vs 2012 em p2

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Class Register Number

Name

4016/02 12/4P2/EM/2

MATHEMATICS PAPER 2

Thursday 16 August 2012 2 hours 30 minutes

VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL VICTORIA SCHOOL




VICTORIA SCHOOL

SECOND PRELIMINARY EXAMINATION

(SECONDARY FOUR) Additional Materials: Answer Paper Graph Paper (1 sheet) Plain Paper (1 sheet)

READ THESE INSTRUCTIONS FIRST 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 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 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 . 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 number of marks for this paper is 100.

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VICTORIA SCHOOL 2012 12/4P2/EM/2

Mathematical Formulae

Compound interest

Total amount = 1100

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= =

2 2 2 2 cosa b c bc A= +

Statistics

Mean = fx

f

Standard deviation =

22fx fx

f f

3


Answer all the questions.

1 (a) (i) Sara, a British tourist, came to Singapore for a holiday. The rate of exchange between British pounds () and Singapore dollars (S$) was 1 = S$1.95.

She bought a diamond ring at S$585 from a shop. A Goods and Services Tax (GST) of 7% was imposed on the ring. Calculate the amount of money, in British pounds, that Sara paid for the ring. [2]

(ii) When Sara returned to England, she discovered that the retail price of the same

ring was 400. Calculate the amount she saved as a percentage of the price paid in Singapore. [2]

(b) The following advertisement was put up by the shop in January 2012.

(i) Calculate the interest payable for the hire purchase scheme. [1]

(ii) Hence calculate the monthly instalment for the hire purchase scheme. [2]

(iii) In February 2012, the interest rate by the hire purchase scheme was increased to

R % while the rest of terms remained unchanged. The new total hire purchase price was $942. Calculate R. [2]

2 Answer the whole of this question on a sheet of plain paper.

(a) Construct triangle ABC where 5.8 cm, 10.4 cm and angle 120 .AB AC ABC= = = [2]

(b) Measure the angle .BAC [1] (c) Construct

(i) the bisector of angle ,ACB [1]

(ii) the perpendicular bisector of AB. [1]

(d) Label the point D where D is the intersection of the bisector of angle ACB and the

perpendicular bisector of AB. Measure AD. [2]

Jamsung Jphone

Cash and carry $888

OR

Hire purchase terms of a cash deposit of $288

followed by monthly instalments at 3% per annum for 2 years

4


3 (a) The table shows the duration, in minutes, that 27 commuters had to wait for their taxi at Taxi Stand A.

(i) Construct a frequency table from the information given above. [1] (ii) Calculate the mean waiting time and standard deviation. [4]

Another group of 27 commuters waited for their taxi at Taxi Stand B. The mean and standard deviation of the waiting time at Taxi Stand B were 9.8 minutes and 3.55 minutes respectively. (b) Compare and comment on the waiting time for the two groups of commuters in two

different ways. [2] (c) The weights of 700 potatoes were measured. Their weights, in grams, are

summarized below. The heaviest potato weighs 210 g. The lightest potato weighs 25 g. The 75th percentile is 170 g. Median weight is 158 g. Interquartile range is 71 g.

The weights of the potatoes measured were represented in a box and whisker diagram as shown.

Using the given information, write down the values of a, b, c, d and e. [2]

13 14 11 13 12 14 13 13 14 13 15 14 13 13 14 12 15 13 14 13 18 12 15 13 13 11 12

a b c d e

Weight (in grams)

5


4 An oil tank is made by joining a hemisphere of radius 4 m to an open cylinder of radius 4 m and height 9 m.

(a) Calculate the total external surface area of the oil tank. [3]

(b) The exterior surface of the oil tank is to be painted. The paint is sold in tins, each of capacity 2.5 litres. One litre of paint covers 6 m2. Find the minimum number of tins that should be purchased. [2]

The oil tank is initially filled with oil to the brim. (c) Calculate the initial volume of oil in the oil tank. [3] (d) Oil is drained from the oil tank by allowing it to flow through a cylindrical outlet

pipe. The inner diameter of this pipe is 15 cm and oil flows out at a rate of 5 metres per second. Find, correct to the nearest minute, the time taken to completely drain the oil tank. [3]

5 (a) Expand and simplify ( )( )2 2

3 2 3 2 8 9y y y yy

+ +

. [3]

(b) (i) Express 2

1 2 3

4 3 1 4 3x x x x

+ as a single fraction in its simplest

form. [4]

(ii) Hence solve the equation 2

1 2 40

4 3 1 4 3x x x x =

+. [3]

9

4

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6 John went on a Malaysian road trip to visit his uncle. He drove a distance of 750 km at an average speed of x km/h.

(a) Write down an expression, in terms of x, the time in hours, that he took to complete

his journey. [1]

(b) On his return journey, his driving speed was 10 km/h slower due to a delay caused by a heavy traffic jam. Write down an expression, in terms of x, the time in hours, he took to complete the return trip. [1]

(c) Given that John took 45 minutes longer on the return trip, form an equation in x and

show that it reduces to 2 10 10 000 0.x x = [3]

(d) Solve the equation 2 10 10 000 0,x x = giving your solutions correct to 2 decimal

places. [2]

(e) Find the average speed for the entire journey. [1]

7 The diagram shows a circle ABCD with centre O. PQ and PR are tangents to the circle.

Angle 140 ,AOC = angle 44ODC = and OA is parallel to CB.

