MATHEMATICS 4016 / 01 - WordPress.com · · 2016-01-04MATHEMATICS 4016 / 01 Paper 1 17 August 2010 ... 4 AHS Prelim 2010 Sec 4 Mathematics Paper 1 ... B and C are on level ground

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ANGLICAN HIGH SCHOOL

Preliminary Examination 2010 Secondary Four

NAME

Class / Index Number 4 /

MATHEMATICS 4016 / 01

Paper 1 17 August 2010

2 hours

Candidates answer on the Question Paper.

READ THESE INSTRUCTIONS FIRST Write in dark blue or black pen in the spaces provided on the Question Paper. 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. Calculators should be used where appropriate. 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 80.

For Examiner’s Use

This Question Paper consists of 14 printed pages.

Write down the model of your calculator used:

Signature of Parent/Guardian & Date

Name of Parent/Guardian 80

Page | 2 AHS Prelim 2010 Sec 4 Mathematics Paper 1

Mathematical Formulae

Compound interest

Total amount = 𝑃 1 +𝑟

100 𝑛

Mensuration

Curved surface area of a cone = πrl

Surface area of a sphere = 4πr2

Volume of a cone = 1

3𝜋𝑟2𝑕

Volume of a sphere = 4

3𝜋𝑟3

Area of triangle ABC = 1

2 ab sinC

Arc length = 𝑟𝜃, where 𝜃 is in radians

Sector area = 1

2𝑟2𝜃, where 𝜃 is in radians

Trigonometry

𝑎

sin𝐴 =

𝑏

𝑠𝑖𝑛𝐵 =

𝑐

𝑠𝑖𝑛𝐶

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos𝐴

Statistics

Mean =

𝑓𝑥

𝑓

Standard deviation = 𝑓𝑥2

𝑓−

𝑓𝑥

𝑓

2

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Answer all the questions.

1 Expressed as the product of prime factors,

360 = 2

3 3

2 5 and

84 = 22 3 7

Use these results to find

(a) the higher common factor of 360 and 84.

(b) the smallest positive integer k, such that 360k is a cube number.

Answer (a) [1]

(b) k = [1]

2 (a) Write the following numbers in order of size, starting with the largest.

648.2 ,13

112 ,648.2 ,684.2

(b) Calculate the simple interest when $ 12500 is invested for 9 months at 0.2 % per

annum.

Answer (a) [1]

(b) $ [1]


3 Given that 94 p and 16 q , find

(a) the greatest possible value of p q ,

(b) the smallest possible value of 𝑝 + 𝑞

𝑞

(c) the smallest possible value of )( 22 qp

Answer (a) [1]

(b) [1]

(c) [1]

4 The rate of exchange between Japanese yen (¥) and Singapore dollars ($) was

¥ 64.1 = $ 1.

(a) Joanne bought a handbag for ¥ 23 000 in Japan. How much is the cost of the

handbag in SGD?

(b) A similar handbag costs $ 414.40 in Singapore. What is the difference in the

price between the two countries in SGD?

[Correct your answers to the nearest dollars]

Answer (a) $ [1]

(b) $ [1]

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5 (a) Write the following numbers correct to four significant figures,

(i) 385.047

(ii) 0.00192287.

Answer (a) (i) [1]

(ii) [1]

(b) Express 9

16 as a percentage.

Answer (b) % [1]

6

(a)

(i) Calculate

2.51− 1.23−7.836 3

20.7 ×96.45+18.01 , showing all the figures on your

calculator display.

(ii) Give your answer correct to 3 decimal places.

Answer (a) (i) [1]

(ii) [1]

(b) Evaluate (2.77 10 –3

) (5.911 10 –7

) ,give your answer in standard form.

Answer (b) [1]


7 Factorise the following expressions completely

(a) ac + 2bc – 3 ad – 6 bd

(b) 2x2 – 32

Answer (a) [1]

(b) [1]

8 A map is drawn to a scale of 1: 250 000. Find

(a) the actual distance, in kilometres, of a highway which is 28 cm in length on the

map.

(b) the area, in cm2, of a garden on the map which has an actual area of 37.5 km

2.

Answer (a) km [1]

(b) cm2

[1]

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9 Liquid is poured into the empty container, as shown in the diagram, at a constant rate.

It filled the container up to a height of d1 in 5 minutes and takes another 5 minutes to

fill up to d2.

On the axes in the answer space, sketch the graph showing the depth (d) of the liquid

varies with time (t).

[2]

Answer

10 The line 𝑥

2−

𝑦

3 = 6 cuts the y–axis at the point P. Find

(a) The coordinates of the point P,

(b) The gradient of the line.

Answer (a) (……. , …….) [1]

(b) [1]

Liquid in

d2

d1

d1

d2

Depth of

liquid (d)

Time (minutes)

0 5 10


11 In the diagram, points A, B and C are on level ground. 86ABC . C is due north

of A and B is in the direction 050 from A. Find

(a) the bearing of B from C.

(b) Find the area of ∆ 𝐴𝐵𝐶 when AC = 5 m and AB = 4 m.

Answer (a) o

[1]

(b) m2

[2]

12 Given that 13

25

3

2

x

xyy

(a) Calculate the value of y when x = −1

8

(b) Express y in terms of x.

Answer (a) y = [2]

(b) [2]

N

B

A

C

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13 The temperature at 0500 is – 12o C

The temperature at 1330 is 24o C

(a) Find the difference between the two temperatures.

