1 Paper 1 1 Based on the given ordered pairs {(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below. (a) The image of 2 is 1. (b) The object of 7 is 6. 2 (a) Let g −1 (7) = y Thus, g(y) = 7 4y − 1 = 7 4y = 8 y = 2 ∴ g −1 (7) = 2 (b) hg(x) = h(4x − 1) = (4x − 1) 2 − 3(4x − 1) + 5 = 16x 2 − 8x + 1 − 12x + 3 + 5 = 16x 2 − 20x + 9 3 (a) The range is {3, 7}. (b) The above relation is a many-to-one relation. Form 4: Chapter 1 (Functions) SPM Flashback Fully-Worked Solutions '3' and '7' are linked to object(s) but '5' and '11' are not linked to any object(s). Therefore, the range is {3, 7}. Element '7' in the codomain is linked to two elements, i.e. '28' and '49' in the domain. Therefore, it is a many-to-one relation.
78
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1
Paper 1 1 Based on the given ordered pairs
{(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below.
(a) The image of 2 is 1. (b) The object of 7 is 6. 2 (a) Let g−1(7) = y
Maximum value = −k 2 −1 But it is given that the maximum value = −r2 − 2k . By comparison,
−r2 − 2k = −k 2 − 1r2 = k 2 − 2k +1r2 = (k −1)2
r = k − 1 (shown)
(b) From f (x) = −(x − 2k )2 − k 2 − 1, the axis
of symmetry is x = 2k . But it is given that the axis of symmetry is x = r2 − 1. By comparison, r2 − 1 = 2k Solve the following simultaneous equations: r = k − 1 … r2 − 1 = 2k … Substitute into :
(k −1)2 −1 = 2kk 2 − 2k +1− 1− 2k = 0
k 2 − 4k = 0k(k − 4) = 0
k = 0 or k = 4k = 0 is not accepted.∴ k = 4When k = 4, r = 4 − 1 = 3
Add and subtract
12
× (−4k )⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
= 4k 2
10
Paper 2 1 4x + y = x2 + x − y = −3
4x + y = −3 …
x2 + x − y = −3 …
From : y = −3− 4x …
Substitute into :
x2 + x − (−3− 4x) = −3x2 + x + 3+ 4x + 3 = 0
x2 + 5x + 6 = 0(x + 2)(x + 3) = 0
x = −2 or −3
From : When x = −2, y = −3 − 4(−2) = 5 When x = −3, y = −3− 4(−3) = 9
Hence, the solutions are x = −2, y = 5
or x = −3, y = 9 . 2 x − y = 1 …
x2 + 3y = 6 …
From : x = 1+ y …
Substituting into , we have:
(1+ y)2 + 3y = 61+ 2y + y2 + 3y − 6 = 0
y2 + 5y − 5 = 0
y =−b ± b2 − 4ac
2a
y =−5 ± 52 − 4(1)(−5)
2(1)
y =−5 ± 45
2(1)y = 0.854 or −5.854
From : When y = 0.854, x = 1+ 0.854 = 1.854 When y = −5.854, x = 1+ (−5.854) = −4.854
Hence, the solutions are x = 1.854, y = 0.854 or x = −4.854, y = −5.854 (correct to 3 decimal places).
From : When x = 1.159, y = 2 − 3(1.159) = −1.477 When x = 0.216, y = 2 − 3(0.216) = 1.352 Hence, the solutions are x = 1.159, y = −1.477 or x = 0.216, y = 1.352 (correct to 3 decimal places).
When x = 0, y = (0 + 3)2 = 9 Hence, the coordinates of point Q are (0, 9) .
