SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 1 Standard Form Paper 1 1 0.009495 = 0.00950 (3 sig. fig.)
4
196 × 1010 ————– = 25 × 104
5 Answer : D
196 × 106 = ————–— 25 14 × 103 = ———– 5 3 2.8 = × 10
2 709 000 = 709 000 = 7.09 × 105 Answer :
196 × 106 ————– 25
B
Answer : A –4
3 0.049 + 3 × 10 = 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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SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 2 Quadratic Expressions and Equations Paper 1 1 6p2 – p(3 – p) = 6 p2 – 3 p + p2 = 7 p2 – 3 p
Paper 2 1
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 2 (–2) – k(–2) – 6 = 0 4 + 2k – 6 = 0 2k – 2 = 0 2k = 2 2 k =— 2 k =1 Answer : A
2
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2
3 p2 + 10p ———— = 3 p +2 3p2 + 10p = 3( p + 2) 3p2 + 10p = 3 p + 6 3p2 + 10p – 3 p – 6 = 0 3p2 + 7 p – 6 = 0 (3p – 2)(p + 3) = 0 3p – 2 = 0 or p + 3 = 0 3p = 2 p = –3 2 p = — 3 2 ∴ p = — or –3 3 3( x2 + 9) ———— = 9 2 x 3( x2 + 9) = 9(2 x) 3 x2 + 27 = 18 x 3 x2 – 18 x + 27 = 0 3( x2 – 6 x + 9) = 0 3( x – 3)( x – 3) = 0 x–3=0 ∴ x = 3
SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 3 Sets Paper 1 1
Paper 2 1
ξ
A
B I
A B I
V II
III
IV III
II
C
(a) A : Regions I and II B : Region I only A’ : Region III only B’ : Regions II and III Shaded region: Region II only
C : II, III, IV A’ A’ : III, IV, V B’ : I, II, III C C
ഫ A’ A’ :
A’ പ ഫ A’
II, III, IV, V B’ : II, III
Answer :
∴ Shaded region is A പ B’ = B’ പ A ↓ ↓ I and II
A
B
II and III
II III
Answer : B
C
2
A
(b)
B ʚ C
ʚ
A
B
C
I
II
III
C B
A I
II
III
IV
A’ A’ : C’ : A’ A’ പ C’ : A’ പ C’ )’ ( A’ )’ :
A’ A’ : II, III, IV B’ : III, IV C’ : IV
∴ A’ A’ പ B’ = B’ ∴ A’ A’ പ C’ = C’
A’ A’ പ B’ : III, IV A’ A’ പ C’ : IV
II, III I, II II I, III
Answer : A
B
C
Answer : A 3
P 4
R
Q 1
2
n(Q’ പ R) = y n(Q’ പ P ) = 4 n(Q’ പ R) – 3 = y – 3 = y = y = ∴ n(R) = 2 + y =2+7 =9
n(Q’ പ P ) 4 4+3 7
y
2 ξ= (a) (b) (c) (d)
{11, 12, 13, 14, 15, 16, 17, 18, 19, 20} P = {19} Since the universal set is less Q={ } than 22 n(Q) = 0 Q’ = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20} P പ Q’ = {19} ∴ n(P പ Q’ ) = 1
Answer : B Weblink
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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) (ii ) Hexagon means a six-sided polygon. polygon. ∴ All hexagons have 6 sides. (b) The converse of ‘If p, then q’ is ‘If q, then p’. p: k Ͼ 4, q: k Ͼ 12 ∴ Converse: If k Ͼ 12, then k Ͼ 4. If k Ͼ 12, then k = 13, 14, 15, … All values greater than 12 are greater than 4 (e.g. 13 Ͼ 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 A is a subset of set B. q: A പ B = A A പ B is not A. ∴ Premise 2: A പ B ≠ A 2 (a)
P Q
Since Q ʚ P , all elements of Q are also elements of P . ∴ Some elements of set Q are elements of set P . False statement (b) 8 – 15 = –4 is false but 6 × 6 = 6 5 × 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.
