Nota math-spm

SPM ZOOM-IN
(Fully-worked Solutions)
Form 4: Chapter 1 Standard Form

Paper 1 196 × 1010 196 × 106
1 0.009495 = 0.00950 (3 sig. fig.) 4 ————– = ————–
25 × 104 25
5 196 × 106
= ————–—
Answer: D 25
14 × 103
2 709 000 = 709 000 = ———–
5
= 7.09 × 105 = 2.8 × 103
Answer: B Answer: A
3 0.049 + 3 × 10–4
= 4.9 × 10–2 + 0.03 × 102 × 10–4
= 4.9 × 10–2 + 0.03 × 10–2
= (4.9 + 0.03) × 10–2
= 4.93 × 10–2
Answer: A

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Suc Math SPM (Zoom IN).indd 1 10/9/2008 1:56:47 PM

SPM ZOOM-IN
Form 4: Chapter 2 Quadratic Expressions and Equations

Paper 1 Paper 2
1 6p2 – p(3 – p) 3p2 + 10p
1 ———— =3
= 6p2 – 3p + p2 p+2
= 7p2 – 3p 3p2 + 10p = 3(p + 2)
Answer: C 3p2 + 10p = 3p + 6
2
3p + 10p – 3p – 6 = 0
2 If p = –2 is a root of the equation 3p2 + 7p – 6 = 0
p2 – kp – 6 = 0, then we substitute p = –2 into (3p – 2)(p + 3) = 0
p2 – kp – 6 = 0 3p – 2 = 0 or p+3=0
2
(–2) – k(–2) – 6 = 0 3p = 2 p = –3
p =— 2
4 + 2k – 6 = 0
3
2k – 2 = 0
∴p= — 2 or –3
2k = 2
3
k =—2
2 3(x2 + 9)
2 ———— = 9
k =1 2x
3(x2 + 9) = 9(2x)
Answer: A 3x2 + 27 = 18x
2
3x – 18x + 27 = 0
3(x2 – 6x + 9) = 0
3(x – 3)(x – 3) = 0
x–3= 0
∴x = 3

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SPM ZOOM-IN
Form 4: Chapter 3 Sets

Paper 1 Paper 2
1 ξ 1 A B
A I V
B II IV
III
I II III
C

(a) C : II, III, IV
A’ : III, IV, V
A : Regions I and II
B’ : I, II, III
B : Region I only
A’ : Region III only
C ഫ A’ : II, III, IV, V
B’ : Regions II and III
C ഫ A’ പ B’ : II, III
Shaded region: Region II only
Answer:
∴ Shaded region is A പ B’ = B’ പ A
↓ ↓ A B

I and II II and III II
III
Answer: B C

2 AʚBʚC (b)
A B C
C
A B I II III
IV
I III
II

A’ : II, III
C’ : I, II
A’ : II, III, IV
A’ പ C’ : II
B’ : III, IV
(A’ പ C’)’ : I, III
C’ : IV
A’ പ B’ : III, IV ∴ A’ പ B’ = B’ Answer:
A’ പ C’ : IV ∴ A’ പ C’ = C’ A B C

Answer: A
3 P Q R

4 1 2 y
2 ξ= {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
(a) P = {19}
n(Q’ പ R) = y (b) Q={ } Since the universal set is less
than 22
n(Q’ പ P) = 4 (c) n(Q) = 0
(d) Q’ = {11, 12, 13, 14, 15, 16, 17, 18, 19,
n(Q’ പ R) – 3 = n(Q’ പ P) 20}
y–3= 4 P പ Q’ = {19}
y= 4+3 ∴ n(P പ Q’) = 1
y = 7
∴ n(R) = 2 + y
=2+7
=9
Answer: B

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SPM ZOOM-IN
Form 4: Chapter 4 Mathematical Reasoning

