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HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 4

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第31～45題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第1～10題答案，請參閱題解Part 1。

如欲查閱第11～20題答案，請參閱題解Part 2。

如欲查閱第21～30題答案，請參閱題解Part 3。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/07/hkdse-solutions-to-hkdse-maths.html）

31. The answer is B.

Let x=1.2.

The given expression= 1.5625

Option A= -4.6875

Option B= 1.5625

Option C= 0.9375

Option D= -0.9375

32. The answer is D.

From the graph,

when x=0, log₅y=2.

y=25

Therefore, 25=ab⁰

a=25

33. The answer is A.

For every digit '1',

The rightmost digit= 2⁰

The 2nd digit= 2¹

The 3rd digit= 2²

…

The nth digit= 2^n-1

Therefore,

the given expression

=2¹²+2¹⁰+2⁷+2³+2⁰

=2¹²+2¹⁰+137

34. The answer is A.

The given expression

=[4ki-(6+ki)]÷i

=(3ki-6)÷i

=3k-6÷i

=3k+6(i²)÷i

=3k+6i

35. The answer is D.

Try (1,3) on different constraints.

It satisfies all the three inequalities.

As (1,3) only lies in triangle ODF but not other triangles,

triangle ODF is the only solution to the inequality.

36. The answer is C.

Arithmetic sequence: same difference between every two consecutive terms

Therefore, the common difference is=(-6-18)÷3=-8

The second term is 18-(-8)=26

37. The answer is C.

The graph of y=g(x) can be obtained by translating y=f(x) 2 units to the left and 3 units downwards.

Translating 2 units to the left: x+2

Translating 3 units downwards: -3

Therefore g(x)=f(x+2)-3

38. The answer is A.

By sine law,

y÷sin(180°-47°-56°)=x÷sin56°

y÷sin77°=x÷sin56°

y=xsin77°÷sin56°

39. The answer is A.

Interest rate p.a.=6%

Interest rate per month=0.5%

Let $x the amount deposited every month.
x(1+0.5%)¹²+x(1.05%)¹¹+...+x(1+0.5%)²+x(1+0.5%)=10000
x=10000÷(1.005+1.005²+1.005³+...+1.005¹²)=806.63

40. The answer is C.

Let X be a point on AC such that EX is perpendicular to AC and DX is perpendicular to AC.

AC²=AB²+BC² (Pyth. Theorem)

AC²=3²+4²

AC=5cm

Area of triangle ADC= AD×DC÷2= DX×AC÷2

AD×DC=DX×AC

4×3=DX×5

DX=2.4cm

tantheta=DE÷DX=6÷2.4=2.5

41. The answer is B.

Let O be the centre of the circle.

OA is perpendicular to TA. (tangent+radius)

As ∠CTA+∠TAO=90°+90°=180°

CT//OA (int. ∠s supp.)

∠TCA=∠OAC=20° (alt. ∠s, CT//OA)

Since OA=OC (radii)

∠ACO=∠OAC=20° (base ∠s, isos. triangle)

∠BCO=20°+20°=40°

OB=OC (radii)

∠CBO=∠BCO (base ∠s, isos. triangle)

∠BOC=180°-40°-40°=100° (∠ sum of triangle)

BC÷sin100°=OB÷sin40° (sine law)

Radius of the circle

=OB

=6÷sin100°×sin40°

=3.9cm

42. The answer is B.

As x=140, y=3.

For option A:

LHS=3, RHS=0.52

For option B:

LHS=3, RHS=3

For option C:

LHS=3, RHS=-0.52

For option D:

LHS=3, RHS=-3

43. The answer is D.

The probability

=P(red, non-red)+P(green, non-green)+P(white, non-white)

=2/9×7/9+3/9×6/9+4/9×1

=68/81

44. The answer is B.

The probability

=(2×₆P₆)÷₇P₇

=2/7

45. The answer is D.

The new mode= (32+3)×2= 70

The new interquartile range

=(Old Q₃+3)×2-(Old Q₁+3)×2

=2×(Old Q₃-Old Q₁)

=2×27

=54

Assume there are only two numbers in the set.

Let M be the old mean.

The new mean= 2M+6

Let X and Y be the old values of the two data.

