Amk Prelim 2009 Em2

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 Paper Graph 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 decimal place. For π, use either your calculator value or 3.142, unless the question requires the answer in terms of π. 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 part question. The total of the marks for this paper is 100.

This document consists of 11 printed pages. [Turn over

2 Mathematical Formulae Compound interest

r   Total amount = P1    100 

n

Mensuration Curve surface area of a cone = πrl Surface area of a sphere = 4πr2 1 2 πr h 3

Volume of a cone =

Volume of a sphere =

Area of triangle ABC =

4 3 πr 3

1 ab sin C 2

Arc length = rθ, where θ is in radians Sector Area =

1 2 r θ, where θ is in radians 2

Trigonometry a b c   sin A sin B sin C

a2 = b 2  c2 – 2bc cos A

Statistics Mean =

Standard deviation =

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

 fx f

 fx f

2

  fx      f 

2

3 Answer all the questions. 1

3a  b 2  , express a in terms of b. 2a  b 5

(a)

Given that

(b)

Express as a single fraction in its simplest form,

[2]

1 1 .  2a  a  1 1  a

[2]

(i)

Factorise 3x2 – 27.

[1]

(ii)

Hence, or otherwise, solve 3x2 – 27 = (x – 3)2.

[3]

2

(c)

2

A bus driver makes a journey from Chicago to Detroit, a journey of 470 km, at an average speed of x 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

[1]

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. (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 equation x2  8x  4512 = 0.

[3]

(e)

Find the average speed for the return journey.

[1]

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

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4 3

(a)

Ahmad needs a loan of $80 000 to buy a new car. Bank A charges an interest rate

[4]

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? (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 EC  600 600 < EC  1 000 1 000 < EC  1 600 1 600 < EC  3 000 EC > 3 000

6-Monthly Road Tax Formulae $200  0.782 [$200 + $0.125 (EC  600)]  0.782 [$250 + $0.375 (EC  1 000)]  0.782 [$475 + $0.75 (EC  1 600)]  0.782 [$1 525 + $1 (EC  3 000)]  0.782

Ahmad paid $1681.30 for a year of road tax. Calculate the engine capacity of his

[3]

car. (c)

Ahmad realised that the cost of petrol he used ($P) was directly proportional to the

[2]

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. (d)

Ahmad’s 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.

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

[2]

5 4

O is the centre of the circle ACBQ of radius 15 cm. APB is an arc of a circle, centre C and angle ACB = 1.2 radians. C 1.2 rad

O A

B

P Q

5

(a)

Find obtuse angle AOB in radians.

[1]

(b)

Show that AC = 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]

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.

Diagram 1

(a)

Diagram 2

(c)

Diagram 4

Find the values of a, b and c. Diagram, n Number of shaded squares, S Number of unshaded squares, U Total number of squares, T

(b)

Diagram 3

[2] 1 1 0 1

2 5 4 9

3 9 16 25

4 13 36 49

5 a b c

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 of n, for T.

[2]

(i)

Write down the number of diagram that has 29 shaded squares.

[1]

(ii)

Find an expression, in terms of n, for S.

[1]

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

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6 6

The diagram below represents a map showing the foot of a cliff Z and two oil rigs X and Y. XYZ is 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 above Z, and the angles of elevation of S from X and Y are 20 and 35 respectively. S N

Z

X

10 km

30

Y (a)

Show that the height of SZ is approximately 7 km.

[2]

(b)

Calculate XYZ.

[3]

(c)

A supply ship is travelling from X to Z. Calculate the shortest distance between the supply ship and the oil rig Y.

[3]

(d)

It is given that the bearing of Z from X is 100. Calculate the bearing of Z from Y.

[2]

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

7 7

In the diagram, X and Y are centres of the two circles. The line ABCD is a tangent to the two circles at B and C. Line GFED is another tangent to the two circles at E and F. P, Q and R are points on the circumference of the circle. Given EDY = 30 and PQR = 70.

G

F E

Q 70

30

R

X

D

Y

C

P B

A

(a)

(b)

Calculate (i)

EYD,

[1]

(ii)

PBR,

[1]

(iii) BPR,

[2]

(iv)

QPR,

[2]

(v)

PXB.

[2]

Show that triangles EYD and BXD are similar.

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

[2]

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8 8

The diagram shows a hemispherical bowl with centre O, inner radius 6 cm and outer radius r cm. Find

r cm

6 cm O

(a)

(i)

the volume of the hemisphere with radius 6 cm,

[2]

(ii)

the value of r if 329 cm3 of ceramic is used to make the bowl.

