MATHEMATICS PAPER 1 - M★th revisemathrevise.weebly.com/uploads/6/2/8/1/62815699/m_2000_01.pdf2000-CE-MATH 1−2 − 1 Page total FORMULAS FOR REFERENCE SPHERE Surface area = 4πr2

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Hong Kong Examinations AuthorityAll Rights Reserved 2000

2000-CE-MATH 1–1 Checker No.

HONG KONG EXAMINATIONS AUTHORITY

HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 2000

MATHEMATICS PAPER 1Question-Answer Book

8.30 am – 10.30 am (2 hours)This paper must be answered in English

1. Write your candidate number, centre number and seatnumber in the spaces provided on this cover.

2. This paper consists of THREE sections, A(1), A(2)and B. Each section carries 33 marks.

3. Attempt ALL questions in Sections A(1) and A(2),and any THREE questions in Section B. Write youranswers in the spaces provided in this Question-Answer Book. Supplementary answer sheets will besupplied on request. Write your Candidate Numberon each sheet and fasten them with string inside thisbook.

4. Write the question numbers of the questions you haveattempted in Section B in the spaces provided on thiscover.

5. Unless otherwise specified, all working must beclearly shown.

6. Unless otherwise specified, numerical answers shouldeither be exact or correct to 3 significant figures.

7. The diagrams in this paper are not necessarily drawnto scale.

Candidate Number

Centre Number

Seat Number

Marker’sUse Only

Examiner’sUse Only

Marker No. Examiner No.

Section AQuestion No. Marks Marks

1–2

3–4

5–6

7–8

9

10

11

12

13

14

Section ATotal

Checker’sUse Only Section A Total

Section BQuestion No.* Marks Marks

Section BTotal

*To be filled in by the candidate.

2000-CEMATHPAPER 1

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FORMULAS FOR REFERENCE

SPHERE Surface area = 24 rπ

Volume = 3

34 rπ

CYLINDER Area of curved surface = rhπ2

Volume = hr 2πCONE Area of curved surface = rlπ

Volume = hr 2

31 π

PRISM Volume = base area × height

PYRAMID Volume =31 × base area × height

SECTION A(1) (33 marks)Answer ALL questions in this section and write your answers in the spaces provided.

1. Let )32(95 −= FC . If C = 30 , find F . (3 marks)

2. Simplify 2

3

xyx −

and express your answer with positive indices. (3 marks)

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Go on to the next page

3. Find the area of the sector in Figure 1. (3 marks)

4. In Figure 2, find a and x . (4 marks)

75°

6 cm

Figure 1

Figure 2

a cm10 cm

7 cm

x°


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5. Solve 15

211 <− x and represent the solution in Figure 3. (4 marks)

6. Let f(x) = 7262 23 −−+ xxx . Find the remainder when f(x) is divided by 3+x . (3 marks)

−1 0−2−3−4−5 1 2 3 4 5

Figure 3


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7. In Figure 4, AD and BC are two parallel chords of the circle. AC and BD (4 marks)intersect at E . Find x and y .

8. On a map of scale 1 : 5 000 , the area of the passenger terminal of the Hong Kong (4 marks)International Airport is 220 cm2 . What is the actual area, in m2 , occupied by theterminal on the ground?

25°

Figure 4

E

D

CB

A

56°

y°

x°


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9. Let L be the straight line passing through (−4, 4) and (6, 0) . (5 marks)

(a) Find the slope of L .

(b) Find the equation of L .

(c) If L intersects the y-axis at C , find the coordinates of C .


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Section A(2) (33 marks)Answer ALL questions in this section and write your answers in the spaces provided.

10. (a) Solve 022910 2 =−+ xx . (2 marks)

(b) Mr. Tung deposited $ 10 000 in a bank on his 25th birthday and $ 9 000 on his 26th birthday.The interest was compounded yearly at r% p.a. , and the total amount he received on his 27thbirthday was $ 22 000 . Find r . (4 marks)


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11. Figure 5 shows the cumulative frequency polygon of the distribution of the lengths of 75 songs.

