Band 5 – Page 47

MATRICES, FUR1 2011 VCAA 5-6 MC

Two politicians, Rob and Anna, are the only candidates for a forthcoming election. At the beginning of the election campaign, people were asked for whom they planned to vote. The numbers were as follows.

During the election campaign, it is expected that people may change the candidate that they plan to vote for each week according to the following transition diagram.

Part 1

The total number of people who are expected to change the candidate that they plan to vote for one week after the election campaign begins is

A. `828`

B. `1423`

C. `2251`

D. `4269`

E. `6891`

Part 2

The election campaign will run for ten weeks.

If people continue to follow this pattern of changing the candidate they plan to vote for, the expected winner after ten weeks will be

A. Rob by about 50 votes.

B. Rob by about 100 votes.

C. Rob by fewer than 10 votes.

D. Anna by about 100 votes.

E. Anna by about 200 votes.

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Students expected to change)`

`= 25text(%) xx 5692 + 24text(%) xx 3450`

`= 2251`

`=> C`

`text(Part 2)`

♦ Mean mark 48%.

`text(After 1 week,)`

`[(0.75,0.24),(0.25,0.76)][(5692),(3450)] = [(5097),(4045)]`

`text(After 10 weeks,)`

`[(0.75,0.24),(0.25,0.76)]^10[(5692),(3450)] = [(4479),(4663)]`

`:.\ text(Anna is ahead by)`

`4663 – 4479 = 184\ text(votes)`

`=> E`

MATRICES, FUR1 2012 VCAA 9 MC

`[(3,4), (1,2)] × [(a),(3)] = [(6,3), (2,-1)] × [(2),(b)]`

Which set of equations below could be used to determine the values of `a` and `b` that are shown in the matrix equation above?

A. `a - b = 2`

`a + b = 0`

B. `a + b = -2`

`a - b = 0`

C. `a + b = 2`

`a - b = 0`

D. `a - b = 8`

`a + b = 2`

E. `a - b = 8`

`a + b = -2`

Show Answers Only

`B`

Show Worked Solution

`text(Finding the matrix product of both sides:)`

`[(3,4),(1,2)][(a),(3)] = [(6,3),(2,−1)][(2),(b)]`

♦ Mean mark 39%.

`3a + 12`	`= 12 + 3b`
`a – b`	`= 0\ \ …\ (1)`

`a + 6`	`= 4 – b`
`a + b`	`=−2\ \ …\ (2)`

`rArr B`

MATRICES, FUR1 2012 VCAA 3 MC

`x + z`	`= 6`
`2y + z`	`= 8`
`2x + y + 2z`	`= 15`

The solution of the simultaneous equations above is given by

Show Answers Only

`A`

Show Worked Solution

`[(1,0,1),(0,2,1),(2,1,2)][(x),(y),(z)] = [(6),(8),(15)]`

♦ Mean mark 41%.

`:. [(x),(y),(z)]`	`= [(1,0,1),(0,2,1),(2,1,2)]^(−1)[(6),(8),(15)]`
	`= [(−3,−1,2),(−2,0,1),(4,1,−2)][(6),(8),(15)]`

`rArr A`

MATRICES, FUR1 2013 VCAA 8 MC

The matrix `S_(n+1)` is determined from the matrix `S_n` using the rule `S_(n+1) = TS_n - C,` where `T, S_0` and `C` are defined as follows.

`T = [(0.5, 0.6), (0.5,0.4)], \ S_0 = [(100), (250)] quad text(and) quad C = [(20), (20)]`

Given this information, the matrix `S_2` equals

A. `[(100), (250)]`

B. `[(148), (122)]`

C. `[(170), (140)]`

D. `[(180), (130)]`

E. `[(190), (160)]`

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 41%.

`S_1`	`= TS_0 – C`
	`= [(0.5,0.6),(0.5,0.4)][(100),(250)] – [(20),(20)]`
	`= [(180),(130)]`

`S_2`	`= TS_1 – C`
	`= [(0.5,0.6),(0.5,0.4)][(180),(130)] – [(20),(20)]`
	`= [(148),(122)]`

`rArr B`

MATRICES, FUR1 2013 VCAA 7 MC

A school has three computer classes, A, B and C. There are 15 students in each class.

Each student is given a mark out of 100 based on their performance in a test.

Matrix `M` below displays the marks obtained by these 45 students, listed by class.

`M = [(56,78,79,43,67,56,80,85,75,89,55,64,95,34,63), (90,45,56,65,76,79,27,45,69,73,70,63,65,34,59), (76,76,89,47,50,66,68,89,88,90,45,67,78,45,87)]{:(A),(B),(C):} quad text (class)`

Two other matrices, `S` and `R`, are defined below.

`S = [(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)] quad text(and) quad R = [(1,1,1)]`

Which one of the following matrix expressions can be used to generate a matrix that displays the mean mark obtained for each class?

A. `1/45M`

B. `1/3R×M`

C. `1/3R×M×S`

D. `1/15M×S`

E. `1/15S×R×M`

Show Answers Only

`D`

Show Worked Solution

`M`	`xx`	`S`	`=`	`MS`
`3 xx 15`		`15 xx 1`		`3 xx 1`

`MS\ text(represents the total scores in each class.)`

`:. 1/15 MS\ text(is the average score in each class.)`

`rArr D`

MATRICES, FUR1 2013 VCAA 6 MC

A worker can assemble 10 bookcases and four desks in 360 minutes, and eight bookcases and three desks in 280 minutes.

If each bookcase takes `b` minutes to assemble and each desk takes `d` minutes to assemble, the matrix `[(b), (d)]` will be given by

A. `[(-1.5, 2), (4, -5)][(360), (280)]`

B. `[(10, 4), (8, 3)][(360), (280)]`

C. `[(3, -4), (-8, 10)][(360), (280)]`

D. `[(5,- 2), (-4, 1.5)][(360), (280)]`

E. `[(10), (4)][360] + [(8), (3)][280]`

Show Answers Only

`A`

Show Worked Solution

`[(10,4),(8,3)][(b),(d)] = [(360),(280)]`

♦ Mean mark 41%.

