Other School 2 EM P2

7/29/2019 Other School 2 EM P2

1/19

Class Candidate Name Register Number

This question paper consists of 10 printed pages. [Turn over

READ THESE INSTRUCTIONS FIRST

Mathematics

Paper 2

Additional Materials: 1. Graph paper (1 sheet)2. Answer Paper ( 6 sheets)

Write your name, class and register number on all the work you hand in.

Write in dark blue or black pen.

You may use a pencil for any diagrams or graphs.

Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all questions.

If working is needed for any question it must be shown with the answer.

Omission of essential working will result in loss of marks.

You are expected to use a scientific calculator to evaluate explicit numericalexpressions.If the degree of accuracy is not specified in the question, and if theanswer is not exact, give the answer to three significant figures. Give answersin degrees to one decimal place.

For , use either your calculator value or 3.142, unless the question requiresthe answer in terms of.

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 orpart question.

The total mark for this paper is 100.

4016/02

2 hours 30 minutes

O-LEVEL PRELIMINARY EXAMINATION


2/19

2

Mathematical Formulae

Compound interest

Total amount = 1 100

nr

P

+

Mensuration

Curved surface area of a cone = rl

Surface area of a sphere = 24 r

Volume of a cone = 21

3r h

Volume of a sphere = 34

3r

Area of triangle ABC =1

sin2

ab C

Arc length = r , where is in radians

Sector area = 21

2r , where is in radians

Trigonometry

sin sin sin

a b c

A B= =

C

A

2 2 2 2 cosa b c bc= +

Statistics

Mean =fx

f

Standard deviation =

22fx fx

f f

O-Level Preliminary Examination Mathematics Paper 2


3/19

3

Answerall the questions

1 (a) Express as a fraction in its simplest form ,5 3

2 3 2y y

+. [3]

(b) Solve the equation5

8

13

7

+=

xx. [2]

.

(c) Given that 312

4

=

m

Tx ,

(i) find the value ofT when 2=m and 1=x , [1]

(ii) express m in terms of xand T . [2]

(d) Given that a2 b2 = 117 and a+ b= 13, find the value of (ab) 2. [2]

2 A group of n people plan to go on a tour and share the total cost of $10 560.

(a) Write down an expression, in terms of n, for the amount each person should

pay. [1]

A few weeks before the holiday, two more people join the tour group. The total cost

now rises to $14 040.

(b) Write down an expression, in terms of n, for the amount each member of the group

should pay now. [1]

(c) If the cost per person is $120 more than was originally expected, form an equation

in n and show that it reduces to:

n [3]0176272

=+ n

(d) Solve the equation n . [3]0176272 =+ n

(e) Hence, find 2 possible amounts each member of the original group was intended

to pay. [2]



4/19

4

3 (a) A shop sold a LCD TV at $5280 after offering a discount of 12%. Despite offering the

discount, the shop is still able to make a profit of 5%.

Calculate

(i) the usual price of the TV before discount, [2]

(ii) the cost price of the TV, giving your answer to the nearest cents. [2]

Another customer bought the same TV on hire purchase by paying a down payment of

10% of the selling price and the rest by a 2 years loan at 4.28 % annual simple interest.

(iii) Calculate the amount required for each monthly instalment, giving your

answer to the nearest dollar. [3]

(b) (i) Rudy borrowed S$5000 at x% per year compound interest. After 3 years,

she owed a total of S$6655. Calculate the value ofx. [2]

(ii) She then decided to change the S$5000 borrowed into euros.

The rate of exchange between S$ and euros was 1 euro = S$2.11 on

a particular day. Calculate the maximum amount in euros, to the nearest

dollar that she could get. [2]

4 The students in two schools took the same standing broad jump test. Information relatingto the results is shown in the tables below.

School A School B

Mean = 110.5

Standard Deviation = 15Jumping distance

(cm)Frequency

10080 < x 20120100 < x 16140120 < x 12160140 < x 2

(a) For School A, calculate(i) the mean , [2](ii) the standard deviation, giving both answers correct to the 1 decimal place. [2]

(b) Compare briefly, the results for the two schools in two different ways. [2]



5/19


5

5

B

A C

G

80 m

32 m

45 m

X

North

o140

North

A, B and C are 3 points on level ground with C due east ofA . GC is a vertical building of

height 45 m. AC = 80 m, BC = 32 m and ACB= 140o.

