2jam bhn per. pmr 2010
TRANSCRIPT
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SEMINAR KECEMERLANGAN PMR • Tahun 2010
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PMR 2012
.SMK SUNGAI PUSU
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HAJI LAMSAH BIN ARBAI @ BAIPPT.,PJK
SMK PENGKALAN PERMATANGKUALA SELANGOR, SELANGOR.
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USUCCESS
PLAN FOR IT.
WILL BE YOURS IF….DREAM FOR IT.
WORK FOR IT.
RACE FOR IT.
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Objectives of the Seminar is to:Objectives of the Seminar is to:
Make sure, you understand the Make sure, you understand the requirements of the PMR requirements of the PMR Mathematics Examination.Mathematics Examination.
Show the simplest steps to solve Show the simplest steps to solve problems with accurate answer.problems with accurate answer.
Understand the marking scheme.Understand the marking scheme.
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GIVE ATTENTION TO:
1. Interpretation of the answer:
(a) (-3)(-2) 6≠(b)
2
3
2
3 ≠−−
(d)0
5
0 ≠
(e)
31
3 ≠(c)
2
3
2
3 ≠−
−
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1. Interpretation of answer:
(f)
2
1
8
4 ≠ Except in certain cases
(g) 5.32
13
2
7 == Except in certain cases
(h)
823 ≠(i)
x
yx
x
yx 32
2
64 −≠−
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THE CORRECT WAY TO WRITE THE FINAL ANSWER.
24
15=
8
5
2. 62
4.
3. xx =1
xx =1
5.
yx
xyx
24
186 2−
=( )
yx
xyx
24
36 −
y
xy
4
3−=
1.
= 36
6. 10 =x
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THE WRONG WAY TO WRITE THE FINAL ANSWER
147 −x1. = ( )27 −x
2.
7
15 no need to change to7
12
3.
−6
4
−
6
4 ( )6,4−≠≠
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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)
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PAPER 1
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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.
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How to get the answer?
• Mental arithmatic
• Do direct calculation • By using the given answer, work
backwards
•The method of calculation The method of calculation need notneed not be shown be shown..
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Objective question answering skills
SUPERSTEP TO
SUCCESS!!!
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Simplify
1. ( – 4x + 5y ) – (8x – 3y) =A. 4x – 8y
B. 8y – 12 x C. 8y + 12x
D. 9x – 5y
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SIMPLIFY BY OPENING THE BRACKET..
( − 4X + 5Y − 8X + 3Y)
−12X + 8Y
REARRANGE
8Y – 12 X
ANSWER B.
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What is the answer?
2. ( 4h – 7)2 =
A . 16h – 49 B . 16h2 + 49 C . 8h2 −12h + 49 D . 16h2 − 56h + 49
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• 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
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3. Simplify to the most simplest form. (2m( m – 5) + 7m)
A. 2m2 + 2m − 5B. 2m2 − 12 mC. 2m2 − 3mD. 2m2 + 2m
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Simplify the multiplication
2m( m – 5) + 7m
Open the bracket …
2m2 – 10m + 7m
2m2 – 3m
The answer is C.
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Simplify this question..
4). ( x – 3y)2 + 7xy =
A. x2 (x – 2) + xyB. x2 +7xy – 9y2 C. x2 + 7y2 + 8xyD. x2 + 9y2 + xy
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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.
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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
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Solution.Determine the median for 2, 2, 2, 3, 4 , 4, 5, 5, 7The median is 4.And mode is 2the difference between median and mode is 2 answer : B
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1. 923 758 becomes 924 000 after it is rounded off to
A oneB tenC hundredD thousand
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923 758
3 000 3 758 4 000
Answer D
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2. Diagram 2 shows some of the factors of 270. The possible value of y is
A 4B 6C 8D 12
2
3y
9 10
270
DIAGRAM 2
15
270
6= 45
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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 0.9B 1.2C 1.9D 2.1
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Sunday = 2.0mMonday = 2.0 –
(25% × 2.0)
= 2.0 – 0.5
= 1.5
Tuesday = 1.5 + (40% × 1.5) = 1.5 + 0.6
= 2.1
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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 43.75B 56.25C 60.00D 62.50
of the girls attended a youth camp.