`:. [(b),(d)]`	`= [(10,4),(8,3)]^(−1)[(360),(280)]`
	`= 1/(10 xx 3 – 4 xx 8)[(3,−4),(−8,10)][(360),(280)]`
	`= −1/2[(3,−4),(−8,10)][(360),(280)]`
	`= [(−1.5,2),(4,−5)][(360),(280)]`

`rArr A`

MATRICES, FUR1 2014 VCAA 6 MC

The order of matrix `X` is `3 xx 2.`

The element in row `i ` and column `j` of matrix `X` is `x_(ij)` and it is determined by the rule

`x_(ij) = i + j`

The matrix `X` is

Show Answers Only

`E`

Show Worked Solution

`[(x_11, x_12), (x_21, x_22), (x_31, x_32)]`

♦ Mean mark 42%.

`:. X = [(2, 3), (3, 4), (4, 5)]`

`=>E`

MATRICES, FUR1 2015 VCAA 9 MC

A fast-food stand at the football sells pies (`P`) and chips (`C`).

Each week, 300 customers regularly buy either a pie or chips, but not both, from this stand.

For the first five weeks, the customers’ choice of pie or chips is expected to change weekly according to the transition matrix `T_1`, where

`{:({:qquadqquadqquad\ text(this week):}),(qquadqquadqquadquadPqquadqquadC),(T_1 = [(0.65,0.25),(0.35,0.75)]{:(P),(C):}qquadtext(next week)):}`

After the first five weeks, due to expected cold weather, the customers’ choice of pie or chips is expected to change weekly according to the transition matrix `T_2`, where

`{:({:qquadqquadqquad\ text(this week):}),(qquadqquadqquadquadPqquadqquadC),(T_2 = [(0.85,0.25),(0.15,0.75)]{:(P),(C):}qquadtext(next week)):}`

In week 1, 150 customers bought a pie and 150 customers bought chips.

Let `S_1` be the state matrix for week 1.

The number of customers expected to buy a pie or chips in week 8 can be found by evaluating

A. `T_2^7S_1`

B. `T_1^8S_1`

C. `T_2^3(T_1^4S_1)`

D. `T_1^3(T_2^4S_1)`

E. `T_1^3(T_2^5S_1)`

Show Answers Only

`C`

Show Worked Solution

`S_2 = T_1S_1`

♦ Mean mark 40%.
MARKER’S COMMENT: Many students didn’t take into account the change in the transition matrix after 5.

`S_3`	`= T_1S_2`
	`= T_1(T_1S_1)`
	`= T_1^2S_1`

`:. S_5 = T_1^4S_1`

`=>\ text(The transition matrix changes to)`

`T_2\ text(after week 5.)`

`S_6`	`= T_2S_5`
	`= T_2(T_1^4S_1)`
	`vdots`
`S_8`	`= T_2^3(T_1^4S_1)`

`=> C`

MATRICES, FUR1 2015 VCAA 8 MC

The order of matrix `X` is `2 xx 3`.

The element in row `i` and column `j` of matrix `X` is `x_(ij)` and it is determined by the rule

`x_(ij) = i - j`

Which one of the following calculations would result in matrix `X`?

A. `[(1,1,1),(2,2,2)] - [(1,2,3),(1,2,3)]`

B. `[(1,2,3),(1,2,3)] - [(1,1,1),(2,2,2)]`

C. `[(2,2,2),(2,2,2)] - [(3,3,3),(3,3,3)]`

D. `[(1,2),(1,2),(1,2)] - [(1,1),(2,2),(3,3)]`

E. `[(1,1),(2,2),(3,3)] - [(1,2),(1,2),(1,2)]`

Show Answers Only

`A`

Show Worked Solution

`x_(ij) = i – j`

♦ Mean mark 43%.

`:. X = [(0,−1,−2),(1,0,−1)]`

`=> A`

MATRICES, FUR1 2015 VCAA 7 MC

Matrix `P` has inverse matrix `P^(−1)`.

Matrix `P` is multiplied by the scalar `w(w ≠ 0)` to form matrix `Q`.

Matrix `Q^(−1)` is equal to

A. `1/w P^(−1)`

B. `1/(w^2)P^(−1)`

C. `wP^(−1)`

D. `w^2P^(−1)`

E. `P^(−1)`

Show Answers Only

`A`

Show Worked Solution

`Q = wP`

♦ Mean mark 43%.

`Q^(−1)`	`= (wP)^(−1)`
	`= 1/w P^(−1)`

`=> A`

MATRICES, FUR1 2006 VCAA 9 MC

A large population of birds lives on a remote island. Every night each bird settles at either location A or location B.

It was found on the first night that the number of birds at each location was the same.

On each subsequent night, a percentage of birds changed the location at which they settled.

This movement of birds is described by the transition matrix

`{:(qquadqquadAqquadB),({: (A), (B):} [(0.8, 0), (0.2, 1)]):}`

Assume this pattern of movement continues.

In the long term, the number of birds that settle at location A will

A. not change.

B. gradually decrease to zero.

C. eventually settle at around 20% of the island’s bird population.

D. eventually settle at around 80% of the island’s bird population.

E. gradually increase.

Show Answers Only

`B`

Show Worked Solution

`text(The transition matrix shows that 20% of birds)`

♦ Mean mark 40%.

`text(that settle at)\ A\ text(each night move to location)`

`B\ text(on the next night, while no birds relocate)`

`text(from)\ B.`

`:.\ text(In the long term, the birds that settle at)\ A`

`text(will gradually reduce to zero.)`

`rArr B`

MATRICES, FUR1 2006 VCAA 7 MC

How many of the following five sets of simultaneous linear equations have a unique solution?

