Band 4 – Page 67

MATRICES, FUR1 2007 VCAA 6 MC

A colony of fruit bats feeds nightly at three different locations, `A, B` and `C.`

Initially, the number of bats from the colony feeding at each of the locations was as follows.

The bats change feeding locations according to the following transition matrix `T.`

`{:(qquadqquadqquadquadtext(this night)),({:qquadqquadqquad\ A\ qquadBquadqquadC:}),(T = [(0.8, 0.1, 0.2), (0.1, 0.6, 0.1), (0.1, 0.3, 0.7)]{:(A), (B), (C):} quad quad text (next night)):}`

If this pattern of feeding continues, the number of bats feeding at location `A` in the long term will be closest to

A. `1254`

B. `1543`

C. `1568`

D. `1605`

E. `1725`

Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ n\ text{large (say}\ n = 50),`

`[(0.8, 0.1, 0.2), (0.1, 0.6, 0.1), (0.1, 0.3, 0.7)]^50[(1568),(1105),(894)]`

`= [(0.45,0.45,0.45),(0.2,0.2,0.2),(0.35,0.35,0.35)][(1568),(1105),(894)]`

`= [(1605.15),(713.4),(1248.45)]`

`=> D`

MATRICES, FUR1 2007 VCAA 4 MC

Consider the following system of three simultaneous linear equations.

`2x+z=5`

`x-2y=0`

`y-z=-1`

This system of equations can be written in matrix form as

A. `[(2, 1), (1, -2), (1, -1)][(x), (y), (z)] = [(5), (0), (-1)]`	B. `[(2,0,1), (1,-2,0), (0,1, -1)][(x), (y), (z)] = [(5), (0), (-1)]`

C. `[(2, 1, 5), (1, -2, 0), (1, -1, -1)][(x), (y), (z)] = [(5), (0), (-1)]`	D. `[(2, 1, 0), (1, -2, 0), (1, -1, 0)][(x), (y), (z)] = [(5), (0), (-1)]`

E. `[(2, 1), (1, -2), (1, -1)][(5), (0), (-1)] = [(x), (y), (z)]`

Show Answers Only

`B`

Show Worked Solution

`=> B`

MATRICES, FUR1 2011 VCAA 7 MC

Each night, a large group of mountain goats sleep at one of two locations, `A` or `B`.

On the first night, equal numbers of goats are observed to be sleeping at each location.

From night to night, goats change their sleeping locations according to a transition matrix `T`.

It is expected that, in the long term, more goats will sleep at location `A` than location `B`.

Assuming the total number of goats remains constant, a transition matrix `T` that would predict this outcome is

Show Answers Only

`A`

Show Worked Solution

`text(Consider option)\ A,`

`text(20% sleeping at)\ A\ text(move to)\ B\ text(the next night,)`

`text(while 40% sleeping at)\ B\ text(move to)\ A\ text(the next)`

`text(night.)`

`=>A`

MATRICES, FUR1 2011 VCAA 4 MC

Matrix `A` is a 3 x 4 matrix.

Matrix `B` is a 3 x 3 matrix.

Which one of the following matrix expressions is defined?

A. `BA^2`

B. `BA - 2A`

C. `A + 2B`

D. `B^2 - AB`

E. `A^-1`

Show Answers Only

`B`

Show Worked Solution

`text(Consider)\ B,`

`B`	`xx`	`A`	`=`	`BA`
`3 xx 3`		`3 xx 4`		`3 xx 4`

`:. BA`	`-`	`2Aqquadtext(is defined)`
`3 xx 4`		`3 xx 4`

`text(All other options can be shown)`

`text(to produce undefined matrices.)`

`=> B`

MATRICES, FUR1 2011 VCAA 3 MC

Each of the following four matrix equations represents a system of simultaneous linear equations.

`[(1,3),(0,2)] [(x),(y)]=[(4),(8)]`

`[(1,1),(2,2)] [(x),(y)]=[(5),(3)]`

`[(1,0),(0,2)] [(x),(y)]=[(4),(8)]`

`[(0,3),(0,2)] [(x),(y)]=[(6),(12)]`

How many of these systems of simultaneous linear equations have a unique solution?

A. 0

B. 1

C. 2

D. 3

E. 4

Show Answers Only

`C`

Show Worked Solution

`text(Consider the 1st system,)`

`Delta = text(det)[(1,3),(0,2)] = 1 xx 2 – 3 xx 0 = 2 != 0`

`:.\ text(Unique solutions exists)`

`text(Similarly for the other systems, we find)`

`text(that)\ Delta != 0\ text{in two (total).}`

`=> C`

MATRICES, FUR1 2014 VCAA 8 MC

Wendy will have lunch with one of her friends each day of this week.

Her friends are Angela (A), Betty (B), Craig (C), Daniel (D) and Edgar (E).

On Monday, Wendy will have lunch with Craig.

Wendy will use the transition matrix below to choose a friend to have lunch with for the next four days of the week.

`{:(qquad qquad qquad qquad qquad text(today)),(\ quad qquad qquad {:(A, B, C, D, E):}),(T = [(0,\ 1,\ 0,\ 0,\ 0), (0,\ 0,\ 0,\ 1,\ 0), (1,\ 0,\ 0,\ 0,\ 0), (0,\ 0,\ 0,\ 0,\ 1), (0,\ 0,\ 1,\ 0,\ 0)] {:(A), (B), (C), (D), (E):} quad text(tomorrow)):}`

The order in which Wendy has lunch with her friends for the next four days is

A. Angela, Betty, Craig, Daniel

B. Daniel, Betty, Angela, Craig

C. Daniel, Betty, Angela, Edgar

D. Edgar, Angela, Daniel, Betty

E. Edgar, Daniel, Betty, Angela

Show Answers Only

`E`

Show Worked Solution

`T xx [(0), (0), (1), (0), (0)] = [(0), (0), (0), (0), (1)] {:(), (), (), (), (E):},\ \ Txx [(0), (0), (0), (0), (1)] = [(0), (0), (0), (1), (0)] {:(), (), (), (D), ():}`

`T xx [(0), (0), (0), (1), (0)] = [(0), (1), (0), (0), (0)] {:(), (B), (), (), ():},\ \ Txx [(0), (1), (0), (0), (0)] = [(1), (0), (0), (0), (0)] {:(A), (), (), (), ():}`

`:.\ text(The order is)\ EDBA`

`=>E`

MATRICES, FUR1 2014 VCAA 5 MC

Students from Year 7 and Year 8 in a school sold trees to raise funds for a school trip.

The number of large, medium and small trees that were sold by each year group is shown in the table below.

The large trees were sold for $32 each, the medium trees were sold for $26 each and the small trees were sold for $18 each.

A matrix product that can be used to calculate the amount, in dollars, raised by each year group by selling trees is

Show Answers Only

`C`

Show Worked Solution

`=>C`

MATRICES, FUR2 2015 VCAA 3

A new model for the number of students in the school after each assessment takes into account the number of students who are expected to leave the school after each assessment.

After each assessment, students are classified as beginner (`B`), intermediate (`I`), advanced (`A`) or left the school (`L`).

Let matrix `T_2` be the transition matrix for this new model.

Matrix `T_2`, shown below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.

`{:(qquadqquadqquadquadtext(before assessment)),(qquadqquadqquadquad{:(B,\ qquadI,qquadA,quadL):}),(T_2 = [(0.30,0,0,0),(0.40,0.70,0,0),(0.05,0.20,0.75,0),(0.25,0.10,0.25,1)]{:(B),(I),(A),(L):}qquadtext(after assessment)):}`

An incomplete transition diagram for matrix `T_2` is shown below.

Complete the transition diagram by adding the missing information. (2 marks)

The number of students at each level, immediately before the first assessment of the year, is shown in matrix `R_0` below.

`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`

Matrix `T_2`, repeated below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.

What percentage of students is expected to leave the school after the first assessment? (1 mark)
How many advanced-level students are expected to be in the school after two assessments

Write your answer correct to the nearest whole number. (1 mark)
After how many assessments is the number of students in the school, correct to the nearest whole number, first expected to drop below 50? (1 mark)

Another model for the number of students in the school after each assessment takes into account the number of students who are expected to join the school after each assessment.

Let `R_n` be the state matrix that contains the number of students in the school immediately after `n` assessments.

Let `V` be the matrix that contains the number of students who join the school after each assessment.

Matrix `V` is shown below.

`V = [(4),(2),(3),(0)]{:(B),(I),(A),(L):}`

The expected number of students in the school after `n` assessments can be determined using the matrix equation

`R_(n + 1) = T_2 xx R_n + V`

where

`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`

Consider the intermediate-level students expected to be in the school after three assessments.

How many are expected to become advanced-level students after the next assessment?

Write your answer correct to the nearest whole number. (2 marks)

Show Answers Only

`17.5 text(%)`
`43`
`5`
`7`

Show Worked Solution

b.	`T_2 R_0`	`= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)] [(20), (60), (40), (0)]`
		`= [(6), (50), (43), (21)]`

`:.\ text(Percentage that leave school)`

♦♦ Mean mark of parts (b)-(e) was 32%.

`= 21/120`

`= 17.5 text(%)`

c. `text(After 2 assessments,)`

`(T_2)^2 R_0`	`= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)]^2 [(20), (60), (40), (0)]`
	`=[(1.8), (37.4), (42.55), (38.25)]`

`:.\ text(43 advanced-level students expected to remain.)`

d. `text(After 4 assessments,)`

`(T_2)^4 R_0\ text(shows 54 students left)`

`text(After 5 assessments,)`

`(T_2)^5 R_0\ text(shows 43 students left.)`

`:.\ text(Numbers drop below 50 after 5 assessments.)`

MARKER’S COMMENT: Understand why the common error `R_3 =` `(T_2)^3R_0 + V` is incorrect!

e.	`R_1`	`= T_2 R_0 + V`
		`= [(6), (50), (43), (21)] + [(4), (2), (3), (0)] = [(10), (52), (46), (21)]`
	`R_2`	`= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(10), (52), (46), (21)] + [(4), (2), (3), (0)] = [(7), (42.4), (48.4), (40.2)]`
	`R_3`	`= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(7), (42.4), (48.4), (40.2)] + [(4), (2), (3), (0)] = [(6.1), (34.48), (48.13), (58.29)]`

`:.\ text(After 3 assessments, the number expected to)`

`text(move from Intermediate to Advanced)`

`= 20text(%) xx 34.48`

`= 6.896`

`= 7\ text{students (nearest whole)}`

MATRICES, FUR2 2012 VCAA 3

When a new industrial site was established at the beginning of 2011, there were 350 staff at the site.

The staff comprised 100 apprentices (`A`), 200 operators (`O`) and 50 professionals (`P`).

At the beginning of each year, staff can choose to stay in the same job, move to a different job at the site or leave the site (`L`).

The number of staff in each category at the beginning of 2011 is given in the matrix

`S_2011 = [(100), (200), (50), (0)]{:(A), (O), (P), (L):}`

The transition diagram below shows the way in which staff are expected to change their jobs at the site each year.

How many staff at the site are expected to be working in their same jobs after one year? (1 mark)

The information in the transition diagram has been used to write the transition matrix `T`.

`{:(qquad qquad qquad qquad qquad qquad text(this year)),((qquad qquad qquad\ A, qquad O, qquad P, qquad L)),(T = [(0.70, 0, 0, 0),(0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)]):} {:(), (), (A), (O), (P), (L):} {:(), (), (qquad text(next year)):}`

Explain the meaning of the entry in the fourth row and fourth column of transition matrix `T`. (1 mark)

If staff at the site continue to change their jobs in this way, the matrix `S_n` will contain the number of apprentices (`A`), operators (`O`), professionals (`P`) and staff who leave the site (`L`) at the beginning of the `n`th year.

Use the rule `S_(n + 1) = TS_n` to ﬁnd
1. `S_2012` (1 mark)
2. the expected number of operators at the site at the beginning of 2013 (1 mark)
3. the beginning of which year the number of operators at the site ﬁrst drops below 30 (1 mark)
4. the total number of staff at the site in the longer term. (1 mark)

Suppose the manager decides to bring 30 new apprentices, 20 new operators and 10 new professionals to the site at the beginning of each year.

The matrix `S_(n + 1)` will then be given by

`S_(n + 1) = T S_n + A` where `S_2011 = [(100), (200), (50), (0)] {:(A), (O), (P), (L):}` and `A = [(30), (20), (10), (0)] {:(A), (O), (P), (L):}`

Find the expected number of operators at the site at the beginning of 2013. (2 marks)

Show Answers Only

`275`
`text(Any worker who has left the)`
`text(site will not return.)`
1. `[(70), (170), (65), (45)]`
2. `143`
3. `2021`
4. `0`
`182`

Show Worked Solution

a. `text(Same jobs after 1 year)`

♦ Mean mark for parts (a)-(d) (combined) was 37%.

`= 0.7 xx 100 + 0.8 xx 200 + 0.9 xx 50`

`= 275`

b. `text(Any worker who has left the site will)`

MARKER’S COMMENT: Most students couldn’t explain the meaning of the 1.0 figure in this context.

`text(not return.)`

c.i.	`S_2012`	`= T S_2011`
		`= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (200), (50), (0)]`
		`= [(70),(150),(65),(45)]`

c.ii.	`S_2013`	`= T S_2012`
		`= [(49), (143), (75.5), (82.5)]`

`:.\ text(143 operators are expected at the site at the)`

`text(start of 2013.)`

c.iii.	`S_2020`	`= T^9 S_2011`
		`= [(0.70, 0, 0, 0), (0.10, 0.8, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)]^9 [(100), (200), (50), (0)] = [(4), (36.2), (78), (231.8)]`
	`S_2021`	`= T^10 S_2011`
		`= [(2.8), (29.4), (73.8), (244)]`

`:.\ text(At the start of 2021, the number of operators)`

MARKER’S COMMENT: “In the 10th year” recieved no marks – make sure you answer with specific years when actual years are used.

`text(drops below 30.)`

c.iv. `text(Consider)\ n\ text(large,)`

`T^100 S_2011 = [(0), (0), (0), (350)]`

`:.\ text(NO staff remain at the site in the long term.)`

d.	`S_2012`	`= T S_2011 + A`
		`= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (200), (50), (0)] + [(30), (20), (10), (0)]`
		`= [(100), (190), (75), (45)]`

	`S_2013`	`= T S_2012 + A`
		`= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (190), (75), (45)] + [(30), (20), (10), (0)]`
		`= [(100), (182), (96.5), (91.5)]`

MARKER’S COMMENT: A common error was `S_2013“=T^2S_2011“+A`. Know why this is not correct!

`:.\ text(182 operators are expected on site at the )`

`text(start of 2013.)`

MATRICES, FUR2 2008 VCAA 2

The following transition matrix, `T`, is used to help predict class attendance of History students at the university on a lecture-by-lecture basis.

`{:(qquad qquad qquad text(this lecture)),(qquad qquad text(attend not attend)),(T = [(0.90 qquad,0.20),(0.10 qquad,0.80)]{:(text(attend)),(text(not attend)):}{:qquad text(next lecture):}):}`

`S_1` is the attendance matrix for the first History lecture.

