further maths 2 2013 solutions final...

This trial examination produced by Insight Publications is NOT an official VCAA paper for the 2013 Further Mathematics 2 written examination. This examination paper is licensed to be printed, photocopied or placed on the school intranet and used only within the confines of the purchasing school for examining their students. No trial examination or part thereof may be issued or passed on to any other party, including other schools, practising or non-practising teachers, tutors, parents, websites or publishing agencies, without the written consent of Insight Publications.

Copyright © Insight Publications 2013

INSIGHT YEAR 12 Trial Exam Paper

2013 FURTHER MATHEMATICS

Written examination 2

Worked solutions

This book presents:

¾�correct solutions with full working ¾�explanatory notes ¾ mark allocations

2


CONTINUES OVER PAGE

3

Core Copyright © Insight Publications 2013

SECTION A Core: Data Analysis Question 1a. Worked solution

Mark allocation: 2 marks

x 1 mark for five-number summary points in the correct position x 1 mark for the correct convention used to draw the box plot

Question 1b. Worked solution The distribution of the weights of the Dorpers is approximately symmetrical (or slightly positively skewed) with an outlier at 43, whereas the Merino weights are negatively skewed. The centre of the Dorper weights as described by the median is 21, compared to the median of the Merinos, which is 31. The Dorper weights are more variable with an interquartile range of 10 and a range of 30, compared to an interquartile range of 8 and a range of 26 for the Merinos.


x 1 mark for comparing the shape of the distributions x 1 mark for comparing the centres (medians) of the distributions x 1 mark for comparing the spreads of the distributions

Note: Values must be used when making comparisons.

Dorpers

0 5 10 15 20 25 30

Merinos

Sheep weights (kg) 35 40 45

4


Question 1c. Worked solution

3upper fence Q (1.5 IQR)upper fence 27 (1.5 10)upper fence 42Since 43 is higher than the upper fence, 42, it is an outlier.

� u � u

Tip

x Values equal to a fence are not considered to be outliers.

Question 2a. Worked solution The graph is non-linear. Question 2b. Worked solution residual actual value predicted value

103 (49.66 3.79 12)103 95.147.868

� � � u � |

Or, alternatively, use the lists and spreadsheet on your calculator to list the residuals. Question 2c. Worked solution log x or y2

5


Question 2d. Worked solution

1Sale price 137.48 454.03age

� u

Question 2e. Worked solution

1sale price 137.48 454.0320

sale price $114.779sale price $115

� u

6


Question 3a. Worked solution June SI 12 (0.8 0.83 0.85 0.92 0.98 1.06 1.10 1.15 1.15 1.12 1.00)

1.04 � � � � � � � � � � �

Question 3b. Worked solution

67July deseasonalised rainfall1.0663.21

Question 3c. Worked solution September 2013 is month number 21. Deseasonalised rainfall 31.72 2.04 (21)

74.56 �

Rainfall predicted for September 2013 74.56 1.15

85.74 mm86 mm

u


x 1 mark for obtaining 74.56 for deseasonalised rainfall x 1 mark for obtaining 86 mm as the annual rainfall prediction for September 2013

7

Module 1: Number patterns Copyright © Insight Publications 2013

SECTION B Module 1: Number patterns Question 1a. Worked solution 500 g for the original 5 bantam hens + 3 110 g for the 3 Isa Browns

500 3 110830 g

u � u

Question 1b. Worked solution 2000 g less 500 g for the bantams leaves 1500 g. If each Isa Brown requires 110 g, Henrietta

has enough food for 1500 13.6110

Isa Browns, but you cannot have 0.6 of a chicken. Therefore,

Henrietta can afford to feed 5 bantams and 13 Isa Browns, which is 18 in total.


x 1 mark for obtaining 13.6 x 1 mark for interpreting this as 18 chickens in total.