(a) Calculate(i) the distance AB, [2]

(ii) CAB, [2]

(iii) the area of . [2]ABC

(b) Find the bearing ofB from A. [1]

Jane stood at the top of building G. She looked at Ken who stood at point A.

(c) Find the angle of depression of Ken from Jane. [2]

Ken started to jog along the path from A to B. X is a point on AB such that CX is the

shortest distance from C to AB.

(d) Calculate the length ofCX. [2]

(e) Calculate the greatest angle of elevation ofG from Ken as he jogged from

A to B. [2]


6/19

6

6 The diagram shows a rectangular pyramid with base ABCD. The diagonals intersect at O.

The mid-point ofBC is X.

The vertex T is 7 cm vertically above O and AB = 8 cm and AD = 6 cm.

O

8 cm

6 cm

A B

D

T

X

7 cm

O

C

Calculate

(a) TX, [1]

(b) the volume of the pyramid, [2]

(c) the total surface area of the pyramid. [3]



7/19

7

7 A shop sells two varieties of chocolates, Black Chocolates and White Chocolates. Each

variety is sold in packets of three different sizes, small, medium and large and of different

prices. The sales in two successive weeks are given in the table below.

First Week Second Week

Size Small Medium Large Small Medium Large

Cost per Chocolate Bar $1.00 $1.20 $1.50 $1.00 $1.20 $1.50

Number of

Black Chocolate sold15 10 12 7 11 9

Number of

White Chocolate sold13 11 14 12 8 17

The information for the first weeks sales can be represented by the matrix, P =

141113

121015

and the cost of each chocolate bar for each size can be represented by the matrix A = .

5.1

2.1

1

The information for the second week sales can be represented by the matrix Q.

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

(b) Calculate S = (P + Q) . [1]

(c) Describe what is represented by the elements in S. [1]

(d) Calculate R =2

1[SA] . [2]

(e) Describe what is represented by the elements ofR . [1]

(f) Calculate and describe what is represented by the elements ofT = SA. [2]( 11 )



8/19


8

8

Q

O

AC

B

D

6834

In the diagram, ACB and BOQ are straight lines. A, C, D and Q lie on the circle and

O is the centre of the circle. O and O .o68 =CA o34 =CB

(a) Find

(i) , [1]AOC

(ii) , [1]AQC

(iii) , [2]COB

(iv) . [2]ACQ

(b) Given that AD = 8 cm and DQ= 11 cm, find AQ and state the reason for your

answer. [2]


9/19

9

9 In ORS, the point P on OR is such that OROP3

2= . M is the midpoint ofPQ.

Q is the midpoint ofOSand RN : NS= 3 : 4 . =OP 2a and =OQ 2b.

R

3

4

PN

2aM

O 2b Q S

(a) Express in terms ofa and/orb,

(i) PM , [1]

(ii) OM , [1]

(iii) RS , [1]

(iv) MR . [1]

(b) (i) Express MN in terms ofa and b. [2]

(ii) State two relationships between O, M and N . [2]

(c) Findarea of

area of

PRM

ORN

. [2]



10/19


10

10 The diagram below is the speed- time graph of a car for the first 18 seconds of a journey.

(a) Calculate

(i) the acceleration of the car during the first 4 seconds, [1](ii) the distance travelled for the first 10 seconds, [2](iii) the speed of the car when t= 16. [2]

Time (s)

2

12

4 10 18

Speed (m/s)

(b) Complete the distance-time graph of the car for the same journey. [2]

4 10 18

Distance (m)

Time (s)


11/19

11

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

The following table is for y= 5 2x 3x2.

x 3 2 1 0 1 2 3

y

16 u 4 5 0 w

28

(a) Calculate the values ofuand w. [2]

(b) Using the scale of 2 cm to 1 unit on the x-axis and a scale of 2 cm to 5 units on

the y-axis, draw the graph ofy= 5 2x 3x2 for 3 x 3. [3]

