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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 1 Standard Form
Paper 1 1 0.009495 = 0.00950 (3 sig. fig.)
Answer: D
2 709 000 = 709 000 = 7.09 × 105
Answer: B
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
4 196 × 1010————–25 × 104
= 196 × 106————–
25
= 196 × 106
————–— 25
= 14 × 103———–
5 = 2.8 × 103
Answer: A
5
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 2 Quadratic Expressions and Equations
Paper 1 1 6p2 – p(3 – p) = 6p2 – 3p + p2
= 7p2 – 3p
Answer: C
2 If p = –2 is a root of the equation p2 – kp – 6 = 0, then we substitute p = –2 into p2 – kp – 6 = 0
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 4 Mathematical Reasoning
Paper 2 1 (a) (i) Only common multiples of 6 and 7
are divisible by 7. All other multiples of 6 are not divisible by 7.
∴ Some multiples of 6 are divisible by 7.
(ii) Hexagon means a six-sided polygon. ∴ All hexagons have 6 sides.
(b) The converse of ‘If p, then q’ is ‘If q, then p’. p: k � 4, q: k � 12 ∴ Converse: If k � 12, then k � 4. If k � 12, then k = 13, 14, 15, … All values greater than 12 are greater
than 4 (e.g. 13 � 4). ∴ The converse is true.(c) This is a form III type of argument. Premise 1: If p, then q. Premise 2: Not q is true. Conclusion: Not p is true. p: Set A is a subset of set B. q: A � B = A ∴ Premise 2: A � B ≠ A
2 (a)
Since Q � P, all elements of Q are also elements of P.
∴ Some elements of set Q are elements of set P.
(b) 8 – 15 = –4 is false but 6 × 6 = 65 × 6–3 is true. ↓ ↓
62 = 65 – 3 = 62
To make a compound statement true from one true and one false statement, the word ‘or’ must be used.
∴ 8 – 15 = –4 or 6 × 6 = 65 × 6–3.(c) The argument is a form III type of argument. Premise 1: If p, then q. Premise 2: Not q is true. Conclusion: Not p is true. ∴ q: n = 0, p: 5n = 0 ∴ Premise 1: If 5n = 0, then n = 0.
3 (a) –8 × (–5) = 40 and –9 � –3 ↓ ↓
‘True’ and ‘False’ is ‘False’. ∴ The statement is false.(b) Implication 1: If p, then q. Implication 2: If q, then p. Statement: p if and only if q.
p: x—y
is an improper fraction, q: x � y.
The required statement is x—y
is an
improper fraction if and only if x > y.(c) Argument form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. ∴ p: x � 7, q: x � 2 ∴ Premise 1: If x � 7, then x � 2.
4 (a) If ‘antecedent’, then ‘consequent’.
∴ If 1% = 1—–
100, then 20% of 200 = 40.
(b) Argument form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. p: cos θ = 0.5, q: θ = 60° Premise 2: cos θθ = 0.5
(c) 34
—–32 = 34 – 2
Let 4 = a and 2 = b.
∴ 3a
—–3b = 3a – b
P
Q
Generalisation
A � B is not A.
False statement
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 5 The Straight Line
Paper 1 1 4x + 3y = 12 3y = –4x + 12
y = – 4—3
x + 4
At y-intercept, x = 0 ∴ y = 4 ∴ y-intercept = 4
Answer: D 2 7x + 4y = 5
4y = –7x + 5
y = – 7—4
x + 5—4
y = mx + c
∴ m = – 7—4
∴ Gradient = – 7—4
Answer: B
3 P(–5, –6), Q(–3, 2), R(1, k) m
PQ = m
PR
2 – (–6)————––3 – (–5)
= k – (–6)
————–1 – (–5)
8—2
= k + 6——–
6
k + 6 = 24 k = 18
Answer: D
Paper 2 1 (a) 4x – 9y + 36 = 0
At G, x = 0, ∴ 4(0) – 9y + 36 = 0 9y = 36 y = 4 ∴ G(0, 4)(b) Let the equation of the straight line JK
be y = mx + c. m
JK = m
GH = 4—
9
y = 4—9
x + c
At J�–4 1—2
, 0�, 0 = 4—9 �– 9—
2 � + c
0 = –2 + c ∴ c = 2
∴ y = 4—9
x + 2
or 9y = 4x + 18 or 4x – 9y + 18 = 0
2 (a) O(0, 0), P(2, 6)
mOP
= 6 – 0——–2 – 0
= 3
The gradient of OP is 3.
