Band 6 – Page 18

GRAPHS, FUR2 2009 VCAA 4

Cheapstar Airlines wishes to find the optimum number of flights per day on two of its most popular routes: Alberton to Bisley and Alberton to Crofton.

Let `x` be the number of flights per day from Alberton to Bisley

`y` be the number of flights per day from Alberton to Crofton

Table 4 shows the constraints on the number of flights per day and the number of crew per flight.

The lines `x + y = 10` and `3x + 5y = 41` are graphed below.

A profit of $1300 is made on each flight from Alberton to Bisley and a profit of $2100 is made on each flight from Alberton to Crofton.

Determine the maximum total profit that Cheapstar Airlines can make per day from these flights. (2 marks)

Show Answers Only

`$17\ 300`

Show Worked Solution

`P = 1300x + 2100y`

♦♦♦ Mean mark 9%.
MARKER’S COMMENT: Many students incorrectly substituted the point of intersection (4.5,5.5) into the equation. Integer values within the feasible region were required.

`text(Within the feasible area, profit is)`

`text(maximised at)\ (2,7).`

`:.\ text(Maximum profit)`

`= 1300 xx 2 + 2100 xx 7`

`= $17\ 300`

CORE*, FUR2 2010 VCAA 4

A home buyer takes out a reducing balance loan of $250 000 to purchase an apartment.

Interest on the loan will be calculated and paid monthly at the rate of 6.25% per annum.

The loan will be fully repaid in equal monthly instalments over 20 years.
1. Find the monthly repayment, in dollars, correct to the nearest cent. (1 mark)
2. Calculate the total interest that will be paid over the 20 year term of the loan.
  Write your answer correct to the nearest dollar. (2 marks)
After 60 monthly repayments have been made, what will be the outstanding principal on the loan?

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

By making a lump sum payment after nine years, the home buyer is able to reduce the principal on his loan to $100 000. At this time, his monthly repayment changes to $1250. The interest rate remains at 6.25% per annum, compounding monthly.

With these changes, how many months, in total, will it take the home buyer to fully repay the $250 000 loan? (1 mark)

Show Answers Only

i. `$1827.32`
ii. `$188\ 557\ \ text{(nearest $)}`
`$213\ 118`
`212\ text(months)`

Show Worked Solution

a.i. `text(By TVM Solver,)`

♦♦ This question (all parts) were generally poorly answered, although exact data unavailable.

`N`	`= 20 xx 12 = 240`
`I(text(%))`	`= 6.25`
`PV`	`= 250\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = −1827.320…`

`:. text(Monthly repayment) = $1827.32`

a.ii. `text(Total of all repayments)`

`= 240 xx 1827.32`

`= $438\ 556.80`

`:.\ text(Total interest paid)`

`= 438\ 556.80 – 250\ 000`

`= 188\ 556.80`

`= $188\ 557\ \ text{(nearest $)}`

b. `text(By TVM Solver:)`

`N`	`= 60`
`I(text(%))`	`= 6.25`
`PV`	`= 250\ 000`
`PMT`	`= −1827.32`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=>FV = −213\ 117.807…`

`:.\ text(After 60 repayments, the loan balance is $213 118.)`

c. `text(By TVM Solver:)`

`N`	`= ?`
`I(text(%))`	`= 6.25`
`PV`	`= 100\ 000`
`PMT`	`= −1250`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 103.75…`

MARKER’S COMMENT: A common error was to forget to add the initial 108 months.

`:.\ text(Total time to repay loan)`

`= 104 + 9 xx 12`

`= 212\ text(months)`

CORE*, FUR2 2010 VCAA 3

Simple Saver is a simple interest investment in which interest is paid annually.

Growth Plus is a compound interest investment in which interest is paid annually.

Initially, $8000 is invested with both Simple Saver and Growth Plus.

The graph below shows the total value (principal and all interest earned) of each of these investments over a 15 year period.

The increase in the value of each investment over time is due to interest

Which investment pays the highest annual interest rate, Growth Plus or Simple Saver?

Give a reason to justify your answer. (1 mark)
After 15 years, the total value (principal and all interest earned) of the Simple Saver investment is $21 800.

Find the amount of interest paid annually. (1 mark)
After 15 years, the total value (principal and all interest earned) of the Growth Plus investment is $24 000.
1. Write down an equation that can be used to ﬁnd the annual compound interest rate, `r`. (1 mark)
2. Determine the annual compound interest rate.
  
  Write your answer as a percentage correct to one decimal place. (1 mark)

Show Answers Only

`text(Simple Saver has the highest annual)`
`text(interest rate because after 1 year,)`
`text(the value of investment is higher.)`
`$920`
1. `24\ 000 = 8000 (1 + r/100)^15`
2. `7.6text{% (1 d.p.)}`

Show Worked Solution

a. `text(Simple Saver has the highest annual)`

♦♦♦ Part (a) was “very” poorly answered although exact data unavailable.
MARKER’S COMMENT: Most students ignored the word “rate” and instead referred to the eventual return of each investment.

`text(interest rate because after 1 year,)`

`text(the value of investment is higher.)`

b. `text(Total interest earned)`

`= 21\ 800 – 8000`

`= $13\ 800`

`:.\ text(Interest paid annually)`

`= (13\ 800)/15`

`= $920`

c.i. `text(Using)\ A = PR^n,`

`24\ 000 = 8000 (1 + r/100)^15`

c.ii.	`(1 + r/100)^15`	`= 3`
	`1 + r/100`	`= 1.0759…`
	`:. r`	`= 0.0759…`
		`= 7.6text{% (1 d.p.)}`

GRAPHS, FUR2 2010 VCAA 3

Let `x` be the number of Softsleep pillows that are sold each week and `y` be the number of Resteasy pillows that are sold each week.

A constraint on the number of pillows that can be sold each week is given by

Inequality 1: `x + y ≤ 150`

Explain the meaning of Inequality 1 in terms of the context of this problem. (1 mark)

Each week, Anne sells at least 30 Softsleep pillows and at least `k` Resteasy pillows.

These constraints may be written as

Inequality 2: `x ≥ 30`

Inequality 3: `y ≥ k`

The graphs of `x + y = 150` and `y = k` are shown below.

State the value of `k`. (1 mark)
On the axes above
1. draw the graph of `x = 30` (1 mark)
2. shade the region that satisﬁes Inequalities 1, 2 and 3. (1 mark)
Softsleep pillows sell for $65 each and Resteasy pillows sell for $50 each.

What is the maximum possible weekly revenue that Anne can obtain? (2 marks)

Anne decides to sell a third type of pillow, the Snorestop.

She sells two Snorestop pillows for each Softsleep pillow sold. She cannot sell more than 150 pillows in total each week.

Show that a new inequality for the number of pillows sold each week is given by

Inequality 4: `3x + y ≤ 150`

where `x` is the number of Softsleep pillows that are sold each week

and `y` is the number of Resteasy pillows that are sold each week. (1 mark)

Softsleep pillows sell for $65 each.

Resteasy pillows sell for $50 each.

Snorestop pillows sell for $55 each.

Write an equation for the revenue, `R` dollars, from the sale of all three types of pillows, in terms of the variables `x` and `y`. (1 mark)
Use Inequalities 2, 3 and 4 to calculate the maximum possible weekly revenue from the sale of all three types of pillow. (2 marks)

Show Answers Only

`text(Inequality 1 means that the combined number of Softsleep)`
`text(and Resteasy pillows must be less than 150.)`
`45`
i. & ii.
`$9075`
`text(See Worked Solutions)`
`R = 175x + 50y`
`$8375`

Show Worked Solution

a. `text(Inequality 1 means that the combined number of Softsleep)`

`text(and Resteasy pillows must be less than 150.)`

b. `k = 45`

c.i. & ii.

d. `text(Checking revenue at boundary)`

`text(At)\ (30,120),`

`R = 65 xx 30 + 50 xx 120 = $7950`

`text(At)\ (105,45),`

`R = 65 xx 105 + 50 xx 45 = $9075`

`:. text(Maximum weekly revenue) = $9075`

e. `text(Let)\ z = text(number of SnoreStop pillows)`

♦♦ Mean mark of parts (e)-(g) (combined) was 24%.

`:. x + y + z <= 150,\ text(and)`

`z = 2x\ \ text{(given)}`

`:. x + y + 2x`	`<= 150`
`3x + y`	`<= 150\ \ …text(as required)`

f.	`R`	`= 65x + 50y + 55(2x)`
		`= 65x + 50y + 110x`
		`= 175x + 50y`

`text(New intersection occurs at)\ (35,45)`

`:.\ text(Maximum weekly revenue)`

`= 175 xx 35 + 50 xx 45`

`= $8375`

GEOMETRY, FUR2 2010 VCAA 4

A ﬂying fox suspension wire begins at `V`, 12.5 metres above `T` as shown in the diagram below. It ends at `S`, 4.5 metres above `K`.

At `P`, the ﬂying fox wire passes over `H`.

The horizontal distances `TH` and `HK` are 95 metres and 65 metres respectively.

Calculate the vertical distance, `PH`, in metres. (2 marks)

Show Answers Only

`7.75\ text(m)`

Show Worked Solution

`Delta VAS\ text(|||)\ Delta PBS\ \ (A A A)`

♦♦♦ Mean mark 11%.

