Pierce Prelim 2009 Em1 Solutions

PEIRCE SECONDARY SCHOOL DEPARTMENT OF MATHEMATICS SECONDARY FOUR EXPRESS / FIVE NORMAL ACADEMIC 4016 MATH 2009 PRELIM 2 PAPER 1 WORKED SOLUTION 1. 2.

3.

1.4 58208 100 68 $85600 Y  120%  X 6 Y X 5 Z  120%  Y 6 Y 5 66  Z   X 55  X 25  Z 36 X : Z = 25 : 36 k y 2 x k 4  2  (1) x k ynew  (2 x) 2

7. (a) (b)

8. (a)

(b)

Z

4.

k  (2) 4x 2 (2)  (1), y new k k  2 2 4 4x x y new k x2  2 4 4x k 1 y new   4 4 1 8 2 8 2    2 x  4 x  2 ( x  2)( x  2) x  2 8 2( x  2)   ( x  2)( x  2) ( x  2)( x  2) 8  2( x  2)  ( x  2)( x  2) 8  2x  4  ( x  2)( x  2) 12  2 x  ( x  2)( x  2) 148 169  173 median  2  171 ynew 

5.

6. (a) (b)

9.

1 goal(s) 0(4)  1(6)  2(3)  3(4)  4(2)  5(1) 20 = 1.85 goal(s) 2 ext.   180  24 15 360 no. of sides   15 sides 24 sum of int.  s  (15  2) 180  2340 also accept: (“their (a)” – 2) × 180° 2 x  1 4 x  2 5x   3 4 6 8(2 x  1) 6(4 x  2) 4(5 x)   24 24 24 8(2 x  1)  6(4 x  2) 20 x  24 24 16 x  8  24 x  12  20 x 20 x  4

mean =

1 5 2 + 6 + 10 + 14 + … + 38 = 200 = 2 × 102 288 S = 2 × (n + 1)2 2 y  3 x  5  (1) x

10. (a) (b) (c) 11.

5 x  6 y  19  (2) From (1), 3x  5 y  (3) 2 sub (3) into (2),  3x  5  5 x  6   19  2  5 x  9 x  15  19

12. (a)

 4x  4 x  1 sub x = –1 into (3), 3(1)  5 y 2  4 2x  3 x   10  2 x 5 2 2x  3 x  5 2 x  10  2 x 2(2 x  3)  5 x 2 x  20  4 x 4 x  6  5 x and 5 x  20 x6 x4 x  6  6  x  4

(b)

20. (a) (b) (c)

4

6 13. (a) (b) (c) 14. (a)

27° BCD  180  97  83 OCD  83  63  20 y

9 5 0

(b) 15. (a)

(b) 16. (a)

x

2

correct shape y−intercept/turning point x=2 1 area of ABGFA  (8  14)(4)  44 cm 2 2 shaded area  44   (1.52 )  36.93  36.9 cm 2 volume  36.93 12  443.16  443 cm 3

QP  OP  OQ

21. (a) 1.44 × 1011 m (b) 700 picometres (c) (i) 4.0 × 10 (ii) 2.25 × 10−2 22. (a) 1 : 250 000 (b) 8 cm (c) area scale = (12) cm2 : (2.52) km2

23. (a) (b)

= 1 cm2 : 6.25 km2 actual area = 6.25 × 7 = 43.75 km2 9, 7, 2, 1

OQ  OP  QP  5   3          3   4  2    1 (b) 17. (a) (b) (c)

18. (a) (b) (c) 19. (a)

QP  32  ( 4) 2  5 units

6.71 2 y 6 2 x 5 y  6  2 x  10 y  2x  4 12 8 84k  (2  3  7)  (2  (3  7))  k  3  7  21 1

2

1

x2  4x 3 4x 3  1 x x2  4x

7 1  3 2 11

(b)

 4x 6 315  81  3 x 315  34  3 x  x  15  4  11

0

(c) 24. (a)

1

2

3

4

5

3 AB 2  AC 2  12 2  5 2  169

BC 2  132  169 2 Since AB + AC2 = BC2, ∆ABC is a rightangled triangle with  CAB = 90°. (b) (i) 12 13 (ii) 12  13

25. (a)

(b)

(c)

1 1 (15)(20)  ( 20  v)(25)  900 2 2 25 150  250  v  900 2 25 v  500 2 v  40 0  40 a 60  40  2 m / s 2  deceleration = 2 m/s2 Distance (m) 1300 900

150 0

15

40 Time, t (s)

Correct shape for t = 0 to 40s. Correct shape for t = 40 to 60s.

60