ADMIRALTY SECONDARY SCHOOL

SUBJECT : Elementary Mathematics

PAPER : 1

CODE : 4016

LEVEL/STREAM : Sec 4 Express / 5 Normal Academic

DATE : 25 August 2011

TIME : 1000 - 1200

DURATION : 2 hours

Instructions to candidates:

1. Write your name, class and index number. 2. Answer ALL questions.

3. Use an electronic calculator to evaluate explicit numerical expressions.

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 , use either your

calculator value or 3.142, unless the question requires the answer in terms of .

4. Essential workings must be shown. Loss of essential workings and

illegible handwriting will lead to loss of marks.

DO NOT TURN OVER THIS PAGE UNTIL YOU ARE TOLD TO DO SO! ______________________________________________________________

This question paper consists of 16 printed pages including this cover page.

PRELIMINARY EXAMINATION 2011

NAME: NO: CLASS:

80

2

Mathematical Formulae

Compound interest

Total Amount = niP )

1001(

Mensuration

Curved Surface area of a cone = rl

Curved surface area of a sphere = 24 r

Volume of a cone = hr 2

3

1

Volume of a sphere = 3

3

4r

Area of triangle ABC = Cab sin2

1

Arc length = r , where is in radians

Sector Area = 2

2

1r , where is in radians

Trigonometry

C

c

B

b

A

a

sinsinsin

bccba 2222 Acos

Statistics

Mean =

f

fx

Standard deviation = 2

2

)(

f

fx

f

fx

3

Answer all the questions 1

(a) Express

25

105 as a percentage.

(b) Express %4

120 as a fraction in its simplest form.

Answer: (a) …………………% [1]

(b) ……………….... [1]

2 Calculate

)04.2467.3(2

759.4

.

Give your answer correct to 2 decimal places.

Answer: ……………………….. [2]

3 Factorise yyx 182 2 completely.

Answer: ………………………... [2]

4

4 Given that 32162 k , find the value of k.

Answer: ……………………… [2]

5 The temperature at the top of a mountain is -20 0C.

It increases by 12 0C at mid-way down the mountain. The temperature between the top and the bottom of the mountain is 520C.

(a) What is the temperature at the bottom of the mountain ? (b) What is the mean temperature at these 3 points of the mountain ?

Answer: (a) …………………0C [1]

(b) …………………0C [2]

6 A tin of biscuits is to be shared amongst 3 boys in the ratio of 3 : 4 : 5.

If, this tin of biscuits is to be shared equally amongst them, the boy that

has the biggest share would have 10 biscuits less.

How many biscuits are there in the tin ?

Answer: ……..…………biscuits [2]

5

7 (a) Factorise 374 2 xx .

(b) Hence, solve 0374 2 xx

Answer: (a) …………………….. [1]

(b)….……or .……….. [1]

8 In the diagram on the right, the points B, C, D, E lie on a circle with centre O. BD is the diameter of the circle, AT is tangent to the circle at B, AED is a straight line

and . Find

(a) (b) (c) (d)

Answer: (a) …………………….. [1]

(b) …………………….. [1]

(c) …………………….. [1]

(d) …………………….. [1]

6

9 (a) Solve the inequality 14523 x .

(b) State the largest integer value of x .

Answer: (a) …………………….. [2]

(b) ………….………….. [1]

10 Written as the product of its prime factors 75328400 24

(a) Express 140 as the product of its prime factors. (b) Hence, write down the smallest integer k such that 140k is a

multiple of 8400.

Answer: (a) …………………….. [1]

(b) …………………….. [1]

7

11 If 3 balls are chosen, one after another, from a bag which consists of 5

yellow balls, 6 green balls and 9 white balls, what is the probability that

(a) all three balls chosen are of the same colours, (b) exactly 2 white balls are chosen ?

Answers: (a) …………………… [1]

(b) …………………… [1]

12 Let 61: xx and given that x is an integer

50:

32:

xxB

xxA

find

(a) ,

(b) BA (c)

Answer: (a) …………………… [1]

(b) …………………… [1]

(c) …………………… [1]

8

13 Given that y is inversely proportional to 2)3( x and that y = 5 when

2x .