`(PB)/(BS)`	`= (VA)/(AS)`
`(PB)/(65)`	`= 8/160`
`:. PB`	`= (8 xx 65)/160`
	`= 3.25\ text(m)`

`:. PH`	`= PB + BH`
	`= 3.25 + 4.5`
	`= 7.75\ text(m)`

`text(Alternative solution)`

`text(In)\ Delta VAS,`

`tantheta`	`= 8/160`
`:. theta`	`= 2.862…^@`

`text(In)\ DeltaPBS,`

`tantheta`	`= (PB)/65`
`PB`	`= 65 xx tan2.862…^@`
	`= 3.25\ text(m)`

`:. PH = 3.25 + 4.5 = 7.75\ text(m)`

CORE*, FUR2 2011 VCAA 4

Tania takes out a reducing balance loan of $265 000 to pay for her house.

Her monthly repayments will be $1980.

Interest on the loan will be calculated and paid monthly at the rate of 7.62% per annum.

i. How many monthly repayments are required to repay the loan? Write your answer to the nearest month. (1 mark)
ii. Determine the amount that is paid off the principal of this loan in the first year. Write your answer to the nearest cent. (1 mark)

Immediately after Tania made her twelfth payment, the interest rate on her loan increased to 8.2% per annum, compounding monthly.

Tania decided to increase her monthly repayment so that the loan would be repaid in a further nineteen years.

Determine the new monthly repayment.

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

Show Answers Only

i. `300`
ii. `$3694.25\ \ text{(nearest cent)}`
`$2265.04`

Show Worked Solution

a.i. `text(By TVM Solver:)`

`N`	`= ?`
`I(text(%))`	`= 7.62`
`PV`	`= 265\ 000`
`PMT`	`= −1980`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 299.573…`

`:.\ text(After 300 months, the loan will be repaid.)`

a.ii. `text(After 12 months, by TVM Solver:)`

`N`	`= 12`
`I(text(%))`	`= 7.62`
`PV`	`= 265\ 000`
`PMT`	`= −1980`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −261\ 305.74…`

`:.\ text(Amount paid off after 1 year)`

`= 265\ 000 – 261\ 305.747…`

`= 3694.252…`

`= $3694.25\ \ text{(nearest cent)}`

b. `text(By TVM Solver:)`

`N`	`= 19 xx 12 = 228`
`Itext(%)`	`= 8.2`
`PV`	`= 261\ 305.75`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = −2265.043`

`:.\ text(New monthly repayment is $2265.04)`

CORE*, FUR2 2012 VCAA 4

Arthur invested $80 000 in a perpetuity that returns $1260 per quarter. Interest is calculated quarterly.

Calculate the annual interest rate of Arthur’s investment. (1 mark)
After Arthur has received 20 quarterly payments, how much money remains invested in the perpetuity? (1 mark)
Arthur’s wife, Martha, invested a sum of money at an interest rate of 9.4% per annum, compounding quarterly.

She will be paid $1260 per quarter from her investment.

After ten years, the balance of Martha’s investment will have reduced to $7000.

Determine the initial sum of money Martha invested.

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

Show Answers Only

`6.3text(%)`
`$80\ 000`
`$35\ 208`

Show Worked Solution

a. `text(Let)\ \ r =\ text(annual interest rate)`

♦♦ Mean mark of all parts (combined) was 26%.

`80\ 000 xx r/(4 xx 100)`	`= 1260`
`:. r`	`= (1260 xx 400)/(80\ 000)`
	`= 6.3text(%)`

b. `$80\ 000`

`text{(The principal invested in a perpetuity}`

`text{remains unchanged.)}`

c. `text(Find)\ PV\ text(using TVM Solver:)`

`N`	`= 4 xx 10 = 40`
`I(text(%))`	`= 9.4`
`PV`	`= ?`
`PMT`	`= 1260`
`FV`	`= 7000`
`text(P/Y)`	`= text(C/Y) = 4`

`=> PV = −35\ 208.002…`

`:.\ text{Martha initially invested $35 208 (nearest $)}`

CORE*, FUR2 2012 VCAA 3

An area of a club needs to be refurbished.

$40 000 is borrowed at an interest rate of 7.8% per annum.

Interest on the unpaid balance is charged to the loan account monthly.

Suppose the $40 000 loan is to be fully repaid in equal monthly instalments over five years.

Determine the monthly payment, correct to the nearest cent. (1 mark)
If, instead, the monthly payment was $1000, how many months will it take to fully repay the $40 000? (1 mark)
Suppose no payments are made on the loan in the first 12 months.
i. Write down a calculation that shows that the balance of the loan account after the first 12 months will be $43 234 correct to the nearest dollar. (1 mark)
ii. After the first 12 months, only the interest on the loan is paid each month.

Determine the monthly interest payment, correct to the nearest cent. (1 mark)

Show Answers Only

`$807.23`
`text(47 months)`
i. `text(See Worked Solutions)`
ii. `$281.02\ \ text{(nearest cent)}`

Show Worked Solution

a. `text(Find monthly payment by TVM solver:)`

♦♦ Mean mark of all parts (combined) 28%.

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 7.8`
`PV`	`= 40\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PMT = −807.232…`

`:.\ text{Monthly payment is $807.23 (nearest cent)}`

b. `text(Find)\ n\ text(when loan fully paid:)`

`N`	`= ?`
`I(text(%))`	`= 7.8`
`PV`	`= 40\ 000`
`PMT`	`= −1000`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N = 46.47…`

`:.\ text(Loan will be fully repaid after 47 months.)`

c.i. `text(Loan balance after 12 months)`

`= 40\ 000 xx (1 + (7.8)/(12 xx 100))^12`

`= 43\ 233.99…`

`= $43\ 234\ \ text{(nearest $) … as required}`

c.ii. `text(Interest paid each month)`

`= 43\ 234 xx (7.8)/(12 xx 100)`

`= 281.021`

`= $281.02\ \ text{(nearest cent)}`

NETWORKS, FUR1 2006 VCAA 8 MC

Euler’s formula, relating vertices, faces and edges, does not apply to which one of the following graphs?

Show Answers Only

`D`

Show Worked Solution

`text(Euler’s formula only applies to)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: Over a third of students incorrectly chose option `A`, unaware that any graph with 4 or less vertices must be planar.

`text(planar graphs.)`

`text(Consider option)\ D,`

`text(Only option)\ D\ text(cannot be redrawn)`

`text(as a planar graph.)`

`rArr D`

NETWORKS, FUR1 2008 VCAA 7 MC

The graph above has

4 faces.
5 faces.
6 faces.
8 faces.
9 faces.

Show Answers Only

`=> C`

Show Worked Solution

`text(Redrawing the graph in planar form,)`

`text(the graph can be seen to have 6 faces.)`

`text(Alternatively, using Euler’s rule:)`

`v + f`	`= e + 2`
`5 + f`	`= 9 + 2`
`:. f`	`= 6`

`=> C`

NETWORKS, FUR1 2010 VCAA 9 MC

The table below shows the time (in minutes) that each of four people, Aiden, Bing, Callum and Dee, would take to complete each of the tasks `U, V, W` and `X.`

If each person is allocated one task only, the minimum total time for this group of people to complete all four tasks is

A. `22\ text(minutes)`

B. `28\ text(minutes)`

C. `29\ text(minutes)`

D. `30\ text(minutes)`

E. `32\ text(minutes)`

Show Answers Only

`B`

Show Worked Solution

`:.\ text(Minimum time)`

`= 9 + 5 + 6 + 8`

`= 28\ text(minutes)`

`=> B`

NETWORKS, FUR1 2010 VCAA 8 MC

A project has 12 activities. The network below gives the time (in hours) that it takes to complete each activity.

The critical path for this project is

A. `ADGK`

B. `ADGIL`

C. `BHJL`

D. `CEGIL`

E. `CEHJL`

Show Answers Only

`D`

Show Worked Solution

`text(Critical path is)\ \ CEGIL`

♦ Mean mark 41%.

`=> D`

NETWORKS, FUR1 2006 VCAA 9 MC

The network below shows the activities and their completion times (in hours) that are needed to complete a project.

The project is to be crashed by reducing the completion time of one activity only.

This will reduce the completion time of the project by a maximum of

A. 1 hour

B. 2 hours

C. 3 hours

D. 4 hours

E. 5 hours

Show Answers Only

`D`

Show Worked Solution

`text(The critical path is)\ BDCEHJ\ text{(19 hours)}`

♦♦♦ Mean mark 17%.
MARKER’S COMMENT: When choosing an activity to crash, take care that a new critical path is not created.

`text(Also,)\ ACEHJ\ text{(15 hours)}`

`:.\ text(Activity)\ B\ text(could be crashed by 4 hours)`

`text(without a new critical path emerging.)`

`rArr D`

MATRICES*, FUR1 2009 VCAA 9 MC

Five soccer teams played each other once in a tournament. In each game there was a winner and a loser.

A table of one-step and two-step dominances was prepared to summarise the results.

One result in the tournament that must have occurred is that

A. Elephants defeated Bears.

B. Elephants defeated Aardvarks.

C. Aardvarks defeated Donkeys.

D. Donkeys defeated Bears.

E. Bears defeated Chimps.

Show Answers Only

`B`

Show Worked Solution

`text(Consider option)\ A,`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: The critical step here is to realise that if Team A beats Team B, then Team B’s one-step dominances become Team A’s two-step dominances.

`text(If elephants defeat bears, elephants)`

`text(will have a 2-step dominance over)`

`text(3 other teams.)`

`:.\ text(Incorrect.)`

`text(Consider option)\ B,`

`text(If elephants defeat aardvarks, elephants)`

`text(will have a 2-step dominance over)`

`text(1 other team.)`

`:.\ text(Correct.)`

`=> B`

NETWORKS, FUR1 2009 VCAA 8 MC

An undirected connected graph has five vertices.

Three of these vertices are of even degree and two of these vertices are of odd degree.

One extra edge is added. It joins two of the existing vertices.

