Answer all the questions. - · PDF file · 2010-09-022010-09-02 · Answer all the questions. 1 ... Assuming no mass loss, find the height of the new cylinder formed, ... 12 seamstresses

[Turn over 3

Answer all the questions.

1 Simplify

)3(

6

3

3

−+

− aaa.

Answer …………….…………………...... [2]

2 (a) Solve the inequality −10 + 2x < 11 – x. (b) Given that x is a real number, state the smallest value of x2.

Answer (a) …….....………………...... [1]

(b) 2x = ….………………...... [1]

[Turn over 4

3 Simplify

)(

)()(52

032323

rs

sr

rs

rs×

−

,

leaving your answer in positive index notation.

Answer …………….…………………...... [2]

4 Factorise completely 22 9324 bbcaca −+− .

Answer …………..……………………...... [2]

[Turn over 5

5 Solve the following pair of simultaneous equations:

235 −= yx

xy 15127 =+

Answer x = ……….…………………......

y = ……………………………...

[2]

6 A shopkeeper sells 80 tins of biscuits for $425. If the shopkeeper bought 100 tins for $430, calculate and express the profit gained from the sale of one tin of biscuits, as a percentage of its cost price.

Answer …………….…………………. % [3]

[Turn over 6

7 The total population of the world was recently estimated to be 6.7 billion. (a) Express 6.7 billion as a number in standard form. (b) The total land surface area of the world is approximately 8106.1 × km 2 . Calculate the average number of people per square kilometre of the land

surface area.

Answer (a) ………………………...... [1]

(b) …….…………… per km2 [2]

8 An interior angle of a regular polygon is five times its exterior angle. Find the number of sides, n, of the polygon.

Answer n = …....….…………………...... [3]

[Turn over 7

9 (a) Express 60 and 825 as products of their prime factors, leaving your answers in

index notation.

(b) Find the smallest positive integer n such that 60n is a multiple of 825.

Answer (a) 60 = ..………………......

825 = ..…………………..

[1]

[1]

(b) n = ..…………………...... [1]

10 Sketch the graph of the function )2)(1( −−−= xxy .

Indicate and mark clearly on the curve, its coordinates of the x – intercept(s), y – intercept(s) and turning point of the curve. [3]

Answer

y

x O

[Turn over 8

11 The diagram below shows the design of an eco-garden.

HOG is a quadrant and the scale of the map is represented as 1 : 5000. All measurements given in the diagram are in centimetres.

Calculate the actual area of the eco-garden in square metres. (Take π = 3.142)

Answer ……………………………......m2 [3]

[Turn over 9

12 On the grid in the answer space lie the positions of the points A, B and C where →

AB = 2a and =→

BC b.

(a) Given that

−=

→

3

1BC , express

→

CD as a column vector.

(b) Find the magnitude of →

AB.

(c) Mark clearly on the grid above, the point E such that =

→

DE a + b. [1]

Answer (a) ………………………...... [1]

(b) ………………..…… units [2]

2a

b

[Turn over 10

13 The diagram shows two similar solid cylinders with their dimensions.

These cylinders are made with the same material.

(a) Find the ratio of the volumes of cylinder J to cylinder K.

(b) Both cylinders are melted and recast into one large cylinder, similar in shape to

the above. Assuming no mass loss, find the height of the new cylinder formed,

giving your answer correct to 3 significant figures.

Answer (a) …....…...………………...…… [1]

(b) ……..……………..…..…... cm [3]

12 cm 21.6 cm

[Turn over 11

14 (a) On average, 12 seamstresses sew 30 dresses in 6 days. Assuming that all the

seamstresses work at the same rate, calculate the number of days needed for 18

seamstresses to sew 15 dresses.

(b) Given that

4

3

34

2=

−

−

yx

yx,

find the value of y

x

5

2.

Answer (a) ………………..…… days [2]

(b) ………………………...... [2]

[Turn over 12

15 A boy runs 5 km in 25 minutes. He then walks a further 10 km at an average speed of 6 km/h. Calculate

(a) his running speed, giving your answer in km/h, (b) the time in minutes when he is walking, (c) his average speed, in km/h, for the whole journey.