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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
112x 100 = 56.25 %
= 28 boys + 35 girls = 63 students
Total number of the student = 112
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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
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A. 412
= 13
B. 416 =
14
C. 420
= 15
D. 824
=13
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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 removedA 42B 85C 425D 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
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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
T11 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
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Area of shaded region
= area of PQT + RSUV Area of PQT
= ½ × 11cm × 8cm = 44cm2
Area of RSUV = 4cm × 5.5cm = 22cmArea of shaded region = 44cm2 + 22cm2
= 66cm2
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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 isA 37 cmB 47 cmC 57 cmD 67 cm
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T
S
R
U
QP
V`
7 cm
5 cm
10cm
10cm
5cm 5cm
5cm
Perimeter = 5cm + 5cm + 5cm + 5cm + 10 + 7 + 10 = 47cm
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9. Given that the mean of 5, 7, 6, 3, p, 8 and 7 is 6, find the value of p is
A. 5B. 6C. 7D. 8
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5 + 7 + 6 + 3 + p + 8 + 7
7= 6
36 + p = 6 x 7
36 + p = 42
p = 42 – 36
p = 6
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53o
32o
yoO
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
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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 0.25B 0.50C 1.00D 2.00
mode = 3median =
2 , 3 , 3 , 4 , 6 , 7 , 8 , 94 + 6
2= 5
median – mode = 5 – 3
= 2
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If you are smart enough, If you are smart enough, nothing can be a problem.nothing can be a problem.
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Table shows the time allocation for Table shows the time allocation for a test.a test.
TestTest Time AllocationTime AllocationPaper 1Paper 1 1 1 ¼ hours¼ hoursBreakBreak 20 minutes20 minutes
Paper 2Paper 2 1 ½ hours1 ½ hoursAll candidates must be in the All candidates must be in the examination hall 10 minutes before examination hall 10 minutes before Paper 1 starts. Paper 2 ends at 1.05 Paper 1 starts. Paper 2 ends at 1.05 p.m. At what time must the p.m. At what time must the candidates be in the hall before candidates 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
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SOLUTIONSOLUTION
Convert 1.05 p.m. Convert 1.05 p.m. in 24-hour system.in 24-hour system.
Find the duration Find the duration of the papers of 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
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In diagram 16, PQRS is a In diagram 16, PQRS is a circle with centre O and PST circle with centre O and PST is a straight line. Find the is a straight line. Find the value of value of xx..AA 1515oo CC 4545oo
BB 3535oo DD 5050oo
85 °
x °
100°
O
TP
Q
S
R
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Knowing that angles subtended at the circumference Knowing that angles subtended at the circumference by an arc are half the one at centre, by an arc are half the one at centre, ∠∠ P PQS QS = = ½½ ∠∠ POSPOS..
Therefore Therefore ∠∠ PQSPQS = = 5050oo..
Since Since PQRS PQRS is cyclic quadrilateral, is cyclic quadrilateral,
∠∠PQRQR = = ∠∠RSTRST = 85 = 85oo
Therefore, Therefore, ∠∠SQR SQR = = ∠∠PQRPQR −− ∠∠PQSPQS
= 85= 85oo −− 50 50oo = = 3535oo
SOLUTIONSOLUTION
Answer: Answer: BB
85 °
x °
100 °
O
TP
Q
S
R
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PAPER 2
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I swear!I swear!I didn't I didn't use the use the
calculator.calculator.
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MathematicsMathematics
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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.
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If a question needs candidates to show the proper solving steps
..,BUT
Only the final answer written in the space provided...
NOMarks will be given, because it is assumed candidates did
not Do calculation.
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• A. Calculate the value of
6(57 ÷ 3 − 4) + 51
Solution:
6(19 − 4) + 51
6(15) + 51 K1 90 + 51
141 N1
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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
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• 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
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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
K1Not complete yet
5−pq = N1
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Paper 2 topics with ‘BIG’ marks !!!!
