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GRAPHS, FUR2 2019 VCAA 1

The graph below shows the membership numbers of the Wombatong Rural Women’s Association each year for the years 2008–2018.
 


 

  1. How many members were there in 2009?  (1 mark)
    1. Show that the average rate of change of membership numbers from 2013 to 2018 was − 6 members per year.  (1 mark)
    2. If the change in membership numbers continues at this rate, how many members will there be in 2021?  (1 mark)
Show Answers Only
  1. `60`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `14`
Show Worked Solution

a.  `60`

 

b.i.   `text(Members in 2013)` `= 62`
  `text(Members in 2018)` `= 32`
  `text(Average ROC)` `= (32 – 62)/5`
    `= -6\ text(members per year.)`

 

b.ii.   `text(Members in 2021)` `= 32 – 3 xx 6`
    `= 14`

GEOMETRY, FUR2 2019 VCAA 3

The following diagram shows a crane that is used to transfer shipping containers between the port and the cargo ship.
 


 

The length of the boom, `BC`, is 25 m. The length of the hoist, `AB`, is 15 m.

    1. Write a calculation to show that the distance `AC` is 20 m.  (1 mark)
    2. Find the angle `ACB`.

        

      Round your answer to the nearest degree.  (1 mark)

  1. The diagram below shows a cargo ship next to a port. The base of a crane is shown at point `Q`.
     

      

      

    `qquad`
     

      

    The base of the crane (`Q`) is 20 m from a shipping container at point `R`. The shipping container will be moved to point `P`, 38 m from `Q`. The crane rotates 120° as it moves the shipping container anticlockwise from `R` to `P`.

      

    What is the distance `RP`, in metres?

      

    Round your answer to the nearest metre.  (1 mark)

  2. A shipping container is a rectangular prism.

      

    Four chains connect the shipping container to a hoist at point `M`, as shown in the diagram below.
     

      

    `qquad` 
     

      

    The shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m.

     

    Each chain on the hoist is 4.4 m in length.

      

    What is the vertical distance, in metres, between point `M` and the top of the shipping container?

      

    Round your answer to the nearest metre.  (2 marks)

Show Answers Only
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `37^@`
  1. `51\ text(m)`
  2. `3\ text(m)`
Show Worked Solution
a.i.   `AC` `= sqrt(BC^2 – AB^2)`
    `= sqrt(25^2 – 15^2)`
    `= sqrt 400`
    `= 20`

 

a.ii.   `tan\  /_ ACB` `= 15/20`
  `/_ ACB` `= tan^(-1) (3/4)`
    `= 36.86…`
    `~~ 37^@`

 

b.  `text(Using cosine rule:)`

`RP^2` `= 20^2 + 38^2 – 2 xx 20 xx 38 xx cos 120^@`
  `= 2604`
`:. RP` `= 51.029…`
  `~~ 51\ text(metres)`

 

c.  

`text(Find)\ h => text(need to find)\ x`

`text(Consider the top of the container)`
 


 

`x` `= sqrt(3^2 + 1.2^2)`
  `~~ 3.2311`

 

`:. h` `= sqrt(4.4^2 – 3.2311^2)`
  `= 2.98…`
  `~~ 3\ text(metres)`

GEOMETRY, FUR2 2019 VCAA 2

A cargo ship travels from Magadan (60° N, 151° E) to Sydney (34° S, 151° E).

  1. Explain, with reference to the information provided, how we know that Sydney is closer to the equator than Magadan.   (1 mark)
  2. Assume that the radius of Earth is 6400 km.

      

    Find the shortest great circle distance between Magadan and Sydney.

      

    Round your answer to the nearest kilometre.  (1 mark)

  3. The cargo ship left Sydney (34° S, 151° E) at 6 am on 1 June and arrived in Perth (32° S, 116° E) at 10 am on 11 June.

      

    There is a two-hour time difference between Sydney and Perth at that time of year.

