2000-CE MATH Marker s Examiner s Use Only Use Only MATHEMATICS PAPER 1 Question-Answer Book Checker s Use Only

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1 000-CE MATH PAPER 1 HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 000 MATHEMATICS PAPER 1 Question-Answer Book 8.0 am 10.0 am ( hours) This paper must be answered in English Candidate Number Centre Number Seat Number Section A Question No. 1 4 Marker s Use Only Marker No. Marks Examiner s Use Only Examiner No. Marks 1. Write your candidate number, centre number and seat number in the spaces provided on this cover.. This paper consists of THREE sections, A(1), A() and B. Each section carries marks.. Attempt ALL questions in Sections A(1) and A(), and any THREE questions in Section B. Write your answers in the spaces provided in this Question- Answer Book. Supplementary answer sheets will be supplied on request. Write your Candidate Number on each sheet and fasten them with string inside this book. 4. Write the question numbers of the questions you have attempted in Section B in the spaces provided on this cover Section A Total 5. Unless otherwise specified, all working must be clearly shown. Checker s Use Only Section A Total 6. Unless otherwise specified, numerical answers should either be exact or correct to significant figures. 7. The diagrams in this paper are not necessarily drawn to scale. Section B Question No.* Marks Marks Section B Total *To be filled in by the candidate. 香港考試局保留版權 Hong Kong Examinations Authority All Rights Reserved 000 Checker s Use Only Section B Total 000-CE-MATH 1 1 Checker No.

2 Page total FORMULAS FOR REFERENCE SPHERE Surface area = 4π r 4 Volume = π r CYLINDER Area of curved surface = π rh Volume = π r h CONE Area of curved surface = π rl Volume = 1 π r h PRISM Volume = base area height PYRAMID Volume = 1 base area height SECTION A(1) ( marks) Answer ALL questions in this section and write your answers in the spaces provided Let C = ( F ). If C = 0, find F. ( marks) 9. Simplify x y and express your answer with positive indices. ( marks) x 000-CE-MATH 1 保留版權 All Rights Reserved 000 1

3 . Find the area of the sector in Figure 1. ( marks) Page total 6 cm 75 Figure 1 4. In Figure, find a and x. (4 marks) 10 cm a cm x 7 cm Figure 000-CE-MATH 1 保留版權 All Rights Reserved 000 Go on to the next page

4 11 x 5. Solve < 1 5 and represent the solution in Figure. Page total (4 marks) Figure 6. Let f(x) = x + 6x x 7. Find the remainder when f(x) is divided by x +. ( marks) 000-CE-MATH 1 4 保留版權 All Rights Reserved 000

5 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. Page total A y 56 B 5 E x D C Figure 4 8. On a map of scale 1 : 5 000, the area of the passenger terminal of the Hong Kong (4 marks) International Airport is 0 cm. What is the actual area, in m, occupied by the terminal on the ground? 000-CE-MATH 1 5 保留版權 All Rights Reserved Go on to the next page

6 Page total 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. 000-CE-MATH 1 6 保留版權 All Rights Reserved 000 5

7 Page total Section A() ( marks) Answer ALL questions in this section and write your answers in the spaces provided. 10. (a) Solve 10x + 9x = 0. ( marks) (b) Mr. Tung deposited $ in a bank on his 5th birthday and $ on his 6th birthday. The interest was compounded yearly at r% p.a., and the total amount he received on his 7th birthday was $ 000. Find r. (4 marks) 000-CE-MATH 1 7 保留版權 All Rights Reserved Go on to the next page

8 Page total 11. Figure 5 shows the cumulative frequency polygon of the distribution of the lengths of 75 songs. The cumulative frequency polygon of the distribution of the lengths of 75 songs Number of songs Length (seconds) (a) Complete the tables below. ( marks) Length ( t seconds) Cumulative frequency Figure 5 Length ( t seconds) Frequency t 0 00 < t 0 t < t 40 1 t < t 60 0 t < t 80 t < t 00 9 (b) Find an estimate of the mean of the distribution. ( 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 0 seconds but not greater than 60 seconds? ( marks) 000-CE-MATH 1 8 保留版權 All Rights Reserved 000 7

9 1. A box contains nine hundred cards, each marked with a different -digit number from 100 to 999. A card is drawn randomly from the box. (a) Find the probability that two of the digits of the number drawn are zero. ( marks) Page total (b) Find the probability that none of the digits of the number drawn is zero. ( marks) (c) Find the probability that exactly one of the digits of the number drawn is zero. ( marks) 000-CE-MATH 1 9 保留版權 All Rights Reserved Go on to the next page