(b) Assuming that the temperature rises at a steady rate, find

(i) the temperature at 0800, correct your answer to one decimal place.

(ii) the time when the temperature is 0o C

Answer (a) o

[1]

(b) (i) o

[2]

(ii) [1]

14 If = {pupils of a school}, F = {female pupils} and S = {Singaporeans}.

(a) Draw a Venn diagram to represent the above information.

[2]

(b) Shade the area that represents male Singaporean pupils.

Answer (a) & (b)

(c) Express each of the following statements in set notation.

(i) Female pupils who are Singaporeans.

(ii) Male pupils who are non Singaporeans.

Answer (c) (i) [1]

(ii) [1]


15 The scale drawing in the answer space below shows a circle with diameter AB and Q is

a point on the circumference of the circle.

(a) Construct the bisector of angle QAB.

(b) Construct a right–angled triangle ABC such that C lie on the bisector of angle

QAB and ABC = 90o. Label the point C.

(c) Measure and write down the size of angle ACB.

Answer

[2]

Answer (c) ACB = o

[1]

A

B

Q

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16 In the diagram, AFDB and CGEB are straight lines. BDE and DEF are two isosceles

triangles and FDE = 36o.

(a) Find FEC.

Answer (a) FEC = o

[2]

(b) By construction, determine the maximum number of isosceles triangles that can

be added in the given diagram such that the vertices of the triangle(s) are on the

two given lines.

Answer (b) [2]

(c) Justify your answer in (b) with reason. [1]

Answer

C

A

36

F

E

B

D


17 It is known that a certain parabola cuts the x–axis at –3 and 4.

(a) Write down the equations of two quadratic curves that fit this description, and

sketch them on the axes in the answer space.

Answer (a) [1]

[1]

[3]

(b) Write down the equation(s) of the line of symmetry of the curves that you have

drawn.

(b) [1]

x

y

O

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18 The points A, B, C and D lie on a circle. A smaller circle, centre C, passes through

points B, E and F and AFC is straight line. ADC = 79o and BAC = 21

o.

Giving your reasons, find

(a) angle CBA,

(b) angle FCB,

(c) angle CFE, if CF is parallel to EB.

Answer (a) CBA = ° [1]

(b) FCB = ° [1]

(c) CFE = ° [2]

79

21

F

C

A

D

B

E


19 ABCD is a parallelogram. 𝐴𝐵 = 𝒂, 𝐵𝐶 = 𝒃, 3BX = 2XC and AE = 3

5𝐴𝑋.

(a) Given that 𝒂 = 25 and 𝒃 =

50 , find

`

(i) 𝐴𝑋

(ii) 𝐵𝐸

(iii) 𝐸𝐷

Answer (a) (i) units [2]

(ii) [1]

(iii) [1]

(iv) Hence, show that B, E and D are collinear.

Answer

[1]

(b) Write down, as a fraction in its simplest form, the value of

𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐸

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝐴𝐵𝐶𝐷

(b) [2]

A

B C

D

X

E a

b

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20 The cumulative frequency shows the time in minutes taken by a group of 32

competitors to complete a Sudoku puzzle.

Using the graph, to estimate

(a) the median,

(b) the interquartile range,

(c) the number of competitors that took 14 minutes or less to complete the game,

(d) the percentage of competitors that took 19.5 minutes or more to complete the

game.

Answer (a) min [1]

(b) min [2]

(c) [1]

(d) % [2]

Cumulative Frequency Curve of Time Taken To Solve Sudoko Puzzle

Number of Competitors

Time (min)

4

0

8

12

16

20

24

28

32

8 10 12 14 16 18 20 22 24 26


21 The first four terms in a sequence of numbers, p1, p2, p3, p4, …, are given below.

p1 = 12 + 2

2 + 2

2 = 9

p2 = 22 + 3

2 + 6

2 = 49

p3 = 32 + 4

2 + 12

2 = 169

p4 = 42 + 5

2 + 20

2 = 441

(a) Write down an expression for p5 and show that p5 = 961.

Answer (a)

[1]

(b) Write down an expression for p6 and evaluate it.

Answer (b)

[1]

(c) Show that pn = n4 + 2n

3 + 3n

2 + 2n + 1.

Answer (c)

[3]

(d) Given that p10 = 102 + 11

2 + r

2 = k, express k as a perfect square in terms of r.

Answer (d) k = [1]

(e) Given that pw = w2 + (w + 1)

2 + r

2 = 5257

2 , find the value of r and of w.

Answer (e) r = [1]

w = [1]

END OF PAPER

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1 (a) HCF of 360 and 84 = 4 3 = 12

(b) k = 3 52 = 75

2 (a) 2.8466 2.84646 2.84615 2.846846

Answer :

648.2

684.2

648.2 13

112

(b) simple interest = 12500 × 9

12 × 0.002

= $ 18.75

3 (a) the greatest possible value of p q = (–4) (–1) = 4

(b) the smallest possible value of 𝑝 + 𝑞

𝑞

= the smallest possible value of ( 𝑝

𝑞+ 1) =

9

−1 + 1 = –8

(c) the smallest possible value of )( 22 qp = 0 2 + (–1)

