan
Dan segala kesedihan
Lalu kujadikan sebuah kenangan
Dan pengalaman

hari inihari ini
Hari ini …
Ku taburkan benih harapan
Ku semaikan dengan baja impian
Ku berdiri dengan semangat perjuangan
Aku tak ‘kan kalah dengan pujukan
Itulah keyakinan di hati
Yang telah ku tanamkan
Kerana ku tahu
Hari ini adalah kenyataan

dan esokdan esok
Esok …
Akanku kutip segala semaian
Akanku buktikan segala lelahan
Yang tak pernah kenal erti kepayahan
Yang ku pasti esok adalah harapan
Harapan yang menuntut
Sebuah kenyataan

REMEMBERREMEMBER
Excellent students make everyone
feels that he or she makes a
difference to the success of the
school.
A student is best, when people
barely know he scores.
A real student faces the music,
even when he doesn’t like the
tune.

SEKIAN TERIMA KASIHSEKIAN TERIMA KASIH
daripadadaripada
Lamsah bin Arbai @ BaiLamsah bin Arbai @ Bai PPT. PJKPPT. PJK
smk pengkalan permatang
45000 kuala selangor
selangor darul ehsan

The allegory of the Frog ...

Lesson of Life N. 1

Once upon a time there was a race ...
of frogs

The goal was to reach the top of a high tower.
Many people gathered to see and support them.

In reality, the people probably didn’t
believe that it was possible that the frogs
reached the top of the tower, and all the
phrases that one could hear were of this
kind :
"What pain !!!
They’ll never make it!"

The frogs began to resign,
except for one who kept on climbing
The people continued :
"... What pain !!! They’ll never make it!..."

And the frogs admitted defeat,
except for the frog who continued to insist.
At the end, all the frogs quit, except the
one who, alone and with and enormous
effort, reached the top of the tower.
The others wanted to know how did he
do it.

And discovered that he...
was deaf!
...Never listen to people who have the
bad habit of being negative...
because they steal the best aspirations
of your heart!

Always remind yourself of the power of the
words that we hear or read.
That’s why, you always have to think
positive
POSITIVE !!!!
Conclusion:
Always be deaf to someone who tells
you that you can’t and won’t achieve
your goals or make come true your
dreams.

• lamsaharbai@yahoo.com
• Tel: 013-3464160

Practice makes perfect
Be the Best
and
Beat the Rest

MATEMATIK
1 Petak ( 2 x 2 ) = 4 = 22
2 Petak ( 1 x 1 ) = 1 = 12
1 Petak ( 3 x 3 ) = 9 = 32
2 Petak ( 2 x 2 ) = 4 = 22
3 Petak ( 1 x 1 ) = 1 = 12
1 Petak ( 4 x 4 ) = 16 = 42
2 Petak ( 3 x 3 ) = 9 = 32
3 Petak ( 2 x 2 ) = 4 = 22
4 Petak ( 1 x 1 ) = 1 = 12
Berapa segiempat sama dalam rajah berikut?
2
5
14
30
= 12
+ 22
+ 32
+ 42
+ 52
= 555 x 5 = ?
?10 x 10 = = 12
+ 22
+ 32
+ 42
+ 52
+ ….. + 102
= ?
?n x n = = 12
+ 22
+ 32
+ 42
+ 52
+ …. + n2
= ?
Pengetahuan sedia ada :
takrif segi empat sama &
kuasa dua

4 DB  =   4 cm
Or  EB  =  5 cm
Cos x  =  4
               5
P1
N2

a) Sin xo
= opp/hip = 6/10
= 3/5 or
=0.6
M
N
Q
R
12 cm
6 cm
8 cmXo
P
10 cm

b) Cos ∠ MRP =⅜ so 3 = 12
8 RP
So length of RP = 4 x 8 = 32 cm.
M
N
Q
R
12 cm
6 cm
8 cmXo
P
10 cm

4. In Diagram 1, AEB is a
straight line and AE = EB.
DIAGRAM 1
A
BC
D
E
8 cm
6 cm
xo
3 cm
Find the value of cos xo
.

4. In Diagram 1, AEB is a
straight line and AE = EB.
DIAGRAM 1
A
BC
D
E
8 cm
6 cm
xo
3 cm
Find the value of cos xo
.
5 cm
4 cm

2ky = 3 + k
k(2y – 1) = 3
2ky – k = 3 1
k
k
y
2
3 +
= 1

4. In Diagram 1. R is the midpoint of the
straight line QRS .Given sin θ = .
Calculate the length, in cm, of QS
5
4
P
Q
R
S
12 cm
θ
Diagram 1

5
4
12 cm
Diagram 1
P
Q
R
Tan X= opp/adj.= PQ/QR.
Sin X= opp/hyp = PQ/PR
Cos X = adj/Hyp = QR/PR.