(a)

Find the value of y when 2

1x

(b) Find the values of x when 45y

Answers: (a) …………………… [2]

(b) …………………… [2]

14 The speed of light is given as sm /1000.3 8 .

(a) Express this speed in kilometres per second.

Give your answer in standard form.

(b) How long does light take to travel half a metre ? Give your answer in nanoseconds.

Answer: (a) ……………………….. [1]

(b) ………………nanosec. [2]

9

15 Given that

01

25X and

14

32Y ,

(a) Evaluate YX 2

(b)

Find the values of p and q if

q

pXY

2

17

Answers: (a)

[1]

(b) …………………………. [2]

16 (a) Simplify

z

x

yz

x

27

12

9

4 3

(b)

Simplify 2)23(

15

)23(

2

x

x

x

Answers: (a) …………………… [2]

(b) …………………… [2]

10

17 The diagram below shows the speed-time graph of an object.

(a) Find the acceleration of the object when t = 12. (b) Find the total distance covered in 16 seconds. (c) Hence, or otherwise, calculate the average speed of the object.

Answers: (a)………………….…ms-2 [1]

(b) ……………….. ..…m [2]

(c) ……………...…… ms-1 [1]

11

18 The box-and-whisker diagram below illustrates the heights, in centimeters, of the students in Class 5N1.

(a) Find the interquartile range. (b) Find the median height. (c) In Class 5N2, the interquartile range and median height is

given as 13.5 and 162 respectively. Compare, briefly, the heights of the students from the two classes in two different ways.

Answers: (a) …………………………………………………cm. [1]

(b) …………………………………………………cm. [1]

(c) ……………………………………………………...

……………………………………………………...

[2]

19 Mr Ang cycles to school at an average speed of 12 km/h every morning from his home. Calculate

(a) the time taken, in minutes, for him to travel to school if his home is 2000 m away from school,

(b) his cycling speed in km/min,

(c) the time that he must leave his home if he is to be in school at 7.15 am.

Answers: (a) ………………...min. [1]

(b) ..……………km/min [1]

(c) …………………… [1]

12

20 (a)

(b)

(i) Sketch the graph of )3)(1( xxy

(ii) Write down the equation of the line of symmetry.

Sketch the graph of 2)1(5 xy

[2]

[2]

Answers: (a) (ii)…………………… [1]

13

21 The diagram below shows a triangle ABC with points A(1,4), B(8,4) and C(10,9).

(a) Find the gradient of AC.

(b) Given that a line parallel to AC passes through a point (0,6), find

the equation of this line.

(c) Calculate the area of triangle ABC.

Answers: (a) ……………………... [2]

(b) ……………………... [2]

(c) ……………….. unit2 [2]

14

22 Rods of equal length are used to make a series of pattern. The first three patterns are shown below.

Pattern No. 1st 2nd 3rd

Pattern formed

No. of rods used 4 10 16

(a) Write down the number of rods required to form the 5th

pattern.

(b) How many rods are needed to form the 10th pattern?

(c) Write down, in terms of n, the number of rods needed to form

the nth pattern.

Answers: (a) …………….….rods [1]

(b) …………….… rods [1]

(c) ……………..…rods [2]

15

23 In the diagram, given that 090 QPSRSP . The point R is the

reflection of T in the line PS and that QT intersects PS at the point U.

(a) State the two triangles that are congruent, and support your

answer with the proof.

(b) Given also that, cmQU 6 and cmRU 3 , calculate the

numerical values of the ratios of :

(i) Area of TSU : Area of QPU

(ii) Area of TSU : Area of trapezium PQRS

Answers: (a) …………and……........ [4]

(b)(i) ……………….……... [1]

(b)(ii) …..…………….…… [2]

16

END OF PAPER !

24 The scale drawing in the answer space below shows the positions

of the 3 estates A, B and C. A is due North of B and C is due east

of B.

(a) Find the bearing of C from A.

(b) Construct the perpendicular bisector of BC. [1]

(c) Construct the angle bisector of angle ABC. [1]

(d) These two bisectors meet at point P, measure the length of

AP.