4
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∴ 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 Ͼ –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 — is an improper fraction, q: x Ͼ y . y The required statement is x — is an y improper fraction if and only if x x > y . (c) Argument form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. ∴ p: x Ͼ 7, q: x Ͼ 2 ∴ Premise 1: If x x Ͼ 7, then x Ͼ 2. 4 (a) If ‘antecedent’, ‘antece dent’, then ‘consequent’. 1 ∴ If 1% = —– , then 20% of 200 = 40 . 100 (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 θ = 0.5 34 (c) —– = 34 – 2 32 Let 4 = a and 2 = b. 3a Generalisation = 3a – b ∴ —– 3b
SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 5 The Straight Line 4 9 or 9 y = 4 x + 18 or 4 x – 9 y + 18 = 0
Paper 1 1 4 x + 3y = 12 3y = –4 x + 12 4 x + 4 y = – — 3
∴
2 (a) O(0, 0), P (2, (2, 6) 6– 0 mOP = ——– 2– 0
At y -intercept, -intercept, x = 0 ∴ y = 4 -intercept -intercept = 4 ∴ y
=3
Answer: D 2 7 x + 4 y = 5 4y = –7 x + 5 7 x + — 5 y = – — 4 4 y = mx + c 7 ∴ m = –— 4 7 ∴ Gradient = – — 4 Answer : B
The gradient of OP is 3. (b) RQ / /OP /OP ∴ m = mOP = 3 RQ Let the equation of the straight line QR be y = 3 x + c.
3 P (–5, (–5, –6), Q(–3, 2), R(1, k) mPQ = mPR P , Q , R are points on
2 – (–6) ————– –3 – (–5) 8 — 2 k+6 k
= = = =
k – (–6) ————– 1 – (–5) k+6 ——– 6 24 18
a straight line.
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 = 3 x – 18. (c) PQ//OR ∴
3 mPQ = mOR = — 7
Let the equation of the straight line PQ 3 x + c. be y = — 7
Answer : D
At point P (2, (2, 6), y = 6 and x = 2.
Paper 2 1 (a) 4 x – 9 y + 36 = 0 At G, x = 0, ∴ 4(0) – 9 y + 36 = 0 9y = 36 y = 4 ∴ G(0, 4) (b) Let the equation of the straight line JK be y = mx + c. 4 m JK = mGH = — 9 4 x + c y = — 9 1 4 –9 + c At J J –4 — , 0 , 0 = — — 2 9 2 0 = –2 + c ∴ c = 2
y = — x + 2
∴
3 (2) + c 6= — 7 6 +c 6 =— 7 36 c = —– 7
The y -intercept -intercept of the straight line 36 PQ is —– . 7
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SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 6 Statistics III Paper 1 1
(b)
Mass (g)
Upper boundary
Cumulative frequency
Number of guidebooks
2
3
4
5
6
Frequency
3
7
8
10
8
580 – 599
599.5
0
Cumulative frequency
3
10 18 28 36
600 – 619
619.5
2
620 – 639
639.5
5
640 – 659
659.5
15
660 – 679
679.5
27
The mode is 5. Answer : C 2
Mode is the value of data with the highest frequency.
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
5
2
The ogive is as shown below. Cumulative frequency
Σ fx
Mean, x = —— Σ f (1 × 5) + (2 × 4) + (3 × 6) + (4 × 3) + (5 × 2) = ———————————— 5+4+6+3+2 53 = —–
40 35 30 25
20
= 2.65
20
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
15 10 5 O
667.5
599.5 599.5 619.5 619.5 639.5 639.5 659.5 659.5 679.5 679.5 699.5 719.5 739.5
Mass (g)
Answer : C Paper 2 1 (a) Mass (g)
Upper boundary
600 – 619
(c) From the ogive, 1 × 40 fish = 20 fish (i) — 2 Hence, the median mass = 667.5 g (ii) The median mass means means that 50% (20) of the fish have masses of less than or equal to 667.5 g .