1
5
4
P
Q
R
S
15 cm12 cm
θ
Diagram 1
Sin θ= PQ/PR
4/5 = 12/15

cm18QS = 2
1
5
4
P
Q
R
S
15 cm12 cm
9 cm9 cm θ
Diagram 1

y
x
1
2
3
4
5
6
0 –1 1 2 3 4 5 6
A’
A
B’
B
D
C’
C
• K
E
•K’

y
x
1
2
3
4
5
6
0 –1 1 2 3 4 5 6
A’
A
B’
B
D
C’
C
• K
E
K’

Transformations
(iii) Reflection
Answer  :
A
B
C D
A’
B’C’
D’
DIAGRAM 3
[2 marks]

(iii) Reflection
Answer  :
A
B
C D
A’
B’C’
D’
DIAGRAM 3
[2 marks]

Diagram 5 shows a polygon.
20m
16m
12m
12m
DIAGRAM 5
Answer : [3 marks]

Diagram 5 shows a polygon.
20m
16m
12m
12m
DIAGRAM 5
Answer :
[3 marks]

Diagram 5 shows trapezium P drawn on a grid squares with sides of 1 unit.
P
DIAGRAM 5
On the grid in the answer space, draw Diagram 5 using the scale 1 : ½ .
The grid has equal squares with sides of 1 unit.
[3 marks]

21. Diagram 2, in the answer space shows a triangle L drawn on a
grid of equal squares.
On the diagram in the answer space, draw the image of triangle
L under an enlargement with centre Q and scale factor 3.
[2 marks]
Q
DIAGRAM 2
L
•

21. Diagram 2, in the answer space shows a
triangle L drawn on a grid of equal
squares.
On the diagram in the answer space,
draw the image of triangle L under
an enlargement with centre Q and scale
factor 3.


[2 marks]

Congruence
Diagram 3 in the answer space shows polygon ABCDEF and a straight line PQ
drawn on a grid of equal squares. Starting from the line PQ, draw polygon
PQRSTU which is congruent to polygon ABCDEF.
Answer :
A
BC
DE
F
P
Q
DIAGRAM 3
[2 marks]

Congruence
Diagram 3 in the answer space shows polygon ABCDEF and a straight line PQ
drawn on a grid of equal squares. Starting from the line PQ, draw polygon
PQRSTU which is congruent to polygon ABCDEF.
Answer :
A
BC
DE
F
P
Q
DIAGRAM 3
[2 marks]
R
S T
U

Solid Geometry
Eg : Diagram 4 shows a right pyramid with a square base.
6 units
5 units
DIAGRAM 4
Draw a full scale the net of the pyramid on the grid in the answer
space. The grid has equal squares with sides of 1 unit.

PMR  2004.
1.   Diagram 6 in the answer space shows four squares, PKJN, KQLJ, NJMS
      and JLRM.  W, X and Y are three moving points in the diagram.
(a)   W moves such that it is equidistant from the straight lines  PS and QR.
        By using the letters in the diagram, state the locus of W
(b)    On the diagram, draw
   (i)    the locus of X such that XJ = JN
   (ii)   the locus of Y such that its distance from point Q and point S are
          the  same.
   (iii)  Hence, mark with the symbol          all the intersections of
               the locus X and he locus Y.                                      [5 marks]
⊗
KP Q
J LN
S M R

2. Diagram 5 in the answer space shows a rectangle PQRS and
SP = 2cm.  W and Y are two points moving in the diagram.
On the diagram, draw
(a)    the locus of point W such that WS = WP,

(b) the locus of point Y such that QY = 1.5 cm,
                                                                                                           [5 marks]

⊗
S R
QP
DIAGRAM 5
(2)
(2)
(1)

