HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
只限教師參閱FOR TEACHERS’ USE ONLY香港考試及評核局HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY香港中學文憑考試HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION練習卷PRACTICE PAPER數學MATHEMATICS必修部分COMPULSORY PART試卷一PAPER 1評卷參考（暫定稿）PROVISIONAL MARKING ��This marking scheme has been prepared by the Hong Kong Examinations andAssessment Authority for teachers’ reference. Teachers should remind theirstudents NOT to regard this marking scheme as a set of model answers. Ourexaminations emphasise the testing of understanding, the practical application ofknowledge and the use of processing skills. Hence the use of model answers, oranything else which encourages rote memorisation, will not help students toimprove their learning nor develop their abilities in addressing and solvingproblems. The Authority is counting on the co-operation of teachers in this regard. 香港考試及評核局保留版權Hong Kong Examinations and Assessment AuthorityAll Rights Reserved 2012PP-DSE-MATH-CP 1 1只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYHong Kong Diploma of Secondary Education ExaminationMathematics Compulsory Part Paper 1General Marking Instructions1.This marking scheme is the preliminary version before the normal standardisation process and some revisionsmay be necessary after actual samples of performance have been collected and scrutinised by the HKEAA.Teachers are strongly advised to conduct their own internal standardisation procedures before applying themarking schemes. After standardisation, teachers should adhere to the marking scheme to ensure a uniformstandard of marking within the school.2.It is very important that all teachers should adhere as closely as possible to the marking scheme. In manycases, however, students will have obtained a correct answer by an alternative method not specified in themarking scheme. In general, a correct answer merits all the marks allocated to that part, unless a particularmethod has been specified in the question. Teachers should be patient in marking alternative solutions notspecified in the marking scheme.3.In the marking scheme, marks are classified into the following three categories:‘M’ marksawarded for correct methods being used;‘A’ marksawarded for the accuracy of the answers;Marks without ‘M’ or ‘A’awarded for correctly completing a proof or arrivingat an answer given in a question.In a question consisting of several parts each depending on the previous parts, ‘M’ marks should be awardedto steps or methods correctly deduced from previous answers, even if these answers are erroneous. However,‘A’ marks for the corresponding answers should NOT be awarded (unless otherwise specified).4.For the convenience of teachers, the marking scheme was written as detailed as possible. However, it is stilllikely that students would not present their solution in the same explicit manner, e.g. some steps would eitherbe omitted or stated implicitly. In such cases, teachers should exercise their discretion in marking students’work. In general, marks for a certain step should be awarded if students’ solution indicated that the relevantconcept/technique had been used.5.Use of notation different from those in the marking scheme should not be penalized.6.In marking students’ work, the benefit of doubt should be given in the students’ favour.7.Marks may be deducted for wrong units (u) or poor presentation (pp).a.The symbol u-1 should be used to denote 1 mark deducted for u. At most deduct 1 mark for u ineach of Section A(1) and Section A(2). Do not deduct any marks for u in Section B.b.The symbol pp-1 should be used to denote 1 mark deducted for pp. At most deduct 1 mark for ppin each of Section A(1) and Section A(2). Do not deduct any marks for