St Gabriels Prelim 2009 Em P2 Solutions

ST GABRIEL’S SECONDARY SCHOOL 2009 PRELIMINARY EXAMINATION Sec 4E/ 5NA/4N1 Mathematics Paper 2 Marking scheme 1 (a) (i) Bearing of A from B = 360° − (180° − 18°) = 198° (B1) 502  110 2  1252 2(50)(110) = − 0.093181818 = 95.3°

(ii) cos ABC =

ABC

(M2 A1)

(b) (i) Let the shortest distance be d m d Sin (180° - 95.3469°) = 50 d = 50 sin (180° - 95.3469°) = 49.8 m (M1 A1) (ii) Let the greatest angle be θ° 20 tan θ° = 49.78246 θ° = 21.9° (M1 A1) _____________________________________________________________ 2 (a) (i) 4y − x2y + 4z − x2z = y(4 − x2) + z(4 − x2) = (4 − x2) (y + z) = (2 + x)(2 −x)(y + z) (M1 A1) (ii) 4t2 − 9 = (2t + 3)(2t − 3) 391 = 400 − 9 = 4(102) − 9 = (20 + 3)(20 − 3) = (23)(17) Hence the prime factors are 23 and 17. (b)

8x+2 = 169 − 3x 23(x +2) = 24(9 − 3x) 3x + 6 = 36 − 12x 15x = 30 x = 2

(B1)

(M1 A1)

(M1 A1)

2(5.4)  6(3)(2.5) 3(5.4)(2.5)  2(3) q = - 0.735

(c) (i) q =

(B1)

2 p  6rs 3 ps  2r q (3ps + 2r) = 2p − 6rs 3pqs + 2rq = 2p − 6rs 2rq + 6rs = 2p − 3pqs 2r(q + 3s) = p(2 − 3qs) p(2  3qs) r = (M2 A1) 2(q  3s ) _____________________________________________________________ 300 3 (a) hrs (B1) x (b) (i) (x − 4) km/h (B1) 308 (ii) hrs (B1) x4 (ii)

(c)

q=

308 300 1 − = x4 x 2

(B1)

616 x − 600(x − 4) = x(x − 4) 616 x − 600 x + 2400 = x2 − 4x x2 − 4x − 16x − 2400 = 0 x2 − 20 x − 2400 = 0 (shown) (d)

x2 − 20 x − 2400 = 0 (x − 60)(x + 40) = 0 x = 60 or x = − 40 (reject)

(B1)

(M1 A1)

(e)

Time taken to travel to Q = (308  56) hr = 5.5 hrs (B1) Arrival time = 0400 (B1) ____________________________________________________________ 4 (a) (i)

ε P 1 9 4

(ii) {4, 6, 8} (iii) 2

3 5 7 6

Q 2

8

(B1)

(B1) (B1)

(iv)

(b) (i)

2 3

(B1)

2 2 3 4 5 5 6 6   7 7 8 8 

(B1)

0.7  1.1  (ii)    2.3     0.5 

2 2 3 4 (iii)  5 5 6 6  7 7 8 8 

(B1)

0.7  12.5  1.1    =  25.8    2.3   35     0.5 

0 1.4 0  (iv) 12.5 25.8 35 0 1.5 0     0 0 1.2 

= 17.5 38.7 42

(M1 A1)

(B1)

(B1)

Q5a Total Rooney earned = $850 + 0.10  $2000 + 0.12  $4000 + 0.20  ($9650  $6000) = $850 + $1410 = $2260 Q5b Amt of profit each employee received (i) = [0.082  0.74 $475 000]  12 = $2401.92 (2dp) (ii)

Amt received in peso = ($2401.917  $4.50)  100 = 53376 peso (correct to the nearest peso)

Q5c In first state : (i) Tax = 0.4  0.0321  $295 000 = $3787.80 In second state : Tax = 0.24  0.0502  $295 000 = $3554.16  Second state charged lower property tax. (ii)

Percentage diff = ($3787.80  $3554.16)  3554.16  100% = 6.57 %

M1 A1 M1 A1 cao M1 A1

M1

A1 M1 A1

Q6a AOC = 64 (i) (ii) OAB = OBA = 64  2 = 32 ( at centre = 2 s at cricumference)  BAE = 90  32 = 58 (tan  rad) Q6b BAD = 180  58 = 122 ACD = 90 + 32= 122 ADC = BDA (common ) ACD = BAD (from above) or CAD = DBA (alt seg thm)  triangle ACD and triangle BAD are similar (NB : only 2 marks awarded if wrong reasons or no reason given and no final statement/conclusion given) Q6c x  4 6  6 4 4x + 16 = 36 4x = 20 x = 5  radius = 2.5 cm (shown) Q7a 1 Area of triangle ABC = 8  10  sin 57 = 33.55 cm2 (i) 2  Vol. of the prism = 33.55  20 cm3 = 671 cm3 (3sf) (NB: No mark awarded if accuracy < 3sf, accuracy > 3sf no penalty) (ii) BC2 = 82 + 102  2(8)(10)cos 57 = 164  160cos 57 BC = 8.767 cm Total surface area Q7b

= 2(33.55) + 1020 + 820 + 8.767  20 = 602 cm2 (3sf)

8.767 10 BMC = 41.241 tan BMC 

BME = 180  2(41.241) = 97.5

B1 M1 A1 M1 M1 B1

M1

A1 M1 A1 B1 (ECF) M1 A1 M1 A1 M1 A1 B1 (ECF)

9

(a)

a = 19, b = 10

(c)

Height = 15 m

(d)

(i)

6.4  0.1 m

(ii)

12  0.2 m

(e)

gradient = 3  0.3

(f)

(i)

5.55  0.1

(f)

(ii)

0.3  0.1 < x < 3.75  0.1