Frequen uency
Cumulative frequency
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
5–9
7
4
28
720 – 739
739.5
2
40
10 – 14
12
7
84
6
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Tally ally
2 (a) Average Average Midpoint Frequency Tally marks (x) (f)
fx
(c)
Frequency
Average Average Midpoint Frequency Tally marks (x) (f)
fx
15 – 19
17
9
153
8
20 – 24
22
8
176
7
25 – 29
27
5
135
30 – 34
32
4
128
35 – 39
37
5
185
3
40 – 44
42
3
126
2
9
6 5
Σ f =
45
Σ fx
= 1015
4
1 0
4.5
Σ fx 1015 5 (b) Mean = —— = ——– = 22 — Σ f 9 45
9.5
14.5 14.5 19.5 24.5 24.5 29.5 29.5 34.5 34.5 39.5 39.5 44.5 44.5
Average marks
(d) Percentage of students who need to attend extra classes 9 + 7 + 4 × 100 = ——–—— 45 4% = 44 — 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
Thus, the table can now be completed, as shown below:
= 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 A
3 Non-gr Non-grad aduat uatee
Male Female Total Total
28
12
6
18
Female
28
4
32
Total Total
40
10
50
4
Number of students
1 – 40
h
Total otal
41 – 70
88
18
71 – 100
8
32 50
Number of graduate teachers = P(graduate teacher) × Total number of teachers 4 = — × 50 5 = 40
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Male
Marks
The information in the above table is given.
8
Total otal
Answer : A
Answer : A
Gradu Graduate ate
NonNon-gr grad adua uate te
Hence, the number of male non-graduate teachers is 6.
8 = 1 P( A) = —– — 2 16
2
Grad Gradua uate te
14 P(marks not more than 70) = —– 15 14 h + 88 ——–——– = —– 15 h + 88 + 8 14 h + 88 ——–— = —– 15 h + 96 15(h + 88) = 14(h + 96) 15h + 1320 = 14h + 1344 15h – 14h = 1344 – 1320 h = 24 Answer : A
Bab 8 tidak ada
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
24 2 tan y ° = —– 7 QS 24 —– = —– QR 7 QS 24 —– = —– 14 7 24 QS = —– × 14 7 QS = 48 cm 1 4 1 QT = — (48) 4 QT = 12 cm
cos x° = –cos
QT PT
= – —–
300° 60°
x
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. tan 90° = 90°
tan 180° = 0
10
PQ2 + QT 2
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O
0° tan 0° = 0 360° tan 360° = 0
y
O
PT = 52 + 122 PT = 13 cm
180°
∞
270° tan 270° = – ∞
QT = — QS
PT =
∠PTQ
Answer : D
90°
180°
270°
360°
x
SPM ZOOM-IN
(Fully-worked Solutions) Form 4: Chapter 10 Angles of Elevation and Depression Paper 1 1
3
Bird R
XW
X
—– = tan 53° 4.2 XW = 4.2 × tan 53°
T
Angle of depression = ∠TRS
YW
53°
Y U
Answer : A CB
2 —– = tan 16° AB CB
= AB × tan 16° = 35 × tan 16° = 10.0361
W
S
Cat
16° B
V
19°
XY = XW – YW
C
35 m
The height of the pole, CB, is 10 m, correct to the nearest integer. Answer : A ∴
4.2 m
—– = tan 19° 4.2 YW = 4.2 × tan 19°
A
= = = =
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
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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 PM and the plane PSTU is ∠ NPM.
S
P
In
R
Q
⌬ NUP ,
NP = N
T
W
U
In
using the Pythagoras’ Theorem,
42 + 62 = 52 = 7.211 7.2111 cm
⌬ NMP ,
V
M
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
∠ NPM
tan
∠ NPM
tan
∠ NPM ∠ NPM
2
NM
= —–– NP
8 = —––– 7.2111 = 1.1094 = 47°58’
M
S
R
P
The angle between the line SM and its orthogonal projection (SN ) is ∠MSN .
Answer :
tan
Q
20 cm D
B
C
4 cm A
2
H
J
B
G
The angle between the plane SABM and the plane SDCR is ∠ ASD.
E
F D
C
A
B
The angle between the plane plane DHGC is ∠ BGC.
HGB
and the
• The line of intersection intersection of the planes 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 ang angle le between between the plane HGB and the plane DHGC is the angle between the lines GC and GB , i.e. ∠BGC .