(a) Complete the tables below. (2 marks)

Length( t seconds)

Cumulativefrequency

Length( t seconds) Frequency

t ≤ 220 3 200 < t ≤ 220 3

t ≤ 240 16 220 < t ≤ 240 13

t ≤ 260 46 240 < t ≤ 260 30

t ≤ 280 260 < t ≤ 280

t ≤ 300 75 280 < t ≤ 300 9

(b) Find an estimate of the mean of the distribution. (2 marks)

(c) Estimate from the cumulative frequency polygon the median of the distribution. (1 mark)

(d) What percentage of these songs have lengths greater than 220 seconds but not greater than 260seconds? (2 marks)

10

20

30

40

50

70

60

80

Num

ber o

f son

gs

0210200 240230220 300280260250

Length (seconds)270 290

Figure 5

The cumulative frequency polygon of the distribution of the lengths of 75 songs


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12. A box contains nine hundred cards, each marked with a different 3-digit number from 100 to 999 . Acard is drawn randomly from the box.

(a) Find the probability that two of the digits of the number drawn are zero. (2 marks)

(b) Find the probability that none of the digits of the number drawn is zero. (2 marks)

(c) Find the probability that exactly one of the digits of the number drawn is zero. (2 marks)


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13. In Figure 6, ABCDE is a regular pentagonand CDFG is a square. BG produced meetsAE at P .

(a) Find ∠ BCG , ∠ ABP and ∠ APB .(5 marks)

(b) Using the fact that APB

ABABP

AP∠

=∠ sinsin

, or otherwise, determine which line segment, AP

or PE , is longer. (3 marks)

A

PG F E

DC

B

Figure 6


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14. An auditorium has 50 rows of seats.All seats are numbered in numericalorder from the first row to the last row,and from left to right, as shown inFigure 7. The first row has 20 seats.The second row has 22 seats. Eachsucceeding row has 2 more seats thanthe previous one.

(a) How many seats are there in the last row? (2 marks)

(b) Find the total number of seats in the first n rows.

Hence determine in which row the seat numbered 2000 is located.(4 marks)

12 19

20

!

!

!

1st row

2122 41

42

4344 65

66

2nd row

3rd row

"

Figure 7


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SECTION B (33 marks)Answer any THREE questions in this section and write your answers in the spaces provided.Each question carries 11 marks.

15. A company produces two brands, A and B , of mixed nuts by putting peanuts and almonds together. Apacket of brand A mixed nuts contains 40 g of peanuts and 10 g of almonds. A packet of brand Bmixed nuts contains 30 g of peanuts and 25 g of almonds. The company has 2 400 kg of peanuts,1 200 kg of almonds and 70 carton boxes. Each carton box can pack 1 000 brand A packets or 800brand B packets.

The profits generated by a box of brand A mixed nuts and a box of brand B mixed nuts are $ 800 and$ 1 000 respectively. Suppose x boxes of brand A mixed nuts and y boxes of brand B mixed nuts areproduced.

(a) Using the graph paper in Figure 8, find x and y so that the profit is the greatest.(8 marks)

(b) If the number of boxes of brand B mixed nuts is to be smaller than the number of boxes of brandA mixed nuts, find the greatest profit. (3 marks)

x

y

0 10 20 30 40 50 60 70

70

60

50

40

30

20

10

Figure 8


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16. In Figure 9, C is the centre of the circle PQS . ORand OP are tangent to the circle at S and Prespectively. OCQ is a straight line and ∠ QOP = 30° .

(a) Show that ∠ PQO = 30° . (3 marks)

(b) Suppose OPQR is a cyclic quadrilateral.

(i) Show that RQ is tangent to circle PQSat Q .

(ii) A rectangular coordinate system isintroduced in Figure 9 so that thecoordinates of O and C are (0, 0) and(6, 8) respectively. Find the equation ofQR .

(8 marks)

C

PO

R

QS

Figure 9

30°


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17. Figure 10 shows a circle with centreO and radius 10 m on a verticalwall which stands on the horizontalground. A , B and C are threepoints on the circumference of thecircle such that A is vertically belowO , ∠ AOB = 90° and ∠ AOC = 20° .A laser emitter D on the groundshoots a laser beam at B . The laserbeam then sweeps through an angleof 30° to shoot at A . The angles ofelevation of B and A from D are60° and 30° respectively.