8 y =
13
u6
y =13
(3x − 6)6
dydx
=63
(3x − 6)5(3)
= 6(3x − 6)5
9 (a) y = 3 + 14x − 2x3
dydx
= 14 − 6x2
When x = 2, dy
dx= 14 − 6(2)2 = −10
(b) δy ≈
dydx
× δx
= (−10) × [(2 + k ) − 2]= −10k
31
Paper 2 1
Using the concept of similar triangles,
r0.5
=h
0.7
r =h
0.7× 0.5
r =57
h
V =13
π r2h
V =13
π 57
h⎛
⎝ ⎜
⎞
⎠ ⎟
2
h
V =25
147πh3
dVdh
=25
147π (3h2 )
dVdh
=2549
π h2
dhdV
=49
25πh2
Rate of change of the height of the water level:
dhdt
=dhdV
×dVdt
dhdt
=49
25πh2× 0.1
dhdt
=49
25π(0.3)2× 0.1
dhdt
=49
25(3.142)(0.3)2× 0.1
= 0.6931 m s−1
2 y = 2x3 − 3x2 − 12x + 11
dydx
= 6x2 − 6x − 12
d 2 ydx2
= 12x − 6
(a) At turning point,
dydx
= 0
6x2 − 6x −12 = 0x2 − x − 2 = 0
(x + 1)(x − 2) = 0x = −1 or 2
When x = −1,
y = 2(−1)3 − 3(−1)2 −12(−1) + 11 = 18 ∴ (−1, 18) is a turning point.
When x = −1,
d 2 ydx2
= 12(−1) − 6 = −18 (negative)
∴ (−1, 18) is a maximum point.
When x = 2,
y = 2(2)3 − 3(2)2 − 12(2) + 11 = −9 ∴ (2, −9) is a turning point.
When x = 2,
d 2 ydx2
= 12(2) − 6 = 18 (positive)
∴ (2, −9) is a minimum point. (b) At point (3, 2),
dydx
= 6(3)2 − 6(3) − 12 = 24
mgradient = 24
∴ mnormal = −124
The equation of normal is
y − 2 = −124
(x − 3)
24( y − 2) = −(x − 3)24y − 48 = −x + 3
24y = −x + 51
At the x-axis, y = 0.
24(0) = −x + 51x = 51
∴ P is point (51, 0).
Rate of increase of the volume of water:
dVdh
= 0.1 m3 s−1
32
Paper 2 1 (a)
Area of ∆PRQ = 4 m2
12
(3)(3)sin∠RPQ = 4
sin∠RPQ =89
∠RPQ = 62.73°
Using the cosine rule,
RQ2 = 32 + 32 − 2(3)(3) cos 62.73°RQ2 = 9.75268RQ = 3.123 m (correct to 4
significant figures)
(b) Step 1 (Find PM where M is the midpoint
of QR)
Area of ∆PRQ = 4 m2
12
× RQ × PM = 4
12
× 3.123× PM = 4
PM = 2.5616 m
Step 2 (Determine ∠VMP )
The angle between the inclined plane VQR and the base PQR is 55°. Therefore, ∠VMP = 55°. Step 3 (Find VM) From ∆VRM , using the Pythagoras' Theorem, VM = 22 −1.56152 = 1.2497 m Step 4 (Calculate PV)
Using the cosine rule,
VP2 = 2.56162 + 1.24972 −2(2.5616)(1.2497) cos 55°
VP = 2.110 m (correct to 4significant figures)
(c) Step 1 (Find ∠VQP)
cos ∠VQP =22 + 32 − 2.1102
2(2)(3)cos ∠VQP = 0.712325cos ∠VQP = 44.58°
Step 2 (Find the area of ∆VPQ ) Area of ∆VPQ
=12
× 2 × 3× sin 44.58°
= 2.106 m2
Form 4: Chapter 10 (Solution of Triangles) SPM Flashback
Fully-Worked Solutions
33
2 (a)
(i) Based on ∆ABC , using the sine rule,
sin ∠ABC15
=sin 30°
9
sin ∠ABC =sin 30°
9×15
sin ∠ABC = 0.83333∠ABC = 56.44°
(ii) Based on ∆ADC , using the cosine
rule,
152 = 102 + 82 − 2(10)(8)cos ∠ADC
cos ∠ADC =102 + 82 −152
2(10)(8)cos ∠ADC = −0.38125
∠ADC = 112.41°
(iii) Based on ∆ABC ,
∠ACB = 180° − 30° − 56.44°= 93.56°
Area of the quadrilateral ABCD
= Area of ∆ADC + Area of ∆ABC
=12
(10)(8) sin 112.41° +
12
(15)(9) sin 93.56°
= 36.9792 + 67.3697= 104.35 cm2
(b) (i) Based on ∆ABC , since the length of BC is shorter than the length of AC and ∠BAC is an acute non-included angle, the ambiguous case will occur. Another triangle (∆A ′ B C) that can be drawn is as shown below.