• The line 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 . S
20 cm
Answer :
D
Paper 2 1
A W
T N
Based on
M
4 cm V
U
6 cm
8 cm P
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⌬SDA,
tan
∠ ASD
tan
∠ ASD
R
S
12
4 cm D
8 cm
Q
∠ ASD
AD
= —–– SD
4 = —– 20 = 11°19’
SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 1 Number Bases Paper 1 1
3 1
1
1
1 1 1 1 1
2
1 1 1 1 1 02 Answer :
12 + 1 2 = 10 2 12 + 1 2 + 1 2 = 11 2 ∴
52
51
50
1
1
3
B
1 1 0 1 0 02 –
1 1 12
0
102 – 1 2 = 1 2 12 – 1 2 = 0
4
51 + 3
×
5°
C
83 + 5 = 1
×
83 + 0
82 + 0
×
83
82
81
80
1
0
0
5
83 + 5 =
10058
×
81 + 5
×
80
10
1 1 0 1 0 02 –
×
1135
52 + 5 + 3 =
Answer :
2
52 + 1
×
1
1 1 1 1 12 +
52 + 5 + 3 = 1
1 1 12 ∴ 0
1 10
10
Answer : A
1 1 0 1 0 0 –
1 1 12 0 12
0
1 1 10 10 10
10
1 1 0 1 0 02 –
5 102 – 1 2 = 1 2 12 – 1 2 = 0
1 ⎧
⎨
110 ⎩
421 ⎧
⎨
⎩
1
⎧
⎨
111 ⎩
⎧
⎨
011 000 00 02 ⎩
421 421 ⎧
⎨
6
⎩
⎧
⎨
⎩
⎧
⎨
⎩
421 4 21 ⎧
7
⎨
3
⎩
⎧
⎨
⎩
421 ⎧
⎨
⎩
0
1 1 12
1 0 1 1 0 12 102 – 1 2 = 1 2
Answer :
∴
1110111011000 1011101100 0 2 =
Answe Answer :
167308
D
C
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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 1 , it is graph III. a= — 2 1 ∴ I: a = 5, II: a = 1, III: a = — 2 Answer : D 18 is a reciprocal graph. 2 y = – —– x B is a quadratic graph. C and D are cubic graphs.
Paper 2 1 (a) Substitute x = –2, y = k into y = 2 x2 – 3 x – 5. k = 2(–2) 2 – 3(–2) – 5 k=8+6–5 k=9
Equation that has to be solved
– ➁: such that 2x 2 + 3 x – 17 = 0 is y = –6 x + 12 rearranged. Draw the straight line y = –6 x + 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). ➀
20
x = 1 y = 5 x + 5
15
Substitute x = 3, y = m into y = 2 x2 – 3 x – 5. m = 2(3)2 – 3(3) – 5 m = 18 – 9 – 5 m=4
16 y = – ––– x
10 y = 5
5 –2.85 –4
y
–3
–2
–1 O –5 –10
y = 2 x 2 – 3 x – 5
30
Graph drawn.
The solution from the graph is x = 2.25. 16 2 (a) Substitute x = –2 into y = – —– , then x –16 —— y = =8 –2 16 Substitute x = 3 into y = – —– , then x –16 y = —— = –5.3 3 y (b), (d)
Answer : A
(b)
y = 2 x2 – 3 x – 5 ......➀ 0 = 2 x2 + 3 x – 17......➁
–15
1
2
2.85 3
4
y = –2 x 16 y = – ––– x
y
= – 6
25
x
+
1 2
20 15 10
–1.5 –2
5 4 –1 O –5
4.35
2.25 1
2
3
4
5
x
(c) From the graph, 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: 14
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16 Graph drawn. (c) y = – —– ......➀ x Equation that 16 has to be solved 0 = – —– + 2 x ......➁ such that x 16 ➀ – ➁: – —– + 2 x = 0 x is rearranged. y = –2 x Draw the straight line y = –2 x 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.