Answers: (a) ………………….. 0. [1]

(d) ………………….cm [1]

A

B C

17

Solutions to Prelim 2011 Paper 1 (ADSS)

1a

1b

420%

400

81

A1

A1

9a

9b

x

x

x

4

28

523

5.4

92

1452

x

x

x

5.44 x

largest integer is 4

M1

A1

A1

2 0.432086

= 0.43 (2 dp)

M1

A1

10a

10b

7522

605322 k

A1

A1

3

)3)(3(2

)9(2 2

xxy

xy

M1

A1

11a

10

1

6840

684)

18

7

19

8

20

9(

)18

4

19

5

20

6()

18

3

19

4

20

5(

A1

4

7,34

22

222

3)4(

34

kk

k

k

M1

A1

11b

95

33

6840

23763)

18

11

19

8

20

9(

A1

5a

5b

52 - 20 = 32

3

11

3

32820

A1

M1

A1

12a

12b

12c

5,4,3,2,1

2,1

B

A

2)(

5,4,3,2,1

6,5,4,3'

BAn

BA

A

A1

A1

A1

6 1 share – 10 biscuits

12 shares – 120 biscuits

M1

A1

13a

5

4

)32/1(

5

)3(

2

2

y

ky

k

x

ky

3

22,

3

13

3

1)3(,

3

1)3(

9

1)3(

)3(45

2

2

xx

xx

x

x

k

M1

A1

M1

A1

7a

7b

(4x-3)(x-1)

1,4

3 xx

A1

A1

13b

8a

8b

8c

8d

36DBE (same segment)

72236 EOD

(angle at center = 2 angle at circ)

549036180 BDE

(angle in semi-circle)

365490180 BAD

(rad. perp. tangent)

A1

A1

A1

A1

18

14a

14b

skm

s

m

s

m

/103

1

10103

1

103

5

188

km8103 -- 1 sec

km5.0 ---

sec3

5

sec10103

5.0 9

8

nano

nano

A1

M1

A1

18a

18b

18c

166 -154 =12 cm

162 cm

The median height of students in

Class 5N1 is smaller than that in

5N2 and the IQR of Class 5N1 is

smaller than that in 5N2.

A1

A1

A2

15a

15b

12

112

14

32

01

252

3,18

32

1718

14

32

01

25

qp

A1

M1

A1

19a

19b

19c

min10min606

1

12

2

min/5

1

min60

12km

km

7.05 am

A1

A1

A1

16a

16b

y

x

x

z

yz

x 23

12

27

9

4

2

2

2

)23(

5

)23(

1546

)23(

15

)23(

2

x

x

x

xx

x

x

x

M1

A1

M1

A1

20a

x = 1

A2

A1

17a

17b

17c

Acceleration, when t =10,

2

8

9

816

312

ms

Average Speed

A1

M1

A1

A1

20b

A2

19

21a

21b

21c

25.17572

1

69

5

6

)0(9

56

9

5

110

49

unitArea

xyEquation

c

c

cmxy

M1

A1

M1

A1

M1

A1

23a

TSU and RSU

RHScongruent

commonsideUSSU

givenRSUTSU

reflectionRUTU

)(

)(90

)(

0

Accept other correct proof

A1

M2

A1

22a

22b

22c

16+6+6=32 rods

4+6(10 – 1)= 58 rods

4+6(n-1) = 6n-2

A1

A1

M1

A1

23b (i) TSU similar to QPU

4

1)

6

3( 2

(ii) Area of TUR = 2(Area of

)UTS

Area of TQR = 3(Area of

)TUR

= 6(Area of

)UTS

Area of PUQ = 4(Area of )UTS

Required ratio = 10

1

64

1

A1

M1

A1

24a 1100 A1 24d 4.6 cm (accept from 4.2 to 5.0 cm) A1

24b

A2 24c

A

B C

P

1

Marking Scheme for Add Math P1 – Preliminary Examination

1.

1......................60151

.........)441(9030151

1...............))21((270))21((90)21(151)631(

)21(

))21(31()631(

1....................45

90)15(3

1...............)......27090151)(31()31)(31(

1................27090151

....)3(3

5)3(

2

5)3(

1

51)31(

2

222

3252

552

2

325

32

325

App

ppppp

Mpppppppp

ppxLet

pppp

A

xofCoeff

Mxxxxxx

Axxx

xxxx

2.

1.....................22

1

8

2

4

1

1.....................7

11

22

53

4

1

1....................7

11

32

52

1......................22

53

4

11

Ayx

y

x

My

x

My

x

MA

3a.