In the resulting graph, it is not possible to have five vertices that are

A. all of even degree.

B. all of equal degree.

C. one of even degree and four of odd degree.

D. three of even degree and two of odd degree.

E. four of even degree and one of odd degree.

Show Answers Only

`E`

Show Worked Solution

`text(Consider an example of the graph)`

♦♦♦ Mean mark 25%.

`text{described (below):}`

`A\ text(is possible – join)\ V\ text(and)\ Z`

`B\ text(is possible – join)\ V\ text(and)\ Z`

`C\ text(is possible – join)\ W\ text(and)\ Y`

`D\ text(is possible – join)\ V\ text(and)\ X`

`E\ text(is NOT possible)`

`=> E`

NETWORKS, FUR1 2011 VCAA 9 MC

An Euler path through a network commences at vertex `P` and ends at vertex `Q`.

Consider the following five statements about this Euler path and network.

• In the network, there could be three vertices with degree equal to one.

• The path could have passed through an isolated vertex.

• The path could have included vertex `Q` more than once.

• The sum of the degrees of vertices `P` and `Q` could equal seven.

• The sum of the degrees of all vertices in the network could equal seven.

How many of these statements are true?

A. `0`

B. `1`

C. `2`

D. `3`

E. `4`

Show Answers Only

`B`

Show Worked Solution

`text(“The path could have included vertex)\ Q`

♦♦ Mean mark 30%.

`text(more than once” is the only true statement.)`

`=> B`

NETWORKS, FUR1 2011 VCAA 7 MC

Andy, Brian and Caleb must complete three activities in total (K, L and M)

The table shows the person selected to complete each activity, the time it will take to complete the activity in minutes and the immediate predecessor for each activity.

All three activities must be completed in a total of 40 minutes.

The instant that Andy starts his activity, Caleb gets a telephone call.

The maximum time, in minutes, that Caleb can speak on the telephone before he must start his allocated activity is

A. 5

B. 13

C. 18

D. 24

E. 34

Show Answers Only

`D`

Show Worked Solution

`text(Maximum speaking time)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students incorrectly answered the earliest starting time, C.

`= 40 – text(duration of)\ M`

`= 40 – 16`

`= 24\ text(minutes)`

`=> D`

NETWORKS, FUR1 2012 VCAA 9 MC

John, Ken and Lisa must work together to complete eight activities, `A, B, C, D, E, F, G` and `H`, in minimum time.

The directed network below shows the activities, their completion times in days, and the order in which they must be completed.

Several activities need special skills. Each of these activities may be completed only by a specified person.

Activities `A` and `F` may only be completed by John.
Activities `B` and `C` may only be completed by Ken.
Activities `D` and `E` may only be completed by Lisa.
Activities `G` and `H` may be completed by any one of John, Ken or Lisa.

With these conditions, the minimum number of days required to complete these eight activities is

A. 14

B. 17

C. 20

D. 21

E. 24

Show Answers Only

`D`

Show Worked Solution

`text(The minimum days require the following path:)`

♦♦♦ Mean mark 18%.

`text(John completes)\ A.`

`text(Lisa completes)\ D\ text(then)\ E\ text{(restriction)}.`

`H\ text(is completed.)`

`text(All other activities can be completed)`

`text(during this time.)`

`text(Minimum days)`	`= 3 + 7 + 6 + 5`
	`= 21`

`rArr D`

NETWORKS, FUR1 2012 VCAA 8 MC

Eight activities, `A, B, C, D, E, F, G\ text(and)\ H`, must be completed for a project.

The graph above shows these activities and their usual duration in hours.

The duration of each activity can be reduced by one hour.

To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is

A. 1

B. 2

C. 3

D. 4

E. 5

Show Answers Only

`C`

Show Worked Solution

`text(Paths)\ AFH\ text(and)\ BCFH => 18\ text(hours)`

♦♦ Mean mark 30%.

`text(Paths)\ AEG\ text(and)\ BCEG => 17\ text(hours)`

`text(Reducing)\ A\ text(and)\ B\ text(by 1 hour each)`

`text(reduces each path above by 1 hour.)`

`:. text(Need to now reduce)\ AFH\ text(and)`

`BCFH\ text(by 1 hour to 16 hours.)`

`text(Reducing)\ F\ text(or)\ H\ text(by 1 hour brings)`

`text(the critical path down to 16 hours.)`

`rArr C`

NETWORKS, FUR1 2012 VCAA 7 MC

Vehicles from a town can drive onto a freeway along a network of one-way and two-way roads, as shown in the network diagram below.

The numbers indicate the maximum number of vehicles per hour that can travel along each road in this network. The arrows represent the permitted direction of travel.

One of the four dotted lines shown on the diagram is the minimum cut for this network.

The maximum number of vehicles per hour that can travel through this network from the town onto the freeway is

A. `310`

B. `330`

C. `350`

D. `370`

E. `390`

Show Answers Only

`C`

Show Worked Solution

`text(Consider each “minimum cut” line,)`

♦♦♦ Mean mark 23%.

`text(Line 1: doesn’t seperate the town and freeway)`

`text(Line 2: 240 + 110 = 350)`

`text(Line 3: 240 + 60 + 90 = 390)`

`text(Line 4: 280 + 90 = 370)`

`:.\ text(Line 2 gives the maximum flow)`

`rArr C`

NETWORKS, FUR1 2013 VCAA 8 MC

The graph above shows five activities, `A, B, C, D\ text(and)\ E`, that must be completed to finish a project.

The number next to each letter shows the completion time, in hours, for the activity.

Each of the five activities can have its completion time reduced by a maximum of one hour at a cost of $100 per hour.

The least cost to achieve the greatest reduction in the time taken to finish the project is

A. $100

B. $200

C. $300

D. $400

E. $500

Show Answers Only

`D`

Show Worked Solution

`text(Clearly, we reduce)\ E\ text(by 1 hour.)`

♦♦♦ Mean mark 23%.

`text(New critical path is)\ B – D – E\ \ text{(23 hours)}`

`text(Next, reduce)\ B and D\ text(by 1 hour each.)`

`text(A new critical path emerges:)`

`A – C – E\ \ text{(23 hours)}`

`:.\ text(Reduce)\ A\ text(or)\ C\ text(by 1 hour so)`

`text(that)\ A – C – E\ \ text{(22 hours) matches}`

`B – D – E\ \ text{(22 hours).}`

`:.\ text(Least cost for maximum time reduction)`

`= 4 xx 100`

`= $ 400`

`=> D`

NETWORKS, FUR2 2007 VCAA 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

The directed network below shows the activities and their completion times in weeks.

Determine the minimum time, in weeks, to complete this project. (1 mark)
Determine the slack time, in weeks, for activity `D`. (2 marks)

The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

Which of these activities, if reduced in time individually, would not result in an earlier completion of the project? (1 mark)

The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is 2 weeks.

Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time. (1 mark)
Determine the minimum additional cost of completing the project in this reduced time. (1 mark)

Show Answers Only

`19\ text(weeks)`
`5\ text(weeks)`
`A, E, G`
`text(15 weeks)`
`$25\ 000`

Show Worked Solution

a. `B-C-F-H-I\ \ text(is the critical path.)`

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

`:.\ text(Minimum time)`	`= 4 + 3 + 4 + 2 + 6`
	`= 19\ text(weeks)`

b.	`text(EST of)\ D`	`= 4`
	`text(LST of)\ D`	`= 9`

`:.\ text(Slack time of)\ D`	`= 9 – 4`
	`= 5\ text(weeks)`

c. `A, E,\ text(and)\ G\ text(are not currently on)`

`text(the critical path, therefore crashing)`

`text(them will not affect the completion)`

`text(time.)`

d. `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks.)`

`text(However, a new critical path)`

`B − E − H − I\ text(takes 16 weeks.)`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum time = 5 weeks)`

e.	`text(Additional cost)`	`= 5 xx $5000`
		`= $25\ 000`

NETWORKS, FUR2 2007 VCAA 3

As an attraction for young children, a miniature railway runs throughout the new housing estate.

The trains travel through stations that are represented by nodes on the directed network diagram below.

The number of seats available for children, between each pair of stations, is indicated beside the corresponding edge.

Cut 1, through the network, is shown in the diagram above.

Determine the capacity of Cut 1. (1 mark)
Determine the maximum number of seats available for children for a journey that begins at the West Terminal and ends at the East Terminal. (1 mark)

On one particular train, 10 children set out from the West Terminal.

No new passengers board the train on the journey to the East Terminal.

Determine the maximum number of children who can arrive at the East Terminal on this train. (1 mark)

Show Answers Only

`43`
`22`
`7`

Show Worked Solution

a. `text(The capacity of Cut 1)`

♦♦ Mean mark for all parts (combined) was 33%.
MARKER’S COMMENT: A common error was counting the edge with “10” in the reverse direction (it should be ignored).

`=14 + 8 + 13 + 8`

`= 43`

`text(Maximum seats)`	`=\ text(minimum cut)`
	`= 6 + 7 + 9`
	`= 22`

c. `text{The path (edge weights) of the train setting out with}`

`text(10 children starts with: 11 → 13.)`

`text(At the next station, a maximum of 7 seats are available)`

`text(which remain until the East Terminal.)`

`:.\ text(Maximum number of children arriving is 7.)`

NETWORKS, FUR1 2014 VCAA 9 MC

A network of tracks connects two car parks in a festival venue to the exit, as shown in the directed graph below.

The arrows show the direction that cars can travel along each of the tracks and the numbers show each track’s capacity in cars per minute.

Five cuts are drawn on the diagram.