x
SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 3 Transformations III Paper 2 1 (a)
2 (a) W
(–2, P (–2,
R T 2) ⎯→ ’(0, ’(0, 2) ⎯→ ⎯→ P ⎯→
(ii)
(–2, P (–2,
T R (–1, (–1, 0) ⎯→ P’ ⎯→ P’’ 2) ⎯→ (0, (0, 1) ⎯→
V
(i)
(4, H (4,
4) ⎯→ H ’(6, ’(6, 1) ⎯→ H ’’’’(0, 1)
(ii)
(4, H (4,
⎯→ 4) ⎯→
V
(i)
(2, H’ (2,
W
⎯→ 4) ⎯→
(4, H’’ (4,
1)
(b) X – Translation 5 3 Y – Anticlockwise rotation of 90° about the point N (7, 10) N (7, (c) (i) Scale factor = 2, Centre = (–1, 8) (ii) Area of ⌬EFG = 2 2 × Area of ⌬ ABC 52 = 4 × Area of ⌬ ABC Area of ⌬ ABC = 13 units 2 ∴ Area of ⌬LMN = Area of ⌬ ABC 2 = 13 units
(b)
(2, P’’ (2,
1)
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 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
(i)
V –
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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) ∴
18 + 3q = 15 ......➀ 3q = 15 – 18 3q = –3 3 q = –— 3 q = –1
3p + (–3q) = 12 3p – 3 q = 12 ......➁
∴
x = 17 and y = 14
2 (a) If no inverse, ad – bc = 0. ∴ (2 × –4) – (4 × d) = 0 –8 – 4d = 0 –4d = 8
8 d = –— 4 d = –2
17 + 3p = 4q =4 7+3=q =4–1 ∴ q = 10 =3 = 3 × 10 = 30
Paper 2
1 (a) Inverse of 4 –5 6 –7 1 = ————————— –7 (4 × –7) – (–5 × 6) –6
1 = ——————– –7 (–28) – (–30) –6
1 = ———— –7 –28 + 30 –6
1 = — –7 2 –6 ∴
16
5 4
5 = k –7 q –6 4 4 5 4
1 k = — and q = 5 2
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(b) Q = 2 –3 if d = –3 4 –4 1 Q–1 = ——————–——– –4 (2 × –4) – (–3 × 4) –4 1 = ———– –4 3 –8 + 12 –4 2 1 = — –4 3 4 –4 2 3 –1 — 4 = 1 –1 — 2 (c) QP = 2 6 2 –3 a = 2 4 –4 b 6 1 –4 3 2 –3 a 1 — = — –4 3 2 4 –4 2 4 –4 b 4 –4 2 6 1 (–4 × 2) + 1 0 a =— 0 1 b 4 (–4 × 2) + 1 –8 + 18 a = — b 4 –8 + 12
Answer : C
×
p + q = 3 + (–1) = 3 – 1 = 2
1+ p ∴ p p p×q
×
×
Answer: A
2
×
34 —– = 2 28 —– 2 = 17 14
Substitute q = –1 into ➁: ∴ 3p – 3(–1) = 12 3p + 3 = 12 3p = 12 – 3 3p = 9 9 p= — 3 p=3 ∴
54 –24 –2) + (5 4) 10 01 xy = —12 (–7 (–6 –2) + (4 4) 14 + 20 xy = —12 12 + 16 1 = — 34 2 28
(b) 4 –5 x = –2 6 –7 y 4 1 –7 5 4 –5 x 1 — = — –7 –6 4 6 –7 y 2 2 –6
5 4
3 2
(3 (2
× ×
6) 6)
ab = —14 140
10 —– 4 1 5 — = 2 1 =
∴
5 or 2 — 1 ,b=1 a=— 2 2
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SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 5 Variations Paper 1 1 Given s
∝
∴ s
When
=
r =
3
1 , —– 2 r k
k is
—– 2 r
s —–
r =
ks , where —–
t 2
t
When
r =
8, s = 2 and
∝
s
=
D
When
r =
27, s = 6 and
6(6) 27 = —–—
r 3 kr 3,
where
k
u
is a constant
When s = 192 and 192 = k(4)3 k= 3 3 ∴ s = 3 r When s = –24, –24 = 3r 3 r 3 = –8 r = –2 Answer :
18
B
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t =
3,
t
2
s
is a constant
k(2)2 8 = —–— 3 24 = 4k k=6 6 s2 ∴ r = —–
2 and s = 5,
r
2
k
a constant.