2 = 1

4 (a) The cost of the handbag in $ = 23 000 64.1 = $ 358.81 $ 359

(b) The difference in the price between the two countries in $

= 414.40 – 358.81 = $ 55.59 $ 56

5 (a) (i) 0.002177359

(ii) 0.002

(b) – 17134913.82 – 17130000

= – 1.713 × 107

6 (a) 1 : 250 000 = 1 cm : 250 000 cm

= 1 cm : 2.5 km

= (1 28) cm : (2.5 28) km

= 28 cm : 70 km

Actual Distance = 70 km

(b) (1cm)2

: (2.5 km)

2 = 1 cm

2 : 6.25 km

2

= (1 37.5

6.25) cm

2 : (6.25

37.5

6.25) km

2

= 6 cm2

: 37.5 km2

map area = 37.5 km2

7 (a) When x = 0, y = –18, P is (0, –18)

(b) y = 3𝑥

2 – 18, gradient = 3/2

8 (a) ac + 2bc – 3 ad – 6 bd = c (a + 2b) – 3d (a + 2b)

= (c – 3d) (a + 2b)

(b) 2x2 – 32 = 2(x

2 – 16) = 2(x – 4)(x + 4)


9

10 (a) Bearing of B from C = 86 + 50 = 136°

(b) Area = 1

2× 5 × 4 × 𝑠𝑖𝑛50° = 7.6604 ≈ 7.66 𝑚2

11 (a) Difference = 24o C – (–12

o C) = 36

o C

(b) (i) the temperature at 0800 = – 12o + 36𝑜 ×

(8 – 5)

(13.5 − 5)

= 0.705 o C

0.7 o C

(ii) the time when the temperature is 0o C

= 0500 + (13.5 – 5) ×12𝑜

36𝑜

= 0500 + 17

6

= 0500 + 170

60

= 0500 + 2 hr 50 min

= 0750

12 (a) & (b)

(c)(i) 𝐹 ∪ 𝑆

(ii) 𝐹 ∪ 𝑆 ′ 𝑜𝑟 𝐹′ ∩ 𝑆′ 13

(a) sub x = −1

8 into

13

25

3

2

x

xyy

1)

8

1(3

)8

1(25

3

2

yy

4

315

4

11

8

11 yy

d1

d1

Depth of liquid

(d cm)

Time (minutes)

0

d2

2 4 6 8 10 12

E

SF

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C

A

B

Q

2

8

131y

131

16y

(b)

13

25

3

2

x

xyy

)25(3)13)(2( xyxy

xyxxy 61526)13(

215)13( yxy

2)163( xy

xor

xy

316

2

163

2

14 (a) & (b)

(c) The point of intersection is the centre of the given circle.

(d) ACB = 90o

15 (a) DBE = 36o/2 = 18

o

DFE = 36o

FEC = DBE + DFE (ext of BEF)

= 18o + 36

o = 54

o

(b)

Construct lines FG and GH

2 more isosceles triangles can be drawn

(c) HFG = (18 + 54)o = 72

o

FHG = 72o

HGC = 18o + 72

o = 90

o

No further triangles can be drawn.

A

C

36

H

G

F

E

B

D


16 (a) y = (x + 3)(x – 4) (There are other possible answers)

and y = – (x + 3)(x – 4)

(b) x = 0.5

17 (a) CBA = 180o – CDA (s in opp segment)

= 180o – 79

o

= 101o

(b) FCB = 180o – BAC – CBA (s sum of ABC)

= (180 – 21 – 79)o

= 80o

(c) CFE = FEB (alt. s, CF // EB)

= ½ FCB ( at centre = 2at circumference)

= 40o

18 (a) (i) 𝐴𝑋 = 𝐴𝐵 + 𝐵𝑋

= 24

+ 1

3

60

= 44

𝐴𝑋 = 42 + 42 = 32 ≈ 5.66 𝑢𝑛𝑖𝑡𝑠

(ii) 𝐵𝐸 = 𝐵𝐴 + 𝐴𝐸

= −2−4

+ 3

4

44 =

1−1

(iii) 𝐸𝐷 = 𝐸𝐴 + 𝐴𝐷

= −3

4

44 +

60

= 3−3

= 3 1−1

Since 𝐵𝐸 𝑎𝑛𝑑 𝐸𝐷 have common vector 1−1

from same point E, therefore B, E and D are

collinear.

x

y

O

y = – (x + 3)(x – 4)

–3 4

– 12.25 y = (x + 3)(x – 4)

12.25

–

12

12

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(b)


𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐷 =

12𝐴𝐵×𝐵𝐸𝑠𝑖𝑛∠𝐴𝐵𝐷

12𝐴𝐵×𝐵𝐷𝑠𝑖𝑛∠𝐴𝐵𝐷

= 𝐵𝐸

𝐵𝐷 =

14𝐵𝐷

𝐵𝐷 =

1

4


𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝐴𝐵𝐶𝐷=


2×𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐷 =

1

8

19 (a)(i) Median =13.5 min

(ii) Interquartile range = 16 – 12 = 4 min

(iii) No of competitors = 18

(iv) Percentage = 4

32× 100 = 12.5%

(b) The lower quartile or median or upper quartile for the 2008 competition is higher than

the 2009 competition. This shows that the quality of the competitors in the 2009 batch

seems to be better than the 2008 batch.

The inter-quartile range for the 2008 competition is wider than the 2009 competition.

This shows that the 2009 competitors are relatively more competitive.