A. 1

B. 2

C. 3

D. 4

E. 5

Show Answers Only

`C`

Show Worked Solution

`text(Consider each set of equations,)`

`text(Set 1: det)[(4,2),(2,1)] = 0\ \ text{(not unique)}`

♦ Mean mark 42%.

`text(Set 2:)\ x = 0, x + y = 6\ \ text{(unique)}`

`text(Set 3: det)[(1,−1),(1,1)] = 2 != 0\ \ text{(unique)}`

`text(Set 4: det)[(2,1),(2,1)] = 0\ \ text{(not unique)}`

`text(Set 5:)\ x = 8, y = 2\ \ text{(unique)}`

`:. 3\ text(sets of equations have a unique solution.)`

`rArr C`

MATRICES, FUR1 2007 VCAA 9 MC

Matrix `M` is a `3 xx 4` matrix.

Matrix `P` has five rows.

`N` is another matrix.

If the matrix product

`M (NP) = [(4, 1, 7, 2), (0, 9, 7, 4), (4, 3, 3, 1)],`

then the order of matrix `N` is

A. `3 xx 5`

B. `5 xx 3`

C. `4 xx 5`

D. `5 xx 4`

E. `5 xx 5`

Show Answers Only

`C`

Show Worked Solution

`M xx (NP) = [(4,1,7,2),(0,9,7,4),(4,3,3,1)]`

♦ Mean mark 50%.

`text{The order of matrices (above) is}`

`(3 xx 4) xx ((a xx b)(5 xx c)) = (3 xx 4)`

`a = 4, b = 5, c = 4`

`:. N\ text{is a (4 × 5) matrix.}`

`=> C`

MATRICES, FUR1 2007 VCAA 8 MC

Kerry sat for a multiple-choice test consisting of six questions.

Each question had four alternative answers, `A`, `B`, `C`, or `D.`

He selected `D` for his answer to the first question.

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

`{:(qquadqquadqquadqquadqquadqquadqquadqquad\ text(This question)),({:qquadqquadqquadqquadqquadqquadqquadqquadqquadAquadBquadCquadD:}),(text(Next question)qquad{: (A), (B), (C), (D):}[(1,0,1,0), (0,0,0,1), (0,1,0,0), (0,0,0,0)]):}`

The answers that he gave to the six test questions, starting with `D`, were

Show Answers Only

`B`

Show Worked Solution

`text(Initial state matrix is)\ [(0),(0),(0),(1)]`

♦ Mean mark 48%.

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

`:.\ text(S)text(econd answer is)\ B`

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

`:.\ text(Third answer is)\ C,\ text(and so on…)`

`=> B`

MATRICES, FUR1 2007 VCAA 7 MC

Each year, a family always goes on its holiday to one of three places; Portland (`P`), Quambatook (`Q`) or Rochester (`R`).

They never go to the same place two years in a row. For example, if they went to Portland one year, they would not go to Portland the next year; they would go to Quambatook or Rochester instead.

A transition matrix that can be used to model this situation is

Show Answers Only

`E`

Show Worked Solution

`text(Transition matrix will have all elements in)`

♦ Mean mark 48%.

`text(the main diagonal equal to 0, and all columns)`

`text(add up to 1.)`

`rArr E`

MATRICES, FUR1 2007 VCAA 5 MC

An international mathematics competition is conducted in three sections – Junior, Intermediate and Senior.

There are money prizes for gold, silver and bronze levels of achievement in each of these sections.

Table 1 shows the number of students who were awarded prizes in each section.

Table 2 shows the value, in dollars, of each prize.

A matrix product that gives the total value of all the Silver prizes that were awarded is

Show Answers Only

`A`

Show Worked Solution

`text(The matrix product of “total” value must be a 1 × 1 matrix.)`

`=> A`

MATRICES, FUR1 2011 VCAA 8 MC

Considering the following matrix `A`.

`A = [(3,k),(-4,-3)]`

`A` is equal to its inverse `A^-1` for a particular value of `k`.

This value of `k` is

A. `– 4`

B. `– 2`

C. `0`

D. `2`

E. `4`

Show Answers Only

`D`

Show Worked Solution

`A = A^(−1)`

♦ Mean mark 49%.

`[(3,k),(−4,−3)]`	`= 1/(−9 – (−4k)) [(−3,−k),(4,3)]`
`:. 3`	`= (−3)/(−9 + 4k)`
`1`	`=(-1)/(4k-9)`
`4k – 9`	`= −1`
`4k`	`= 8`
`k`	`= 2`

`=> D`

MATRICES, FUR1 2014 VCAA 9 MC

`A` and `B` are square matrices such that `AB = BA = I`, where `I` is an identity matrix.

Which one of the following statements is not true?

A. `ABA = A`

B. `AB^2A = I`

C. `B\ text(must equal)\ A`

D. `B\ text(is the inverse of)\ A`

E. `text(both)\ A\ text(and)\ B\ text(have inverses)`

Show Answers Only

`C`

Show Worked Solution

`A\ text(is the inverse of)\ B\ text(in this example, but)`

♦ Mean mark 40%.

`text(this does not mean that)\ B=A.`

`=>C`

MATRICES, FUR2 2015 VCAA 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Matrix `V` is shown below.

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

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

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

where

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

`= 21/120`

`= 17.5 text(%)`

c. `text(After 2 assessments,)`

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

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

d. `text(After 4 assessments,)`

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

`text(After 5 assessments,)`

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

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

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

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

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

`text(move from Intermediate to Advanced)`

`= 20text(%) xx 34.48`

`= 6.896`

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

MATRICES, FUR2 2012 VCAA 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

`= 275`

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

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

`text(not return.)`

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

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

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

`text(start of 2013.)`

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

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

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

`text(drops below 30.)`

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

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

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

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

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

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

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

`text(start of 2013.)`

MATRICES, FUR2 2008 VCAA 2

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

b. `S_n = T^(n – 1)S_1`

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

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

`text(with less than 400 students.)`

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

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

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

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

`text(the long run.)`

GRAPHS, FUR2 2012 VCAA 3

A company repairs phones and laptops.