(b) RQ//OP ∴ m
RQ = m
OP = 3
Let the equation of the straight line QR be y = 3x + c.
At point R(7, 3), y = 3 and x = 7. ∴ 3 = 3(7) + c 3 = 21 + c c = –18 The equation of the straight line QR is
y = 3x – 18.
(c) PQ//OR
∴ mPQ
= mOR
= 3—7
Let the equation of the straight line PQ
be y = 3—7
x + c.
At point P(2, 6), y = 6 and x = 2.
∴ 6 = 3—7
(2) + c
6 = 6—7
+ c
c = 36—–7
The y-intercept of the straight line
PQ is 36—–7
.
P, Q, R are points on a straight line.
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 6 Statistics III
Paper 1
1 Number of guidebooks 2 3 4 5 6
Frequency 3 7 8 10 8
Cumulative frequency 3 10 18 28 36
The mode is 5. Answer: C
2 Score Frequency
1 5
2 4
3 6
4 3
5 2
Mean, x = Σfx
——Σf
=
(1 × 5) + (2 × 4) + (3 × 6) + (4 × 3) + (5 × 2)
———————————— 5 + 4 + 6 + 3 + 2
= 53—–20
= 2.65
The scores higher than the mean (2.65) are 3, 4 and 5 with the frequencies 6, 3 and 2 participants respectively.
Hence, the number of participants getting scores higher than the mean score is 6 + 3 + 2 = 11
Answer: C
Paper 2
1 (a)
Mass (g) Upper boundary Tally Frequency Cumulative
frequency
600 – 619 619.5 2 2
620 – 639 639.5 3 5
640 – 659 659.5 10 15
660 – 679 679.5 12 27
680 – 699 699.5 7 34
700 – 719 719.5 4 38
720 – 739 739.5 2 40
(b) Mass (g) Upper boundary
Cumulative frequency
580 – 599 599.5 0
600 – 619 619.5 2
620 – 639 639.5 5
640 – 659 659.5 15
660 – 679 679.5 27
680 – 699 699.5 34
700 – 719 719.5 38
720 – 739 739.5 40
The ogive is as shown below.
Mode is the value of data with the highest frequency.
Cumulative frequency
O
40
35
30
25
20
15
10
5
Mass (g)599.5 619.5 639.5 659.5 679.5
667.5
699.5 719.5 739.5
(c) From the ogive,
(i) 1—2
× 40 fish = 20 fish
Hence, the median mass = 667.5 g (ii) The median mass means that 50%
(20) of the fish have masses of less than or equal to 667.5 g.
2 (a)
Average marks
Midpoint (x)
TallyFrequency
(f)fx
5 – 9 7 4 28
10 – 14 12 7 84
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Frequency
9
Averagemarks
8
7
6
5
4
3
2
1
09.54.5 19.514.5 24.5 29.5 34.5 39.5 44.5
Average marks
Midpoint (x)
TallyFrequency
(f)fx
15 – 19 17 9 153
20 – 24 22 8 176
25 – 29 27 5 135
30 – 34 32 4 128
35 – 39 37 5 185
40 – 44 42 3 126
Σf = 45 Σfx = 1015
(b) Mean = Σfx
——Σf
= 1015——–
45 = 22 5—
9
(c)
(d) Percentage of students who need to attend extra classes
= 9 + 7 + 4——–——
45 × 100
= 44 4—9
%
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 7 Probability I
Paper 1 1 S = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30} n(S) = 16
A = Event that the sum of digits of the number on the chosen card is even
A = {15, 17, 19, 20, 22, 24, 26, 28} n(A) = 8
P(A) = 8—–16
= 1—2
Answer: A
2 Graduate Non-graduate Total
Male 18
Female 28 4 32
Total 50
The information in the above table is given.
Number of graduate teachers = P(graduate teacher) × Total number of teachers
= 4—5
× 50
= 40
Thus, the table can now be completed, as shown below:
Graduate Non-graduate Total
Male 12 6 18
Female 28 4 32
Total 40 10 50
Hence, the number of male non-graduate teachers is 6.