TOPIC MARKS
1. Algebraic Expressions (Form1,2 & 3)
i. Expansion ii. Factorizationiii. Algebraic Fractions iv. Algebraic formulae
2 marks3 marks3 marks3 marks
2. Loci in Two Dimensions (Form 2)
5 marks
3. Transformations1 or 2 questions
4 to 6 marks
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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
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FRACTIONSFRACTIONS
4 basic operations
Simplifying
Involving LCM
Transferring the answer
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1. Calculate the value of
+÷
52
31
153
2
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1. Calculate the value of
+÷
52
31
153
2
+÷=52
34
513
÷=
1526
513
N1
K1
211=2/3
130195
2615
513 =
×=
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2. Calculate the value of
−×−
4
3
5
1
3
21
and express the answer as a fraction in its lowest term.
[2 marks]
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−×−
20
11
3
5
12
11
K1
N1
![Page 66: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/66.jpg)
Calculate the value ofCalculate the value of
532
52
312 ÷
−
and express the answer and express the answer as a fraction in its as a fraction in its lowest term.lowest term.
![Page 67: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/67.jpg)
135
156
1535 ×
−
532
52
312 ÷
− 5
1352
37 ÷
−
STEPS TO BE SHOWNSTEPS TO BE SHOWN
135
1529 ×
3929
=
=
=
=
1
3
K1
N1
![Page 68: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/68.jpg)
4
11
164
515 −÷
1.Calculate 15 ÷ + ( −16)
5
415 × − 16
12 − 16
− 4
K1
N1
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DECIMALSDECIMALS
Place value
4 basic operations
Change decimal to fractions and vice versa
Bracket operations
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)4
31(2.004.6 −−÷−1. Calculate
75.12.0
04.6 +−
75.12
4.60 +−
75.12.30 +−
45.28−
K1
N1
![Page 71: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/71.jpg)
)5
21(3.043.5 −−÷3. Calculate
4.1)3.043.5( +÷
4.11.18 +
19.5
K1
N1
![Page 72: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/72.jpg)
Calculate the value and express the answer correct to two decimal places
0.45661 12 ×−−
) 0.076 - ( - 12=0.07612 +=
N1
K112.076=12.08=
![Page 73: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/73.jpg)
Calculate the value and express the answer correct to two decimal places
( ) 0.33.1252
2 ÷−−
0.33.122.4 ÷+=10.42.4 +=
N1
K1
12.80=
![Page 74: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/74.jpg)
3
4(6 − 0.24) ÷ 2. Calculate
Correct to two decimal places
5.76 × 4
3 K1
1.44 × 3
= 4.32 N1
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SQUARES, SQUARE ROOTS
• CUBES AND CUBES ROOTS
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1. a. Find the value
9
45
9
49
3
7or
3
12 P1
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1. b. Calculate the value 3 064.0 - (-1)2
14.0 − K1
6.0− N1
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2. a Find the value of 36.06.0= P1
b. Calculate 23 )275( −+2)]3(5[( −+
4K1
N1
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3a. Find the value of 30.3
0.30.30.3 ××=0.027= P1
![Page 80: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/80.jpg)
64)2( 3 ×−3.a. find the value of
88×−64− P1
b. Calculate
9
714 −
3
44 −
3
8
K1
N1
![Page 81: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/81.jpg)
INDICES
• MARKING SCHEME
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1. a. Simplify 2435 )()( qq ÷−−
815−q7q= P1
b. 2739 2 =× −x
322 333 =× −x
322 =−+ x
3=x
K1
N1
![Page 83: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/83.jpg)
2. a. Simplify
542 )( mm ÷58−m
3m P1
b. 22216 =× m
24 222 =× m
24 =+ m2−=m
K1
N1
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3. a .simplify
10
4433
4
)()(16
xy
yx
10
169
4
16
xy
yx=
684 yx P1
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b.
xx
5
1255 2 =−
xx
5
55
32 =−
xx −− = 32 55xx −=− 32
52 =x2
5=x
K1
N1
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ALGEBRAIC EXPRESSIONS
• EXPANSION
• FACTORIZATION
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8. Simplify
(a) 2(n + 5) − 3= 2n + 10 − 3
= 2n + 7 N1
(b) 3(4m – 3k) – (5k – m)
12m – 9k – 5k + m13m – 14k
K1
N1
![Page 88: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/88.jpg)
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 +++−
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qrpq 64 +
)5(4 ++ xx
2, Factorise completely a.
b.
)32(2 rpq + P1
xx 54 2 ++452 ++ xx
)4)(1( ++ xx
K1
N1
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)2(45 −−− kk
)1(4)2( 2 −+− yy
3. Simplify to its simplest form a.
b.