1................62

1..................0)6)(2(

0124

1.................0)3)(1(4)(

04

1...............0)3(

2

2

2

2

Am

Mmm

mm

Mmm

acb

Mmmxx

2

3b.

solutionnohasy

My

Mx

2

1................8

11

1..............0)2

5( 2

4.

1................4.198,4.18

1..................3

1tan

cossin3

1............0cossin3

0det

Ax

Mx

xx

Mxx

oo

5.

1.....................5.309,5.103

92.372,08.17643.63

1..............20

1)43.63sin(

1)43.63sin(20

1..................43.63

2

4tan

1.............20

A

M

M

MR

oo

o

6a.

1...).........(1

1................26.1

43

4

1............0)1)(4(

043

3

1.......043.33

349

2

2

1

ANAu

or

Ax

u

Muu

uu

ulet

M

x

x

xx

xx

3

b.

1................1

98

1.............38

1..............28

log

2log2

1)8(log

1.........log2

1

2

3

33

3

Ax

xx

Mx

x

Mx

x

xx

Ax

c.

1..........36

11

1..........4

3210

1...........33

1..........

)3(

3)3(

27

39

4

3

210

4

1

3

152

4

15

Ax

Mxx

M

M

xx

xx

xx

7. cosine graph……….B1

1.....3

4B

xy

4

i.

1.........2sec

1..........1cos2

2

1cos

AtionInter

Mx

x

ii.

1...........1sec

1.........3

4cos2

2cos3

AtionInter

Mx

x

xx

8a.

ptstationaryNo

Mdx

dy

Me

Me

e

e

dx

dy

x

x

x

x

1.....0

1............0

1......1

1

2

2

2

2

b.

1...........2

1

2

1

1...........2

1

2

11

1,0

1...........2

1...........1

2

2

2

Aexy

Mc

c

yxwhen

Mce

xy

Mdxey

x

x

x

5

9a.

1............2

1

1.........)2

1(

)21(

1(

1..........)2(

)2(

1

1

0

1

1

0 2

A

M

Mx

dxx

b.

1............707.0

1...........2

)42

sin(

2

)4

2sin(

1............2

)4

2sin(

)4

2cos(

4

4

A

M

M

x

dxx

10.

One interior angle of polygon = 1206

180)26(

……….M1

One angle in triangle = 60o. Thus, it is equilateral triangle.

Area = )60sin2

1(6 2x ……….M1

= 1.......32

3 2 Mx

6

1...........6.15

1..).........5.0(33

5.0

1.........33

A

Mx

dt

dx

dx

dA

dt

dA

dt

dx

Mxdx

dA

11

Correct intersection with x axis…….M1

Correct intersection with y axis………M1

Sketch of modulus……A1

a. when x=-3,

1..............05

53)1(2

,1

33)3(2

Ay

y

xwhen

y

b.

when y=-2,

1........2

5

322

1..........2

1

322

322

Ax

x

or

Ax

x

x

12.

1...........25.02

1...........0)14)(2(

1..........0274

2)74(

2

Axx

Mxx

Mxx

xx

M1

y

x

(0,-3)

(-1.5,0)

(-3,-3)

(1,-5)

7

13i.

1.......331

17MmBC

ii.

1...........3

21

3

1

)2(3

11

1.......3

1

Axy

xy

MmAL

iii.

x=1………..A1

iv.

1...).........2,1(

23

21

3

1

AH

y

v.

?

1

2

2

1

ABCH

ABCH

AB

CH

mism

mm

m

m

M1

vi.

1.......5.2,5.2

.tan

3

21

3

1

1.........103

)1(37

Myx

EqneousSolveSimul

xALofEqn

Mxy

xy

BCofEqn

1.........75.3

1.............72

11

5.27

5.21

2

1

A

MArea

A1

ADMIRALTY SECONDARY SCHOOL

SUBJECT : Elementary Mathematics

PAPER : 2

CODE : 4016

LEVEL/STREAM : Sec 4 Express / 5 Normal Academic

DATE : 13 Sept 2011

TIME : 0750 - 1020

DURATION : 2 h 30 min

Instructions to candidates:

1. Write your name, class and index number. 2. Answer ALL questions. Tie your writing papers separately from the question

paper using the thread provided.