pp in Section B.At most deduct 1 mark in each of Section A(1) and Section A(2).In any case, do not deduct any marks in those steps where students could not score any marks.c.d.8.In the marking scheme, ‘r.t.’ stands for ‘accepting answers which can be rounded off to’ and ‘f.t.’ stands for‘follow through’. Steps which can be skipped are shaded whereas alternative answers are enclosed withrectangles . All fractional answers must be simplified.PP-DSE-MATH-CP 1 2只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱SolutionFOR TEACHERS’ USE ONLYMarksRemarks(m 5 n 2 ) 61.m 4 n 3m 30 n 12 m 4 n 3m30 4n12 3m 26 1Mfor (ab) p a pb p or (a p )q a pq1Mforapap1p q aor aq aq paq1An9----------(3)2.5 b 3b1 a5 b 3b(1 a)5 b 3b 3ab3ab 2b 52b 5a 3b5 b 3b1 a5 b 3b(1 a)5 b3b3b (5 b)a 3b2b 5a 3ba 1 1Mfor 3b(1 a)1Mfor putting a on one side1Aor equivalent1Mfor 3b(1 a)1Mfor putting a on one side1Aor equivalent----------(3)3.(a)9 x 2 42 xy 49 y 2 (3x 7 y ) 2(b)1Aor equivalent1Mfor using (a)9 x 2 42 xy 49 y 2 6 x 14 y (3x 7 y ) 2 6 x 14 y2 (3x 7 y ) 2(3x 7 y ) (3x 7 y )(3 x 7 y 2)1Aor equivalent----------(3)PP-DSE-MATH-CP 1 3只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution4.MarksLet x be the marked price of the chair.x (1 20%) 360 (1 30%)360(1.3)x 0.8x 585Thus, the marked price of the chair is 585 .pp 1 for undefined symbol1M 1M 1A1M for x (1 20%) 1M for 360 (1 30%)1Au 1 for missing unitThe marked price of the chair360 (1 30%) 1 20% 5855.Remarks1M 1M 1A1M for 360 (1 30%) 1M for dividing by (1 20%)1Au 1 for missing unit----------(4)Let x litres and y litres be the capacities of a bottle and a cup respectively. x 4 y 3 7 x 9 y 11 pp 1 for undefined symbol1A 1A 3x So, we have 7 x 9 11 . 4 4Solving, we have x .5Thus, the capacity of a bottle is1Mfor getting a linear equation in x or y only1A0.84litre.5Let x litres be the capacity of a bottle. 3x 7 x 9 11 4 4Solving, we have x .54Thus, the capacity of a bottle islitre.5u 1 for missing unitpp 1 for undefined symbol1A 1M 1A1A1A for y 3x 1M for 7 x 9 y 1140.8u 1 for missing unit----------(4)PP-DSE-MATH-CP 1 4只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱Solution6.(a)(b) AOC 337 157 180 Thus, A , O and C are collinear.MarksRemarks1Mfor considering AOC1Af.t.Note that BO AC .The area of ABC1(13 15)(14)2 196 7.FOR TEACHERS’ USE ONLYNote that BCD 90 .Also note that CBD 180 90 36 54 .Further note that BAC BDC 36 .Since AB AC , we have ACB ABC .180 36 .2Therefore, we have ABC 72 .1M1A----------(4)1A1MSo, we have ABC 1M ABD ABC CBD 72 54 18 1ANote that BAC BDC 36 .Since AB AC , we have ACB ABC .180 36 .2Therefore, we have ACB 72 .Also note that BCD 90 .So, we have ACB u 1 for missing unit1M1M1A ACD 90 72 18 ABD ACD 18 1Au 1 for missing unit----------(4)PP-DSE-MATH-CP 1 5只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution8.(a)(b)MarksRemarksThe coordinates of A′ (3 , 4)1App–1 for missing ‘(’ or ‘)’The coordinates of B ′ (5 , 2)1App–1 for missing ‘(’ or ‘)’Let ( x , y ) be the coordinates of P .( x 3) 2 ( y 4) 2 ( x 5) 2 ( y ( 2)) 21M 1Ax 2 6 x 9 y 2 8 y 16 x 2 10 x 25 y 2 4 y 44 x 12 y 4 0Thus, the required equation is x 3 y 1 0 .1Aor equivalentThe coordinates of the mid-point of A′B ′ 3 5 4 ( 2) , 2 2 (4 , 1)1MThe slope of A′B′4 ( 2) 3 5 31A1So, the required equation is y 1 ( x 4) .3Thus, the required equation is x 3 y 1 0 .9.(a)1Aor equivalent----------(5)The least possible value of the inter-quartile range of the distribution 5 5 or 2 2 01M1Aeither oneThe greatest possible value of the inter-quartile range of the distribution 5 2 3(b)1ASince r 9 and the median of the distribution is 3 ,we have 9 8 12 s .Therefore, we have s 5 .So, we have s 1 , 2 , 3 or 4 .Thus, there are 4 possible values of s .1M1Af.t.