dunman high school 2009 maths d paper 2 answer key

8/8/2019 Dunman High School 2009 Maths D Paper 2 Answer Key

1/19

DHS 2009 Sec 4 SAP Preliminary Exam Mathematics Paper 2

1 (a) Given that2

2

p q p

q

= , express q in terms ofp. [3]

(b) Express as a fraction in its lowest terms,

2

3 2

5 6 3

x x

x x x

+ . [3]

Answer:

1 (a)2 2

4

p q p

q

=

2 2

4

4

q

p p=

+

2

2

4

4

pq

p=

+

(b)( ) ( )

( )( )23 2 23 2

5 6 3 2 3

x x xx x

x x x x x

+ =

+

( )( )

( ) ( )

1 3

2 3

x x

x x

=

1

2

x

x

=

2

1st 2nd 3rd

pattern pattern patte

In the diagram above, each pattern is made up of dots, lines and small triangles. In the

1st pattern, there are 9 dots, 15 lines and 7 small triangles.


2/19

(a) How many small triangles are there in the

(i) 4th pattern,

(ii) n th pattern? [2]

(b) How many lines are there in the n

th

pattern? [1](c) If there are ddots, l lines and Ttriangles in one of these patterns, write down

an equation connecting d, l and T. [2]

2 (a) (i) 16

(ii) 3 4n +

(b) 6 9n +

(c) ( ) ( ) (6 9 3 6 1 3 4n n n+ + + = + )

1l d T + =

3 A cylindrical container which has an internal diameter of 60 cm and an internal height

of 1.05 m weighs 7 kg when empty.

(a) Find the weight of the container when it is full of oil, if the density of oil is

37 g/cm9

.

(b) How many times will the oil in the container fill a hemispherical bowl of

internal diameter of 7 cm? [Take22

7= ] [5]

(c) Find the internal surface area of the hemispherical bowl in contact with the

oil. [2]

3 (a) Volume of the cylindrical container

2

3

2230 105

7

297 000 cm

=

=

Weight of the cylindrical container

77 297

9

7 231 238 kg

= +

= + =


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3

(b) Volume of a hemispherical bowl

3

3

1 4 223.5

2 3 7

539

cm6

=

=

Number of times the oil will fill the bowl

539297 000

6

63306

49

=

=

(c) Internal surface area in contact with oil

2

2

1 224 3

2 7

77 cm

=

=

.5

4 In May 2007, the Credit Bureau Singapore released the following data on

Singaporeans home loans/ mortgages for the period from March 2005 to March

2007.

No of Singaporeans with: March 2005 March 2006 March 2007

2 or more home loans 19901 25977 41078

2 or more home loans valued at

a total of more than S$1 million

1416 1962 2925

More than S$1 million

in home loans

2381 2381 4291

The information for those Singaporeans with 2 or more home loans over this period of

comparison can be represented by the matrix P = .19901

25977

41078

The information for those Singaporeans with 2 or more home loans valued at a total

of more than S$1 million over this period of comparison is represented by a matrix Q.

(i) Write down the matrix Q. [1]

(ii) Calculate the matrix ( )P Q . [1]

[Turn over


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(iii) Describe what is represented by the elements of ( )P Q . [1]

The information for those Singaporeans with home loans in 2005 is represented by the

matrix A = .( )19901 1416 2381

Information for those Singaporeans with home loans in 2007 is represented by the

matrix B.

(iv) Write down the matrix B. [1]

(v) Show that the matrix C, in terms of A and/ or B, which has its elements

showing the increase of each category over the period of 2005 to 2007 is

. [1]( )21177 1509 1910

(vi) A matrix D is given by

1 0 019901

10

1416

10 0

2381

0 )

)

. Evaluate ( , rounding

off each element to the nearest whole number. [1]

100CD

(vii) Describe what is represented by the elements of the matrix . [2](100CD

4 (i) Q =

1416

1962

2925

(ii) =( )P Q

19901 1416

25977 1962

41078 2925

=

18485

24015

38153

(iii) The elements of ( )P Q represent the information for those

Singaporeans with 2 or more home loans valued at a total ofless than or equalto S$1 million over this period of comparison.