Answer (a) …………………...…...…km/h [1]

(b) …………………....… minutes [1]

(c) ……………….……….....km/h [2]

[Turn over 13

16 The points J(-4, 5), K(2, -3) and L(5, -3) are shown in the diagram below.

Find

(a) the length of JK, (b) the gradient of the line JK,

(c) the equation of the line passing through (1, 3) which has the same gradient as the line JK,

(d) the value of cos ∠JKL.

Answer (a) …………...………… units [1]

(b) ………………………..….. [1]

(c) …………….……………... [1]

(d) …………………………… [1]

x

K (2, -3)

O

L (5, -3)

J (-4, 5)

y

[Turn over 14

17 Let ε = {x : x is an integer between 10 and 20 inclusive}.

A = {x : x is a factor of 45} and B = {x : x is an odd number}.

(a) List the elements in the set 'BA ∩ .

(b) State n ( ) .'

BA ∪

(c) Represent the above given sets on a clearly labelled Venn diagram, showing

clearly the elements contained in each set. [2]

Answer (a) ………………….………... [1]

(b) …………………………….. [1]

[Turn over 15

18 In the diagram, POQ is a sector of a circle with centre O and radius 10 cm.

Given that QR is perpendicular to OP and PR = 4 cm, find

(a) ∠POQ in radians,

(b) the area of the shaded region.

Answer (a) ………………..…….…...radians [2]

(b) ………………............…….. cm2 [2]

[Turn over 16

19 The first four terms of a number sequence are 5, 12, 19 and 26.

(a) Find the values of p and q if the nth term of the number sequence can be

expressed as pn + q.

(b) Hence, find the 100th term of the number sequence.

(c) From the original number sequence above, deduce the nth term of a sequence

that has 8, 15, 22 and 29 as its first four terms.

Answer (a) p = …..………………......

q = ………………………

[1]

[1]

(b) …………………………... [1]

(c) ………….……………...... [1]

[Turn over 17

20 The stem-and-leaf diagram shows the marks obtained during a 50-marks class test.

Stem Leaf

0 6 8 1 1 4 6 8 9

2 0 1 3 5 3 1 2 3 6 6

4 2 3 4 6

Find (a) the number of pupils who took the test (b) the median mark obtained. (c) the mean mark attained. (d) the standard deviation.

Answer (a) …...……………..…….……... [1]

(b) ………………….....………..... [1]

(c) ……………………………….. [2]

(d) ……………………………….. [2]

Key: 2|3

means 32 marks

[Turn over 18

21 The diagram shows the speed-time graph of a lorry over a 10-second interval.

(a) Write down the initial speed of the lorry. (b) How long was the lorry travelling at constant speed? (c) Calculate the acceleration of the lorry during the first 4 seconds. (d) Find total distance covered by the lorry during 10-second interval. (e) Sketch the corresponding distance-time graph.

Answer (e) [2]

Answer (a) …………...………..…. m/s [1]

(b) ………...…...……....……s [1]

(c) ……..………….…… m/s 2 [2]

(d) ………………………… m [1]

distance (m)

O time (s)

|

2 |

4 |

6 |

8 |

10

[Turn over 19

22 Two corners, A and B, of a horizontal triangular field are 240 m apart. The diagram below is part of a scale drawing of the field and AB measures 12 cm.

(a) Find the scale of the drawing in the form 1 : n.

(b) Using appropriate measurement, find the bearing of A from B.

The third corner, C, of the field which lies below the line AB, is 220 m from A and 170 m from B.

(c) Using ruler and compass only, find and label the position of C, using an accurate scale drawing.

(d) A tree, T, in the field is equal in distances from the three corners A, B and C. (i) By constructing perpendicular bisectors, find and label the position of the

tree, T. (ii) Calculate the actual distance of the tree from the corner of the field A.