845 +−− kkk98 − P1
44442 −++− yyy2y
K1
N1
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77 Factorise completely Factorise completely (a) 2y + 6(a) 2y + 6
(b) 12 – 3x(b) 12 – 3x22
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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 + 6 7 (a) 2y + 6
= 2(y + 3)= 2(y + 3)P1
K1
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
N1
![Page 93: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/93.jpg)
88 Expand Expand (a) q(2 + p)(a) q(2 + p)
(b) (3m – n)(b) (3m – n)22
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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 + n22K1
N1
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
8 (a) q(2 + p)8 (a) q(2 + p)
= 2q + pq = 2q + pq P1
![Page 95: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/95.jpg)
Factorize completelyFactorize completely
55hh((k k – 3) – 2(3 – – 3) – 2(3 – kk))
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55hh((kk – 3) – 2(3 – – 3) – 2(3 – kk))
55hh((kk – 3) – 6 + 2 – 3) – 6 + 2kk
55hh((kk – 3) + 2 – 3) + 2k k – 6– 6
55hh((kk – 3 – 3) + 2() + 2(k k – 3– 3))
(5(5hh + 2)( + 2)(kk – 3) – 3)
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
K1
N1
![Page 97: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/97.jpg)
qq 4
1
3
4 −1.
qq 12
3
12
16 −
q12
13
K1
N1
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2.
)2(3
4
2
1
qpqp ++
+
)2(3
43
qp ++
)2(3
7
qp +
K1
N2
![Page 99: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/99.jpg)
Express Express
as a single fraction as a single fraction in its simplest form. in its simplest form.
n
w
n 6
32
3
5 −+
![Page 100: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/100.jpg)
( )( ) 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 −=
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
1
2
K1
K1
N1
![Page 101: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/101.jpg)
17. Express as a single fraction in its simplest form
6b2 b
3b2 −−
6bb - 6=
K1
K1
N1
b 62b
b 3 x 2 4 −−=
6b 2 b - 4 +=
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17. Express as a single fraction in its simplest form.
9abb 2
3a2 −−
9ab) b- 2 ( - 3b2×=
9abb 2 - 6b +=
9ab2 - b 7 =
N1
K1
K1
![Page 103: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/103.jpg)
LINEAR EQUATION
• SOLVE THE VALUE
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pp =−− 6)3(4
5
732
kk
−=
1. a.
b.
pp =−− 6124183 =p6=p
kk 7310 −=317 =k
17
3=k
K1
N1
K1
N1
![Page 105: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/105.jpg)
432 −= nn
34
23 −=−n
n
2. a
2.b.
4−=− n4=n
P1
12423 −=− nn21243 +−=− nn
10=n
K1
N1
![Page 106: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/106.jpg)
6435 +=− xx
31)46(2
3 −=−+ ff
3 a.
3.b.
3645 +=− xx9=x
6212182 −=−+ ff186210 −−=− f
8010 =f8=f
P1
K1
N1
![Page 107: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/107.jpg)
kk −−= 14
9)1(3 =+− pp
4. a.
b.
142 −=k7−=k
P1
933 =−− pp122 =− p
6−=p
K1
N1
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LOCUS IN TWO DIMENSION
• DRAW THE LOCUS
• MARK THE INTERSECTION
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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.
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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.
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LOCI IN TWO DIMENSIONS .
A B
i). Always equidistant from two points (AB)
ii). Always equidistant from PR
P Q
RS
1) e.g
![Page 112: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/112.jpg)
Always equidistant from a point
e.g
i) The locus of X such that XJ = JM
J K
LM
X
2)
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ii) The locus of Y such that QY = 4cm
4cm
6 cm
P Q
RS
Y
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Always equidistant from a line or two lines.
e.g.
i) Locus K is 3 cm from AB
3.
A
B
3 cm 3 cmK
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ii) Always equidistant 2 lines.
eg. Locus W is always equidistant from PQ and RS or PW = WS
P Q
RS
W
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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)
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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,
(c) mark with all the possible points of intersection between locus W and locus Y.
[5 marks]
⊗
S R
QP
DIAGRAM 5
W
Y
⊗
(2)(2)
(1)
3cm
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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.