----------(5)PP-DSE-MATH-CP 1 6只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution10. (a)MarksNote that when f ( x) is divided by x 1 , the remainder is 4 .Remarks1Mcan be absorbed1Mfor ( x 1)(6 x 2 17 x 2) rf ( x) ( x 1)(6 x 2 17 x 2) 4 6 x 3 11x 2 19 x 6f ( 3) 6( 3) 3 11( 3) 2 19( 3) 6 0(b)1A----------(3)f ( x) ( x 3)(6 x 2 7 x 2) ( x 3)(2 x 1)(3 x 2)11. (a)1M 1A1A----------(3)Let C a bx 2 , where a and b are non-zero constants.22So, we have a (20 )b 42 and a (120 )b 112 .Solving, we have a 40 and b 1.200The required cost1 40 (50 2 )200 52.5(b)40 1M for ( x 3)(ax 2 bx c)1 2x 582001A1Mfor either substitution1Afor both correct1Au 1 for missing unit----------(4)1Mx 2 3 600x 60Thus, the required length is 60 cm .1Au 1 for having unit----------(2)PP-DSE-MATH-CP 1 7只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution12. (a)MarksThe required duration 63 32 31 minutesRemarks1M1Au 1 for missing unit----------(2)(b) Suppose Ada and Billy meet at a place which is at a distance of x kmfrom town P .x12 78 120x 7 .8Thus, Ada and Billy meet at a place which is at a distance of 7.8 kmfrom town P .1M 1A1M for ratio 78 : 1201A----------(3)(c)The average speed of Ada12 2 6 km/hThe average speed of Billy16 2 2 7 km/hNote that 7 6 .Thus, Billy runs faster.1Meither one1ADuring the period, Ada runs 12 km while Billy runs 14 km .Note that 14 12 .So, the average speed of Billy is higher than that of Ada.Thus, Billy runs faster.f.t.1M1Af.t.----------(2)PP-DSE-MATH-CP 1 8只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution13. (a)MarksLet n be the number of students in the group.6 3 n 20n 40pp 1 for undefined symbol1Mk 40 6 11 5 10 8(b) (i)Remarks1M1A----------(3)The required angle5 (360 )40 45 1M1A(ii) Let m be the number of new students.Assume that the angle of the sector representing that the mostfavourite fruit is orange will be doubled.5 m (45)(2) 40 m36020 4m 40 m3m 20Since 20 is not a multiple of 3 , the angle of the sector representingthat the most favourite fruit is orange will not be doubled.u 1 for missing unitpp 1 for undefined symbol1Mfor considering1Af.t.----------(4)PP-DSE-MATH-CP 1 9只限教師參閱FOR TEACHERS’ USE ONLY5 mn m
只限教師參閱FOR TEACHERS’ USE ONLYSolution14. (a)Marks2A----------(2) BCD OA D(b) (i)RemarksLet (0 , h) be the coordinates of C .1M216 CD By (b), we have . AD45 1Mfor using similarity1Mfor either AD or CD2 12 h 16 22 45 6 12 h 2 24h 80 0h 4 or h 20 (rejected)Thus, the coordinates of C are (0 , 4) .1App–1 for missing ‘(’ or ‘)’(ii) Note that AC is a diameter of the circle OABC .So, the coordinates of the centre of the circle are (3 , 2) .Also, the radius of the circle is1M1Meither one(3 0) 2 (4 2) 2 13 .Thus, the equation of the circle OABC is ( x 3) 2 ( y 2) 2 13 .1ASuppose that the equation of the circle OABC isx 2 y 2 k1 x k 2 y k3 0 , where k1 , k 2 and k3 are constants.1M 0 2 0 2 k1 (0) k 2 (0) k3 0 22 6 0 k1 (6) k 2 (0) k3 0 22 0 4 k1 (0) k 2 (4) k 3 0Solving, we have k1 6 , k 2 4 and k 3 0 .1M22Thus, the equation of the circle OABC is x y 6 x 4 y 0 .x 2 y 2 6x 4 y 0for solving system of equations1A----------(7)PP-DSE-MATH-CP 1 10只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution15. (a)MarksLet s be the standard deviation of the scores in the test.36 48 2ss 6Remarks1Meither oneThe standard score of John in the test66 48 6 31A----------(2)(b) Note that the score of David is equal to the mean of the scores.So, the mean of the scores remains unchanged.The sum of squares of the deviations of the scores in the test remainsunchanged while the number of students decreases by 1 .Therefore, the standard deviation increases.Hence, the standard score of John decreases.Thus, the standard score of John will change.1M1Af.t.