1

Paper 1

1. 0.005429=0.00543(3significantfigures) Thetwozerosarenotcountedassignificantfigures.

Answer: C

2. 0.000248 = 2.48 × 10−4

Answer: C

3. Thenumberofsmallbottlesofshampoo

= 4500 × 103ml––––––––––––15ml

= 3 × 105

Answer: D

4. 3.7 × 10−6−0.00000012= 3.7 × 10−6−0.12×10−6=(3.7−0.12)×10−6= 3.58 × 10−6

Answer: B

5. 1011112

5 7Hence,1011112 = 578

Answer: C

6. 10112 + 11112––––––––– 110102–––––––––Answer: B

7.

115°

108°36°

72°y°F

E

L K

JHGx°

y = 72x=180−115 =65Hence,x + y=72+65 = 137

Answer: C

8. ∠FJH = 40°∠FOJ =180°−40°−40° = 100°

Answer: D

9.

A

D

C

B

II

I

III

R

TrianglesA,BandCareimagesofRunderreflectioninlinesI,IIandIIIrespectively.

Answer: D

10. P

L MK J

G H

E

N

Scalefactor=EP–––EK

= 2CentreofenlargementisE.

Answer: A

11. T

R Q

13 cm12 cm

5 cm

x °

P

SPM Model Paper

2

RQ = 10–––2

=5cm

TR = 132−52

=12cm

sin x ° = sin ∠TQR

= 12–––13

Answer: B

12. R

Q

10 cm x °

O

QO = QS–––2

= 16–––2

=8cm

OR = 102−82

=6cm

tanx° = 8—6

= 4—3

Answer: D

13. sin 0° = 0sin 90° = 1sin 180° = 0sin270°=−1

Answer: B

14.

SV

T W

R

QP

U

Answer: B

15. E

F

G R

T

Answer: C

16.

Y

X

Q31°

50 m

tan31°= XY–––50

XY=50×tan31° =30.0m

Answer: B

17.

G

H

60°

North

Answer: D

18. LatitudeofR=(60°−40°)S = 20°S

LongitudeofR=(180°−130°)W =50°WHence,thepositionofR=(20°S,50°W)

Answer: A

19. 5x−x(2−3x) =5x−(2x−3x2) = 5x−2x + 3x2

= 3x2 + 3x

Answer: D

3

20. (4p2−2pq)×p

–––––––––q2(2p−q)

= 2p(2p−q)×p

–––––––––q2(2p−q)

= 2p2

––––q2

Answer: C

21. ef + 1 = 4e−3f––––––

3

3(ef+1)=4e−3f 3ef + 3 = 4e−3f 3ef + 3f = 4e−3 f (3e+3)=4e−3

f = 4e−3––––––3e + 3

Answer: D

22. 7—2 + 2m =−4(1−m)

7—2 + 2m =−4+4m

7—2 + 4 = 4m−2m

7—2 + 8—

2 = 2m

2m = 15–––2

m = 15–––4

Answer: D

23. 213–––––––7

1—2 × 34

2 = 73 × 33

–––––––7

1—2 × 34

2

= 76 × 36–––––––71 × 38

= 76−1 × 36−8

= 75 × 3−2

Answer: A

24. x + 1 1—2 x , x 1—

4 (x+6)

2x + 2 x , 4x x+6 2x−x −2 , 4x−x 6 x −2 , 3x 6 x 2

Hence,x=−2,−1,0,1

Answer: C

25. Median= 20 + 1––––––2 thobservation

=10.5thobservation

Median= 2 + 3–––––2

= 2.5

Answer: B

26. Thenumberofcandrinkssoldin3rdand4thweek=4050−(12×150)= 2250

Letthenumberofcandrinkssoldinthe4thweekbex.2x + x = 2250 3x = 2250 x = 750

Hence,thenumberofcandrinkssoldinthe3rdweek= 750 × 2= 1500

Answer: B

27. 5,5,5,k,k,10 ↑Median=6

5 + k–––––2 =6

k = 7

Mean= 5 + 5 + 5 + 7 + 7 + 10 + 3 + 4––––––––––––––––––––––––––8

= 46–––8

= 5.75

Answer: C

28. y=8−2x2 isaquadraticfunctionwithamaximumpointandy-intercept=8.