![Page 119: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/119.jpg)
lociP Q
R
M
S
DIAGRAM 8
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15.a. W is a moving point such that it is equidistant from points S and Q.b. X is a moving point such thatit 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
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15. a. State the locus of Wby 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
N1K1
![Page 122: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/122.jpg)
TRANSFORMATIONS
• TRANSLATION
• ROTATION
• REFLECTION
• ENLARGEMENT
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TRANSFORMATIONSTRANSFORMATIONS
Describe the transformations
Writing technique
Spelling
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12
10
8
6
4
2
0 2 4 6 8 10 12
P’
P
P’P’ is the image of is the image of P P under transformation under transformation MM. . Describe in full transformation M.Describe in full transformation M.
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12
10
8
6
4
2
0 2 4 6 8 10 12
PP
P’
TranslationTranslation
−54
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
![Page 126: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/126.jpg)
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
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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]
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Q
P
T
R
P’
Q’
R’
![Page 129: 2jam bhn per. pmr 2010](https://reader033.vdocuments.net/reader033/viewer/2022042518/559cc30f1a28ab83788b46ed/html5/thumbnails/129.jpg)
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
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Diagram 4 in the answer spaceshows triangle PQR drawn on a grid of equal squares.
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5. Draw and label the image
M
N
K1
D
C
B
N1
A
A’
B’ C’
D’
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6. Draw and label the image under reflection on the line MN
K1
M
N
P
P’N1
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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]
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(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]
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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
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Question 6
K’
J’
M’ L’P2
J
K
LM
K’
L’M’
J’
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Question 6
K’
J’
M’ L’
P
P2
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y
x1
2
3
45
6
0 –1 1 2 3 4 5 6
A’ A
B’
B
D
C’
C• K
E
•K’
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On the grid, draw the image of triangle PQR Under an enlargement with scale factor 2 at centre M
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K1
N1
7. Draw an enlargement image
P’
Q
R’
Q’
RM
P
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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]
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Q
DIAGRAM 2
L
•
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Q
DIAGRAM 2
L
•
L’
•
• •
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CONSTRUCTIONS
• CONSTRUCT THE DIAGRAM
• FOLLOW THE INSTRUCTIONS
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Set square and protractor cannot be used to answer this Set square and protractor cannot be used to answer this questionquestion
Diagram 5 in the answer space shows a straight line Diagram 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 with with PQPQ = 8 cm, = 8 cm, PRPR = 5 cm and = 5 cm and ∠∠RPQRPQ = 60 = 60oo
ii.ii. bisector of bisector of ∠∠ PQRPQR
b)b) Based on the diagram constructed, measure the distance Based on the diagram constructed, measure the distance between the point between the point QQ and the straight line and the straight line PRPR..
QP
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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 line Diagram 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 with with PQPQ = 8 cm, = 8 cm, PRPR = 5 cm and = 5 cm and ∠∠RPQRPQ = 60 = 60oo
ii.ii. bisector of bisector of ∠∠ PQRPQR
b)b) Based on the diagram constructed, measure the distance Based on the diagram constructed, measure the distance between the point between the point QQ and the straight line and the straight line PRPR..
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
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A
B CD
6 cm 8 cm
Diagram 6
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19. a (i). Construct triangle ABC (ii). Construct AD lineb (i). Measure the angle of ABC (ii). measure AD
11
1
11
1
6 cm 8 cm
AD = 5.9±0.1 cm
A
B CD
ABC = 740 ±1
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STATISTICS
• CONSTRUCT PIE CHART
• CONSTRUCT BAR CHART
• CONSTRUCT LINE GRAPH
• CONSTRUCT PICTOGRAM
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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
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First, find the value of angles that represents the number of stamps according to the table.
Abu 0360240
30 ×⇒
Bakar
Chua
David
Ellan
0360240
50 ×⇒
0360240
20 ×⇒
0360240
40 ×⇒
0360240
60 ×⇒
Fadil 0360240
40 ×⇒
= 45o
= 75o
= 30o
= 60o
= 90o
= 60o
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Using a protractor, draw the angles and Using 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
DavidChua
Title : Number of stamps each student have.
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Grade A B C D E
Number ofStud 25 38 12 15 10
2.
