Download Answer Technique MM 2

ANSWERING TECHNIQUES:
SPM MATHEMATICS
Paper 2
Section A
Simultaneous Linear equation (4 m)
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Simultaneous linear equations with two unknowns can be solved
by (a) substitution or (b) elimination.
Example: (SPM07-P2) Calculate the values of p and q that
satisfy the simultaneous :
g + 2h = 1
4g  3h = 18
g + 2h = 1

4g  3h = 18 
 :
g = 1  2h  1
 into : 4(1  2h)  3h = 18 1
4  8h 3h = 18
 11h = 22

h=2

When h = 2, from :
g = 1  2(2)
g=14
 g = 3

Hence,
and
h=2
g = 3
2
Simultaneous Linear equation
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
Simultaneous linear equations with two unknowns can be solved
by (a) elimination or (b) substitution.
Example: (SPM04-P2) Calculate the values of p and q that
satisfy the simultaneous :
½p – 2q =13
3p + 4q = 2
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  2:
 + :

½p – 2q =13 
3p + 4q = 2 
p – 4q = 26  1
4p = 24
1
p=6
When p = 6, from :
½ (6) – 2q = 13
2q = 3 – 13

2q = - 10
q=-5

Hence,
and
p=6
q=-5
2
Solis geometry (4 marks)
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Include solid geometry of cuboid, prism, cylinder, pyramid, cone and
sphere.
Example : (SPM04-P2) The diagram shows a solid formed by joining
a cone and a cylinder. The diameter of the cylinder and the base of the
cone is 7 cm. The volume of the solid is 231 cm3. Using  = 22/7,
calculate the height , in cm of the cone.

Let the height of the cone be t cm.
Radius of cylinder = radius of cone= 7/2 cm (r)
Volume of cylinder = j2t2
 22  7 
    4 = 154 cm3

Hence volume of cone = 231 – 154 = 77 cm3
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 7  2 
1
2
2
t = 6 cm
1
1  22  7 
   t
= 77 2
3  7  2 
7
2
t = 77  3    
22  7 
t cm
4 cm
7/
2
cm
Rujuk rumus yang
diberi dalam kertas
soalan.
Perimeters & Areas of circles (6 m)

Usually involve the calculation of both the arc and area of
part of a circle.

Example : (SPM04-P2) In the diagram, PQ and RS are the
arc of two circles with centre O. RQ = ST = 7 cm and PO =
14 cm.
Using  = 22/7 , calculate
S
R
2
(a) area, in cm , of the shaded region,
Q
T
60
(b) perimeter, in cm, of the whole diagram.
O
P
Perimeters & Areas of circles
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(a) Area of shaded region
= Area sector ORS –Area of DOQT
1
= 1  22  212  2  14  14 2
4 7
= 346½ – 98
= 248½ cm2
1
Formula
given in exam
paper.
S
R
Q
T
60
O
P
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(b) Perimeter of the whole diagram
= OP + arc PQ + QR + arc RS + SO
1
22
60
22
= 14 + 4  2  7  21 + 7 + 360  2  7  14
= 346½ – 98
= 248½ cm2
1
2
+ 21
Formula given
.
Mathematical Reasoning (5 marks)
(a) State whether the following compound statement is true or false
5  125 and  6  7
3
Ans: False
1
Mathematical Reasoning
(b) Write down two implications based on the following compound
statement.
x  64 if and only if x  4.
3
Ans: Implication I : If x3 = -64, then x = -4
Implication II : If x = -4, then x3 = -64
2
(c) It is given that the interior angle of a regular polygon of n sides

2
is 1  n  180
Make one conclusion by deduction on the size of the size of the
interior angle of a regular hexagon.
2

Ans: The interior angle of a regular hexagon  1   180
 6
2
 120
The Straight Line ( 5 or 6 marks)
Diagram shows a trapezium PQRS drawn on a Cartesian plane. SR
is parallel to PQ.
Find
(a) The equation of the
straight line SR.
Ans: mPQ  5  1 1
3  11
4