[3]

B A

C

O D

A

V

C

O D

V

A solid pyramid with square base ABCD and height OV, 6 cm, is placed in the bowl. The points V, A, B, C and D touches the inner surface of the hemispherical bowl. (b)

(i)

Show that AB  6 2 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.

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

[3]

9 9

(a)

The age of 60 members of a club was recorded. The cumulative frequency curve

Cumulative Frequency

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< x 24 24< x 29 29< x 34 34< x 39 39< x 44 44< x 49 49< x 54 54< x 59 Freq [2]

(ii)

Using your grouped frequency table, calculate an estimate of (a)

the mean age of the members in the club,

[2]

(b)

the standard deviation.

[2]

(iii) The ages of another club of 60 members also have the same median but a smaller standard deviation. Describe how the cumulative frequency curve will differ from the given

[1]

curve. AMKSS 4E/5N Math Prelim Exam 2009 4016/02

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10 (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 the other 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.

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

[2]

11 10 Answer the whole of this question on a sheet of graph paper The table below shows values of x and y (correct to 1 decimal place) connected by the equation

x y

0.7 5.5

y  2 x

4 . x2

0.8 3.5

0.9 2.0

1.0 1.0

1.2 0.4

1.4 p

1.6 2.0

1.8 2.6

2.0 q

2.2 3.4

2.4 3.7

(a)

Find the values of p and q, rounding off your answers to 1 decimal place.

(b)

Using a scale of 2 cm to represent 0.5 units on the x-axis and 2 cm to represent 1 4 unit on the y-axis, draw the graph of y  2  x  2 for 0.7  x  2.4. x 4  0. x2

(c)

Use your graph to solve the equation 2  x 

(d)

(i)

On the same axes, draw a suitable line to solve the equation 3x3  x2  4 = 0.

(ii)

State the range of values of x for which x 

(e)

4  1  2x . x2

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.

End of Paper

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

[2]

[3]

[1] [3] [1] [2]

12 Ang Mo Kio Secondary School Preliminary Examination 2009 Secondary 4 Express / 5 Normal Academic Paper 2 Qn 1a

Solutions 5 ( 3a + b ) = 2 ( 2a – b) 15a + 5b = 4a – 2b 15a – 4a = –2b – 5b 11a = –7b a = –7b/11

Marks

M1 A1

.

1b

1ci 1cii

1 1  2a  1a  1 1  a 1 12a  1   2a  1a  1 a  12a  1 2a  2  2a  1a  1 3 ( x2 – 9 ) =3(x–3)(x+3) 3x2 – 27 = ( x – 3 )2 3 ( x – 3 ) ( x + 3 ) = ( x – 3 )2 3 ( x – 3 ) ( x + 3 ) – ( x – 3 )2 = 0 (x–3)[3(x+3)–(x–3)]=0 x – 3 = 0 OR 3x + 9 – x + 3 = 0 x = 3 OR x = –6

M1

A1 B1

M1

A1, A1

OR 3x2 – 27 = x2 – 6x + 9 2x2 + 6x – 36 = 0 2 ( x2 + 3x – 18 ) = 0 2(x–3)(x+6)=0 x = 3 OR x = –6 2a 2b 2c

Time = 470/x Time = 470/x – 8 470 470 50   x 8 x 60 470(6)( x)  470(6)( x  8)  5( x  8)( x)

M1 A1, A1 B1 B1 M1 M1

2820 x  2820 x  22560  5 x 2  40 x 5 x 2  40 x  22560  0 5( x 2  8 x  4512)  0 x 2  8 x  4512  0

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

A1

13 2d x

  8 

 82  41 4512 21

2e

8  18112 2 x = 71.29041537 or –63.29041537 x = 71.3 or –63.3 Speed = 71.3 – 8 = 63.3 km/h

3a

Bank A:

M1

x

A1, A1 B1

125

3b

3c

3d

4a

4b

4c

4d

 2.45  12  800001   100    $90414.24 Bank B: 80000  2.65  5 80000  100  $90600.00 He should borrow from Bank A Half year: 1681.30 ÷ 2 = 840.65 [475 + 0.75 (EC – 1600)] × 0.782 = 840.65 475 + 0.75 EC – 1200 = 1075 0.75 EC = 1075 + 1200 – 475 0.75 EC = 1800 EC = 2400 cc 1600  320 500  320/1600 × 500  $100 3840 × 0.8 × 1.075 = $3302.40 AOB = 1.2 × 2 ( at center = 2 ×  at circumference) = 2.4 rad AC 15  sin(  1.2) sin 0.6 15 AC  sin   1.2  sin 0.6 AC  24.76 Perimeter = 15 (2.4) + 24.76006845 (1.2) = 65.71208214 = 65.7