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GRADE
Number of Students
5
10
15
20
25
30
35
40
A B C D E
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GRAPH OF FUNCTION
• DRAW THE AXIS
• MARK THE LOCATIONS
• JOIN THE POINTS
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1. Use the real graph paper not a square grid 1. Use the real graph paper not a square grid paper.paper.2.2. Always remember that the big square Always 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 of Identify 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 and Label the x-axis and the y-axis correctly and
uniformly. uniformly. 5.5. Relate the number on the value table with Relate 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 small Then mark all the points correctly using small cross or dots.cross or dots. 7.7. Draw smooth curve that passes through all the Draw smooth curve that passes through all the
correct points.correct points.
GRAPH OF FUNCTIONS.
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Use graph paper provided to answer this Use graph paper provided to answer this question.question.
Table 2 shows the values of two Table 2 shows the values of two variables variables xx and and 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 function Draw the graph of the function using a scale of 2 cm to 1 unit at using a scale of 2 cm to 1 unit at the the xx-axis and 2 cm to 10 units -axis and 2 cm to 10 units at the at the y y –axis.–axis.
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X
y
10
-10
20
30
40
-20
-30
-40
- 3 -2 -1 0 1 2 3
⊗
⊗
⊗
⊗
⊗
⊗
⊗
1.
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-3 -2 -1 0 1 32 x
y
×
× ×
××
×
× -10
-20
10
-30
20
30
40
-40
SSTTEEPPSS TTOO BBEE SSHHOOWWNN
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x -3 -2 -1 0 1 2.5 4 5
y 15 5 -1 -3 -1 9.5 29 47
2.
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-3 -2 -1 0 1 32 4 5 x
×
×
×
×
××
y
35
10
5
15
-5
20
25
40
-10
30
45
50
× ×
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X
Y
-4
-27
-2.5
-7.5
-1
3
0 1 2 3 4 5
5 3 -3 -13 -27 -45
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y
-10
-35
-40
-30
-45
-25
-20
-5
-50
-15
5
10
-3 -2 -1 0 1 32 4 5 x-4-5
×
×
×
××
×
×
×
×
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-3 - 2 -1 0 1 2 3 4
y
4
8
12
×
× ×
×
× ×
×
×
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SOLID GEOMETRYSOLID GEOMETRY
Net and layout
Surface area
Volume of solid geometry and combinations
Basic characteristics
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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 units8 units
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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 units8 units
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7. Draw the net of the prism
8 unit
3 unit 5 unit4 unit
K2
N1
3 unit
3 unit
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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
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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.
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daripadadaripada
Lamsah bin Arbai @ BaiLamsah bin Arbai @ Bai PPT. PJKPPT. PJK
603 - 32893611
013 - 3464160
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semalam,semalam, hari ini hari ini dan esokdan esok
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semalam,semalam,
Semalam …
Telahku campakkan benih kekecewaan
Ku lontarkan segala kedukaan
Dan segala kesedihan
Lalu kujadikan sebuah kenangan
Dan pengalaman
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hari inihari ini
Hari ini …Ku taburkan benih harapan
Ku semaikan dengan baja impianKu berdiri dengan semangat perjuangan
Aku tak ‘kan kalah dengan pujukanItulah keyakinan di hatiYang telah ku tanamkan
Kerana ku tahuHari ini adalah kenyataan
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dan esokdan esok
Esok …Akanku kutip segala semaian
Akanku buktikan segala lelahanYang tak pernah kenal erti kepayahanYang ku pasti esok adalah harapan
Harapan yang menuntutSebuah kenyataan
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REMEMBERREMEMBERExcellent 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.
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USUCCESS
PLAN FOR IT.
WILL BE YOURS IF….DREAM FOR IT.
WORK FOR IT.
RACE FOR IT.
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SEKIAN TERIMA KASIHSEKIAN TERIMA KASIH
daripadadaripada
Lamsah bin Arbai @ BaiLamsah bin Arbai @ Bai PPT. PJKPPT. PJK
smk pengkalan permatang45000 kuala selangorselangor darul ehsan
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The allegory of the Frog ...
Lesson of Life N. 1
Once upon a time there was a race ...
of frogs
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The goal was to reach the top of a high tower.
Many people gathered to see and support them.