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

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

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

The constraints on the company are as follows.

Constraint 1 `x ≥ 0`

Constraint 2 `y ≥ 0`

Constraint 3 `35x + 50y ≤ 1750`

Constraint 4 `y ≤ 4/5 x`

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

Use this constraint to complete the following sentence.

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

`text(minutes on any given day.)`

b. `y <= 4/5x`

`text(When)\ x = 10,`

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

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

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

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

`text(of laptops below the intersection)`

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

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

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

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

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

`text(In the feasible region, maximum)`

`text(profits occur at the intersection.)`

`:. 18\ text(laptops)`

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

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

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

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

`= 60 xx 24 + 100 xx 18`

`= $3240`

GRAPHS, FUR2 2012 VCAA 2

The cost, `C`, and revenue, `R`, in dollars, for making and selling `n` laptops respectively is given by

cost	`C = 320n + 125\ 000`
revenue	`R = 600n`

What is the minimum number of laptops that should be made and sold in order to obtain a profit? (1 mark)

The cost of making each laptop increases by $50.

The selling price of each laptop will need to increase to offset this cost increase.

Find the new selling price of each laptop so that the break-even point occurs when 400 laptops are made and sold. (1 mark)

Show Answers Only

`447`
`682.50`

Show Worked Solution

a. `text(Breakeven occurs when)`

♦ Mean mark for all parts combined was 39%.

`600n`	`= 320n + 125\ 000`
`280n`	`= 125\ 000`
`:. n`	`= (125\ 000)/280`
	`= 446.42…`

`:.\ text(A minimum of 447 laptops need)`

`text(to be sold to make a profit.)`

b. `text(If the cost price increases by $50,)`

MARKER’S COMMENT: A common error was to round to the nearest whole rather than round up.

`C = 370n + 125\ 000`

`text(If breakeven occurs when)\ n = 400,`

`text(Let new selling price = $)P`

`:. 400 xx P`	`= 370 xx 400 + 125\ 000`
`P`	`= (273\ 000)/400`
	`= $682.50`

GEOMETRY, FUR2 2012 VCAA 4

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

Triangle `OAB` is a right-angled triangle.

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

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

Calculate the length, `OA`.

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

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

The area of triangle `OBC` is 35 m².

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

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

Show Answers Only

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

Show Worked Solution

a. `text(In)\ DeltaOAB,`

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

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

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

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

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

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

d. `text(In)\ DeltaBCO,`

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

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

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

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

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

GEOMETRY, FUR2 2012 VCAA 3

A tree is growing near the block of land.

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

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

Calculate the height of the tree.

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

Show Answers Only

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

Show Worked Solution

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

`text(Using the sine rule:)`

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

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

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

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

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

CORE*, FUR2 2013 VCAA 3

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

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

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

Determine the monthly instalment that Hugo will pay.

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

♦ Mean mark of all parts combined was 37%.

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

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

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

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

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

`= 7500 (1 – 0.15)^5`

`= 7500(0.85)^5`

`= 3327.789…`

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

CORE*, FUR2 2013 VCAA 2

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

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

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

Show Answers Only

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

Show Worked Solution

a. `text(Monthly interest rate)`

♦ Mean mark for all parts combined was 39%.

`= 4.8/12`

`= 0.4text(%)`

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

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

`= 5000(1.004)^12`

`= 5245.351…`

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

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

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

`FV = 7698.8614…`

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

b.ii. `text(Total interest)`

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

`= 7698.86 – (5000 + 12 xx 200)`

`= 7698.86 – 7400`

`= $298.86`

GRAPHS, FUR2 2013 VCAA 4

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

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

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

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

inequality 1 `x ≤ 5`

inequality 2 `y ≤ 10`

There are 48 students to accommodate in total.

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

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

inequality 3 `ax + by ≥ 48`

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

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

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

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

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

School regulations require boys and girls to be accommodated separately.

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

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

Show Answers Only

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

Show Worked Solution

a. `a = 6`

`b = 4`

b. `text(Inequality 4)`

`y ≥ 2x`

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

GRAPHS, FUR2 2013 VCAA 3

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

Each student who participates will pay $24.

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

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

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

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

Show Answers Only

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

Show Worked Solution

a. `text(Revenue = 24)n`

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

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

b. `text(Profit) >= 500`

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

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

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

GEOMETRY, FUR2 2014 VCAA 3

The chicken coop contains a circular water dish.

Water flows into the dish from a water container.

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

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

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

The hemisphere has a radius of 5 cm.

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

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

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

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

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

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

The surface area of the water container is 628 cm².

What is the surface area of the feed container?

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

`=1178.09… + 261.79…`

`=1439.8…`

`=1440\ text{cm³ (nearest cm³)}`

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

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

`=628 xx 1.2114…`

`=760.7…`

`=761\ text(cm²)`

GEOMETRY, FUR2 2013 VCAA 3

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

`= 7966 + 4983.91…`

`= 12\ 949.91…\ text(m²)`

`:.\ text(Volume of rubber)`

♦ Mean mark of part (c) 34%.

`=\ text(Area of track × depth)`

`= (12\ 949.91 – 9549) xx 0.1`

`= 340.09`

`= 340\ text{m³ (nearest m³)}`

GEOMETRY, FUR2 2013 VCAA 2

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

The staircase is 1.6 m high and 3.0 m deep.

Its cross-section comprises identical rectangles.

One of these rectangles is shaded in the diagram below.

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

The concrete staircase is 2.5 m wide.