k 5 = —– 22 k = 20 20 ∴ s = —– 2 Answer :
2
r ∝
r =
4,
u u
216 = —–– 27 =8
Answer :
C
t = u,
SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 6 Gradient and Area Under a Graph Paper 2 1 (a) Average Average speed of the lorry for the whole journey from point P to point Q
Total Total distance travelled = ——————————— Total Total time taken 300 = —— 16 3 = 18 — m s–1 4 (b) Speed of the car for the whole journey = Gradient of the straight line ABC Vertical Vertical axis = – ——————— Horizontal axis 300 – 0 = – ———— 10 – 0 = –30 m s–1
Hence, the speed of the car for the whole journey from point Q to point P is 30 m s–1. (c) The point on the distance–time graph graph when the lorry and the car meet is the intersection point of the graph OBD and the graph ABC, i.e. point B. Hence, the distance from point Q when the lorry and the car meet is 300 – 60 = 240 m
2 (a) Total distance travelled by the particle for the whole journey is 310 m. Total area under the graph = 310 1 1 (4 × 25) + — (25 + 40)(4) + — (t – 8)(40) = 310 2 2 100 + 130 + 20( t – 8) = 310 20(t – 8) = 80 t – 8 = 4 t = 12
(b) Rate of speed of the particle from the 4th second to the 8th second = Gradient of the graph from the 4th second to the 8th second 40 – 25 = ———— 8– 4 3 = 3 — m s–2 4 (c) Average Average speed of the particle in the first first 8 seconds Total Total distance = ——————– Total Total time Area under the graph in the first 8 s = ——————–—————————— 8 From (a) 100 + 130 = ————— 8 3 = 28 — m s–1 4
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SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 7 Probability II Paper 1 1 Let R = Event of obtaining a round biscuit Sq = Event of obtaining a square biscuit T = Event of obtaining a triangular biscuit S = Sample space
P(T ) = 1 – P( R) – P(Sq) 3 1 P(T ) = 1 – — – — 7 4 9 P(T ) = —– 28 n(T ) 9 —–— = —– n(S) 28 36 9 —–— = —– n(S) 28 9 × n(S) = 36 × 28 36 × 28 n(S) = ———— 9 n(S) = 112 n(R) + n(Sq) + n(T ) n(R) + n(Sq) + 36 n(R) + n(Sq) n(R) + n(Sq)
= 112 = 112 112 = 112 112 – 36 = 76
Answer : C
2 Let G = Event of obtaining a green disc B = Event of obtaining a blue disc S = Sample space
6 = 5 P(B) = 1 – P( G) = 1 – —– —– 11 11 n(B) 5 —–— = —– n(S) 11 30 5 —–— = —– n(S) 11
5
∴
×
n(S) = 30 × 11 30 × 11 n(S) = ———— 5 n(S) = 66
n(G) = n(S) – n(B) = 66 – 30 = 36
Answer : A
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3 Let B = Event of drawing a blue ball R = Event of drawing a red ball S = Sample space
5 Given P(R) = — , 8 n(R) 5 —–— = — 8 n(S) n(R) 5 —–— = — 8 32 5 n(R) = — 8
×
32
n(R) = 20
Let the number of blue balls added = h Therefore, Therefore, n(S) = 32 + h 5 P(R) = — New value of P(R ) 9 n(R) 5 —–— = — 9 n(S) 20 5 ——— =— 32 + h 9 5(32 + h) = 180 160 + 5 h = 180 5h = 20 h =4 Hence, the number of new blue balls that have to be added to the bag is 4. Answer : D
Paper 2 1 (a) P(letter M M) n( M) = —–— n(S) 2+5 = —————— 2+5+3+4 7 = —– 14 1 =— 2
(b) P(both the cards drawn are cards with the letter N N ) After 1 card with the letter 7 × 6 N is taken out, it is left with = —– —– 6 cards with the letter N out 14 13 of the balance of 13 cards.
Initially, there are 7 cards with the letter N out of 14 cards.
(c) P(both the cards drawn drawn are of different colours) G – Green = P(GY or YG) Y – Yellow = P(GY ) + P(YG) 5 × —– 9 + —– 9 × —– 5 = —– 14 13 14 13 Initially, there are 5 green cards out of 14 cards.
1 P( Z ) = — 5 Number of male students from school Z 1 —————————————— = — 5 Total Total number of male students from all the three schools 10 1 —————– =— k + 22 + 10 5 k + 32 = 50 k = 18
3 = —– 13
2 (a)
After 1 green card is taken out, it is left with 13 cards and so there are 9 yellow cards out of the 13 cards.