Question 2a. Worked solution a = 2, b = 50, c = 100


x 2 marks for all three values correct OR

x 1 mark for two out of three values correct

Question 2b. Worked solution t1 = 100, t2 = 250, t3 = 550, t4 = 1150, t5 = 2350, 250 100 550 250, therefore not arithmetic

250 550 , therefore not geometric100 250

� z �

z


x 1 mark for the first 5 terms x 1 mark for proof that the sequence is neither arithmetic nor geometric

8



11

1

11

1 1

1

( 1)( 1)

50(2 1)2 100(2 1)

100 2 50 2 50

150 2 50

nn

n

nn

n

n nn

nn

b at a ta

t

tt

��

��

� �

�

� �

�

� u �

�

u � u �

u �


x 1 mark for substituting the values of a, b and t1 into the given equation (line 2) x 1 mark for simplifying correctly


1

67

7

7

150 2 50

150 2 50150 64 509550

nnt

ttt

� u �

u � u �

9


Question 3a. Worked solution

32

1 2

300 450200 3001.5

ttrt t

r

r

Question 3b. Worked solution Generate the sequence 200, 300, 450, 675, 1012.5, 1518.75, 2278.125,… These are the duck numbers at the start of each year. At the start of the 7th year she will have 2278 ducks. Therefore during the 6th year she will first exceed 2000 ducks.


10

10

10

( 1)( 1)

200(1.5 1)(1.5 1)

22 666

n

na rS

r

S

S

�

�

�

�

Question 3d. Worked solution a = 1.5 and b = 0


x 1 mark for obtaining a = 1.5 x 1 mark for obtaining b = 0

10

Module 2: Geometry and trigonometry Copyright © Insight Publications 2013

Module 2: Geometry and trigonometry Question 1a. Worked solution The height of B above the ground is the opposite side of the triangle (with respect to 20°) and the sloping edge is the hypotenuse. opposite hypotenuse sin

height 3 sin 20height 3 0.3420height 1.026height 1.0 m

u T u q u |

Question 1b. Worked solution The horizontal distance is the adjacent side of the triangle (with respect to 20°) and the sloping edge is the hypotenuse.

adjacent hypotenuse coshorizontal 3 coshorizontal 3 0.9397horizontal 2.8191horizontal 2.8 m

u T u T u |

11


Question 2a. Worked solution The three angles of a triangle add up to 180°. angle at B = 180º – 30º – 47º angle at B = 103°

Question 2b. Worked solution To find this length, use the sine rule.

sin sin5

sin 30 sin 475 sin 30

sin 473.418 m3.42 m or 342 cm

a cA C

BC

BC

BCBC

q q

u qq

|


x 1 mark for substituting the values correctly (line 2) x 1 mark for obtaining BC = 3.42 m


1area of triangle sin21area of triangle 5 8.45 sin 302

area of triangle 10.5625 marea of triangle 7.04

1.5

ab C

u u u q

Therefore, Bill can fit a maximum of 7 steers.


x 1 mark for finding the correct area of the triangle x 1 mark for obtaining a maximum of 7 steers

12



ABC� is 35 + 60 = 95 Using the cosine rule to find a side length.

2 2 2

2 2 2

2 cos120 100 2 120 100cos95162.76 m

b a c ac Bb

CA

� �

� � u u q


x 1 mark for correct substitution into the cosine rule x 1 mark for obtaining CA = 162.76 m


Find ACB� using the sine rule.

1

sin sin

sin 95sin 100162.76

sin 0.612sin 0.612 37.7389

38

C Bc b

ACB

ACB

ACB

�

u

q� | q

Therefore the bearing of A from C is 270°െ8°, which is 262°T.


x 1 mark for correct substitution into sine rule x 1 mark for the correct bearing

13



front 2 30 6 2 360end 2 10 6 2 120

roof slant 2 30 6 2 360

u u uu u u u u u

2

roof end 2 ( )( )( ) 2

roof end 2 11(11 10)(11 6)(11 6) 2roof end 2 33.17

total surface area 873.17 m

s s a s b s cu � � � u

u � � � u

u


3

volume area of end 30

volume (60 11(11 10)(11 6)(11 6) 30volume 76.58 30volume 2297.4 m

u

� � � � u

u

Question 4c. Worked solution 30 metres : 0.3 metres

100 :1

Question 4d. Worked solution Area ratio is length ratio squared.