`S_1 = [(540),(36)]{:(text(attend)),(text(not attend)):}`

`S_1` indicates that 540 History students attended the first lecture and 36 History students did not attend the first lecture.

Use `T` and `S_1` to
1. determine `S_2` the attendance matrix for the second lecture (1 mark)
2. predict the number of History students attending the fifth lecture. (1 mark)
Write down a matrix equation for `S_n` in terms of `T`, `n` and `S_1`. (1 mark)

The History lecture can be transferred to a smaller lecture theatre when the number of students predicted to attend falls below 400.

For which lecture can this first be done? (1 mark)
In the long term, how many History students are predicted to attend lectures? (1 mark)

Show Answers Only

1. `[(493.2),(82.8)]`
2. `421`
`S_n = T^(n – 1)S_1`
`text(8th lecture)`
`384`

Show Worked Solution

a.i.	`S_2`	`= TS_1`
		`= [(0.9, 0.2), (0.1, 0.8)] [(540), (36)]`
		`= [(493.2), (82.8)]`

ii.	`S_5`	`= TS_4`
		`= [(0.9, 0.2), (0.1, 0.8)]^4 [(540), (36)]`
		`= [(421.46), (154.54)]`

`:. 421\ text(History students attend the 5th lecture.)`

b. `S_n = T^(n – 1)S_1`

c. `S_8 = [(396.847),(179.15)]`

`:.\ text(The 8th lecture is predicted to be the 1st one)`

`text(with less than 400 students.)`

d. `text(Consider)\ n = 51 and n = 52,`

`S_51 = T^50 S_1 = [(384.00), (191.99)]`

`S_52 = T^51 S_1 = [(384.00), (191.99)]`

`:. 384\ text(students are predicted to attend in)`

`text(the long run.)`

GRAPHS, FUR2 2012 VCAA 3

A company repairs phones and laptops.

Let `x` be the number of phones repaired each day

`y` be the number of laptops repaired each day.

It takes 35 minutes to repair a phone and 50 minutes to repair a laptop.

The constraints on the company are as follows.

Constraint 1 `x ≥ 0`

Constraint 2 `y ≥ 0`

Constraint 3 `35x + 50y ≤ 1750`

Constraint 4 `y ≤ 4/5 x`

Explain the meaning of Constraint 3 in terms of the time available to repair phones and laptops. (1 mark)
Constraint 4 describes the maximum number of phones that may be repaired relative to the number of laptops repaired.

Use this constraint to complete the following sentence.

For every ten phones repaired, at most _______ laptops may be repaired. (1 mark)

The line `y = 4/5 x` is drawn on the graph below.

Draw the line `35x + 50y = 1750` on the graph. (1 mark)
Within Constraints 1 to 4, what is the maximum number of laptops that can be repaired each day? (1 mark)
On a day in which exactly nine laptops are repaired, what is the maximum number of phones that can be repaired? (1 mark)

The profit from repairing one phone is $60 and the profit from repairing one laptop is $100.

1. Determine the number of phones and the number of laptops that should be repaired each day in order to maximise the total profit. (2 marks)
2. What is the maximum total profit per day that the company can obtain from repairing phones and laptops? (1 mark)

Show Answers Only

`text(Max total repair time is 1750 minutes.)`
`8`
`text(See Worked Solutions)`
`18`
`37`
1. `(24,18)`
2. `$3240`

Show Worked Solution

a. `text(Constraint 3 means that the maximum time)`

`text(available to repair phones and laptops is 1750)`

`text(minutes on any given day.)`

b. `y <= 4/5x`

`text(When)\ x = 10,`

`y`	`<= 4/5 xx 10`
	`<= 8`

`:.\ text(At most, 8 laptops may be repaired)`

d. `text(From the graph, the highest number)`

♦♦ Mean mark of parts (c) – (f) combined was 32%.

`text(of laptops below the intersection)`

`text{(in the feasible region) is 18.}`

e. `text(When)\ y = 9,\ text(constraint 3 requires)`

`35x + (50 xx 9)`	`<= 1750`
`35x`	`<= 1300`
`:. x`	`<= 37.14…`

`:.\ text(The maximum number of phones = 37.)`

f.i. `text(Profit)\ = 60x + 100y`

`text(In the feasible region, maximum)`

`text(profits occur at the intersection.)`

`:. 18\ text(laptops)`

`text(When)\ y = 18,\ text(constraint 3 requires)`

`35x + (50 xx 18)`	`<= 1750`
`35x`	`<= 850`
`x`	`<= 24.28…`

`:. text(Maximum profit occurs at)\ (24,18)`

f.ii. `text(Maximum daily profit)`

`= 60 xx 24 + 100 xx 18`

`= $3240`

GRAPHS, FUR2 2012 VCAA 1

The cost, `C`, in dollars, of making `n` phones, is shown by the line in the graph below.

1. Calculate the gradient of the line, `C`, drawn above. (1 mark)
2. Write an equation for the cost, `C`, in dollars, of making `n` phones. (1 mark)
The revenue, `R`, in dollars, obtained from selling `n` phones is given by `R = 150n`
.
1. Draw this line on the graph above. (1 mark)
2. How many phones would need to be sold to obtain $54 000 in revenue? (1 mark)
Determine the number of phones that would need to be made and sold to break even. (1 mark)

Show Answers Only

1. `50`
2. `C = 20\ 000 + 50n`
1. `text(See Worked Solutions)`
2. `360`
`200`

Show Worked Solution

a.i. `text{Using (0, 20 000) and (300, 35 000)}`

`text(Gradient)`	`= (y_2 – y_1)/(x_2 – x_1)`
	`= (35\ 000 – 20\ 000)/(300 – 0)`
	`= 50`

a.ii. `C = 50n + 20\ 000`

b.i.

b.ii.	`54\ 000`	`= 150n`
	`:. n`	`= (54\ 000)/150`
		`= 360`

`:. 360\ text(planes need to be sold.)`

c. `text(Breakeven occurs when)`

`text(Revenue)`	`=\ text(C)text(osts)`
`150n`	`= 50n + 20\ 000`
`100n`	`= 20\ 000`
`:. n`	`= 200\ text(phones)`

GEOMETRY, FUR2 2012 VCAA 4

`OABCD` has three triangular sections, as shown in the diagram below.

Triangle `OAB` is a right-angled triangle.

Length `OB` is 10 m and length `OC` is 14 m.

Angle `AOB` = angle `BOC` = angle `COD` = 30°

Calculate the length, `OA`.

Write your answer, in metres, correct to two decimal places. (1 mark)
Determine the area of triangle `OAB`.

Write your answer, in m², correct to one decimal place. (1 mark)
Triangles `OBC` and `OCD` are similar.

The area of triangle `OBC` is 35 m².

Find the area of triangle `OCD`, in m². (2 marks)
Determine angle `CDO`.

Write your answer, correct to the nearest degree. (2 marks)

Show Answers Only

`8.66\ text{m (2 d.p.)}`
`21.7\ text{m² (1 d.p.)}`
`68.6\ text(m²)`
`43^@\ \ text{(nearest degree)}`

Show Worked Solution

a. `text(In)\ DeltaOAB,`

`cos30^@`	`= (OA)/10`
`:. OA`	`= 10 xx cos30`
	`= 8.660…`
	`= 8.66\ text{m (2 d.p.)}`

b. `text(Using the sine rule,)`

`text(Area)\ DeltaAOB`	`= 1/2 ab sinC`
	`= 1/2 xx 8.66 xx 10 xx sin30^@`
	`= 21.65…`
	`= 21.7\ text{m² (1 d.p.)}`

c. `text(Linear scale factor) = 14/10 = 1.4`

`:. (text(Area)\ DeltaOCD)/(text(Area)\ DeltaOBC)`	`= 1.4^2`

`:. text(Area)\ DeltaOCD`	`= 35 xx 1.4^2`
	`= 68.6\ text(m²)`

d. `text(In)\ DeltaBCO,`

`BC^2`	`= 10^2 + 14^2 – 2 xx 10 xx 14 xx cos30^@`
	`= 53.51…`
`:. BC`	`= 7.315…\ text(m)`

`text(Using the sine rule in)\ DeltaBCO,`

`(sin angleBCO)/10`	`= (sin30^@)/(BC)`
`sin angleBCO`	`= (10 xx sin30)/(7.315…)`
	`= 0.683…`
`:. angleBCO`	`= 43.11^@`

`text(S)text(ince)\ DeltaBCO\ text(|||)\ DeltaCDO,`

`angleCDO`	`= angleBCO`
	`= 43^@\ \ text{(nearest degree)}`

GEOMETRY, FUR2 2012 VCAA 3

A tree is growing near the block of land.

The base of the tree, `T`, is at the same level as the corners, `P` and `S`, of the block of land.

Show that, correct to two decimal places, distance `ST` is 41.81 metres. (1 mark)
From point `S`, the angle of elevation to the top of the tree is 22°.

Calculate the height of the tree.

Write your answer, in metres, correct to one decimal place. (1 mark)

Show Answers Only

`text(See Worked Solutions)`
`16.9\ text(m)`

Show Worked Solution

a.	`∠STP`	`= 180 − (72 + 47)`
		`= 61^@`

`text(Using the sine rule:)`

`(ST)/(sin47^@)`	`= 50/(sin61^@)`
`ST`	`= (50 xx sin47^@)/(sin61^@)`
	`= 41.809…`
	`= 41.81\ text{m (to 2 d.p.) … as required}`

b. `text(Let)\ \ X\ text(be the top of the tree)`

♦♦ Exact data unavailable although this part was highlighted as “very poorly answered”.

`text(In)\ DeltaSXT, XT\ text(is height of tree.)`

`tan22^@`	`= (XT)/(41.81)`
`:. XT`	`= 41.81 xx tan22^@`
	`= 16.892…`
	`= 16.9\ text{m (1 d.p.)}`

GEOMETRY, FUR2 2012 VCAA 1

A rectangular block of land has width 50 metres and length 85 metres.

Calculate the area of this block of land.

Write your answer in m². (1 mark)

In order to build a house, the builders dig a hole in the block of land.

The hole has the shape of a right-triangular prism, `ABCDEF`.

The width `AD` = 20 m, length `DC` = 25 m and height `EC` = 4 m are shown in the diagram below.

Calculate the volume of the right-triangular prism, `ABCDEF`.

Write your answer in m³. (1 mark)

Once the right-triangular prism shape has been dug, a fence will be placed along the two sloping edges, `AF` and `DE`, and along the edges `AD` and `FE`.

Calculate the total length of fencing that will be required.

Write your answer, in metres, correct to one decimal place. (1 mark)

Show Answers Only

`4250\ text(m²)`
`1000\ text(m³)`
`90.6\ text(m)`

Show Worked Solution

a.	`text(Area)`	`= 50 xx 85`
		`= 4250\ text(m²)`

b.	`V`	`= Ah`
	`A`	`= 1/2 xx 4 xx 25`
		`= 50\ text(m²)`

`:. V`	`= 50 xx 20`
	`= 1000\ text(m³)`

c. `text(Using Pythagoras in)\ DeltaECD,`

`ED`	`= sqrt(25^2 + 4^2)`
	`= sqrt(641)`
	`= 25.317…`

MARKER’S COMMENT: A poorly done question with many students failing to include the two 20 metre sections of fencing.

`:.\ text(Total length of fencing)`

`= 2 xx 20 + 2 xx 25.317…`

`= 90.635…`

`= 90.6\ text(m)`

CORE*, FUR2 2013 VCAA 3

Hugo paid $7500 for a second bike under a hire-purchase agreement.

A flat interest rate of 8% per annum was charged.

He will fully repay the principal and the interest in 24 equal monthly instalments.

Determine the monthly instalment that Hugo will pay.

Write your answer in dollars, correct to the nearest cent. (2 marks)
Find the effective rate of interest per annum charged on this hire-purchase agreement.

Write your answer as a percentage, correct to two decimal places. (1 mark)
Explain why the effective interest rate per annum is higher than the flat interest rate per annum. (1 mark)
The value of his second bike, purchased for $7500, will be depreciated each year using the reducing balance method of depreciation.

One year after it was purchased, this bike was valued at $6375.

Determine the value of the bike five years after it was purchased.

Write your answer, correct to the nearest dollar. (2 marks)

Show Answers Only

`$362.50`
`15.36text(%)`
`text(Simple interest does not take the)`
`text(reducing balance over the life)`
`text(of the loan into account.)`
`$3328\ \ text{(nearest dollar)}`

Show Worked Solution

a.	`text(Total interest)`	`= (PrT)/100`
		`= (7500 xx 8 xx 2)/100`
		`= $1200`

`:.\ text(Total to repay)`	`= 7500 + 1200`
	`= $8700`

`:.\ text(Repayments)`	`= 8700/24`
	`= $362.50`

♦ Mean mark of all parts combined was 37%.

b.	`text(Effective Rate)`	`= (2n)/(n + 1) xx text(flat rate)`
		`= (2 xx 24)/(24 + 1) xx 8`
		`= 15.36text(%)`

c. `text(Simple interest does not take the reducing)`

`text(balance over the life of the loan into account.)`

d.	`text(Depreciation Rate)`	`= 1 – 6375/7500`
		`= 0.15`
		`= 15text(%)`

`:.\ text(Value after 5 years)`

`= 7500 (1 – 0.15)^5`

`= 7500(0.85)^5`

`= 3327.789…`

`= $3328\ \ text{(nearest dollar)}`

CORE*, FUR2 2013 VCAA 2

Hugo won $5000 in a road race and invested this sum at an interest rate of 4.8% per annum compounding monthly.

What is the value of Hugo’s investment after 12 months?

Write your answer in dollars, correct to the nearest cent. (1 mark)
1. Suppose instead that at the end of each month Hugo added $200 to his initial investment of $5000.
  
  Find the value of this investment immediately after the 12th monthly payment of $200 is made.
  
  Write your answer in dollars, correct to the nearest cent. (1 mark)
2. Assume Hugo follows the investment that is described in part b.i.
  
  Determine the total interest he would earn over the 12-month period.
  
  Write your answer in dollars, correct to the nearest cent. (1 mark)

Show Answers Only

`$5245.35`
1. `$7698.86`
2. `$298.86`

Show Worked Solution

a. `text(Monthly interest rate)`

♦ Mean mark for all parts combined was 39%.

`= 4.8/12`

`= 0.4text(%)`

`:.\ text(Value after 12 months)`

`= 5000(1 + 0.4/100)^12`

`= 5000(1.004)^12`

`= 5245.351…`

`= $5245.35\ \ text{(nearest cent)}`

b.i. `text(Using TVM Solver,)`

`N`	`= 12`
`I(text(%))`	`= 4.8`
`PV`	`= −5000`
`PMT`	`= −200`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`FV = 7698.8614…`

`:. text(Value of interest) = $7698.86`

b.ii. `text(Total interest)`

`=\ text{Final value − (original + payments)}`

`= 7698.86 – (5000 + 12 xx 200)`

`= 7698.86 – 7400`

`= $298.86`

CORE*, FUR2 2013 VCAA 1

Hugo is a professional bike rider.

The value of his bike will be depreciated over time using the flat rate method of depreciation.

The graph below shows his bike’s initial purchase price and its value at the end of each year for a period of three years.