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

Show Answers Only

`0.048\ text(m²)`
`6.6\ text(m³)`

Show Worked Solution

a. `text(Height of rectangle)`

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

`= 1.6/10`

`= 0.16\ text(m)`

`text(Length of rectangle)`

`= 3.0/10`

`= 0.3\ text(m)`

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

b. `V = Ah`

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

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

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

CORE*, FUR2 2014 VCAA 3

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

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

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

Interest on the account had been calculated and paid quarterly.

What was the annual rate of interest for this investment?

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

a. `text(Interest earned)`

`= text(End value − original investment)`

`= 60\ 000 – 45\ 550`

`= $74\ 450`

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

b. `text(By TVM Solver:)`

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

`I(text(%)) = 6.948…`

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

c.i. `text(account balance)`

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

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

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

`FV = 75\ 443.014…`

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

CORE*, FUR2 2014 VCAA 2

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

The annual interest from this perpetuity is $750.

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

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

Show Answers Only

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

Show Worked Solution

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

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

b. `$20\ 000`

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

`text{constant.)}`

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

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

`= 750 xx 1.03`

`= $772.50`

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

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

GRAPHS, FUR2 2014 VCAA 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.)`

GRAPHS, FUR2 2014 VCAA 1

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

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

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

Arthur is a farmer who grows tomatoes.

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

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

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

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

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

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

Each kilogram of Booster contains 0.06 kg of potassium.

Each kilogram of Fastgrow contains 0.04 kg of potassium.

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

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

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

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

(Answer on the graph above.)

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

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

(Answer on the graph above.)

Show Answers Only

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

Show Worked Solution

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

`=0.02× 2`

`= 0.04\ text(kg)`

b. `text(Total Phosphorous)`

`=0.05×100 + 0.05× 400`

`= 25\ text(kg)`

c. `text(Inequality 5,)`

`0.06x + 0.04y >= 180`

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

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

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

`0.02x + 0.06y = 120`

d.ii.

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

`x+y=4000`

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

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

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

CORE*, FUR2 2015 VCAA 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)`

CORE*, FUR2 2015 VCAA 4

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

They each choose a different investment strategy.

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

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

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

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

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

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

Interest is compounded at the end of each quarter.

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

`=4072.542…`

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

b. `text(Annual compounding rate)`

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

`= 3.2text(%)`

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

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

`FV=6664.629…`

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

CORE*, FUR2 2015 VCAA 2

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

a. `text(Annual depreciation)`

`=(3800 – 3150)/ 2`

`= $325\ …\ text(as required)`

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

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

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

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

GRAPHS, FUR2 2015 VCAA 4

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

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

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

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

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

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

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

(Answer on the graph above.)

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

Show Answers Only

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

Show Worked Solution

a. `$250`

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

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

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

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

`text(occurs when)`

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

Calculus, MET2 2014 VCAA 5

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

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

Show Answers Only

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

Show Worked Solution

a. `text(Solution 1)`

`g(x) = x^4 – 8x`

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

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

`text(Solution 2)`

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

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

♦ Mean mark part (b) 37%.

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

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

♦♦♦ Mean mark 7%.

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

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

`:. d ∈ [1,3)`

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

♦♦♦ Mean mark part (c)(ii) 19%.

`text(right.)`

`:. d < 1`

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

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

♦♦♦ Mean mark 17%.

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

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

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

♦♦ Mean mark 28%.

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

e.ii. `u^3 + v^3 = 4\ …\ (1)`

♦♦♦ Mean mark 10%.

`u + v = 1\ …\ (2)`

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

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

f.i. `text(Solution 1)`

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

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

`text(Solution 2)`

♦♦ Mean mark (f)(i) 22%.

`y = 4(p^3 – 2)x – 3p^4`

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

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

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

♦♦♦ Mean mark 14%.

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

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

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

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

Probability, MET2 2014 VCAA 4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

♦ Mean mark 43%.

`X ∼\ text(N)(14,4^2)`

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

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

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

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

`= 0.10565… xx 2000`

`= 211\ text(plants)`

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

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

`= 0.15\ \ text(for)\ \ a ∈ (0,50)`

♦♦ Mean mark 33%.

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

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

♦♦ Mean mark 30%.

`Y ∼\ text(Bi) (n,0.2)`

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

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

Calculus, MET2 2014 VCAA 3

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

`t=2/3`

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

b.i.

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

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

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

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

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

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

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

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

♦ Mean mark part (c)(ii) 38%.

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

d. `text(Solution 1)`

♦ Mean mark 46%.

`text(Equations from given information:)`

`n(1/2) = 0.74`

`0.74 = 1/2Ae^(−1/2k)\ …\ (1)`

`n′ (1/2) = 0`

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

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

`text(Solution 2)`

`n(1/2) = 0.74`

`0.74 = 1/2Ae^(−1/2k)\ …\ (1)`

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

`n′ (1/2) = 0`

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

`:.A`

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

Calculus, MET2 2014 VCAA 2

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculus, MET2 2015 VCAA 5

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

`:. c = 8`

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

a.iii.

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

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

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

`:. d = 20/11`

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

♦♦ Mean mark part (c)(i) 28%.

`d=20/3,`

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

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

♦♦ Mean mark part (c)(ii) 27%.

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

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

♦♦♦ Mean mark part (d) 15%.

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

`= 0`

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

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

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

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

Calculus, MET2 2015 VCAA 4

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

`:. a = 4`

♦ Mean mark part (a) 36%.

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

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

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

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

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

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

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

`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`

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

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

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

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

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

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

`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`

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

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

Probability, MET2 2015 VCAA 3

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

`Y ∼\ text(Bi)(3,11/6)`

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

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

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

`L ∼\ N(74,9^2)`

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

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

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

`W ∼\ text(Bi)(4,0.03)`

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

`text(Solution 2)`

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

d.ii. `W ∼\ text(Bi)(n,0.03)`

♦ Mean mark 44%.

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

GRAPHS, FUR2 2006 VCAA 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`

Calculus, MET2 2015 VCAA 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

`text(At)\ \ x=-40,\ \ f′(x)=3`

♦ Mean mark part (a) 40%.