Answer (c) [2] (d)(i) [1]

Answer (a) .……….………………...... [1]

(b) ………..………………….. [1]

(d) (ii) …………………..… m [2]

End of Paper 1

North

B

A

Presbyterian High School 4E/5N Preliminary Examination Elementary Math P1 Answer Scheme

1

Qn. Solutions

1

)3(

6

3

3

−+

− aaa

= )3(

6

)3(

3

aaa −−

− (o.e.)

= )3(

63

aa

a

−

− (o.e)

= )3(

)2(3

aa

a

−

− or =

)3(

)2(3

−

−

aa

a (o.e)

2(a) −10 + 2x < 11 – x 3x < 21 x < 7

2(b) Smallest value of x2 = 0

3

)(

)(

)(

)(52

032323

rs

sr

rs

rs×

−

= 63

69

rs

rs−

= 12

6

r

s (showing ans. in positive index notation)

4 22 9324 bbcaca −+−

= bcacba 3294 22+−−

= )32()32)(32( bacbaba −−−+

= )32)(32( cbaba −+−

5 ,235 −= yx xy 15127 =+

5=x , 9=y

6 Total sale of 80 tins = $425

Sale of 1 tin = 3125.5$

80

425=

Cost of 1 tin = $4.30 Profit made from 1 tin = $5.3125 - $ 4.30 = $1.0125

Percentage profit = %5.23%10030.4

0125.1=×

7(a) 9107.6 × 9107.6 ×= (no rounding off to 3 s.f.)

7(b) Average = 875.41

106.1

107.68

9

=×

× per km 2 or 41.9 per km2

8 Let x be the size of an exterior angle. 1 interior angle = 5x

Alt. angles on a st.line = 6x = 180°

x = 30°

Sum of exterior angles = 360°


2

360=⇒ nx °

⇒ 30°n = 360°

n = 12 sides

9(a) 60 = 22 53××

825 = 3 5×2 11×

9(b) (trick is to pick the largest power from each prime factor)

LCM = 22 53××2 11×

1153260 22×××=∴ n

= 5560×

55=∴n (o.e. workings for n = 55 obtained)

10

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

x-intercepts at (1, 0) and (2, 0) y-intercept at (0, -2) Turning point at (1.5, 0.25)

11 Scale of map = 1 cm : 50 m

∴area on map : actual area

= 1 cm2 : 50 × 50 m2

= 1 cm2 : 2500 m2

Ratio = 1 : 2500

Area of eco-garden on the map ( ) ( ) ( )

××+×+×+×=

21142.34

1112453

= 24.7855 cm2 ---------------------------------------------------------------

∴actual area 25007855.24 ×=

= 61963.75 m2 62000≈ m2

12(a)

−

−=

2

4CD

12(b) Magnitude of

→

AB = 20 units or 4.47 units.


3

12(c)

13(a) Reduced to lowest term.

Ratio = 33 9:5 = 125 : 729

13(b) Let H be the height of the new formed cylinder. 3

12.

.

=

H

cylindersmallofvol

cylindernewofvol (o.e)

3

12125

125729

=

+ H

H = 22.8 cm (3 s.f.)

14(a) 18 seamstresses took 4 days to sew 30 dresses. 18 seamstresses took 2 days to sew 15 dresses.

14(b)

4

3

34

2=

−

−

yx

yx

2

1

5

2

15

4

45

91248

=

=

=

−=−

y

x

y

x

xy

yxyx

15(a) Running speed = 12

6025

5= km/h

15(b) Time in min = 60

6

10× = 100 min

15(c) Total ave. speed =

60100

6025

510

+

+

= 7.2 km/h

16(a) Length of JK = 22 )35()24( ++−−

= 10100 = units

E x

2a

b


4

16(b) Gradient =

3

4

24

)3(5−=

−−

−−

16(c) Grad. =

3

4−

Subst (1, 3) : c = 3

13

Eqn: 3

13

3

4+−= xy

Or 3y = - 4x +13

16(d) Length of JK = 10 (s.o.i. use of Pythagoras Theorem)

cos ∠JKL = 5

3

10

6−=−

17(a) 'BA ∩ = { } award mark if and only if ans. represented in curly bracket

17(b) n ( ) ( ) {6'''=∩=∪ BAnBA 10, 12, 14, 16, 18, 20}

{'=A 11, 12, 13, 14, 16, 17, 18, 19, 20}

{'=B 10, 12, 14, 16, 18, 20}

17(c) Refer to diagram attached

18(a) Length of OR = 6 cm

∠POQ = ∠ROQ =

−

10

6cos 1

= 0.927 rad (3s.f.)