(b) P(Two students from school Y are of the same gender) = P ( MM or FF ) = P ( MM) + P (FF ) 22 × 21 + 18 × 17 = —– —– —– —– 40 39 40 39 32 = —– 65
45 = —– 91
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SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 8 Bearing Paper 1 1
Let the bearing of point K from point H be θ . θ = 360° – 35° – 40° = 285°
North
North
B
60°
60° A
Answer : C
Bearing of A from B
120°
3 Label the east and north direction and write down all the provided information onto the diagram.
Bearing of A from B = 180° + 60° = 240°
North
North
Answer : C
Q
30° P
30°
60°
30°
2 Label the north direction and write down all the provided information onto the diagram. Bearing of F from K = 065° North
Alternate angles are equal.
60° North
Alternate angles are equal.
R
F North
65° 35° 65° 40°
35° 40°
K ∠FHK =
22
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East
180° – 100° —————– 2
From the above diagram, the bearing of 030°. point Q from point P is 030°.
H θ
Answer : A
SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 9 Earth as a Sphere (i) The latitude of point R is 35°S. (ii) The longitude longitude of point L is (180 – 70)°E = 110°E (b) Distance of ML ML (along the parallel of latitude) = 180 × 60 × cos 35° = 8846.8 n.m. (c) Distance of LMR (via the North Pole) = 180 × 60 = 10 800 n.m. (a)
Paper 1 1
N D
O
50° 40°
32°
148°
148°E
0°
S
Based on the above diagram:
Average speed Distance = ———— Speed 10 800 = ——— 600 = 18 hours
The latitude of point D = (90 – 50)°N = 40°N The longitude of point D = (180° – 32°)E = 148°E Hence, the location of point D is (40°N, 148°E). Answer : A
Hence, the time the aeroplane reached point R is 0500 + 1800 = 2300.
2
N 0°
2
N J
C
30°N L O R
40° 50°
100°
80°
K
P Q
S
The latitude of point R = (90 – 50)°S = 40°S The longitude of point R = (180 – 80)°W = 100°W Hence, the position of point R is (40°S, 100°W). Answer : B
Paper 2 1
30°
O 30°
M 35°E
S
(a) The position of point Q is (30°S, (180 – 35)°W) = (30°S, 145°W) (b) (i) JK = 3300 n.m. ∠ JOK × 60 = 3300 3300 ∠ JOK = ——– 60 ∠ JOK = 55°
30°N
N
J
L 35° N
M
55°
180° 35° 35°
Equator 0°
O x °S °S
R 70°W
S
K
110°E
∴ x
= 55 – 30 = 25 Weblink
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(ii) JL = 4936 n.m. ∠ JCL × 60 × cos 30° = 4936 4936 ∠ JCL = —————– 60 × cos 30° ∠ JCL
y °W y °W
0°
35°E
L
G
J
95°
∴
= 95°
Total Total distance travelled (c) Time taken = ——————————— Speed JL + LM = ———— 600 4936 + [(30 + 40) × 60] = ——————————— 600 4936 + 4200 = —————— 600 9136 = ——— 600 = 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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SPM ZOOM-IN
(Fully-worked Solutions) Form 5: Chapter 10 Plans and Elevations Paper 2 1 (a)
2 (a) 4 cm
K/Q
C/D
B/A
2 cm
L/P
J/I
5 cm 4 cm
A/D
3 cm
B/C
4 cm
K/L J/G
G/H
3 cm
Elevation as viewed from Y
M/N Plan
(b)
F/E
(b)
(i)
(i) 4 cm
A
B
M/A/H
2 cm H/J/K
4 cm
N/R/S
W/Q/T
1 cm
1 cm
P/B/G
3 cm
M/L D/I/E
6 cm
C/J/F
4 cm 4 cm D/G/Q
4 cm
N/D/P
4 cm
E/F/C
L
K
8 cm
Elevation as viewed from X
Plan
(ii)
(ii)
N
M
W
B/A
2 cm 2 cm
H
5 cm
M/J
R
D/A
Q
C/B
L/K
2 cm I
4 cm
P
J
4 cm 3 cm 4 cm
E/D
5 cm
F/N/G
C/D
Elevation as viewed from Y
1 cm P/Q
L/E/H
S
6 cm
1 cm T
K/F/G
Elevation as viewed from X
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