(a) Show that AC bisects OCB . [2] (b) Calculate

(i) angle ,BAC [2]

(ii) angle ,ODA [2]

(iii) angle ,ADP [2]

(iv) angle .APD [2]

B

140

44

O

Q

C

A

P

R D

7


8

In the diagram, OAB is a triangle. C is a point on AB such that AC : CB = 2 : 1. The side OB is produced to the point D such that OB : BD = 3 : 2.

It is given that =OA a and =OB b.

(a) Express, as simply as possible, in terms of a and b .

(i) AB

, [1]

(ii) AC

. [1]

(b) Show that 1

3CD = b a

. [4]

(c) It is given that E is the point on OA such that 5

9OE = a

.

Express ED

, as simply as possible, in terms of a and b . [1]

(d) (i) Show that, ED kCD=

, where k is a constant. [1]

(ii) Write down one fact about E, C and D. [1]

(e) Find the exact value of

(i) area of triangle

,area of triangle

AEC

OEC [1]

(ii) area of triangle

area of triangle

AEC

ODC. [2]

A

b

D

C

a

B

E O

8


9 The diagram shows the map of a park, ABCD, and a path AC. D is due north of A.

76 m andAD = 182 m.AC = The bearing of C from D is 065 .

(a) Calculate (i) angle DAC, [3] (ii) the bearing of A from C. [2]

(b) The area of triangle ABC is 217 973 m and angle 61 .ACB = Calculate AB. [3]

(c) A boy stands at C and notices a viewing tower at A. He then jogs along CD produced

to a point E, where the angle of depression from the top of the tower to E is the greatest.

(i) Find AE. [2] (ii) If the height of the viewing tower is 48 m, calculate the angle of depression

from the top of the tower to E. [2]

N

C

B

A

D

76

182

61

9


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

The variables x and y are connected by the equation2 5

12

xy

x= + . Some corresponding

values of x and y are given in the following table.

x 0.5 0.75 1 1.2 1.5 2 2.5 3 4

y 9.125 5.95 4.5 3.89 3.46 3.5 p 5.17 8.25

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

(b) Using a scale of 4 cm to represent 1 unit, draw a horizontal x-axis for 0 4.x

Using a scale of 2 cm to represent 1 unit, draw a vertical y-axis for 0 10.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 two solutions of 2 5 1

62 5

x

x+ = in the range 0 4.x [2]

(d) By drawing a tangent, find the gradient of the curve at the point ( )3, 5.17 . [2]

(e) On the same axes, draw the graph of 4 7 30y x+ = for 0 4.x [1]

(f) (i) Write down the x-coordinate of the points where the two graphs intersect. [2]

(ii) These values of x are the solutions of the equation 3 2 0.x Ax Bx C+ + + = Find the value of A, the value of B and the value of C. [1]

End of Paper

This document is intended for internal circulation in Victoria School only. No part of this document may be reproduced, stored in a retrieval

system or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior permission of the

Victoria School Internal Exams Committee.

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25

170 71

99

158

170

210

a

b

c

d

e

=

=

=

=

=

=

2012 Prelim 2 Mathematics Paper 2 Answer Key

1(a)(i) 321 4(a) 327 m2

1(a)(ii) 24.6 % 4(b) 22

1(b)(i) $36 4(c) 586 m3

1(b)(ii) $26.50 4(d) 111 min

1(b)(iii) 4.5 5(a) 481y

2(b) 31 5(b)(i) ( )

( )( )

5 2

4 3 1

x

x x

+

+

2(d) AD = 3 cm 5(b)(ii) 3

24

x =

3(a)(i)

Waiting time (minutes)

11 12 13 14 15 18

frequency 2 4 11 6 3 1

6(a) 750

x

3(a)(ii) 13.3 min; 1.39 min 6(b) 750

10x

3(b)

The commuters at Taxi Stand B had a shorter waiting time as compared to those at Taxi Stand A as the mean waiting time at Taxi Stand B is lesser. The commuters at Taxi Stand B had a wider/bigger spread of waiting time as compared to those at Taxi Stand A as the standard deviation at Taxi Stand B is higher.

6(d) x =105.12 or 95.12

3(c)

6(e) 99.9 km/h

11


7(b)(i) 50 9(a)(i) 42.8

7(b)(ii) 26 9(a)(ii) 222.8

7(b)(iii) 64 9(b) 210 m

7(b)(iv) 52 9(c)(i) 68.9 m

8(a)(i) b a 9(c)(ii) 34.9

8(a)(ii) ( )2

3b a 10(a) p = 4.125

8(c) ( )5

39

b a 10(c) x = 0.85 or 3.025

8(d)(i) 5

3ED CD=

10(d) 2.49

8(d)(ii)

E, C and D are collinear or lie on the same straight line. or CD is parallel to ED.

10(f)(i) x = 0.725 or 2.2

8(e)(i) 4

5 10(f)(i)

13

2

17

10

A

B

C

=

=

=

8(e)(ii) 8

15

vs 2012 em p2

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