18(b) Shaded area = ( ) 222

610)6(2

1)927295.0(10

2

1−−

= 22.36 cm2

4.22≈ cm2

Or

Shaded area

×××−

××= 927.0sin106

2

1927.010

2

1 2

= 22.4 cm2 (correct to 3 sig. fig.)

19(a) When 1=n , 5=+ qp ------- (1)

When 2=n , 122 =+ qp ------- (2) (o.e.)

(2) − (1): 7=p

Substituting 7=p into (1),

57 =+ q

2−=q

Or Recognising this number sequence as Arithmetic Progression (AP)

= dn )1(5 −+ (where d = 7)

Simplify: 5 + 7n – 7 = 7n – 2 ( 7=p , 2−=q )

19(b) nth term 27 −= n


5

100th term ( ) 21007 −= = 698

19(c) Adding 3 to each term in the number sequence 5, 12, 19, 26, we get a new sequence 8, 15, 22, 29

∴nth term for new sequence 327 +−= n 17 += n

20(a) 20

20(b) Median – 24

20(c) Mean =

20

464443423636333231232120191816141186 ++++++++++++++++++

= 26.2

20(d) std. deviation = 2.12155.122.26

20

16684 2≈=−

21(a) 5 m/s

21(b) 10 − 4 = 6 s The lorry was travelling at constant speed for 6 s.

21(c) Acceleration

4

515 −=

2

12= ms-2

21(d) Total distance covered ( ) ( )( )410

2

11015 −=

= 150 − 20

= 130 m

21(e) Refer to diagram attached

22(a) 12 cm rep 240 m 12 cm rep 24000 cm 1 cm rep 2000 cm ∴ scale = 1 : 2000

22(b) Measured angle North to the line AB anticlockwise = 72° The required angle = 360°- 72° = 288° Bearing = 288° Final ans: 288° ± 1° (accept also 290°)

22(c) Refer to diagram to only award 1 mark for visible arc seen

22(d) (i) Refer to diagram attached


6

(ii) AT = BT = CT = 6.2 cm ( ± 0.1 cm)

∴ Distance of tree from corner A = 20002.6 × = 12 400 cm = 124 m (ans. is the same from corners B and C) Final ans between 122 m and 126 m (inclusive both)

Qn 21(e)

________________________________________________________________________________

This question paper consists of 10 printed pages including this page.

PRESBYTERIAN HIGH SCHOOL

MATHEMATICS 4016/02

2 September 2010 Thursday 2 hours 30 minutes

PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL





SECONDARY 4 EXPRESS / FIVE NORMAL PRELIMINARY EXAMINATION TIME 2 hours 30 minutes

READ THESE INSTRUCTIONS FIRST

Write your name, index number and class on all the work you hand in. Write in dark blue or black ink in the foolscap paper provided. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Omission of essential working will result in loss of marks. You are expected to 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 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.

Setter: Mr Tan Chee Wee

FOR EXAMINER’S USE

100

[Turn over

Sec Four Express/ Five Normal Preliminary Examination 2010 Mathematics Paper 2

Page 2 of 10

Mathematical Formulae

Compound Interest

Total amount = n

rP

+100

1

Mensuration

Curved surface area of a cone = lrπ

Surface area of a sphere = 24 rπ

Volume of a cone = hr2

3

1 π

Volume of a sphere = 3

3

4rπ

Area of triangle ABC = Cab sin2

1

Arc length = θr , where θ is in radians

Sector area = θ2

2

1r , where θ is in radians

Trigonometry

C

c

B

b

A

a

sinsinsin==

Abccba cos2222 −+=

Statistics

Mean = ∑

∑

f

fx

Standard deviation =

22

−

∑

∑

∑

∑

f

fx

f

fx


Page 3 of 10

ANSWER ALL QUESTIONS.