Thus, ∠ ′ A BD = 180° − 45° − ∠D ′ A B= 180° − 45° −106.07°= 28.93°
First quadrant
Second quadrant
35
Paper 2 1
Item Price in 2000
Price in 2002
Price index
for 2002
based on 2000 (I)
Weekly expenses(weigh-tage, w)
Iw
P RMp RM1.75 140 12 1680 Q RM2.00 RM2.30 115 28 3220 R RM4.00 RM4.80 q = 120 20 2400 S RM6.00 RM7.50 125 30 3750 T RM2.50 RMr 110 10 1100
w =∑
100
Iw =∑12 150
(a) (i) For item P,
I2002 =P2002
P2000×100
140 =1.75
p×100
p = 1.25
(ii) For item R,
I2002 =P2002
P2000×100
q =4.804.00
×100
q = 120
(iii) For item T,
I2002 =P2002
P2000×100
110 =r
2.50×100
r = 2.75
(b)
I =Iw∑w∑
=12 150
100= 121.5
(c)
I =P2002
P2000×100
121.5 =P2002
RM500×100
P2002 = RM607.50
Hence, the corresponding total monthly expenses in the year 2002 was RM607.50.
(d)
Hence, I 2004 (based on the year 2000)
=100 + 20
100× I 2002
=120100
×121.5
= 145.8
Form 4: Chapter 11 (Index Numbers) SPM Flashback
Fully-Worked Solutions
+ 21.5% + 20% Year 2000 Year 2002 Year 2004
36
2 (a) (i) For item S,
I2004 =P2004
P2002×100
110 =1.50P2002
×100
P2002 =1.50110
×100
P2002 = RM1.36
Hence, the price of item S in the year 2002 was RM1.30.
(ii) For item P,
It is given that:
I2002 (based on 2000) = 105P2002
P2000×100 = 105
∴ P2000 =
P2002 ×100105
…
From the table, we can see that:
I2004 (based on 2002) = 115P2004
P2002×100 = 115
∴ P2004 =
115 × P2002
100 …
I2004 (based on 2000)
=P2004
P2000×100
=
115 × P2002
100P2002 ×100
105
×100
=115100
×105100
×100
= 120.75
Hence, the price index of item P for the year 2004 based on the year 2000 is 120.75.
(b) (i) I = 110
Iw∑w∑
= 110
(115 × 20) + (10x) + (105 × 40) + (110 × 30)
20 + 10 + 40 + 30= 110
9800 + 10x100
= 110
9800 + 10x = 11 00010x = 1200
x = 120
(ii) I 2004 =
P2004
P2002× 100
110 =22
P2002×100
P2002 =22
110×100
P2002 = 20.00
Hence, the price of a box of icecream in the year 2002 was RM20.00.