What was the initial purchase price of the bike? (1 mark)
1. Show that the bike depreciates in value by $1500 each year. (1 mark)
2. Assume that the bike’s value continues to depreciate by $1500 each year.
  
  Determine its value five years after it was purchased. (1 mark)

The unit cost method of depreciation can also be used to depreciate the value of the bike.

In a two-year period, the total depreciation calculated at $0.25 per kilometre travelled will equal the depreciation calculated using the flat rate method of depreciation as described above.

Determine the number of kilometres the bike travels in the two-year period. (1 mark)

Show Answers Only

`$8000`
1. `text(See Worked Solutions)`
2. `$500`
`12\ 000\ text(km)`

Show Worked Solution

a. `$8000`

b.i. `text(Value after 1 year) = $6500`

`:.\ text(Annual depreciation)`	`= 8000 – 6500`
	`= $1500`

b.ii. `text(Value after)\ n\ text(years)`

`= 8000 – 1500n`

`:.\ text(After 5 years,)`

`text(Value)`	`= 8000 – 1500 xx 5`
	`= $500`

c. `text(After 2 years,)`

`text(Depreciation)`	`= 2 xx 1500`
	`= $3000`

`:.\ text(Distance travelled)`

`= 3000/0.25`

`= 12\ 000\ text(km)`

GRAPHS, FUR2 2013 VCAA 4

The school group may hire two types of camp sites: powered sites and unpowered sites.

Let `x` be the number of powered camp sites hired

`y` be the number of unpowered camp sites hired.

Inequality 1 and inequality 2 give some restrictions on `x` and `y`.

inequality 1 `x ≤ 5`

inequality 2 `y ≤ 10`

There are 48 students to accommodate in total.

A powered camp site can accommodate up to six students and an unpowered camp site can accommodate up to four students.

Inequality 3 gives the restrictions on `x` and `y` based on the maximum number of students who can be accommodated at each type of camp site.

inequality 3 `ax + by ≥ 48`

Write down the values of `a` and `b` in inequality 3. (1 mark)

School groups must hire at least two unpowered camp sites for every powered camp site they hire.

Write this restriction in terms of `x` and `y` as inequality 4. (1 mark)

The graph below shows the three lines that represent the boundaries of inequalities 1, 3 and 4.

On the graph above, show the points that satisfy inequalities 1, 2, 3 and 4. (1 mark)
Determine the minimum number of camp sites that the school would need to hire. (1 mark)
The cost of each powered camp site is $60 per day and the cost of each unpowered camp site is $30 per day.
1. Find the minimum cost per day, in total, of accommodating 48 students. (1 mark)

School regulations require boys and girls to be accommodated separately.

The girls must all use one type of camp site and the boys must all use the other type of camp site.

Determine the minimum cost per day, in total, of accommodating the 48 students if there is an equal number of boys and girls. (1 mark)

Show Answers Only

`a = 6\ text(and)\ b = 4`
`y ≥ 2x`
`text(11 camp sites)`
1. `$390`
2. `$480`

Show Worked Solution

a. `a = 6`

`b = 4`

b. `text(Inequality 4)`

`y ≥ 2x`

d. `text(From the graph, minimum number)`

♦♦ Mean mark of parts (c)-(e) combined was 22%.

`text(of camps is 11, which occurs at)\ (2,9)`

`text(and)\ (3,8).`

e.i. `text(Test)\ (2,9)\ text(and)\ (3,8)\ text(for minimum cost.)`

`text(At)\ (2,9),`

`text(C)text(ost) = 60 xx 2 + 30 xx 9 = $390`

`text(At)\ (3,8),`

`text(C)text(ost) = 60 xx 3 + 30 xx 8 = $420`

`:. text(Minimum cost per day) = $390.`

e.ii. `text(If equal boys and girls at each site.)`

`text(Powered sites) = 24/6 = 4`

`text(Unpowered sites) = 24/4 = 6`

`:. x >= 4\ text(and)\ y = 8`

`:.\ text(Minimum cost)`	`= 60 xx 4 + 30 xx 8`
	`= $480`

GRAPHS, FUR2 2013 VCAA 3

A rock-climbing activity will be offered to students at the camp on one afternoon.

Each student who participates will pay $24.

The organisers have to pay the rock-climbing instructor $260 for the afternoon. They also have to pay an insurance cost of $6 per student.

Let `n` be the total number of students who participate in rock climbing.

Write an expression for the profit that the organisers will make in terms of `n`. (1 mark)
The organisers want to make a profit of at least $500.

Determine the minimum number of students who will need to participate in rock climbing. (1 mark)

Show Answers Only

`text(Profit) = 18n − 260`
`43\ text(students)`

Show Worked Solution

a. `text(Revenue = 24)n`

`text(C)text(osts) = 260 + 6n`

`:.\ text(Profit)`	`= 24n – (260 + 6n)`
	`= 18n – 260`

b. `text(Profit) >= 500`

`:. 18n – 260`	`>= 500`
`18n`	`>= 760`
`n`	`>= 42.22…`

`:. 43\ text(students are needed to make)`

`text(a profit) >= $500.`

GEOMETRY, FUR2 2014 VCAA 3

The chicken coop contains a circular water dish.

Water flows into the dish from a water container.

The water container is in the shape of a cylinder with a hemispherical top.

The water container and the dish are shown in the diagrams below.

The cylindrical part of the water container has a diameter of 10 cm and a height of 15 cm.

The hemisphere has a radius of 5 cm.

What is the surface area of the hemispherical top of the water container?

Write your answer, correct to the nearest square centimetre. (1 mark)
What is the maximum volume of water that the water container can hold?

Write your answer, correct to the nearest cubic centimetre. (2 marks)

The eating space of the chicken coop also has a feed container.

The feed container is similar in shape to the water container.

The volume of the water container is three-quarters of the volume of the feed container.

The surface area of the water container is 628 cm².

What is the surface area of the feed container?

Write your answer, correct to the nearest square centimetre. (2 marks)

Show Answers Only

`157\ text(cm)²`
`1440\ text(cm)³`
`761\ text(cm)²`

Show Worked Solution

a.	`text(Area)`	`= 1/2 xx (4pi xx r^2)`
		`= 1/2 xx (4pi xx 5^2)`
		`= 157\ text{cm² (nearest cm²)}`

b.	`text(Volume of Cylinder)`	`=pi r^2 h`
		`=pi xx 5^2 xx 15`
		`=1178.09…\ text(cm³)`
	`text(Volume of Hemisphere)`	`=1/2 xx 4/3 pi r^3`
		`=1/2 xx 4/3 xx pi xx 5^3`
		`=261.79…\ text(cm³)`

♦ Mean mark of all parts of question (combined) was 44%.

`:.\ text(Maximum volume of the container)`

`=1178.09… + 261.79…`

`=1439.8…`

`=1440\ text{cm³ (nearest cm³)}`

c.	`text(Volume Factor)`	`=4/3`
	`text(Linear Factor)`	`=root3(4/3)`
	`text(Area Factor)`	`=(root3(4/3))^2=1.2114…`

`:.\ text(S.A. of Feed Container)`

`=628 xx 1.2114…`

`=760.7…`

`=761\ text(cm²)`

GEOMETRY, FUR2 2013 VCAA 3

A grassed region in the athletics ground is shown shaded in the diagram below.

The perimeter of the grassed region comprises two parallel lines, `BA` and `CD`, each 100 m in length, and two semi-circles, `BC` and `AD`.

In total, the perimeter of the grassed region is 400 m.

The diameter of the semi-circle `AD` is 63.66 m, correct to two decimal places.

Show how this value could be obtained. (1 mark)
Determine the area of the grassed region, correct to the nearest square metre. (1 mark)

A running track, shown shaded in the diagram below, surrounds the grassed region. This running track is 8 m wide at all points.

The running track is to be resurfaced with special rubber material that is 0.1 m deep.

Find the volume of rubber material that is needed to resurface the running track.

Write your answer, correct to the nearest cubic metre. (2 marks)

Show Answers Only

`63.66\ text{m (2d.p.)}`
`9549\ text{m² (nearest m²)}`
`340\ text{m³ (nearest m³)}`

Show Worked Solution

a. `text(Total perimeter = 400 m)`

`text(Perimeter)`	`= AB + CD + 2\ text(semi-circles)`
`400`	`= 100 + 100 + 2 xx (1/2 xx pi xx AD)`
	`= 200 + pi xx AD`
`:. AD`	`= 200/pi`
	`= 63.66\ text{m (2d.p.)}`

b.	`text(Grassed Area)`	`= 100 xx 63.66 + 2 xx 1/2 xx pi xx (63.66/2)^2`
		`= 6366 + 3182.90…`
		`= 9548.90…`
		`= 9549\ text{m² (nearest m²)}`

c. `text(Area within the outer boundary)`

`= 100 xx (63.66 + 16) + 2 xx 1/2 xx pi xx ((63.66 + 16)/2)^2`

`= 7966 + 4983.91…`

`= 12\ 949.91…\ text(m²)`

`:.\ text(Volume of rubber)`

♦ Mean mark of part (c) 34%.

`=\ text(Area of track × depth)`

`= (12\ 949.91 – 9549) xx 0.1`

`= 340.09`

`= 340\ text{m³ (nearest m³)}`

GEOMETRY, FUR2 2013 VCAA 2

A concrete staircase leading up to the grandstand has 10 steps.

The staircase is 1.6 m high and 3.0 m deep.

Its cross-section comprises identical rectangles.

One of these rectangles is shaded in the diagram below.

Find the area of the shaded rectangle in square metres. (1 mark)

The concrete staircase is 2.5 m wide.

Find the volume of the solid concrete staircase in cubic metres. (2 marks)

Show Answers Only

`0.048\ text(m²)`
`6.6\ text(m³)`

Show Worked Solution

a. `text(Height of rectangle)`

MARKER’S COMMENT: Keep units in metres as many students made mistakes converting cm² to m².

`= 1.6/10`

`= 0.16\ text(m)`

`text(Length of rectangle)`

`= 3.0/10`

`= 0.3\ text(m)`

`:.\ text(Area)`	`= 0.16 xx 0.3`
	`= 0.048\ text(m²)`

b. `V = Ah`

`text(S)text(ince there are 55 rectangles)`

`A`	`= 55 xx 0.048`
	`= 2.64\ text(m²)`

`:. V`	`= 2.64 xx 2.5`
	`= 6.6\ text(m³)`

GEOMETRY, FUR2 2013 VCAA 1

A spectator, `S`, in the grandstand of an athletics ground is 13 m vertically above point `G`.

Competitor `X`, on the athletics ground, is at a horizontal distance of 40 m from `G`.

Find the distance, `SX`, correct to the nearest metre. (1 mark)

Competitor `X` is 40 m from `G` and competitor `Y` is 52 m from `G`.

The angle `XGY` is 113°.

1. Calculate the distance, `XY`, correct to the nearest metre. (1 mark)
2. Find the area of triangle `XGY`, correct to the nearest square metre. (1 mark)
Determine the angle of elevation of spectator `S` from competitor `Y`, correct to the nearest degree.
Note that `X`, `G` and `Y` are on the same horizontal level. (1 mark)

Show Answers Only

`42\ text{m (nearest m)}`
1. `77\ text{m (nearest m)}`
2. `957\ text{m² (nearest m²)}`
`14^@\ \ text{(nearest degree)}`

Show Worked Solution

a. `text(Using Pythagoras:)`

`SX^2`	`= 40^2 + 13^2`
	`= 1769`
`:. SX`	`= 42.05…`
	`= 42\ text{m (nearest m)}`

b.i. `text(Using the cosine rule:)`

`XY^2`	`= 40^2 + 52^2 – 52^2 xx 40 xx 52cos113^@`
	`= 5929.44…`
`:. XY`	`= 77.00…`
	`= 77\ text{m (nearest m)}`

b.ii. `text(Using the sine rule:)`

`text(Area)`	`= 1/2ab sinC`
	`= 1/2 xx 40 xx 52 xx sin113^@`
	`= 957.32…`
	`= 957\ text{m² (nearest m²)}`

`theta`	`= text(angle of elevation of)\ S\ text(from)\ Y`
`tantheta`	`= 13/52`
`:. theta`	`= tan^(−1)\ 1/4`
	`= 14.036…`
	`= 14^@\ \ text{(nearest degree)}`

CORE*, FUR2 2014 VCAA 3

The cricket club had invested $45 550 in an account for four years.

After four years of compounding interest, the value of the investment was $60 000.

How much interest was earned during the four years of this investment? (1 mark)

Interest on the account had been calculated and paid quarterly.

What was the annual rate of interest for this investment?

Write your answer, correct to one decimal place. (1 mark)

The $60 000 was re-invested in another account for 12 months.

The new account paid interest at the rate of 7.2% per annum, compounding monthly.

At the end of each month, the cricket club added an additional $885 to the investment.

1. The equation below can be used to determine the account balance at the end of the first month, immediately after the $885 was added.
  
  Complete the equation by filling in the boxes. (1 mark)
2. What was the account balance at the end of 12 months?
  
  Write your answer, correct to the nearest dollar. (1 mark)

Show Answers Only

`14\ 450`
`text(6.9%)`
1. `text(account balance)\ = 60\ 000 xx (1 + (7.2)/(12 xx 100)) + 885`
2. `75\ 443`

Show Worked Solution

a. `text(Interest earned)`

`= text(End value − original investment)`

`= 60\ 000 – 45\ 550`

`= $74\ 450`

♦ Mean mark of all parts (combined) was 40%.

b. `text(By TVM Solver:)`

`N`	`= 4 xx 4 = 16`
`I(text(%))`	`= ?`
`PV`	`= −45\ 550`
`PMT`	`= 0`
`FV`	`= 60\ 000`
`text(P/Y)`	`= text(C/Y) = 4`

`I(text(%)) = 6.948…`

`:.\ text(Annual interest rate = 6.9%)`

c.i. `text(account balance)`

`= 60\ 000 xx (1 + 7.2/(12 xx 100)) + 885`

c.ii. `text(By TVM Solver:)`

`N`	`= 12`
`I(text(%))`	`= 7.2`
`PV`	`= −60\ 000`
`PMT`	`= −885`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`FV = 75\ 443.014…`

`:. text(Balance after 12 months) = $75\ 443`

CORE*, FUR2 2014 VCAA 2

A sponsor of the cricket club has invested $20 000 in a perpetuity.

The annual interest from this perpetuity is $750.

The interest from the perpetuity is given to the best player in the club every year, for a period of 10 years.

What is the annual rate of interest for this perpetuity investment? (1 mark)
After 10 years, how much money is still invested in the perpetuity? (1 mark)
The average rate of inflation over the next 10 years is expected to be 3% per annum.
1. Michael was the best player in 2014 and he considered purchasing cricket equipment that was valued at $750.
  
  What is the expected price of this cricket equipment in 2015? (1 mark)
2. What is the 2014 value of cricket equipment that could be bought for $750 in 2024?
  