1 (a) It is given that g

lT

π2= .

(i) Find T when l = 26 and g = 9.81. [1] (ii) Express l in terms of T and g. [2]

(b) (i) Factorise completely 249 q− . [1]

(ii) Simplify 249

214

q

q

−−

. [2]

(c) Solve the equation .9

7

52

6

xx=

+ [2]

2 In July 2008, John and Winnie decided to buy a private condominium worth $878 000. The

couple needed to pay a down payment of 20% of the list price, of which John came out with

70% and Winnie paid the remaining amount.

The outstanding amount was met using a bank loan at a simple interest of 2.5% per annum on a

20 year loan tenure.

(a) Calculate the amount Winnie paid for the down payment. [1]

(b) At the end of the first year, how much interest would the couple have paid to the bank?

[2]

(c) At the end of the second year, this property's worth has risen by 15%. Taking into

account of the couple's down payment, the bank loan and the interest paid, how much

profit did they make if they sell the property at the end of the second year? [2]

(d) John told Winnie that if they would decide to sell the property at the end of the second

year, he would be entitled to two-third of the profit.

Express their individual profit as a percentage of the amount they put in initially. [3]


Page 4 of 10

3 A shopkeeper bought some porcelain toys for $990. He paid $x for each toy. (a) Find, in terms of x, an expression for the number of toys he bought. [1] (b) During transportation, 8 toys were broken. He sold each of the remaining toys for $2

more than the amount he paid for it. Write down an expression (no need to simplify), in terms of x, for the sum of money he received. [1] (c) He made a profit of $388. (i) Write down an equation in terms of x and show that it reduces to

.04951012 2 =−+ xx [3]

(ii) Solve the equation .04951012 2 =−+ xx Explain why you would reject one of the two solutions. [3]

(d) Find the number of toys the shopkeeper sold. [1]

4 A rice distributor supplies rice to 3 different supermarkets in Singapore. The number of bags of rice supplied per delivery to each supermarket, the sizes and the number of deliveries are shown in the table below.

Number of bags of rice per

delivery

Number of

deliveries in one

month

Size of each bag 1 kg 5 kg 10 kg

Giant 35 45 15 4

NTUC 85 65 35 12

Sheng Siong 45 55 30 6

The costs of 1 kg, 5 kg and 10 kg of rice are $3, $10 and $18 respectively.

(a) If ( )154535=P evaluate 4P. Describe what is represented by the elements of

4P. [2]

(b) (i) Given that

=305545

356585

154535

Q , write down another matrix R, such that the

product RQ gives the total number of each size of bag the distributor need to supply to the three supermarkets in one month. Evaluate this product. [3]

(ii) Using the product from part (b)(i) and multiplying by another matrix, evaluate the total amount of money the distributor will receive from the three supermarkets in one month. [2]


Page 5 of 10

5

In the diagram, TA and TB are tangents to a circle centre O.

Given that °=∠ 62ATB , by stating the reasons clearly, calculate

(a) ,BAT∠

(b) ,AOB∠

(c) ,ACB∠

(d) ,ADB∠

(e) ,EBS∠

(f) the radius of the circle given that the length of BT is 20 cm. [8]

T

62°

S

U

C

A

D

O

B

E


Page 6 of 10

6 Diagram I Diagram II Diagram I represents a road bridge. The horizontal road, ABC, consists of two equal parts, AB and CB, each of length 48 metres,

which can be rotated round A and C respectively, to allow tall boats to pass through the bridge. Vertical towers, AD and CE, each of height 64 metres, stand at the ends of the bridge. Cables, DB and EB, join the tops of the towers to the ends of the road. Diagram II shows the road when it has been raised.