From
From
37
3 (a) I2004 (based on 2002)
=
P2004
P2002×100
For material K,
p =
1.751.40
×100 = 125
For material M,
q2
×100 = 140
q =140 × 2
100q = 2.80
For material N,
2.40r
×100 = 80
r =2.40 ×100
80r = 3.00
(b) (i)
Material I2004 (based on the year
2002)
Angle of pie chart (degrees)
w
K 125 75 15 L 150 40 8 M 140 155 31 N 80 90 18
I 2004 (based on 2002)
=Iw∑w∑
=(125 × 15) + (150 × 8) + (140 × 31) + (80 × 18)
15 + 8 + 31 + 18
=885572
= 122.99
(ii) I 2004 (based on 2002) = 122.99
P2004
P2002×100 = 122.99
RM9P2002
×100 = 122.99
P2002 =RM9 ×100
122.99= RM7.32
(c)
I 2002+ 22.99%⎯ → ⎯ ⎯ ⎯ I 2004
+ 20%⎯ → ⎯ ⎯ I 2006
(100) (122.99) (?)
I 2006 (based on 2002)
=100 + 20
100×122.99
= 147.59
38
4 (a) For component U,
I2006 = 120P2006
P2004×100 = 120
h50
×100 = 120
h = 60
(b) For component S,
I2006 = 125P2006
P2004×100 = 125
mk
×100 = 125
100m = 125k
4m = 5k … P2006 = 20 + P2004
m = 20 + k … Substituting into :
4(20 + k ) = 5k80 + 4k = 5k
k = 80
From : m = 20 + 80 = 100
(c) (i) I = 132
P2006
P2004×100 = 132
1716P2004
×100 = 132
P2004 = RM1300
(ii)
Component I2006 w U 120 1
R 4025
×100 = 160 3
S 125 4
T 4440
×100 = 110 p
I = 132(120 × 1) + (160 × 3) + (125 × 4) + 110p
1 + 3 + 4 + p= 132
1100 + 110p8 + p
= 132
1100 + 110p = 1056 + 132p44 = 22pp = 2
39
Paper 1 1 (a) k + 1, k + 5, 2k + 6 , … arithmetic progression
d = T2 − T1 = k + 5 − (k + 1) = 4d = T3 − T2 = 2k + 6 − (k + 5) = k + 1
Since common difference is always the same, k +1 = 4 ⇒ k = 3
(b) When k = 3, we have 4, 8, 12, …
S8 =
82
[2(4) + 7(4)] = 144
2 (a) T4 = 24
ar3 = 2481r3 = 24
r3 =2481
=827
r =23
(b)
S∞ =a
1− r=
81
1− 23
= 243
3 Since k, 3, 9
k, m are four consecutive terms
of a geometric progression,
3k
=m9k
3k
=mk9
mk 2 = 27
m =27k 2
4 The arithmetic progression 5, 9, 13, … has a common difference of 4.
S3 = 5732
[2a + (3 − 1)(4)] = 57
32
(2a + 8) = 57
3a +12 = 57a = 15
Hence, the three consecutive terms which sum up to 57 are 15, 19 and 23.
+4 +4 5 Volumes of water in litres:
410, 425, 440, … Volume of water at the end of the 8th day
= T8
= a + 7d= 410 + 7(15)= 515 litres
6 0.848484K
= 0.84 + 0.0084 + 0.000084 +K
=0.84
1− 0.01
=0.840.99
=8499
=2833
Form 5: Chapter 12 (Progressions) SPM Flashback
Fully-Worked Solutions
S∞ =
a1− r
These three terms are not the first three terms but any three consecutive terms with a common difference of 4. Therefore, a new value of a (first term) has to be determined.
40
7 (a) If 3, k, 48 are in an arithmetic progression, then
k − 3 = 48 − k2k = 51
k = 25.5
(b) If 3, k, 48 are in a geometric progression,
then
k3
=48k
k 2 = 144k = 12
8 (a) The common difference of the arithmetic
progression 2, 5, 8, … is 5 − 2 = 3 . (b) The sum of all the terms from the 4th term
to the 23rd term
= S23 − S3
=232
[2(2) + (23− 1)(3)]−32
[2(2) + (3− 1)(3)]
= 805 −15= 790
9 (a) The common ratio of the geometric
progression 2, 6, 18, K is 62
= 3 .
