2007 Ce Math Solution

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

p=

2.

3q + 7 2

m6 m 6 n5 n5 = = 3 m9 n −5 m9 m

3.

(a) r 2 + 10r + 25 = (r + 5) 2 (b) r 2 + 10r + 25 − s 2 = (r + 5) 2 − s 2 = (r + 5 − s)(r + 5 + s)

4.

Median = 67 kg Range = 75 − 50 = 25 kg S.D. = 7.65 kg

5.

Δ<0

142 − 4k < 0 196 k> 4 k > 49 6.

(a) Answer = 400 × 80% = $320 (b) Cost = 320 − 70 = 250 Percentage Profit =

7.

70 × 100% = 28% 250

Let x be the number of elderly patients consulted. 120 x + 160(67 − x) = 9000 120 x + 10720 − 160 x = 9000 40 x = 1720 x = 43

8.

x = 180 − 110 = 70 y = 110 − 90 = 20 Let ∠CBA = a a = y = 20 ∴ z = a = 20 1

9.

(a) Let ∠AOB = θ 2π ⋅ 40 ⋅

θ 360

= 16π

θ = 72 (b) Area = π (40) 2 ⋅

72 = 320π cm 2 360

Section A2

10. (a) The least possible length of the metal mire = 5 − (b) (i)

Length of the wire ≤ 2.0 +

1 = 4.5 cm 2

0.1 = 2.05 m = 205 cm 2

∴ It is impossible that the actual length of this metal wire exceeds 206 m 1 = 4.5 cm 2 ∵ 4.5 × 46 = 207 cm, ∴ the answer is impossible.

(ii) Maximum length of each shorter wire = 5 −

1 11. (a) Volume of the cone = π ⋅182 ⋅ 24 = 2592π cm 3 3 3

⎛ 8 ⎞ ⎛1⎞ Volume of water = 2592π × ⎜ ⎟ = 2592π × ⎜ ⎟ ⎝ 24 ⎠ ⎝3⎠

(b) (i)

= 96π cm 3 Let the slant height of the cone be

3

,

= 182 + 242 = 30

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

⎛1⎞ Wet area = 540π ⎜ ⎟ = 60π cm 2 ⎝3⎠ (ii) Let the slant height of the bigger cone be s , s = 302 + 27 2 = 45

Curved surface area of the bigger cone = π ⋅ 27 ⋅ 45 = 1215π 1 Volume of the larger cone = π (27) 2 ⋅ 36 = 8748π 3 Let A be the required wet area By similar solid,

1215π 3 8748π = = 4.5 A 96π

2

∴A=

12. (a)

1215π = 60π cm 2 2 4.5

k 63 = 17 153

∴ k = 17 ×

63 =7 153

(b) Number of students = 17 × (c) Prob. =

360 = 40 153

4 1 = 40 10

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

13. (a)

y −3 4 =− 3 x − 10 3 y − 9 = −4 x + 40 4 x + 3 y − 49 = 0 AB :

(b) 4 ⋅ 4 + 3h − 49 = 0 h = 11 (c) (i)

k + 10 =4 2 ∴ k = −2

(ii) ΔABC =

1 (12)(11 − 3) = 48 2

AC = 62 + 82 = 10

10 ⋅ BD = 48 2 ∴ BD = 9.6

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Section B 14. (a) (i)

f ( −3) = 0

4(−3)3 + k (−3) 2 − 243 = 0 −108 + 9k − 243 = 0 k = 39 (ii) f ( x) = 4 x 3 + 39 x 2 − 243 4

39

−3

0 −81

−12

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

(b) (i)

−243 243

0

C = k1 x 3 + k2 x 2 , where k1 and k2 are constants. 7381 = k1 (5.5)3 + k2 (5.5) 2 3

2

⎛ 11 ⎞ ⎛ 11 ⎞ 7381 = k1 ⎜ ⎟ + k2 ⎜ ⎟ , ∴ 488 = 11k1 + 2k2 .................(1) ⎝2⎠ ⎝2⎠ 9072 = k1 ⋅ 63 + k2 ⋅ 62 , ∴ 252 = 6k1 + k2 ...................(2) 2 × (2) − (1) :16 = k1 ∴ k2 = 156 ∴ C = 16 x 3 + 156 x 2 (ii) 972 = 16 x3 + 156 x 2 4 x 3 + 39 x 2 − 243 = 0 9 ∴ x = cm (As x > 0 ) 4

15. (a) (i)

Prob. =

48 3 = 80 5

(ii) Prob. =

12 3 = 80 20

(iii) Prob. =

48 + 4 52 13 = = 80 80 20

(iv) Let L = Large size shirt , and B = Boy 3 P( L and B) 20 3 5 1 P( L | B) = = = × = 3 20 3 4 P( B) 5 4

(b) (i)

16 ⋅15 C216 3 Prob. = 80 = 2 = 80 ⋅ 79 79 C2 2

(ii) P(Same Size) =

C228 + C236 + C216 28 ⋅ 27 + 36 ⋅ 35 + 16 ⋅15 = C280 80 ⋅ 79

2256 141 = = 0.357 6320 395 ∴ P(Different Sizes) = 1 − 0.357 = 0.643 =

∴ P(Different Sizes) > P(Same Sizes)

16. (a)

s=

5+6+9 = 10 2

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

Volume of ABCDEF = 20 × 200 + (b)

1 200 × 3 = 210 2 cm 3 3

DE = 62 + 32 = 45 FE = 9

DF = 32 + 52 = 34

Let ∠DEE = θ 34 + 81 − 45 cos θ = = 0.6669389422 2 34 ⋅ 9

θ = 48.1688 = 48.2 (3 sig. fig.) Let the shortest distance be L L = DF sin θ = 34 sin 48.1688 = 4.344714 = 4.34 cm (3 sig. fig.) (c) Area of the rectangle = 4 × 5 = 20 1 FE ⋅ L = 19.6 cm 2 2 As 20 > 19.6 , ∴ the answer is impossible.

Area of ΔFDE =

17. (a) (i)

∠ABG = ∠DBG ( I is the In-center) BG is common

AB = BD (Given) ∴ΔABG ≅ ΔDBG (SAS) (ii) ∠ABE = 90 ( ∠ in semi-circle) ∠AGB = 90 (Isos. Δ ) ∴∠AGI = ∠ABE 5

Also ∠IAG = ∠BAE (In-center) ∴ΔAIG ~ ΔAEB (AAA) ∴ (b) (i)

GI BE = AG AB

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

⎛ 11 − 25 ⎞ , 0 ⎟ = (−7, 0) Also G is the mid-point of AD , ∴ G = ⎜ ⎝ 2 ⎠ (ii)

BE 1 = AD 2 By (a)(ii),

GI 1 = AG 2

1 (18) = 9 2 ∴ I = (−7,9) ∴ GI =

Radius of the required circle = GI = 9 ∴ Equation of the inscribed circle: ( x + 7) 2 + ( y − 9) 2 = 92

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CE 2007 Math II Key: 1. 11. 21. 31. 41. 51.

ADAAD DDABC DDBDC DAACB ACBCC CCBD

6. 16. 26. 36. 46.

BBDAB AABAD BDBCC CCDBA DBAAC

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