----------(2)PP-DSE-MATH-CP 1 11只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution16. (a)MarksRemarksThe required probability C418C4301Mfor numerator or denominator 686091Ar.t. 0.1121Mfor1Ar.t. 0.112The required probability 18 17 16 15 30 29 28 27 68 609 r r 1 r 2 r 3 n n 1 n 2 n 3 , r n----------(2)(b)The required probability 1 68 C412 609 C 4305306091Mfor 1 – (a) – p11Ar.t. 0.8701Mfor considering 3 cases1Ar.t. 0.8701Mfor 1 – (a) – p21Ar.t. 0.8701Mfor considering 14 cases1Ar.t. 0.870The required probability C118C312 C 218C212 C318C112C430530609The required probability68 12 11 10 9 1 609 30 29 28 27 530 609The required probability 18 12 11 10 18 17 12 11 18 17 16 12 4 6 4 3029282730292827 30 29 28 27 530609----------(2)PP-DSE-MATH-CP 1 12只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution17. (a)Marks11 2i 1 1 2i 1 2i 1 2i 1 2 i5 5Remarks1M1A----------(2)(b) (i)1010 2 4i and 2 4i .1 2i1 2iThe sum of roots1010 1 2i 1 2i (2 4i ) (2 4i ) 4Note thatThe product of roots 10 10 1 2i 1 2i 20Thus, we have p 4 and q 20 .1Meither1Aeither1Afor both correct(ii) When the equation x 2 4 x 20 r has real roots, we have 0 .So, we have ( 4) 2 4(1)(20 r ) 0 .Thus, we have r 16 .1M1A----------(5)PP-DSE-MATH-CP 1 13只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution18. (a)MarksBy cosine formula,AB 2 AC 2 BC 2 2( AC )( BC ) cos ACB22Remarks1M2AB 20 12 2(20)(12) cos 60 AB 4 19 cm1Ar.t. 17.4 cmAB 17.43559577 cm----------(2)(b) By sine formula,sin BAC sin ACB BCABsin BAC sin 60 124 19 BAC 36.58677555 Let Q be the foot of the perpendicular from C to AB .CQsin BAC ACCQ 20 sin 36.58677555 CQ 11.92079121 cmSince ABC ABD , the required angle is CQD .1CD CQD 2sin 2CQ CQDsin 0.5872093452 CQD 71.91844786 CQD 71.9 Thus, the angle between the plane ABC and the plane ABD is 71.9 .By sine formula,sin ABC sin ACB ACABsin ABC sin 60 204 19 ABC 83.41322445 Let Q be the foot of the perpendicular from C to AB .CQsin ABC BCCQ 12 sin 83.41322445 CQ 11.92079121 cmSince ABC ABD , the required angle is CQD .1CD CQD 2sin 2CQ CQDsin 0.5872093452 CQD 71.91844786 CQD 71.9 Thus, the angle between the plane ABC and the plane ABD is 71.9 .1M1M1Mfor identifying the angle1Ar.t. 71.9 1M1M1Mfor identifying the angle1Ar.t. 71.9 PP-DSE-MATH-CP 1 14只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolutionMarksBy sine formula,sin BAC sin ACB BCABsin BAC sin 60 124 19 BAC 36.58677555 Let Q be the foot of the perpendicular from C to AB .CQsin BAC ACCQ 20 sin 36.58677555 CQ 11.92079121 cmBy symmetry, we have DQ CQ .DQ 11.92079121 cmSince ABC ABD , the required angle is CQD .22Remarks1M1M1Mfor identifying the angle1Ar.t. 71.9 2CD CQ DQ 2(CQ)( DQ) cos CQD22214 11.92079121 11.92079121 2(11.92079121)(11.92079121) cos CQD CQD 71.91844786 CQD 71.9 Thus, the angle between the plane ABC and the plane ABD is 71.9 .The area of ABC1 ( AC )( BC ) sin ACB21 (20)(12) sin 60 21M 60 3 cm 2Let Q be the foot of the perpendicular from C to AB .1( AB)(CQ ) 60 321(4 19 )(CQ ) 60 32CQ 11.92079121 cmSince ABC ABD , the required angle is CQD .1CD CQD 2sin 2CQ CQDsin 0.5872093452 CQD 71.91844786 CQD 71.9 Thus, the angle between the plane ABC and the plane ABD is 71.9 .1M1Mfor identifying the angle1Ar.t. 71.9 ----------(4)(c)Let Q be the foot of the perpendicular from C to AB .1CD CPD 2Note that sin .2CPSince CP CQ , we have CPD CQD .Thus, CPD increases as P moves from A to Q and decreases as Pmoves from Q to B .1M1Af.t.----------(2)PP-DSE-MATH-CP 1 15只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYSolution19. (a)Marks4 000 000(1 r %)3 1 048 5761 048 576(1 r %)3 4 000 0001 r % 0.64r 36(b) (i)1M1A----------(2)Let n be the number of years needed for the total revenue made bythe firm to exceed 9 000 000 .2 000000 2 000000(1 20%) L 2 000000(1 20%)n 1 9 0000002 000 000 (1 (0.8) n ) 9 000 0001 0 .8(0.8) n 0.1n log 0.8 log 0.1log 0.1log 0.8n 10.31885116Thus, the least number of years needed is 11 .n (ii)RemarksThe total revenue made by the firm 2 000000 2 000000(1 20%) 2 000000(1 20%)2 L2 000 000 1 0.8 10 000 000Thus, the total revenue made by the firm will not exceed 10 000 000 .