3. Use an electronic calculator to evaluate explicit numerical expressions. 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 , use either your

calculator value or 3.142, unless the question requires the answer in terms of

.

4. Essential workings must be shown. Loss of essential workings and illegible handwriting will lead to loss of marks.

DO NOT TURN OVER THIS PAGE UNTIL YOU ARE TOLD TO DO SO! ______________________________________________________________

This question paper consists of 9 printed pages including this cover page.

PRELIMINARY EXAMINATION 2011

NAME: NO: CLASS:

100

2

Mathematical Formulae

Compound interest

Total Amount = niP )

1001(

Mensuration

Curved Surface area of a cone = rl

Curved surface area of a sphere = 24 r

Volume of a cone = hr 2

3

1

Volume of a sphere = 3

3

4r

Area of triangle ABC = Cab sin2

1

Arc length = r , where is in radians

Sector Area = 2

2

1r , where is in radians

Trigonometry

C

c

B

b

A

a

sinsinsin

bccba 2222 Acos

Statistics

Mean =

f

fx

Standard deviation = 2

2

)(

f

fx

f

fx

3

Answer all the questions.

1.

The diagram above shows a plot of land, ABCD. BD and BM are two

paths that cut across the land. It is also given that AB = 15 m,

BC = 18 m, AD = 20 m, BD = 25 m and 0105CBD .

(a) Calculate

(i) length of CD, [2]

(ii) length of BM, given that it is the shortest distance of B from

CD.

[3]

(b) (i) Show that BAD is a right angle. [2]

(ii) Hence, or otherwise, find the area of the plot of land. [2]

2 (a)

Solve the equation x

x 14

5

2

[3]

(b) Express p as a subject, given that

32

47

q

pqp

[3]

(c) Given that

9

5

2

3

ba

ba, find the value of

b

a

[2]

(d) Solve 0

3

2

9

12

xx

.

[3]

M

4

3. The cash price of a sports car is $420,000.

(a) Mr Ben plans to buy it on hire purchase terms. He pays a deposit of

70% of the cash price follows by 24 monthly installments at an interest

rate of 7% per annum. Calculate his monthly installment.

[3]

(b) Mr John plans to buy it on simple interest loan terms. He pays a down

payment of $250 000 and the balance at the end of 2 years with a

simple interest rate of 10% per year. Calculate how much more Mr

John has to pay for the car in simple interest loan terms compared to

cash terms.

[3]

(c) Mr Lionel buys it on compound interest loan terms. He pays a down

payment $300 000 and the balance at the end of 5 years with a

compound interest rate of 5% per annum. Calculate the amount that

Mr Lionel has to pay at the end of 5 years.

[3]

4 (a) Given that

5

15,

6

4XYOX , find

(i) XY [2]

(ii) OY [2]

(b) Given that

kXZ

8 and OXhXZ , find the value of h and k.

[3]

(c) The points X, Y and Z are vertices of a triangle, find the equation of

the line XZ.

[3]

5

5. (a) Mrs Lee cycled 80 km from X to Y at an average speed of x km/h.

Write down an expression, in terms of x, for the number of hours taken

to travel from X to Y.

[1]

(b) She returned from Y to X by the same route. Her average speed for

the return journey was 5 km/h more than on the outward journey.

Write down an expression, in terms of x, for the number of hours taken

to travel from Y to X.

(b)

[1]

(c) She took 2 hours more on the outward journey than on the return

journey. Write down an equation in x, and show that it reduces to

020052 xx .

[3]

(d) Solve the equation 020052 xx

[3]

(e) Find the total time she was cycling, giving your answer in minutes and

seconds.

[3]

6.

In the diagram, PMNQ is a trapezium in which PM is parallel to QN, PM = 8

cm and QN = 5 cm. Arc MX is drawn using P as centre and arc NX is drawn

using Q as centre.

(a) Show that XQN is approximately 1.8 radian. [2]

(b) Calculate

(i) the length of arc XN, [2]

(ii) the area of the shaded region. [4]

6

7. The diagram shows two triangular fields, ABC and ACD. DAB is a straight line,

AC = 66 m, 060CAB and 075ABC .

(a) Calculate the length of BC. [3]

(b) Given that AD = 88 m, calculate the length of CD. [3]

(c) A vertical tree is growing at C. The angle of elevation of the top of the

tree from D is 150. Calculate the height of the tree.