The maximum number of cars per minute that will reach the exit is given by the capacity of

A. Cut A

B. Cut B

C. Cut C

D. Cut D

E. Cut E

Show Answers Only

`D`

Show Worked Solution

`text(Cut)\ A and B\ text(don’t separate both)`

♦♦♦ Mean mark 24%.
COMMENT: Note that the “4” is not included in Cut D as it is flowing in the opposite direction.

`text(car parks from the exit.)`

`text(Cut)\ D\ text(has the minimum cut of)`

`text(the three cuts remaining.)`

`=> D`

NETWORKS, FUR1 2014 VCAA 6-7 MC

Consider the following four graphs.

Part 1

How many of these four graphs have an Eulerian circuit?

A. `0`

B. `1`

C. `2`

D. `3`

E. `4`

Part 2

How many of these four graphs are planar?

A. `0`

B. `1`

C. `2`

D. `3`

E. `4`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(An Euler circuit cannot exist if any vertices)`

`text(have an odd degree.)`

`=> B`

♦♦♦ Mean mark part 2: 7%!
MARKER’S COMMENT: A majority of students chose option A, not understanding that a graph with intersecting edges can still be planar.

`text(Part 2)`

`=> E`

NETWORKS, FUR2 2013 VCAA 3

The rangers at the wildlife park restrict access to the walking tracks through areas where the animals breed.

The edges on the directed network diagram below represent one-way tracks through the breeding areas. The direction of travel on each track is shown by an arrow. The numbers on the edges indicate the maximum number of people who are permitted to walk along each track each day.

Starting at `A`, how many people, in total, are permitted to walk to `D` each day? (1 mark)

One day, all the available walking tracks will be used by students on a school excursion.

The students will start at `A` and walk in four separate groups to `D`.

Students must remain in the same groups throughout the walk.

1. Group 1 will have 17 students. This is the maximum group size that can walk together from `A` to `D`.
  
  Write down the path that group 1 will take. (1 mark)
2. Groups 2, 3 and 4 will each take different paths from `A` to `D`.
  
  Complete the six missing entries shaded in the table below. (2 marks)

Show Answers Only

`37`
1. `A − B − E − C − D`
2. `text{One possible solution is:}`

Show Worked Solution

a.	`text(Maximum flow)`	`=\ text(minimum cut through)\ CD and ED`
		`= 24 + 13`
		`= 37`

♦ Mean mark of all parts (combined) was 41%.

`:.\ text(A maximum of 37 people can walk)`

`text(to)\ D\ text(from)\ A.`

b.i. `A − B − E − C − D`

b.ii. `text(One solution using the second possible largest)`

`text(group of 11 students and two groups from the)`

`text(remaining 9 students is:)`

NETWORKS, FUR2 2009 VCAA 4

A walkway is to be built across the lake.

Eleven activities must be completed for this building project.

The directed network below shows the activities and their completion times in weeks.

What is the earliest start time for activity `E`? (1 mark)
Write down the critical path for this project. (1 mark)
The project supervisor correctly writes down the float time for each activity that can be delayed and makes a list of these times.

Determine the longest float time, in weeks, on the supervisor’s list. (1 mark)

A twelfth activity, `L`, with duration three weeks, is to be added without altering the critical path.

Activity `L` has an earliest start time of four weeks and a latest start time of five weeks.

Draw in activity `L` on the network diagram above. (1 mark)
Activity `L` starts, but then takes four weeks longer than originally planned.

Determine the total overall time, in weeks, for the completion of this building project. (1 mark)

Show Answers Only

`7`
`BDFGIK`
`H\ text(or)\ J\ text(can be delayed for)`
`text(a maximum of 3 weeks.)`
`text(25 weeks)`

Show Worked Solution

a. `7\ text(weeks)`

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

b. `BDFGIK`

c. `H\ text(or)\ J\ text(can be delayed for a maximum)`

`text(of 3 weeks.)`

e. `text(The new critical path is)\ BLEGIK.`

`L\ text(now takes 7 weeks.)`

`:.\ text(Time for completion)`

`= 4 + 7 + 1 + 5 + 2 + 6`

`= 25\ text(weeks)`

NETWORKS, FUR2 2011 VCAA 4

Stormwater enters a network of pipes at either Dunlop North (Source 1) or Dunlop South (Source 2) and flows into the ocean at either Outlet 1 or Outlet 2.

On the network diagram below, the pipes are represented by straight lines with arrows that indicate the direction of the flow of water. Water cannot flow through a pipe in the opposite direction.

The numbers next to the arrows represent the maximum rate, in kilolitres per minute, at which stormwater can flow through each pipe.

Complete the following sentence for this network of pipes by writing either the number 1 or 2 in each box. (1 mark)

Determine the maximum rate, in kilolitres per minute, that water can flow from these pipes into the ocean at Outlet 1 and Outlet 2. (2 marks)

A length of pipe, show in bold on the network diagram below, has been damaged and will be replaced with a larger pipe.

The new pipe must enable the greatest possible rate of flow of stormwater into the ocean from Outlet 2.

What minimum rate of flow through the pipe, in kilolitres per minute, will achieve this? (1 mark)

Show Answers Only

`text(Storm water from Source 2 cannot reach Outlet 1)`
`text(Outlet 1: 700 kL/min)`
`text(Outlet 2: 700 kL/min)`
`text(300 kL per min)`

Show Worked Solution

a. `text(Storm water from Source 2 cannot reach Outlet 1)`

♦ Mean mark of all parts (combined) was 35%.

`text(The minimum cut includes the 200 kL/min pipe from Source 1.)`

`:.\ text(Maximum rates are)`

`text(Outlet 1: 700 kL/min)`

`text(Outlet 2: 700 kL/min)`

c. `text(The next smallest cut in the lower pipe system is 800.)`

`:.\ text(The minimum flow through the new pipe that will achieve)`

`text(this is 300 kL/min.)`

NETWORKS, FUR2 2014 VCAA 4

To restore a vintage train, 13 activities need to be completed.

The network below shows these 13 activities and their completion times in hours.

Determine the earliest starting time of activity `F`. (1 mark)

The minimum time in which all 13 activities can be completed is 21 hours.

What is the latest starting time of activity `L`? (1 mark)
What is the float time of activity `J`? (1 mark)

Just before they started restoring the train, the members of the club needed to add another activity, `X`, to the project.

Activity `X` will take seven hours to complete.

Activity `X` has no predecessors, but must be completed before activity `G` starts.

What is the latest starting time of activity `X` if it is not to increase the minimum completion time of the project? (1 mark)

Activity `A` can be crashed by up to four hours at an additional cost of $90 per our.

This may reduce the minimum completion time for the project, including activity `X`.

Determine the least cost of crashing activity `A` to give the greatest reduction in the minimum completion time of the project. (1 mark)

Show Answers Only

`text(7 hours)`
`text(18 hours)`
`text(2 hours)`
`text(4 hours)`
`$270`

Show Worked Solution

a. `5 + 2 = 7\ text(hours)`

♦ Mean mark for all parts (combined) was 42%.

b. `text(Latest starting time of)\ L`

`= text(Length of critical path – duration of)\ L`

`= 21 – 3`

`= 18\ text(hours)`

c. `text(Float time of)\ J`

`=\ text(LST – EST)`

`= 13 – 11`

`= 2\ text(hours)`

d. `X\ text(precedes)\ G`

`text(EST of)\ G = 11`

`:. text(LST of)\ X = 11`

`text(EST)\ text(of)\ X`

`= text(LST of)\ X – text(duration of)\ X`

`= 11 – 7`

`= 4\ text(hours)`

e. `text(Longer paths are)`

`A – C – G – K = 21\ text{hours (critical path)}`

`A – D – E – H – K = 20\ text(hours)`

`A – D – F – J – M = 19\ text(hours)`

`A – D – E – I – M = 18\ text(hours)`

`B – E – H – K = 18\ text(hours)`

`B – F – J – M = 17\ text(hours)`

`:.\ text(Reduce)\ \ A – C – G – K\ \ text(by 3 hours to get)`

`text{to 18 hours (equals}\ \ B – E – H – K)`

`:.\ text(Least cost)`	`= 3 xx 90`
	`= $270`

MATRICES, FUR2 2006 VCAA 3

Market researchers claim that the ideal number of bookshops (`x`), sports shoe shops (`y`) and music stores (`z`) for a shopping centre can be determined by solving the equations

`2x + y + z = 12`

`x-y+z=1`

`2y-z=6`

Write the equations in matrix form using the following template. (1 mark)

`qquad[(qquadqquadqquad),(),()][(quad),(quad),(quad)] = [(quad),(quad),(quad)]`
Do the equations have a unique solution? Provide an explanation to justify your response. (1 mark)
Write down an inverse matrix that can be used to solve these equations. (1 mark)
Solve the equations and hence write down the estimated ideal number of bookshops, sports shoe shops and music stores for a shopping centre. (1 mark)

Show Answers Only

`[(2,1,1),(1,-1,1),(0,2,-1)][(x),(y),(z)] = [(12),(1),(6)]`
`text(Yes. See worked solutions.)`
`[(2,1,1),(1,-1,1),(0,2,-1)]^(-1) = [(-1,3,2),(1,-2,-1),(2,-4,-3)]`
`text(3 bookshops, 4 sports shoe shops, 2 music stores.)`

Show Worked Solution

a.	`[(2,1,1),(1,-1,1),(0,2,-1)][(x),(y),(z)] = [(12),(1),(6)]`

♦ Mean mark 35% for all parts (combined).

b.	`text(det)\ [(2,1,1),(1,-1,1),(0,2,-1)] = 1 != 0`

`:.\ text(A unique solution exists.)`

c. `text(By CAS,)`

`[(2,1,1),(1,-1,1),(0,2,-1)]^(-1) = [(-1,3,2),(1,-2,-1),(2,-4,-3)]`

d.	`[(x),(y),(z)]`	`= [(-1,3,2),(1,-2,-1),(2,-4,-3)][(12),(1),(6)]`
		`= [(3),(4),(2)]`

`:.\ text(Estimated ideal numbers are:)`

`text(3 bookshops)`

`text(4 shoe shops)`

`text(2 music stores)`

MATRICES, FUR2 2007 VCAA 2

To study the life-and-death cycle of an insect population, a number of insect eggs (`E`), juvenile insects (`J`) and adult insects (`A`) are placed in a closed environment.