(b) Sn = 6560
a(r n −1)r − 1
= 6560
2(3n −1)3 − 1
= 6560
3n −1 = 65603n = 65613n = 38
∴ n = 8
10 138 += kT
a + 7d = 3k + 1a + 7(4) = 3k + 1
a = 3k − 27 …
S8 = 13k + 682
(2a + 7d ) = 13k + 6
8a + 28d = 13k + 68a + 28(4) = 13k + 6
8a = 13k −106 … Substituting into ,
8(3k − 27) = 13k − 10624k − 216 = 13k − 106
11k = 110k = 10
11 (a) T3 = 10
ar2 = 10 …
T3 + T4 = 1510 + T4 = 15
T4 = 5
ar3 = 5 …
: ar3
ar2=
510
r =12
From :
a 12
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 10
a 14
⎛
⎝ ⎜
⎞
⎠ ⎟ = 10
a = 40
(b)
S∞ =40
1− 12
= 80
41
Paper 2 1 (a)
T1 = Area of the first triangle
=12
bh
T2 = Area of the second triangle
=12
b2
⎛
⎝ ⎜
⎞
⎠ ⎟
h2
⎛
⎝ ⎜
⎞
⎠ ⎟ =
18
bh
T3 = Area of the third triangle
=12
b4
⎛
⎝ ⎜
⎞
⎠ ⎟
h4
⎛
⎝ ⎜
⎞
⎠ ⎟ =
132
bh
T2
T1=
18
bh
12
bh=
14
T3
T2=
132
bh
18
bh=
14
Since
T2
T1=
T3
T2=
14
(a constant), the
areas of the triangles form a geometric
progression with a common ratio of 14
.
(b) (i)
T1 =
12
(160)(80) = 6400
T2 =18
(160)(80) = 1600
T3 =1
32(160)(80) = 400
Tn = 25ar n−1 = 25
6400 14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
= 25
14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
=1
256
14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
=14
⎛
⎝ ⎜
⎞
⎠ ⎟
4
n − 1 = 4n = 5
Hence, the 5th triangle has an area of 25 cm2.
(ii) Sum to infinity,
S∞ =a
1− r
=6400
1− 14
= 8533 13
cm2
42
2 (a) The number of cubes in each storey forms a geometric progression 1, 4, 16, 64, …, where a = 1 and r = 4.
2 sin2 θ + 3 sin θ +1 = 0(2 sin θ +1)(sin θ +1) = 0
sin θ = −
12
or sin θ = −1
When sin θ = −12
,
Basic ∠ = 30°θ = 210°, 330°
When sin θ = −1,θ = 270°
∴θ = 210°, 270°, 330°
Form 5: Chapter 16 (Trigonometric Functions) SPM Flashback
Fully-Worked Solutions
tan θ = t
62
5 15 cos2 x − 7 cos x = 4 cos 60°
15 cos2 x − 7 cos x = 4(0.5)15 cos2 x − 7 cos x = 2
15 cos2 x − 7 cos x − 2 = 0(3 cos x − 2)(5 cos x +1) = 0
cos x =
23
or cos x = −15
When cos x =23
Basic ∠ = 48.19°∴ x = 48.19°, 311.81°
When cos x = −15
Basic ∠ = 78.46°∴ x = 101.54°, 258.46°
∴ x = 48.19°, 101.54°, 258.46°, 311.81°
63
Paper 2 1 (a) LHS
= tan x2+ cot x
2
=sin x
2
cos x2
+cos x
2
sin x2
=sin2 x
2+ cos2 x
2
sin x2
cos x2
=1
sin x2
cos x2
=2
2 sin x2
cos x2
=2
sin x= 2 cosec x= RHS
(b) (i)
(ii) sin 3
2x =
34π
x −1
2 sin 3
2x =
32π
x − 2
The solutions to the equation
2 sin 3
2x =
32π
x − 2 are given by the
x-coordinates of the intersection
points of the graphs of y = 2 sin 3
2x
and y =
32π
x − 2.