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 9 Trigonometry II
Paper 1 1 tan θ = –1.7321 Basic ∠ = 60° θ = 360° – 60° θ = 300°
Answer: C
2 tan y° = 24—–7
QS—–QR
= 24—–7
QS—–14
= 24—–7
QS = 24—–7
× 14 QS = 48 cm
QT = 1—4
QS
QT = 1—4
(48)
QT = 12 cm
PT = PQ2 + QT 2
PT = 52 + 122
PT = 13 cm
300°
60°x
cos x° = –cos ∠PTQ
= –QT—–PT
= – 12—–13
Answer: B
3 The information on special angles of the unit circle is used to draw the graph of y = tan x°. Therefore, the graph of y = tan x° is D.
Answer: D
tan 90° = ∞
tan 180° = 0 tan 0° = 0tan 360° = 0
tan 270° = –∞
180°
90°
0°360°
270°
O
y
O 90° 180° 270° 360°x
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 10 Angles of Elevation and Depression
Paper 1 1
Answer: A 2 CB—–
AB = tan 16°
CB = AB × tan 16° = 35 × tan 16° = 10.0361
∴ The height of the pole, CB, is 10 m, correct to the nearest integer.
Answer: A
3 XW—–4.2
= tan 53°
XW = 4.2 × tan 53°
YW—–4.2
= tan 19°
YW = 4.2 × tan 19°
XY = XW – YW = 4.2(tan 53°) – 4.2(tan 19°) = 4.2(tan 53° – tan 19°) = 4.127 m = 4.1 m (correct to one decimal place)
Answer: B
R T
US
Bird
Cat
Angle of depression = ∠TRS
AB
C
35 m16°
X
Y
W V4.2 m
53°
19°
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SPM ZOOM-IN(Fully-worked Solutions)
Form 4: Chapter 11 Lines and Planes in 3-Dimensions
Paper 1 1
The angle between the line SM and the plane PTWS is ∠MSN,
where MN – Normal to the plane PTWS SN – Orthogonal projection on the
plane PTWS
Answer: B
2
The angle between the plane HGB and the plane DHGC is ∠BGC.
Answer: D
Paper 2 1
The angle between the line PM and the plane PSTU is ∠NPM.
In �NUP, using the Pythagoras’ Theorem,
NP = 42 + 62 = 52 = 7.2111 cm
In �NMP, tan ∠NPM = NM—––NP
tan ∠NPM = 8—–––7.2111
tan ∠NPM = 1.1094 ∠NPM = 47°58’
2
The angle between the plane SABM and the plane SDCR is ∠ASD.
Based on �SDA,
tan ∠ASD = AD—––SD
tan ∠ASD = 4—–20
∠ASD = 11°19’
P
Q R
S
T
U V
WN
M
A B
CDE F
GHJ
• The line of intersection of the planes HGB and DHGC is HG.
• The line that lies on the plane DHGC and is perpendicular to the line of intersection HG is GC.
• 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.
S
P Q
RS
T
U V
W
N M
6 cm
4 cm
8 cm
8 cm
A B
CD
P Q
RS M
20 cm
4 cm
• The line of intersection of the planes SABM and SDCRM is SM.
• The line that lies on the plane SABM and is perpendicular to the line of intersection (SM) is SA.
• 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 SDCR is the angle between the lines SA and SD, i.e. ∠ASD.
A D4 cm
20 cm
The angle between the line SM and its orthogonal projection (SN) is ∠MSN.
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SPM ZOOM-IN(Fully-worked Solutions)
Form 5: Chapter 1 Number Bases
Paper 1 1
Answer: B
2 1 1 0 1 0 02
– 1 1 12
0 10
1 1 0 1 0 02
– 1 1 12
0 10 10
1 1 0 1 0 0 – 1 1 12
0 12
0 10 10 10 10
1 1 0 1 0 02
– 1 1 12
1 0 1 1 0 12
Answer: C
1 1 1 1
1 1 1 1 12
+ 1 1 1 1 12
1 1 1 1 1 02
12 + 12 = 102
12 + 12 + 12 = 112
102 – 12 = 12
102 – 12 = 12
102 – 12 = 12
12 – 12 = 0
52 51 50
1 1 3
3 52 + 5 + 3 = 1 × 52 + 1 × 51 + 3 × 5°
∴ 52 + 5 + 3 = 1135
Answer: C
4 83 + 5 = 1 × 83 + 0 × 82 + 0 × 81 + 5 × 80
∴ 83 + 5 = 10058
Answer: A
5 1 110 111 011 0002
421 421 421 421 421
1 6 7 3 0
∴ 11101110110002 = 167308
Answer: D
83 82 81 80
1 0 0 5
⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩
⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩1 1
1
12 – 12 = 0
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16y = – ––– x
SPM ZOOM-IN(Fully-worked Solutions)
Form 5: Chapter 2 Graphs of Functions II
Paper 1 1 y = ax2
The greater the value of ‘a’, the graph will be closer to the y-axis.