20 (a) p5 = 52 + 6

2 + 30

2 = 25 + 36 + 900 = 961

(b) k = (s + 1)2

(c) r = 5256

w (w + 1) = 5256 = 72 × 73

w = 72

(d) pn = n2 + (n + 1)

2 + [n(n +1)]

2

= n2 + n

2 + 2n + 1 + n

2(n + 1)

2

= 2n2 + 2n + 1 + n

2(n

2 + 2n + 1)

= 2n2 + 2n + 1 + n

4 + 2n

3 + n

2

= n4 + 2n

3 + 3n

2 + 2n + 1.

Anglican High School Preliminary Examinations 2010

Name ( ) Class 4 ________

Thursday 19 August 2010 2 hours 30 minutes

Additional Materials: 8 pieces of writing paper and 1 piece of graph paper. INSTRUCTIONS TO CANDIDATES

Answer all questions. Write your answers on the writing papers provided. Omission of essential working will result in loss of marks. Calculators should be used where appropriate. 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 .

INFORMATION FOR CANDIDATES

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.

Question 1 2 3 4 5 6 7 8 9 10 11

Marks

This question paper consists of 9 printed pages.

For Examiner’s Use

100

ANGLICAN HIGH SCHOOL

Preliminary Examination Secondary Four

MATHEMATICS 4016/02

Assessment noted by :

_____________________________ _____________________________ _____________________________

Name of Parent/Guardian Signature of Parent/Guardian Date

Anglican High School Preliminary Examinations 2010 Mathematics Paper 2

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

3

4r

Area of triangle ABC = Cabsin2

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

Abccba cos2222

Statistics

Mean = f

fx

Standard deviation =

22

f

fx

f

fx


3

Answer all the questions.

1 (a) A train travelled a distance of 375 km in 5 hours from Station A to Station B. After

taking a break of 45 minutes, it travelled a further distance of 157.5 km at an average

speed of 90 km/h.

Calculate:

i) its average speed for the first part of the journey, [1]

ii) the time taken, in hours and minutes, for the second part of the journey, [1]

iii) its average speed for the whole journey. [3]

(b) Solve the equation

32−𝑥 = 1

35𝑥+13 [2]

(c) Express the following as a single fraction in its simplest form.

5 + 𝑦

25 − 𝑦2−

3

𝑦 − 5 [3]

2 (a) A soccer club has 150 members. X is the set of strikers. Y is the set of defenders.

The letters a, b and c in the Venn diagram represent the number of members in

each of the subsets of X and Y. The letter d represents the numbers of members

who are neither strikers nor defenders. Given that n(X) = 90 and n(Y) = 68, find

(i) the value of b if d = 0, [1]

(ii) the value of d if b = c, [1]

(iii) the largest possible number of members who are neither strikers nor

defenders. [1]

X

Y

a b

c

d


4

2 (b) The table below shows the number of pens own by each student in a class.

Number of pens 0 1 2 3 4 5

Number of students 1 3 x 7 5 2

(i) If the mode of the distribution is 3, find the range of values of x. [1]

(ii) If the median is 3, what is the largest value of x. [1]

(iii) If the mean is 2.9, find the value of x. [2]

3 (a)

The diagram shows the arcs AB and EF of a circle, centre O, with radius r cm.

BCX, OP and EDY are straight lines that are perpendicular to the line segment

AXPYF. CDYX is a rectangle with CX = 4 cm and AX =XY =YF = 5cm.

i) Show that r = 8.5 cm [1]

ii) Calculate the arc AB. [4]

(b) The cost price of a particular model of camera is $300. Shop A advertises its sale

as “usual price $450 now $360”. Shop B sells the same model of camera at $420

after a 25% discount.

Calculate

i) the discount given by shop A as a percentage, [2]

ii) the original marked price of the camera at shop B, [1]

iii) the profit gained by shop B as a percentage. [2]

B E

C D

A X P Y F

O


5

4 (a) It takes 3 hours and 10 min for 5 workers to paint a hall. Given that all the workers

work at the same rate, how long will 2 workers take to paint the same hall?

[Leave your answer in hours and minutes] [2]

(b) Mr Hai takes x minutes to carve a swan out of a block of ice and Mr Li takes 25

more minutes to carve an identical sculpture. If both of them work together, it only

takes 30 minutes to complete carving an identical sculpture.

i) Form an equation involving x, and show that it simplifies to

x2 – 35x – 750 = 0. [2]

ii) Determine the time taken by Mr Li to carve the sculpture on his own.

[3]

(c) i) Solve – 4 < 2x – 5 13 [2]

ii) Write down all the prime numbers which satisfy this condition. [1]

5 The diagram shows a solid in which ABCD, DCFE and ABFE are rectangles. G is the

foot of perpendicular from D to AE. Given that AB = 5 cm, AD = 8 cm, DE = 5.5 cm

and angle DAE is 25°.

Find,

(i) the length of DG, [1]

(ii) angle DCG, [2]

(iii) the length of AE, [4]

C

A

B F

D

E G

5 cm 8 cm 5.5 cm

25o


6

6 Two shops A and B sell the same brand of compact discs (CDs) and printing paper. Each

box of CDs is sold at $3.70 and $3.80 at Shop A and Shop B respectively. Each ream of

paper is sold at $6.80 and $6.50 at Shop A and Shop B respectively. The prices of the

items are represented by the matrix P,

𝑷 = 3.70 3.806.80 6.50

Andrew plans to buy 6 boxes of CDs and 2 reams of paper. Betty plans to buy 4 boxes of

CDs and 3 reams of paper. Charles plans to buy 5 boxes of CDs and 1 ream of paper.

