Dunman High School 2009 Maths D Paper 1 Answer Key

DHS 2009 Sec 4 (SAP) Preliminary Examination Mathematics Paper 1

DHS 2009 Sec 4 SAP Preliminary Exam Mathematics Paper 1 1 The diameter of a H1N1 flu virus in a form of a circle is measured by a team of health

officers from WHO as 0.000 131 01 mm. Find the area of the H1N1 flu virus, giving your answer in square metres and express this answer in standard form correct to 3 significant figures.

Answer:

1

8

0.000 131 01 10 100Radius2

6.5 10 m−

÷ ÷=

= ×

( )28 14

14

area of H1N1 flu virus is π 6.5 10 1.348 10

1.35 10 m (to 3sf)

− −

−

∴ × × ≈ ×

= ×

M1 A1

Correct conversion of mm to m cao

2 (i) Simplify ( )( )2a a b a b− − + .

(ii) Hence evaluate 231.01 28.01 34.01− × . Answer:

2(ai) ( )( ) ( )2 2 2 2

2 2 2

2

a a b a b a a b

a a bb

− − + = − −

= − +

=

B1

(aii) ( ) ( )2 2

2

Hence, 31.01 28.01 34.01 31.01 31.01 3 31.01 3

39

− × = − − × +

==

B1

Award this mark only if result from (ai) is used


3. (i) Find the prime factorisation of 4950 and 5880. (ii) Hence find

(a) the highest common factor of 4950 and 5880, (b) the smallest positive integer value of n for which 4950 5880 n× × is a perfect

square. Answer:

3(i) 2 2

3 2

4950 2 3 5 115880 2 3 7 5

= × × ×

= × × ×

B1 B1

cao cao

(iia) ( )HCF 4950,5880 2 3 530

= × ×

=

B1

cao

(iib) Smallest positive integer value of n for which 4950 5880 n× × is a perfect square is 3 5 11 165× × =

B1 cao

4. Given that 3x4 < 2 + < 29 , find (i) the least prime number of x, (ii) the greatest natural number of 2x. Answer:

4(i) 32 1

12 the least prime number of is 2.

xx

x

x

4 < 2 +>

>

∴

B1

(ii) 32

the greatest natural number of 2 is 25.

xx

x

2 + < 29< 26∴

B1


5 Solve the simultaneous equations:

20 18 7312 10 45

x yx y

− =− =

Answer:

5 20 18 7312 10 45

15, 12

x yx y

x y

− =− =

= =

M1 A1 + A1

By any valid method ft using c’s answer for 1st

variable answer

6 Simplify 1 2 15

318

.512

32 125x

x

−⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

.

Answer:

6 161 2 31 355 65

318

6 6

6

.512 5

32 2 8125

564 84164

x x xx

x x

x

− ⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎢ ⎥+ = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= +

=

M1 A1

Award the mark for either correct simplified expression of part of the sum cao


7 A polygon has n sides. Two of its exterior angles are 88° and 103°. The remaining ( )2n − exterior angles are each 13° , Calculate the value of n.

Answer:

7 ( )( )

88 103 2 13 360

13 2 360 191169213

13 215

n

n

n

n

° + ° + − × ° = °

− = −

− =

= +=

M1 A1

8 Sum of interior angles in a regular polygon is 2520°. Find the order of rotational symmetry

of the polygon. Answer:

8 ( )2 180 25202520 2180

16Hence, the order of rotational symmetry is 16.

n

n

− × ° = °

= +

=

M1 A1


9 A map of the downtown district of Singapore is drawn to a scale of 1 : 2000.

The diagram below shows a section of the Marina’s Floating Platform.

(i) Measure the radius of the audience seating area on the map with a ruler and write down the actual radius in metres.

(ii) The organizer wanted to light up a section of the audience seating area denoted by arc BC and the perimeter of the whole floating stage EFGH in the upcoming 44th National Day Parade.

By using a protractor and a ruler, measure and calculate the actual length, in metres, of wire required.

(iii) If the holding area for all the participants of the 44th National Day Parade contingents (not shown in this section of map) covers an area of 1.5 square kilometers. Find, in square centimeters, the area representing the holding area on the same map.