(c) From the graph, solve the equation 5 2x 3x2 = 0 . [2]

(d) By drawing a suitable straight lineon the same axes, solve the equation

10 - 4x 3x2 = 0. [2]

(e) By drawing a tangent, find the gradient of the curve at the point where x= -2. [2]


~~~~~End of Paper 2~~~~~


12/19

12

Mathematics Paper 2

Essential Steps Marks Alternative Steps/Remarks

1(a) 5 3

2 3 2y y + = ( )( ) ( )5( 2) 3(2 3 )

2 3 2 2 3 ( 2)

y y

y y y y

+ + +

=( ) ( )5 10 6 9

2 3 2

y y

y y

+ +

+

=( ) ( )

14 4

2 3 2

y

y y

+

+or

( )( )2(7 2)

2 3 2

y

y y

+

+

M1

M1

A1

(b)

5

8

13

7

+=

xx

17

92or

17

43

1743

724835

824357)13(8)5(7

=

=

=+

=+

=+

x

x

xx

xxxx

M1

A1

(c)(i)

4

3=T B1

(ii)

33

33

3

3

42

2

)12(4

12

4

xTmx

xmx

mxT

m

T

x

+=

=

=

=

3

3

2

4 xTm

+= or

2

123

+=T

m

M1

A1

(d) (a+ b) (ab) = 11713 (ab) = 117

13

117

= ba

( ) 8113

1172

2 =

= ba

M1

A1

10

2(a) 10560

n

B1



13/19

13


(b) 14040

2n+ B1

(c) 14040 10560- 120

2n n

=

+

14040 10560( 2)=120

( 2)

n n

n n

+

+

214040 10560 21120 =120( 2 )n n n + n

2 27 176 0n n + =

M1

M1

A1

(d) 2 27 176 0n n + = ( 16)( 11) 0n n =

16n= or 11n=

M1

A1, A1

(e)

When n = 11 , amount = 11

10560

= 960

When n = 16, amount =16

10560= 660

A1

A1

103(a)(i)

6000$

%10088

5280

=

M1

A1

(ii)

57.5028$

%100105

5280

=

M1

A1 No mark awarded ifanswer not given to nearestcents

(iii) Down payment = 10% x 5280 = $528

Amount of loan = 5280 528 = $4752

Total amount plus interest = %28.4247524752 +

= $5158.77

Monthly instalment =

94.214$

24

77.5160

=

= $ 215.00

M1

M1

A1

Alternative:Amt loan = 90% x 5280

= 4752

No mark awarded ifanswer not given to nearestdollar

(b)(i)

6655100

15000

3

=

+

x M1



14/19

14


x= 10 A1(ii)

11.2

5000= 2369.69 euro

= 2369 euro ( maximum euros)

M1

A1

No mark awarded ifanswer is rounded up to

2370 euros11

4(a)(i)

Mean =fx

f

= 108.4 cm

M1

A1

Standard deviation =

22fx fx

f f

= 17.8 cm

M1

A1

(b) School Bs students have a higher mean than School As.

Hence, School Bs scored better in the test.

Also, School Bs lower standard deviation indicates that theirperformance is more consistent than students in School A.

B1

B1

6

5(a)(i) 2 2 280 32 2(80)(32) cos140AB = + 0

106.518AB =

m ( 3 s.f.s.)107=

M1

A1

(ii)518.106

140sin32

sin0

=CAB

0134.11=CAB

(to 1 dec. pls.)011.1=

M1

A1

(iii)Area = 0140sin3280

2

1

= 822.768

= 823 m2

(3 s.f.s.)

M1

A1

(b) Bearing = 0 090 11.1 101.1= + = 0 B1

(c) 45tan

80x=

029.35 29.4x= = 0 (to 1 dec. pls.)

M1

A1



15/19

15


(d)0sin11.134

80

CX=

15.44 15.4CX = = (to 3 s.f.s)

M1

A1

Alternative :

1106.518 822.768

2CX =

CX= 15.4

(e) 45tan

15.44y=

071.06 71.1y= = 0 (to 1 dec. pls.)

M1

A1

13

6(a) TX= 22 47 + = 8.0623