(i) Represent the quantities of CDs and papers to be bought by the 3 persons as

a matrix Q. [1]

(ii) Evaluate the product, R = QP [2]

(iii) State what the elements of R represent. [1]

(iv) Evaluate T = (1 1 1) R. [1]

(v) State what the elements of T represent. [1]

(vi) If Andrew, Betty and Charles combine their purchase, which shop will give them a

better deal. [1]

7 A bowl of sweets contains 12 chocolate nuggets, 13 mints and 15 toffees. Mr Chan takes

two sweets at random and eats them.

(a) Draw a tree diagram to represent all the possible outcome. [3]

(b) Calculate the probability, as a fraction in its simplest form, that

i) a chocolate nugget and a mint was taken. [1]

ii) none of the two sweets chosen were toffees. [1]

(c) A third sweet was randomly taken from the bowl. Calculate the probability that the

first two sweets drawn were of the same type and the third was a chocolate nugget.

[2]


7

8. The diagram shows two regular octagons that meet at points A and B.

(a) Find the size of each interior angle of the octagons.

[1]

(b) Find PAQ.

[2]

(c) Show that

𝐴𝐵

𝐴𝐶 = 1.85, correct to 3 significant figures.

[3]

(d) Hence or otherwise, find the area of the smaller octagon if the area of the bigger

octagon is 36.2 cm2.

[3]

9. The diagram shows a circle, centre O and radius 3 cm.

An inscribed triangle ABC is

drawn with A and B on the

y-axis. AC is extended to

meet the x-axis at D and

OD meets BC at E.

(a) Prove that triangles BOE and DOA are similar.

[3]

(b) If OE = 1 cm,

(i) find OD,

(ii) show that AC = 3 10

5 cm,

(iii) hence find the exact length of DC.

[2]

[3]

[2]

y

xE

C

A

B

OD

Q

C

B

P

A


8

10

O, P , W and R are points on a horizontal plane. A vertical memorial tower, OT, is due

north of P. The angle of elevation of a man at P to the top of the tower is 15°. He walks a

distance of 200 metres to point W. Given that the bearing of W from P is 040° and the

bearing of O from W is 280°, calculate

i) ∠PWO, [2]

ii) OP, [2]

.

iii) the height of the tower OT, [2]

iv) the angle of depression of W from T. [2]

From W, the man walks 150 metres to a point R on a bearing of 𝜃° on the horizontal

plane. At R, the angle of elevation of the man to the top of the tower is 10°.

Calculate

v) angle OWR, [3]

vi) the value of 𝜃. [1]

N

N

R

T

O

P 150 m 200 m

15O

40O

280O

W


9

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

The variables x and y are connected by the equation

𝑦 = 3 2𝑥 − 4.

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

x –2 –1 0 1 2 3 4

y a –2.5 –1 2 8 20 44

(a) Find the value of a.

[1]

(b) Using a scale of 2 cm to 1 unit, draw a horizontal x–axis for –2 ≤ x ≤ 4.

Using a scale of 2 cm to 5 units, draw a vertical y–axis for –5 ≤ y ≤ 45.

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 solution of 2𝑥 = 9.

[2]

(d) By drawing a tangent, find the gradient of the curve at the point (3, 20).

[2]

(e) Use your graph to find the solutions of the equation 3 2𝑥 − 10𝑥 − 8 = 0.

[3]

END OF PAPER


10

BLANK PAGE


11

AHS 2010 Prel EM P2 Answers

1ai) 75 km/h 1aii) 1 h 45 min 1aiii) 71 km/h 1b) –3 ½ 1c) 4/(5 – y)

2ai) 8 2aii) 26 2aiii) 60

2bi) 0 ≤ 𝑥 ≤ 6 2bii) 9 2biii) 2

3aii) 15.0 cm 3bi) 20% 3bii) $ 560

3biii) 40% 4a) 7 hrs 55 min 4bii) 1 hr 15 min

4ci) ½ < x 9 4cii) 2, 3, 5, 7 5i) 3.38 cm

5ii) 34.1o 5iii) 11.6 cm 6vi) Shop B

6i)

6 24 35 1

6ii)

35.80 35.8035.20 34.7025.30 25.50

6iv) 96.30 96.00

6iii) The elements in R represent the total cost of buying the CDs and paper at shop A or shop

B for Andrew, Betty and Charles respectively.

6v) The elements in T represent the total spent at Shop A or Shop B by Andrew, Betty and

Charles.

7bi)) 1/5 7bii) 5/13 7c) 119/1235

8a) 135o

8b) 112.5o

8d) 10.6 cm2

9bi) 9 cm 9biii) 12 10

5 cm

10i) 60o

10ii) 176 m 10iii) 19.8o

10iv) 144.5o

10v) 64.5o 11a) –3.25 11c) 3.25

11d) 16.7 11e) –0.6 , 4


12

AHS 2010 Prel EM P2 Solution

1 (a) A train travelled a distance of 375 km in 5 hours from Station A to Station B. After

taking a break of 45 minutes, it then travelled a further distance of 157.5 km at an

average speed of 90 km/h.

Calculate:

i) its average speed for the first part of the journey, [1]

ii) the time taken, in hours and minutes, for the second part of the journey, [1]

iii) its average speed for the whole journey. [3]

(b) Solve the equation

32−𝑥 = 1

35𝑥+13 [2]

(c) Express the following as a single fraction in its simplest form.