Answer:

(i) Actual map measurement of radius = 4.9 cm Actual radius = 4.9 2000 9 800× = cm = 98 m

B1 B1

s.o.i. and 0.1± cm cao and 2± m

Audience seating area

radius

Floating stage

E

G

B C

F

H


(ii) Arc length BC of the audience area (sector) = 2π360

rθ× ×

°

= 75 2π 98360

× ×

= 540 π 1286

≈ m

(to 3 s.f.) Actual length of the stage = 3.9 × 2000 cm = 7800 cm = 78 m Actual width of the stage = 5.9 × 2000 cm = 11 800 cm = 118 m

Required length = ( )540 π 2 78 1186

+ +

= 520 m (to 3 s.f.)

M1 B1 B1

Accept if using equivalent angle in radian to calculate ( )1± ° s.o.i., and 0.1± cm or 2± m, both must be correct at the same time to get this mark. cao, accept correct answers derived from those within acceptable range

(iii) Ratio of length = 1 : 2000 Therefore, Ratio of areas = 1: 4 000 000

So the area on the map = 1.54 000 000

km2

= 1.5 (100 000 100 000)4 000 000

× ×

= 3 750 cm2

B1 M1 A1

s.o.i. show of correct conversion from km2 to cm2

cao


10 Two containers bought from “EKIA” shown in the diagram are geometrically similar. The

cost of painting the base area of the smaller container is 1649

of the cost of painting the base

area of the larger container. (a) The top of the larger container has a circumference of 24.5 cm, calculate the

circumference of the top of the smaller container. (b) The containers are completely filled with liquid. Given that the smaller container

holds 1.28 litres of liquid, calculate the number of litres of liquid the larger container holds.

Answer:

(a) 21649

16 49

47 the circumference of the top of the smaller container is

4 24.5 14 cm7

s

l

s

l

ll

ll

⎛ ⎞=⎜ ⎟

⎝ ⎠

=

=

∴

× =

M1 A1

Showing ratio of areas = square of ratio of lengths cao

(b) 347

1.28 64 343 the number of litres of liquid the larger container holds is

343 1.28 6.8664

s

l

l

VV

V

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

∴

× =

M1 A1

Showing ratio of vols = cube of ratio of lengths cao


11 For the 44th National Day Parade (NDP) held at the Marina Floating Platform (MFP), it is estimated that a total of 50 000 people gathered at the MFP, Marina and Collyer Quay area to view and bask in the atmosphere of the Parade and its celebrations.

(a) Given that the holding capacity of the MFP for audience is 42.7 10× pax, calculate the percentage of people gathered at the Marina and Collyer Quay area who celebrated the National Day outside the MFP.

(b) If in addition to the 50 000 present at the MFP, Marina and Collyer Quay area, an estimated TV and internet viewership of 3.2 million was also recorded all around the world. Moreover, 89% of these two groups of people took the National pledge at 20 22 at the same time, calculate the number of people who watched the parade and took the pledge that night.

Answer:

(a) Number of people celebrated outside the platform = 450 000 2.7 10− × = 23 000 Percentage of people celebrated outside the platform

= 23 000 100%50 000

× = 46%

M1 A1

Indication of subtraction cao

(b) Total who watched the NDP = 50 000 3 200 000+ = 3 250 000 Number of people who took the pledge that night

= 89 3 250 000100

× = 2 892 500

B1 B1

cao o.e. in standard form etc cao o.e. in standard form etc


12 Yvonne started her journey from home for the ‘Danzage’ performance at 18 24. Due to a massive traffic jam, she did not manage to reach Dunman High’s Performing Arts Centre (PAC) until 20 02.

(i) Calculate, expressing your answer in exact form and in hours, the time taken by her to reach her destination.

(ii) Find the distance, in kilometers, between her home and the PAC if her taxi travelled at an average speed of 19.92 km/h.

Answer:

(i) Time taken by her = 20 02 – 18 24 = 1 hours and 38 mins

= 19130

hours

B1

cao, no mark if not in fraction in lowest term

(ii) Distance = 1919.92 130

×

= 32.536 km

M1 A1

ft with c’s result from (i) ft with c’s result from (i) ; if 32.536 is rounded off, this mark is lost.


13 In a bid to encourage Singaporeans to go Green for the environment, Singapore Power Grid Pte Ltd conducted a survey to find out the extent of Singaporeans’ readiness in this drive.

Out of a total of 200 Singaporeans surveyed, 130 claimed that they used energy-efficient electrical appliances while 80 claimed that they had replaced using air-conditioners with electric fans. The survey also revealed that 30 neither used energy-efficient electrical appliances nor replaced air-conditioners with electric fans.

If ξ, E and A represent the sets of Singaporeans surveyed, and those who used energy-efficient electrical appliances and those who replaced using air-conditioners with electric fans respectively,

(i) draw a Venn diagram to illustrate the above information.