The initial state of this population can be described by the column matrix

`S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`

A row has been included in the state matrix to allow for insects and eggs that die (`D`).

What is the total number of insects in the population (including eggs) at the beginning of the study? (1 mark)

In this population

- eggs may die, or they may live and grow into juveniles
- juveniles may die, or they may live and grow into adults
- adults will live a period of time but they will eventually die.

In this population, the adult insects have been sterilised so that no new eggs are produced. In these circumstances, the life-and-death cycle of the insects can be modelled by the transition matrix

`{:(qquadqquadqquadqquadquadtext(this week)),((qquadqquadqquadE,quad\ J,quadA,\ D)),(T = [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)]{:(E),(J),(A),(D):}):}`

What proportion of eggs turn into juveniles each week? (1 mark)
1. Evaluate the matrix product `S_1 = TS_0` (1 mark)
2. Write down the number of live juveniles in the population after one week. (1 mark)
3. Determine the number of live juveniles in the population after four weeks. Write your answer correct to the nearest whole number. (1 mark)
4. After a number of weeks there will be no live eggs (less than one) left in the population.
  When does this first occur? (1 mark)
5. Write down the exact steady-state matrix for this population. (1 mark)
If the study is repeated with unsterilised adult insects, eggs will be laid and potentially grow into adults.

Assuming 30% of adults lay eggs each week, the population matrix after one week, `S_1`, is now given by

`qquad S_1 = TS_0 + BS_0`

where `B = [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]` and `S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
1. Determine `S_1` (1 mark)
  
  This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
  
  `qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
2. Determine the number of live eggs in this insect population after two weeks. (1 mark)

Show Answers Only

`700`
`50text(%)`

1. `[(160),(280),(180),(80)]{:(E),(J),(A),(D):}`
2. `280`
3. `56`
4. `text(7th week)`
5. `[(0),(0),(0),(700)]`
1. `[(190),(280),(180),(80)]`
2. `130`

Show Worked Solution

a. `400 + 200 + 100 + 0 = 700`

b. `50text(%)`

c.i.	`S_1`	` = TS_0`
		`= [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)][(400),(200),(100),(0)]`
		`= [(160),(280),(180),(80)]{:(E),(J),(A),(D):}`

c.ii. `280`

c.iii.	`S_4`	` = T^4S_0`
		`= [(10.24),(56.32),(312.96),(320.48)]{:(E),(J),(A),(D):}\ \ \ text{(by graphics calculator)}`

`:. 56\ text(juveniles still alive after 4 weeks.)`

c.iv.	`text(Each week, only 40% of eggs remain.)`
	`text(Find)\ \ n\ \ text(such that)`

`400 xx 0.4^n`	`< 1`
`0.4^n`	`<1/400`
`n`	`> 6.5`

`:.\ text(After 7 weeks, no live eggs remain.)`

c.v. `text(Consider)\ \ n\ \ text{large (say}\ \ n = 100 text{)},`

`[(0.4, 0, 0, 0), (0.5, 0.4, 0, 0), (0, 0.5, 0.8, 0), (0.1, 0.1, 0.2, 1)]^100 [(400), (200), (100), (0)] ~~ [(0), (0), (0), (700)]`

d.i.	`S_1`	`= TS_0 + BS_0`
		`= [(160),(280),(180),(80)] + [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(400),(200),(100),(0)]`
		`= [(190),(280),(180),(80)]`

♦♦ Mean mark for part (d) was 30%.

d.ii.	`S_2`	`= TS_1 + BS_1`
		`= [(130), (207), (284), (163)]`

`:.\ text(There are 130 live egss after 2 weeks.)`

MATRICES, FUR2 2011 VCAA 3

A breeding program is started in the wetlands. It is aimed at establishing a colony of native ducks.

The matrix `W_0` displays the number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) that are introduced to the wetlands at the start of the breeding program.

`W_0 = [(32),(64)]{:(J),(A):}`

In total, how many female ducks are introduced to the wetlands at the start of the breeding program? (1 mark)

The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year 1 of the breeding program is determined using the matrix equation

`W_1 = BW_0`

In this equation, `B` is the breeding matrix

`{:((qquadqquadqquad\ J,qquadA)),(B = [(0,2),(0.25,0.5)]{:(J),(A):}):}`

Determine `W_1` (1 mark)

The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year `n` of the breeding program is determined using the matrix equation

`W_n = BW_(n - 1)`

The graph below is incomplete because the points for the end of Year 3 of the breeding program are missing.

1. Use the matrices to calculate the number of juvenile and the number of adult female ducks expected in the colony at the end of Year 3 of the breeding program.
  
  Plot the corresponding points on the graph. (2 marks)
2. Use matrices to determine the expected total number of female ducks in the colony in the long term.
  
  Write your answer correct to the nearest whole number. (1 mark)

The breeding matrix `B` assumes that, on average, each adult female duck lays and hatches two female eggs for each year of the breeding program.

If each adult female duck lays and hatches only one female egg each year, it is expected that the duck colony in the wetland will not be self-sustaining and will, in the long run, die out.

The matrix equation

`W_n = PW_(n - 1)`

with a different breeding matrix

`{:((qquadqquadqquad\ J,qquadA)),(P = [(0,1),(0.25,0.5)]{:(J),(A):}):}`

and the initial state matrix

`W_0 = [(32),(64)]{:(J),(A):}`

models this situation.

During which year of the breeding program will the number of female ducks in the colony halve? (1 mark)

Changing the number of juvenile and adult female ducks at the start of the breeding program will also change the expected size of the colony.

Assuming the same breeding matrix, `P`, determine the number of juvenile ducks and the number of adult ducks that should be introduced into the program at the beginning so that, at the end of Year 2, there are 100 juvenile female ducks and 50 adult female ducks. (2 marks)

Show Answers Only

`96`
`W_1 = [(128),(40)]`
2. `144`
`text(year 5)`
`400\ text(juvenile ducks and 0 adult)`
`text(ducks should be introduced.)`

Show Worked Solution

a. `text(Total female ducks introduced)`

`= 32 + 64`

`= 96`

b.	`W_1`	`= BW_0`
		`= [(0,2),(0.25,0.5)][(32),(64)]`
		`= [(128),(40)]`

c.i.

`W_n = BW_(n – 1)`

`W_2 = [(0,2),(0.25,0.5)][(128),(40)] = [(80),(52)]`

`W_3 = [(0,2),(0.25,0.5)][(80),(52)] = [(104),(46)]`

`:.\ text{Plot (3, 104) and (3, 46)}`

c.ii. `text(Consider)\ n = text(50 and 51,)`

`W_50 = [(0,2),(0.25,0.5)]^49[(32),(64)] = [(96),(48)]`

`W_51 = [(0,2),(0.25,0.5)]^50[(32),(64)] = [(96),(48)]`

`:.\ text(Total female ducks in the long term)`

`= 96 + 48`

`= 144`

d. `text(Initial female ducks = 96)`

♦♦♦ Mean mark of parts (d)-(e) was 20%.

`W_n = PW_(n – 1)`

`W_4 = [(0,1),(0.25,0.5)]^3[(32),(64)] = [(28),(33)]`

`:. 51\ text(ducks at end of year 4.)`

`W_5 = [(0,1),(0.25,0.5)]^4[(32),(64)] = [(23),(18.5)]`

`:. 41.5\ text(ducks at end of year 5.)`

`:.\ text{Numbers halve (drop below 48) in year 5.}`

e.	`text(Let )`	`a = text(initial juvenile ducks)`
		`b = text(initial adult ducks)`

`text(Find)\ a\ text(and)\ b\ text(such that,)`

`[(0,1),(0.25,0.5)]^2[(a),(b)]= [(100),(50)]`

`[(0.25,0.5),(0.125,0.5)][(a),(b)] = [(100),(50)]`

`:. [(a),(b)]`	`= [(0.25,0.5),(0.125,0.5)]^(-1)[(100),(50)]`
	`= [(400),(0)]`

`:. 400\ text(juvenile ducks and 0 adult ducks)`

`text(should be introduced.)`

MATRICES, FUR2 2013 VCAA 2

10 000 trout eggs, 1000 baby trout and 800 adult trout are placed in a pond to establish a trout population.

In establishing this population

- eggs (`E`) may die (`D`) or they may live and eventually become baby trout (`B`)
- baby trout (`B`) may die (`D`) or they may live and eventually become adult trout (`A`)
- adult trout (`A`) may die (`D`) or they may live for a period of time but will eventually die.

From year to year, this situation can be represented by the transition matrix `T`, where

`{:(qquadqquadqquadqquadqquadtext(this year)),((qquadqquadqquadE,quad\ B,quad\ A,\ D)),(T = [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]):}{:(),(),(E),(B),(A),(D):}{:(),(),(qquadtext(next year)):}`

Use the information in the transition matrix `T` to
1. determine the number of eggs in this population that die in the first year (1 mark)
2. complete the transition diagram below, showing the relevant percentages. (2 marks)

The initial state matrix for this trout population, `S_0`, can be written as

`S_0 = [(10\ 000),(1000),(800),(0)]{:(E),(B),(A),(D):}`

Let `S_n` represent the state matrix describing the trout population after `n` years.