  Write your answer, correct to the nearest dollar. (1 mark)

Show Answers Only

`3.75`
`$20\ 000`
1. `$772.50`
2. `$558\ \ text{(nearest dollar)}`

Show Worked Solution

a.	`20\ 000 xx r`	`= 750`
	`:. r`	`= 750/(20\ 000)`
		`= 0.0375`

`:.\ text(Annual interest rate = 3.75%)`

b. `$20\ 000`

`text{(A perpetuity’s balance remains}`

`text{constant.)}`

c.i. `text(Expected price in 2015)`

`= 750 xx (1 + 3/100)`

`= 750 xx 1.03`

`= $772.50`

c.ii. `text(Value in 2014) xx (1.03)^10 = 750`

`:.\ text(Value in 2014)\ `	`= 750/((1.03)^10)`
	`= 558.07…`
	`= $558\ \ text{(nearest dollar)}`

GRAPHS, FUR2 2014 VCAA 2

The cost, `C`, in dollars, of producing `n` kilograms of tomatoes is given by

`C = 1.25n + 36\ 000qquadqquad0 ≤ n ≤ 40\ 000`

The revenue, `R`, in dollars, from selling `n` kilograms of tomatoes is given by

`R = 3.5nqquadqquadqquad0 ≤ n ≤ 40\ 000`

The cost, `C`, for the production of `n` kilograms of tomatoes is graphed below.

On the graph above, draw the revenue equation line, `R = 3.5n`. (1 mark)

(Answer on the graph above.)

What profit will Arthur make if he sells a total of 20 000 kg of tomatoes? (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`$9000`

Show Worked Solution

b. `text(Profit)`

`=\ text(Revenue – Costs)`

`= 3.5 xx 20\ 000 − (1.25 xx 20\ 000 + 36\ 000)`

`=70\ 000 – 61\ 000`

`=$9000`

GRAPHS, FUR2 2014 VCAA 1

Fastgrow and Booster are two tomato fertilisers that contain the nutrients nitrogen and phosphorus.

The amount of nitrogen and phosphorus in each kilogram of Fastgrow and Booster is shown in the table below.

How many kilograms of phosphorus are in 2 kg of Booster? (1 mark)
If 100 kg of Booster and 400 kg of Fastgrow are mixed, how many kilograms of nitrogen would be in the mixture? (1 mark)

Arthur is a farmer who grows tomatoes.

He mixes quantities of Booster and Fastgrow to make his own fertiliser.

Let `x` be the number of kilograms of Booster in Arthur’s fertiliser.

Let `y` be the number of kilograms of Fastgrow in Arthur’s fertiliser.

Inequalities 1 to 4 represent the nitrogen and phosphorus requirements of Arthur’s tomato field.

Inequality 1	`x`	`≥ 0`
Inequality 2	`y`	`≥ 0`
Inequality 3 (nitrogen)	`0.05x + 0.05y`	`≥ 200`
Inequality 4 (phosphorus)	`0.02x + 0.06y`	`≥ 120`

Arthur’s tomato field also requires at least 180 kg of the nutrient potassium.

Each kilogram of Booster contains 0.06 kg of potassium.

Each kilogram of Fastgrow contains 0.04 kg of potassium.

Inequality 5 represents the potassium requirements of Arthur’s tomato field.

Write down Inequality 5 in terms of `x` and `y`. (1 mark)

The lines that represent the boundaries of Inequalities 3, 4 and 5 are shown in the graph below.

1. Using the graph above, write down the equation of line `A`. (1 mark)
2. On the graph above, shade the region that satisfies Inequalities 1 to 5. (1 mark)

(Answer on the graph above.)

Arthur would like to use the least amount of his own fertiliser to meet the nutrient requirements of his tomato field and still satisfy Inequalities 1 to 5.

1. What weight of his own fertiliser will Arthur need to make? (1 mark)
2. On the graph above, show the point(s) where this solution occurs. (2 marks)

(Answer on the graph above.)

Show Answers Only

`0.04\ text(kg)`
`25\ text(kg)`
`0.06x + 0.04y ≥ 180`
1. `0.02x + 0.06y = 120`
2. `text(See Worked Solutions)`
1. `4000\ text(kg)`
2. `text(See Worked Solutions)`

Show Worked Solution

a. `text(Phosphorous in 2 kg of Booster)`

`=0.02× 2`

`= 0.04\ text(kg)`

b. `text(Total Phosphorous)`

`=0.05×100 + 0.05× 400`

`= 25\ text(kg)`

c. `text(Inequality 5,)`

`0.06x + 0.04y >= 180`

d.i. `text(Line)\ A\ text(is the boundary of Inequality 4.)`

♦♦ Mean mark for parts (d) and (e) combined was 32%.
MARKER’S COMMENT: A majority of students didn’t identify Inequality 4 as relevant to Line A.

`:.\ text(It can be expressed)`

`0.02x + 0.06y = 120`

d.ii.

e.i. `text(The minimum amount is on the boundary)`

`x+y=4000`

`:.\ text(Arthur will have to make at least 4000 kg.)`

e.ii. `text(All points on the bold line are solutions,)`

`text{including (1000, 3000) and (3000, 1000).}`

CORE*, FUR2 2015 VCAA 4

As their business grows, Jane and Michael decide to invest some of their earnings.

They each choose a different investment strategy.

Jane opens an account with Red Bank, with an initial deposit of $4000.

Interest is calculated at a rate of 3.6% per annum, compounding monthly.

Determine the amount in Jane’s account at the end of six months.

Write your answer correct to the nearest cent. (1 mark)

Michael decides to open an account with Blue Bank, with an initial deposit of $2000.

At the end of each quarter, he adds an additional $200 to his account.

Interest is compounded at the end of each quarter.

The equation below can be used to determine the balance of Michael’s account at the end of the first quarter.

account balance = 2000 × (1 + 0.008) + 200

Show that the annual compounding rate of interest is 3.2%. (1 mark)
Determine the amount in Michael’s account, after the $200 has been added, at the end of five years.

Write your answer correct to the nearest cent. (1 mark)

Show Answers Only

`$4072.54`
`3.2text(%)`
`$6664.63`

Show Worked Solution

a. `text(Amount in account after 6 months)`

`= 4000 xx (1 + 3.6/(12 xx 100))^6`

`=4072.542…`

`=$4072.54\ \ text{(nearest cent)}`

b. `text(Annual compounding rate)`

`=0.008 xx text(100%) xx 4`

`= 3.2text(%)`

c. `text(By TVM Solver, after 5 years)`

`N`	`=5 xx 4 = 20`
`I (%)`	`=3.2`
`PV`	`= – 2000`
`PMT`	`=200`
`FV`	`=?`
`text(P/Y)`	`= text(C/Y) = 4`

`FV=6664.629…`

`:.\ text(The amount in Michael’s account is $6664.63)`

CORE*, FUR2 2015 VCAA 3

Jane and Michael decide to set up an annual music scholarship.

To fund the scholarship, they invest in a perpetuity that pays interest at a rate of 3.68% per annum.

The interest from this perpetuity is used to provide an annual $460 scholarship.

Determine the minimum amount they must invest in the perpetuity to fund the scholarship. (1 mark)
For how many years will they be able to provide the scholarship? (1 mark)

Show Answers Only

`12\ 500`
`text(It will last forever.)`

Show Worked Solution

a. `text(Let)\ \ $A\ =\ text(Perpetuity amount)`

`3.68text(%) xx A`	`= 460`
`:.A`	`=460/0.0368`
	`= $12\ 500`

b. `text(It will last forever.)`

CORE*, FUR2 2015 VCAA 2

The sound system used by the business was initially purchased at a cost of $3800.

After two years, the value of the sound system had depreciated to $3150.

Assuming the flat rate method of depreciation was used, show that the value of the sound system was depreciated by $325 each year. (1 mark)
The value of the sound system will continue to depreciate by $325 each year.

How many years will it take, after the initial purchase, for the sound system to have a value of $550? (1 mark)
The recording equipment used by the business was initially purchased at a cost of $2100.

After five years, the value of the recording equipment had depreciated to $1040 using the reducing balance method.

Find the annual percentage rate by which the value of this recording equipment depreciated.

Write your answer correct to two decimal places. (1 mark)

Show Answers Only

`$325`
`10`
`13.11text(%)`

Show Worked Solution

a. `text(Annual depreciation)`

`=(3800 – 3150)/ 2`

`= $325\ …\ text(as required)`

b. `text(Let)\ t =\ text(number of years.)`

`3800 – 325 xx t`	`= 550`
`325t`	`= 3250`
`:. t`	`=3250/325`
	`=10\ text(years)`

c. `text(Let)\ r =\ text(annual depreciation rate)`

`2100(1 – r)^5`	`= 1040`
`(1-r)^5`	`= 1040/2100`
`(1-r)`	`=0.868…`
`:. r`	`=0.13111…`
	`=13.11 text{% (2 d.p.)}`

CORE*, FUR2 2015 VCAA 1

Jane and Michael have started a business that provides music at parties.

The business charges customers $88 per hour.

The $88 per hour includes a 10% goods and services tax (GST).

Calculate the amount of GST included in the $88 hourly rate. (1 mark)

Jane and Michael’s first booking was a party where they provided music for four hours.

Calculate the total amount they were paid for this booking. (1 mark)

After six months of regular work, Jane and Michael decided to increase the hourly rate they charge by 12.5%.

Calculate the new hourly rate (including GST). (1 mark)

Show Answers Only

`$8`
`$352`
`$99`

Show Worked Solution

a. `text(Let)\ \ x=\ text(hourly rate before GST)`

`x +\ text(GST)`	`= 88`
`x+0.1x`	`= 88`
`x`	`=88/1.1`
	`=80`

`:.\ text(GST)\ = 88-80 = $8`

b.	`text(Total amount)`	`=88 xx 4`
		`=$352`

c.	`text(New hourly rate including GST)`
	`= 88 + 88 xx 12.5 text(%)`
	`= 88 xx 1.125`
	`=$99`

GRAPHS, FUR2 2015 VCAA 5

When Ben is in Japan, he will study at a Japanese school.

Some of his lessons will be in English and some of his lessons will be in Japanese.

Let `x` be the number of lessons in English that he will attend each week.

Let `y` be the number of lessons in Japanese that he will attend each week.

There are 35 lessons each week.

It is a condition of his exchange that Ben attends at least 24 lessons each week.

It is also a condition that Ben attends no more than two lessons in English for every lesson in Japanese.

This information can be represented by Inequalities 1, 2 and 3.

Inequality 1 `x + y ≤ 35`

Inequality 2 `x + y ≥ 24`

Inequality 3 `y ≥ x/2`

There is another constraint given by

Inequality 4 `y ≥ 10`

Describe Inequality 4 in terms of the lessons that Ben must attend. (1 mark)
The graph below shows the lines that represent the boundaries of Inequalities 1 to 4.

On the graph below, shade the region that contains the points that satisfy these inequalities. (1 mark)

Determine the maximum number of lessons in English that Ben can attend. (1 mark)

Show Answers Only

`text(Ben must attend at least 10 lessons in Japanese.)`
`23`

Show Worked Solution

a. `text(Ben must attend at least 10 lessons in Japanese.)`

c. `text{The maximum integral value of}\ x\ text(in)`

`text(the shaded feasible region is 23.)`

`:.\ text(Maximum lessons in English)\ = 23`

GRAPHS, FUR2 2015 VCAA 4

The airline that Ben uses to travel to Japan charges for the seat and luggage separately.

The charge for luggage is based on the weight, in kilograms, of the luggage.

If the luggage is paid for at the airport, the graph below can be used to determine the cost, in dollars, of luggage of a certain weight, in kilograms.

Find the cost at the airport for 23 kg of luggage. (1 mark)

If the luggage is paid for online prior to arriving at the airport, the equation for the relationship between the online cost, in dollars, and the weight, in kilograms, would be

`text(online cost) = {{:(75),(22.5 xx text(weight) - 375):} qquad {:(\ \ 0 < text(weight) <= 20),(20 < text(weight ≤ 40)):} :}`

Find the online cost for 30 kg of luggage. (1 mark)
On the graph above, sketch a graph of the online cost of luggage for `0 <` weight `≤ 40`. Include the end points. (2 marks)

(Answer on the graph above.)

Determine the weight of luggage for which the airport cost and online cost are the same.
Write your answer correct to one decimal place. (1 mark)

Show Answers Only

`$250`
`$300`
`27.8\ text(kg)`

Show Worked Solution

a. `$250`

b.	`text(Online cost)`	`= 22.5 xx 30 – 375`
		`= 300\ text(dollars)`

c.	`text{Online cost (40 kg)}`	`= 22.5 xx 40 – 375`
		`= 525`

`:.\ text(End points are)\ \ (0, 75) and (40, 525)`

d. `text(From the graph, the intersection`

`text(occurs when)`

`22.5 xx text(weight) – 375`	`= 250`
`text(weight)`	`= 625/22.5`
	`= 27.77…`
	`= 27.8\ text(kg)`

GRAPHS, FUR2 2015 VCAA 3

The graph below shows the relationship between the yen and the dollar on the same day at a different currency exchange agency.

The points `(100, 8075)` and `(200, 17\ 575)` are labelled.

The equation for the relationship between the yen and the dollar is

yen = 95 × dollars – k

Use the point `(100, 8075)` to show that the value of `k` is 1425. (1 mark)
1. Determine the intercept on the horizontal axis. (1 mark)
2. Interpret the intercept on the horizontal axis in the context of converting dollars to yen. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
1. `$15`
2. `text(There is a $15 fixed charge for)`
  
  `text(any conversion of dollars into yen.)`

Show Worked Solution

a.	`text(Substituting)\ \ (100, 8075)\ \ text(into equation)`
	`8075`	`= 95 xx 100 – k`
	`:. k`	`= 9500 – 8075`
		`= 1425\ …\ text(as required)`

b.i.	`text(Intercept when yen) = 0`
	`0`	`= 95 xx text(dollars) – 1425`
	`text(dollars)`	`= 1425/95`
		`= 15`

`:.\ text(Intercept is at $15)`

b.ii.	`text(There is a $15 fixed charge for)`
	`text(any conversion of dollars into yen.)`

Calculus, 2ADV C4 SM-Bank 1 MC

The diagram below is the graph of `y = x^2 - x - 6`

What is the correct expression for the area bounded by the `x`-axis and the graph `y = x^2 - x - 6` between `0 <= x <= 4`?