8
1
1
1

Equation of the straight line SR, y  9   x  8
1
2
2
1
y 9   x  4
2
1
y   x  13
2
1
The Straight Line
Diagram shows a trapezium PQRS drawn on a Cartesian plane. SR
is parallel to PQ.
Find
(b) The y-intercept of
the straight line SR
Ans: The y-intercept
of SR is 13.
1
Graphs of Functions (6 marks)
Diagram shows the speed-time graph for the movement of a particle
for a period of t seconds.
Graphs of Functions
(a) State the uniform speed, in m s-1, of the particle.
Ans: 20 m s-1
1
(b) Calculate the rate of change of speed, in m s-1, of the particle
in the first 4 seconds.
Ans:
20  12 1
The rate of change of speed of the particle in the first 4 seconds 
4
 2ms 2 1
(c) The total distance travelled in t seconds is 184 metres.
Calculate the value of t.

Ans:  1
2







12

20

4

20

t

4

184
 2

64  20t  80  184
20t  200
t  10seconds 1
Probability (5 or 6 marks)
Diagram shows three numbered cards in box P and two cards
labelled with letters in box Q.
2
3
P
6
Y
R
Q
A card is picked at random from box P and then a card is picked at
random from box Q.
Probability (5 or 6 marks)
By listing the sample of all the possible outcomes of the event,
find the probability that 1
(a) A card with even number and the card labeled Y are picked,
P(Even number and Y card)  P(Even number)  P(Y card)
2 1
1
 
3 2
1
1

3
(b) A card with a number which is multiple of 3 or the card
labeled R is picked.
P(multiple of 3 or R card)  P(multiple of 3)  P(R card)  P(multiple of 3  R card)
2 1 1
1
  
3 2 3
5

1
6
Lines and planes in 3-Dimensions(3m)
Diagram shows a cuboid. M is
the midpoint of the side EH and
AM = 15 cm.
E
M
F
H
G
(a) Name the angle between the
line AM and the plane ADEF.
1
Ans: EAM
(b) Calculate the angle between
the line AM and the plane
ADEF.
Ans:
M
15 cm
4 cm
D
E
A
C
8 cm
B
θ
4
sin  
15
  1528'
EAM  1528'
A
1
1
Matrices
This topic is questioned both in Paper 1 &
Paper 2
 Paper 1: Usually on addition, subtraction
and multiplication of matrices.
 Paper 2: Usually on Inverse Matrix and the
use of inverse matrix to solve simultaneous
equations.
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Matrices (objective question)
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  2
 
 4 
Example 1: (SPM03-P1)
 5 1   2 

 
 3 4  4 
  10  4 

 
  6  16 
  6
  
 10 
5 1


3 4
5(-2) + 14
3(-2) + 44
Matrices (6 or 7 marks)
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Example 2: (SPM04-P2)
(a) Inverse Matrix for
is
 6
m
5
3  4


5  6
p

3
3  4


5  6
Hence, m = ½ , p = 4.
2
1
Inverse matrix
formula is given
in the exam
paper.
  6  (4) 
1



3 
3  (6)  (4)  5   5
1   6 4

 
2   5 3
1
Matrices
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Example 2: (SPM04-P2) (cont’d)
(b) Using the matrix method , find the value of x and y that
satisfy the following matrix equation:
3x – 4y =  1
5x – 6y = 2
Change the simultaneous equation into matrix equation:
 3  4  x    1

    
 5  6  y   2 
1
Solve the matrix equation:
1
1
 3  4   3  4  x   3  4    1

 
   
  
 5  6   5  6  y   5  6   2 
 x  1   6 4   1
   
 
 y  2   5 3  2 
1
 x  1  (6)  (1)  4  2 
   

 y  2  (5)  (1)  3  2 
 x  1 14 
    
 y  2  11 
 x   7 
   1
 y   5 2 
Maka, x = 7, y = 5½
2
1
Paper 2
Section B
Graphs of functions (12 marks)
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This question usually begins with the calculation of two to
three values of the function.( Allocated 2-3 marks)
Example: (SPM04-P2)
y = 2x2 – 4x – 3
Using calculator, find the values of k and m:
When x = - 2, y = k.
hence, k = 2(-2)2 – 4(-2) – 3
Usage of calculator:
= 13
Press 2 ( - 2 ) x2 - 4
( - 2 ) - 2 = .
When x = 3, y = m.
Answer 13 shown on
hence, m = 2(3)2 – 4(3) – 2
screen.
=3
2
To calculate the next
value, change – 2 to 3.
Graphs of functions
To draw graph
(i) Must use graph paper.
(ii) Must follow scale given
in the question.
(iii) Scale need to be
uniform.
(iv) Graph needs to be
smooth with regular
shape.
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
Example: (SPM04-P2)
y = 2x2 – 4x – 3