Using the rule `S_n = T S_(n – 1)`, determine each of the following.
1. `S_1` (1 mark)
2. the number of adult trout predicted to be in the population after four years (1 mark)
The transition matrix `T` predicts that, in the long term, all of the eggs, baby trout and adult trout will die.
1. How many years will it take for all of the adult trout to die (that is, when the number of adult trout in the population is first predicted to be less than one)? (1 mark)
2. What is the largest number of adult trout that is predicted to be in the pond in any one year? (1 mark)
Determine the number of eggs, baby trout and adult trout that, if added to or removed from the pond at the end of each year, will ensure that the number of eggs, baby trout and adult trout in the population remains constant from year to year. (2 marks)

The rule `S_n = T S_(n – 1)` that was used to describe the development of the trout in this pond does not take into account new eggs added to the population when the adult trout begin to breed.

To take breeding into account, assume that 50% of the adult trout lay 500 eggs each year.

The matrix describing the population after one year, `S_1`, is now given by the new rule

`S_1 = T S_0 + 500\ M\ S_0`

where `T=[(0,0,0,0),(0.40,0,0,0),(0,0.25,0.50,0),(0.60,0.75,0.50,1.0)], M=[(0,0,0.50,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]\ text(and)\ S_0=[(10\ 000),(1000),(800),(0)]`
1. Use this new rule to determine `S_1`. (1 mark)

This pattern continues so that the matrix describing the population after `n` years, `S_n`, is given by the rule

`S_n = T\ S_(n – 1) + 500\ M\ S_(n – 1)`

Use this rule to determine the number of eggs in the population after two years (2 marks)

Show Answers Only

1. `6000`
2. `text(See Worked Solutions)`
1. `text(See Worked Solutions)`
2. `text(See Worked Solutions)`
1. `text(13 years)`
2. `text(Largest number of adult trout is 1325.)`
`text(Add 10 000 eggs, remove 3000 baby trout and add 150)`

`text(150 adult trout to keep the population constant.)`
1. `text(See Worked Solutions)`
2. `text(See Worked Solutions)`

Show Worked Solution

a.i. `text(60% of eggs die in 1st year,)`

`:.\ text(Eggs that die in year 1)`

`= 0.60 xx 10\ 000`

`= 6000`

MARKER’S COMMENT: A 100% cycle drawn at `D` was a common omission. Do not draw loops and edges of 0%!

a.ii.

b.i.	`S_1`	`= TS_0`
		`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(10\ 000),(1000),(800),(0)]`
		`= [(0),(4000),(650),(7150)]`

b.ii.	`S_4`	`= T^4S_0`
		`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]^4[(10\ 000),(1000),(800),(0)]`
		`= [(0),(0),(331.25),(11\ 468.75)]`

`:. 331\ text(trout is the predicted population)`

`text(after 4 years.)`

c.i.

`S_12 = T^12S_0 = [(0),(0),(1.29),(11\ 791)]`

`S_13 = T^13S_0 = [(0),(0),(0.65),(11\ 799)]`

`:.\ text{It will take 13 years (when the trout}`

`text{population drops below 1).}`

c.ii.

`S_1 = TS_0 = [(0),(4000),(650),(7150)]`

`text(After 1 year, 650 adult trout.)`

`text(Similarly,)`

`S_2 = T^2S_0 = [(0),(0),(1325),(10\ 475)]`

`S_3 = T^3S_0 = [(0),(0),(662.5),(11\ 137.5)]`

`S_4 = T^4S_0 = [(0),(0),(331),(11\ 469)]`

`:.\ text(Largest number of adult trout = 1325.)`

d.	`S_0 – S_1 = [(10\ 000),(1000),(800),(0)] – [(0),(4000),(650),(7150)] = [(10\ 000),(−3000),(150),(−7150)]`

`:.\ text(Add 10 000 eggs, remove 3000 baby trout)`

`text(and add 150 adult trout to keep the)`

`text(population constant.)`

e.i.	`S_1`	`= TS_0 + 500MS_0`
		`= [(0),(4000),(650),(7150)] + 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(10\ 000),(1000),(800),(0)]`
		`= [(0),(4000),(650),(7150)] + 500[(400),(0),(0),(0)]`
		`= [(200\ 000),(4000),(650),(7150)]`

e.ii.

`S_2`

`= TS_1 + 500MS_1`

`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(200\ 000),(4000),(650),(7150)]`

`+ 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(200\ 000),(4000),(650),(7150)]`

`= [(162\ 500),(80\ 000),(1325),(130\ 475)]`

MATRICES, FUR1 2008 VCAA 7-9 MC

A large population of mutton birds migrates each year to a remote island to nest and breed. There are four nesting sites on the island, A, B, C and D.

Researchers suggest that the following transition matrix can be used to predict the number of mutton birds nesting at each of the four sites in subsequent years. An equivalent transition diagram is also given.

`{:(qquad qquad qquad qquad {:text(this year):}), (qquad qquad quad quad \ {:(A,\ \ B,\ \ C,\ D):}), (T = [(0.4, 0, 0.2, 0),(0.35, 1, 0.15, 0), (0.15, 0, 0.55, 0), (0.1, 0, 0.1, 1)] {:(A), (B), (C), (D):} quad {:text(next year):}):}`

Part 1

Two thousand eight hundred mutton birds nest at site C in 2008.

Of these 2800 mutton birds, the number that nest at site A in 2009 is predicted to be

A. `560`

B. `980`

C. `1680`

D. `2800`

E. `3360`

Part 2

This transition matrix predicts that, in the long term, the mutton birds will

A. nest only at site A.

B. nest only at site B.

C. nest only at site A and C.

D. nest only at site B and D.

E. continue to nest at all four sites.

Part 3

Six thousand mutton birds nest at site B in 2008.

Assume that an equal number of mutton birds nested at each of the four sites in 2007. The same transition matrix applies.

The total number of mutton birds that nested on the island in 2007 was

A. `6000`

B. `8000`

C. `12\ 000`

D. `16\ 000`

E. `24\ 000`

Show Answers Only

`text(Part 1:) A`

`text(Part 2:) D`

`text(Part 3:) D`

Show Worked Solution

`text(Part 1:)`

`text(20% of birds at site)\ A\ text(in 2008 are predicted)`

`text(to move to site)\ C.`

`:.\ text(Number of birds)`

`= 20 text(%) xx 2800`

`= 560`

`=> A`

`text(Part 2:)`

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

`T^50 = [(0, 0, 0, 0), (0.75, 1, 0.66, 0), (0, 0, 0, 0), (0.25, 0, 0.33, 0)]`

`=> D`

`text(Part 3:)`

`text(Let)\ \ x = text(mutton birds at each site in 2007)`

`text(In 2008, 6000 birds nest at)\ B.`

♦♦♦ Mean mark 25%.

`text(Using the diagram,)`

`x + 0.35x + 0.15x`	`= 6000`
`1.5x`	`= 6000`
`x`	`= 4000`

`:.\ text(Total number nested in 2007)`

`= 4 xx 4000`

`= 16\ 000\ text(birds)`

`=> D`

MATRICES, FUR1 2010 VCAA 9 MC

Robbie completed a test of four multiple-choice questions.

Each question had four alternatives, A, B, C or D.

Robbie randomly guessed the answer to the first question.

He then determined his answers to the remaining three questions by following the transition matrix

`{:(qquad qquad qquad {:text(this question):}), (qquad qquad quad \ {:(A, B, C, D):}), (T = [(1,\ 0,\ 0,\ 0), (0,\ 0,\ 1,\ 0), (0,\ 0,\ 0,\ 1), (0,\ 1,\ 0,\ 0)] {:(A), (B), (C), (D):} quad {:text(next question):}):}`

Which of the following statements is true?

A. It is impossible for Robbie to give the same answer to all four questions.

B. Robbie would always give the same answer to the first and fourth questions

C. Robbie would always give the same answer to the second and third questions.

D. If Robbie answered A for question one, he would have answered B for question two

E. It is possible that Robbie gave the same answer to exactly three of the four questions.

Show Answers Only

`B`

Show Worked Solution

`text(Consider when the first guess is)\ \ C,`

♦♦ Mean mark 29%.

`text(i.e. the state matrix is)\ \ [(0), (0), (1), (0)]`

`text(Using the transition matrix,)`

`[(1, 0, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 1, 0, 0)] [(0), (0), (1), (0)] = [(0), (1), (0), (0)]`

`:.\ text(The 2nd answer is)\ \ B`

`[(1, 0, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 1, 0, 0)] [(0), (1), (0), (0)] = [(0), (0), (0), (1)]`

`:.\ text(The 3rd answer is)\ \ D`

`[(1, 0, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 1, 0, 0)] [(0), (0), (0), (1)] = [(0), (0), (1), (0)]`

`:.\ text(The 4th answer is)\ \ C`

`text(Other guesses for the first answer can)`

`text(be similarly followed to show)\ B\ text(is true.)`

`=> B`

MATRICES, FUR1 2010 VCAA 6 MC

Vince, Nev and Rani all service office equipment.

The matrix `T` shows the time that it takes (in minutes) for each of Vince (V), Nev (N) and Rani (R) to service a photocopier (P) a fax machine (F) and a scanner (S).

`{:(qquad qquad qquad {:(V,\ N,\ R):}), (T = [(12, 15, 14),(8, 7, 8), (20, 19, 17)] {:(P), (F), (S):}):}`

The matrix `U` below displays the number of photocopiers, fax machines and scanners to be serviced in three schools, Alton (A), Borton (B) and Carlon (C).