`A = int_0^4 x^2 - x - 6\ dx`
`A = |\ int_0^3 x^2 - x - 6\ dx\ | + int_3^4 x^2 - x - 6\ dx`
`A = int_0^3 x^2 - x - 6\ dx + |\ int_3^4 x^2 - x - 6\ dx|`
`A = |\ int_0^4 x^2 - x - 6\ dx\ |`

Show Answers Only

`=> B`

Show Worked Solution

`y`	`= x^2 – x – 6`
	`= (x – 3)(x + 2)`

`text(Graph intersects the)\ xtext(-axis at)\ (3,0)`

`:. A = |\ int_0^3 x^2 – x – 6\ dx\ | + int_3^4 x^2 – x – 6\ dx`

`=> B`

Calculus, MET2 2014 VCAA 5

Let `f: R -> R, \ \ f (x) = (x - 3)(x - 1)(x^2 + 3) and g: R-> R, \ \ g (x) = x^4 − 8x.`

Express `x^4 - 8x` in the form `x(x - a) ((x + b)^2 + c)`. (2 marks)
Describe the translation that maps the graph of `y = f (x)` onto the graph of `y = g (x)`. (1 mark)
Find the values of `d` such that the graph of `y = f (x + d)` has
i. one positive `x`-axis intercept (1 mark)
ii. two positive `x`-axis intercepts. (1 mark)
Find the value of `n` for which the equation `g (x) = n` has one solution. (1 mark)
At the point `(u, g(u))`, the gradient of `y = g(x)` is `m` and at the point `(v, g(v))`, the gradient is `-m`, where `m` is a positive real number.
i. Find the value of `u^3 + v^3`. (2 marks)
ii. Find `u` and `v` if `u + v = 1`. (1 mark)
i. Find the equation of the tangent to the graph of `y = g(x)` at the point `(p, g(p))`. (1 mark)
ii. Find the equations of the tangents to the graph of `y = g(x)` that pass through the point with coordinates `(3/2, text(−12))`. (3 marks)

Show Answers Only

`x(x- 2)((x + 1)^2 + 3)`
`text(See Worked Solutions)`
i. `[1,3)`
ii. `d < 1`
`−6 xx 2^(1/3)`
i. `4`
ii. `u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`
i. `y = 4(p^3 – 2)x – 3p^4`
ii. `y = −8x;quady = 24x – 48`

Show Worked Solution

a. `text(Solution 1)`

`g(x) = x^4 – 8x`

`= x(x – 2)(x^2 + 2x + 4)\ \ \ text([by CAS])`

`= x(x- 2)((x + 1)^2 + 3)`

`text(Solution 2)`

`x^4 – 8x`	`=x(x^3-2^3)`
	`=x(x-2)(x^2 +2x+4)`
	`=x(x-2)(x^2 +2x+1+3)`
	`=x(x-2)((x+1)^2+3)`

b.	`f(x + 1)`	`= ((x + 1) – 3)((x + 1) – 1)((x + 1)^2 + 3)`
		`= x(x – 2)((x + 1)^2 + 3)`
		`= g(x)`

♦ Mean mark part (b) 37%.

`:.\ text(Horizontal translation of 1 unit to the left.)`

c.i. `text(Consider part of the)\ \ f(x)\ text(graph below:)`

♦♦♦ Mean mark 7%.

`text(For one positive)\ xtext(-axis intercept, translate at least)`

`text(1 unit left, but not more than 3 units left.)`

`:. d ∈ [1,3)`

c.ii. `text(Translate less than 1 unit left, or translate)`

♦♦♦ Mean mark part (c)(ii) 19%.

`text(right.)`

`:. d < 1`

d. `text(If)\ \ g(x)=n\ \ text(has one solution, then it)`

`text(will occur when)\ \ g′(x)=0 and x>0.`

♦♦♦ Mean mark 17%.

`g′(x)`	`=4x^3-8`
`4x^3`	`=8`
`x`	`=2^(1/3)`

`n`	`=g(2^(1/3))`
	`=2^(4/3)-8xx2^(1/3)`
	`=-6 xx 2^(1/3)`

e.i. `gprime(u) = mqquadgprime(v) = −m`

♦♦ Mean mark 28%.

`gprime(u)`	`= −gprime(v)`
`4u^3 – 8`	`= −(4v^3 – 8)`
`4u^3 + 4v^3`	`= 16`
`:. u^3 + v^3`	`= 4`

e.ii. `u^3 + v^3 = 4\ …\ (1)`

♦♦♦ Mean mark 10%.

`u + v = 1\ …\ (2)`

`text(Solve simultaneous equation for)\ \ u > 0:`

`:. u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`

f.i. `text(Solution 1)`

`text(Using the point-gradient formula,)`

`y-g(p)`	`=g′(p)(x-p)`
`y-(p^4-8p)`	`=(4p^3-8)(x-p)`
`y`	`=(4p^3-8)x -3p^4`

`text(Solution 2)`

♦♦ Mean mark (f)(i) 22%.

`y = 4(p^3 – 2)x – 3p^4`

`text([CAS: tangentLine)\ (g(u),x,p)]`

f.ii. `text(Sub)\ (3/2,−12)\ text(into tangent equation,)`

`text(Solve:)\ \ −12 = 4(p^3 – 2)(3/2) – 3p^4\ \ text(for)\ p,`

♦♦♦ Mean mark 14%.

`p = 0\ \ text(or)\ \ p = 2`

`text(When)\ \ p = 0:`	`y`	`= 4(−2)x`
		`= −8x`

`text(When)\ \ p = 2:`	`y`	`= 4(2^3 – 2)x – 3(2)^4`
		`= 24x – 48`

`:. text(Equations are:)\ \ y = −8x, \ y = 24x – 48`

Probability, MET2 2014 VCAA 4

Patricia is a gardener and she owns a garden nursery. She grows and sells basil plants and coriander plants.

The heights, in centimetres, of the basil plants that Patricia is selling are distributed normally with a mean of 14 cm and a standard deviation of 4 cm. There are 2000 basil plants in the nursery.

Patricia classifies the tallest 10 per cent of her basil plants as super.

What is the minimum height of a super basil plant, correct to the nearest millimetre? (1 mark)

Patricia decides that some of her basil plants are not growing quickly enough, so she plans to move them to a special greenhouse. She will move the basil plants that are less than 9 cm in height.

How many basil plants will Patricia move to the greenhouse, correct to the nearest whole number? (2 marks)

The heights of the coriander plants, `x` centimetres, follow the probability density function `h(x)`,

`h(x) = {(pi/100 sin ((pi x)/50), 0 < x < 50), (\ \ \ \ \ \0, text(otherwise)):}`

State the mean height of the coriander plants. (1 mark)

Patricia thinks that the smallest 15 per cent of her coriander plants should be given a new type of plant food

Find the maximum height, correct to the nearest millimetre, of a coriander plant if it is to be given the new type of plant food. (2 marks)

Patricia also grows and sells tomato plants that she classifies as either tall or regular. She finds that 20 per cent of her tomato plants are tall.

A customer, Jack, selects `n` tomato plants at random.

Let `q` be the probability that at least one of Jack’s `n` tomato plants is tall.

Find the minimum value of `n` so that `q` is greater than 0.95. (2 marks)

NB. Parts (f) - (g) are no longer in the syllabus.

Show Answers Only

`191\ text(mm)`
`211\ text(plants)`
`25\ text(cm)`
`127\ text(mm)`
`14\ text(plants)`

Show Worked Solution

a. `text(Let)\ \ X = text(plant height,)`

♦ Mean mark 43%.

`X ∼\ text(N)(14,4^2)`

`text(Pr)(X > a)`	`= 0.1`
`a`	`= 19.1\ text(cm)quadtext([CAS: invNorm)\ (.9,14,4)]`
	`=191\ text{mm (nearest mm)}`

`:.\ text(Min super plant height is 191 mm.)`

b. `text(Pr)(X < 9) = 0.10565…\ qquadtext([CAS: normCdf)\ (−∞,9,14,4)]`

`:.\ text(Number moved to greenhouse)`

`= 0.10565… xx 2000`

`= 211\ text(plants)`

c.	`text(E)(X)`	`= int_0^50 (x xx pi/100 sin((pix)/50))dx`
		`= 25\ text(cm)`

`text(Solve:)\ \ int_0^a h(x)\ dx`

`= 0.15\ \ text(for)\ \ a ∈ (0,50)`

♦♦ Mean mark 33%.

`:.a`	`=12.659…\ text(cm)`
	`=127\ text{mm (nearest mm)}`

e. `text(Let)\ \ Y =\ text(Number of tall plants,)`

♦♦ Mean mark 30%.

`Y ∼\ text(Bi) (n,0.2)`

`text(Pr)(Y >= 1)`	`> 0.95`
`1 – text(Pr)(Y = 0)`	`> 0.95`
`0.05`	`> 0.8^n`
`n`	`> 13.4\ \ text([by CAS])`

`:. n_text(min) = 14\ text(plants)`

Calculus, MET2 2014 VCAA 3

In a controlled experiment, Juan took some medicine at 8 pm. The concentration of medicine in his blood was then measured at regular intervals. The concentration of medicine in Juan’s blood is modelled by the function `c(t) = 5/2 te^(-(3t)/2), t >= 0`, where `c` is the concentration of medicine in his blood, in milligrams per litre, `t` hours after 8 pm. Part of the graph of the function `c` is shown below.

What was the maximum value of the concentration of medicine in Juan’s blood, in milligrams per litre, correct to two decimal places? (1 mark)
i. Find the value of `t`, in hours, correct to two decimal places, when the concentration of medicine in Juan’s blood first reached 0.5 milligrams per litre. (1 mark)
ii. Find the length of time that the concentration of medicine in Juan’s blood was above 0.5 milligrams per litre. Express the answer in hours, correct to two decimal places. (2 marks)
i. What was the value of the average rate of change of the concentration of medicine in Juan’s blood over the interval `[2/3, 3]`? Express the answer in milligrams per litre per hour, correct to two decimal places. (2 marks)
ii. At times `t_1` and `t_2`, the instantaneous rate of change of the concentration of medicine in Juan's blood was equal to the average rate of change over the interval `[2/3, 3]`.
Find the values of `t_1` and `t_2`, in hours, correct to two decimal places. (2 marks)

Alicia took part in a similar controlled experiment. However, she used a different medicine. The concentration of this different medicine was modelled by the function `n(t) = Ate^(-kt),\ t >= 0` where `A` and `k in R^+`.

If the maximum concentration of medicine in Alicia’s blood was 0.74 milligrams per litre at `t = 0.5` hours, find the value of `A`, correct to the nearest integer. (3 marks)

Show Answers Only

`0.61\ text(mg/L)`
i. `0.33\ text(hours)`
ii. `0.86\ text(hours)`
i. `−0.23\ text(mg/L/h)`
ii. `t = 0.90\ text(or)\ t = 2.12`
`4`

Show Worked Solution

a. `text(Solve:)\ \ cprime(t)=0\ text(for)\ t >= 0`

`t=2/3`

`c(2/3)= 0.61\ text(mg/L)`

b.i.

`text(Solve:)\ \ c(t)`

`= 0.5\ text(for)\ t >= 0`

`t=0.33\ \ text(or)\ \ t = 1.19`

`:. text(First reached 0.5 mg/L at)\ \ t = 0.33\ text(hours)`

MARKER’S COMMENT: In part (b)(ii), work to sufficient decimal places to ensure your answer has the required accuracy.

b.ii.	`text(Length)`	`= 1.18756… – 0.326268…`
		`= 0.86\ text(hours)`

c.i.	`text(Average ROC)`	`= (c(3) – c(2/3))/(3 – 2/3)`
		`= (0.083 – 0.613)/(7/3)`
		`=-0.227…`
		`= −0.23\ text{mg/L/h (2 d.p.)}`

c.ii. `text(Solve:)\ \ cprime(t) = −0.22706…\ text(for)\ \ t >= 0`

♦ Mean mark part (c)(ii) 38%.

`:. t = 0.90\ \ text(or)\ \ t = 2.12\ text{h (2 d.p.)}`

d. `text(Solution 1)`

♦ Mean mark 46%.

`text(Equations from given information:)`

`n(1/2) = 0.74`

`0.74 = 1/2Ae^(−1/2k)\ …\ (1)`

`n′ (1/2) = 0`

`text(Solve simultaneously for)\ A and k\ \ text([by CAS])`

`:.A`	`= 4.023`
	`= 4\ \ text{(nearest integer)}`

`text(Solution 2)`

`n(1/2) = 0.74`

`0.74 = 1/2Ae^(−1/2k)\ …\ (1)`

`n′(t)`	`=Ae^(- k/2) – 1/2 Ake^(- k/2)`
	`=Ae^(- k/2) (1- k/2)\ …\ (2)`

`n′ (1/2) = 0`

`:. k=2\ \ text{(using equation (2))}`

`:.A`

`= 4\ \ text{(nearest integer)}`

Calculus, MET2 2014 VCAA 2

On 1 January 2010, Tasmania Jones was walking through an ice-covered region of Greenland when he found a large ice cylinder that was made a thousand years ago by the Vikings.

A statue was inside the ice cylinder. The statue was 1 m tall and its base was at the centre of the base of the cylinder.

The cylinder had a height of `h` metres and a diameter of `d` metres. Tasmania Jones found that the volume of the cylinder was 216 m³. At that time, 1 January 2010, the cylinder had not changed in a thousand years. It was exactly as it was when the Vikings made it.

Write an expression for `h` in terms of `d`. (2 marks)
Show that the surface area of the cylinder excluding the base, `S` square metres, is given by the rule `S = (pi d^2)/4 + 864/d`. (1 mark)

Tasmania found that the Vikings made the cylinder so that `S` is a minimum.

Find the value of `d` for which `S` is a minimum and find this minimum value of `S`. (2 marks)
Find the value of `h` when `S` is a minimum. (1 mark)

NB. Parts (e) - (h) have been removed from the syllabus.

Show Answers Only

`h = 864/(pi d^2)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`12/(pi^(1/3))\ text(m);quad108pi^(1/3)\ text(square metres)`
`6/(root(3)(pi))m`
`V = pih^3`
No longer in course
No longer in course
`2031`

Show Worked Solution

a.	`V`	`= pir^2h`
	`216`	`= pi(d/2)^2h`
	`:. h`	`= (864)/(pi d^2)`

b.	`S`	`= A_text(top) + A_text(curved surface)`
		`=pi xx (d/2)^2 + pi xx d xx h`

`text(Substitute)\ \ h = 864/(pid^2),`

`:. S`	`= (pid^2)/4 + pi(d)(864/(pid^2))`
	`= (pid^2)/4 + 864/d`

c. `(dS)/(dd) = (pi d)/2 -864/d^2`

`text(Stationary point when)\ \ (dS)/(dd)=0,`

`text(Solve:)\ \ (pi d)/2 -864/d^2=0\ \ text(for)\ d`

`d=12/(pi^(1/3))\ text(m)`

`:. S_text(min)=S(12/pi^(1/3)) = 108pi^(1/3)\ text(m²)`

d. `text(Substitute)\ \ d=12/(pi^(1/3))\ \ text{into part (a):}`

♦ Mean mark part (d) 42%.
MARKER’S COMMENT: An exact answer required here!

`h= 864/(pi(12/(pi^(1/3)))^2)`

`= 6/(root(3)(pi))\ \ text(m)`

Algebra, MET2 2014 VCAA 1

The population of wombats in a particular location varies according to the rule `n(t) = 1200 + 400 cos ((pi t)/3)`, where `n` is the number of wombats and `t` is the number of months after 1 March 2013.

Find the period and amplitude of the function `n`. (2 marks)
Find the maximum and minimum populations of wombats in this location. (2 marks)
Find `n(10)`. (1 mark)
Over the 12 months from 1 March 2013, find the fraction of time when the population of wombats in this location was less than `n(10)`. (2 marks)

Show Answers Only

`text(Period) = text(6 months);\ text(Amplitude) = 400`
`text(Max) = 1600;\ text(Min) = 800`
`1000`
`1/3`

Show Worked Solution

a. `text(Period) = (2pi)/n = (2pi)/(pi/3) = 6\ text(months)`

MARKER’S COMMENT: Expressing the amplitude as [800,1600] in part (a) is incorrect.