(iv) B = ( ) 41078 2925 4291

(v) C = ( )41078 19901 2925 1416 4291 2381

= [shown]( )21177 1509 1910


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5

(vi) =( )100CD ( )

10 0

19901

1100 21177 1509 1910 0 0

14161

0 02381

= ( )106 106 80

(vii) 106 represents the percentage increase in number of Singaporeanshaving 2 or more home loans over the period of March 2005 to March 2007.106 represents the percentage increase in number of Singaporeans having 2 ormore home loans valued at more than S$1 million over the period of March2005 to March 2007.

80 represents the percentage increase in number of Singaporeans with home

loans of more than S$1 million over the period of March 2005 to March 2007.

5 In Singapore, the rate for the usage of water for the month of July in 2009 is as

follows:

Water used : $1.17 per m3

Water borne fee : $0.28 per m3

Sanitary Appliance fee : $2.80 per fitting

Water Conservation tax : 30% of the amount payable for water used

Goods and Services tax (GST): 7% of all the above fees/ tax

(i) In July, the GST payable for water used only by a Pasir Ris 5-room household

is $3.11.

Calculate the amount, excluding GST, paid for water used in July by this

household. [2]

(ii) Show that the amount of water used by this household in July, is

approximately 38.0 m3. [1]

(iii) Hence, find the overall water bill if this household has 2 sanitary fittings. [2]

(iv) If the national average of water usage per month for a typical 5-room HDB flat

in Singapore is 19.1 m3,

(a) how many percent above average is the water usage for this

household? [2]

[Turn over


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(b) what is the average water usage per day for a typical 5-room HDB flat

in Singapore for the month of July? [1]

5(i) Amount paid for water used only =

$3.11 100

7

= $44.43 (to nearest cent)

(ii) 344.43

37.97 38.0 m (to 3 sig. fig.)1.17

=

(iii) ( )38.0 1.17 1.3 38.0 0.28 2 2.80 1.07 + +

= $79.17 (to nearest cent)

(iv) (a)38.0 19.1

100 49.7% (to 3 sig. fig.)38.0

=

(b) 319.1

0.616 m (to 3 sig. fig.)

31

=

6

C

PD

B

The pointsD,H,R and P lie on the circumference of a circle.DR is a diameter of the

circle,DA is a tangent to the circle atD, CBHis a straight line and .DRH 42= (a) Find, with reason,

AR H

42


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7

(i) , (ii)DHR RDH,

(iii) , (iv)DAR RPH. [4]

(b) Given also that DBH 107= , find

(i) RCH, (ii) . [2]DHC

(c) Show that the trianglesDHR andAHD are similar. [2]

6(a) (i) 90 ( in a semicircle)DHR =

(ii) 90 42 (complementary s, )

48

RDH DHR=

=

(iii)

90 (tangent radius)

90 42 (complementary s, )48

RDA

DAR RDA

=

= =

(iv) ( s in the same segment)

48

RPH RDH=

=

(b) (i)

(ext. sum of int. opp. s)

107 90

17

RCH CDB DBH

RCH

+ = =

=

=

(ii)

(ext. sum of int. opp. s)

48 17

31

DHC RCH RDHDHC

+ = =

=

=

(c)

90 ( s on a straight line)

48 ((a)(ii)&(iii))

(3rd s of s)

Since there are 3 pairs of equal corresponding s,

triangles and are similar. (Shown)

DHR AHD

RDH DAH

DRH ADH

DHR AHD

= =

= =

=

Q7

SR

CBA

P

[Turn over


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The diagram shows three semicircles each of radius 18 cm with centres at A,B and C

in a straight lines shown above. A fourth circle centre at P and with radius rcm is

drawn to touch the other three semicircles. Given thatBPQ is a straight line which is

tangential to the two semicircles with centresA and Cat pointB,

(a) show that cm, [3]4.5r=

(b) Find the value of in radians, [2]PAC

(c) Calculate the area of the shaded region. [3]

7

(a)

( ) ( )

( ) ( )

2 2 2

2 2 2 2

: 18 18 18

18 2 18 18 2 18 18

4.5

ABP r r

r r r r

r

+ = +

+ + = + +

=

2

(b)

13.5tan18

0.644 rad

PAC

PAC

=

=

(c) Area of shaded region

= ( )2 2

1 1 1 2 18 13.5 18 0.644 4.5 0.6442 2 2 2

= 15.7 cm2


9/19

9

8P Q

0.874 km

1.3 kmR

In the diagram, STrepresents the northward-bound MRT line. The quadrilateral PQRS

formed the fence that boarded a carnival for the F1 Night Race in September. The

point P is due west ofS and PS is parallel to QR. Given that PRTis a straight line,

km, km, and . Find0.874QR = 1.3PS = 26.3RST= 90SRT=

(i) the bearing ofR from T, [1]