The x-axis and y-axis in Diagram 4 in the answer space is
drawn on a square grid of sides 1 unit.
On the diagram,
(a) Construct locus of point P which always moves at a

      distance of 2 units from the point  (1,2),
(a) Consruct locus of point R which always moves at
an equidistant from points K and L,
(b)    Mark with
⊗ all the possible points of intersections of
locus P and locus R.
[5 marks]

Answer :
1
2
3
4
5
6
0 21 3 4 5–1–2–3–4
y
x
DIAGRAM 4
L •
• K
⊗
⊗
–1
–2
–3
•

2. Bar Chart
i) Label both axes uniformly
and correctly.
ii) Draw all bars correctly.

P Q R
T
S
xo
yo
10. TRIGONOMETRY
In Diagram 3, PQR is a straight line and T is the midpoint
of the straight QTS.
(a) Given that tan xo
= 1, calculate the length of QTS.
(b) State the value of cos yo
.
[3 marks]
4cm
15cm

Table 1 shows the number of cans of orange juice
consumed by students of a school on three consecutive
weeks in a month.
Week Number of cans
1 80
2 160
3 140
TABLE 1
The above information is shown in a pictograph in the
answer space. Complete the pictograph to represent all the
information in Table 1
[3 marks]

Week 1
Week 2
Week 3
Key :
Represents ______ cans of orange juice

8. Statistics Bar Chart
Month Number of
Lorries
January 850
February 1 280
March 1 050
April 1 500
TABLE 2
Table 2 shows the productions of lorries by an
automobile factory in the first four months of 2004.
On the square grid provided, consruct a bar chart to
represent all the information given in the table.
[4 marks]
Eg :

Month
Example 1.
Number of
Lorries
200
400
600
800
1000
1200
1400
1600
Jan Feb March April

STATISTICS
1. Line Graph
i) Label both axes uniformly and
correctly.
ii) Mark all points correctly.
iii) Join all points with straight line.
iv) Use a ruler to join all the points.
v) Do not start from 0.

Rainfall(mm)
M A M J J Month
30
60
90
120
150
180
210
240
270
300
x
x
x
x
x
2.

3. Pie Chart
i) Convert all the given
values/quantities to angle
of sectors.
Use proportion rule.
ii) Measure angle of sectors
correctly (using a
protractor).

Size of sports shirt Number of Students
S 5
M 20
L 15
XL 10
3.
On the circle in the answer space, with X as the centre of the circle,
construct a pie chart to represent all the information given in the table
[4 marks]

Size of sports shirt Number of Students
S 5
M 20
L 15
XL 10
00
36360
50
5
=×
On the circle in the answer space, with X as the centre of the circle,
construct a pie chart to represent all the information given in the table
S =
M =
00
144360
50
20
=×
L =
00
108360
50
15
=×
XL =
00
72360
50
10
=×
S
XL
L
M
5 = 36
20 = 36 x 4
= 144

1.   Diagram 6 shows rhombus PQRS and T lies on the straight line

      PQ.
      (a) Starting with triangle PQR in Diagram 7 in the answer space,
construct
(i) rhombus PQRS,
(ii)    QRT =          such that point T lies on the straight
line PQ to form Diagram 6
      (b) Based on the diagram drawn in (a) , measure
(i)              PRT,
(ii) the distance between points T and S.
∠ 0
30∠
GEOMETRICAL CONSTRUCTION

Answer :
R
P QT
(b) (i) 300
(ii) 5.8 cm
S

GEOMETRICAL CONSTRUCTION
2.   Diagram 7 in the answer space shows a straight line AB.
(a)  Using only a ruler and a pair of compasses,
      (i)   construct a triangle ABC beginning from the straight line AB
             such that BC = 6 cm and AC = 5 cm
      (ii) hence, construct the perpendicular line to the straight lne AB
        which passes through the point C.
(b)Based on the diagram constructed in (a), measure the
perpendicular  distance between point C and the straight line AB.

Answer:
A B
C
6 cm
5 cm
(b) 4.6cm

Graph of functions
Use the graph paper provided to answer this question.
x – 4 – 2.5 – 1 0 1 2 3 4 5
y – 27 – 7.5 3 5 3 –3 –13 –27 –45
Table 1 represents the table of values for certain function.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to
5 units on y-axis, draw the graph.
Table 1

2jam bhn per. pmr 2010

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