[2]

(d) To provide more shades at the field, a tree is to be planted every 10 m

along CD. Find the minimum number of trees needed to be planted.

[Assuming the first tree to be planted starts from point D.]

[2]

7

8. A metallic container as shown in the diagram below is used to collect rain

water samples by a group of students. The container is made by extracting a

metallic circular cone of base diameter 5 cm from a solid hemisphere of

diameter 15 cm.

(a) Given that the volume of the extracted cone is

16

1 of the volume of the

hemisphere, find the height of the cone.

[3]

(b) The metallic container needs to be coated with a layer of zinc to prevent

it from rusting. Find the total surface area of the metallic container to be

coated, giving your answer correct to 1 decimal place.

[3]

(c) The metallic cone extracted from the hemisphere above is melted and

mould into five identical metallic cubes. Calculate the length of one side

of each cube, giving your answer correct to 1 decimal place.

[4]

5 cm

15 cm

8

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

The following is the table of values for the graph of 3234 xxy .

x -2 -1 0 1 2 3 4

y 24 8 4 p 8 4 -12

(a)

Calculate the value of p.

[1]

(b) Using a scale of 2 cm to represent 1 unit on the horizontal axis and 2 cm

to represent 5 units on the vertical axis, draw the graph of

3234 xxy for 42 x .

[3]

(c) Use your graph to solve the equation 037 32 xx . [2]

(d) The equation kxx 3234 has 3 solutions for 42 x . State the

range of values of k.

[2]

(e) Find the x-coordinate of a point on the curve, where x > 0, such that the

gradient is -12.

[2]

(f) By drawing a suitable line graph on the same graph, solve the equation

0623 23 xxx

[3]

9

10. The waiting time, in minutes, for 50 patients at two dental clinics are given as follow.

Hospital A

Time (minutes)

Number of patients

6 12 21 8 3

Hospital B

Mean = 29.0

Standard Deviation = 3.5

(a) For Hospital A, calculate

(i) the mean waiting time,

(ii) the standard deviation.

[1]

[2]

(b) Compare, briefly, the waiting time for the two dental clinics. [2]

(c) Find the probability that one of the patient chosen has to wait for more than

28 minutes to see the doctor.

[2]

(d) Two patients are chosen at random from the group of 50.

Calculate the probability that both waited less than or equal to 24 minutes.

[2]

END OF PAPER !

10

Solutions to ADSS_EMath_Prelim_Paper 2

1a (i)

)3(4.34379.34

937141.1181

105cos)25)(18(22518

2

222

sfmCD

CD

CD

M1

A1

(ii)

Area of BCD

33.217

105sin25182

1

)3(6.12643.12

33.2172

1

sfmMB

BMCD

M1

M1

A1

1b (i)

2

2

22

22

25

2015

BD

ADAB

, by Pythagoras theorem, angle BAD is right angle.

M1

A1

(ii)

233.367

33.217)20152

1(

m

M1

A1

2a

43.9,43.7

2

)70(422

0702

702

14

5

2

2

2

2

xx

x

xx

xx

x

x

M1

A2

b

)12(

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)23(7)12(

14212

4211432

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2

2

q

qqp

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M1

M1

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c

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ab

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ba

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072

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1

x

x

xx

x

xxx

xxx

M1

M1

A1

3a Balance to pay

126000

420000100

30

Interest for 2 years =0.07x126000x2 = $17640

Monthly instalment = 24

17640126000

= $5985

M1

M1 A1

3b Balance to pay = 420000 -250000 = 170000 Total to pay at end of year 2

204000

170000)2170000100

10(

Difference = (250000+204000) – 420000 = $34000

M1

M1

A1

3c Balance to pay = 420000 -300000 =120000 Total to pay at end of 5 years

79.153153$

)100

51(120000 5

M1

M1

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4a (i)

8.15515 22 XY M1 A1

(ii)

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sec604.0min689

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80

86.11

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s

s

hours

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M1

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M1

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M1 A1

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9a p = 6 B1

9b Correct scale used All points plotted Smooth curve drawn

M1 M1 M1

9c y = -3, from graph, x=3.68 M1 A1

9d 4<k<8 A2

9e Draw line with gradient -12 X = 3.3 (accept 3.1 to 3.5)

M1 A1

9f

1,.

