hai sing c 2012 prelim 4e5n p2 marking scheme.pdf0

8/14/2019 Hai Sing C 2012 Prelim 4E5N P2 Marking Scheme.pdf0

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2

HSCS/4016/2

Mathematical Formulae

Compound interest

Total amount =

nr

P

1001

Mensuration

Curved surface area of a cone = rl

Surface area of a sphere = 24 r

Volume of a cone = hr 2

3

1

Volume of a sphere = 3

34 r

Area of triangle ABC = C absin2

1

Arc length = , r where is in radians

Sector area = ,2

1 2 r where is in radians

Trigonometry

C

c

B

b

A

a

sinsinsin

Abccba cos2222

Statistics

Mean =

f

fx

Standard deviation =

22

f

fx

f

fx



3

HSCS/4016/2

1 (a) Solve the equation

3

54

2

17 mm

[2]

(b) Given that a = b

b

3

85 , express b in terms of a [2]

(c) Express as a single fraction in its simplest form [2]

p p

p

3

4

92

2

(d) Simplify

y x

x43

2

[2]

_______________________________________________________________________________________

2 In the figure, the 2 circles XABY and XCDY intersect at X and Y . CYB and DYA are straight lines. XA and

YB are parallel.

It is given that 30YAX , 100 ABY and 120YDC . Stating your reasons clearly, find

(a) XYA [2]

(b) BXC [2]

(c) BCX [1]

(d) DCY [2]

(i)

D

X

BY

C

A

E



4

HSCS/4016/2

3 (a) In 2011, Ms Lim received a salary of $6 500 every month. She donated 1% of her income to charity

monthly. She was entitled for an annual relief amount of $10 500 for looking after her father and

her handicapped sister.

[Taxable income refers to the annual gross income less all donations and relief for the same year]

(i) Calculate Ms Lim’s taxable income for 2011. [2]

(ii) Taxable income is taxed according to the rates shown in the table below:

Calculate the amount of income tax that Ms Lim had to pay for 2011. [2]

(b) Cheryl invested $50 000 with a finance company which earned an annual interest of 4%. The

compound interest is compounded twice a year for 3 years. Calculate the amount of money,

correct to the nearest cent, in her account after 3 years. [2]

(c) The cost of making a particular car is divided between materials, wage and other miscellaneous

costs in the ratio 5:4:3. The material cost used for a car is $15 000.

(i) What is the total cost of making a car? [2](ii) Due to a new minimum wage regulation, wage was raised by 10%.

Express, correct to one decimal place, the new wage as a percentage of the new total cost. [3]

_________________________________________________________________________________________

Taxable income Rate (%) Gross Tax Payable ($)



5

HSCS/4016/2

4 As part of his triathlon training, Jeffrey cycles from A to C in a rural part of Australia. He travels for 50

km at a constant speed of x km/h, until he reaches a point B, where his bicycle chain breaks.

(a) Write down the time, in hours, taken for his cycled journey from A to B. [1]

In order to find a bicycle repair shop, he then walks the remaining 8 km from B to C at a constant speed

of ( x – 20) km/h.

(b) Write down the time, in hours, taken for his walk from B to C . [1]

(c) Given that the total time f or Jeffrey’s whole journey from A to C is 5 hours, write down an equation

in x and show that it reduces to

010001585 2 x x [3]

(d) Solve the equation 010001585 2 x x , giving your answer correct to 2 decimal places. [3]

(e) Find the time, in hours, minutes and seconds, Jeffrey would have taken if he had completed the

entire journey by bicycle at his original constant speed. [2]

_______________________________________________________________________________________



6

HSCS/4016/2

N

R

Q

H

63 km69 km

P

65 108 47 km

5 (a) A

The diagram above represents a map showing a harbour H and three oil production platforms, P , Q and R, along the Gulf of Mexico, where R is due east of H . HPQ is a straight line which lies on a

bearing of 065 from H and angle HPR = 108. It is given that R is 47 km from P , HP = 69 km and RQ = 63 km.