(a) Let A be h m above the ground.

(i) Express AD and BD in terms of h .

(ii) Find h .(7 marks)

(b) Another laser emitter E on the ground shoots a laser beam at A with angle of elevation 25° .The laser beam then sweeps through an angle of 5° to shoot at C . Find ∠ ACE .

(4 marks)

20°

Verticalwall

30°

30°60°

A

B

C

D

O

Figure 10


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P

Q

Figure 11.1 Figure 11.2

h cm

18. Figure 11.1 shows a solid hemisphere of radius 10 cm . It is cut into two portions, P and Q , along aplane parallel to its base. The height and volume of P are h cm and V cm3 respectively.

It is known that V is the sum of two parts. One part varies directly as 2h and the other part varies

directly as 3h . π3

29=V when 1=h and π81=V when 3=h .

(a) Find V in terms of h and π . (3 marks)

(b) A solid congruent to P is carved away from the top of Q to form a container as shown inFigure 11.2 .

(i) Find the surface area of the container (excluding the base).

(ii) It is known that the volume of the container is π3

4001 cm3 . Show that 030030 23 =+− hh .

Using the graph in Figure 11.3 and a suitable method, find the value of h correct to 2decimal places.

(8 marks)

Figure 11.3

The graph of 23 30xxy −= for 0 ≤ x ≤ 5

x

−100

−200

−300

−400

−500

−600

y0

21 3 4 5


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END OF PAPER


2000

Mathematics 1Section A(1)

1. 86

2. 5xy

3. 23.6 cm2

4. a = 51

x ≈ 45.6

5. 3>x

6. −1

7. x = 25

y = 74

8. 550 000 m2

9. (a)52−

(b) 01252 =−+ yx

(c) )5

12,0(


Section A(2)

10. (a)1011or 2−=x

(b) 22000%)1(9000%)1(10000 2 =+++ rr

022%)1(9%)1(10 2 =−+++ rr

From (a), 1.1%1 =+ rr = 10

11. (a) Missing value in 1st table = 66Missing value in 2nd table = 20

(b) An estimate of the mean

= 75

92902027030250132303210 ×+×+×+×+× seconds

≈ 255 seconds

(c) Median ≈ 254 seconds

(d) Percentage required = %10075

3013 ×+ ≈ 57.3%

12. (a) Probability required = 101

101 × =

1001

(b) Probability required = 109

109 × =

10081

(c) Probability required = 10081

10011 −− =

509


13. (a) Size of each interior angle of the pentagon = 5

180)25( °×− = 108°

∠BCG = 108° − 90° = 18°

∠CBG = 2

18180 °−° = 81°

∠ABP = 108° − 81° = 27°∠APB = 180° − 27° −108° = 45°

(b)°

=° 45sin27sin

ABAP

∴ AP = AB°°

45sin27sin = AE

°°

45sin27sin ≈ AE642.0

PE ≈ (1−0.642)AE ≈ 0.358 AE∴ AP is longer than PE .

14. (a) Number of seats in the last row = )150(220 −+ = 118

(b) Total number of seats in the first n rows = )]1(2202[2

−+× nn = nn 192 +

If 2000192 =+ nn , then02000192 =−+ nn

2)2000(41919 2 −−±−

=n

n ≈ 36.2 or −55.2

∴ The seat numbered 2000 can be found in the 37th row.


Section B

15. (a) x and y satisfy the following conditions:2400000)30(800)40(1000 ≤+ yx or 30035 ≤+ yx1200000)25(800)10(1000 ≤+ yx or 1202 ≤+ yx

70≤+ yxx , y are non-negative integers

x

y

0 10 20 30 40 50 60 70

70

60

50

40

30

20

10

5x + 3y = 300x + y = 70

x + 2y = 120

x = y


Let $P(x, y) be the profit generated by x boxes of brand A mixednuts and y boxes of brand B mixed nuts. Then P(x, y) = 800x + 1000y

= 200(4x + 5y)

By drawing parallel lines of 4x + 5y = 0 ,P(x, y) attains its maximum at (20, 50).∴ The profit is the greatest when x = 20 and y = 50 .

