Download - Amk Prelim 2009 Em2
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Class Index Number Name
ANG MO KIO SECONDARY SCHOOL
PRELIMINARY EXAMINATION 2009
SECONDARY FOUR EXPRESS/FIVE NORMAL ACADEMIC
Mathematics 4016/02
Setter: Mr Desmond Tong
Thursday 27 Aug 2009 2 hours 30 minutes
Additional Materials: Answer PaperGraph Paper (1 sheet)
READ THESE INSTRUCTIONS FIRST
Write your name, index number and class on all the work you hand in.Write in dark blue or black pen on both sides of the paper.You may use a pencil for any diagrams or graphs.Do not use highlighters, glue or correction fluid.
Answer all questions.If working is needed for any questions it must be shown with the answer.Omission of essential working will result in loss of marks.Calculators should be used where appropriate.
If the degree of accuracy is not specified in the question, and if the answer is not exact,give the answer to three significant figures. Give answers in degrees to one decimalplace.For , use either your calculator value or 3.142, unless the question requires theanswer in terms of .
At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or partquestion.The total of the marks for this paper is 100.
This document consists of 11 printed pages.
[Turn over
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Mathematical Formulae
Compound interest
Total amount =
nr
P
1001
Mensuration
Curve surface area of a cone = rl
Surface area of a sphere = 4r2
Volume of a cone =3
1r
2h
Volume of a sphere =
3
4r
3
Area of triangleABC=2
1ab sin C
Arc length = r, where is in radians
Sector Area =2
1r2, where is in radians
Trigonometry
C
c
B
b
A
a
sinsinsin
a2 = b2c2 2bc cosA
Statistics
Mean =
f
fx
Standard deviation =
22
f
fx
f
fx
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AMKSS 4E/5N Math Prelim Exam 2009 4016/02 [Turn over
Answer all the questions.
1 (a) Given that5
2
2
3
ba
ba, express a in terms ofb. [2]
(b) Express as a single fraction in its simplest form,
aaa
1
1
12
12
. [2]
(c) (i) Factorise 3x2 27. [1]
(ii) Hence, or otherwise, solve 3x2 27 = (x 3)2. [3]
2 A bus driver makes a journey from Chicago to Detroit, a journey of 470 km, at an average
speed ofx km/h.
(a) Write down an expression for the number of hours taken for the journey. [1]
(b) On his return journey from Detroit to Chicago, his average speed for the journey is
reduced by 8 km/h due to heavy traffic along the highway. Write down an
expression for the number of hours taken for the return journey.
[1]
(c) If the return journey takes 50 minutes longer. Write down an equation to represent
this information, and show that it simplifies to
x2 8x 4512 = 0. [3]
(d) Solve the equationx2 8x 4512 = 0. [3]
(e) Find the average speed for the return journey. [1]
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3 (a) Ahmad needs a loan of $80 000 to buy a new car. Bank A charges an interest rate
of 2.45% p.a. compounded monthly. Bank B charges a simple interest rate of
2.65% p.a. Which bank should he borrow from if he were to take a five year loan?
[4]
(b) The road tax is charged based on the engine capacity of the car. The road tax for
cars is shown in the table below:
Engine Capacity (EC) in cc 6-Monthly Road Tax Formulae
EC 600 $200 0.782600 < EC 1 000 [$200 + $0.125 (EC 600)] 0.782
1 000 < EC 1 600 [$250 + $0.375 (EC 1 000)] 0.7821 600 < EC 3 000 [$475 + $0.75 (EC 1 600)] 0.782
EC > 3 000 [$1 525 + $1 (EC 3 000)] 0.782
Ahmad paid $1681.30 for a year of road tax. Calculate the engine capacity of his
car.
[3]
(c) Ahmad realised that the cost of petrol he used ($P) was directly proportional to the
distance (D km) he travelled. At the end of a certain month, he found that he had
travelled a distance of 1 600 km at a cost of $320. Find the cost of petrol for a
journey of 500 km.
[2]
(d) Ahmads petrol bill for 2007 was $3 840. In 2008, the price of petrol increased by
7.5% and his petrol consumption decreased by 20%. Find his petrol bill for 2008.
[2]
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AMKSS 4E/5N Math Prelim Exam 2009 4016/02 [Turn over
4 O is the centre of the circleACBQ of radius 15 cm.APB is an arc of a circle, centre Cand
angleACB = 1.2 radians.
(a) Find obtuse angleAOB in radians. [1]
(b) Show thatAC= 24.76 cm correct to 2 decimal places. [2]
(c) Calculate the perimeter of the shaded region. [2]
(d) Calculate the area of the shaded region. [4]
5 A series of diagrams of shaded and unshaded small squares is shown below. The shaded
squares are those which lie on the diagonals of the diagram.