Hence, the equation of the straight line for solving the equation
sin 3
2x =
34π
x −1 is y =
32π
x − 2 .
Number of solutions = Number of intersection points = 3
This is a y = sin θ graph
with 1 1
2 cycles
because θ = 1 1
2x.
x 0 2π y −2 1
64
2 (a) The sketch of the graph of y = cos 2x for 0 ≤ x ≤ π is as shown below:
(b) 2 sin2 x = 2 − x
π
1− cos 2x = 2 − xπ
−cos 2x = 1− xπ
cos 2x =xπ−1
The straight line that has to be drawn is
y = x
π− 1.
x 0 π y −1 0
Hence, the number of solutions to the
equation 2 sin2 x = 2 − x
π for 0 ≤ x ≤ π
= Number of intersection points = 2
3 (a) LHS = −2 cos2 x + cosec2 x − cot2 x
= −2 cos2 x +1
= −(2 cos2 x −1)= −cos 2x= RHS
(b) (i) The sketch of the graph of
y = −cos 2x is as shown below.
(ii) 2(−2 cos2 x + cosec2 x − cot2 x) = x
π−1
2(−cos 2x) = xπ−1
−cos 2x =x
2π−
12
The straight line that has to be
sketched is y =
x2π
−12
.
x 0 2π
y −
12
12
Number of solutions = Number of intersection points = 4
If cot2 x +1 = cosec2 x , then cosec2 x − cot2 x = 1.
cos 2x = 1− 2 sin2 x∴ 2 sin2 x = 1− cos 2x
65
4 (a), (b)
sin 2x + 2 cos x = 0
sin 2x = −2 cos x
Number of solutions = Number of intersection points = 2
66
Paper 1 1 Number of codes that can be formed
= 6P3 × 4P2 = 120×12 = 1440. 2 (a) Number of teams that can be formed if
each team consists of 3 boys (and 5 girls)
= 7C3 × 6C5
= 35× 6= 210
(b) Available: 7 boys and 6 girls
Needed: 7 boys and 1 girl or 8 boys and 0 girls Hence, the number of teams that can be formed if each team consists of not more than 1 girl = 7C7 × 6C1 = 6
3 (a) Number of arrangements = 5! = 120 (b) Step 1
If the letter 'O ' and 'E ' have to be side by side, they will be counted as 1 item. Together with the letters 'P ', 'W ' and 'R ', there are altogether 4 items, as illustrated below:
Number of arrangements = 4! Step 2 But the letters 'O ' and 'E ' can also be arranged among themselves in their group.
Number of arrangements = 2!
Step 3 Hence, using the rs multiplication principle, the number of arrangements of all the letters of the word 'POWER ' in which the letters O and E have to be side by side = 4!× 2!= 24 × 2 = 48
4 (a) If there is no restriction, the number of
ways the dancing groups can be formed
= 16C6
= 8008
(b) If the group of dancers must consist of 1
Form Five student and exactly 2 Form Six students, the number of Form Four students required is 3. Hence, the number of ways the dancing groups can be formed
= 7C3 × 5C1 × 4C2
= 1050
5 Number of arrangements
N 7P3
T 7P3
G 7P3
R 7P3
L 7P3
Hence, the total number of arrangements = 7P3 × 5 = 1050
Form 5: Chapter 17 (Permutations and Combinations) SPM Flashback
Fully-Worked Solutions
7C7 × 6C1 Impossible
7 Form Four + 5 Form Five + 4 Form Six students
67
Paper 1
1 P(Blue) =
25
k6 + k
=25
5k = 12 + 2k3k = 12k = 4
2 P(both students participate in the same game)
= P(TT or BB or HH )= P(TT ) + P(BB) + P( HH )
=5
14×
413
⎛
⎝ ⎜
⎞
⎠ ⎟ +
614
×5
13⎛
⎝ ⎜
⎞
⎠ ⎟ +
314
×2
13⎛
⎝ ⎜
⎞
⎠ ⎟
=4
13
3 Let A − Azean D − Dalilah N − Nurur P(only one of them will qualify)
= P( A D N ) + P( A D N ) + P( A D N )
=13
×25
×47
⎛
⎝ ⎜
⎞
⎠ ⎟ +
23
×35
×47
⎛
⎝ ⎜
⎞
⎠ ⎟ +
23
×25
×37
⎛
⎝ ⎜
⎞
⎠ ⎟
=8
105+
835
+435
=44
105
Form 5: Chapter 18 (Probability) SPM Flashback
Fully-Worked Solutions
Initially, there are 5 students playing table tennis out of 14 students.