5 + 𝑦

25 − 𝑦2−

3

𝑦 − 5 [3]

Answer:

1a i) average speed for the first part of the journey = 375 5

= 75 km/h

ii) time taken for the second part of the journey = 157.5 90

= 1¾ h

= 1 h 45 min

iii) average speed for the whole journey

=

4

31

4

35

5.157375

= 5.75.532 = 71 km/h

Q1b) 32−𝑥 = 1

35𝑥+13

32−𝑥 = 3−(5𝑥

3 +

1

3)

2 – x = −5𝑥

3−

1

3

2𝑥

3= −

7

3

x = −7

2


13

Q1c) 5+𝑦

25 − 𝑦2 − 3

𝑦 − 5 =

5+𝑦

5−𝑦 (5+𝑦)+

3

5 − 𝑦

= 5 + 𝑦 + 3(5+𝑦)

5−𝑦 (5+𝑦)

= 20 + 4𝑦

5−𝑦 (5+𝑦)

= 4(5 + 𝑦)

5−𝑦 (5+𝑦)

= 4

5−𝑦

2 (a) A soccer club has 150 members. X is the set of strikers. Y is the set of defenders.

The letters a, b and c in the Venn diagram represent the number of members in

each of the subsets of X and Y. The letter d represents the numbers of members

who are neither strikers nor defenders. Given that n(X) = 90 and n(Y) = 68, find

(i) the value of b if d = 0, [1]

(ii) the value of d if b = c, [1]

(iii) the largest possible number of members who are neither strikers nor

defenders. [1]

2a i)

When d = 0, 90+ 68 – b = 150

b = 8

ii) When b = c, b = c = (68÷ 2) = 34

Therefore, a – b + b + c + d = 150

90 + 34 + d =150

d = 26

iii) Largest possible value of members who are neither strikers nor

defenders is when Y becomes a proper subset of X.

Therefore d + a = 150

Largest value of d = 150 – 90 = 60

X Y

a b

c

d


14

2 (b) The table below shows the number of pens own by each student in a class.

Number of pens 0 1 2 3 4 5

Number of students 1 3 x 7 5 2

(i) If the mode of the distribution is 3, find the range of values of x. [1]

(ii) If the median is 3, what is the largest value of x. [1]

(iii) If the mean is 2.9, find the value of x. [2]

54 + 2𝑥 = 52.2 + 2.9𝑥

2bi) 0 ≤ 𝑥 ≤ 6

ii) 1+3+𝑥+7+5+2+1

2 ≤ 7+5+2

19 + 𝑥 ≤ 28 𝑥 ≤ 9

Therefore, largest value of x is 9.

iii) 0+3+2𝑥+21+20+10

1+3+𝑥+7+5+2 = 2.9

54+2𝑥

18+𝑥 = 2.9

0.9𝑥 = 1.8

𝑥 = 2


15

3 (a) The diagram shows the arcs AB and EF

of a circle, centre O, with radius r cm.

BCX, OP and EDY are straight lines that

perpendicular to the line segment AXPYF.

Given that AX =XY =YF = 5cm,

CDYX is a rectangle with CX = 4 cm.

i) Show that r = 8.5 cm [1]

ii) Calculate the arc AB. [4]

Solution

(a) i) r = 𝐴𝑃2 + 𝑂𝑃2

= (5 + 2.5)2 + 42

= 8.5 cm (shown)

ii) COA = OAP

= tan – 1 4

7.5

= 28.07o

BOC = cos –1

2.5

8.5

= 72.89o

AOB = AOC + COB

= 28.07o + 72.89

o

= 100.96o

Arc AB = 100.96𝑜

360𝑜 2 (8.5)

= 14.97

15.0 cm

B E

C D

A X P Y F

O


16

3 (b) The cost price of a particular model of camera is $300. Shop A advertises its sale

as “usual price $450 now $360”. Shop B sells the same model of camera at $420

after 25% discount.

Calculate

i) the discount given by shop A as a percentage, [2]

ii) the original marked price of the camera at shop B, [1]

iii) the profit gained by shop B as a percentage. [2]

3b) i) Discount = 450−360

450 100%

= 20 %

ii) Original marked price = 420

75%

= $ 560

iii) Percentage of profit = 420−300

300 100%

= 40%

4 (a) It takes 3 hours and 10 min for 5 workers to paint a hall. Given that all the workers

work at the same rate, how long will 2 workers take to paint the same hall?

[Leave your answer in hours and minutes] [2]

4(a) No. of workers : No. of hours

= 5 : 3 1

6

= 5 2

5 :

19

6

5

2

= 2 : 95

12

= 2 : 475

60

= 2 : 7 hours 55 min


17

3(b) Mr Hai takes x minutes to carve a swan out of a block ice and Mr Li takes 25 more

minutes to carve an identical sculpture. If both of them work together, it only takes

30 minutes to complete carving an identical sculpture.

i) Form an equation involving x, and show that it simplifies to

x2 – 35x – 750 = 0. [2]

ii) Determine the time taken by Mr Li to carve the sculpture on his own.