      

    How many hours did it take the cargo ship to travel from Sydney to Perth?  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `10\ 500\ text(km)`
  3. `246\ text(hours)`
Show Worked Solution
a.  

`text{Sydney’s angle with the equator (34°S) is smaller than Magadan’s}`

`text{(60°N) resulting in a shorter distance to the equator.)`

 

b.   `text(Distance)` `= (34 + 60)/360 xx 2pi xx 6400`
    `~~ 10\ 500\ text(km)`

 

c.    `text(Perth is 2 hours behind Sydney)`

`text(6 am Sydney = 4 am Perth)`

`:.\ text(Travel time)` `= 4\ text{am (1 Jun)} – 10\ text{am (11 Jun)}`
  `= 6\ text(hours) + 10\ text(days)`
  `= 6 + 10 xx 24`
  `= 246\ text(hours)`

GEOMETRY, FUR2 2019 VCAA 1

The following diagram shows a cargo ship viewed from above.
 

 
The shaded region illustrates the part of the deck on which shipping containers are stored.

  1. What is the area, in square metres, of the shaded region?  (1 mark)

Each shipping container is in the shape of a rectangular prism.

Each shipping container has a height of 2.6 m, a width of 2.4 m and a length of 6 m, as shown in the diagram below.
 

  1. What is the volume, in cubic metres, of one shipping container?  (1 mark)
  2. What is the total surface area, in square metres, of the outside of one shipping container?  (1 mark)
  3. One shipping container is used to carry barrels. Each barrel is in the shape of a cylinder.

      

    Each barrel is 1.25 m high and has a diameter of 0.73 m, as shown in the diagram below.

      

    Each barrel must remain upright in the shipping container
     
     

      

    `qquad qquad`
     
    What is the maximum number of barrels that can fit in one shipping container?  (1 mark)

Show Answers Only
  1. `6700\ text(m²)`
  2. `37.44\ text(m³)`
  3. `72.48\ text(m²)`
  4. `48`
Show Worked Solution
a.   `text(Area)` `= 160 xx 40 + 12 xx 25`
    `= 6700\ text(m²)`

 

b.   `text(Volume)` `= 6 xx 2.4 xx 2.6`
    `= 37.44\ text(m³)`

 

c.   `text(S.A.)` `= 2(6 xx 2.6) + 2 (6 xx 2.4) + 2(2.4 xx 2.6)`
    `= 72.48\ text(m²)`

 

d.   `6 ÷ 0.73 = 8.21` `=> \ text(8 barrels along length)`
  `2.4 ÷ 0.73 = 3.28` `=> \ text(3 barrels along width)`
  `2.6 ÷ 1.25 = 2.08` `=> \ text(2 vertical layers)`

 

`:.\ text(Maximum barrels)` `= 2 xx 8 xx 3`
  `= 48`

Networks, STD2 N3 2019 FUR2 3

Fencedale High School is planning to renovate its gymnasium.

This project involves 12 activities, `A` to `L`.

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


 

The minimum completion time for the project is 35 weeks.

  1. Identify the critical path and state how many activities are on it?  (2 marks)

  2. Determine the latest start time of activity `E`.  (1 mark)

  3. Which activity has the longest float time?  (1 mark)

It is possible to reduce the completion time for activities `C, D, G, H` and `K` by employing more workers.

  1.  The completion time for each of these five activities can be reduced by a maximum of two weeks.

      

    What is the minimum time, in weeks, that the renovation project could take?  (1 mark)

Show Answers Only
  1. `8\ text(activities)`
  2. `12\ text(weeks)`
  3. `text(Activity)\ J`
  4. `29\ text(weeks)`
Show Worked Solution

a.  `text(Scanning forwards and backwards:)`
 

 

 

`text(Critical path:)\ ABDFGIKL`

`:. 8\ text(activities)`
 

b.  `text(LST for activity)\ E = 12\ text{weeks  (i.e. start of 13th week)}`
 

c.   `text(Consider float times of all activities not on critical path.)`

`J-5, H-1, E-1, C-1`

`:.\ text(Activity)\ J\ text(has the largest float time.)`
 

d.   `text(Critical path after reducing)\ CDGHK\ text(by 2 weeks is)`

`ABDFGIKL.`
 

`:.\ text(Minimum time)` `= 2 + 4 + 7 + 1 + 2 + 2 + 5 + 6`
  `= 29\ text(weeks)`

NETWORKS, FUR2 2019 VCAA 1

Fencedale High School has six buildings. The network below shows these buildings represented by vertices. The edges of the network represent the paths between the buildings.
 