The road has been rotated through 52° and its ends are at M and N. Calculate

(a) the width of the gap, MN, between the ends of road sections, [3]

(b) the length of DM, [3]

(c) the reduction in the length of each cable. [2]

64 64 64 64

48 48 48 48 A B C A B C

D E D E

M N

52° 52°


Page 7 of 10

7 (a) The cumulative frequency table below shows the number of SMS sent out using

handphones by a group of 120 pupils in June.

(i) Use your graph to find

(a) the median, [1] (b) the 20th percentile, [1]

(c) the interquartile range. [2]

(ii) If 30% of the pupils are considered frequent users, use the graph to find the

minimum number of SMS these pupils make. [1]

Jemmy conducted a survey to find the number of SMS these 120 pupils made in July. He

represented the data using a box and whisker plot shown below.

(iii) Compare the number of SMS sent in June and July in two ways. [2]

(b) The weight, in kilogram of eleven adults are given below. 67, 53, 47, 75, 49, 58, 64, 72, 57, 60, 80

Find (i) the mean, (ii) the standard deviation. [3]

0 200 400 600 800

200 400

20

600 800

40

60

80

100

120

0 No. of SMS

Cumulative Frequency


Page 8 of 10

8 The given solid in Figure A is made up of two identical hemispheres and a right cylinder. The radii of the hemispheres are 12 cm each and the cylinder has a radius of 4 cm and a height of 20 cm.

(a) Find the volume of the solid in terms of π. [2] (b) Find the total surface area of the solid. [3]

(c) The solid was melted to form a frustum as shown in Figure B. (A frustum is the resultant figure when a right circular cone has been cut out from a larger right circular cone.) Given that the radii of the top and the base of the frustum are 0.12 m and 0.28 m respectively, calculate the height. [2]

9

Three points, A, B and C lie on a horizontal field. ∠ BAC = 58o, ∠ BCA = 43o, AB = 78 m and the bearing of C from A is 281o. (a) Find

(i) the bearing of B from A, [1] (ii) AC, [2] (iii) area of triangle ABC, [2] (iv) the bearing of C from B. [2]

(b) A boy is standing at B, flying a kite, K, which is vertically above A. The string BK,

attached to the kite makes an angle of 32o with the horizontal. Calculate the angle of elevation of the kite when view at C. [3]

Figure B

0.12 m

0.28 m

Figure A

N

C

A

B

43°

78

58°

12 cm

20 cm

4 cm


Page 9 of 10

10

(a) In the diagram, OA = 3a, OB = 4b, BC is parallel to OA and BC = 3

5OA. X is the point on OC

such that OX = 3

2XC. Y is the midpoint of BC.

(i) Express in terms of a and/or b,

(a) AB ,

(b) OC ,

(c) OX ,

(d) AX . [4]

(ii) Does B lies on AX produced? Explain your answer. [1]

(iii) Find the numerical value of

(a) OAC

OAX

of Area

∆ of Area

∆,

(b) of Area

∆ of Area

OBY

OAY

∆. [2]

(b) The letters in the word “HIPPOPOTAMI ” are printed on 11 cards.

(i) If one card is randomly chosen, find the probability that (a) the letter “O ” or “P ” is printed on it, [1] (b) a consonant is printed on it. [1]

(ii) Suppose 3 cards are randomly drawn one at a time without replacement. Find the

probability that (a) the card printed with the letter “P” will only be drawn in the 3rd draw,

[2] (b) at least one of the cards is printed with the letter “P”. [2]

C

X

O B

Y

A

3a

4b


Page 10 of 10

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

The variables x and y are connected by the equation ( )262

1xxy −= .