2

2 2

Area ratio is 10 000 :1.873.17 marea of the model

10 000area of the model 0.087317 m or 873.17 cm

Alternatively, calculate the surface area of the model (as in the solution to part a) but using the model dimensions instead of the actual dimensions; e.g. length of 30 m becomes 30 cm, etc.

Given that 2

a b cs � �

14

Module 3: Graphs and relations Copyright © Insight Publications 2013

Module 3: Graphs and relations Question 1a. Worked solution R = 900n

Question 1b. Worked solution C = 1500 + 500n

Question 1c. Worked solution P = 900n – 1500 – 500n P = 400n – 1500

Question 1d. Worked solution 400 1500 0

400 1500154

3.75

nn

n

n

�

Therefore, Stacey will make a profit after she makes four or more surfboards.

15


CONTINUES OVER PAGE

16


Question 2 Worked solution 80 50 800040 60 6000

x yx y� d� d


x 1 mark for each correct constraint Question 3a. Worked solution If Jake can hire out Malibus no more than 100 times per month (20 days), he must have only five Malibus available for hire.


y

200

100

50 100 x

17



Question 3d. Worked solution P = 50x + 70y

Question 3e. Worked solution P = 50x + 70y Test vertices: (0, 0) P = $0; (0, 200) P = $14 000; (50, 200) P = $16 500; (100, 100) P = $12 000; (100, 0) P = $5000 Therefore, the maximum profit Jake can earn in a month is $16 500.

100

200

50 100

y

x

18


Question 3f. Worked solution 50 Malibus and 200 short boards (working shown in part e).

Question 3g. Worked solution

New constraint will be 300y d . Any higher than 300 and it will not be a boundary of the feasible region. This means Jake can hire out any extra short boards 100 times per month; divide this by 20 days and it means he should purchase an extra 5 short boards.


x 1 mark for obtaining 300y d x 1 mark for an extra 5 short boards

Question 3h. Worked solution P = 50x + 70y P = 50(0) + 70(300) P = $21 000

19

Module 4: Business-related mathematics Copyright © Insight Publications 2013

Module 4: Business-related mathematics Question 1a. Worked solution

21 125 100 2.5%845 000

u

Question 1b. Worked solution Using the TiNspire calculator page:

Tony must pay $85 500 in Capital Gains Tax.

20


Question 2a. Worked solution interest paid total repayments amount borrowedinterest paid 120 24 2400interest paid $480

� u �


(interest 100)=(principal time)(480 100)=(2400 2)

=10%

f

f

f

R

R

R

uu

uu


2, where is equal to the total number of repayments.

( 1)481025

19.2%

fe

e

e

R nR n

n

R

R

u

�

u

Question 2d. Worked solution Interest is charged at the flat interest rate (10%) on the initial principal ($2400) for the duration of the loan (24 months), even though this principal is being reduced during the loan. The effective interest rate (19.2%) takes into account the fact that the principal is being reduced while the interest charges remain unchanged and, therefore, the effective rate is higher.

21


Question 3a. Worked solution investment principal interest

(55 000 4.75 5)investment 55 000100

investment 55 000 13 062.50investment $68 062.50

�u u

�

�


� �5 121 4.255500012 100

67996.60

Au�§ · u¨ ¸u© ¹

Or use TVM solver:

Question 3c. Worked solution Use TVM solver:

Monthly deposit required is $152.26.

22


Question 4a. Worked solution Using TVM solver:

Quarterly instalment is $13 822.79.