[3]

4(b) 1

𝑥+

1

𝑥 + 25=

1

30

2𝑥+25

𝑥(𝑥 + 25)=

1

30

60 x + 750 = x2 + 25x

x2 – 35x – 750 = 0 (shown)

(x + 15)( x – 50) = 0

x = – 15 (rej) or x = 50

Time taken by Mr Li = 50 + 25

= 75 min

= 1 hr 15 min

4 (c) i) Solve – 4 < 2x – 5 13 [2]

ii) Write down all the prime numbers which satisfies this condition. [1]

4 (c) i) – 4 < 2x – 5 13

1 < 2x 18

½ < x 9

ii) Prime numbers are 2, 3, 5 & 7


18

5 The diagram shows a solid in which ABCD, DCFE and ABFE are rectangles. G is the

foot of perpendicular from D to AE. Given that AB = 5 cm, AD = 8 cm, DE = 5.5 cm

and angle DAE is 25°.

Find,

(i) the length of DG, [1]

(ii) angle DCG, [2]

(iii) the length of AE, [4]

5 (i) 𝐷𝐺

8 = sin 25

o

𝐷𝐺 = 8 × 𝑠𝑖𝑛25°

= 3.3809

≈ 𝟑.𝟑𝟖 𝒄𝒎

ii) In ΔDCG, tan DCG = 𝐷𝐺

𝐷𝐶

= 3.3809

5

∠𝐷𝐶𝐺 = 34.065𝑜

≈ 𝟑𝟒.𝟏°

C

A

B F

D

E G

5 cm 8 cm 5.5 cm

25o


19

iii) AG = 82 − 3.38092

= 7.250 cm

GE = 5.52 − 3.38092

= 4.338 cm

AE = AG + GE

= 7.250 + 4.338

= 11.588

11.6 cm

6 Two shops A and B sell the same brand of compact discs (CDs) and printing paper. Each

box of CDs is sold at $3.70 and $3.80 at Shop A and Shop B respectively. Each ream of

paper is sold at $6.80 and $6.50 at Shop A and Shop B respectively. The prices of the

items are represented by the matrix P,

𝑷 = $3.70 $3.80$6.80 $6.50

Andrew plans to buy 6 boxes of CDs and 2 reams of paper. Betty plans to buy 4 boxes of

CDs and 3 reams of paper. Charles plans to buy 5 boxes of CDs and 1 ream of paper.

(i) Represent the quantities of CDs and papers to be bought by the 3 persons as

a matrix Q. [1]

(ii) Evaluate the product, R = QP [2]

(iii) State what the elements of R represent. [1]

(iv) Evaluate T = (1 1 1) R. [1]

(v) State what the elements of T represent. [1]

(vi) If Andrew, Betty and Charles combine their purchase, which shop will give them a

better deal. [


20

6i) Q = 6 24 35 1

ii) 𝑹 = 6 24 35 1

3.70 3.806.80 6.50

= 6 × 3.7 + 2 × 6.8 6 × 3.8 + 2 × 6.54 × 3.7 + 3 × 6.8 4 × 3.8 + 3 × 6.55 × 3.7 + 1 × 6.8 5 × 3.8 + 1 × 6.5

= 35.80 35.8035.20 34.7025.30 25.50

iii) The elements in R represent the total cost of buying the CDs and paper at shop A

or shop B for Andrew, Betty and Charles respectively.

iv) 𝑻 = 1 1 1 35.80 35.8035.20 34.7025.3 25.5

= 96.30 96.00

v) The elements in T represent the total spent at Shop A or Shop B by Andrew, Betty

and Charles.

vi) Shop B

7 A bowl of sweets contains 12 chocolate nuggets, 13 mints and 15 toffees. Mr Chan takes

two sweets at random and eats them.

(a) Draw a tree diagram to represent all the possible outcome. [3]

(b) Calculate the probability, as a fraction in its simplest form, that

i) a chocolate nugget and a mint was taken. [1]

ii) none of the two sweets chosen were toffees. [1]

(c) A third sweet was randomly taken from the bowl. Calculate the probability that the

first two sweets drawn were of the same type and the third was a chocolate nugget.

[2]


21

7a)

bi) P(that to get chocolate and mint) =12

40×

13

39 +

13

40×

12

39 =

1

5

bii) P(none of the two sweets are toffees) = 12

40×

24

39+

13

40×

24

39 =

5

13

(c) P(two sweets of the same type and the third a chocolate nugget)

=12

40×

11

39 ×

10

38+

13

40×

12

39×

12

38+

15

40×

14

39×

12

38

= 119

1235

14

39

13

39

12

39

15

39

12

39

13

39

15

39

12

39

11

39

T

M

C

T

M

C

T

M

C

15

40

13

40

12

40

T

M

C


22

8. The diagram shows two regular octagons meet at points A and B.

(a) Find the size of each interior angle of the octagons.

[1]

(b) Find PAQ.

[2]

(c) Show that

𝐴𝐵

𝐴𝐶 = 1.85, correct to 3 significant figures.

[3]

(d) Hence or otherwise, find the area of the smaller octagon if the area of the bigger

octagon is 36.2 cm2.

[3]

8 (a) Interior angle = 180o – 360

o/8 = 135

o

(b) CAB = (180 – 135)

o/2 = 22.5

o

PAQ = 360o – 135

o – (135

o – 22.5

o)

= 112.5o

Q

C

B

P

A


23

9. The diagram shows a circle, centre O and radius 3 cm.

An inscribed triangle ABC is

drawn with A and B are on

the y-axis.

AC is extended to meet the

x-axis at D and OD meets

BC at E.

(a) Prove that triangles BOE and DOA are similar.

[3]

(b) If OE = 1 cm,

(i) find OD,

(ii) show that AC = 3 10

5 cm,

(iii) hence find the exact length of DC.