After 1 student is chosen, it is left with 4 students playing table tennis out of 13 students.
68
Paper 1 1
P(Z > k ) = 0.5 − 0.2580 = 0.2420
2 X − Number of students who passed
X ~ B(7, 0.6)
P( X = 5)= 7C5(0.6)5 (0.4)2
= 0.2613
3 (a) X ~ N (4.8, 1.22 )
Z =X − µ
σ
Z =4.2 − 4.8
1.2Z = −0.5
(b) P(4.8 < X < 5.2)
= P 4.8 − 4.81.2
< Z <5.2 − 4.8
1.2⎛
⎝ ⎜
⎞
⎠ ⎟
= P(0 < Z < 0.333)= 0.5 − 0.3696= 0.1304
4 X − Volume, in ml X ~ N(650, 252)
(a) Z =
25
X − 65025
=25
X − 650 = 10X = 660
Hence, the volume which is equivalent to
the standard score of 25
is 660 ml .
(b) P( X > 620)
= P Z >620 − 650
25⎛
⎝ ⎜
⎞
⎠ ⎟
= P(Z > −1.2)= 1− 0.1151= 0.8849
Hence, the percentage of bottles of soy sauce that have volumes of more than 620 ml
= (0.8849 ×100)%= 88.49%
5 P(Z > m) = 0.5 − 0.1985
P(Z > m) = 0.3015
∴ m = 0.52
Form 5: Chapter 19 (Probability Distributions) SPM Flashback
Fully-Worked Solutions
69
Paper 2 1 (a) X − Number of children
X ~ B(8, 0.4)
(i) P( X ≥ 2)
= 1− P( X = 0) − P( X = 1)= 1− 8C0 (0.4)0 (0.6)8 − 8C1(0.4)1 (0.6)7
= 1− 0.01680 − 0.08958= 0.8936
(ii) Variance = 192
npq = 192n(0.4)(0.6) = 192
n = 800
Hence, the population of the housing estate is 800.
We have to "plus 24" here because at the beginning of the motion, object Q is at 24 m to the right of point A.
74
4 (a) (i) v = t 2 − 6t + k
When t = 0, v = 8
8 = 02 − 6(0) + kk = 8
(ii) When v < 0,
t2 − 6t + 8 < 0(t − 2)(t − 4) < 0
Hence, the range of values of t when the particle moves to the left is 2 < t < 4 .
(iii) a =
dvdt
= 2t − 6
When a < 0,2t − 6 < 0
2t < 6t < 3
Hence, the range of values of t when the particle decelerates is 0 ≤ t < 3 .