`text(A)text(mplitude) = 400`

b. `text(Max:)\ 1200 + 400 = 1600\ text(wombats)`

`text(Min:)\ 1200 – 400 = 800\ text(wombats)`

c. `n(10) = 1000\ text(wombats)`

`text(Solve)\ n(t)`

`= 1000\ text(for)\ t ∈ [0,12]`

`t= 2,4,8,10`

`text(S)text(ince the graph starts at)\ \ (0,1600),`

♦ Mean mark 48%.

`=> n(t) < 1000\ \ text(for)`

`t ∈ (2,4)\ text(or)\ t ∈ (8,10)`

`:.\ text(Fraction)`	`= ((4 – 2) + (10 – 8))/12`
	`= 1/3\ \ text(year)`

Calculus, MET2 2015 VCAA 5

Let `S(t) = 2e^(t/3) + 8e^((-2t)/3)`, where `0 <= t <= 5`.
i. Find `S(0) and S(5)`. (1 mark)
ii. The minimum value of `S` occurs when `t = log_e(c)`.
State the value of `c` and the minimum value of `S`. (2 marks)
iii. On the axes below, sketch the graph of `S` against `t` for `0 <= t <= 5`.
Label the end points and the minimum point with their coordinates. (2 marks)
iv. Find the value of the average rate of change of the function `S` over the interval `[0, log_e(c)]`. (2 marks)

Let `V = [0, 5] -> R,\ \ \ V(t) = de^(t/3) + (10 - d)e^(-(2t)/3)`, where `d` is a real number and `d` in `(0, 10)`.

If the minimum value of the function occurs when `t = log_e (9)`, find the value of `d`. (2 marks)
i. Find the set of possible values of `d` such that the minimum value of the function occurs when `t = 0`. (2 marks)
ii. Find the set of possible values of `d` such that the minimum value of the function occurs when `t = 5`. (2 marks)
If the function `V` has a local minimum `(a, m)`, where `0 <= a <= 5`, it can be shown that `m = k/2 d^(2/3) (10 - d)^(1/3)`.
Find the value of `k`. (2 marks)

Show Answers Only

1. `S(0) = 10quadS(5) = 2e^(5/3) + 8e^(-10/3)`
2. `c = 8; S = 6`
4. ` – 4/(log_e 8)`
`20/11`
1. `[20/3 ,10)`
2. `(0, 20/(2 + e^5)]`
`3 xx 2^(1/3)`

Show Worked Solution

a.i.	`S(0)`	`=2e^0+8e^0`
		`=10`
	`S(5)`	`=2e^(5/3)+8e^(-10/3)`

a.ii. `text(When)\ \ Sprime(t)=0,`

`t`	`= 3ln(2)`
	`= log_e(2^3)`

`:. c = 8`

`:. S_text(min) = S(log_e(8)) = 6`

a.iii.

a.iv.	`text(Average ROC)`	`= (S(log_e(8)) – S(0))/(log_e(8) – 0)`
		`=(6-10)/(log_(8))`
		`= (−4)/(log_e(8))`

b. `V(t) = de^(t/3) + (10 – d)e^(-(2t)/3)`

`text(Solve:)\ \ Vprime(log_e(9)) = 0\ text(for)\ d,`

`:. d = 20/11`

c.i. `text(Solve:)\ \ Vprime(0) = 0\ \ text(for)\ \ d,`

♦♦ Mean mark part (c)(i) 28%.

`d=20/3,`

`:. 20/3 <= d <10`

c.ii. `text(Solve:)\ \ Vprime(5) = 0\ \ text(for)\ \ d,`

♦♦ Mean mark part (c)(ii) 27%.

`d=20/(2+e^5),`

`:. 0< d <=20/(2 + e^5)`

♦♦♦ Mean mark part (d) 15%.

`text(Local min when)\ Vprime(a)`

`= 0`

`text(Solve)\ \ Vprime(a)=0\ \ text(for)\ a`

`a= log_e(20/d -2)`

`text(Solve)\ \ V(log_e(20/d-2))=k/2 d^(3/2)(10-d)^(1/3)\ \ text(for)\ k,`

`k= 3 xx 2^(1/3)`

Calculus, MET2 2015 VCAA 4

An electronics company is designing a new logo, based initially on the graphs of the functions

`f(x) = 2 sin (x) and g(x) = 1/2 sin (2x),\ text(for)\ 0 <= x <= 2 pi`

These graphs are shown in the diagram below, in which the measurements in the `x` and `y` directions are in metres.

The logo is to be painted onto a large sign, with the area enclosed by the graphs of the two functions (shaded in the diagram) to be painted red.

The total area of the shaded regions, in square metres, can be calculated as `a int_0^pi sin(x)\ dx`.

What is the value of `a`? (1 mark)

The electronics company considers changing the circular functions used in the design of the logo.

Its next attempt uses the graphs of the functions `f(x) = 2 sin(x) and h(x) = 1/3 sin (3x),\ text(for)\ \ 0 <= x <= 2 pi`.

On the axes below, the graph of `y = f(x)` has been drawn.

On the same axes, draw the graph of `y = h(x)`. (2 marks)
State a sequence of two transformations that maps the graph of `y = f (x)` to the graph of `y = h(x)`. (2 marks)

The electronics company now considers using the graphs of the functions `k(x) = m sin(x) and q (x) = 1/n sin (nx)`, where `m` and `n` are positive integers with `m >= 2 and 0<= x <= 2pi`.

i. Find the area enclosed by the graphs of `y = k(x)` and `y = q(x)` in terms of `m` and `n` if `n` is even.
Give your answer in the form `am + b/n^2`, where `a` and `b` are integers. (2 marks)
ii. Find the area enclosed by the graphs of `y = k(x)` and `y = q(x)` in terms of `m` and `n` if `n`is odd.
Give your answer in the form `am + b/n^2`, where `a` and `b` are integers. (2 marks)

Show Answers Only

`4`
`text(See Worked Solutions)`
1. `4m + 0/n^2`
2. `4m + (−4)/(n^2)`

Show Worked Solution

a.	`text(Area)`	`= 2 xx int_0^pi (2 sin (x))\ dx`
		`= 4 xx int_0^pi sin (x)\ dx`

`:. a = 4`

♦ Mean mark part (a) 36%.

c. `text(Find sequence that takes)\ f(x) -> h(x)`

`text(A dilation by factor of)\ 1/6\ text(from the)\ xtext(-axis.)`

`text(A dilation by factor of)\ 1/3\ text(from the)\ ytext(-axis.)`

d.i. `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 20%.
MARKER’S COMMENT: Note that `cos(npi)=1` for `n` even.

`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`

`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`

`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`

`= 2(2m+(cos(npi)-1)/n^2)`

`= 4m + 0/(n^2)`

d.ii. `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 14%.
MARKER’S COMMENT: Note that `cos(npi)=-1` for `n` odd.

`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`

`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`

`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`

`= 2(2m+(cos(npi)-1)/n^2)`

`= 4m + (-4)/(n^2)`

Probability, MET2 2015 VCAA 3

Mani is a fruit grower. After his oranges have been picked, they are sorted by a machine, according to size. Oranges classified as medium are sold to fruit shops and the remainder are made into orange juice.

The distribution of the diameter, in centimetres, of medium oranges is modelled by a continuous random variable, `X`, with probability density function

`f(x) = {(3/4(x - 6)^2(8 - x), 6 <= x <= 8), (\ \ \ \ \ \ \ 0, text(otherwise)):}`

1. Find the probability that a randomly selected medium orange has a diameter greater than 7 cm. (2 marks)
2. Mani randomly selects three medium oranges.
  Find the probability that exactly one of the oranges has a diameter greater than 7 cm.
  Express the answer in the form `a/b`, where `a` and `b` are positive integers. (2 marks)
Find the mean diameter of medium oranges, in centimetres. (1 mark)

For oranges classified as large, the quantity of juice obtained from each orange is a normally distributed random variable with a mean of 74 mL and a standard deviation of 9 mL.

What is the probability, correct to three decimal places, that a randomly selected large orange produces less than 85 mL of juice, given that it produces more than 74 mL of juice? (2 marks)

Mani also grows lemons, which are sold to a food factory. When a truckload of lemons arrives at the food factory, the manager randomly selects and weighs four lemons from the load. If one or more of these lemons is underweight, the load is rejected. Otherwise it is accepted.

It is known that 3% of Mani’s lemons are underweight.

1. Find the probability that a particular load of lemons will be rejected. Express the answer correct to four decimal places. (2 marks)
2. Suppose that instead of selecting only four lemons, `n` lemons are selected at random from a particular load.
  
  Find the smallest integer value of `n` such that the probability of at least one lemon being underweight exceeds 0.5 (2 marks)

Show Answers Only

1. `11/16`
2. `825/4096`
`36/5\ text(cm)`
`0.778`
1. `0.1147`
2. `23`

Show Worked Solution

a.i.	`text(Pr)(X > 7)`	`= int_7^8 f(x)\ dx`
		`= int_7^8 (3/4(x – 6)^2(8 – x))\ dx`
		`= 11/16`

a.ii. `text(Let)\ \ Y =\ text(number with diameter > 7cm)`

`Y ∼\ text(Bi)(3,11/6)`

`text(Pr)(Y = 1)`	`= ((3),(1))(11/16)^1 xx (5/16)^2`
	`= 825/4096`

b.	`text{E(X)}`	`= int_6^8 (x xx f(x))\ dx`
		`= 36/5`
		`=7.2\ text(cm)`

c. `text(Let)\ \ L = text(Large juice quantity)`

`L ∼\ N(74,9^2)`

`text(Pr)(L < 85 \| L > 74)`	`= (text(Pr)(L < 85 ∩ L>74))/(text(Pr)(L > 74))`
	`= (text(Pr)(74 < L < 85))/(text(Pr)(L > 74))`
	`= (0.3891…)/0.5`
	`= 0.7783…`
	`=0.778\ \ text{(3 d.p.)}`

d.i. `text{Solution 1 [by CAS]}`

`text(Let)\ \ W =\ text(number of lemons under weight)`

`W ∼\ text(Bi)(4,0.03)`

`text(Pr)(W >= 1) = 0.1147qquad[text(CAS: binomCdf)\ (4,0.03,1,4)]`

`text(Solution 2)`

`text(Pr)(W>=1)`	`=1-text(Pr)(W=0)`
	`=1-(0.97)^4`
	`=0.11470…`
	`=0.1147\ \ text{(4 d.p.)}`

d.ii. `W ∼\ text(Bi)(n,0.03)`

♦ Mean mark 44%.

`text(Pr)(W >= 1)`	`> 1/2`
`1 – text(Pr)(W = 0)`	`> 1/2`
`1/2`	`> (0.97)^n`
`n`	`> 22.8`
`:. n_text(min)`	`= 23`

GRAPHS, FUR2 2006 VCAA 2

In one particular week, Harry began with 50 litres of fuel in the tank of his van.

After he had travelled 160 km there were 30 litres of fuel left in the tank of his van.

The amount of fuel remaining in the tank of Harry's van followed a linear trend as shown in the graph below.

Determine the equation of the line shown in the graph above. (2 marks)

Assume this linear trend continues and that Harry does not add fuel to the tank of his van.

How much further will he be able to travel before the tank is empty? (1 mark)

Harry stopped to refuel his van when there were 12 litres of fuel left in the tank.

He completely filled the tank in 3½ minutes when fuel was flowing from the pump at a rate of 18 litres per minute.

How much fuel does the tank hold when it is completely full?

Write your answer in litres. (1 mark)

Show Answers Only

`f = -1/8 d + 50`
`240\ text(km)`
`75\ text(litres)`

Show Worked Solution

a. `y text(-intercept) = 50`

`text(Using)\ (0, 50) and (160, 30)`

`text(Gradient)`	`= (y_2 – y_1)/(x_2 – x_1)`
	`= (30 – 50)/(160 – 0)`
	`= -1/8`

`:.\ text(Equation is:)\ \ \ f = -1/8 d + 50`

b. `text(Find)\ d\ text(when)\ \ f = 0`

`0`	`= -1/8 d + 50`
`1/8 d`	`= 50`
`:. d`	`= 400\ text(km)`

`:.\ text(Further distance to travel)`

MARKER’S COMMENT: Parts (b) and (c) saw many students miss easy marks by not reading the question carefully.

`= 400 – 160`

`= 240\ text(km)`

c. `text(Fuel added to tank)`

`= 3.5 xx 18`

`= 63\ text(litres)`

`text(T) text(ank capacity)`	`= 63 + 12`
	`= 75\ text(litres)`

Calculus, MET2 2015 VCAA 2

A city is located on a river that runs through a gorge.

The gorge is 80 m across, 40 m high on one side and 30 m high on the other side.

A bridge is to be built that crosses the river and the gorge.

A diagram for the design of the bridge is shown below.

The main frame of the bridge has the shape of a parabola. The parabolic frame is modelled by `y = 60 - 3/80x^2` and is connected to concrete pads at `B (40, 0)` and `A (– 40, 0).`

The road across the gorge is modelled by a cubic polynomial function.

Find the angle, `theta`, between the tangent to the parabolic frame and the horizontal at the point `(– 40, 0)` to the nearest degree. (2 marks)

The road from `X` to `Y` across the gorge has gradient zero at `X (– 40, 0)` and at `Y (40, 30)`, and has equation `y = x^3/(25\ 600) - (3x)/16 + 35`.

Find the maximum downwards slope of the road. Give your answer in the form `-m/n` where `m` and `n` are positive integers. (2 marks)

Two vertical supporting columns, `MN` and `PQ`, connect the road with the parabolic frame.

The supporting column, `MN`, is at the point where the vertical distance between the road and the parabolic frame is a maximum.

Find the coordinates `(u, v)` of the point `M`, stating your answers correct to two decimal places. (3 marks)

The second supporting column, `PQ`, has its lowest point at `P (– u, w)`.

Find, correct to two decimal places, the value of `w` and the lengths of the supporting columns `MN` and `PQ`. (3 marks)

For the opening of the bridge, a banner is erected on the bridge, as shown by the shaded region in the diagram below.

Find the `x`-coordinates, correct to two decimal places, of `E` and `F`, the points at which the road meets the parabolic frame of the bridge. (3 marks)
Find the area of the banner (shaded region), giving your answer to the nearest square metre. (1 mark)

Show Answers Only

`72^@`
`−3/16`
`M (2.49,34.53)`
`w = 35.47\ text(m);quadMN = 25.23\ text(m);quadPQ = 24.30\ text(m)`
`x_E = -23.71;quadx_F = 28.00`
`870\ text(m²)`

Show Worked Solution

a.	`f(x)`	`=60 – 3/80x^2`
	`f′(x)`	`=- 3/40 x`

`text(At)\ \ x=-40,\ \ f′(x)=3`

♦ Mean mark part (a) 40%.