[2]

[3]

[2]

(c)

(d)

𝐴𝐵

sin 135𝑜 = 𝐴𝐶

sin 22.5𝑜

𝐴𝐵

𝐴𝐶 =

sin 135𝑜

sin 22.5𝑜

= 1.8477

1.85

The two octagons are similar objects,

area of the larger octagon

area of the smaller octagon = (

𝐴𝐵

𝐴𝐶 )2

= 1.852

area of the smaller octagon = 36.2

1.852

= 10.577

10.6 cm

y

xE

C

A

B

OD


24

Solution

9.

9

(a) BOE = DOA = 90o (axes are perpendicular)

ACB = 90o ( rt in a semicircle)

ABC = 180o – 90

o – BAC ( sum of )

= 90o – BAC

ADO = 180o – 90

o – BAC

= 90o – BAC

ABC = ADO

By AAA property, triangles BOE and DOA are similar.

(b) (i) Since triangles BOE and DOA are similar,

𝑂𝐷

𝑂𝐵=

𝑂𝐴

𝑂𝐸

𝑂𝐷

3=

3

1

OD = 9 cm

(ii) ABC is similar to EBO,

𝐴𝐶

𝐸𝑂=

𝐴𝐵

𝐸𝐵

𝐴𝐶

1=

6

1+9

AC = 6 10

10

= 3 10

5 cm

DC = AD – AC

= 9 + 81 – 3 10

5

= 3 10 – 3 10

5

= 12 10

5 cm


25

10

O, P , W and R are points on a horizontal plane. A vertical memorial tower, OT, is due

north of P. The angle of elevation of a man at P to the top of the tower is 15°. He walks a

distance of 200 metres to point W. Given that the bearing of W from P is 040° and the

bearing of O from W is 280°, calculate

i) ∠PWO, [2]

ii) OP, [2]

iii) the height of the tower OT, [2]

iv) the angle of depression of W from T. [2]

From W, the man walks 150 metres to a point R on a bearing of 𝜃° on the horizontal

plane. At R, the angle of elevation of the man to the top of the tower is 10°.

Calculate

v) angle OWR, [3]

vi) the value of 𝜃. [1]

Solution

N

N

R

T

O

P 150 m 200 m

15O

40O

280O

W


26

10 i) ∠𝑂𝑊𝑁 = 360 − 280 = 80°

∠𝑃𝑂𝑊 = 80° 𝑎𝑙𝑡 𝑎𝑛𝑔𝑙𝑒 ∠𝑃𝑊𝑂 = 180 − 80 − 40 = 60°

ii) 𝑂𝑃

sin 60 =

200

𝑠𝑖𝑛80

𝑂𝑃 = 𝑠𝑖𝑛60 ×200

𝑠𝑖𝑛80

= 175.87 ≈ 𝟏𝟕𝟔𝒎

iii) Height of Tower = 𝑂𝑃 𝑡𝑎𝑛15°

= 47.124 ≈ 𝟒𝟕.𝟏𝒎

iv) 𝑂𝑊

𝑠𝑖𝑛40𝑜 =200

sin 80𝑜

𝑂𝑊 = 200 × 𝑠𝑖𝑛40𝑜 ÷ sin 80𝑜

≈ 130.54

𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑒𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑓 𝑊 𝑓𝑟𝑜𝑚 𝑇 = 𝑡𝑎𝑛−1 47.124

130.54

= 19.849° ≈ 19.8°

v) 47.124

𝑂𝑅 = tan 10o

𝑂𝑅 = 47.124 ÷ 𝑡𝑎𝑛10°

267.25

cos∠𝑂𝑊𝑅 =130.542 + 1502 − 267.252

2 × 130.54 × 150

∠𝑂𝑊𝑅 = 144.50 𝑜 ≈ 𝟏𝟒𝟒.𝟓°

vi) Value of = 144.50o – 80

o

= 64.50o

64.5o


27

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

The variables x and y are connected by the equation

𝑦 = 3 2𝑥 − 4.

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

x –2 –1 0 1 2 3 4

y a –2.5 –1 2 8 20 44

(a) Find the value of a.

[1]

(b) Using a scale of 2 cm to 1 unit, draw a horizontal x–axis for –2 ≤ x ≤ 4.

Using a scale of 2 cm to 5 units, draw a vertical y–axis for –5 ≤ y ≤ 45.

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 solution of 2𝑥 = 9.

[2]

(d) By drawing a tangent, find the gradient of the curve at the point (3, 20).

[2]

(e) Use your graph to find the solutions of the equation 3 2𝑥 − 10𝑥 − 8 = 0.

[3]


28

11. (a) 𝑎 = 3 2−2 − 4 = –3.25

(b)

(c) 2𝑥 = 9

3(2𝑥) − 4 = 3 9 − 4

3(2𝑥) − 4 = 23

From the graph, when y=23, x= 3.25

(d) gradient of the curve at the point (3, 20) 16.7

(e) 3 2𝑥 − 10𝑥 + 8 = 0

3 2𝑥 = 10𝑥 + 8

3 2𝑥 − 4 = 10𝑥 + 4

Insert 𝑦 = 10𝑥 + 4,

From the graph,

x – 0.6, 4

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MATHEMATICS 4016 / 01 - WordPress.com · · 2016-01-04MATHEMATICS 4016 / 01 Paper 1 17 August 2010 ... 4 AHS Prelim 2010 Sec 4 Mathematics Paper 1 ... B and C are on level ground