Answer: A

29. Element‘0’isnotinthesetF.

Answer: B

4

30. E

F'

ξ FE

E'

ξ F

=

E

E' � F'

ξ F

Answer: A

31. n(P Q)=n(R′) 4 + 2k=8+4+7−k 4 + 2k=19−k 2k + k=19−4 3k = 15 k = 5

Answer: A

32. 2y + 3x−5=0 2y=−3x + 5

y=− 3—2 x + 5—

2

Hence,y-intercept= 5—2

Answer: A

33. 4y + mx−28=0Substitute(−8,1)intotheequation.4(1)+m(−8)−28=0 −8m−24=0 8m=−24 m=−3

Answer: B

34. Totalnumber=13Thenumberdivisibleby2are4,6,8,10,14,16.Theprobabilitythatthenumberisdivisibleby2

= 6–––13

Answer: B

35. Totalnumberofpens=6+4+x= 10 + x

P(choosingaredpen)= 4––––––10 + x

1—6 = 4––––––

10 + x 10 + x = 24

x = 14

Answer: D

36. y ∝ 1–––t 2

y = k–––t 2

4 = k–––32

k = 32 × 4 =36

Hence,y = 36–––t 2

Wheny=9, 9=36–––t 2

t2 = 4 t = 2

Answer: A

37. H ∝ x3

H = kx3

3 = k 1—2

3

k = 24Hence,H = 24x3

Answer: D

38. w ∝ p

–––q

w = kp

–––q

1—3 = k 8–––

4

k = 1–––12

Hence,w = 1–––12

p–––

q

1—4 =

1–––12

x––––25

1—4 =

1–––12 x—

5 x = 15

Answer: D

Drawtheline

Shadecorrectly

Correctequation

Factorisecorrectly

Correct answers

Correctworking

Correctanswer

Correct method

Correctanswer

Correctanswers

Equationwithone unknown

Correctanswers

Correct method

Identifyanglecorrectly

5

39. 4–2

+ 3 –12

− –58

= 4–2

+ –36

− –58

= 4 – 3 + 5–2+6–8

= 6– 4

Answer: A

40. EF=(–24) 1–3

= (–2 × 1 + 4 ×(–3)) =(–14)

Answer: A

Paper 2

Importantstep 1.

y = 5

0

y

x

32y = – x – 4 x + y = 5

2. 2x(x−6) =x + 7 2x2−12x = x + 7 2x2−12x−x−7 =0 2x2−13x−7 =0 (2x+1)(x−7) =02x+1=0 or x−7=0 x=− 1—

2 or x = 7

3.

FE

C

NDA M

B 8 cm

6 cm

12 cm

The angle between lineEB and the baseABCD is ∠EBN. BN 2 = BC 2 + CN 2 = 82+62

BN = 100 =10cm

tan∠EBN = EN––––BN

= 9–––10

∠EBN = 41.99°

4. Volumeoftheremainingsolid=489

1—2 (8+12)×9×8−πr2×6=489

720− 22–––7 × r2×6=489

720− 132––––7 r2 = 489

132––––7 r2=720−489

r2 = 231 × 7––––132

r = 12.25 =3.5cm

5. (a) (i) 2isaprimenumberor 32=6.

(ii) (−4)(−2)=8and−4 −2.

(b) Numberpisanevennumber.

(c) 2n3+1,n=1,2,3,…

6. x + 4y=−12 .........................1

3—4 x−y = 11 ............................2

2×4,3x−4y = 44 .............. 31 + 3, 4x = 32 x = 8

Substitutex=8into1.8 + 4y =−12 4y =−12−8

y =− 20–––4

y =−5

7. (a) P(choosingaboy) =n(boys)–––––––

n(S)

= 4—5

Correctanswers

Correctmethod

Addalltheedges

Subtracttofindtheareaoftheshadedregion

Substitutecorrectly

Correctmethod

Correctmethod

Correctanswers

Correctanswers

Correctmethod

Correctanswers

6

(b) LetBbeboyandGbegirl.

1stpupil 2ndpupil Outcomes

B1

B2 B1B2

B3 B1B3

G B1G

B2

B1 B2B1

B3 B2B3

G B2G

B3

B1 B3B1

B2 B3B2

G B3G

GB1 GB1

B2 GB2

B3 GB3

n(S)=12 P(thetwootherpupilsareofdifferentgender)

= 6–––12

= 1—2

8. (a) QT=7cm LengthofarcPQR

= 120°––––360°

× 2 × 22–––7 × 21

=44cm

Hence,theperimeterofthesectorOPQR = 44 + 21 + 21 =86cm

(b) AreaofthesectorOPQR

= 120°––––360°

× 22–––7 × 212

=462cm2

Areaofthecircle

= 22–––7 × 72

=154cm2

Hence,theareaoftheshadedregion =462−154 =308cm2

9. (a) 5x + 2y−12=0 2y=−5x + 12

y=− 5—2 x+6

Using m=− 5—2 xandG(2,−4),

y = mx + c −4=− 5—

2 (2)+c

−4=−5+c c=−4+5 = 1

Hence, the equation of the straight line passes

throughGandparalleltoEF is y =− 5—2 x + 1.