10 Page total 1. In Figure 6, ABCDE is a regular pentagon and CDFG is a square. BG produced meets AE at P. B G A P F E (a) Find BCG, ABP and APB. (5 marks) C Figure 6 D (b) Using the fact that or PE, is longer. AP AB = sin ABP sin APB, or otherwise, determine which line segment, AP ( marks) 000-CE-MATH 1 10 保留版權 All Rights Reserved 000 9

11 14. An auditorium has 50 rows of seats. " All seats are numbered in numerical! order from the first row to the last row, and from left to right, as shown in! Figure 7. The first row has 0 seats The second row has seats. Each 66 rd row! 41 succeeding row has more seats than 1 the previous one. 1 4 nd row st row Figure 7 (a) How many seats are there in the last row? ( marks) Page total (b) Find the total number of seats in the first n rows. Hence determine in which row the seat numbered 000 is located. (4 marks) 000-CE-MATH 1 11 保留版權 All Rights Reserved Go on to the next page

12 Page total SECTION B ( 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. A packet of brand A mixed nuts contains 40 g of peanuts and 10 g of almonds. A packet of brand B mixed nuts contains 0 g of peanuts and 5 g of almonds. The company has 400 kg of peanuts, 1 00 kg of almonds and 70 carton boxes. Each carton box can pack brand A packets or 800 brand B packets. The profits generated by a box of brand A mixed nuts and a box of brand B mixed nuts are $ 800 and $ respectively. Suppose x boxes of brand A mixed nuts and y boxes of brand B mixed nuts are produced. (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 brand A mixed nuts, find the greatest profit. ( marks) y Figure 8 x 000-CE-MATH 1 1 保留版權 All Rights Reserved

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14 Page total 16. In Figure 9, C is the centre of the circle PQS. OR and OP are tangent to the circle at S and P respectively. OCQ is a straight line and QOP = 0. R (a) Show that PQO = 0. ( marks) (b) Suppose OPQR is a cyclic quadrilateral. (i) Show that RQ is tangent to circle PQS at Q. S C Q (ii) A rectangular coordinate system is introduced in Figure 9 so that the coordinates of O and C are (0, 0) and (6, 8) respectively. Find the equation of QR. (8 marks) O 0 P Figure CE-MATH 1 14 保留版權 All Rights Reserved 000 1

15 Page total 000-CE-MATH 1 15 保留版權 All Rights Reserved Go on to the next page

16 Page total 17. Figure 10 shows a circle with centre O and radius 10 m on a vertical wall which stands on the horizontal ground. A, B and C are three points on the circumference of the circle such that A is vertically below O, AOB = 90 and AOC = 0. A laser emitter D on the ground shoots a laser beam at B. The laser beam then sweeps through an angle of 0 to shoot at A. The angles of elevation of B and A from D are 60 and 0 respectively. Vertical wall C O 0 A B (a) Let A be h m above the ground. Figure 10 D (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 5. The laser beam then sweeps through an angle of 5 to shoot at C. Find ACE. (4 marks) 000-CE-MATH 1 16 保留版權 All Rights Reserved

17 Page total 000-CE-MATH 1 17 保留版權 All Rights Reserved Go on to the next page

18 18. Figure 11.1 shows a solid hemisphere of radius 10 cm. It is cut into two portions, P and Q, along a plane parallel to its base. The height and volume of P are h cm and V cm respectively. Page total P h cm Q Figure 11.1 Figure 11. It is known that V is the sum of two parts. One part varies directly as directly as h. V = 9 π when h = 1 and V = 81π when h =. h and the other part varies (a) Find V in terms of h and π. ( marks) (b) A solid congruent to P is carved away from the top of Q to form a container as shown in Figure 11.. (i) Find the surface area of the container (excluding the base) (ii) It is known that the volume of the container is π cm. Show that h 0h =. Using the graph in Figure 11. and a suitable method, find the value of h correct to decimal places. (8 marks) The graph of y = x 0x for 0 x 5 y x Figure CE-MATH 1 18 保留版權 All Rights Reserved

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20 Page total END OF PAPER 000-CE-MATH 1 0 保留版權 All Rights Reserved

21 000 Mathematics 1 Section A(1) y 5 x..6 cm 4. a = 51 x x > x = 5 y = m 9. (a) 5 (b) x + 5y 1 = 0 (c) 1 (0, ) 5 保留版權 All Rights Reserved 000

