Pierce Prelim 2009 Em2 Solutions

Answer Scheme for 4E/5A Prelim II Mathematics 2009 Paper 2 4016

1a) AC = AX (tangents from ext. point are equal) CAX  BAX AX is common length ABX is congruent to ACX Case : SAS Alternatively : AXB  AXC  90 AX is common AB  AC

ABC  ACX RHS OR BX  CX AX is common AB  AC ABC  ACX SSS 1b) Since ABX is congruent to ACX , X is midpoint of chord BC. BXO  90 (perp bisector of chord passes through O) BCE  90 (right  in semi circle) XBO  CBE (common angle)

 80 2a) A   70 80  90 

80 60 90 90

70  90  60  80 

 80 80 70  1 BA  (1 1 1 1)  70 60 90  80 90 60 2b) 4  90 90 80      80 80 75  It represents the average score of the 3 categories of the admission criteria for the four applicants. 2c)  0.25    X   0.6   0.15     80 AX   70 80  90 

80 60 90 90

70  0.25  78.5  90   0.6    67  60   0.15   83    88.5  80  

3a)(i)  C

M 11

11

9

9

3a)(ii) C  M ' 3a)(iii) n( M  C ' ) = 9 The number of people who only possess motorcycle licenses. 3b)  X

Y

Answer Scheme for 4E/5A Prelim II Mathematics 2009 Paper 2 4016

4a) x 2  x  12 x3 x 2  x  12 x3  4 x 2    2( x 2  16) x3  4 x 2 2( x 2  16) x3 ( x  3)( x  4) x 2 ( x  4)   2( x  4)( x  4) x 3 

x2 2

4b) y  3

x x9

 x  y2  9    x9 ( x  9) y 2  9 x xy 2  9 y 2  9 x xy 2  9 x  9 y 2 x( y 2  9)  9 y 2 x

9 y 2 9 y2 or y2  9 9  y2

4c) 2n 3  mn 7 7(2n)  3(m  n)

14n  3m  3n 11n  3m m 11 2  or 2 n 3 3

4  60 240  x x 3.5  60 210 5b)  x2 x2 5c) 240 210 5(  )  500 x x2 240 210   100 x x2 240( x  2)  210 x  100 x( x  2) 240 x  480  210 x  100 x 2  200 x 100 x 2  250 x  480  0 10 x 2  25 x  48  0 (proven)

5a)

5d) 10 x 2  25 x  48  0 (25)  (25)2  4(10)(48) x 2  10 x  3.772 or  1.272

210  36 3.772+2 Number of patients = 37 also acceptable Number of patients 

6a)(i)

Area =

1 2 11 9   2 9

99 2  155.5  156 cm 2 6a) (ii) 99  ( XA)  9  2 XA  5.5 cm 6a)(iii) OX 2  92  5.52 OX  9 2  5.52 OX  7.1239 

1  (5.50) 2 (7.1239) 3  225.6689517  226 cm 3 Volume =

Answer Scheme for 4E/5A Prelim II Mathematics 2009 Paper 2 4016

(b)

2

2  (30) 2 h   (30)3  255000 3  (30) 2 h  198451.3322 h  70.1878  70.2 cm

7a) (i)  



AC  AD  DC 

2

(3q)  p

3  2q  p

   7a (ii ) HC  HA  AC  3q  2q  p  p  q    7a (iii ) HF  HA  AF

3  3q  (2q  p ) 4 ’ 3 3  3q  q  p 2 4 3 3  p q 4 2    7a (iv ) DE  DA  AE

1  2q  (2q  p ) 2 1  pq 2  3 HF  ( p  2q) 4  1 DE  ( p  2q) 2 Fact 1 : HF and DE are parallel Fact 2 . Ratio of HF : DE = 3:2

A1

Area of ADE  2  4 7(c)(i)    Area of AHF  3  9 Area of ADE AE 1 (ii)   Area of ADC EC 2 Area of ADE 1 1   Area of ABCD 1  1  1  1 4 8) Refer to graph paper 9(a) Chargeable income = $95000 -$15297 = $79703 First $40000 : tax = $900 Next $39703 : tax = $39703 × 8.5% = $3374.76 Total = $900 + $3374.76 = $4274.76 9(b) Loan = 80%  $1200000 = $960000 Year 1 : interest : 3%  $960000 = $28800 Year 2 : interest : 3.5%  $960000 = $33600 Year 3 – 30 : Interest : 5%  $960000  28 = $1344 000 Total value of loan (including interest) = $$960000 + $28800 + $33600 + $1344 000 = $2366400 Monthly Installment = $2366400 ÷ (30  12) = $6573.33 9(c) (i) Principal Amount = $200000 ÷ 1.216 = AUD 164473.68 AUD (c)(ii) $164473.68 × (1 + 5.5÷100)10 = AUD 280944.805 In SGD : 280944.805 × 1.345 = SGD 377870.77

Answer Scheme for 4E/5A Prelim II Mathematics 2009 Paper 2 4016

2

2

2

10 a ) RT  9.8  13.5  2(9.8)(13.5) cos 55 2

RT  126.5216749 RT  11.248  11.2 cm 10b)

sin RTS

sin 55



9.8

sin RTS  0.713699 





11(d) The average weight of 4A2 is lower than that of 4A1. But the weight distribution of 4A2 is less consistent than class 4A1. 



10c )SRT  180  45.5  55  79.5 



Bearing of R from T  360  (79.5 )  280.5





10 d ) tan 40 

QR 9.8

QR  9.8  tan 40

11(a) 25% percentile = 64 kg 75% percentile = 72.5 kg M1 Interquartile range = 72.5 – 64 = 8.5 kg 11(b) 62 ≤ x < 66 : p = 9 70 ≤ x < 74 : q = 14 11(c) Mean = 68 kg Standard Deviation = 5.727  5.73 kg



11.248

RTS  45.53  45.5





QR  8.223  8.22m



21 40 19 18 17 (ii)1  (   ) 40 39 38 469  520

11e(i )

Question 8

10e )(i ) 

sin 55 

RX 9.8 

RX  sin 55  9.8 RX  8.02769  8.03 8.223

(ii) tan QXR 

8.02769 QXR  45.6885



 45.7 (a) q = 6.1 (c)(i) y = 4.3 to 4.4 (ii) 2  x  4 2

( e )3 x  10 x  8  0 3 x  10 

8

0

x 8

(d) Gradient = -1 x  2  x  2 x  8  0 x2

8

 8  2x

x Draw y  8  2 x x  1.4 and 2.0

Answer Scheme for 4E/5A Prelim II Mathematics 2009 Paper 2 4016