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DSE-MAT-15-C2S

2012012012015555 HKDSE Mathematics CoreHKDSE Mathematics CoreHKDSE Mathematics CoreHKDSE Mathematics Core Paper Paper Paper Paper 2222

Suggested SolutionsSuggested SolutionsSuggested SolutionsSuggested Solutions Prepared by Andy LaiPrepared by Andy LaiPrepared by Andy LaiPrepared by Andy Lai

HKDSE HKDSE HKDSE HKDSE MathematicsMathematicsMathematicsMathematics TeaTeaTeaTeachercherchercher

(100% M.C. Right!)(100% M.C. Right!)(100% M.C. Right!)(100% M.C. Right!) Enrollment Hotline: 6772 3001 Website: www.andylai.hk MSN: mrandylai@hotmail.com

Address: Rm706, Prosper Commercial Building, 9 Yin Chong Street, Mong Kok, Kowloon, Hong Kong.

MCABC Grade,

2015 HKDSE Mathematics Paper 2 Suggested Answers

1.

D 2. D 3. A 4. D 5. B

6. A 7. A 8. D 9. B 10. D

11. C 12. D 13. B 14. C 15. C

16. B 17. D 18. A 19. C 20. C

21. B 22. A 23. C 24. A 25. B

26. C 27. A 28. B 29. A 30. B

31. C 32. A 33. A 34. B 35. C

36. D 37. D 38. B 39. D 40. C

41. B 42. A 43. C 44. D 45. B

MCABC Grade,

Andy's predicted M.C. Grade boundaries:

5**: 43 / 45 5*: 40 / 45 5: 38 / 45

4: 34 / 46 3: 30 / 45 2: 26 / 45

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 1

Section A

1. D Sir :

? 1221)1)(1(232232+++=+++++=+++xxxxxxxxxxx

2. D

Sir :

? 22 2244

2462733

3)3(yyy

yy==

3. A

Sir :

1. Method 1: Elimination of Simultaneous equation in two unknowns

29543
qpqp

29520155

qpqp ? 186=q ? 3=q

4)3(3=+p ? 5-=p

2. Method 2: Program in calculator!

5-=p and 3=q

4. D

Sir :

? 0.0023456789 = 0.00235 (correct to 5 decimal places) ? 0.0023456789 = 0.002346 (correct to 6 decimal places) ? 0.0023456789 = 0.00234568 (correct to 6 significant figures) ? Therefore, the answer is D sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 2

5. B

Sir :

? 6))(4(2+-+≡++mxxnmxx mxmxnmxx46)4(22-+-+≡++ ? Comparing coefficients and constants on both sides gives mm=-4 and nm=-46 ? 2=m and 2-=n

6. A

Sir :

? 4718>+x or 325<-x

714->x or 22

-->x

2->x or 1>x

? Therefore, 2->x

7. A

Sir :

? 01542=--xx and β is the root of the equation. ? Therefore,

01542=--ββ ? 021082=--ββ ? 581072=-+ββ

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 3

8. D

Sir :

? 2)(bxay+= ? Vertex form: 0)]([2+--=bxay ? Min Point: (-b, 0) ? From the graph,

0<-b ? 0>b.

? From the graph, the parabola's opening is upward ? 0>a ? Therefore,

0>a and 0>b

9. B

Sir :

? 68.0)6.01(7.1=-× %3232.0168.0-=-=- ? Therefore, percentage change = -32%

10. D

Sir :

? 597819.780,59)4%61(5000043==+× (correct to the nearest dollar)

11. C

Sir :

? a : c = 5 : 3 b : c = 3 : 2 a : b : c = 10 : 9 : 6 ? Let a = 10k, b = 9k and c = 6k ? 15:1615:16)(:)(==++kkcbca sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 4

12. D

Sir :

? 23ykxz= ? 23)1()2(14k= ? 47=k

189)2(34723=-=z

13. B

Sir :

? 1st Term: 5 dots, 2nd Term: 9 dots, 3rd Term: 13 dots, etc.. ? 4 th Term = 13 + 4 = 17 dots, 5th Term = 17 + 4 = 21 dots ? Therefore, 6 th Term = 21 + 4 = 25 dots.