Area of sector OAQB

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

M1 M1

M1 A1 M1 M1

A1

M1 A1 M1 A1

B1 M1

A1 M1 A1

14 = 0.5 × 152 × 2.4 = 270 Area of sector CAPB = 0.5 × (24.76…)2 × 1.2 = 367.8365937 Area of triangle OAC = 0.5 × 152 × sin (π – 1.2) = 104.8543972 Area of shaded region = 270 – [ 367.8365937 – 2(104.8543972) ] = 111.8722006 = 112 cm2 5a

a = 17 b = 64 c = 81

5bi 5bii 5ci 5cii

192 = 361 T = ( 2n – 1 )2 8 S =4(n–1)+1 = 4n – 4 + 1 = 4n – 3

6a

6b

6c

6d

7ai

tan 35° = SZ/10 SZ = 10 tan 35° SZ = 7.00 (shown) tan 20 = SZ/XZ XZ = SZ/tan 20 XZ = 19.238044 sin XYZ sin 30  19.238044 10 sin 30 sin XYZ   XZ 10 XYZ  74.13368338 XYZ  74.1 XZY = 180° – 30° – 74.13368338° XZY = 75.86631662° sin 75.86631662° = d/10 d = 10 × sin 75.86631662° d = 9.697286299 d = 9.70 km 180° – 100° – 30° = 50° Bearing = 74.13368338° – 50° Bearing = 024.1°

EYD = 180° – 30° – 90°

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

M1

M1

M1

A1 B1 for 1 correct answer B2 for 3 correct answers B1 B2 B1

B1 M1 A1

M1 M1

A1 M1 M1

A1 M1 A1

15 7aii 7aiii

7aiv 7av

7b

8ai

8aii

8bi

B1

EYD = 60° PBR = 180°  70° (s in opp segment) PBR = 110° RXB = 60° BPR = 60°/2 (s at center = 2 ×  at circumference) BPR = 30° QPR = 180° – 70°  90° QPR = 20° XBR = 60° (equal ) XBP = 110°  60° = 50° PXB = 180°  2 (50°) (isos ) PXB = 80° EDY = BDX = 30° YED = XBD = 90° Triangle EYD and BXD are similar 2 of their corresponding s are equal

B1 M1 A1 M1 A1 M1 A1 M1 A1

Vol = 1/2 × 4/3 × π × 6 3 = 904.7786842 = 452.3893421 = 452 ( 1/2 × 4/3 × π × r3 ) – 452.3893421 = 329 2 /3 × π × r3 = 781.3893421 r3 = 373.0859288 r = 7.198957725 r = 7.20 2

AB  6  6

M1

A1 M1 M1 A1 M1

2

AB  36  2 8bii

8c

9ai

9aiia

9aiib

A1

AB  6 2 2 1 Vol   6 2  6 3 Vol  144 Total = 144 × 30 = 4320 4320 ÷ 1200 = 3.6 No of packets = 4

M1

 

A1 M1 M1 A1

Age

0< x 24

24< x 29

29< x 34

34< x 39

39< x 44

44< x 49

49< x 54

54< x 59

Freq

4

5

7

9

11

11

8

5

Mean = [4(12)+5(26.5)+7(31.5)+9(36.5)+11(41.5)+11(46.5) +8(51.5)+5(56.5)] ÷ 60 = 39.86666667 = 39.9 SD = 11.23308605 SD = 11.2

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

B1 for 4 correct answers B2 for 8 correct answers M1

A1 or B2 B2

16 9iii

9bi

bii

biii

10a 10b 10c 10d

The new curve has a steeper slope. OR The new curve has a smaller interquartile range. 60  27 59  27 P  60 59 88 P 295  60  49 5  P  2   59   60 11 P 354 56 55 54 53 P    60 59 58 57 P  0.7532068043 P  0.753

B1

p = 1.4 (must be 1 dp) q = 3.0 (must be 1 dp) Refer to graph x = 1.05 to 1.20 3x 3  x 2  4  0

B1 B1

3 x3 x 2 4   0 x 2 x 2 x2 4 3x  1  2  0 x 4 x  2  2  3  2x x y  3  2x 10di 10dii 10e

x = 1.15 to 1.30 x < (di) (2, 3)

AMKSS 4E/5N Math Prelim Exam 2009 4016/02

B1 M1 A1 M1

A1

B1

B1 B1 B1 B1

17 10b

AMKSS 4E/5N Math Prelim Exam 2009 4016/02