Paper 2 ∴ 8 – 15 = –4 or 6 × 6 = 65 × 6–3.
1 (a) (i) Only common multiples of 6 and 7 (c) The argument is a form III type of argument.
are divisible by 7. All other multiples Premise 1: If p, then q.
of 6 are not divisible by 7. Premise 2: Not q is true.
∴ Some multiples of 6 are Conclusion: Not p is true.
divisible by 7. ∴ q: n = 0, p: 5n = 0
(ii) Hexagon means a six-sided polygon. ∴ Premise 1: If 5n = 0, then n = 0.
∴ All hexagons have 6 sides. 3 (a) –8 × (–5) = 40 and –9 Ͼ –3
(b) The converse of ‘If p, then q’ is ‘If q, then p’. ↓ ↓
p: k Ͼ 4, q: k Ͼ 12 ‘True’ and ‘False’ is ‘False’.
∴ Converse: If k Ͼ 12, then k Ͼ 4. ∴ The statement is false.
If k Ͼ 12, then k = 13, 14, 15, … (b) Implication 1: If p, then q.
All values greater than 12 are greater Implication 2: If q, then p.
than 4 (e.g. 13 Ͼ 4). Statement: p if and only if q.
∴ The converse is true. x
p: — is an improper fraction, q: x Ͼ y.
(c) This is a form III type of argument. y
x
The required statement is — is an
Premise 1: If p, then q. y
Premise 2: Not q is true. improper fraction if and only if x > y.
Conclusion: Not p is true. (c) Argument form II
p: Set A is a subset of set B. Premise 1: If p, then q.
q: A പ B = A Premise 2: p is true.
∴ Premise 2: A പ B ≠ A A പ B is not A. Conclusion: q is true.
∴ p: x Ͼ 7, q: x Ͼ 2
2 (a) P
∴ Premise 1: If x Ͼ 7, then x Ͼ 2.
Q
4 (a) If ‘antecedent’, then ‘consequent’.
1
∴ If 1% = —– , then 20% of 200 = 40.
100
Since Q ʚ P, all elements of Q are also
(b) Argument form II
elements of P.
Premise 1: If p, then q.
∴ Some elements of set Q are elements
Premise 2: p is true.
of set P. False statement
Conclusion: q is true.
(b) 8 – 15 = –4 is false but
p: cos θ = 0.5, q: θ = 60°
6 × 6 = 65 × 6–3 is true.
Premise 2: cos θ = 0.5
↓ ↓ 34
62 = 65 – 3 = 62 (c) —– = 34 – 2
32
To make a compound statement true Let 4 = a and 2 = b.
from one true and one false statement, 3a
the word ‘or’ must be used. ∴ —– = 3a – b Generalisation
3b

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SPM ZOOM-IN
Form 4: Chapter 5 The Straight Line

Paper 1 4
∴ y = —x + 2
1 4x + 3y = 12 9
3y = –4x + 12 or 9y = 4x + 18
4
y = –—x + 4 or 4x – 9y + 18 = 0
3
2 (a) O(0, 0), P(2, 6)
At y-intercept, x = 0 6–0
∴y=4 mOP = ——–
2–0
∴ y-intercept = 4
=3
Answer: D
2 7x + 4y = 5 The gradient of OP is 3.
4y = –7x + 5
(b) RQ//OP
7 5
y = – —x + — ∴ mRQ = mOP = 3
4 4
y = mx + c Let the equation of the straight line QR
∴ m = –— 7 be y = 3x + c.
4
7 At point R(7, 3), y = 3 and x = 7.
∴ Gradient = – —
4 ∴ 3 = 3(7) + c
Answer: B 3 = 21 + c
c = –18
3 P(–5, –6), Q(–3, 2), R(1, k) The equation of the straight line QR is
mPQ = mPR P, Q, R are points on y = 3x – 18.
a straight line.
2 – (–6) k – (–6) (c) PQ//OR
————– = ————–
–3 – (–5) 1 – (–5) 3
8 k+6 ∴ mPQ = mOR = —
— = ——– 7
2 6
k+6 = 24 Let the equation of the straight line PQ
k = 18 3
be y = — x + c.
7
Answer: D
At point P(2, 6), y = 6 and x = 2.
Paper 2 3
1 (a) 4x – 9y + 36 = 0 ∴ 6 = — (2) + c
7
At G, x = 0, 6
∴ 4(0) – 9y + 36 = 0 6 =—+c
7
9y = 36 36
y=4 c = —–
7
∴ G(0, 4)
(b) Let the equation of the straight line JK The y-intercept of the straight line
be y = mx + c. 36
4 PQ is —– .
mJK = mGH = — 7
9
4
y = —x + c
9
1
2΂ ΃4 9
At J –4 — , 0 , 0 = — – — + c
9 2 ΂ ΃
0 = –2 + c
∴c=2