(ii) the length ofPR, [1]

Hence, show that 0.54PQ = km, [2]

(iii) . [1]QPR

(iv) The Singapore Flyer is built at the point Q. If the angle of depression of P

from the highest point of the wheel is , find the height, in metres, of the

entire flyer. [1]

8

(v) A man walked from P along PS and reached a point Xsuch that the angle of

elevation of the highest point of the wheel is a maximum. Find the angle of

elevation, (you may ignore the height of the man). [3]

8 (i) 90 26.3

63.7 (complementary angles)

RTS =

=

Hence, the bearing ofR from Tis 180 63.7 243.7+ =

(ii) : 26.3PST SPT =

cos26.3

1.3

1.17 km (3 sig. fig.)

PR

PR

=

=

26.3

S

T

North

[Turn over


10/19

( ) ( )( )

22 2 0.874 2 0.874 cos 26.3

0.54 km (Shown)

PQ PR PR

PQ

= +

=

(iii)

sin sin 26.3

0.874

QPR

PQ=

45.4 (to 1 dec. pl.)QPR =

(iv) tan8h

PQ=

Height of the entire flyer is 76.4 m (to 3 sig. fig.)

(v)( )sin 45.4 26.3

0.5164 (to 4 sig. fig.)

XQ PQ= +

=

Let the angle of elevation be

tanh

XQ =

8.4 (to 1 dec. pl.) =


11/19

11

9 According to the Straits Times, a check on a random selection of basic goods at

several supermarkets in Singapore revealed an increase in the prices since the

beginning of the year. In particular, a pack of fresh chicken (between 1 to 1.3 kg) now

cost 70 cents more than its original cost at the beginning of the year.In 2008, Yusof budgeted $234 for fresh chicken to be used during his wedding

reception in January 2009.

(i) Ifx represents the number of packs of fresh chicken (between 1 to 1.3 kg)

which Yusof could buy at the beginning of 2009, write down an expression, in

terms ofx, for the original cost of a pack of fresh chicken (between 1 to 1.3

kg). [1]

(ii) Yusof found that he would get 7 packs of fresh chicken (between 1 to 1.3 kg)

less than that at the beginning of the year if he decided to delay the wedding

reception till September 2009.

Write down an expression, in terms ofx, for the current cost of a pack of fresh

chicken (between 1 to 1.3 kg). [1]

(iii) Write down an equation inx, and show that it reduces to . [3]2 7 2340 0x x =

(iv) Solve the equation 2 7 2340 0x x = . [2]

(v) Calculate the percentage increase in the price of a pack of fresh chicken(between 1 to 1.3 kg). [2]

9 (i) The original cost of a pack of fresh chicken (between 1 to 1.3 kg) = $234

x

(iii)234 234 7

7 1x x =

0

234 234( 7) 7

( 7) 10

x x

x x

=

[ ] 210 234 234 1638 7 49 x x x + = x

x =

27 49 16380 0x x =

[shown]2 7 2340 0x x =

(iv) 2 7 2340 0x x =

( )( )52 45 0x x + =

(rejected)52 or 45x =

(v) GKC could buy 52 7 = 45 packs now.

[Turn over


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(vi) original price = $234

52= $4.50

Percentage increase in price per pack =0.70

100%4.50

=5

15 %9

10 In a recent Olympic diving event, a male participant stood on a platform and

performed a dive into the water.

During the dive, the horizontal distance of the participant away from the platform,

x m, and the corresponding vertical distance of the participant above the platform,

y m, are related by the equation

210

13 2xxy = .

Some corresponding values ofx andy are given in the table below.

x 0 1 2 3 4 5 6

y 0 0.8 0.6 0.6 2.8 6 p

(a) Find the value ofp. [1]

(b) Using a scale of 2 cm to 1 unit, draw a horizontal x-axis for .60 x

Using a scale of 2 cm to 1 unit, draw a vertical y-axis for .111 y

On your axes, plot the points given in the table and join them with a smooth

curve. [3]

(c) Use your graph to find the distance(s) the participant was from the platform the

when he was 0.5 m above the platform. [2](d) Use your graph to find the maximum height above the platform reached by the

participant.