(a) A ship carrying supplies for the crew members is travelling from H to R. Calculate

(i) the distance between H and R, [2]

(ii) the shortest distance between this supply ship and the oil platform P . [3]

(b) Calculate the bearing of R from Q. [3]

(c) A compliant tower, 610 m high, stands at oil platform P . Find the greatest angle of elevation ofthe top of the tower when viewed from any point along HR. [2]

(d) A helicopter, C , is hovering at a point vertically above P . If the angle of depression of

R from the helicopter is 33, find the height of the helicopter. [2]

_______________________________________________________________________________________



7

HSCS/4016/2

6 In the diagram, the base diameter and the height of solid cone PXQ are 40 cm and 32 cm respectively.

The base diameter of another solid cone AXB is 16 cm.

The smaller cone AXB is cut off from the bigger cone PXQ.

A cylindrical hole ABCD is then drilled into the frustum as shown.

The curved surface area of the cylindrical hole is given to be 256 cm2

.

(All the dimensions shown are in cm)

(a) Find

(i) the height of the cone AXB, [1]

(ii) the height of the cylinder, BC , [2]

(iii) the slant length PA. [3]

(iv) the total surface area of the cone AXB (correct to 3 significant figures) [2]

(b) If the cost of paint needed to paint the curved surface area of cone PXQ is $40.

Calculate the cost of paint needed to paint the curved surface area of cone AXB. [2]

(c) Calculate the volume of the remaining solid in terms of π . [3]

_______________________________________________________________________________________

X

P Q

D

B

C

A

40

32

16



8

HSCS/4016/2

7 (a) The nth term of a sequence is given by Tn = 2n(n + 2).

(i) Use the formula to find T50. [1]

(ii) Which term of the sequence has the value 15 840? [1]

(b) Consider the pattern shown below:

Term Sum Number of odd integers

1 1 + 3 4 2

2 1 + 3 + 5 9 3

3 1 + 3 + 5 + 7 16 4

4 1 + 3 + 5 + 7 + 9 25 5

: : : :

n 1 + 3 + 5 + 7 + 9 + … + k S

(i) Is it possible to have a sum of 400? Explain your answer. [1]

(ii) Express k in terms of n. [1]

(iii) Find the formula relating S and n. [1]

(iv) How many odd integers are there from 1 to 99? [1]

8



9

HSCS/4016/2

In the diagram, OP = p and OQ= q . The points K and M lie on PQ and OQ respectively such that

3

1

KQ

PK and

5

2

MQ

OM . The lines OK and PM intersect at the point X .

(a) Express PM in terms of p and q . [1]

(b) Show that4

1OK (3p + q ). [2]

(c) Given that PM h PX , show that hOX 1 p + h

7

2q . [2]

(d) Given also that OK OX 13

8 , find the value of h. [2]

(e) Find the numerical value of

(i)OPK

OPX

of Area

Δof Area [1]

(ii)OPM

OPK

of Area

Δof Area [2]

_______________________________________________________________________________________

O

M

Q

P

K

X

q

p



10

HSCS/4016/2

9 (a)(i) The cumulative frequency curve below shows the time spent on homework and revision each

day by a group of 140 Sec 4 Express students in Hai Sing Catholic School.

Use the curve to estimate

(a) the median number of minutes spent on revision and homework, [1]

(b) the interquartile range, [2]

(c) the probability that when two students are selected at random from the group, onewould have spent at most 50 minutes while the other would have spent at least 84minutes on revision each day. [2]

0

20

40

60

80

100

120

140

160

0 20 40 60 80 100 120 140 160

Time (minutes)

C u m u l a t i v e F

r e q u e n c y



11

HSCS/4016/2

Time (minutes)

800 20 40 60 140120100

(ii) The box and whisker diagram shows the time spent on homework and revision each day by agroup of 140 Sec 5N(A) students from Hai Sing Catholic School.

(a) Use the diagram to find the median and interquartile range of the distribution. [2]

(b) Teachers feel that students from Sec 5 N(A) are more hardworking.

Do you agree? Give two reasons to justify your answer. [2]

(b) The Parkway Health Medical Group measured and recorded the masses of 60 newborns. The table below shows the distribution of their masses.

