[spmsoalan]-Skema-SPM-2014-Mathematics

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 = 140E 

(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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