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SPM ZOOM-IN
Form 4: Chapter 6 Statistics III

Paper 1 Upper Cumulative
(b) Mass (g)
1 Number of guidebooks 2 3 4 5 6 boundary frequency
Frequency 3 7 8 10 8 580 – 599 599.5 0
Cumulative frequency 3 10 18 28 36 600 – 619 619.5 2
The mode is 5. Mode is the value of data 620 – 639 639.5 5
with the highest frequency.
Answer: C 640 – 659 659.5 15
660 – 679 679.5 27
2 Score Frequency
1 5 680 – 699 699.5 34

2 4 700 – 719 719.5 38

3 6 720 – 739 739.5 40
4 3 The ogive is as shown below.
5 2
Cumulative frequency
Σfx
Mean, x = ——
Σf 40
(1 × 5) + (2 × 4) + (3 × 6)
+ (4 × 3) + (5 × 2)
= ————————————
35

5+4+6+3+2 30
53
= —–
20 25
= 2.65 20
The scores higher than the mean (2.65) are
15
3, 4 and 5 with the frequencies 6, 3 and
2 participants respectively. 10

Hence, the number of participants getting 5
667.5
scores higher than the mean score is
O Mass (g)
6 + 3 + 2 = 11 599.5 619.5 639.5 659.5 679.5 699.5 719.5 739.5

Answer: C

Paper 2 (c) From the ogive,
1
(i) — × 40 fish = 20 fish
1 (a) 2
Upper Cumulative Hence, the median mass = 667.5 g
Mass (g) Tally Frequency (ii) The median mass means that 50%
boundary frequency
600 – 619 619.5 2 2 (20) of the fish have masses of
less than or equal to 667.5 g.
620 – 639 639.5 3 5
640 – 659 659.5 10 15 2 (a)

660 – 679 679.5 12 27
Average Midpoint Frequency
Tally fx
680 – 699 699.5 7 34 marks (x) (f)
700 – 719 719.5 4 38 5–9 7 4 28
720 – 739 739.5 2 40 10 – 14 12 7 84

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Average Midpoint Frequency (c) Frequency
Tally fx
marks (x) (f) 9
15 – 19 17 9 153 8

20 – 24 22 8 176 7
6
25 – 29 27 5 135
5
30 – 34 32 4 128 4
35 – 39 37 5 185 3

40 – 44 42 3 126 2

1
Σf = 45 Σfx = 1015
0 Average
4.5 9.5 14.5 19.5 24.5 29.5 34.5 39.5 44.5 marks

Σfx 1015 5 (d) Percentage of students who need to
(b) Mean = —— = ——– = 22 —
Σf 45 9 attend extra classes
9+7+4
= ——–—— × 100
45
4%
= 44 —
9

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SPM ZOOM-IN
Form 4: Chapter 7 Probability I

Paper 1 Thus, the table can now be completed, as
1 S = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, shown below:
25, 26, 27, 28, 29, 30}
n(S) = 16 Graduate Non-graduate Total
Male 12 6 18
A = Event that the sum of digits of the
Female 28 4 32
number on the chosen card is even
A = {15, 17, 19, 20, 22, 24, 26, 28} Total 40 10 50
n(A) = 8 Hence, the number of male non-graduate
teachers is 6.
8 1
P(A) = —– = —
16 2 Answer: A