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

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

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

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

`gprime(0) = −3/16`

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

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

♦ Mean mark part (c) 39%.

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

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

`:. M (2.49,34.53)`

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

d. `P(-2.49, w)`

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

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

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

e. `text(Intersection occurs when:)`

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

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

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

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

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

GEOMETRY, FUR2 2006 VCAA 3

A closed cylindrical water tank has external diameter 3.5 metres.

The external height of the tank is 2.4 metres.

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

`= 1/2 xx (3.5 – 2 xx 0.25)`

`= 1.5\ text(m)`

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

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

GEOMETRY, FUR2 2006 VCAA 2

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

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

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

Distance `SQ` is 125 metres.

Distance `TQ` is 98 metres.

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

GEOMETRY, FUR2 2006 VCAA 1

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

Boundary `AB` is 251 metres.

Boundary `AC` is 142 metres.

Angle `BAC` is 45°.

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

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

`= 78 – 45`

`= 033^@`

`text(Using the cosine rule,)`

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

e. `text(Using sine rule,)`

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

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

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

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

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

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

g. `/_ AMN = 180 – (90 + 45) = 45^@`

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

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

`text(Area)\ Delta AMN`	`= 1/2 xx AN xx MN`
`3200`	`= 1/2 xx MN xx MN\ (AN = MN)`
`MN^2`	`= 6400`
`:. MN`	`= 80\ text(m)`

GEOMETRY, FUR2 2007 VCAA 1-3

Question 1

Tessa is a student in a woodwork class.

The class will construct geometrical solids from a block of wood.

Tessa has a piece of wood in the shape of a rectangular prism.

This prism, `ABCDQRST`, shown in Figure 1, has base length 24 cm, base width 28 cm and height 32 cm.

On the front face of Figure 1, `ABRQ`, Tessa marks point `W` halfway between `Q` and `R` as shown in Figure 2 below. She then draws line segments `AW` and `BW` as shown.

Determine the length, in cm, of `QW`. (1 mark)
Calculate the angle `WAQ`. Write your answer in degrees, correct to one decimal place. (1 mark)
Calculate the angle `AWB` correct to one decimal place. (1 mark)
What fraction of the area of the rectangle `ABRQ` does the area of the triangle `AWB` represent? (1 mark)

Question 2

Tessa carves a triangular prism from her block of wood.

Using point `V`, halfway between `T` and `S` on the back face, `DCST`, of Figure 1, she constructs the triangular prism shown in Figure 3.

Show that, correct to the nearest centimetre, length `AW` is 34 cm. (1 mark)
Using length `AW` as 34 cm, find the total surface area, in cm², of the triangular prism `ABCDWV` in Figure 3. (2 marks)

Question 3

Tessa's next task is to carve the right rectangular pyramid `ABCDY` shown in Figure 4 below.

She marks a new point, `Y`, halfway between points `W` and `V` in Figure 3. She uses point `Y` to construct this pyramid.

Calculate the volume, in cm³, of the pyramid `ABCDY` in Figure 4. (1 mark)
Show that, correct to the nearest cm, length `AY` is 37 cm. (2 marks)
Using `AY` as 37 cm, demonstrate the use of Heron's formula to calculate the area, in cm², of the triangular face `YAB`. (2 marks)

Show Answers Only

Question 1

`12\ text(cm)`
`20.6^@`
`41.2^@\ text{(1 d.p.)}`
`1/2`

Question 2

`34\ text(cm)`
`3344\ text(cm²)`

Question 3

`7168\ text(cm³)`
`37\ text(cm)`
`420\ text(cm²)`

Show Worked Solution

`text(Question 1)`

a.	`QW`	`= 1/2 xx QR`
		`= 1/2 xx 24`
		`= 12\ text(cm)`

`tan\ /_ WAQ`	`= 12/32`
`:. /_ WAQ`	`= tan^-1\ 3/8`
	`= 20.55…`
	`= 20.6^@\ text{(1 d.p.)}`

c.	`/_ AWB`	`= 2 xx /_ WAG`
		`= 2 xx 20.6`
		`= 41.2^@\ text{(1 d.p.)}`

d.	`text(Area)\ ABRQ`	`= 24 xx 32`
		`= 768\ text(cm²)`
	`text(Area)\ Delta AWB`	`= 2 xx text(Area)\ Delta AQW`
		`= 2 xx 1/2 xx 12 xx 32`
		`= 384\ text(cm²)`

`:.\ text(Fraction of area)`

`= 384/768`

`= 1/2`

`text(Question 2)`

`text(Using Pythagoras in)\ Delta AWX,`

`AW`	`= sqrt (12^2 + 32^2)`
	`= sqrt 1168`
	`= 34.176…`
	`= 34\ text{cm (nearest cm) … as required.}`

b. `text(Total S.A. of triangular prism)`

`= text(area of base) + text(area of sides) + text(area of ends)`

`= 24 xx 28 + 2 xx (34 xx 28) + 2 xx (1/2 xx 24 xx 32)`

`= 672 + 1904 + 768`

`= 3344\ text(cm²)`

`text(Question 3)`

a.	`V`	`= 1/3 xx b xx h`
		`= 1/3 xx 24 xx 28 xx 32`
		`= 7168\ text(cm³)`

`text(Using Pythagoras in)\ Delta ABC,`

`AC`	`= sqrt (24^2 + 28^2)`
	`= 36.878…`

`text(Let)\ Z\ text(be the midpoint of)\ AC`

`:. AZ = (36.878…)/2 = 18.439…`

`text(Using Pythagoras in)\ Delta AYZ,`

`AY`	`= sqrt (32^2 + (18.439…)^2)`
	`= sqrt (1364)`
	`= 36.9…`
	`= 37\ text{cm (nearest cm) … as required}`

c. `text(Using Heron’s formula,)`

`s`	`= (37 + 37 + 24)/2=49`

`:.\ text(Area of)\ Delta YAB`

`= sqrt (49 (49 – 24) (49 – 37) (49 – 37))`

`= sqrt (49 xx 25 xx 12 xx 12)`

`= 420\ text(cm²)`

MATRICES, FUR2 2008 VCAA 3

The bookshop manager at the university has developed a matrix formula for determining the number of Mathematics and Physics textbooks he should order each year.