`tan theta`	`= 3`
`:. theta`	`= tan^(−1)(3)`
	`=71.56…`
	`= 72^@`

b.	`g(x)`	`= (x^3)/(25\ 600) – 3/16 x + 35`
	`g′(x)`	`=(3x^2)/(25\ 600) – 3/16`

`text(S)text(ince)\ \ 3x^2>=0\ \ text(for all)\ \ x,`

`text(Max downwards slope occurs at)\ \ x= 0.`

`gprime(0) = −3/16`

c. `text(Let)\ \ V=\ text(distance)\ MN`

`V`	`= (60 – 3/80x^2) – ((x^3)/(25\ 600) – 3/16x + 35)`
`(dV)/(dx)`	`=-3/40 x – (3x^2)/(25\ 600) + 3/16`

♦ Mean mark part (c) 39%.

`(dV)/(dx)`	`= 0quadtext(for)\ x ∈ [−40,40]`
`x`	`= 2.490…`

`text(When)\ \ x=2.490…,\ \ g(2.490…) = 34.533…`

`:. M (2.49,34.53)`

♦♦ Mean mark part (d) 29%.
MARKER’S COMMENT: Many students didn’t work to a sufficient number of decimal places.

d. `P(-2.49, w)`

`text(When)\ \ x=-2.490…,\ \ g(-2.490…) = 35.47…`

`:. w = 35.47\ \ text{(2 d.p.)}`

`V_(MN)`	`=(60 – 3/80(2.49…)^2) – (((2.49…)^3)/(25\ 600) – 3/16(2.49…) + 35)`
	`=25.23\ text{m (2 d.p.)}`
`V_(PQ)`	`=(60 – 3/80(-2.49…)^2) – (((-2.49…)^3)/(25\ 600) – 3/16(-2.49…) + 35)`
	`=24.30\ text{m (2 d.p.)}`

e. `text(Intersection occurs when:)`

`f(x) = g(x)quadtext(for)\ x ∈ (−40,40)`

`60 – 3/80x^2=(x^3)/(25\ 600) – 3/16x + 35\ \ text([by CAS])`

`x_E = -23.7068… = -23.71\ text{(2 d.p.)}`

`x_F = 27.9963… = 28.00\ text{(2 d.p.)}`

f.	`text(Area)`	`= int_-23.71^28.00 (f(x) – g(x))dx`
		`=int_-23.71^28.00 (60 – 3/80x^2 – ((x^3)/(25\ 600) – 3/16x + 35))\ dx`
		`=869.619…\ \ text([by CAS])`
		`= 870\ text(m²)`

Calculus, MET2 2015 VCAA 1

Let `f: R -> R,\ \ f(x) = 1/5 (x - 2)^2 (5 - x)`. The point `P(1, 4/5)` is on the graph of `f`, as shown below.

The tangent at `P` cuts the y-axis at `S` and the x-axis at `Q.`

Write down the derivative `f prime (x)` of `f (x)`. (1 mark)
1. Find the equation of the tangent to the graph of `f` at the point `P(1, 4/5)`. (1 mark)
2. Find the coordinates of points `Q` and `S`. (2 marks)
Find the distance `PS` and express it in the form `sqrt b/c`, where `b` and `c` are positive integers. (2 marks)

Find the area of the shaded region in the graph above. (3 marks)

Show Answers Only

`−3/5(x – 4)(x – 2)`
1. `y = – 9/5x + 13/5`
2. `S (0, 13/5)`
  
  `Q (13/9, 0)`
`sqrt 106/5`
`108/5\ text(units²)`

Show Worked Solution

a.	`f(x)`	`= 1/5 (x – 2)^2 (5 – x)`
	`fprime(x)`	`=1/5 xx 2(x-2)(5-x) – 1/5 (x-2)^2`
		`= −3/5(x – 4)(x – 2)`

b.i. `text(Solution 1)`

`y = −9/5x + 13/5qquad[text(CAS:tangentLine)\ (f(x),x,1)]`

`text(Solution 2)`

`m_text(tan) = -9/5,\ \ text(through)\ \ (1, 4/5)`

`y-4/5`	`=-9/5 (x-1)`
`y`	`=-9/5 x +13/5`

b.ii. `text(At)\ S,\ \ x=0`

`y = −9/5 xx 0 + 13/5 = 13/5`

`:. S(0,13/5)`

`text(At)\ Q,\ \ y=0`

`0`	`= −9/5x + 13/5`
`x`	`=13/5 xx 5/9=13/9`

`:. Q(13/9,0)`

c. `P(1, 4/5),\ \ S(0,13/5)`

`text(dist)\ PS`	`=sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`
	`= sqrt((1 – 0)^2 + (4/5 – 13/5)^2)`
	`= (sqrt106)/5`

d. `text(Find intersection pts of)\ SQ\ text(and)\ f(x),`

MARKER’S COMMENT: Many students complicated their answer by splitting up the area.

`text{Solve (using technology):}`

`1/5 (x – 2)^2 (5 – x) = −9/5x + 13/5`

`x = 1,7`

`:.\ text(Area)`	`= int_1^7(f(x) – (−9/5x + 13/5))dx`
	`= 108/5\ text(u²)`

GEOMETRY, FUR2 2006 VCAA 3

A closed cylindrical water tank has external diameter 3.5 metres.

The external height of the tank is 2.4 metres.

The walls, floor and top of the tank are made of concrete 0.25 m thick.

What is the internal radius, `r`, of the tank? (1 mark)
Determine the maximum amount of water this tank can hold.

Write your answer correct to the nearest cubic metre. (2 marks)

Show Answers Only

`1.5\ text(m)`
`13\ text{m³ (nearest m³)}`

Show Worked Solution

a. `text(Internal radius)\ (r)`

♦ Mean mark of both parts (combined) was 50%.

`= 1/2 xx (3.5 – 2 xx 0.25)`

`= 1.5\ text(m)`

b.	`text(Height)`	`= 2.4 – (2 xx 0.25)`
		`= 1.9\ text(m)`

`:.\ text(Volume)`	`= pi r^2 h`
	`= pi xx 1.5^2 xx 1.9`
	`= 13.43…`
	`= 13\ text{m³ (nearest m³)}`

GEOMETRY, FUR2 2006 VCAA 2

An allotment of land contains a communications tower, `PQ`.

Points `S`, `Q` and `T` are situated on level ground.

From `S` the angle of elevation of `P` is 20°.

Distance `SQ` is 125 metres.

Distance `TQ` is 98 metres.

Determine the height, `PQ`, of the communications tower.

Write your answer, in metres, correct to one decimal place. (1 mark)
Determine the angle of depression of `T` from `P`.

Write your answer, in degrees, correct to one decimal place. (1 mark)

Show Answers Only

`45.5\ text{m (1 d.p.)}`
`24.9^@\ text{(1 d.p.)}`

Show Worked Solution

a. `text(In)\ Delta PQS,`

♦ Mean mark of both parts (combined) was 50%.

`tan 20^@`	`= (PQ)/125`
`:. PQ`	`= 125 xx tan 20^@`
	`= 45.49…`
	`= 45.5\ text{m (1 d.p.)}`

`theta = text(angle of depression of)\ T\ text(from)\ P`

`tan theta`	`= 45.5/98`
	`= 0.464…`
`:. theta`	`= 24.90…`
	`= 24.9^@\ text{(1 d.p.)}`

GEOMETRY, FUR2 2006 VCAA 1

A farmer owns a flat allotment of land in the shape of triangle `ABC` shown below.

Boundary `AB` is 251 metres.

Boundary `AC` is 142 metres.

Angle `BAC` is 45°.

A straight track, `XY`, runs perpendicular to the boundary `AC`.

Point `Y` is 55 m from `A` along the boundary `AC`.

Determine the size of angle `AXY`. (1 mark)
Determine the length of `AX`.

Write your answer, in metres, correct to one decimal place. (1 mark)
The bearing of `C` from `A` is 078°.

Determine the bearing of `B` from `A`. (1 mark)
Determine the shortest distance from `X` to `C`.

Write your answer, in metres, correct to one decimal place. (2 marks)
Determine the area of triangle `ABC` correct to the nearest square metre. (1 mark)

The length of the boundary `BC` is 181 metres (correct to the nearest metre).

1. Use the cosine rule to show how this length can be found. (1 mark)
2. Determine the size of angle `ABC`.
  Write your answer, in degrees, correct to one decimal place. (1 mark)

A farmer plans to build a fence, `MN`, perpendicular to the boundary `AC`.

The land enclosed by triangle `AMN` will have an area of 3200 m².

Determine the length of the fence `MN`. (2 marks)

Show Answers Only

`45^@`
`77.8\ text(m)`
`033^@`
`102.9\ text(m)`
`12\ 601\ text(m²)`
1. `text(Proof)\ \ text{(See Worked Solutions)}`
2. `33.7^@`
`80\ text(m)`

Show Worked Solution

a.	`/_ AXY`	`= 180 – (90 + 45)`
		`= 45^@`

b. `text(In)\ Delta AXY,`

`cos 45^@`	`= 55/(AX)`
`:. AX`	`= 55/(cos 45^@`
	`= 77.78…`
	`= 77.8\ text{m (1 d.p.)}`

`text(Bearing of)\ B\ text(from)\ A`

`= 78 – 45`

`= 033^@`

`text(Using the cosine rule,)`

`XC^2`	`= 77.8^2 + 142^2 – 2 xx 77.8 xx 142 xx cos 45^@`
	`= 10\ 593.17…`
`:. XC`	`=sqrt (10\ 593.17…)`
	`= 102.92…`
	`= 102.9\ text{m (1 d.p.)}`

e. `text(Using sine rule,)`

♦ Mean mark of parts (e)-(g) (combined) was 40%.

`text(Area)\ Delta ABC`	`= 1/2 bc sin A`
	`= 1/2 xx 142 xx 251 xx sin 45^@`
	`= 12601.34…`
	`= 12\ 601\ text{m² (nearest m²)}`

f.i. `text(Using the cosine rule,)`

`BC^2`	`= 142^2 + 251^2 – 2 xx 142 xx 251 xx cos 45^@`
	`= 32\ 759.60…`
`:. BC`	`= 180.99…`
	`= 181\ text{m (nearest m) … as required}`

f.ii. `text(Using the cosine rule,)`

`cos /_ ABC`	`= (251^2 + 181^2 – 142^2)/(2 xx 251 xx 181)`
	`= 0.832…`
`:. /_ ABC`	`= 33.69…`
	`= 33.7^@\ text{(1 d.p.)}`

g. `/_ AMN = 180 – (90 + 45) = 45^@`

MARKER’S COMMENT: Part (g) was “very poorly answered” with many failing to realise the equality of `AN` and `MN`.

`:. Delta AMN\ text(is isosceles.)`

`text(Area)\ Delta AMN`	`= 1/2 xx AN xx MN`
`3200`	`= 1/2 xx MN xx MN\ (AN = MN)`
`MN^2`	`= 6400`
`:. MN`	`= 80\ text(m)`

GEOMETRY, FUR2 2007 VCAA 1-3

Question 1

Tessa is a student in a woodwork class.

The class will construct geometrical solids from a block of wood.

Tessa has a piece of wood in the shape of a rectangular prism.

This prism, `ABCDQRST`, shown in Figure 1, has base length 24 cm, base width 28 cm and height 32 cm.

On the front face of Figure 1, `ABRQ`, Tessa marks point `W` halfway between `Q` and `R` as shown in Figure 2 below. She then draws line segments `AW` and `BW` as shown.

Determine the length, in cm, of `QW`. (1 mark)
Calculate the angle `WAQ`. Write your answer in degrees, correct to one decimal place. (1 mark)
Calculate the angle `AWB` correct to one decimal place. (1 mark)
What fraction of the area of the rectangle `ABRQ` does the area of the triangle `AWB` represent? (1 mark)

Question 2

Tessa carves a triangular prism from her block of wood.

Using point `V`, halfway between `T` and `S` on the back face, `DCST`, of Figure 1, she constructs the triangular prism shown in Figure 3.

Show that, correct to the nearest centimetre, length `AW` is 34 cm. (1 mark)
Using length `AW` as 34 cm, find the total surface area, in cm², of the triangular prism `ABCDWV` in Figure 3. (2 marks)

Question 3

Tessa's next task is to carve the right rectangular pyramid `ABCDY` shown in Figure 4 below.

She marks a new point, `Y`, halfway between points `W` and `V` in Figure 3. She uses point `Y` to construct this pyramid.

Calculate the volume, in cm³, of the pyramid `ABCDY` in Figure 4. (1 mark)
Show that, correct to the nearest cm, length `AY` is 37 cm. (2 marks)
Using `AY` as 37 cm, demonstrate the use of Heron's formula to calculate the area, in cm², of the triangular face `YAB`. (2 marks)

Show Answers Only

Question 1

`12\ text(cm)`
`20.6^@`
`41.2^@\ text{(1 d.p.)}`
`1/2`

Question 2

`34\ text(cm)`
`3344\ text(cm²)`

Question 3

`7168\ text(cm³)`
`37\ text(cm)`
`420\ text(cm²)`

Show Worked Solution

`text(Question 1)`

a.	`QW`	`= 1/2 xx QR`
		`= 1/2 xx 24`
		`= 12\ text(cm)`

`tan\ /_ WAQ`	`= 12/32`
`:. /_ WAQ`	`= tan^-1\ 3/8`
	`= 20.55…`
	`= 20.6^@\ text{(1 d.p.)}`

c.	`/_ AWB`	`= 2 xx /_ WAG`
		`= 2 xx 20.6`
		`= 41.2^@\ text{(1 d.p.)}`

d.	`text(Area)\ ABRQ`	`= 24 xx 32`
		`= 768\ text(cm²)`
	`text(Area)\ Delta AWB`	`= 2 xx text(Area)\ Delta AQW`
		`= 2 xx 1/2 xx 12 xx 32`
		`= 384\ text(cm²)`

`:.\ text(Fraction of area)`

`= 384/768`

`= 1/2`

`text(Question 2)`

`text(Using Pythagoras in)\ Delta AWX,`

`AW`	`= sqrt (12^2 + 32^2)`
	`= sqrt 1168`
	`= 34.176…`
	`= 34\ text{cm (nearest cm) … as required.}`

b. `text(Total S.A. of triangular prism)`

`= text(area of base) + text(area of sides) + text(area of ends)`

`= 24 xx 28 + 2 xx (34 xx 28) + 2 xx (1/2 xx 24 xx 32)`

`= 672 + 1904 + 768`

`= 3344\ text(cm²)`

`text(Question 3)`

a.	`V`	`= 1/3 xx b xx h`
		`= 1/3 xx 24 xx 28 xx 32`
		`= 7168\ text(cm³)`

`text(Using Pythagoras in)\ Delta ABC,`

`AC`	`= sqrt (24^2 + 28^2)`
	`= 36.878…`

`text(Let)\ Z\ text(be the midpoint of)\ AC`

`:. AZ = (36.878…)/2 = 18.439…`

`text(Using Pythagoras in)\ Delta AYZ,`

`AY`	`= sqrt (32^2 + (18.439…)^2)`
	`= sqrt (1364)`
	`= 36.9…`
	`= 37\ text{cm (nearest cm) … as required}`

c. `text(Using Heron’s formula,)`

`s`	`= (37 + 37 + 24)/2=49`

`:.\ text(Area of)\ Delta YAB`

`= sqrt (49 (49 – 24) (49 – 37) (49 – 37))`

`= sqrt (49 xx 25 xx 12 xx 12)`

`= 420\ text(cm²)`

MATRICES, FUR2 2008 VCAA 3

The bookshop manager at the university has developed a matrix formula for determining the number of Mathematics and Physics textbooks he should order each year.