The new value of X= 2X+6

The new value of Y= 2Y+6

The old variance= [(M-X)²+(M-Y)²]÷2= 25

The new variance

={[(2M+6)-(2X+6)]²+[(2M+6)-(2Y+6)]²}÷2

=[(2M-2X)²+(2M-2Y)²]÷2

=[4(M-X)²+4(M-Y)²]÷2

=4[(M-X)²+(M-Y)²]÷2

=4×25

=100

相關文章：香港中學文憑試模擬試題、HKDSE數學必修部分模擬試題卷二連答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 3

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第11～20題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第1～10題答案，請參閱題解Part 1。

如欲查閱第11～20題答案，請參閱題解Part 2。

如欲查閱第31～45題答案，請參閱題解Part 4。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/06/hkdse-solutions-to-hkdse-maths_19.html）

21. The answer is B.

Join BD.

∠ADB=90° (∠ in semi-circle)

In triangle BCD,

∠BCA=90° (∠ in semi-circle)

∠BCD=90°+28°=118°

∠CBD=∠CDB (base ∠s, isos triangle)

∠BCD+∠CBD+∠CDB=180° (∠sum of triangle)

∠BCD+2∠CDB=180°

118°+2∠CDB=180°

∠CDB=31°

∠ADC=∠ADC+∠CDB=90°+31°=121°

22. The answer is C.

Join DA.

∠ABE=∠ADE=30° (∠s in the same segment)

∠ADC=105°-30°=75°

∠AOC=2∠ADC=75°×2=150° (∠ at centre= 2∠ at circumference)

23. The answer is B.

TriangleAFB~TriangleDFE~TriangleCBE (AAA)

CD:DE=AF:FD=2:1

AD=BC (opp. sides, //gram)

AF:BC=AF:AD=AF:(AF+FD)=2:(1+2)=2:3

24. The answer is D.

∠ADB=20°=∠DAB (ext. ∠ of triangle)

Therefore, AB=BD=10 cm (base ∠s, isos. triangle)

As BD=CD (given),

∠DBC=∠DCB=40° (base ∠s, isos. triangle)

∠CDB+∠DBC+∠DCB=180° (∠ sum of triangle)

∠CDB=180°-40°-40°=100°

CB÷sin100°=BD÷sin40° (sine law)

CB=10÷sin40°×sin100°

CB=15cm

25. The answer is A.

The figure on the right can be obtained by rotating the figure on the left by 90 degrees clockwise.

26. The answer is D.

As the point is rotated by 180 degrees, the MAGNITUDE of the x-coordinate and the y-coordinate DO NOT CHANGE.

Point (-4,3) is in Quadrant II. After rotating about the point of origin by 180 degrees, the image will be locate in Quadrant IV. So the image is at (4, -3).

27. The answer is D.

The lower quartile= 50 marks (The left end of the box)

So 25% of the students got less than 50 marks.

Passing rate=1-25%=75%

28. The answer is B.

There are 23 staff members.

As (23+1)÷2=12, the height of the 12th shortest staff member is the median.

The median= 165cm.

29.

Mean of the first set= [(a-7)+(a-1)+a+(a+2)+(a+4)+(a+8)]÷6=a+1

Mean of the second set= a+1/6

So I is wrong.

Median of the first set= [a+(a+2)]÷2=a+1

Median of the second set= [(a-1)+(a+3)]÷2= a+1

So II is wrong.

Standard deviation of the first set= 5.05…

Standard deviation of the second set= 7.06…

So III is wrong.

30.

I is correct: Only 2 people out of 950 are selected.

II is correct: Only the chairpersons of the Student Union are selected.

III is correct: Other students have no opportunity to be selected.

相關文章：香港中學文憑試模擬試題、HKDSE數學必修部分模擬試題卷二連答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 2

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第11～20題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第1～10題答案，請參閱題解Part 1。

如欲查閱第21～30題答案，請參閱題解Part 3。

如欲查閱第31～45題答案，請參閱題解Part 4。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/06/hkdse-solutions-to-hkdse-maths.html。）

11. The answer is C.

a₃=4+5=9

a₄=5+9=14

a₅=9+14=23

a₆=14+23=37

a₇=23+37=60

a₈=37+60=97

a₉=60+97=157

a₁₀=97+157=254

12. The answer is B.

Let h be the length and w be the width of the rectangle.

Original area of the rectangle= hw

New area of the rectangle= (1+50%)(hw)= (1+20%)h(1+x%)w

Therefore, 150%hw= (120%)(1+x%)hw

150%= (120%)(1+x%)

125%=1+x%

x=25

13. The answer is D.

2x=3y => x:y=3:2=6:4

x=2z => x:z=2:1=6:3

Therefore, x:y:z=6:4:3

(x+z):(x+y)=(6+3):(6+4)=9:10

14. The answer is C.

Let z=kx÷y.