Answer: A 3 Marks Number of students
1 – 40 h
2 Graduate Non-graduate Total 41 – 70 88
Male 18 71 – 100 8
Female 28 4 32 14
P(marks not more than 70) = —–
15
Total 50 h + 88 14
——–——– = —–
h + 88 + 8 15
The information in the above table is given. h + 88 14
——–— = —–
h + 96 15
Number of graduate teachers
= P(graduate teacher) × Total number of 15(h + 88) = 14(h + 96)
teachers 15h + 1320 = 14h + 1344
4 15h – 14h = 1344 – 1320
= — × 50 h = 24
5
= 40 Answer: A

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SPM ZOOM-IN
Form 4: Chapter 9 Trigonometry II

Paper 1 cos x° = –cos ∠PTQ
1 tan θ = –1.7321 QT
300° = – —–
Basic ∠ = 60° PT
θ = 360° – 60°
x
60° 12
= – —–
θ = 300° 13
Answer: C Answer: B
24 3 The information on special angles of the
2 tan y° = —–
7 unit circle is used to draw the graph of
QS —– 24 y = tan x°. Therefore, the graph of
—– =
QR 7 y = tan x° is D.
QS —– 24 tan 90° = ∞
—– =
14 7 90°
24
QS = —– × 14
7 tan 180° = 0
180° 0° tan 0° = 0
O 360° tan 360° = 0
QS = 48 cm
1
QT = — QS
270°
tan 270° = – ∞
4
1 y
QT = — (48)
4
QT = 12 cm x
O 90° 180° 270° 360°
PT = PQ2 + QT 2
PT = 52 + 122
PT = 13 cm Answer: D

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SPM ZOOM-IN
Form 4: Chapter 10 Angles of Elevation and Depression

Paper 1 X XW
Bird 3 —– = tan 53°
1 4.2
R T XW = 4.2 × tan 53°
Angle of depression
= ∠TRS YW
53° —– = tan 19°
Y 4.2
U
W V YW = 4.2 × tan 19°
S 4.2 m 19°
Answer: A Cat

CB XY = XW – YW
2 —– = tan 16° C
AB = 4.2(tan 53°) – 4.2(tan 19°)
CB = AB × tan 16° = 4.2(tan 53° – tan 19°)
16°
= 35 × tan 16° 35 m = 4.127 m
B A
= 10.0361 = 4.1 m (correct to one decimal place)
∴ The height of the pole, CB, is 10 m, Answer: B
correct to the nearest integer.
Answer: A

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SPM ZOOM-IN
Form 4: Chapter 11 Lines and Planes in 3-Dimensions

Paper 1 The angle between the line PM and the
1
S
plane PSTU is ∠NPM.
P
In ⌬NUP, using the Pythagoras’ Theorem,
R
Q
NP = 42 + 62 = 52 = 7.2111 cm
NM
T
N W In ⌬NMP, tan ∠NPM = —––
NP
8
tan ∠NPM = —–––
U M V
7.2111
The angle between the line SM and the tan ∠NPM = 1.1094
plane PTWS is ∠MSN, ∠NPM = 47°58’
where
MN – Normal to the plane PTWS 2
SN – Orthogonal projection on the S M R
plane PTWS
P Q
The angle between the line SM
and its orthogonal projection
20 cm
(SN) is ∠MSN.
D
Answer: B 4 cm
C

A B
2 J H G
The angle between the plane SABM and the
plane SDCR is ∠ASD.
E F
D C

• The line of intersection of the planes SABM and
A SDCRM is SM.
B
• The line that lies on the plane SABM and is
The angle between the plane HGB and the perpendicular to the line of intersection (SM) is
SA.
plane DHGC is ∠BGC. • The line that lies on the plane SDCR and is
perpendicular to the line of intersection (SM) is
SD.
• The angle between the plane SABM and the plane
• The line of intersection of the planes HGB and SDCR is the angle between the lines SA and SD,
DHGC is HG. i.e. ∠ASD.
• The line that lies on the plane DHGC and is
perpendicular to the line of intersection HG is GC.
S
• The line that lies on the plane HGB and is
perpendicular to the line of intersection HG is GB.
• Hence, the angle between the plane HGB and the
plane DHGC is the angle between the lines GC
and GB, i.e. ∠BGC.
20 cm
Answer: D