`{:(qquad qquad qquad {:(P, F, S):}), (U = [(5,\ 3,\ 2),(4,\ 4,\ 3), (6,\ 1,\ 2)] {:(A), (B), (C):}):}`

A matrix that displays the time that it would take each of Vince, Nev and Rani, working alone, to service the photocopiers, fax machines and scanners in each of the three schools is

A.	`[(17, 18, 16), (12, 11, 11), (26, 20, 19)]`	B.	`[(204, 110, 97), (116, 60, 53), (278, 153, 131)]`

C.	`[(124, 134, 128), (140, 145, 139), (120, 135, 126)]`	D.	`[(7, 12, 12), (4, 3, 5), (14, 18, 15)]`

E.	`[(60, 15, 28), (32, 35, 24), (120, 19, 34)]`

Show Answers Only

`C`

Show Worked Solution

`text(Matrix)\ \ UT\ \ text(gives the information.)`

♦♦♦ Mean mark 19%.
MARKER’S COMMENT: Most students incorrectly calculated matrix `TU`. Make sure you know why this is incorrect!

`UT`	`= [(5, 3, 2), (4, 4, 3), (6, 1, 2)] [(12, 15, 14), (8, 7, 8), (20, 19, 17)]`
	`= [(124, 134, 128), (140, 145, 139), (120, 135, 126)]`

`=> C`

MATRICES, FUR1 2010 VCAA 3 MC

The total cost of one ice cream and three soft drinks at Catherine’s shop is $9.

The total cost of two ice creams and five soft drinks is $16.

Let `x` be the cost of an ice cream and `y` be the cost of a soft drink

The matrix `[(x), (y)]` is equal to

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

B. `[(1, 3), (2, 5)] [(9), (16)]`

C. `[(1, 2), (3, 5)] [(9), (16)]`

D. `[(– 5, 2), (3, – 1)] [(9), (16)]`

E. `[(– 5, 3), (2, – 1)] [(9), (16)]`

Show Answers Only

`E`

Show Worked Solution

`[(1, 3), (2, 5)][(x), (y)] = [(9), (16)]`

♦♦ Mean mark 35%.

`:. [(x), (y)]`	`= [(1, 3), (2, 5)]^-1 [(9), (16)]`
	`= [(– 5, 3), (2, – 1)] [(9), (16)]`

`=> E`

MATRICES, FUR1 2013 VCAA 9 MC

`P, Q, R` and `S` are matrices such that the matrix product `P = QRS` is defined.

Matrix `Q` and matrix `S` are square, non-zero matrices for which `Q + S` is not defined.

Which one of the following matrix expressions is defined?

A. `R - S`

B. `Q + R`

C. `P^2`

D. `R^-1`

E. `P × S`

Show Answers Only

`E`

Show Worked Solution

`text(If)\ Q(n xx n)\ text(and)\ S(m xx m)\ text(are square)`

♦ Mean mark 39%.

`text(matrices and)\ Q + S\ text(is not defined,)`

`m != n.`

`text(S)text(ince)\ QRS\ text(is defined, we can deduce)`

`text(that)\ R\ text(is)\ n xx m,\ text(and)\ P\ text(is)\ n xx m.`

`text(Consider)\ E,`

`P xx S\ text(is defined because the number)`

`text(of columns in)\ P(m)\ text(equals the number)`

`text(of rows in)\ S(m).`

`text(All other options can be shown to not)`

`text(be defined.)`

`rArr E`

MATRICES, FUR1 2006 VCAA 6 MC

If `A = [(1,3), (6,4), (0,0)]` and the matrix product `XA = [(4,1),(1,4), (3,5)],` then the order matrix `X` is

A. `(2 xx 2)`

B. `(2 xx 3)`

C. `(3 xx 1)`

D. `(3 xx 2)`

E. `(3 xx 3)`

Show Answers Only

`E`

Show Worked Solution

`X`	`xx`	`A`	`=`	`XA`
`a xx b`		`3 xx 2`		`3 xx 2`

♦♦♦ Mean mark 23%.

`b = 3\ \ text{(columns in}\ X = text(rows in)\ A)`

`a = 3\ \ text{(rows in}\ X = text(rows in)\ XA)`

`:. XA\ text(is order)\ 3 xx 3.`

`rArr E`

MATRICES, FUR1 2011 VCAA 9 MC

Matrix `A` is a 3 x 3 matrix. Seven of the elements in matrix `A` are zero.

Matrix `B` contains six elements, none of which are zero.

Assuming the matrix product `AB` is defined, the minimum number of zero elements in the product matrix `AB` is

A. `0`

B. `1`

C. `2`

D. `4`

E. `6`

Show Answers Only

`C`

Show Worked Solution

`text(Matrix)\ B\ text(is)\ 3 xx 2`

`text(If non-zero elements of matrix)\ A`

`text(are in the same row,)`

`[(a,0,b),(0,0,0),(0,0,0)][(c,d),(e,f),(g,h)] = [(x,y),(0,0),(0,0)]`

`text(where)\ a,b,c …, x, y != 0`

`text(If non-zero elements of matrix)\ B`

`text(are not in the same row,)`

`[(a,0,0),(b,0,0),(0,0,0)][(c,d),(e,f),(g,h)] = [(ac,ad),(bc,bd),(0,0)]`

`:.\ text(Minimum number of zero elements)`

`text(in matrix)\ AB\ text(is 2.)`

`=> 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.)`

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`

CORE*, FUR2 2013 VCAA 4

Hugo took out a reducing balance loan of $25 000 to compete in road races overseas.

Interest was charged at a rate of 12% per annum compounding quarterly.

His loan is to be repaid fully in four years with equal quarterly payments.

After two years, how much of the $25 000 will Hugo have repaid?

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

Show Answers Only

`$11\ 029\ \ text{(nearest dollar)}`

Show Worked Solution

`text(Find quarterly payment by TVM Solver,)`

♦♦ Mean mark 25%.
MARKER’S COMMENT: Very few students showed any TVM input or other working in this question. Labelling steps to organise thoughts is a feature of the most successful answers.

`N`	`= 4 xx 4 = 16`
`I(text(%))`	`= 12`
`PV`	`= 25\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= 12`
`text(C/Y)`	`= 12`

`:. text(Quarterly payment) = $1990.2712…`

`text(Find principal remaining after 2 years)`

`N`	`= 8`
`Itext(%)`	`= 12`
`PV`	`= 25\ 000`
`PMT`	`= 1990.2712…`
`FV`	`=?`
`text(P/Y)`	`= 4`
`text(C/Y)`	`= 4`

`FV= −13\ 971.09…`

`:.\ text(Amount repaid)`

`=\ text(Loan − principal left)`

`= 25\ 000 – 13\ 971.09…`

`= 11\ 028.90…`

`= $11\ 029\ \ text{(nearest dollar)}`

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)}`

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`

GEOMETRY, FUR2 2013 VCAA 5

Daniel threw a javelin a distance of 68.32 m on a bearing of 057° on his first throw.

On his second throw from the same point, he threw the javelin a distance of 72.51 m.

The second throw landed at a point on a bearing of 125°, measured from the point where the first throw landed.

Determine the distance between the point where Daniel’s first throw landed and the point where his second throw landed.

Write your answer in metres, correct to one decimal place. (2 marks)

Show Answers Only

`9.7\ text{m (1 d.p.)}`

Show Worked Solution

`text(Using the sine rule,)`

`(sin∠ABT)/68.32`	`= (sin∠TAB)/72.51`
`sin∠ABT`	`= (68.32 xx sin112^@)/72.51`
	`= 0.873…`
`:. ∠ABT`	`= 60.88…^@`
`:. ∠ATB`	`= 180 – (112 + 60.88…)`
	`= 7.11…^@`

`text(Using the cosine rule,)`

`AB^2`	`= 68.32^2 + 72.51^2 – 2 xx 68.32 xx 72.51 xx cos7.11…^@`
	`= 93.74…`
`:. AB`	`= 9.68…`
	`= 9.7\ text{m (1 d.p.)}`

GEOMETRY, FUR2 2014 VCAA 4

A chicken escapes into a neighbouring field through an open gate.

The chicken’s owner is 50 m due north of the gate, searching for the chicken.

The chicken is 40 m from the gate on a bearing of 295°.

What is the bearing of the chicken from its owner?

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

Show Answers Only

`228^@`

Show Worked Solution

`text(Using the cosine rule,)`

♦♦ Mean mark 28%.

`OC`	`= sqrt(40^2 + 50^2 − 2 xx 40 xx 50 xx cos65^@)`
	`=49.08…\ text(m)`

`text(Using the sine rule,)`

`(sin theta)/40`	`= (sin65^@)/49`
`sin theta`	`=(40 xx sin65^@)/49`
	`=0.739…`
`:.theta`	`=47.7…^@`
	`=48^@\ \ text{(nearest degree)`

`:.\ text(The bearing of)\ C\ text(from)\ O`

`=180^@ + 48^@`

`= 228^@`

GEOMETRY, FUR2 2013 VCAA 4

Competitors in the intermediate division of the discus use a smaller discus than the one used in the senior division, but of a similar shape. The total surface area of each discus is given below.

By what value can the volume of the intermediate discus be multiplied to give the volume of the senior discus? (2 marks)

Show Answers Only

`1.728`

Show Worked Solution

`k^2 = 720/500 = 36/25`

♦♦ Mean mark well below 50% (exact data unavailable.
MARKER’S COMMENT: A common error was to use a ratio where `k<1` which was obviously wrong since the volume needed to be scaled up.

`:. k`	`= sqrt(36/25) = 1.2`
`:. k^3`	`= 1.2^3 = 1.728`

`:.\ text(Volume of intermediate discuss can)`

`text(be multiplied by 1.728)`

CORE*, FUR2 2014 VCAA 4

The cricket club borrowed $400 000 to build a clubhouse.

