[spmsoalan]-Skema-SPM-2014-Mathematics

Prepared by Lembaga Peperiksaan Malaysia Answer scheme prepared by SPM Soalan

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MATHEMATICS PAPER 1 & 2 SPM 2014



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SKEMA JAWAPAN KERTAS 1 & 2

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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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.

Suggested answer for SPM 2014 Mathematics Paper 2




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




(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=

90




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) (2,B) (3,B) (4,B) (5,B) (6,B)

(1,R) (2,R) (3,R) (4,R) (5,R) (6,R)

(1,G) (2,G) (3,G) (4,G) (5,G) (6,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)}




12 (a)

X -2 3

y -11 9 (b)

(c) (i) when x=0.5, y=11.5 (ii) when y=-18, x=-2.5

(d)

(1)-(2) Y=-2x+4

X 0 2

y 4 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.

-30

-25

-20

-15

-10

-5

0

5

10

15

20

-4 -2 0 2 4 6

Axi

s Ti

tle

Axis Title

Chart Title

Series1

series2




(ii) Area of image PQRS

Area of shaded region = 120-30

14 (a)(i) Body mass index Frequency Midpoint Upper boundary Cumulative

Frequency 10 – 14 0 12 14.5 0

15 – 19 3 17 19.5 3

20 – 24 6 22 24.5 9 25 – 29 8 27 29.5 17

30 – 34 14 32 34.5 31

35 – 39 12 37 39.5 43 40 – 44 5 42 44.5 48

(ii) 30 – 34

(b)

Frequency, f Midpoint, x fx

0 12 0

3 17 51 6 22 132

8 27 216

14 32 448 12 37 444

5 42 210

(c)




(d) Number of students = 48 – 24 = 24

15 (a)

(b)(i)

0 3

9

17

31

43

48

0

10

20

30

40

50

60

0 10 20 30 40 50

Series1

F/E G/H

A/D B/C

6 cm

9 cm

6 cm

3 cm

3 cm

3 cm

3 cm

L/Q

M/R

D

E/D H/C

N/A B

F/P G




END OF ANSWER SCHEME

(ii)

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.

B/A/R C/D

H/E N/M

9cm

6cm




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