(iii)1Mfor left side1Mfor sum of geometric sequence1Mfor solving inequality1A1M1Af.t.The total revenue made by the firm minus the total amount ofinvestment in the first m years2 000 000(1 (0.8) m ) 4 000 000(1 (0.64) m ) 1 0 .81 0.6410 10 000 000 (1 (0.8) m ) (1 (0.64) m ) 9 10 10 000 000 (1 (0.8) m ) (1 (0.8) 2m ) 9 10 000 000m 2m 10 ((0.8) ) 9(0.8) 1910 000 00010 (0.8) m 1 (0.8) m 1 9Note that (0.8) m 0 and (0.8) m 1 for any positive integer m . (())(1M1Mfor quadratic expression1Mfor either one)Therefore, we have 10(0.8) m 1 0 and (0.8) m 1 0 .2 000 000 (1 (0.8) m ) 4 000 000 (1 (0.64) m ) 0 .1 0.81 0.64Thus, the claim is disagreed.So, we have1Af.t.---------(10)PP-DSE-MATH-CP 1 16只限教師參閱FOR TEACHERS’ USE ONLY
只限教師參閱FOR TEACHERS’ USE ONLYHONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITYHONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATIONPRACTICE PAPERMATHEMATICSCOMPULSORY PARTPAPER 2Question No.1.2.3.4.5.KeyACADDQuestion .29.30.DBABC只限教師參閱FOR TEACHERS’ USE ONLY
pp-dse-math-cp 1 1 只限教師參閱 for teachers’ use only 香港考試及評核局 hong kong examinations and assessment authority 香港中學文憑考試 hong kong diploma of secondary education examination 練習卷 practice paper 數學 必修部分 試卷一
Ho Chi Minh City SB1 19 Ho Chi Minh City SE 4 Hong Kong CR1 6 Hong Kong DV 10 Hong Kong F2B 1 Hong Kong F3 6 Hong Kong F4 2 Hong Kong FX 2 Hong Kong IR1 20 Page 12 of 30. Post Visa Class Issuances Immigrant Visa Issuances by Post
Occurrence and significance of thunderstorms in Hong Kong Choi E.C.C.a, Mok H. Y. b. & Lee T. C. b a Department of Building and Construction, City University of Hong Kong, Hong Kong bHong Kong Observatory, Hong Kong ABSTRACT: Hong Kong is situ
to Hong Kong and meet with Hong Kong officials, legislators, civil society, and business representatives in the territory and when they visit the United States. They should also continue to express support for freedom of expression and rule of law in Hong Kong. Introduction In 2019, Hong Kong Chief Executive Carrie Lam directed her ad-
institution incorporated in Hong Kong by or under the Companies Ordinance (Cap. 32) or any other Ordinance and any reference to a bank incorporated in Hong Kong, a deposit-taking company incorporated in Hong Kong or a restricted licence bank incorporated in Hong Kong shall be construed accordingly; “authorized institution incorporated outside .
Sep 03, 2019 · Hong Kong ranked as the fifth-largest market for U.S. fruit exports and the United States was the third-largest supplier for Hong Kong by value in 2018. The United States exported USD347 million worth of fresh fruits to Hong Kong. Though U.S. fresh fruit exports to Hong Kong remain
Jun 29, 2018 · 33 Garden Road Central, Hong Kong Food Retailing Industry In 2017, Hong Kong’s retail food sector sales rose 1.82 percent to US 12.28 billion. The Hong Kong food retail market is made up of supermarkets, convenience stores, and traditional il
Core-competencies for registered nurse grade (Internal official document). Hong Kong SAR: Author. The College of Nursing, Hong Kong. (1993). Standards for nursing practice. Hong Kong: Author. The Nursing Council of Hong Kong (2004). Core-competencies for registered nurses (general). Retrieved from
The Hong Kong Institute of Chartered Secretaries (Incorporated in Hong Kong with limited liability by guarantee) 3/F, Hong Kong Diamond Exchange Building, 8 Duddell Street, Central, Hong Kong Tel: (852) 2881 6177 Fax: (852) 2881 5050 Email: ask@hkics.org.hk (general) ecpd@hkics.org.hk (professional development)