(b) (i) t (s) 0 1 2 3 4
v (m s−1) 8 3 0 −1 0
(ii) Total distance travelled in the first 4 s
= Area under the v − t graph
= v dt0
2∫ + v dt2
4∫= (t 2 − 6t + 8) dt
0
2∫ + (t 2 − 6t + 8) dt2
4∫
=t 3
3− 3t 2 + 8t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
2
+t 3
3− 3t 2 + 8t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
4
=23
3− 3(2)2 + 8(2) − 0 +
43
3− 3(4)2 + 8(4) −
23
3− 3(2)2 + 8(2)
⎛
⎝ ⎜
⎞
⎠ ⎟
=203
+163
−203
=203
+ −43
=203
+43
= 8 m
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Paper 2 1 (a) The total mass of seafood is not less than
20 kg. The inequality is x + y ≥ 20.
x 0 20 y 20 0
The mass of prawns is at most three times that of squids. The inequality is x ≤ 3 y .
x 0 30 y 0 10
The allocation is RM250. The inequality is 10x + 5y ≤ 250 ⇒ 2x + y ≤ 50
x 0 25 y 50 0
(b)
(c) If the restaurant buys 15 kg of squids,
y = 15. From the graph, when y = 15, the minimum value of x = 5. Therefore, if the restaurant buys 15 kg of squids, the minimum mass of prawns it can buy is 5 kg. Total expenditure = 10x + 5y When x = 5 and y = 15, the total expenditure = 10(5) + 5(15) = RM125 Hence, the maximum amount of money that could remain from its allocation is RM250 − RM125 = RM125.
Form 5: Chapter 21 (Linear Programming) SPM Flashback
Fully-Worked Solutions
76
2 (a) The total number of participants is at least 30. The inequality is x + y ≥ 30 .
x 0 30 y 30 0
The number of Mathematics participants is at most twice that of Science. The inequality is y ≤ 2x .
x 0 30 y 0 60
The expenditure for a Science participant and a Mathematics participant are RM80 and RM60 respectively. The maximum allocation is RM3600. The inequality is 80x + 60y ≤ 3600 ⇒
4x + 3y ≤ 180.
x 0 45 y 60 0
(b)
(c) (i) When y = 12, from the graph,
xmin = 18 and xmax = 36 . Hence, when the number of Mathematics participants is 12, the minimum and maximum number of Science participants are 18 and 36 respectively.
(ii) Cost = 80x + 60y
Draw the straight line 80x + 60y = 480
x 0 6 y 8 0
For minimum cost, from the graph, xmin = 10 and ymin = 20. Hence, the minimum cost
= 80x + 60y= 80(10) + 60(20)= RM2000
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3 (a) For the constraint "the total number of balls bought should not be more than 80", the inequality is x + y ≤ 80 .
x 0 80 y 80 0
For the constraint "the number of footballs bought should not be more than 4 times the number of volleyballs bought", the inequality is y ≤ 4x .
x 0 20 y 0 80
For the constraint "the number of footballs bought should exceed the number of volleyballs bought by at least 15", the inequality is y − x ≥ 15.
x 0 40 y 15 55
(b)
(c) (i) From the graph, if the number of volleyballs bought is 25 (x = 25), the range of the number of footballs bought is 40 ≤ y ≤ 55 .
(ii) Cost = 60x + 80y
Draw the straight line 60x + 80y = 4800 .
x 0 80 y 60 0
From the graph, the optimal point is (16, 64). Hence the maximum cost = 60(16) + 80(64) = RM6080.
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4 (a) The maximum time for the use of machine P is 12 hours:
60x + 20y ≤ 12 × 603x + y ≤ 36
x 0 12 6 y 36 0 18
The use of machine Q is at least 8 hours:
30x + 40y ≥ 8 × 603x + 4y ≥ 48
x 0 16 y 12 0
The ratio of the number of 'Premier' pewter plates produced to the number of 'Royal' pewter plates produced is at least 1 : 3:
xy
≥13
3x ≥ yy ≤ 3x
x 0 6 y 0 18
(b)
(c) (i) From the graph, if y = 12, xmax = 8
(ii) Profits = 100x + 140y Draw the straight line 100x +140y = 1400
x 0 14 y 10 0
From the graph, the optimal point = (6, 18) Hence, the maximum profit = 100(6) +140(18) = RM3120
(0, 36) is out of the graph paper. So, another point (6, 18) has to be determined.