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
Graphs of functions


Example: (SPM04-P2)
Draw y = 2x2 – 4x – 3

4
To solve equation
2x2 + x – 23 = 0,
2x2 + x + 4x – 4x – 3 -20 = 0
2x2 – 4x – 3 = - 5x + 20
y = - 5x + 20 1
 Hence, draw straight line
y = - 5x + 20 1
From graph find values of x

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2

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


Plans & Elevations (12 marks)
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NOT ALLOW to sketch.
Labelling not important.
The plans & elevations can be drawn from any angle.
(except when it becomes a reflection)
Points to avoid:
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Inaccurate drawing e.g. of the length or angle.
Solid line is drawn as dashed line and vice versa.
The line is too long.
Failure to draw plan/elevation according to given scale.
Double lines.
Failure to draw projection lines parallel to guiding line
and to show hidden edges.
Plans & Elevations (3/4/5 marks)
N
H
M
J
3 cm
L
K
G
X
F
6 cm
D
4 cm
E
Statistics (12 marks)
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Use the correct method to draw ogive, histogram and
frequency polygon.
Follow the scale given in the question.
Scale needs to be uniform.
Mark the points accurately.
The ogive graph has to be a smooth curve.
Example (SPM03-P2) The data given below shows the
amount of money in RM, donated by 40 families for a
welfare fund of their children school.
Statistics
Amount
(RM)
40
23
27
30
20
24
28
35
34
32
17
33
45
31
29
30
33
21
37
26
22
39
38
40
32
26
34
22
32
22
35
39
27
14
38
19
28
35
28
44
Frequency Cumulative Upper
Frequency boundary
0
10.5
11 - 15
1
1
15.5
16 - 20
3
4
20.5
21 - 25
6
10
25.5
26 - 30
10
20
30.5
31 - 35
11
31
35.5
36 - 40
7
38
40.5
41 - 45
2
40
45.5
To draw an ogive,
•Show the Upper
boundary column,
•An extra row to indicate
the beginning point.
3
Statistics
Kekerapan Longgokan
The ogive drawn is
a smooth curve.
Ogif bagi wang yang didermakan
45
40
35
30
25
20
15
10
5
0
0
10
20
30
Wang (RM)
40
Q3
50
4
d) To use value from graph to solve question given (2 m)
Combined Transformation


(SPM03-P2)
(a) R – Reflection in the line y = 3,
T – translasion   2 
 4
 

Image of H under
(i) RT
2
(ii) TR
y
P
L
8

6
2
4
D
G
A

H
2
M
N
C

B
F
E
-6
-4
-2
O
2
4
K
6
J
8
x
10
Combined Transformation (12 marks)
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



(SPM03-P2)
(b) V maps ABCD to ABEF
V is a reflection in the line AB.
W maps ABEF
to GHJK.
W is a reflection
in the line x = 6.
2
y
P
L
8
6
4
D
G
A
H
2
2
M
N
C
B
F
E
-6
-4
-2
O
2
4
K
6
J
8
x
10
Combined Transformation
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

(SPM03-P2)
(b) (ii) To find a transformation that is equivalent to two
successive transformations WV.
Rotation of 90 anti clockwise about point (6, 5). 3
y
8
6
4
D
G
A
H
2
C
B
K
-6
-4
-2
O
2
4
6
J
8
x
10
Combined Transformation





(SPM03-P2)
(c) Enlargement which maps ABCD to LMNP.
Enlargement centered at point (6, 2) with a scale factor of 3.
3
Area LMNP
y
2
= 325.8 unit
L
P
8
Hence,
6
Area ABCD

1
3
2
 325.8
4
1
= 36.2 unit2
2
M
N
D
C
A
B
1
-6
-4
-2
O
x
2
4
6
8
10
THE END
GOD BLESS
&
Enjoy teaching