(a) Find the values ofa,b andc. [2]
Diagram,n 1 2 3 4 5
Number of shaded squares, S 1 5 9 13 aNumber of unshaded squares, U 0 4 16 36 bTotal number of squares, T 1 9 25 49 c
(b) By considering the number patterns, without further diagrams,
(i) write down the total number of squares in diagram 10, [1]
(ii) find an expression, in terms ofn, for T. [2]
(c) (i) Write down the number of diagram that has 29 shaded squares. [1]
(ii) Find an expression, in terms ofn, for S. [1]
Diagram 1Diagram 2
Diagram 3Diagram 4
A B
C
O
P
Q
1.2 rad
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6 The diagram below represents a map showing the foot of a cliffZand two oil rigs
Xand Y.XYZis a triangle lying on a horizontal plane with YXZ= 30 and YZ= 10 km.
S is the top of a lighthouse situated on the top of the cliff vertically aboveZ, and the
angles of elevation ofS fromXand Yare 20 and 35 respectively.
(a) Show that the height ofSZis approximately 7 km. [2]
(b) Calculate XYZ. [3]
(c) A supply ship is travelling fromXtoZ. Calculate the shortest distance between thesupply ship and the oil rig Y.
[3]
(d) It is given that the bearing ofZfromXis 100. Calculate the bearing ofZfrom Y. [2]
X
Y
Z
S
30 10 km
N
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AMKSS 4E/5N Math Prelim Exam 2009 4016/02 [Turn over
7 In the diagram,Xand Yare centres of the two circles. The lineABCD is a tangent to the
two circles atB and C. Line GFED is another tangent to the two circles atEand F. P, Q
andR are points on the circumference of the circle. Given EDY= 30 and PQR = 70.
(a) Calculate
(i) EYD, [1]
(ii) PBR, [1]
(iii) BPR, [2]
(iv) QPR, [2]
(v) PXB. [2]
(b) Show that trianglesEYD andBXD are similar. [2]
D
G
X
B
C
E
F
Y
P
Q
R
A
30
70
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8 The diagram shows a hemispherical bowl with centre O, inner radius 6 cm and outer
radius rcm. Find
(a) (i) the volume of the hemisphere with radius 6 cm, [2]
(ii) the value ofrif 329 cm3 of ceramic is used to make the bowl. [3]
A solid pyramid with square baseABCD and height OV, 6 cm, is placed in the bowl. The
points V, A,B, CandD touches the inner surface of the hemispherical bowl.
(b) (i) Show that 26AB cm. [2]
(ii) Calculate the volume of the pyramid. [2]
Modelling clay is used to make 30 such pyramids. The clay is sold in packets, each
containing 1200 cm3 of clay.
(c) Calculate the number of packets of clay required. [3]
O
rcm
6 cm
OA
C
D
V
A
CO
B
D
V
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AMKSS 4E/5N Math Prelim Exam 2009 4016/02 [Turn over
9 (a) The age of 60 members of a club was recorded. The cumulative frequency curve
below shows the distribution of the ages.
Age distribution
(i) Copy and complete the grouped frequency table of the age distribution of the
members of the club.
Age 0
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(b) Use the cumulative frequency curve from (a) to solve the following questions:
(i) Two members are selected at random in succession without replacement.Find the probability that both members are above the age of 40.
[1]
(ii) Two members are selected at random in succession without replacement.Find the probability that one of the member is above the age of 50 and theother is below the age of 25.
[2]
(iii) Four members are selected at random in succession without replacement.Find the probability none of them are above the age of 55.
[2]
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10 Answer the whole of this question on a sheet of graph paper
The table below shows values ofx andy (correct to 1 decimal place) connected by the
equation
2
42
xxy .
x 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4
y 5.5 3.5 2.0 1.0 0.4 p 2.0 2.6 q 3.4 3.7
(a) Find the values ofp and q, rounding off your answers to 1 decimal place. [2]
(b) Using a scale of 2 cm to represent 0.5 units on thex-axis and 2 cm to represent 1
unit on they-axis, draw the graph of2
42
xxy for 0.7 x 2.4. [3]
(c) Use your graph to solve the equation 04
22
x
x . [1]
(d) (i) On the same axes, draw a suitable line to solve the equation 3x3x2 4 = 0. [3]
(ii) State the range of values ofx for which xx
x 214
2 . [1]
(e) By drawing a suitable straight line on the graph, write down the coordinates of the
point on the curve at which the gradient of the tangent is 2.
[2]
End of Paper
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Ang Mo Kio Secondary School
Preliminary Examination 2009
Secondary 4 Express / 5 Normal Academic Paper 2
Qn Solutions Marks
1a 5 ( 3a + b ) = 2 ( 2a b)15a + 5b = 4a 2b15a 4a = 2b 5b11a = 7ba = 7b/11.
