WATP Sem 2 2016 Applications units 3&4 Marking Guide PDF

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SEMESTER TWO

MATHEMATICS APPLICATIONS UNITS 3 & 4 2016

SOLUTIONS

Applications Units 3&4

2

2016 Solutions

Calculatorfree Solutions 1.

(a)

(b)

2.

(a)

(i)

T3 = T2 – 5 = 95



(ii)

T1 = 105



(iii)

T11 = 105 + 10 ( - 5) = 55



(i)

Tn+1 = Tn + 5, T1 = 15



(ii)

Linear



(iii)

5



(iv)

10



(v)

Tn = 10 + 5n



V = 7, F = 6, E = 11

 (b) Vertex Odd or Even (c)

3.



V+F–E=2 A Even

[9]

 B Odd

C Even

D Odd

E Even

F Odd

G Odd

A path which starts at an odd vertex, travels along every edge, once only, and finishes at the other odd vertex. This network contains 4 odd vertices. Semi-Eulerian trails can have only 2 odd vertices. So, Peter is not correct. [7]

(a)



(b)

Minimum Completion Time = 18.



(c)

t=4



(d)

7 hours slack 

He could start at 3 pm on Saturday.

© WATP

 [7]

2016 Solutions

4.

(b)

(c)

V = 30 000 0 After 1 year: V = 30 000 (0·8) = 24 000 1 1 (i) After 2 years: V = 24 000 (0·90) = 21 600 2 2 (ii) After 3 years: V = 24 000 (0·90) = 19 440 3 (iii) After n years: V = 24 000 (0·90)n–1 n Let x be the original amount spent. x(0·9) . (1·1) = 0·99x This represents a 1% reduction in spending

   

  [6]

(a) A 24 20

A B C D E F

6.

Applications Units 3 & 4

Original Value: (a)

5.

3

B 24 26 22 21

C 26 28 -

D 28 25 27

E 22 25 23

F 20 21 27 23  

(b)

Yes

(c)

121 minutes



(d)

AFDEBC



(e)

AFBEDC is 116 minutes



(f)

Hamiltonian Path or Hamiltonian Circuit



(a)

Not just Saturday mornings. More answer options. Questions about times of day when shopping is done.



(b)

Men might do shopping on Saturday mornings to give their partners a break Some men are stay at home “mums” and always do the shopping. Women might do their shopping at night, if they work during the day. 

(c)

30



(d)

3 x 40 = 120



© WATP

[8]

[6]

Applications Units 3 & 4

7.

(a)

4

V = 8, F = 6, E = 12 8 + 6 – 12 = 2 Verified.

  

Because BF crosses AD



(b)

(c)

Solutions 2016



(d)

© WATP

[7]

2016 Solutions

5

Applications Units 3 & 4

Calculatorassumed Solutions 8.

(a) (b)

4% (i)

 

1.04(60) = 62.4

(ii) (c)

9.

(a) (b)



Because if n = 14, the maximum of 100 marks.

(i) (ii)

which is more than  [7]

There are 5 ways of getting from A to A in 2 steps.

 



(c)

10.

(i)

X=



(ii)

X2 =



(a)

W decreases



(b)

X and W because |  0.95 | > | 0.63 |



(c)

(i) (ii)



(d)

0.3969 It says that 39.69 % of the variation in one is associated with the change in the other

X, since the correlation is stronger.

[6]

  [7]

© WATP

Applications Units 3 & 4

11.

6

Solutions 2016

(a)

 (b)

12.

13.

14.

(i) (ii) (iii)

  

0.94 Y

(c)

x = 2 means y = 52.6



(d)

It is an extrapolation, so likely to be unreliable.



(e)

14.72 g per kg.



(f)

High value of r does not mean a cause/effect relationship.



(a)

Path is ACEFH with a length of 12.



Shows a systematic method , or trial and error,on the diagram



(b)

x=2

 [5]

(a)

T1 = a = 5 and T24 = ar23 = 60 r23 = 12 r = 1.114 Rate is 11.4% p.a.

 

(b)

The correct model is (i)



(c)

5(1.114)45 = $6.44 So $2.70 is much less than expected.

 

(a)

a = $15.15, b = $1530.20, c = $2030.20



(b)

4 years = 48 months. So 48 lines required.



(c)

A1 = 500, A2 = 1005, A3 = 1515.05 A49 = 31 417.42

 

(d)

31 417.42 – (49 x 500)

 © WATP

[10]



[6]

2016 Solutions

7

Applications Units 3 & 4

= $6 917.42

15.

(a)

17.



(b)

  

(c)

(ii) SC for April = t = 17

   

(d)

= 5.34 – 0.1t = 3.64 3.64 (0.756) = 2.75 million bags August 2014 has a value of 3.2 Deseasonalised value = = 4.23 million Gradient = -0.1 Negative value indicates a decreasing trend.

   [13]

(a)

(i) (ii) (iii)

   

(b)

The pattern indicates that the linear model may not be appropriate. So the geometric( exponential) may be better.

 

ACG = 8, ACEG = 5, ACFG = 1, ABEG = 4, ADFG = 5, ADCFG = 4 Total is 27 Marking of the diagram

  

(i) (ii)

 

(a)

(b) 18.

[8]

b = 6.1 (i) x = 300 – ( 84 + 76 ) = 140 %

(e) 16.



15.5 15.5 Both are reliable, since correltion is strong and 8 is an interpolation.

CG or CF or DF or CE Increase by 1

(a)

A – H1 and B – H2



(b)

$350



(c)

Because 4 lines will cover all the zeros.



(d)

4 appropriate zeros marked. H1 H2 H3 100 60 0 170 0 10 0 0 0 0 0 20 A – H3, B – H4, C – H1, D – H2 Or A – H4, B – H2, C – H3, D – H1

A B C D (e) (f)

160 + 280 + 140 + 290 = $870 Or 200 + 200 + 220 + 250 = $870

© WATP

[6]

[5]

 H4 0 0 10 30  

[6]

Applications Units 3 & 4

19.

20.

8

(a)

Solutions 2016



(b)

BD, DC, DA



(c)

103 km



(a)

Tn+1 = (1.08) Tn, T0 = 15 000



n 0 1 2 3 He owes $18 895.68 Table or calculator

 

[5]

(b)

(c)

21.

22.

e

Tn 15 000 16 200 17 496 18 895.68

= = 0.082999 = 8.30% p.a.

  

(a)

x = 0, y = 40%



(b)

247



(c)

48.6%



(d)

England (1)



(e)

( 0.04 x 120 ) + ( 0.05 x 40 ) + ( 0.3 x 16 ) + ( 0.1 x 20 ) = 14 students

 

(a)

A2 =



(b)

By solving 4An = 3An or by interrogating a table, steady state solution is An = 0

= 3.75, A3 =

= 2.8125

© WATP



[7]

[6]

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