MATHEMATICS PAPER 1 & 2 SPM 2014
SKEMA JAWAPAN KERTAS 1 & 2 Prepared by Lembaga Peperiksaan Malaysia Answer scheme prepared by SPM Soalan
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Suggested Answer for SPM 2014 Mathematics Paper 1
1
D
21
C
2
D
22
B
3
A
23
B
4
A
24
A
5
C
25
A
6
A
26
A
7
A
27
C
8
B
28
C
9
C
29
C
10
A
30
D
11
B
31
D
12
D
32
D
13
B
33
D
14
A
34
C
15
B
35
C
16
B
36
D
17
C
37
C
18
B
38
C
19
D
39
B
20
B
40
A
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Suggested answer for SPM 2014 Mathematics Paper 2 1
(a)
(b)
2
Sub x=4 into (1), 4 + y = 11 y=7
Price of 1 kg durian is RM4, price of 1kg jackfruit is RM 7.
3 When the rocket hit ground, h=0,
The water rocket hits the ground at t=2.
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4
(a)
(b)
5
’
(a) Since, PO parallel to JK, therefore, gradient JK = R(2,5) : sub x=2, y=5, Y=mx + c
C=6
(b) Let y=0,
6
(a)(i) m=150-70 =80 n=64+26 =90
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90
(ii)
7
(a)(i) False (ii) (b) (c) Premise 2:3 is an odd number (d)
8 (a) Inverse matrix =
—
= (b)
=
Thus, x=5, y=
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9
PQ=4cm 10 (a)
=11cm
=16.5cm Perimeter = 11+16.5+3.5+3.5 =34.5cm (b)
Area of shaded region =38.5 + 28.875 – 6.125
11 (a) (1,B) (1,R) (1,G)
(2,B) (2,R) (2,G)
(3,B) (3,R) (3,G)
(4,B) (4,R) (4,G)
(5,B) (5,R) (5,G)
(b) (i) {(1,R), (2,R), (3,R), (4,R), (5,R)}
(ii) {(4,B), (5,B), (6,B), (4,R), (5,R), (6,R), (4,G), (5,G), (6,G), (1,G), (2,G), (3,G)}
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(6,B) (6,R) (6,G)
12 (a) X y
-2 -11
3 9
(b)
Chart Title 20 15 10 Axis Title
5 -4
0 -2
-5 0
2
4
6
-10
Series1 series2
-15 -20 -25 -30
Axis Title
(c) (i) when x=0.5, y=11.5 (ii) when y=-18, x=-2.5 (d)
(1)-(2) Y=-2x+4 X y
0 4
2 0
x =-0.5, x=4.5 13 (a)(i) (ii) (b)(i)(a) N is a reflection in the line y=4 (b) M is an enlargement from the centre S(1,5) and a scale factor of 2.
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(ii) Area of image PQRS
Area of shaded region = 120-30 14 (a)(i) Body mass index 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 35 – 39 40 – 44
Frequency
Midpoint
Upper boundary
0 3 6 8 14 12 5
12 17 22 27 32 37 42
14.5 19.5 24.5 29.5 34.5 39.5 44.5
(ii) 30 – 34 (b) Frequency, f 0 3 6 8 14 12 5
Midpoint, x 12 17 22 27 32 37 42
(c)
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fx 0 51 132 216 448 444 210
Cumulative Frequency 0 3 9 17 31 43 48
60 50
48 43
40 31
30 20
Series1
17
10
9
0
0 0
10
3 20
30
40
50
(d) Number of students = 48 – 24 = 24 15 (a)
F/E
G/H
9 cm A/D
B/C 6 cm
(b)(i) E/D
H/C
3 cm F/P
L/Q
3 cm
G
3 cm M/R D
3 cm
N/A
6 cm
B
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(ii) N/M
H/E
9cm
B/A/R
6cm
C/D
16 (a) Longitude of R = (180-40) E = 140E
(b) = 5969 n.m. (c)
Latitude of V = (74-34)N = 40 N (d) Total distance = QP + PV = 5969 + 4440 = 10409 n.m.
END OF ANSWER SCHEME
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