22 Section A() 10. (a) x = or (b) 10000(1 + r%) (1 + r%) = (1 + r%) + 9(1 + r%) = 0 From (a), 1 + r% = 1. 1 r = (a) Missing value in 1st table = 66 Missing value in nd table = 0 (b) An estimate of the mean = seconds seconds (c) Median 54 seconds (d) Percentage required = % % 1. (a) Probability required = (b) Probability required = = = (c) Probability required = 1 = 保留版權 All Rights Reserved 000

23 1. (a) ( 5 ) 180 Size of each interior angle of the pentagon = 5 BCG = = CBG = = 81 ABP = = 7 APB = = 45 = 108 (b) AP AB = sin 7 sin 45 sin 7 sin 7 AP = AB = AE sin 45 sin 45 PE (1 0.64)AE 0.58 AE AP is longer than PE. 0.64AE 14. (a) Number of seats in the last row = 0 + (50 1) = 118 n (b) Total number of seats in the first n rows = [ 0 + ( n 1)] If n + 19n = 000, then n + 19n 000 = 0 19 ± 19 4( 000) n = n 6. or 55. = n + 19n The seat numbered 000 can be found in the 7th row. 保留版權 All Rights Reserved 000

24 Section B 15. (a) x and y satisfy the following conditions: 1000(40x) + 800(0y) or 5x + y (10x) + 800(5y) or x + y 10 x + y 70 x, y are non-negative integers y 70 x + y = 70 5x + y = x = y x + y = x 保留版權 All Rights Reserved 000

25 Let $P(x, y) be the profit generated by x boxes of brand A mixed nuts and y boxes of brand B mixed nuts. Then P(x, y) = 800x y = 00(4x + 5y) By drawing parallel lines of 4x + 5y = 0, P(x, y) attains its maximum at (0, 50). The profit is the greatest when x = 0 and y = 50. (b) In addition to the conditions in (a), x, y should also satisfy y < x. By considering lines parallel to 4x + 5y = 0 P(x, y) attains its maximum at (6, 4). The greatest profit is $6800. 保留版權 All Rights Reserved 000

26 16. (a) Join CP. OPC = 90 (tangent radius) PCO = = 60 ( sum of ) 1 PQO = PCO = 0 ( at centre twice at circumference) (b) (i) ROQ = QOP = 0 (tangents from ext. pt.) PQO = 0 (proved) RQP + POR = 180 (opp. s of cyclic quad.) CQR = = 90 Hence RQ is tangent to circle PQS at Q. (conv. of tangent radius) (b) (ii) Slope of OC = 4 Slope of QR = 4 OC = = 10 CQ = CP = OC sin0 = 5 Let the coordinates of Q be (x, y). OC : CQ = 10 : 5 = : 1 x + 1(0) y + 1(0) = 6 and = 8 x = 9 and y = 1 Hence the equation of QR is y 1 = x 9 4 x + 4y 75 = 0 保留版權 All Rights Reserved 000

27 17. (a) (i) AD = BD = h sin 0 h + 10 sin 60 m = h m m = ( h + 10) m = ( h + 10) m (ii) AB = ( ) m By cosine law, AB = AD + DB ( AD)( DB) cos ADB = h h + 10 h h cos sin 0 sin 60 sin 0 sin = 4h + ( h + 10) 4h( h + 10) h 10h 50 = 0 h or.660 h 1.7 or.66 (rejected) (b) AC = (10sin10 ) m.4796 m h AE = m. m sin 5 AE sin 5 By sine law, sin ACE = AC h sin 5 0sin10 sin ACE = 54. or 16 保留版權 All Rights Reserved 000

28 18. (a) Let V = ah + bh where a, b are non-zero constants. 9 π = a + b a + b = 9 π...(1) or 81π = 9a + 7b a + b = 9π...() () (1) gives b = π π Hence b = and a = 10π π V = 10πh h (b) (i) Surface area = π 10 cm 68 cm (ii) Volume of hemisphere = π 10 cm π 10 V = 1400 π π 1400 π 10 (10πh h ) = π π (1000 0h + h h 0h + 00 = 0 700) = 0 From the graph in Figure 11.,. < h <.4 Let f( h ) = h 0h + 00, then f(.) > 0 and f(.4) < 0. Using the method of bisection, Interval mid-value (m) f(m). < h <.4.5 +ve (0.904).5 < h <.4.75 ve (.754).5 < h <.75.6 ve ( 1.58).5 < h <.6.57 ve ( 0.519).5 < h < ve (0.507).54 < h < ve ( 0.084).54 < h < ve (0.08).55 < h <.56 h.6 (correct to decimal places) 保留版權 All Rights Reserved 000