Paper 2 A 4 cm D

1 W
T
N M
Based on ⌬SDA,
4 cm AD
U V tan ∠ASD = —––
R
SD
S
4
8 cm
6 cm
tan ∠ASD = —–
20
∠ASD = 11°19’
P 8 cm Q

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SPM ZOOM-IN
Form 5: Chapter 1 Number Bases

Paper 1 3 52 + 5 + 3 = 1 × 52 + 1 × 51 + 3 × 5°
1 1 1 1 1

1 1 1 1 12 52 51 50
12 + 12 = 102
+ 1 1 1 1 12 1 1 3
12 + 12 + 12 = 112
1 1 1 1 1 02
∴ 52 + 5 + 3 = 1135
Answer: B
Answer: C
2 1 1 0 1 0 02
102 – 12 = 12
– 1 1 12
12 – 12 = 0
4 83 + 5 = 1 × 83 + 0 × 82 + 0 × 81 + 5 × 80

0 10
83 82 81 80
1 1 0 1 0 02 1 0 0 5
– 1 1 12
∴ 83 + 5 = 10058
1
0 10 10
11010 0 Answer: A
– 1 1 12
5 1 110 111 011 0002
0 12
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
102 – 12 = 12
421 421 421 421 421
1 1
12 – 12 = 0
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
⎧
⎨
⎩
0 10 10 10 10
1 1 0 1 0 02 1 6 7 3 0
– 1 1 12
∴ 11101110110002 = 167308
1 0 1 1 0 12

102 – 12 = 12
Answer: D

Answer: C

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SPM ZOOM-IN
Form 5: Chapter 2 Graphs of Functions II

Paper 1 y = 2x2 – 3x – 5 ......➀ Graph drawn.

1 y = ax2 0 = 2x2 + 3x – 17......➁
The greater the value of ‘a’, the graph will be Equation that has to be solved
➀ – ➁: such that 2x 2 + 3x – 17 = 0 is
closer to the y-axis.
y = –6x + 12 rearranged.
∴ When a = 5, it is graph I, Draw the straight line y = –6x + 12 by
a = 1, it is graph II and plotting the following points:
1
a = — , it is graph III. When x = 0, y = –6(0) + 12 = 12.
2 ∴ Plot (0, 12).
1
∴ I: a = 5, II: a = 1, III: a = — When x = 1, y = –6(1) + 12 = 6.
2 ∴ Plot (1, 6).
Answer: D
When x = 2, y = –6(2) + 12 = 0.
18
2 y = – —– is a reciprocal graph. ∴ Plot (2, 0).
x
B is a quadratic graph. The solution from the graph is x = 2.25.
C and D are cubic graphs. 16
2 (a) Substitute x = –2 into y = – —– , then
x
Answer: A –16
y = —— = 8
–2
Paper 2 16
Substitute x = 3 into y = – —– , then
1 (a) Substitute x = –2, y = k x
–16
into y = 2x2 – 3x – 5. y = —— = –5.3
3
k = 2(–2)2 – 3(–2) – 5 (b), (d) y
k=8+6–5
20
k=9 x=1
15 y = 5x + 5
Substitute x = 3, y = m
into y = 2x2 – 3x – 5. 16
y = – –––
10
x y=5
m = 2(3)2 – 3(3) – 5 5
m = 18 – 9 – 5 –2.85 2.85
x
m=4 –4 –3 –2 –1 O 1 2 3 4
–5
(b) y
–10 y = –2x
y = 2x2 – 3x – 5 16
y = – –––
30 –15 x
y
=

25
–6

16
x

(c) y = – —– ......➀ Graph drawn.
+

20
x
12

Equation that
15 16 has to be solved
0 = – —– + 2x ......➁
10
x such that
16
➀ – ➁: – —– + 2x = 0
x
5 4 y = –2x is rearranged.
–1.5 2.25 4.35
–2 –1 O 1 2 3 4 5
x Draw the straight line y = –2x by
–5 plotting the following points:
When x = 0, y = –2(0) = 0.
∴ Plot (0, 0).
(c) From the graph, When x = 1, y = –2(1) = –2.
(i) when x = –1.5, y = 4, ∴ Plot (1, –2).
(ii) when y = 20, x = 4.35. When x = –1, y = –2(–1) = 2.
(d) To find the equation of the suitable ∴ Plot (–1, 2).
straight line to be drawn, do the From the graph, the solutions are
following: x = 2.85 and x = –2.85.