(e) By drawing a tangent, find the gradient of the curve at the point (3, 0.6).

What can be said about the movement of the participant at this instant? [3]

(f) The participant entered the water when he was 4.4 m away from the platform

horizontally. Use your graph to determine the height of the platform above the

water. [1]


13/19

13

(g) Is the graph useful in finding the position of the participant beyond a

horizontal distance of 4.4 m? Justify your answer. [1]

10

(a) p = 10.2

(b)

(b) Correct axes --- B1Points plotted correctly --- B1

Shape --- B1

[Turn over


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(c) x = 0.45 or 2.15

(d) Maximum height = 0.85 m

(e) Drawing of correct tangent line

Gradient = 1.7

The participant is moving downwards andaway from the platform at this point.

(f) Distance the platform is above the water = 4 m

(g) No, because beyond 4.4 m, the participant has entered the water, and afterentering the water, the water will slow down his movement.


15/19

15

11 A bag holds some coloured balls. There are 15 red, 3 blue and 2 white balls. Two

balls are picked from the bag at random, without replacement. The tree diagram

below shows the possible outcomes and some of their probabilities.Second Pick

Red

First Pick

Red Blue

White

Red

Blue Blue

White

Red

White Blue

White1

19

d

15

19

c

2

19

2

19

15

19

3

19

b

a

3

20

3

4

(a) State the values ofa, b, c and d. [2]

(b) Expressing your answers as a fraction in its lowest terms, find the probability

that

(i) both balls are white, [1]

(ii) at least one ball is red. [2]

11 (a)1

10a = ,

14

19b = ,

2

19c = and

3

19d=

(b) (i) P (both white) =1 1

10 19

[Turn over


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1

190= .

(ii) P (at least 1 ball is red) =

3 14 3 3 3 2 182 24 19 4 19 4 19 19

+ + =

12 In a bid to make our society more environmentally friendly, a survey was conducted

and the cumulative frequency curve shown illustrates the number of plastic bags used,

by 200 Singaporeans in a week.

(a) Use the graph to find

(i) the median number of plastic bags used, [1]

(ii) the lower quartile, [1]

(iii) the interquartile range, [1]

(b) A person is considered to be a reddie if he uses more than 18 plastic bags in

a week. A Singaporean is chosen at random. Calculate, leaving your answer

as a fraction in its lowest term, the probability of getting a reddie. [2]

(c) Given that 19.5% of Singaporean surveyed are green crusaders, use the

graph to find the minimum number of plastic bags used by a Singaporean who

is not a green crusader. [2]

(d) The frequency table for this set of data is given below. Showing your method

clearly, prove that the values are as shown in the table. [2]

Number of plastic

bags used per week

Number of Singaporeans

surveyed

0 4x< 10

4 8x< 29

8 1x< 2 52

12 16x< 75

16 20x< 30

20 24x< 4

(e) Calculate,

(i) the mean, [3]

(ii) the standard deviation. [2]


17/19

17

(f) A similar survey was also conducted in Hong Kong and the table below shows

the results of the processed data.

Mean 11.96 Compare, briefly, the results for the

two countries.Standard Deviation 2.90 [1]

[Turn over


18/19

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

180

190

200

0 5 10 15 20 25

Cumulative FrequencyCurve showing thedistribution of number of

plastic bags used by 200

Singaporeans in a week

Cumulative Frequency

Number of plastic bags used in a week


19/19

19

[T

12

(ai) Median = 12.5 plastic bags(aii) Lower Quartile = 9 plastic bags(aiii) Upper Quartile = 15 plastic bags

Interquartile range = 15 9 = 6 plastic bags

(b) 200 190 = 10 reddies who used more than 18 plastic bags in a week.

Probability of getting a reddie = 10 1200 20

=

(c) From the graph, there are19.5

200 39100

= green crusaders who used 8

or less plastic bags in a week.So, the minimum number of plastic bags used by a non-green crusader in aweek = 8 + 1= 9

(d)

Number of plastic bags

used per week

Number of Singaporeans

surveyed,fMid-

values,x0 4x< 10 0 = 10 24 8x< 39 10 = 29 68 12x 91 39 = 52 10

dunman high school 2009 maths d paper 2 answer key

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