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The race began.
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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!"
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The frogs began to resign, except for one who kept on climbing
The people continued : "... What pain !!! They’ll never make it!..."
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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.
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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!
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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.
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Practice makes perfect
Be the Best
and
Beat the Rest
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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
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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
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Practice makes perfect
Be the Best
and
Beat the Rest
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4 DB = 4 cmOr EB = 5 cm
Cos x = 4 5
P1
N2
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a) Sin xo = opp/hip = 6/10
= 3/5 or =0.6
M
N Q
R
12 cm6 cm
8 cmXo
P
10 cm
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b) Cos ∠ MRP =⅜ so 3 = 12
8 RP
So length of RP = 4 x 8 = 32 cm.
M
N Q
R
12 cm6 cm
8 cmXo
P
10 cm
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4. In Diagram 1, AEB is a straight line and AE = EB.
DIAGRAM 1
A
BC
D
E
8 cm
6 cmxo
3 cm
Find the value of cos xo.
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4. In Diagram 1, AEB is a straight line and AE = EB.
DIAGRAM 1
A
BC
D
E
8 cm
6 cmxo
3 cm
Find the value of cos xo.
5 cm
4 cm
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2ky = 3 + k
k(2y – 1) = 32ky – k = 3 1
k
ky
2
3+= 1
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4. In Diagram 1. R is the midpoint of the straight line QRS .Given sin θ = . Calculate the length, in cm, of QS
54
P
QR
S
12 cm
θ
Diagram 1
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4. In Diagram 1. R is the midpoint of the straight line QRS .Given sin θ = . Calculate the length, in cm, of QS
54
12 cm
Diagram 1
P
QR
Tan X= opp/adj.= PQ/QR.
Sin X= opp/hyp = PQ/PR
Cos X = adj/Hyp = QR/PR.
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4. In Diagram 1. R is the midpoint of the straight line QRS .Given sin θ = . Calculate the length, in cm, of QS
1
54
P
QR
S
15 cm12 cm
θ
Diagram 1
Sin θ= PQ/PR
4/5 = 12/15
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4. In Diagram 1. R is the midpoint of the straight line QRS .Given sin θ = . Calculate the length, in cm, of QS
cm 18QS = 2
1
54
P
QR
S
15 cm12 cm
9 cm9 cm θ
Diagram 1
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y
x1
2
3
45
6
0 –1 1 2 3 4 5 6
A’ A
B’
B
D
C’
C• K
E
•K’
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y
x1
2
3
45
6
0 –1 1 2 3 4 5 6
A’ A
B’
B
D
C’
C
• K
E
K’
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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]
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(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]
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Diagram 5 shows a polygon. 20m
16m
12m
12m
DIAGRAM 5
Answer : [3 marks]
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Diagram 5 shows a polygon. 20m
16m
12m
12m
DIAGRAM 5
Answer :[3 marks]
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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]
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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]
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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
•
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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]
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Q
DIAGRAM 2
L
•
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Q
DIAGRAM 2
L
•
L’
•
• •
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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]
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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
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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
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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
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Solid GeometryEg : Diagram 4 shows a right pyramid with a square base.
6 units
5 units
DIAGRAM 4Draw 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.
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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
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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)
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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]
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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
•
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2. Bar Chart
i) Label both axes uniformly and correctly.
ii) Draw all bars correctly.
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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
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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]
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Week 1
Week 2
Week 3
Key :
Represents ______ cans of orange juice
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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 :
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Month
Example 1.
Number of Lorries
200
400
600
800
1000
1200
1400
1600
Jan Feb March April
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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.
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Rainfall(mm)
M A M J J Month
30
60
90
120
150
180
210
240
270
300
x
x
x
x
x
2.
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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).
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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]
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Size of sports shirt Number of Students
S 5
M 20
L 15
XL 10
00 3636050
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 14436050
20 =×
L =00 108360
50
15 =×
XL =00 72360
50
10 =×
SXL
LM
5 = 36
20 = 36 x 4
= 144
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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.
∠ 030∠
GEOMETRICAL CONSTRUCTION
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Answer :
R
P QT
(b) (i) 300
(ii) 5.8 cm
S
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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.
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Answer:
A B
C
6 cm5 cm
(b) 4.6cm
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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