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SPM ZOOM-IN
Form 5: Chapter 3 Transformations III

Paper 2 2 (a) (i) P(–2, 2) ⎯→ P’(0, 2) ⎯→ P’’(2, 1)
R T

W V
1 (a) (i) H(4, 4) ⎯→ H’(6, 1) ⎯→ H’’(0, 1) (ii) P(–2, 2) ⎯→ P’(0, 1) ⎯→ P’’(–1, 0)
T R

V W
(ii) H(4, 4) ⎯→ H’(2, 4) ⎯→ H’’(4, 1) (b) (i) V – Reflection in the straight line
y=x
(b) X – Translation 5
3΂΃ W – Enlargement with centre
Y – Anticlockwise rotation of 90° (4, –1) and a scale factor of 3
about the point N(7, 10) (ii) Area of ⌬DEF = 32 × Area of ⌬LMN
(c) (i) Scale factor = 2, Centre = (–1, 8) 54 = 9 × Area of ⌬LMN
(ii) Area of ⌬EFG = 22 × Area of ⌬ABC Area of ⌬LMN = 6 units2
52 = 4 × Area of ⌬ABC ∴ Area of the shaded region
Area of ⌬ABC = 13 units2 = Area of ⌬DEF – Area of ⌬LMN
∴ Area of ⌬LMN = Area of ⌬ABC = 54 – 6
= 13 units2 = 48 units2

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SPM ZOOM-IN
Form 5: Chapter 4 Matrices

Paper 1
1 3(6 p) + q(3 –3) = (15 12)
(18 3p) + (3q –3q) = (15 12)
΂
6 –7 y ΃΂ ΃
(b) 4 –5 x = –2
4 ΂ ΃
∴ 18 + 3q = 15 ......➀ ΂ ΃΂
1 –7 5 4 –5 x
—
2 –6 4 6 –7 y ΃΂
1
= — –7 5 –2
2 –6 4 4 ΃ ΂ ΃΂ ΃
3q = 15 – 18
1 0 x = — (–7 × –2) + (5 × 4)
3q = –3
0 1 y΂ ΃΂ 1
΃ ΂
2 (–6 × –2) + (4 × 4) ΃
3
q = –—
q = –1
3 ΂
1
x = — 14 + 20
y 2 12 + 16 ΃ ΂ ΃
3p + (–3q) = 12
1
= — 34
2 28 ΂ ΃
3p – 3q = 12 ......➁

΂ ΃
34
—–
Substitute q = –1 into ➁: = 2
28
∴ 3p – 3(–1) = 12 —–
2
3p + 3 = 12
3p = 12 – 3 = 17
14 ΂ ΃
3p = 9
9 ∴ x = 17 and y = 14
p= —
3
p=3 2 (a) If no inverse, ad – bc = 0.
∴ (2 × –4) – (4 × d) = 0
∴ p + q = 3 + (–1) = 3 – 1 = 2 –8 – 4d = 0
Answer: A –4d = 8
8
d = –—
7 ΂΃ ΂΃ ΂΃
2 1 + p = 4
3 q d = –2
4

