This document provides fully worked solutions to exam questions from Form 4 mathematics chapters on standard form, quadratic expressions and equations, sets, mathematical reasoning, the straight line, and statistics. The solutions include: 1) Detailed working to obtain the answers for multiple choice and structured questions. 2) Explanations of mathematical concepts and reasoning such as determining gradients, interpreting graphs, and identifying argument forms. 3) Step-by-step derivations to find equations of lines from given points and gradients.

1. 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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2. SPM ZOOM-IN
(Fully-worked Solutions)
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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3. SPM ZOOM-IN
(Fully-worked Solutions)
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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4. SPM ZOOM-IN
(Fully-worked Solutions)
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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5. SPM ZOOM-IN
(Fully-worked Solutions)
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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6. SPM ZOOM-IN
(Fully-worked Solutions)
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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7. 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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8. SPM ZOOM-IN
(Fully-worked Solutions)
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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9. Bab 8 tidak ada

10. SPM ZOOM-IN
(Fully-worked Solutions)
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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11. SPM ZOOM-IN
(Fully-worked Solutions)
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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12. SPM ZOOM-IN
(Fully-worked Solutions)
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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13. SPM ZOOM-IN
(Fully-worked Solutions)
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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14. SPM ZOOM-IN
(Fully-worked Solutions)
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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15. SPM ZOOM-IN
(Fully-worked Solutions)
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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16. SPM ZOOM-IN
(Fully-worked Solutions)
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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17. ΂ 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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Suc Math SPM (Zoom IN).indd 17 10/9/2008 1:58:32 PM

18. SPM ZOOM-IN
(Fully-worked Solutions)
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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19. SPM ZOOM-IN
(Fully-worked Solutions)
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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20. SPM ZOOM-IN
(Fully-worked Solutions)
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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21. (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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22. SPM ZOOM-IN
(Fully-worked Solutions)
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
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23. SPM ZOOM-IN
(Fully-worked Solutions)
Form 5: Chapter 9 Earth as a Sphere
Paper 1 (a) (i) The latitude of point R is 35°S.
1 N (ii) The longitude of point L is
D
(180 – 70)°E = 110°E
(b) Distance of ML (along the parallel of
50° latitude)
O 40°
32° 148° = 180 × 60 × cos 35°
= 8846.8 n.m.
(c) Distance of LMR (via the North Pole)
0°
S
148°E
= 180 × 60
Based on the above diagram: = 10 800 n.m.
The latitude of point D Average speed
= (90 – 50)°N = 40°N Distance
= ————
Speed
The longitude of point D
= (180° – 32°)E = 148°E 10 800
= ———
600
Hence, the location of point D is
= 18 hours
(40°N, 148°E).
Answer: A Hence, the time the aeroplane reached point
R is 0500 + 1800 = 2300.
2 N
0°
2 N
C J
30°N
L
O 40° O 30°
50°
30°
K
R 80° P
100°
Q M
S
35°E
The latitude of point R = (90 – 50)°S S
= 40°S
(a) The position of point Q is
The longitude of point R = (180 – 80)°W
(30°S, (180 – 35)°W) = (30°S, 145°W)
= 100°W
(b) (i) JK = 3300 n.m.
Hence, the position of point R is ∠JOK × 60 = 3300
(40°S, 100°W). 3300
∠JOK = ——–
60
Answer: B ∠JOK = 55°
Paper 2
1 30°N J
N
M L 35° N
55°
180°
35° Equator 0°
35° O
x°S K
R
110°E
S
70°W
∴ x = 55 – 30 = 25
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24. (ii) JL = 4936 n.m. Total distance travelled
(c) Time taken = ———————————
∠JCL × 60 × cos 30° = 4936 Speed
4936 JL + LM
∠JCL = —————– = ————
60 × cos 30° 600
4936 + [(30 + 40) × 60]
∠JCL = 95° = ———————————
600
4936 + 4200
y°W 0° 35°E
= ——————
600
9136
L G J = ———
600
95°
= 15.23 hours
= 15 hours 14 minutes
∴ y = 95 – 35 = 60
0.23 hours
= 0.23 × 60
= 13.8
= 14 (correct to the nearest minute)
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25. SPM ZOOM-IN
(Fully-worked Solutions)
Form 5: Chapter 10 Plans and Elevations
Paper 2 2 (a) C/D B/A
1 (a)
K/Q 4 cm L/P 2 cm
J/I
5 cm
4 cm 3 cm
A/D B/C
K/L 4 cm F/E 3 cm G/H
Elevation as viewed from Y
J/G M/N
Plan
(b) (i)
(b) (i)
4 cm N/R/S W/Q/T
A B P/B/G
M/A/H 1 cm
1 cm
2 cm
H/J/K 4 cm 3 cm
M/L
6 cm D/I/E C/J/F
4 cm
4 cm
D/G/Q 4 cm N/D/P 4 cm E/F/C
L 8 cm K
Elevation as viewed from X Plan
(ii)
(ii) M N W P
B/A
2 cm
2 cm R Q
D/A C/B
5 cm M/J
H L/K 2 cm
I J
4 cm 4 cm
3 cm
4 cm 1 cm 6 cm 1 cm
E/D 5 cm F/N/G C/D P/Q L/E/H S T K/F/G
Elevation as viewed from Y Elevation as viewed from X
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