Mass (kg) 2.0 x < 2.5 2.5 x < 3.0 3.0 x < 3.5 3.5 x < 4.0

Number of

newborns6 14 30 10

(i) Find

(a) the mean mass, [1]

(b) the standard deviation. [1]

(ii) The Raffles Medical Group also measured and recorded the masses of its newborns. Themean mass and the standard deviation obtained were 3.31 kg and 1.521 kg respectively.Compare and comment on the masses of babies born in the two hospitals. [2]

_______________________________________________________________________________________



12

HSCS/4016/2

10 Answer the whole of this question on a sheet of graph paper.

The following is a table of values for the graph of ).3)(1(2 x x x y

x 0 0.2 0.5 0.8 1 1.5 2 2.2 2.5 2.8 3

y 0 0.9 1.25 0.7 a -2.3 -4 -4.2 b -2.0 0

(a) Calculate the values of a and b . [2]

(b) Using scale of 4 cm to 1 unit on both the x-axis and the y-axis,

draw the graph of )3)(1(2 x x x y for .30 x [3]

(c) By drawing a tangent, find the gradient of the curve at the point where

x = 1. [2]

(d) Use your graph to find the two values of x which satisfy the equation.1)3)(1(2 x x x [2]

(e) By drawing a suitable straight line on the same axes, use your graph to

find the three solutions of .12)3)(1(2 x x x x [4]

___________________________________________________________________________________

End of Paper



13

HSCS/4016/2

HSCS Prel im Maths P2 2012 Answer Scheme

1a) 3

54

2

17 mm

1

1111

108321)54(2)17(3

m

m

mmmm

b)

3ab + 8b = 5

b(3a + 8) = 5 [1]

b =

83

5

a

c) p p

p

3

4

92

2

)3)(3(

124

)3)(3(

)3(4

3

4

)3)(3(

3

4

)3)(3(

2

2

2

2

p p

p p

p p

p p

p p p

p

p p p

p

)3)(3(

)2)(6(

p p

p p

[1]

[1]

[1]

[1]

[1]



14

HSCS/4016/2

d) y x

x43

2

=

xy

x y x

43

2

= xy

x y

x

432

= x y

xy

x 43

2

= x y

y

43

2

2(a)

cyc.quad)of angles(opp60

120180

CXY

[1]

segment)samein(angles70

30100segment)samein(angles30

ABX XYA XAY XBY

[1]

(b)

triangle)of sum(angle50

303070180

YXB

[1]

110

5060 BXC

[1]

(c)triangle)of sum(angle40

30110180

BCX [1]

(d)

30120180

angles)(vert.opp.30

DCY

AYBCYD

[1]

triangle)of sum(angle30 [1]

3(a)(i) Taxable income = (12 x $6 500) – 0.01 x ($78 000) - $10 500 [1] = $66 720 [1]

(ii) Amount of income tax payable = $550 + [0.07 x ($66 720 - $40 000)] [1] = $2 420.40(nearest cent) 1]

(b) A = P(1 +100

r )n

= 50 000(1 +100

2)6 [1]

= 56 308.12 [1]

[1]

[1]

Alternative solution:

(b) triangle)of sum(angle50

303070180

YXB

130

5080 BXC

(c) triangle)of sum(angle20

30130180

BCX

(d)

50

angle(vert.opp.30

DCY

AYBCYD

triangle)of sum(angle



15

HSCS/4016/2

(c)(i) 5 units - $15 000

1 unit - $3 000

12 units – 12 x $ 3000 [1] = $36 000 -> Total cost of making a car = $36 000 [1]

(ii) Required percentage = %5.3548.351003720013200100

)300041.0(36000300041.1

4(a)

x

50 h A1

(b)

20

8

x h A1

(c)5

20

850

x x

M1

20582050 x x x x M1

(shown)010001585

10058100050

2

2

x x

x x x x

A1

(d)

10

10005424964158 x M1

75.8or85.22

10

4964158

A1, A1

(e) Time taken

846.2258

M1

20smin32h2

h539.2

A1

[1]

[1]

[1]



16

HSCS/4016/2

5(a)(i) Distance HR

108cos476924769 22 M1

km7.94

7327.94

A1

Alternative Solution 1:

km106

77.105

25sin

47

108sin

HR

HR

Alternative Solution 2:

km7.89

728.89

47sin

69

108sin

HR

HR

M1

A1

(ii) Are of triangle PHR

2km138.1542

108sin47692

1

M1

Let the shortest distance be d

558.32

138.15427327.942

1

d

d

M1

km6.32 A1

Alternative solution 1:

km2.29

161.29

6925sin

:2SolutioneAlternativ

km4.34

374.34

4743cos

d

d

d

d



17

HSCS/4016/2

(b)

