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CE 2007 Math I Solution: Section A1 1. 5 7 3( )p p q− = + 5 7 3 3p p q− = + 2 3 7p q= +

3 72

qp +=

2. 6 6 5 5

9 5 9 3

m m n nm n m m− = =

3. (a) 2 210 25 ( 5)r r r+ + = + (b) 2 2 2 210 25 ( 5)r r s r s+ + − = + − ( 5 )( 5 )r s r s= + − + + 4. Median = 67 kg Range = 75 50 25− = kg S.D. 7.65= kg 5. 0Δ < 214 4 0k− <

1964

k >

49k > 6. (a) Answer 400 80% $320= × = (b) Cost 320 70 250= − =

Percentage Profit 70 100% 28%250

= × =

7. Let x be the number of elderly patients consulted. 120 160(67 ) 9000x x+ − = 120 10720 160 9000x x+ − = 40 1720x = 43x = 8. 180 110 70x = − = 110 90 20y = − = Let CBA a∠ = 20a y= = 20z a∴ = =

2

9. (a) Let AOB θ∠ =

2 40 16360θπ π⋅ =⋅

72θ =

(b) Area 2 72(40) 320360

π π= ⋅ = cm 2

Section A2

10. (a) The least possible length of the metal mire 15 4.52

= − = cm

(b) (i) Length of the wire 0.12.0 2.052

≤ + = m 205= cm

∴It is impossible that the actual length of this metal wire exceeds 206 m

(ii) Maximum length of each shorter wire 15 4.52

= − = cm

4.5 46 207× =∵ cm, ∴ the answer is impossible.

11. (a) Volume of the cone 21 18 24 25923π π= ⋅ ⋅ = cm 3

Volume of water 3 38 12592 2592

24 3π π⎛ ⎞ ⎛ ⎞= × = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

96π= cm 3 (b) (i) Let the slant height of the cone be ,

2 218 24 30= + =

Curved surface area of the cone 18 30 540π π= ⋅ ⋅ = cm 2

Wet area 21540 60

3π π⎛ ⎞= =⎜ ⎟⎝ ⎠

cm 2

(ii) Let the slant height of the bigger cone be s ,

2 230 27 45s = + =

Curved surface area of the bigger cone 27 45 1215π π= ⋅ ⋅ =

Volume of the larger cone 21 (27) 36 87483π π= ⋅ =

Let A be the required wet area

By similar solid, 31215 8748 4.5

96Aπ π

π= =

3

2

1215 604.5

A π π∴ = = cm 2

12. (a) 6317 153k=

6317 7153

k∴ = × =

(b) Number of students 36017 40153

= × =

(c) Prob. 4 140 10

= =

(d) For the Bar Chart: The "Number of students" need modification. For the Pie Chart: No modification is required

13. (a) 3 4:10 3

yABx−

= −−

3 9 4 40y x− = − + 4 3 49 0x y+ − = (b) 4 4 3 49 0h⋅ + − = 11h =

(c) (i) 10 42

k +=

2k∴ = −

(ii) 1 (12)(11 3) 482

ABCΔ = − =

2 26 8 10AC = + =

10 482BD⋅

=

9.6BD∴ =

4

Section B 14. (a) (i) ( 3) 0f − = 3 24( 3) ( 3) 243 0k− + − − = 108 9 243 0k− + − = 39k = (ii) 3 2( ) 4 39 243f x x x= + − 4 39 0 243−

3− 12− 81− 243 4 27 81− 0 2( ) ( 3)(4 27 81)f x x x x∴ = + + − ( 3)( 9)(4 9)x x x= + + −

(b) (i) 3 21 2C k x k x= + , where 1k and 2k are constants.

3 21 27381 (5.5) (5.5)k k= +

3 2

1 211 1173812 2

k k⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, 1 2488 11 2k k∴ = + .................(1)

3 21 29072 6 6k k= ⋅ + ⋅ , 1 2252 6k k∴ = + ...................(2)

12 (2) (1) :16 k× − = 2 156k∴ = 3 216 156C x x∴ = + (ii) 3 2972 16 156x x= + 3 24 39 243 0x x+ − =

94

x∴ = cm (As 0x > )

15. (a) (i) Prob. 48 380 5

= =

(ii) Prob. 12 380 20

= =

(iii) Prob. 48 4 52 1380 80 20+

= = =

(iv) Let L =Large size shirt , and B =Boy

3( and ) 3 5 120( | ) 3( ) 20 3 4

5

P L BP L BP B

= = = × =

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(b) (i) Prob. 162802

16 1532

80 79 792

CC

⋅

= = =⋅

(ii) P(Same Size) 28 36 162 2 2

802

28 27 36 35 16 1580 79

C C CC

+ + ⋅ + ⋅ + ⋅= =

⋅

2256 141 0.3576320 395

= = =

∴ P(Different Sizes) 1 0.357 0.643= − = ∴P(Different Sizes) > P(Same Sizes)

16. (a) 5 6 9 102

s + += =

10(10 9)(10 5)(10 6) 200 10 2ABCΔ = − − − = =

Volume of 120 200 200 3 210 23

ABCDEF = × + × = cm 3

(b) 2 26 3 45DE = + =

9FE =

2 23 5 34DF = + =

Let DEE θ∠ =

34 81 45cos 0.66693894222 34 9

θ + −= =

⋅

48.1688θ = 48.2= (3 sig. fig.) Let the shortest distance be L sin 34 sin 48.1688 4.344714L DF θ= = = 4.34= cm (3 sig. fig.) (c) Area of the rectangle 4 5 20= × =

Area of 1 19.62

FDE FE LΔ = ⋅ = cm 2

As 20 19.6> , ∴ the answer is impossible. 17. (a) (i) ABG DBG∠ = ∠ ( I is the In-center) BG is common AB BD= (Given) ABG DBG∴Δ ≅ Δ (SAS)

(ii) 90ABE∠ = (∠ in semi-circle) 90AGB∠ = (Isos. )Δ AGI ABE∴∠ =∠

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Also IAG BAE∠ = ∠ (In-center) ~AIG AEB∴Δ Δ (AAA)

GI BEAG AB

∴ =

(b) (i) O is the mid-point of AC ( 25,0)A∴ = −

Also G is the mid-point of AD , 11 25 ,0 ( 7,0)2

G −⎛ ⎞∴ = = −⎜ ⎟⎝ ⎠

(ii) 12

BEAD

=

By (a)(ii), 12

GIAG

=

1 (18) 92

GI∴ = =

( 7,9)I∴ = − Radius of the required circle 9GI= = ∴ Equation of the inscribed circle: 2 2 2( 7) ( 9) 9x y+ + − =

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CE 2007 Math II Key: 1. ADAAD 6. BBDAB 11. DDABC 16. AABAD 21. DDBDC 26. BDBCC 31. DAACB 36. CCDBA 41. ACBCC 46. DBAAC 51. CCBD