31 Model[A Math CD]
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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