(b) Wheny=0, 0 =− 5—2 x + 1

5—2 x = 1

x = 2—5

Hence,thex-intercept= 2—5

10. (a) Lengthoftimewhichthecarisstationary =55−30 =25minutes

(b) Speedofthecarinthefirst30minutes

= 40–––30

= 4—3 kmperminute

(c) (i) Averagespeedofthemotorcycle

= 60–––80

=0.75kmperminute

(ii) Distance travelled by the motorcycle in 30minutes

= 0.75 × 30 =22.5km

Distancetravelledbythecarin30minutes =40km

Hence,thedistancebetweenthecarandthemotorcycle

=40−22.5 =17.5km

11. (a) E = 4 –13 t

WhenEhasnoinverse, 4t−(−1)(3) =0 4t =−3

t =− 3—4

Correctmatrixequation

Correct method

Correctanswers

Correctmethod

Correctinversematrix

Correctanswers

7

(b) E = 4 –13 2

(i) E−1 = 1–––––––––––––(4)(2)−(–1)(3)

2 1–3 4

= 1–––11

2 1–3 4

(ii) 4x−y = 17 3x + 2y=−1

4 –13 2

xy

= 17–1

xy =

1––11

2 1–3 4

17–1

= 1––11 34 – 1

–51 – 4

=

1––11 × 33

1––11 ×(–55)

= 3–5

Hence,x=3,y=−5

12. (a)

40

–2

2

2 8

(8, –2)

(3, 1)

(6, 2)

(1, 5)

6

4

y

x

(i) (1,5)→T (3,1) (ii) (1,5)→R (6,2) (i) (1,5)→R (6,2)→T (8,−2)

(b) y

x

AJ K

LM

H

E F

G

D

CB

4

2

–2

2 4 6 8 100

(i) (I) W is a reflection in the line y = 0 (orthex-axis).

(II) V is an enlargement about the centre (5,2)withscalefactorof3.

Correcttransformations Correctdescriptions

(ii) AreaofJKLM = 32×areaofADCB = 9 × 15.4 =138.6m2

Areaoftheshadedregion =138.6−15.4 =123.2m2

13. (a)x −1 2y 10 −11

(b)

5

7

1.75 2.3–2.45

–5

–1 10 2 3

2 cm

2 cm

x

y

y = –4 – 2x

y = 9 – 4x – 3x2

–2–3

–10

–15

–20

–25

–30

–35

–40

10

Correctaxesanduniformscale Allpointscorrectlyplotted Smoothcurve

(c) Fromthegraph, (i) whenx=−1.7,y = 7 (ii) wheny=−16,x = 2.3

Correctreadings

Correctmethod

Correctanswers

8

(d) y=9−4x−3x2 ........................1 0=−13+2x + 3x2 ..................2

1 + 2,y=−4−2x

Drawthecorrectstraightline

\ x=−2.45or1.75Correctanswers

14. (a) Savings (RM) Midpoint Upper

boundaryCumulative frequency

6−10 8 10.5 311−15 13 15.5 916−20 18 20.5 1721−25 23 25.5 2826−30 28 30.5 3631−35 33 35.5 40

(b) Estimatedmeansavings

=

(3×8)+(6×13)+(8×18)+(11×23)+(8×28)+(4×33)

–––––––––––––––––––––––––––––––––––3+6+8+11+8+4

= 855––––40

= RM21.38

(c)

0

5

10

15

20

25

30

35

40

26.5

5.5 10.5 15.5 20.5 25.5 30.5 35.5

Cumulative frequency

Savings (RM)

2 cm

2 cm

Extendcurvetozero

(d) (i) Thethirdquartile=RM26.50.

(ii) There are 10pupilswho savedmore thanRM26.50permonth.

15. (a)G/H

F/E

A/D B/C

L/KM/J

3 cm

3 cm

6 cm

4 cm

4 cm

(b) (i)F/A

2 cm

G/M

4 cm

L/B4 cm

U/H/J/P

E/D

T/S

K/C

3 cm

3 cmQ R

Correctmethod

Correctmethod

9

(ii)TU

K/J

R/SB/A

L/M

F

G

Q/C/P/D

H

E 8 cm

3 cm

3 cm

1 cm

2 cm

4 cm

Correctforms Correctmeasurements Anglesatcornerare90°

16. (a) LongitudeofB =(60°−38°)E = 22°E

(b)

θ0°D

B

O

N

S

q = 3420–––––60

= 57

LatitudeofD =(57°−52°)S = 5°S Hence,thepositionofD=(5°S,22°E)

(c) (i) DistancetravelledbyaeroplaneX =180×60×cos52° =6649nauticalmiles

(ii) DistancetravelledbyaeroplaneY =[180−2(52)]×60 =4560nauticalmiles TimeofflightofaeroplaneX

= 6649–––––600

=11.08hours TimeofflightofaeroplaneY

= 4560–––––600

=7.6hours

Hence,thedifferenceinarrivaltime =11.08−7.6 =3.48hours

Correctanswers