(b) In addition to the conditions in (a), x , y should also satisfy xy < .By considering lines parallel to 4x + 5y = 0P(x, y) attains its maximum at (36, 34).∴ The greatest profit is $62800.


16. (a) Join CP .∠OPC = 90° (tangent ⊥ radius)∠PCO = 180° − 90° − 30° = 60° (∠ sum of ∆)

∠PQO = PCO∠21 = 30° (∠ at centre twice ∠ at circumference)

(b) (i) ∠ROQ = ∠QOP = 30° (tangents from ext. pt.)∠PQO = 30° (proved)

∠RQP + ∠POR = 180° (opp. ∠s of cyclic quad.)∴ ∠CQR = 180° − 3×30° = 90°Hence RQ is tangent to circle PQS at Q. (conv. of tangent ⊥ radius)

(b) (ii) Slope of OC = 34

∴ Slope of QR = 43−

OC = 22 86 + = 10CQ = CP = OC sin30° = 5

Let the coordinates of Q be (x, y).OC : CQ = 10 : 5 = 2 : 1

∴ 63

)0(12=

+x and 83

)0(12=

+y

x = 9 and y = 12

Hence the equation of QR is

43

912

−=−−

xy

07543 =−+ yx


17. (a) (i) AD = °30sin

h m = 2h m

BD = °

+60sin10h m = )10(

32 +h m = )10(3

32 +h m

(ii) )1010( 222 +=AB m2

By cosine law,ADBDBADDBADAB ∠−+= cos))((2222

°

°+

°−

°++

°= 30cos

60sin10

30sin2

60sin10

30sin200

22 hhhh

)10(4)10(344200 22 +−++= hhhh

050102 =−− hhh ≈ 13.660 or −3.660h ≈ 13.7 or −3.66 (rejected)

(b) AC = )10sin10(2 ° m ≈ 3.47296 m

AE = °25sin

h m ≈ 32.3 m

By sine law, sin∠ACE = AC

AE °5sin

≈ °°

°25sin10sin20

5sinh

≈ 0.8112∴ ∠ACE = 54.2° or 126°


18. (a) Let 32 bhahV += where a , b are non-zero constants.

+=

+=

ba

ba

279813

29

π

πor

=+

=+

)2(..........93)1(..........

329

ππ

baba

(2) − (1) gives π322 −=b

Hence 3π−=b and π10=a

∴ 32

310 hhV ππ −=

(b) (i) Surface area = 2102 ×π cm2 ≈ 628 cm2

(ii) Volume of hemisphere = 31032 ×π cm3

∴ ππ3

140021032 3 =−× V

ππππ3

1400)3

10(21032 323 =−−× hh

0)700301000(32 32 =−+− hhπ

030030 23 =+− hh

From the graph in Figure 11.3, 3.3 < h < 3.4

Let 30030)f( 23 +−= hhh , then f(3.3) > 0 and f(3.4) < 0.Using the method of bisection,

Interval mid-value (m) f(m)3.3 < h < 3.4 3.35 +ve (0.9204)

3.35 < h < 3.4 3.375 −ve (−3.2754)3.35 < h < 3.375 3.363 −ve (−1.2583)3.35 < h < 3.363 3.357 −ve (−0.2519)3.35 < h < 3.357 3.354 +ve (0.2507)

3.354 < h < 3.357 3.356 −ve (−0.0843)3.354 < h < 3.356 3.355 +ve (0.0832)

∴ 3.355 < h < 3.356h ≈ 3.36 (correct to 2 decimal places)


Documents

MATHEMATICS PAPER 1 - M★th revisemathrevise.weebly.com/uploads/6/2/8/1/62815699/m_2000_01.pdf2000-CE-MATH 1−2 − 1 Page total FORMULAS FOR REFERENCE SPHERE Surface area = 4πr2