The corresponding pairs of values of x and y, are given in the table below.

x –3 –2 –1 0 1 2 3

y 4.5 –2 –2.5 0 2.5 2 p

(a) Find the value of p . [1]

(b) Using a scale of 2 cm to 1 unit, draw a horizontal x-axis for 33 ≤≤− x . Using a scale of 2 cm to 1 unit, draw a vertical y-axis for 55 ≤≤− 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 least value of x for which ( ) 164

1 2 =−xx . [2]

(d) By drawing a tangent, find the gradient of the curve at the point (–1, –2.5). [2]

(e) On the same axes, draw the graph of 425 −=+ yx for 33 ≤≤− x . [2]

(f) (i) Write down the x coordinates of the point where the two graphs intersect. [1]

(ii) This value of x is the solution of the equation 023 =+++ CBxAxx . Find the value

of A , the value of B and the value of C . [1]

END OF PAPER

MATHEMATICS PRELIMINARY EXAMINATION 2010 SEC 4 EXPRESS/5 NORMAL

SUGGESTED MARKING SCHEME FOR PAPER 2

1

1

g

lT

π2=

a 08.4

81.9

)26(2 == πT (3 s.f.)

B1

ii

g

lT

π2=

g

lT

π22 = M1

lgT π22 =

l

gT =π2

2

A1

bi )7)(7(49 2 qqq +−=− B1

ii 249

214

q

q

−−

)7)(7(

)7(2

qq

q

−+−=

M1

q+=

7

2

A1

c .

9

7

52

6

xx=

+

351454 += xx M1

3540 =x

8

7=x

A1

2a The amount Winnie paid 8780002.03.0 ××=

52680$= B1

b Interest 025.08780008.0 ××= M1

17560$= A1

c New price of the property 100970087800015.1 =×=

Profit 1756028780001009700 ×−−= M1

96580$= A1

d

Winnie percentage profit %10052680

965803

1

××

=

%1.61= B1

MATHEMATICS PRELIMINARY EXAMINATION 2010 SEC 4 EXPRESS/5 NORMAL


2

John’s percentage profit %1008780002.07.0

965803

2

×××

×=

M1

%4.52= A1

3a Number of toys

x

990=

B1

b Sum of money received )2(8

990 +

−= xx

or xx

81980

974 −+ B1

ci 388990)2(8

990 +=+

− xx

M1

1378168

1980990 =−−+ x

x

0

19804048 =−+

xx

M1

019804048 2 =−+ xx A1

ii 04951012 2 =−+ xx cao

0)55)(92( =+− xx M1

5.4=x or 55−=x (rejected) A1

55−=x is rejected because number of toys buy cannot be negative B1

d Number of toys sold = 8

5.4

990 −

212= B1

4a ( )15453544 =P

( )601801404 =P B1

The elements represent the number of bags of each size of bag delivered to Giant in

one month

A1

bi ( )6124=R B1

( )

=305545

356585

154535

6124RQ

( )306351215455665124544568512354 ×+×+××+×+××+×+×=RQ M1

( )66012901430=RQ A1

ii

( )

18

10

3

66012901430

M1

)29070(

The amount is $29070 A1

MATHEMATICS PRELIMINARY EXAMINATION 2010

SEC 4 EXPRESS/5 NORMAL


3

5a

2

62180 −=∠BAT (base angle of isosceles triangle)

°=∠ 59BAT B1

b °−°−°−°=∠ 629090360AOB (angle sum of a quad)

°=∠ 118AOB B1

c

2

118=∠ACB (angles at centre is twice angle at circumference)

°=∠ 59ACB or BATACB =∠ (alternate segment theorem) B1

d 59180 −=∠ADB (Angles in opposite segment)

°=∠ 121ADB B1

e °=∠ 90ABE (angle in a semi circle)

°=∠=∠ 59BATABT

°−°−°=∠ 5990180EBS (angle sum of a triangle)

°=∠ 31EBS B1

Give 1 mark for correct reasons stated.