14. C

Sir :

? Upper limit of the weight of the white sugar in the bad = 215+ = 5.5 kg ? Lower limited of the weight of white sugar in each packet =

2110- = 9.5 g

? n

15. C

Sir :

? By Pyth. Theorem, 1251322=-=AN and 861022=-=CN ? Since

222BCABAC+= and 12162022=-=AB

? Therefore, Area of

296)16)(12(21cmABC==Δ

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 5

16. B

Sir :

? Similar figures ? 3 2 1 21
=ll VV 3 2 1 412)
=VV ? 12721 =VV ? Volume of the Biggest circular cone = ( ) ( )ππ324129)(312= ? Volume of the upper smaller circular cone =

ππ1227324=

? Therefore, the volume of the frustum =

331212324cmπππ=-

17. D

Sir :

? Given DE : EC = 2 : 3 ? Area of ADEΔ : Area of BECΔ = 2a : 3a (Same height ? Base ratio = Area ratio)

DEFΔ~CEBΔ (AAA) ?

2 32
38)
((=a ? 6=a

AEFΔ~GEBΔ (AAA) ?

2 32
328)
Xaa ? 227cmX= ? Do you know how to calculate the area of

ABEΔ?

230cm Why?

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 6

18. A

Sir :

? αcos=ACAD and βtan=ABAC ? Therefore,

βαtancos=×ABAC

ACAD

19. C

Sir :

? )270sin(1360cos )90sin(1180cosθθ+++++°° cos11

20. C

Sir :

? °=?=?90ACDABD (Angle in semi circle) ? BC = CD ? DACBAC?=? (Equal chord, Equal angle) ? °°==?29258BAC °°°=--=?329058180BCA (opp angles, cyclic quadrilateral)

°°°°=--=?2923290180CBD

? Therefore,

°°°=--=?619029180AEC

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 7

21. B

Sir :

? Area of 27290sin)224)(224(21cmAEB==Δ°

22. A

Sir :

? )360(5)180)(2( nnn =- ? 12=n ? Therefore, each interior angle of the polygon =

°°=-15012)180)(212(

? The number of diagonals of the polygon =

541212

2=-C ? The number of folds of rotational symmetry of the polygon is 12.

23. C

Sir :

? Reflection about y-axis ? No change in y-coordinate ? Reflection about y-axis ? Negative value of the x-coordinate ? The rectangular coordinates of its image is )1,3(-- ? By

31tan=θ ? 31tan=θ ? °=30θ and 231=+=r

? Therefore, polar coordinates of its image is (2,

°210)

sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 8

24. A

Sir :

? A moving point P is always equidistant from two points A and B. ? The locus of P is perpendicular bisector of AB.

Locus of P: Perpendicular bisector of AB

25. B

Sir :

? 1=ax ? ax1= ? 011=+cybx ? cxcby1+-= ? 0>-cb and 01>c ? 0>c and 0

011< ? Label the diagram below: sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 9

26. C

Sir :

? Standard form of circle: 222)()(rkyhx=-+- ? Coordinates of the center of circle: ),(kh and radius = r

2222)33()04()3())4((-+--=-+--yx

16)3()4(22=-++?yx

169616822=+-+++?yyxx

096822=+-++?yyxx

27. A

Sir :

? Expected value = 11$6363036366=×+×

28. B

Sir :

? 112 448

44816128==++++

29. A

Sir :

? Inter-quartile range = 75%th datum - 25%th datum = 45 - 25 = 20 sir DSE-MAT-15-C2S

2015 HKDSE Mathematics Core Paper II Suggested Solutions

©2015 A Lai learning Center. ALL RIGHTS RESERVED. www.andylai.hk Prepared by Andy Lai P. 10

30. B

Sir :

? Total numbers of data = 15 ? The median = 8th datum = q ? Mode = r = 3

15109886553333322++++++++++++++=mp ? 1570mp+=

quotesdbs_dbs46.pdfusesText_46

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