For 2009, the starting point for the formula is the column matrix `S_2008`. This lists the number of Mathematics and Physics textbooks sold in 2008.

`S_2008 = [(456),(350)]{:(text(Mathematics)),(text(Physics)):}`

`O_2009` is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for 2009.

`O_2009` is given by the matrix formula

`O_2009 = A\ S_2008 + B` where `A = [(0.75, 0),(0, 0.68)]` and `B = [(18),(12)]`

Determine `O_2009` (1 mark)

The matrix formula above only allows the manager to predict the number of books he should order one year ahead. A new matrix formula enables him to determine the number of books to be ordered two or more years ahead.

The new matrix formula is

`O_(n + 1) = C O_n - D`

where `O_n` is a column matrix listing the number of Mathematics and Physics textbooks to be ordered for year `n`.

Here, `C = [(0.8, 0),(0, 0.8)]` and `D = [(40),(38)]`

The number of books ordered in 2008 was given by

`O_2008 = [(500),(360)]{:(text(Mathematics)),(text(Physics)):}`

Use the new matrix formula to determine the number of Mathematics textbooks the bookshop manager should order in 2010. (2 marks)

Show Answers Only

`O_2009 = [(360),(250)]`
`248`

Show Worked Solution

a.	`O_2009`	`= [(0.75, 0), (0, 0.68)] [(456), (350)] + [(18), (12)]`
		`= [(342), (238)] + [(18), (12)]`
		`= [(360), (250)]`

b.	`O_2009`	`= [(0.8, 0), (0, 0.8)] [(500), (360)] – [(40), (38)]`
		`= [(400), (288)] – [(40), (38)]`
		`= [(360), (250)]`
	`O_2010`	`= [(0.8, 0), (0, 0.8)] [(360), (250)] – [(40), (38)]`
		`= [(248), (162)]`

`:. 248\ text(Maths books should be ordered in 2010.)`

GEOMETRY, FUR2 2008 VCAA 3

A tree, 12 m tall, is growing at point `T` near a shed.

The distance, `CT`, from corner `C` of the shed to the centre base of the tree is 13 m.

Calculate the angle of elevation of the top of the tree from point `C`.

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

`N` and `C` are two corners at the base of the shed. `N` is due north of `C`.

The angle, `TCN`, is 65°.

Show that, correct to one decimal place, the distance, `NT`, is 12.6 m. (1 mark)
Calculate the angle, `CNT`, correct to the nearest degree. (1 mark)
Determine the bearing of `T` from `N`. Write your answer correct to the nearest degree. (1 mark)
Is it possible for the tree to hit the shed if it falls?

Explain your answer showing appropriate calculations. (2 marks)

Show Answers Only

`42.7^@`
`text(See Worked Solution)`
`69^@`
`111^@`
`text(See Worked Solutions)`

Show Worked Solution

`text(Let)\ \ theta`	`=\ text(angle of elevation)`
	`= 12/13`
`:. theta`	`= 42.709…`
	`= 42.7^@\ text{(1 d.p.)}`

`text(Using the cosine rule:)`

MARKER’S COMMENT: Show the squareroot step in the calculations as per the solution.

`NT^2`	`= 10^2 + 13^2 – 2 xx 10 xx 13 xx cos65^@`
	`= 159.11…`
`:. NT`	`= sqrt(159.11…)`
	`= 12.61…`
	`= 12.6\ text{m (1 d.p.) … as required}`

c. `text(Using the sine rule:)`

♦♦ Mean mark of parts (c)-(e) (combined) was 23%.

`(sin /_ CNT)/13`	`= (sin 65^@)/12.6`
`sin /_ CNT`	`= (13 xx sin 65^@)/12.6`
	`= 0.935…`
`:. /_ CNT`	`= 69.24…`
	`= 69^@\ text{(nearest degree)}`

`text(Bearing of)\ T\ text(from)\ N`

`= 180 – 69`

`= 111^@`

`ST\ text(is the shortest distance between the)`

MARKER’S COMMENT: Many students had significant difficulty in understanding the 3-dimensional diagram.

`text(tree and the shed.)`

`text(In)\ \ Delta NST,`

`sin 69^@`	`= (ST)/12.6`
`:. ST`	`= 12.6 xx sin 69^@`
	`= 11.76…\ text(m)`

`:.\ text(S) text(ince the tree is 12 m, and 12 m) > 11.76\ text(m),`

`text(it could hit the shed if it falls.)`

GEOMETRY, FUR2 2008 VCAA 1

A shed is built on a concrete slab. The concrete slab is a rectangular prism 6 m wide, 10 m long and 0.2 m deep.

Determine the volume of the concrete slab in m³. (1 mark)
On a plan of the concrete slab, a 3 cm line is used to represent a length of 6 m.
1. What scale factor is used to draw this plan? (1 mark)

The top surface of the concrete slab shaded in the diagram above has an area of 60 m².

What is the area of this surface on the plan? (1 mark)

Show Answers Only

`12\ text(m³)`
1. `1/200`
2. `text(15 cm²)`

Show Worked Solution

a.	`V`	`= lbh`
		`= 10 xx 6 xx 0.2`
		`= 12\ text(m³)`

b.i. `text(Scale Factor)`

`= 3/600`

`= 1/200`

♦ A poorly answered question, although exact data unavailable.
MARKER’S COMMENT: The most successful answers established the scale factor and then squared it, as shown in the solution.

ii.	`text(Plan Area)`	`= text(Slab Area) xx (1/200)^2`
		`= 60 xx (1/200)^2`
		`= 0.0015\ text(m²)`
		`= 15\ text(cm²)`

GEOMETRY, FUR2 2015 VCAA 4

Wires support the communications tower, as shown in the diagram below.