∴ When a = 5, it is graph I,
a = 1, it is graph II and
a = 1—2
, it is graph III.
∴ I: a = 5, II: a = 1, III: a = 1—2
Answer: D
2 y = – 18—–x
is a reciprocal graph.
B is a quadratic graph.C and D are cubic graphs.
Answer: A
Paper 2 1 (a) Substitute x = –2, y = k
into y = 2x2 – 3x – 5. k = 2(–2)2 – 3(–2) – 5 k = 8 + 6 – 5 k = 9
Substitute x = 3, y = m into y = 2x2 – 3x – 5. m = 2(3)2 – 3(3) – 5 m = 18 – 9 – 5 m = 4
(b)
(c) From the graph, (i) when x = –1.5, y = 4, (ii) when y = 20, x = 4.35.
(d) To find the equation of the suitable straight line to be drawn, do the following:
y = 2x2 – 3x – 5 ......➀ 0 = 2x2 + 3x – 17......➁
➀ – ➁: y = –6x + 12 Draw the straight line y = –6x + 12 by
plotting the following points: When x = 0, y = –6(0) + 12 = 12. ∴ Plot (0, 12). When x = 1, y = –6(1) + 12 = 6. ∴ Plot (1, 6). When x = 2, y = –6(2) + 12 = 0. ∴ Plot (2, 0).
The solution from the graph is x = 2.25.
2 (a) Substitute x = –2 into y = – 16—–x
, then
y = –16——–2
= 8
Substitute x = 3 into y = – 16—–x
, then
y = –16——3
= –5.3
(b), (d)
(c) y = – 16—–x
......➀
0 = – 16—–x
+ 2x ......➁
➀ – ➁: y = –2x Draw the straight line y = –2x by
plotting the following points: When x = 0, y = –2(0) = 0. ∴ Plot (0, 0). When x = 1, y = –2(1) = –2. ∴ Plot (1, –2). When x = –1, y = –2(–1) = 2. ∴ Plot (–1, 2). From the graph, the solutions are
x = 2.85 and x = –2.85.
30
25
20
15
10
5
x
y
4
–2 –1
–1.5
O 1 2
2.25
3 4
4.35
5
–5
y = –6x + 12
y = 2x2 – 3x – 5
Graph drawn.
Equation that has to be solved such that 2x 2 + 3x – 17 = 0 is rearranged.
x
y
20
15
10
5
–4 –3
–2.85
–2 –1 O 1 22.85
3 4
–5
–10
–15 16y = – ––– x
x = 1y = 5x + 5
y = 5
y = –2x
Graph drawn.
Equation that has to be solved such that –
16—–x + 2x = 0
is rearranged.
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SPM ZOOM-IN(Fully-worked Solutions)
Form 5: Chapter 3 Transformations III
Paper 2
1 (a) (i) H(4, 4) ⎯→ H’(6, 1) ⎯→ H’’(0, 1)
(ii) H(4, 4) ⎯→ H’(2, 4) ⎯→ H’’(4, 1)
(b) X – Translation 5� � 3
Y – Anticlockwise rotation of 90° about the point N(7, 10)
(c) (i) Scale factor = 2, Centre = (–1, 8) (ii) Area of �EFG = 22 × Area of �ABC 52 = 4 × Area of �ABC Area of �ABC = 13 units2
∴ Area of �LMN = Area of �ABC = 13 units2
W V
V W
2 (a) (i) P(–2, 2) ⎯→ P’(0, 2) ⎯→ P’’(2, 1)
(ii) P(–2, 2) ⎯→ P’(0, 1) ⎯→ P’’(–1, 0)
(b) (i) V – Reflection in the straight line y = x
W – Enlargement with centre (4, –1) and a scale factor of 3
(ii) Area of �DEF = 32 × Area of �LMN 54 = 9 × Area of �LMN Area of �LMN = 6 units2
∴ Area of the shaded region = Area of �DEF – Area of �LMN = 54 – 6 = 48 units2
R T
T R
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