M1
A1
1b
11222
121
121
112
1
1
1
112
1
aa
a
aa
a
aa
aaa
M1
A1
1ci 3 (x2 9 )
= 3 (x 3 ) (x + 3 ) B11cii 3x2 27 = (x 3 )2
3 (x 3 ) (x + 3 ) = (x 3 )23 (x 3 ) (x + 3 ) (x 3 )2 = 0(x 3 ) [ 3 (x + 3 ) (x 3 ) ] = 0
x 3 = 0 OR 3x + 9 x + 3 = 0x = 3 OR x = 6
OR
3x2 27 =x2 6x + 92x2 + 6x 36 = 02 (x2 + 3x 18 ) = 02 (x 3 ) (x + 6 ) = 0
x = 3 OR x = 6
M1
A1, A1
M1A1, A1
2a Time =470
/x B12b Time = 470/x 8 B1
2c
045128
0)45128(5
0225604054052256028202820
))(8(5)8)(6(470))(6(470
60
50470
8
470
2
2
2
2
xx
xx
xxxxxx
xxxx
xx
M1
M1
A1
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2d
2
181128
12
451214882
x
x
x = 71.29041537 or 63.29041537x = 71.3 or 63.3
M1
A1, A1
2e Speed = 71.3 8 = 63.3 km/h B1
3a Bank A:
24.90414$
100
1245.2180000
512
Bank B:
00.90600$
100
565.28000080000
He should borrow from Bank A
M1M1
M1A1
3b Half year: 1681.30 2 = 840.65[475 + 0.75 (EC 1600)] 0.782 = 840.65475 + 0.75EC 1200 = 10750.75EC= 1075 + 1200 4750.75EC= 1800
EC= 2400 cc
M1M1
A1
3c 1600 320500
320/1600 500
$100M1A1
3d 3840 0.8 1.075
= $3302.40
M1
A1
4a AOB= 1.2 2 ( at center = 2 at circumference)= 2.4 rad B1
4b
76.24
6.0sin
152.1sin
6.0sin
15
)2.1sin(
AC
AC
AC
M1
A14c Perimeter
= 15 (2.4) + 24.76006845 (1.2)= 65.71208214= 65.7
M1
A1
4d Area of sector OAQB
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= 0.5 152 2.4= 270Area of sector CAPB= 0.5 (24.76)2 1.2= 367.8365937Area of triangle OAC= 0.5 152 sin ( 1.2)= 104.8543972Area of shaded region= 270 [ 367.8365937 2(104.8543972) ]= 111.8722006= 112 cm2
M1
M1
M1
A1
5a a = 17b = 64c = 81
B1 for 1correctanswerB2 for 3correctanswers
5bi 192
= 361 B15bii T= ( 2n 1 )2 B2
5ci 8 B1
5cii S= 4 ( n 1 ) + 1= 4n 4 + 1= 4n 3 B1
6a tan 35 = SZ/10SZ= 10 tan 35SZ= 7.00 (shown)
M1
A1
6b tan 20 = SZ/XZ
XZ= SZ/tan 20XZ= 19.238044
1.74
13368338.74
10
30sinsin
10
30sin
238044.19
sin
XYZ
XYZ
XZXYZ
XYZ
M1
M1
A16c XZY= 180 30 74.13368338
XZY= 75.86631662
sin 75.86631662 =
d
/10d= 10 sin 75.86631662d= 9.697286299d= 9.70 km
M1M1
A1
6d 180 100 30 = 50Bearing = 74.13368338 50Bearing = 024.1
M1A1
7ai EYD = 180 30 90
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EYD = 60 B17aii PBR = 180 70 (s in opp segment)
PBR = 110 B1
7aiii RXB = 60BPR = 60/2 (s at center = 2 at circumference)BPR = 30
M1A1
7aiv QPR = 180 70 90
QPR = 20
M1
A17av XBR = 60 (equal )
XBP = 110 60 = 50PXB = 180 2 (50) (isos )PXB = 80
M1
A1
7b EDY= BDX= 30YED = XBD = 90TriangleEYD andBXD are similar
2 of their corresponding s are equal
M1
A1
8ai Vol
=1
/2 4
/3 63
= 904.7786842= 452.3893421= 452
M1
A1
8aii ( 1/2 4/3 r
3 ) 452.3893421 = 3292/3 r
3= 781.3893421
r3 = 373.0859288r= 7.198957725r= 7.20
M1
M1
A1
8bi
26
236
66 22
AB
AB
AB
M1
A1
8bii
144
6263
1 2
Vol
Vol
M1
A1
8c Total = 144 30 = 43204320 1200 = 3.6No of packets = 4
M1M1A1
9ai Age 0