Sub. x=3, y=4 and z=18,

18=(k)(3)÷4

k=24

Therefore, when x=2 and z=8,

8=(24)(2)÷y

y=6

15. The answer is B.

Maximum absolute error of the length of a side= 0.5cm

Maximum absolute error of the perimeter= 0.5+0.5+0.5= 1.5cm

Percentage error= 1.5÷(15+24+25)×100%= 2.3%

16. The answer is B.

cos∠DOA= OA:OD= 10:20= 0.5

∠DOA= 60°

tan∠BOA= AB:OA= 10:10= 1

∠BOA= 45°

Area of the shaded area

=Area of sector DOC- Area of triangle DOB

=20²π×105°÷360°-(DA+AB)(OA)÷2

=20²π×105°÷360°-(20sin60°+10)(10)÷2

=230cm²

17. The answer is A.

a= (2rπ)(r÷2)= r²π

b= 4πr²÷2= 2πr²

c= πrl= πr√r²+(2r²)=  √5πr²

Therefore, a<b<c

18. The answer is B.

Area of ABCD= CD×DT

36=9×DT

DT=4cm

Since AT:TB=1:2,

AT:AB=1:3

AT:9=1:3

AT=3cm

AD²=AT²+DT² (Pyth. Theorem)

AD²=3²+4²

AD=5cm

Perimeter of ABCD= 5+9+5+9= 28cm

19. The answer is A.

Refer to:

20. The answer is A.

Horizontal distance= AB+CD+EF= 1+2+3= 6cm

Vertical distance= BC+DE= 2+2= 4cm

Therefore, AF²= 6²+4² (Pyth. Theorem)

AF=7.2cm

相關文章：香港中學文憑試模擬試題、HKDSE數學必修部分模擬試題卷二連答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 1

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第1～10題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第11～20題答案，請參閱題解Part 2。

如欲查閱第21～30題答案，請參閱題解Part 3。

如欲查閱第31～45題答案，請參閱題解Part 4。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/06/hkdse-hkdse-maths-compulsory-part-paper_11.html。）

1. The answer is C.

(3a)²(a³)

=9a²(a³)

=9a⁵

2. The answer is C.

5-3m=2n

5-2n=3m

m=(5-2n)÷3

3. The answer is C.

a²-b² +2b-1

=a²-(b² -2b+1)

=a²-(b-1)²

=[a+(b-1)][a-(b-1)]

=(a-b+1)(a+b-1)

4. The answer is A.

L.H.S.=x²+px+(5p+q)

R.H.S.=x²+3x-10

p=3, 5p+q=-10

Therefore, 5(3)+q=10

q=-25

5. The answer is C.

f(-2)

=(-2)³+2(-2)²-7(-2)+3

=-8+8+14+3

=17 (Remainder Theorem)

6. The answer is D.

(x-a)(x-a-1)=(x-a)

(x-a)[(x-a-1)-1]=0

(x-a)(x-a-2)=0

x-a=0 or x-a-2=0

x=a or x=a+2

7. The answer is D.

x²-6x=2-k

x²-6x+(k-2)=0

Discriminant

=(-6)²-4(1)(k-2)

=36-4(k-2)

=36-4k+8

=44-4k<0

Therefore,

-4k<-44

k>11

8. The answer is A.

III is wrong because the mid-point of A and B should be where the axis of symmetry passes through. As the x-coordinate of the mid-point is (1+7)÷2=4, the axis of symmetry should be x=4. So III is wrong and the answer is A.

9. The answer is C.

5-2x<3

-2x<-2

x>1

4x+8>0

4x>-8

x>-2

The result is x>1 and x>-2.

The range that satisfies both domains will be x>1.

10. The answer is A.

Total cost of the handbags= 240÷(1+20%)+240÷(1-20%)= $500

Total selling price= 240(2)= $480

So the girl has lost $20.

相關文章：香港中學文憑試模擬試題、HKDSE數學必修部分模擬試題卷二連答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

HKDSE數學必修部分模擬試題卷二答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

經過多日鑽研，筆者已完成HKDSE數學必修部分模擬試題的卷二部分。以下是筆者的答案（共45題，由左至右、上至下順序）：


CCCAC DDACA
CBDCB BABAA
BCBDA DDBDB
BDAAD CCAAC
BBDBD

讀者亦可在這裏下載試卷的中文版及英文版。另外，筆者已上載試題的部分詳細題解，敬請留意！

相關文章：
HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 1

香港中學文憑試模擬試題
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香港中學文憑試模擬試題

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相關文章：
HKDSE數學必修部分模擬試題卷二連答案 (HKDSE Maths Compulsory Part Paper 2 (Sample Paper) with Answers)

HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 1