f

2031tan

r= M1

31tan20=r

0.12=r cm A1

6a length of AM = 48cm

Let M’ be the foot of AB from M

48

'52cos

AM=°

°= 52cos48'AM M1

°×−+= 52cos4824848MN M1

9.36=MN m A1

b °=−=∠ 385290DAM

°−+= 38cos)48)(64(24864 22'2DM M1,

M1

5.39=DM m A1

Or

48

'52sin

MM=°

°= 52sin48'MM

°−= 52sin4864'DD M1

°== 52cos48'' AMMD

22'2 'MDDDDM +=

22'2 '55.2918.26 +=DM M1

5.39=DM m A1

c 222 6448 +=DB

80=DB M1

Reduction in length 48.3980 −=

52.40= A1

7ia Medium = 520 B1




4

b 370 B1

c 630-410 M1

= 220 A1

ii 610

iii The average number of SMS sent in June is more than that of July B1

The spread of SMS sent in June is smaller than that of July. B1

bi Mean weight

11

8060577264584975475367 ++++++++++=

Mean weight 62

11

682 ==

B1

ii Standard deviation

222222222222

6211

8060577264584975475367 −++++++++++=

2.10= kg

or

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

11

62806260625762726264

62586249627562476253626722222

222222

−+−+−+−+−+

−+−+−+−+−+−

=

M1 A1 or B2

8a volume ( )32 12

3

22)20()4( ππ ×+=

M1

32624 cmπ= A1

b Total surface area ( ) ( ) 222)4(21221222)20)(4(2 ππππ −+×+= M1,

M1

23120cm= A1

c Let the height of the big cone be h,

then the height of the small cone hh7

3

28.0

12.0 =×

ππ 2624

7

31228

3

1 22 =

×− hh M1

×−=

7

31228

3

1

2624

22π

πh

899.10=h

Therefore height of frustrum 23.6899.10

7

4 =×= cm or 0.0623m A1




5

9ai Reflex °=+=∠ 33958281NAB

Bearing of B from A °= 339 B1

ii °=−−=∠ 794358180ABC

°=

° 43sin

78

79sin

AC

M1

°°=

43sin

79sin78AC

3.112=AC A1

iii Area of °×××=∆ 58sin27.11278

2

1ABC

M1

3710= A1

iv °=−−=∠ 2158281360NAB

2382179180 =−+=∠NBC M1

Bearing of C from B °= 238 A1

b Let h be the height of the kite

7832tan

h=

74.4832tan78 ==h m M1

3.112

74.48tan =∠CAK

M1

°=∠ 5.23CAK A1

10

aia OAOBAB −=

ab 34 −=AB B1

b BCBOOC +=

ab 3

3

54 ×+=OC

ab 54 +=OC B1

c OCOX

5

2=

)54(

5

2ab +=OX

B1

d OAOXAX −=

aab 32

5

8 −+=AX




6

ab −=

5

8AX

B1

ii No. AX is not parallel to BA B1

iiia

5

2

5.0

5.0

OAC triangleof area

OAX triangleof area =××××=

hOC

hOX

B1

b

55.0

3

'5.0

'5.0

OBY triangleof area

OAY triangleof area

×=

××××=

hBY

hOA

2.1

OBY triangleof area

OAY triangleof area = (accept 5

6)

B1

bia P(“O ” or “P ”)

11

5= B1

b P(a consonant) = P(“H” , “M” , “P”or “T”)

11

6= B1

iia P(1st “P” in the 3rd draw) = P(first 2 cards not “P”and 3rd card is “P”)

9

3

10

7

11

8 ××= M1

165

28= A1

b P(at least one “P” in all 3 draws) −= 1 P(none is “P” in all 3 draws)

9

6

10

7

11

81 ××−=

M1

165

109= A1




7

11

(a) 5.4−=p B1

(b) All points plotted correctly P2

1-2 incorrect points P1

Smooth Curve C1

(c) ( ) 164

1 2 =−xx ⇒ ( ) 262

1 2 =−xx

Draw the line 2=y M1

74.2−=x accept -2.8 – -2.7 B1

(d) Drawing tangent M1

5.1 accept 1.35 – 1.65 A1

(e) Drawing correct straight line B2

(f) -0.36 accept -0.4 – -0.3 B1

xxx2

52)6(

2

1 2 −−=−

xxx 54)6( 2 −−=−

xxx 546 3 −−=−

4110 3 −−= xx

Therefore, B = 0, C = -11 , D = -4 B1

Documents

Answer all the questions. - · PDF file · 2010-09-022010-09-02 · Answer all the questions. 1 ... Assuming no mass loss, find the height of the new cylinder formed, ... 12 seamstresses