Interest is calculated at the rate of 4.5% per annum, compounding monthly.

The cricket club will make monthly repayments of $2500.

After a number of monthly repayments, the balance of the loan will be reduced to $143 585.33.

What percentage of the next monthly repayment will reduce the balance of the loan?

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

Show Answers Only

`text(78%)`

Show Worked Solution

`text(Loan balance) = $143\ 585.33`

♦♦♦ Mean mark 10%.

`text(Interest in the next repayment)`

`= 4.5/(12 xx 100) xx 143\ 585.33`

`= 538.44…`

`:.\ text(Principal amount of next repayment)`

`= 2500 – 538.44`

`= 1.961.55…`

`:.\ text(% of repayment reducing loan)`

`= (1961.55…)/2500 xx 100text(%)`

`= 78.46…`

`= 78text{% (nearest %)}`

GRAPHS, FUR2 2014 VCAA 3

A shop owner bought 100 kg of Arthur’s tomatoes to sell in her shop.

She bought the tomatoes for $3.50 per kilogram.

The shop owner will offer a discount to her customers based on the number of kilograms of tomatoes they buy in one bag.

The revenue, in dollars, that the shop owner receives from selling the tomatoes is given by

`text(revenue)\ {{:(qquadqquad5.4n),(10.8 + 4(n - 2)),(a + 2(n - 10)):}qquadqquad{:(0 < n <= 2),(2 < n <= 10),(10 < n < 100):}`

where `n` is the number of kilograms of tomatoes that a customer buys in one bag.

What is the revenue that the shop owner receives from selling 8 kg of tomatoes in one bag? (1 mark)
Show that `a` has the value 42.8 in the revenue equation above. (1 mark)
Find the maximum number of kilograms of tomatoes that a customer can buy in one bag, so that the shop owner never makes a loss. (2 marks)

Show Answers Only

`$34.8`
`text(See Worked Solutions)`
`15.2`

Show Worked Solution

a. `text(Revenue from selling 8 kg)`

♦♦ Mean mark of Q3 (all parts) was 25%.

`= 10.8 + 4(8 – 2)`

`=$34.80`

b. `text(When)\ n =10,`

`10.8 + 4(n – 2)`	`= a + 2(n – 10)`
`:. a`	`= 10.8 + 4(10 – 2)`
	`= 42.8`

c. `text(A loss occurs when costs > revenue.)`

`3.5n`	`>42.8+2(n-10)`
`3.5n`	`>42.8 +2n -20`
`1.5n`	`>22.8`
`n`	`>15.2`

`:.\ text(The maximum number of kg without)`

`text(loss is 15.2.)`

CORE*, FUR2 2015 VCAA 5

Jane and Michael borrow $50 000 to expand their business.

Interest on the unpaid balance is charged to the loan account monthly.

The $50 000 is to be fully repaid in equal monthly repayments of $485.60 for 12 years.

Determine the annual compounding rate of interest.

Write your answer correct to two decimal places. (1 mark)
Calculate the amount that will be paid off the principal at the end of the first year.

Write your answer correct to the nearest dollar. (1 mark)
Halfway through the term of the loan, at the end of the sixth year, Jane and Michael make an additional one-off payment of $3500.

Assume no other changes are made to their loan conditions.

Determine how much time Jane and Michael will save in repaying their loan.

Give your answer correct to the nearest number of months. (2 marks)

Show Answers Only

`text(5.91%)`
`$2951`
`10\ text(months)`

Show Worked Solution

a. `text(By TVM Solver, after 12 years,)`

`N`	`=12 xx 12 = 144`
`I (%)`	`=?`
`PV`	`= 50\ 000`
`PMT`	`= – 485.60`
`FV`	`=0`
`text(P/Y)`	`=text(C/Y) = 12`

`=> I=5.9100…`

`:.\ text(The annual compounding interest rate = 5.91%)`

b. `text(By TVM Solver,)`

MARKER’S COMMENT: Showing solver “input” allowed a possible error mark in part (b) even if part (a) was incorrect.

`N`	`=12`
`I (%)`	`=5.91`
`PV`	`= 50\ 000`
`PMT`	`= – 485.60`
`FV`	`=?`
`text(P/Y)`	`=12`
`text(C/Y)`	`=12`

`=> FV= – 47\ 048.7077…`

`:.\ text(Paid off)`	`= 50\ 000 – 47\ 048.7077…`
	`= $2951.29…`
	`=$2951\ \ text{(nearest dollar)}`

c. `text(By TVM Solver, after 6 years,)`

`N`	`=12 xx 6=72`
`I (%)`	`=5.91`
`PV`	`= 50\ 000`
`PMT`	`= – 485.60`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV= – 29\ 376.045…`

`:.\ text(New)\ PV = 29\ 376.05 – 3500 = 25\ 086.05`

`text(Find)\ \ N\ \ text(such that)\ \ FV=0,`

`N`	`=?`
`I (%)`	`=5.91`
`PV`	`= 25\ 086.05`
`PMT`	`= – 485.60`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> N=61.96…`

`:.\ text(Time saving)`	`= 72 – 61.96…`
	`= 10.03…`
	`=10\ text(months)`

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`

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)`

GRAPHS, FUR2 2006 VCAA 3

Harry offers dog washing and dog clipping services.

Let `x` be the number of dogs washed in one day

`y` be the number of dogs clipped in one day.

It takes 20 minutes to wash a dog and 25 minutes to clip a dog.

There are 200 minutes available each day to wash and clip dogs.

This information can be written as Inequalities 1 to 3.

Inequality 1: `x ≥ 0`

Inequality 2: `y ≥ 0`

Inequality 3: `20x + 25y ≤ 200`

Draw the line that represents `20x + 25y = 200` on the graph below. (1 mark)

In any one day the number of dogs clipped is at least twice the number of dogs washed.

Write Inequality 4 to describe this information in terms of `x` and `y`. (1 mark)
1. On the graph on page 18 draw and clearly indicate the boundaries of the region represented by Inequalities 1 to 4. (2 marks)
2. On a day when exactly five dogs are clipped, what is the maximum number of dogs that could be washed? (1 mark)

The profit from washing one dog is $40 and the profit from clipping one dog is $30.

Let `P` be the total profit obtained in one day from washing and clipping dogs.

Write an equation for the total profit, `P`, in terms of `x` and `y`. (1 mark)
1. Determine the number of dogs that should be washed and the number of dogs that should be clipped in one day in order to maximise the total profit. (1 mark)
2. What is the maximum total profit that can be obtained from washing and clipping dogs in one day? (1 mark)

Show Answers Only

`y >= 2x`
2. `2`
`P = 40x + 30y`
1. `text(2 washes and 6 clips.)`
2. `$260`

Show Worked Solution

b. `text(Inequality 4:)\ \ y >= 2x`

c.i.

c.ii. `text(When)\ y =5,\ text(the maximum value of)\ x`

♦♦ Mean mark of parts (c)-(e) (combined) was 27%.

`text{(whole number) in the feasible region is 2.)`

`:. 2\ text(dogs can be washed.)`

d. `P = 40x + 30y`

e.i. `text(Test points for maximum profit)`

`text(At)\ (0, 8),`

`text(Profit) = 40 xx 0 + 30 xx 8 = $240`

`text(At)\ (1, 7),`

`text(Profit) = 40 xx 1 + 30 xx 7 = $250`

`text(At)\ (2, 6),`

`text(Profit) = 40 xx 2 + 30 xx 6 = $260`

`:.\ text(2 washes and 6 clips produce maximum profit.)`

e.ii. `text(Maximum profit)`

`= 40 xx 2 + 30 xx 6`

`= $260`

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 4

Tessa has a task that involves removing the top 24 cm of the height of her right regular pyramid below.

The shape remaining is shown in Figure 5 below. The top surface, `JKLM`, is parallel to the base, `ABCD`.

What fraction of the height of the pyramid has Tessa removed to produce Figure 5? (1 mark)
What fraction of the volume of the pyramid remains in Figure 5? (2 marks)

Show Answers Only

`3/4`
`37/64`

Show Worked Solution

a. `text(Fraction of the height removed)`

`= 24/32`

`= 3/4`

b. `text(Linear scale factor)\ =3/4`

`text(Volume scale factor) = (3/4)^3 = 27/64`

♦♦ Mean mark of both parts (combined) was 18%.

`:.\ text(Fraction of pyramid remaining)`

`= 1 – 27/64`

`= 37/64`

MATRICES, FUR2 2008 VCAA 4

By the end of each academic year, students at the university will have either passed, failed or deferred the year. Experience has shown that

88% of students who pass this year will also pass next year
10% of students who pass this year will fail next year
2% of students who pass this year will defer next year

52% of students who fail this year will pass next year
44% of students who fail this year will fail next year
4% of students who fail this year will defer next year

65% of students who defer this year will pass next year
10% of students who defer this year will fail next year
25% of students who defer this year will defer next year.

Twelve hundred and thirty students began a business degree in 2007.

By the end of the 2007 academic year, 880 students had passed, 230 had failed, while 120 had deferred the year.

No students have dropped out of the business degree permanently.

Use this information to predict the number of business students who will defer the 2009 academic year. (2 marks)

Show Answers Only

`42`

Show Worked Solution

`T = [(0.88,0.52,0.65),(0.10,0.44,0.10),(0.02,0.04,0.25)]`

♦♦ Exact data unavailable although “few” students were able to write down the correct transition matrix.

`N = [(880),(230),(120)]`

`S_2009`	`= T^2 N`
	`= [(996.9), (191.4), (41.7)]`

`:.\ text(42 students are predicted to defer in 2009.)`

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.)`