63

72sin

47

sin

72

108180

PQR

QPR

196.45 PQR M1

804.19196.4565 M1

Bearing of R from Q

804.19180

8.199 A1

Also Accept:

205.5QfromR of bearing

5.39 PQR

(c)

558.32

61.0tan

M1

6(a)(i)5

2

40

16

32

h

h = 12.8 cm [1]

(ii) Curved surface area = 256π

25682 BC [1]

16CF cm [1]

(iii) 736.3714242032 22 PX [1]

40

16

1424

AX

094.15 AX [1]

cm PA 6.2264.22094.151424 (3 s. f.) [1]



18

HSCS/4016/2

(iv) Total surface area of cone AXB = π (8)(15.094) + π (8)2 [1]

= 580.42 (580 2cm 3 s.f.) [1]

(b)25

4

5

2

40

16 22

2

1

A

A

[1]

Cost = 40.6$40$25

4

[1]

7(a)(i) T50 = 2(50)(50 + 2) = 5 200 [1]

(ii) 2n(n + 2) = 15 840

2n2 + 4n – 15 840 = 0

n = 88 or -90(inadmissible)

n = 88 [1]

(b)(i) Yes, It is a perfect square. [1]

(ii) k = 2n + 1 [1]

(iii) S = (n + 1)2 [1]

(iv) 50 [1]

8(a)

q p

OM PO PM

7

2

[1]

(b)

q p

OQ PO

PQ PK

4

1

4

1

4

1

[1]



19

HSCS/4016/2

q p

q p

q p p

PK OP OK

34

1

4

1

4

3

4

1

[1]

(c)

q ph p

PX OP OX

q ph PX

7

2

7

2

[1]

hq ph

hqhp p

7

21

72

[1]

(d)

q p

q p

q pOX

13

2

13

6

313

2

34

1

13

8

[1]

13

7

13

61

7

21

13

2

13

6

h

h

hq phq p

[1]

(e)(i)

13

8

of Area

of Area13

8

OPK

OPX

OK OX

[1]



20

HSCS/4016/2

(ii)

104 : 91 : 56

-------------------

13 : 8

13: 7

:: OPM OPK OPX

8

7

104

91

of Area

of Area

OPM

OPK

[1]

[1]

9(a)(i)(a) 50% students70140

Median = 68 min

(b) Q1 = 56 min

Q2 = 81 min

Interquartile range = 81 – 56

= 25 min

(c) No. of students that spend at most 50 min = 22

No. of students that spend at least 84 min = 140 – 110 = 30

Probability =139

22

140

30

139

30

140

22

973

66

(ii)(a) Median = 74 min

Interquartile range = 98 – 28

= 70 min

(b) Yes, pupils from Sec 5N(A) are more hardworking as(a) they have a higher median, i.e. more than half of the students spent more timerevising their work,(b) the upper quartile is higher, i.e. the top 25% of pupils are more hardworking.No, pupils from Sec 4 Express are generally more hardworking as(a) their lower quartile is higher,(b) the minimum (20 min) and maximum (140 min) time spent by Sec 4 Express

students are higher.



21

HSCS/4016/2

(b)(i)(a) Mean

kg12.3

60

187

60

1075.33025.31475.2625.2

(b) Standard deviation

kg427.0

60

12.375.31012.325.33012.375.21412.325.262222

10 (a) y = 2(1)(1 - 1)(1 - 3) = 0 a = 0 [1]

y = 2(2.5)(2.5 – 1)(2.5 – 3) = -3.75 a = -3.75 [1]



22

HSCS/4016/2

(b)

(b) Axes marked and named – 1 mark

Smooth curve plotted –

1 markEqn of curve written – 1 mark



23

(c) gradient = -

= 3.466 3.47 [2]

(d) x = 1.2 [1] or 2.9 [1] (0.1)

(e) Plot y = -2x + 1 [1]

x = 0.15 or 1.35 or 2.45 [3]

hai sing c 2012 prelim 4e5n p2 marking scheme.pdf0

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