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

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