1+p=4 7+3=q
∴p=4–1 ∴ q = 10 ΂
(b) Q = 2 –3 if d = –3
4 –4 ΃
p=3 1
p × q = 3 × 10 = 30 Q = ——————–——– –4
–1
(2 × –4) – (–3 × 4) –4 ΂ 3
2 ΃
1
Answer: C = ———– –4 3
–8 + 12 –4 2 ΂ ΃
Paper 2
1
= — –4 3
4 –4 2 ΂ ΃
1 (a) Inverse of 4 –5 ΂ ΃
΂ ΃
3
6 –7 –1 — 4
=
1
΂ ΃
1
= ————————— –7 5 –1 —
(4 × –7) – (–5 × 6) –6 4 2
1
= ——————– –7 ΂ 5
΃
(c) QP = ΂΃2
6
(–28) – (–30) –6 4
1 ΂ ΃΂ ΃
2 –3 a = 2
4 –4 b 6 ΂΃
= ———— –7 ΂ 5
΃
–28 + 30 –6 4 — ΂ ΃΂
1 –4 3 2 –3 a
4 –4 2 4 –4 b ΃΂ 1
΃
= — –4 3 2
4 –4 2 6 ΂ ΃΂ ΃
1
= — –7
2 –6 ΂ 4 ΃ ΂
5 = k –7
–6
q
4 ΃ 1 0 a = — (–4 × 2) +
΂ ΃΂
0 1 b
1
΃
4 (–4 × 2) + ΂ (3 × 6)
(2 × 6) ΃
1
∴ k = — and q = 5
2
΂ 1
΃
a = — –8 + 18
b 4 –8 + 12 ΂ ΃
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΂ a ΃ = — ΂ 10 ΃
b
1
4 4

΂ ΃
10
—–
= 4
1

΂ ΃
5
—
= 2
1
5 1
∴ a = — or 2 — , b = 1
2 2

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SPM ZOOM-IN
Form 5: Chapter 5 Variations

Paper 1 s2
3 r ∝ —–
1 t
1 Given s ∝ —– ,
r2 ks2 , where k is a constant
r = —–
k t
∴ s = —– k is a constant.
r2 When r = 8, s = 2 and t = 3,
When r = 2 and s = 5, k(2)2
8 = —–—
5 = —–k 3
22 24 = 4k
k = 20 k=6
20 6s2
∴ s = —– ∴ r = —–
r2 t
Answer: D When r = 27, s = 6 and t = u,
6(6)2
2 s ∝ r3 27 = —–—
u
s = kr3, where k is a constant 216
u = —––
When s = 192 and r = 4, 27
192 = k(4)3 u=8
k=3 Answer: C
∴ s = 3r3
When s = –24,
–24 = 3r3
r3 = –8
r = –2
Answer: B

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SPM ZOOM-IN
Form 5: Chapter 6 Gradient and Area Under a Graph

Paper 2 2 (a) Total distance travelled by the particle
1 (a) Average speed of the lorry for the whole for the whole journey is 310 m.
journey from point P to point Q Total area under the graph = 310
Total distance travelled 1 1
= ——————————— (4 × 25) + — (25 + 40)(4) + — (t – 8)(40) = 310
Total time taken 2 2
300 100 + 130 + 20(t – 8) = 310
= —— 20(t – 8) = 80
16
3 t–8=4
= 18 — m s–1 t = 12
4
(b) Speed of the car for the whole journey (b) Rate of speed of the particle from the
= Gradient of the straight line ABC 4th second to the 8th second
= Gradient of the graph from the 4th
΂ ΃
Vertical axis
= – ——————— second to the 8th second
Horizontal axis
40 – 25
= – ————΂
300 – 0
10 – 0 ΃ = ————
8–4
3
= –30 m s–1 = 3 — m s–2
4
Hence, the speed of the car for the (c) Average speed of the particle in the first
whole journey from point Q to point P 8 seconds
is 30 m s–1. Total distance
(c) The point on the distance–time graph = ——————–
Total time
when the lorry and the car meet is the
intersection point of the graph OBD and Area under the graph in the first 8 s
= ——————–——————————
the graph ABC, i.e. point B. 8
Hence, the distance from point Q when 100 + 130 From (a)
= —————
the lorry and the car meet is 8
300 – 60 = 240 m 3
= 28 — m s–1
4