(ii) Find the number of Singaporeans who used energy-efficient electrical appliances and had replaced using air-conditioners with electric fans.

(iii) Find the number of Singaporeans who had started using energy-efficient electrical appliances but still had not replaced air-conditioners with electric fans.

Answer:

(i)

B2

Deduct 1 mark for any missing relevant information

(ii) 130 80 30 200x+ − + = 40x =

M1 A1

ft c’s Venn diagram cao

(iii) Number of Singaporeans who had started using energy-efficient electrical appliances but still had not replaced air-conditioners with electric fans = 130 – x = 90

B1

cao

ξ (200) E (130) A (80)

x 30

130 - x 80 - x


14 Given that A = 2 1 34 5 15 8 0

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

and B = ( )2 2 4− .

Evaluate (i) BA (ii) AB.

Answer:

(i)

BA = ( )2 1 3

2 2 4 4 5 15 8 0

−⎛ ⎞⎜ ⎟− × −⎜ ⎟⎜ ⎟⎝ ⎠

= ( )16 20 8

B1

cao

(ii)

AB = ( )2 1 34 5 1 2 2 45 8 0

−⎛ ⎞⎜ ⎟− × −⎜ ⎟⎜ ⎟⎝ ⎠

= not possible

B1

Or equivalent


15 The diagram shows a line l, which makes an angle of θ with the x-axis shown in the diagram.

(i) Write down the value of tanθ .

(ii) On the grid, draw the line n, which is perpendicular to l and passing through the point ( )4 , 2− .

(iii) Write down the equation of the line n.

Answer:

(i) tanθ = 12

B1 cao

(ii)

B1 cao

4

2

-2

-4

-5 5

n

l

θ

4

2

-2

-4

-5 5

θ

l


(iii) Equation of n: 2 6y x= − −

B1 cao

16 In the figure below, ABC is a triangle with °= 90ˆCAB . AB = 12 cm and 1715ˆcos =CBA .

D is a point on AB such that BD = 9 cm.

(i) Find the length of AC.

(ii) Find the value of CDB ˆcos .

Answer:

(i) 1715ˆcos =CBA

12 15

17BC=

13.6 cmBC∴ =

By Pythagoras’ Theorem, in ABCΔ ,

2 2 213.6 12AC = −

2 40.96AC =

6.4 cmAC∴ =

M1 A1

correct use of trigonometric ratio

(ii) AD = 12 – 9

= 3 cm

By Pythagoras’ Theorem, in ADCΔ ,

2 2 23 6.4CD = +

2 49.96CD =

C

A

B

D 12 cm

9 cm


49.96 cmCD∴ =

CDACDB ˆcosˆcos −=∴

3

49.960.424

= −

= −

(to 3 s.f.)

M1 A1

Correct use of Pythagoras’ Thm o.e.


17 In each of the axes provided below, the point with coordinates (1, 1) has been marked.

(a) (i) Sketch the graph of 223 xxy −= , indicating clearly the coordinates of the x-

intercept(s) and y-intercept, if any.

(ii) State the coordinates of the turning point of the curve.

(b) Sketch the graph of 1 2yx

= − + , indicating clearly the coordinates of the x-

intercept(s) and y-intercept, if any.

Answer (ai)

[2]

y

x

● (1, 1)

O


Answer (b)

[2]

Answer:

(ai)

B1 B1

Correct Downward Parabola Passes through O,

(1, 1) and (23 , 0)

(aii) Coordinates of turning point = (89,

43 ) B1 o.e. but must be

exact

y

x

● (1, 1)

O

23

y

x

● (1, 1)

O


(b)

B1 B1

Correct shape and orientation Passing through O, (1, 1) and indication of asymptote

18 In the diagram, a, b and c are three vectors as shown.

(a) Express the vector c in terms of a and/ or b.

(b) Given that XY = 2a – b,

(i) draw and label the vector XY on the grid provided below,

(ii) find XY .

(c) Z is a point such that 21

YZ−⎛ ⎞

= ⎜ ⎟−⎝ ⎠.

(i) Mark the point Z clearly with a cross on the grid below.

(ii) Write down one fact about X, Y and Z.

(d) Points P and Q are as shown on the grid below. If QP = b – d, express d as a column vector.

y

x

● (1, 1)

O

1 2yx

= − +2

0.5


Answer:

(a) c = a + 12

b B1 cao

(bi) See answer on grid B1 cao (bii) XY = 2 26 3 45 6.71+ = = units (to 3 s.f.) B1 ft c’s answer for

(bi) (ci) See answer on grid B1 ft c’s answer for

(bi) – i.e. use c’s point Y to draw the vector correctly.