cm06.8

B1

(b)

Volume = 7)68(3

1

=112 cm 3

M1

A1

(c) Total Surface Area =

++

+ 22 378

2

120623.86

2

1286

= 157.29998

157 cm 2

M2

A1

M1 for any 2 correctareas

6

7(a)

=

17812

9117Q B1

(b) S = (P + Q)

=

+

17812

9117

141113

121015

=

311925

212122

B1

(c) Total number of black and white chocolates of different sizes(each size) sold in 2 weeks.

B1

(d)

][2

1SAR = =

5.1

2.1

1

311925

212122

2

1=

3.94

7.78

2

1

=

15.47

35.39

M1

A1

(e) The averagesum received from the sales ofeach type of B1 No mark awarded if any of



16/19

16


chocolates in 2 weeks. the underlined words (ordifferent words with samemeaning) is not mentioned.

(f)( ) ( ) (173

3.94

7.781111 =

== SAT )

It represents the total collection from the sales of both types ofchocolates in 2 weeks.

B1

B1

8

8(a)(i) 000 44)68(2180 ==COA B1

(ii)0

0

222

44==CQA B1

(iii)

0

0000

34

443468180

=

=BOC

M1

A1

(iv) 0 0 0180 180 34 146COQ BOC = = = 0

000

172

146180

2

180=

=

=

COQOCQ

000 511768 === OCQOCAQCA

M1

A1

(b) AQ = 22 811 = 7.5498

cm55.7Angle in a semi-circle is a right angle and by Pythagoras Theorem.

B1

B1

8

9(a)(i)== PQPM

2

1b - a B1

(ii)

ba

aba

+=

+= 2OM

B1

(iii) b4a3 +=RS B1

(iv) =MR 3a ( a+ b) = 2ab or MO+uuuur

OR =uuur

-(a+b) +3a= 2a - b B1

(b)(i)

7

3=RN RS =

7

3(-3a+ 4b)

MN =uuuur

MRuuur

+ RNuuur

= 2a b +

7

3(-3a+ 4b)

=7

5(a+ b)

M1

A1

(ii) O, M and N are collinear.5

7MN OM=uuuur uuuru

B1

B1



17/19

17


(c) area of 1

area of 3

PRM

ORM

=

---(1)

area of 7

area of 12

ORM

ORN

=

---(2)

From (1) and (2) ,3 area of 7

area of 12

PRM

ORN

=

area of 7

area of 36

PRM

ORN

=

M1

A1

10

10(a)(i)

2

12

04

212=

m/s2 B1

(ii)( ) 6124122

2

1++

= 100 m

M1

A1(iii)

Deceleration =2

11

8

12= m/s2

Speed =2

11612 =3m/s

M1

A1

(b)

B2

1 mark awarded if any 2out of 3 parts of thegraph is correct, with thecorrect value on thedistance axis.

7

11(a) u= -3

w= -11

B1

B1(b) Points plotted correctly

Scale is correctCurve is smooth

B1B1B1

(c) x= 1x= - 1.7

B1B1 x= -1.7 0.1

(d) 10 - 4x 3x2 = 0 No mark awarded if onlyone solution is correct

4 10 18

Distance (m)

Time (s)

100

28

148



18/19

18


5 - 2x 3x2 = 2x -5

x= 1.3 , x= -2.6Draw y = 2x 5

M1A1

x= 1.3 0.1x= -2.6 0.1

(e) gradient 10.2 M1,A1

10.2 0.2

11

(b)AO = AC

2

1= 22 68

2

1+

= 5

5

7tan =TAO

4623.54=TAO

5.54

M1

A1



19/19


19

(c)(i)

BOC

AOC

ofarea

ofarea=

hBC

2

1

hAC2

1

=AC2

AC=

2

1

M1

A1

(ii) area = 23

57 = 34 cm2 B1

(iii)MN=

7

5(a+ b) =

7

5OM

5

7. =

MN

OM B1

(d)(i) 2 2x 3x2 = 05 2x 3x2 = 3

Draw y = 3x= -1.2 , x= 0.6

M1B1

No mark awarded if

only one solution iscorrectx= -1.2 0.1x= 0.6 0.1

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