Question 4b. Worked solution total repayments 13 822.79 80

$1 105 823.20interest paid $1 105 823.20 $520 000

$585 823.20

u �


x 1 mark for obtaining correct value for total repayments x 1 mark for obtaining the correct value for interest paid

23


Question 5a. Worked solution yearly depreciation 0.7 40 000

$28 000 u

Question 5b. Worked solution book value new value depreciationbook value 125 000 3 0.24 125 000book value $35 000

� � u u


Therefore, after 6 years Fred’s truck will be valued at less than $40 000.

24

Module 5: Networks and decision mathematics Copyright © Insight Publications 2013

Module 5: Networks and decision mathematics Question 1a. Worked solution ABEFG = 35 + 11 + 11 + 9 = 66 m

Question 1b. Worked solution A minimum spanning tree.



Question 1e. Worked solution F – E – H – G – D – A – B – C – F length = 219 m


x 1 mark for a correct Hamilton circuit x 1 mark for correct length

A C

D

E

F

G

H

B

25


Question 2a. Worked solution Yes. Start at classroom A and finish at E, or vice versa.

Question 2b. Worked solution Between A and E. Euler circuit.


x 1 mark for A and E x 1 mark for Euler circuit

26


Question 3a. Worked solution 4 weeks because the earliest start time for H is 8, which means we must be waiting 4 weeks for F to be completed.


Task EST LST A 0 1

B 0 3

C 0 0

D 3 4

E 3 5

F 4 4

G 8 9

H 8 8

I 12 12

J 10 11

K 4 11 Question 3c. Worked solution float = LST – EST, so 1 week Question 3d. Worked solution 14 weeks. Critical path is CFHXI.

27


Question 3e. Worked solution New minimum completion time is 13 weeks. Minimum extra cost is $10 000. The new critical path is ADI. Path CFHXI, if not reduced, takes 14 weeks, so take 1 week off C to make it another critical path. Any further reduction of C or any reduction of B has no reducing effect on the shortest completion time, and, so, would be a waste of money.


x 1 mark for 13 weeks x 1 mark for $10 000

28

Module 6: Matrices Copyright © Insight Publications 2013

Module 6: Matrices Question 1a. Worked solution 4 × 1 (4 rows and 1 column)

Queston 1b. Worked solution

The matrix product NP gives the total sales. N is a 1 ൈ 4 matrix and P is a 4 ൈ 1 matrix resulting in a 1 ൈ 1 matrix, the element of which represents total sales. PN results in a 4 ൈ 4 matrix for which, in the context of this question, the elements have no meaning.


> @

55085

(203) (567) (45) (32)180120

N

ª º« »« » u« »« »¬ ¼

> @(171 785)N Total sales at market = $171 785

29



27 250



150 20 15 114025 30 25 98545 0 25 925

151210

bgp

bgp

�ª º ª º ª º« » « » « » u« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼ª º ª º« » « » « » « »« » « »¬ ¼ ¬ ¼

Beef is $15, goat $12 and pork $10.


x 1 mark for matrix solution (line 2) x 1 mark for stating the prices of each meat

30


Question 3a. Worked solution 250 + 375 + 120 = 745

Question 3b. Worked solution Since the columns of the transition matrix add up to 1, the total number of shoppers each week remains constant at 745; i.e. the total number of shoppers stays the same each week.

Question 3c. Worked solution Accept 0.1 or 10%.


1 0

1

1

0.7 0.0 0.1 2500.1 0.9 0.3 3750.2 0.1 0.6 120

187398.5159.5

S TS

S

S

ª º ª º« » « » u« » « »« » « »¬ ¼ ¬ ¼ª º« » « »« »¬ ¼

Question 3e. Worked solution S1 is the second week since S0 is the first week.

Question 3f. Worked solution 0.1 × 375 = 37.5, therefore 38 people switched.

31


Question 3g. Worked solution S5 will be the 6th week of the study. S5 = T5S0 485 people will shop at Myway in the 6th week.

Question 3h. Worked solution Find the steady state matrix. 53 at Safeless, 532 at Myway and 160 at Coldstore.

END OF SOLUTIONS BOOK

further maths 2 2013 solutions final...

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