(cii) X, Z and Y are collinear points or XZ is parallel to ZY or equivalent or mentioning about the correct ratio of lengths of line segments formed by these 3 points

B1 cao

c

P

Q b

a


(d) QP = b – d

d = 4 10 6

1 7 6− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

B1

cao

c

P

XY

2a

– b

X

Y

Z

Q

(bi)

(ci)


19 The shoe sizes of the 17 pupils from 4A were recorded and represented in the dot diagram shown below.

(i) Write down the modal shoe size.

(ii) Find the median shoe size.

(iii) Find the interquartile range for the above data.

Answer:

(i) Modal shoe size = 8 B1 cao (ii) Median = 7 B1 cao (iii) Q1 = 6

Q3 = 8 9 8.52+

=

Interquartile range = 8.5 – 6 = 2.5

B1 M1

cao – for Q3 ft using c’s answers

4 5 6 8 7 9 10 Shoe size

Dot diagram showing the distribution of shoe sizes of 17 4A pupils


20 With the arrival of the haze, the mean PSI index of Singapore over the past 20 days were recorded as shown below:

41 66 65 85

89 53 71 76

56 79 47 48

50 69 45 51

68 63 70 55

Represent the data in a stem-and-leaf diagram on the space provided below.

Answer:

(a) Stem-and-Leaf diagram showing the distribution of PSI for 20 days

S L 4 1 5 7 8 5 0 1 3 5 6 6 3 5 6 8 9 7 0 1 6 9 8 5 9

Key: 4 | 2 means 42.

B3

Deduct 1 mark for any mistake


21 In the Swim-suit segment of the 2009 Miss World Pageant, the mean scores of the 11 semi-finalists were as shown below in ascending order.

7.7 7.7 x 8.0 8.0 y 8.6 8.7 9.0 9.4 z

The same data is represented in the box-and-whisker diagram shown below:

(i) State the values of x, y and z.

(ii) The mean scores of these 11 semi-finalists in the Interview segment were represented in another box-and-whisker diagram shown below:

Mean score of Swim- Suit Segment

Mean score of Interview Segment


Which segment of the 2009 Miss World Pageant was more competitive amongst the 11 semi-finalists? Explain your answer.

Answer:

(i) x = 7.9

y = 8.4

z = 9.5

B2

Deduct 1 mark for any mistake

(ii) Since the range of scores of the middle 50% of the semi-finalists is much smaller in the Interview Segment as indicated by its smaller inter-quartile range, it shows that the Interview Segment was more competitive.

B2 1 mark – key word smaller interquartile range 1 mark – key word Interview segment more competitive


22 Hafiz left home at 17 00 and took a bus to the National Stadium to witness the opening ceremony of the Asian Youth Games. He reached the National Stadium at 18 30. The speed-time graph below shows the bus journey that Hafiz took.

(a) Calculate the retardation of the bus journey.

(b) Find the average speed for the whole journey.

Answer:

(a) Retardation = gradient

= 600.5

= 120 km/h2

B1

Speed (km/h)

Time

16 30 17 00 17 30 18 00 18 30 19 00

0

20

30

40

50

60

70

10


(b) Total distance travelled

= ( )1 15 1 30 1 3 140 40 60 602 60 2 60 2 4 4

⎛ ⎞× × + × + × + × + ×⎜ ⎟⎝ ⎠

= 60 km

∴Average speed for whole journey

609060 ÷=

= 40 km/h

M1 A1

concept of finding total distance


22 (c) The distance-time graph below shows the journey of Hafiz’s friend, Yiming, who

also travelled along the same path as Hafiz, but in a different vehicle.

(i) Draw the distance-time graph for Hafiz’s journey in the same graph from 16 30

to 19 00.

(ii) Estimate, to the nearest minute, the time at which they first meet.

Distance from Hafiz’s home

(km)

Time 16 30 17 00 17 30 18 00 18 30 19 00

0

20

30

40

50

60

70

10


Answer:

(ci)

From 16 30 to 17 45

From 17 45 to 19 00

B1 B1

ft fr (i)’s answer

Distance from Hafiz’s home

(km)

Time

16 30 17 00 17 30 18 00 18 30 19 00

0

20

30

40

50

60

70

10


(cii) From the distance-time graph,

they first meet at around 17 55.

B1

accept between 17 51 to 17 58

Dunman High School 2009 Maths D Paper 1 Answer Key

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