The shortest wire is 31 m long.

The shortest wire makes an angle of 38° with the communications tower.

The longest wire is 37 m long.

The longest wire is attached to the communications tower `x` metres above the shortest wire.

What is the value of `x`?

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

Show Answers Only

`7.3\ text(m)`

Show Worked Solution

`text(Using the sine rule:)`

♦ Mean mark sub 50% (exact data not available).

`(sin theta)/31`	`= sin142^@/37`
`sin theta`	`= (31 xx sin142^@)/37`
`:. theta`	`=31.05…^@`

`:.phi`	`=180 – 142 – 31.05…`
	`= 6.94…^@`

`text(Using the cosine rule:)`

`x^2`	`=37^2+31^2 – 2 xx 31 xx 37 xx cos 6.95^@`
	`=52.85…`
`:. x`	`=7.27…`
	`=7.3\ text(m)`

GEOMETRY, FUR2 2015 VCAA 3

Cabins are being built at the camp site.

The dimensions of the front of each cabin are shown in the diagram below.

The walls of each cabin are 2.4 m high.

The sloping edges of the roof of each cabin are 2.4 m long.

The front of each cabin is 4 m wide.

The overall height of each cabin is `h` metres.

Show that the value of `h` is 3.73, correct to two decimal places. (1 mark)

Each cabin is in the shape of a prism, as shown in the diagram below.

All external surfaces of one cabin are to be painted, excluding the base.

What is the total area of the surface to be painted?

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

Show Answers Only

`3.73`
`76\ text(m²)`

Show Worked Solution

a.	`h`	`= 2.4 + sqrt(2.4^2 – 2^2)`
		`= 3.726…`
		`=3.73\ text{m (2 d.p.)}`

b. `text(Area of walls and roof)`

♦♦ “Few students answered this question correctly” (exact data unavailable).
MARKER’S COMMENT: In extended calculations, clearly label each step and draw supporting diagrams where applicable.

`=2xx(2.4 xx 5.4) + 2xx(2.4 xx 5.4) `

`=25.92+25.92`

`=51.84`

`text(Area of front and back)`

`=2xx(2.4 xx 4 + 1/2 xx 4 xx 1.33)`

`=2xx12.26`

`=24.52`

`:.\ text(Total area to be painted)`

`=51.84+24.52`

`=76.36`

`=76\ text{m² (nearest m²)}`

GEOMETRY, FUR2 2009 VCAA 4

The ferry has two fuel filters, `A` and `B`.

Filter `A` has a hemispherical base with radius 12 cm.

A cylinder of height 30 cm sits on top of this base.

Calculate the volume of filter `A`. Write your answer correct to the nearest cm³. (2 marks)

Filter `B` is a right cone with height 50 cm.

Originally filter `B` was full of oil, but some was removed.

If the height of the oil in the cone is now 20 cm, what percentage of the original volume of oil was removed? (2 marks)

Show Answers Only

`17\ 191\ text(cm³)`
`text(93.6%)`

Show Worked Solution

a. `text(Volume of cylinder)`

♦ Mean mark of all parts (combined) 36%.

`= pi r^2 h`

`= pi xx 12^2 xx 30`

`= 13\ 571.68…\ text(cm³)`

`text(Volume of hemisphere)`

`= 1/2 xx 4/3 pi r^3`

`= 1/2 xx 4/3 xx pi xx 12^3`

`= 3619.11…\ text(cm³)`

`:.\ text(Volume of filter)\ A`

`= 13\ 571.68… + 3619.11…`

`= 17\ 190.79…`

`= 17\ 191\ text(cm³)`

b. `text(Linear Factor) = 20/50 = 2/5`

`:.\ text(Volume Factor) = (2/5)^3 = 8/125`

`:.\ text(Percentage of original volume removed)`

`= (1 – 8/125) xx 100 text(%)`

`= 93.6text(%)`

GEOMETRY, FUR2 2015 VCAA 2

There are plans to construct a series of straight paths on the flat top of the mountain.

A straight path will connect the cable car station at `C` to a communications tower at `T`, as shown in the diagram below.

The bearing of the communications tower from the cable car station is 060°.

The length of the straight path between the communications tower and the cable car station is 950 m.

How far north of the cable car station is the communications tower? (1 mark)

Paths will also connect the cable car station and the communications tower to a camp site at `E`, as shown below.

The length of the straight path between the cable car station and the camp site is 1400 m.

The angle `TCE` is 40°.

1. What will be the length of the straight path between the communications tower and the camp site?
  
  Write your answer correct to the nearest metre. (1 mark)
2. Use the cosine rule to find the bearing of the camp site from the communications tower.
  
  Write your answer correct to the nearest degree. (2 marks)

Show Answers Only

`475\ text(metres)`
1. `908\ text(m)`
2. `142^@`

Show Worked Solution

a.	`sin 30°`	`=x/950`
	`:. x`	`=950 xx sin 30°`
		`=475\ text(m)`

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

`TE`	`= sqrt(950^2 + 1400^2 – 2 xx 950 xx 1400 xx cos40^@)`
	`=908.196…`
	`=908\ text{m (nearest m)}`

b.ii.

`text(Let)\ \ alpha= angleCTE`

`cos alpha`	`= (950^2 + 908.2^2 – 1400^2)/(2 xx 950 xx 908.196…)`
	`=-0.1348…`
`:. alpha`	`=97.7…°`

`text(Let point)\ \ B\ \ text(be due south of)\ \ T`

`∠BTE`	`= 97.7… -60`
	`=37.7…°`

`:. text(Bearing of)\ E\ text(from)\ T`

`= 180 – 37.7…`

`=142.2…`

`=142°\ \ text{(nearest °)}`