102

0102)43(

0623

23

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xgraphfrom

xy

xxx

xxx

M2

A1

10a (i)

2.2950

1460

A1

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50

43496 2 M1 A1

10b Hospital B has a shorter average waiting time and is more consistent in the waiting time as SD(B) < SD(A)

A2

10c

25

16

50

3821

M1 A1

10d

245

3

490

6

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5

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6

M1 A1

1

Marking Scheme for Add Math P2 – Preliminary Exam

1i.

1........2

112

1.............)2

1(5)

2

1(4)

2

1(4)

2

1(

544)(

23

23

Ac

cc

Mcf

cxxxxf

ii.

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1..........

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03544

12544

2

23

23

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Axxx

Mxxxxf

Mdivisionlong

Mf

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2.

15,2

5

15,2

5

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kwhen

kwhen

Mk

k

M

b.

3

1

3

7

A1

M1

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1..........03

1

3

15

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1

2510104)52)(52(

1.........3

15

10)(25252

2 Axx

M

M

3. Long division – M1

1............)1(

1

)3(

214

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21374

1............1,2

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1.............)1()3()1)(3(

5

1..........)1)(3(

514

)1)(3(

21374

23

23

Axx

xxx

xxx

MBA

xBxAx

Mx

B

x

A

xx

x

Mxx

xx

xx

xxx

4i. Initial mass of radioactive element when t=0………B1

ii.

1.............

,0

1...........

ANdt

dN

twhen

MeNdt

dN

o

t

o

3

iii

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4

1...............2

1

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int4

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1........2

)2

1(3

3

3

3

3

3

A

e

N

A

NN

dt

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1)......3,2(

)2

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(

A

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A

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r

i.

when y=0,

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1.........01242

A

PQLength

Mx

Mxx

4

ii. Using similar triangle, Alt Method Using Midpoint,

1).........7,1(

8

1.........10

54

6

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53

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y

My

x

Mx

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7

2

13

1

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52

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y

y

x

x

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6i.

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2

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1.........)2cos1(

)2sin2(2sin)2cos2)(2cos1(

2cos1

2sin

2

2

22

2

Ak

Mx

Mx

x

x

xxx

Mx

xxxx

dx

dy

x

xy

ii.

1..............6

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1........0cos1

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6

1

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2sin

6

1

1...........)2cos1(

2

6

1

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1

4

0

4

0

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0

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M

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cb

ma

baxxy

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Mxx

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1

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cossin

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22

2

b.

1............24

11,

8

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17,

6

13,

6

5,

6122

2

1)

122sin(

Ax

Mx

x

9.

1.............)2(2

2

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1(

1...........)2(

)2

1)(2(

2

2

2

2

2

1

Axx

x

dx

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Mx

xxx

dx

dy

Mx

xxx

dx

dy

x

xy

6

when x=4,

1..........2983

1..).........4(3

81

,1,4

1........3

8

1...........8

3

Axy

Mxy

yxwhen

Mm

Mdx

dy

normal

10i.

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1..........306

72303

2

2

Asmatwhen

Mtdt

dv

ttv

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1.........0)4)(6(

1.........072303

0

2

Mst

Mtt

Mtt

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iii.

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6

4

23

6

4

2

M

Mttt

Mdttts

Dist moved = 4m………A1

7

v.

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1...........112

7215

72303

4

0

23

4

0

2

M

ttt

dttts

Between t=4-6, s=-4

Between t=6-7,

1...........4

7215

72303

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6

23

7

6

2

M

ttt

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Total distance = 112+4+4

=120m………..A1

11a.

In ,ABC

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1

1......

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1.).........(//

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4

32

2

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2

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xx

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Mx

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Mxxxyx

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ii.

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50,14

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MNAx

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Mxx

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1,2

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20

1..........)1(3

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3

3

2

2

2

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yx

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xx

Mxx

xy

xy

yx

M1

9

ii.

when y=0, x=1……..M1

1.........6

51

2

1

3

12

1...............2

1

112

1

1........3

12

3

1

1.........12

3

2

23

3

2

2

A

AreaShaded

M

ofArea

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xxx

MdxxxcurveunderArea