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SPM ZOOM-IN
Form 5: Chapter 7 Probability II

Paper 1 3 Let
1 Let B = Event of drawing a blue ball
R = Event of obtaining a round biscuit R = Event of drawing a red ball
Sq = Event of obtaining a square biscuit S = Sample space
T = Event of obtaining a triangular biscuit
5
S = Sample space Given P(R) = — ,
8
P(T) = 1 – P(R) – P(Sq) n(R) 5
—–— = —
3 1 n(S) 8
P(T) = 1 – — – —
7 4
9 n(R) 5
P(T) = —– —–— = —
28 32 8
5
n(T)
—–— = —–
9 n(R) = — × 32
n(S) 28 8
36 9 n(R) = 20
—–— = —–
n(S) 28
9 × n(S) = 36 × 28 Let the number of blue balls added = h
36 × 28
n(S) = ————
Therefore, n(S) = 32 + h
9 5
P(R) = — New value of P(R)
n(S) = 112 9
n(R) 5
n(R) + n(Sq) + n(T) = 112 —–— = —
n(S) 9
n(R) + n(Sq) + 36 = 112
20 5
n(R) + n(Sq) = 112 – 36 ——— = —
32 + h 9
n(R) + n(Sq) = 76
5(32 + h) = 180
Answer: C 160 + 5h = 180
5h = 20
2 Let h =4
G = Event of obtaining a green disc Hence, the number of new blue balls that
B = Event of obtaining a blue disc have to be added to the bag is 4.
S = Sample space
Answer: D
6 5
P(B) = 1 – P(G) = 1 – —– = —–
11 11
Paper 2
n(B) 5 1 (a) P(letter M)
—–— = —–
n(S) 11 n(M)
= —–—
30 5 n(S)
—–— = —– 2+5
n(S) 11 = ——————
2+5+3+4
5 × n(S) = 30 × 11 7
30 × 11 = —–
n(S) = ———— 14
5 1
n(S) = 66 =—
2

∴ n(G) = n(S) – n(B) = 66 – 30 = 36

Answer: A

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(b) P(both the cards drawn are cards with 1
2 (a) P(Z) = —
the letter N) 5
After 1 card with the letter Number of male students
7
= —– × —– 6 N is taken out, it is left with from school Z 1
14 13 6 cards with the letter N out —————————————— = —
of the balance of 13 cards. Total number of male students 5
from all the three schools
Initially, there are 7 10 1
cards with the letter —————– = —
N out of 14 cards. k + 22 + 10 5
k + 32 = 50
3
= —– k = 18
13
(b) P(Two students from school Y are of the
(c) P(both the cards drawn are of different same gender)
colours) = P(MM or FF)
G – Green
= P(GY or YG) Y – Yellow = P(MM) + P(FF)
= P(GY) + P(YG)
5
΂ 9
= —– × —– + —– × —–9
΃ ΂
5
΃ ΂ 22
40
21
39 ΃ ΂ 18
= —– × —– + —– × —–
40
17
39 ΃
14 13 14 13 32
= —–
After 1 green card is taken 65
out, it is left with 13 cards
Initially, there are 5 and so there are 9 yellow
green cards out of cards out of the 13 cards.
14 cards.

45
= —–
91

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SPM ZOOM-IN
Form 5: Chapter 8 Bearing

Paper 1 Let the bearing of point K from point H be
1 North
θ.
θ = 360° – 35° – 40° = 285°
North
B Answer: C
60°
60°
3 Label the east and north direction and write
A Bearing of A
120°
from B down all the provided information onto the
diagram.
Bearing of A from B Alternate angles
are equal.
= 180° + 60° North

= 240° North

Q
Answer: C 30°
P 30°
60°
30°
2 Label the north direction and write down all
the provided information onto the diagram.
60°
Bearing of F Alternate angles
from K = 065° North are equal. R

North F
North
65° 35°
East
65°
40° 35°
K 40° From the above diagram, the bearing of
H point Q from point P is 030°.
180° – 100° θ
∠FHK = —————–
2 Answer: A

22 Weblink


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