For 2009, the starting point for the formula is the column matrix `S_2008`. This lists the number of Mathematics and Physics textbooks sold in 2008.

`S_2008 = [(456),(350)]{:(text(Mathematics)),(text(Physics)):}`

`O_2009` is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for 2009.

`O_2009` is given by the matrix formula

`O_2009 = A\ S_2008 + B` where `A = [(0.75, 0),(0, 0.68)]` and `B = [(18),(12)]`

Determine `O_2009` (1 mark)

The matrix formula above only allows the manager to predict the number of books he should order one year ahead. A new matrix formula enables him to determine the number of books to be ordered two or more years ahead.

The new matrix formula is

`O_(n + 1) = C O_n - D`

where `O_n` is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for year `n`.

Here, `C = [(0.8, 0),(0, 0.8)]` and `D = [(40),(38)]`

The number of books ordered in 2008 was given by

`O_2008 = [(500),(360)]{:(text(Mathematics)),(text(Physics)):}`

Use the new matrix formula to determine the number of Mathematics textbooks the bookshop manager should order in 2010. (2 marks)

Show Answers Only

`O_2009 = [(360),(250)]`
`248`

Show Worked Solution

a.	`O_2009`	`= [(0.75, 0), (0, 0.68)] [(456), (350)] + [(18), (12)]`
		`= [(342), (238)] + [(18), (12)]`
		`= [(360), (250)]`

b.	`O_2009`	`= [(0.8, 0), (0, 0.8)] [(500), (360)] – [(40), (38)]`
		`= [(400), (288)] – [(40), (38)]`
		`= [(360), (250)]`
	`O_2010`	`= [(0.8, 0), (0, 0.8)] [(360), (250)] – [(40), (38)]`
		`= [(248), (162)]`

`:. 248\ text(Maths books should be ordered in 2010.)`

MATRICES, FUR2 2008 VCAA 1

Two subjects, Biology and Chemistry, are offered in the first year of a university science course.

The matrix `N` lists the number of students enrolled in each subject.

`N = [(460),(360)]{:(text(Biology)),(text(Chemistry)):}`

The matrix `P` lists the proportion of these students expected to be awarded an `A`, `B`, `C`, `D` or `E` grade in each subject.

`{:((qquadqquadqquadA,qquad\ B,qquadquadC,qquad\ D,qquadE)),(P = [(0.05,0.125,0.175,0.45,0.20)]):}`

Write down the order of matrix `P`. (1 mark)
Let the matrix `R = NP`.
1. Evaluate the matrix `R`. (1 mark)
2. Explain what the matrix element `R_24` represents. (1 mark)
Students enrolled in Biology have to pay a laboratory fee of $110, while students enrolled in Chemistry pay a laboratory fee of $95.
1. Write down a clearly labelled row matrix, called `F`, that lists these fees. (1 mark)
2. Show a matrix calculation that will give the total laboratory fees, `L`, paid in dollars by the students enrolled in Biology and Chemistry. Find this amount. (1 mark)

Show Answers Only

`1 xx 5`
1. `[(23,57.5,80.5,207,92),(18,45,63,162,72)]`
2. `text(See Worked Solutions)`
1. `text(See Worked Solutions)`
2. `text(See Worked Solutions)`

Show Worked Solution

a. `1 xx 5`

b.i.	`R`	`= NP`
		`= [(460), (360)] [0.05 quad 0.125 quad 0.175 quad 0.45 quad 0.20]`
		`= [(23,57.5,80.5,207,92),(18,45,63,162,72)]`

b.ii. `R_24\ text(represents the number of chemistry students)`

`text(expected to get a)\ D.`

c.i.

`{:(qquad qquad qquad B quad qquad C),(F = [(110, 95)]):}`

c.ii. `text(Total laboratory fees)`

`L = [(110, 95)] [(460), (360)] = [84\ 800]`

GEOMETRY, FUR2 2008 VCAA 3

A tree, 12 m tall, is growing at point `T` near a shed.

The distance, `CT`, from corner `C` of the shed to the centre base of the tree is 13 m.

Calculate the angle of elevation of the top of the tree from point `C`.

Write your answer, in degrees, correct to one decimal place. (1 mark)

`N` and `C` are two corners at the base of the shed. `N` is due north of `C`.

The angle, `TCN`, is 65°.

Show that, correct to one decimal place, the distance, `NT`, is 12.6 m. (1 mark)
Calculate the angle, `CNT`, correct to the nearest degree. (1 mark)
Determine the bearing of `T` from `N`. Write your answer correct to the nearest degree. (1 mark)
Is it possible for the tree to hit the shed if it falls?

Explain your answer showing appropriate calculations. (2 marks)

Show Answers Only

`42.7^@`
`text(See Worked Solution)`
`69^@`
`111^@`
`text(See Worked Solutions)`

Show Worked Solution

`text(Let)\ \ theta`	`=\ text(angle of elevation)`
	`= 12/13`
`:. theta`	`= 42.709…`
	`= 42.7^@\ text{(1 d.p.)}`

`text(Using the cosine rule:)`

MARKER’S COMMENT: Show the squareroot step in the calculations as per the solution.

`NT^2`	`= 10^2 + 13^2 – 2 xx 10 xx 13 xx cos65^@`
	`= 159.11…`
`:. NT`	`= sqrt(159.11…)`
	`= 12.61…`
	`= 12.6\ text{m (1 d.p.) … as required}`

c. `text(Using the sine rule:)`

♦♦ Mean mark of parts (c)-(e) (combined) was 23%.

`(sin /_ CNT)/13`	`= (sin 65^@)/12.6`
`sin /_ CNT`	`= (13 xx sin 65^@)/12.6`
	`= 0.935…`
`:. /_ CNT`	`= 69.24…`
	`= 69^@\ text{(nearest degree)}`

`text(Bearing of)\ T\ text(from)\ N`

`= 180 – 69`

`= 111^@`

`ST\ text(is the shortest distance between the)`

MARKER’S COMMENT: Many students had significant difficulty in understanding the 3-dimensional diagram.

`text(tree and the shed.)`

`text(In)\ \ Delta NST,`

`sin 69^@`	`= (ST)/12.6`
`:. ST`	`= 12.6 xx sin 69^@`
	`= 11.76…\ text(m)`

`:.\ text(S) text(ince the tree is 12 m, and 12 m) > 11.76\ text(m),`

`text(it could hit the shed if it falls.)`

GEOMETRY, FUR2 2014 VCAA 2

The chicken coop has two spaces, one for nesting and one for eating.

The nesting and eating spaces are separated by a wall along the line `AX`, as shown in the diagrams below.

`DX = 3.16\ text(m), ∠ADX = 45^@ and ∠AXD = 60^@.`

Write down a calculation to show that the value of `theta` is 75°. (1 mark)
The sine rule can be used to calculate the length of the wall `AX`.

Fill in the missing numbers below. (1 mark)
What is the length of `AX`?

Write your answer in metres, correct to two decimal places. (1 mark)
Calculate the area of the floor of the nesting space, `ADX`.

Write your answer in square metres, correct to one decimal place. (1 mark)

The height of the chicken coop is 1.8 m.

Wire mesh will cover the roof of the eating space.

The area of the walls along the lines `AB, BC and CX` will also be covered with wire mesh.

What total area, in square metres, will be covered by wire mesh?

Write your answer, correct to the nearest square metre. (2 mark)

Show Answers Only

`75^@`
`(AX)/(sin 45^@) = (3.16)/(sin 75^@)`
`2.31`
`3.2\ text(m)^2`
`17\ text(m)^2`

Show Worked Solution

a.	`theta`	`= 180^@ − (45^@ + 60^@)`
		`= 75^@`

b. `(AX)/(sin 45^@) = (3.16)/(sin 75^@)`

c.	`(AX)/(sin 45^@)`	`= (3.16)/(sin 75^@)`

	`:. AX`	`= (3.16 xx sin45^@)/(sin 75^@)`
		`= 2.313…`
		`=2.31\ text{m (2 d.p.)}`

d.	`text(Area of)\ \ ΔADX`	`= 1/2 xx b xx h`
		`= 1/2 xx 3.16 xx 2`
		`= 3.2\ text{m² (1 d.p.)}`

MARKER’S COMMENT: An excellent example where many students do not read the question carefully and give away easy marks.

`text(Roof area)`	`=1/2 xx BC xx (AB + XC)`
	`= 1/2 xx 2 xx (3 + 1.84)`
	`= 4.84`

`text(Wall area)`	`= (1.8 xx 3) + (1.8 xx 2) + (1.8 xx 1.84)`
	`= 12.312`

`:.\ text(Total area)`	`= 4.84 + 12.312`
	`=17.152`
	`=17\ text{m² (nearest m²)}`

GEOMETRY, FUR2 2014 VCAA 1

The floor of a chicken coop is in the shape of a trapezium.

The floor, `ABCD`, and the chicken coop are shown below.

`AB = 3\ text(m), BC = 2\ text(m and)\ \ CD = 5\ text(m.)`

What is the area of the floor of the chicken coop?

Write your answer in square metres. (1 mark)
What is the perimeter of the floor of the chicken coop?

Write your answer in metres, correct to one decimal place. (1 mark)

Show Answers Only

`8\ text(m²)`
`12.8\ text(m)`

Show Worked Solution

a.	`A`	`=1/2 h (a+b)`
		`= 1/2 xx 2 xx (3 + 5)`
		`= 8\ text(m²)`

`text(Using Pythagoras,)`

`AD^2`	`=sqrt(2^2+2^2)`
	`=2.82…\ text(m)`

`:.\ text(Perimeter)`	`= 3 + 2 + 5 + 2.82…`
	`= 12.8\ text{m (1 d.p.)}`

GEOMETRY, FUR2 2008 VCAA 2

A shed has the shape of a prism. Its front face, `AOBCD`, is shaded in the diagram below. `ABCD` is a rectangle and `M` is the mid point of `AB`.

Show that the length of `OM` is 1.6 m. (1 mark)
Show that the area of the front face of the shed, `AOBCD`, is 18 m². (1 mark)
Find the volume of the shed in m³. (1 mark)
All inside surfaces of the shed, including the floor, will be painted.
1. Find the total area that will be painted in m². (2 marks)
  
  One litre of paint will cover an area of 16 m².
2. Determine the number of litres of paint that is required. (1 mark)

Show Answers Only

`text(See Worked Solution)`
`text(See Worked Solution)`
`180`
1. `208`
2. `13\ text(litres)`

Show Worked Solution

`text(In)\ Delta AOM,\ text(using Pythagoras:)`

`OM`	`= sqrt (3.4^2 – 3^2)`
	`= sqrt 2.56`
	`= 1.6\ text(m … as required)`

b. `text(Area of front face of shed)`

MARKER’S COMMENT: “The “Show that” directive requires students to include all mathematical steps, as shown in the solution. Short cuts will not gain full marks!

`= text(Area)\ Delta AOB + text(Area)\ ABCD`

`= 1/2 xx 1.6 xx 6 + 2.2 xx 6`

`= 18\ text(m² … as required.)`

c.	`V`	`= Ah`
		`= 18 xx 10`
		`= 180\ text(m³)`

d.i.	`text(Roof areas)`	`= 2(1/2 xx 1.6 xx 6 + 3.4 xx 10)`
		`= 2 (4.8 + 34)`
		`= 77.6\ text(m²)`
	`text(Wall areas)`	`= 2 (2.2 xx 6 + 2.2 xx 10)`
		`= 2 (13.2 + 22)`
		`= 70.4\ text(m²)`
	`text(Floor area)`	`= 60\ text(m²)`

`:.\ text(Total Area to be painted)`

`= 77.6 + 70.4 + 60`

`= 208\ text(m²)`

ii. `text(Litres of paint required)`

`= 208/16`

`= 13\ text(litres)`

GEOMETRY, FUR2 2008 VCAA 1

A shed is built on a concrete slab. The concrete slab is a rectangular prism 6 m wide, 10 m long and 0.2 m deep.

Determine the volume of the concrete slab in m³. (1 mark)
On a plan of the concrete slab, a 3 cm line is used to represent a length of 6 m.
1. What scale factor is used to draw this plan? (1 mark)

The top surface of the concrete slab shaded in the diagram above has an area of 60 m².

What is the area of this surface on the plan? (1 mark)

Show Answers Only

`12\ text(m³)`
1. `1/200`
2. `text(15 cm²)`

Show Worked Solution

a.	`V`	`= lbh`
		`= 10 xx 6 xx 0.2`
		`= 12\ text(m³)`

b.i. `text(Scale Factor)`

`= 3/600`

`= 1/200`

♦ A poorly answered question, although exact data unavailable.
MARKER’S COMMENT: The most successful answers established the scale factor and then squared it, as shown in the solution.

ii.	`text(Plan Area)`	`= text(Slab Area) xx (1/200)^2`
		`= 60 xx (1/200)^2`
		`= 0.0015\ text(m²)`
		`= 15\ text(cm²)`

GEOMETRY, FUR2 2015 VCAA 3

Cabins are being built at the camp site.

The dimensions of the front of each cabin are shown in the diagram below.

The walls of each cabin are 2.4 m high.

The sloping edges of the roof of each cabin are 2.4 m long.

The front of each cabin is 4 m wide.

The overall height of each cabin is `h` metres.

Show that the value of `h` is 3.73, correct to two decimal places. (1 mark)

Each cabin is in the shape of a prism, as shown in the diagram below.

All external surfaces of one cabin are to be painted, excluding the base.

What is the total area of the surface to be painted?

Write your answer correct to the nearest square metre. (2 marks)

Show Answers Only

`3.73`
`76\ text(m²)`

Show Worked Solution

a.	`h`	`= 2.4 + sqrt(2.4^2 – 2^2)`
		`= 3.726…`
		`=3.73\ text{m (2 d.p.)}`

b. `text(Area of walls and roof)`

♦♦ “Few students answered this question correctly” (exact data unavailable).
MARKER’S COMMENT: In extended calculations, clearly label each step and draw supporting diagrams where applicable.

`=2xx(2.4 xx 5.4) + 2xx(2.4 xx 5.4) `

`=25.92+25.92`

`=51.84`

`text(Area of front and back)`

`=2xx(2.4 xx 4 + 1/2 xx 4 xx 1.33)`

`=2xx12.26`

`=24.52`

`:.\ text(Total area to be painted)`

`=51.84+24.52`

`=76.36`

`=76\ text{m² (nearest m²)}`

A. `[(24, -14), (13, -7.5)]`	B. `[(-4.25, -5.5), (9.75, 12.5)]`

C. `[(-3.75, 7), (-6.5, 12)]`	D. `[(25, 11), (-19.5, -8.5)]`

E. `[(0.625, 1.5), (1.6, 3.333)]`

`AX`	`= [(5,6),(8,10)]`
`X`	`= [(8,4),(5,3)]^(−1)[(5,6),(8,10)]`
	`= [(0.75,−1),(−1.25,2)][(5,6),(8,10)]`
	`= [(−4.25,−5.5),(9.75,12.5)]`