 

  1. Which building in the school can be reached directly from all other buildings?   (1 mark)

  2. A school tour is to start and finish at the office, visiting each building only once.
     i.     What is the mathematical term for this route?   (1 mark)

  3. ii. Draw in a possible route for this school tour on the diagram below.   (1 mark)

 

Show Answers Only
  1. `text(Office)`
  2.  i.  `text(Hamiltonian cycle)`
    ii. `text(See Worked Solutions)` 
Show Worked Solution

a.  `text(Office)`
 

b.i.   `text(Hamiltonian cycle)`

`text{(note a Hamiltonian path does not start and finish}`

  `text{at the same vertex.)}`
 

b.ii.   `text(One of many solutions:)`
 

MATRICES, FUR2 2019 VCAA 3

On Sunday, matrix `V` is used when calculating the expected number of visitors at each location every hour after 10 am. It is assumed that the park will be at its capacity of 2000 visitors for all of Sunday.

Let `L_0` be the state matrix that shows the number of visitors at each location at 10 am on Sunday.

The number of visitors expected at each location at 11 am on Sunday can be determined by the matrix product

 
`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad  \ A qquad quad F qquad \  G \ quad quad W),({:V xx L_0 qquad text(where) qquad L_0 = [(500), (600), (500), (400)]{:(A),(F),(G),(W):}, qquad text(and):} qquad V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Safety restrictions require that all four locations have a maximum of 600 visitors.
  2. Which location is expected to have more than 600 visitors at 11 am on Sunday?   (1 mark) 
  3. Whenever more than 600 visitors are expected to be at a location on Sunday, the first 600 visitors can stay at that location and all others will be moved directly to Ground World `(G)`.
  4. State matrix `R_n` contains the number of visitors at each location `n` hours after 10 am on Sunday, after the safety restrictions have been enforced.
  5. Matrix `R_1` can be determined from the matrix recurrence relation
  6. `qquad qquad qquad R_0 = [(500),(600),(500),(400)]{:(A),(F),(G),(W):}, qquad qquad R_1 = V xx R_0 + B_1`
  7. where matrix `B_1` shows the required movement of visitors at 11 am.
    1. Determine the matrix `B_1`.   (1 mark)

    2. State matrix `R_2` can be determined from the new matrix rule
    3. `qquad qquad R_2 = VR_1 + B_2`
    4. where matrix `B_2` shows the required movement of visitors at 12 noon.
    5. Determine the state matrix `R_2`.   (1 mark)

Show Answers Only
  1. `text(Location)\ A`
  2.  i. `B_1 = [(-210),(0),(210),(0)]`
    ii. `R_2 = [(600),(288),(512),(600)]`
Show Worked Solution
a.   `text{By inspection (higher decimal values in row 1)}`
  `=>\ text(test location)\ A`
  `text(Visitors at)\ A\ text{(11 am)}` `= 0.3 xx 500 + 0.4 xx 600 + 0.6 xx 500 + 0.3 xx 400`
    `= 810`

 
`text(Location)\ A\ text(will have more than 600 visitors.)`

 

b.i.   `VR_0` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(500),(600),(500),(400)]=[(810),(300),(310),(580)]`
     
  `R_1` `= V xx R_0 + B_1`
    `= [(810),(300),(310),(580)] + [(-210),(0),(210),(0)]`
  `:. B_1` `= [(-210),(0),(210),(0)]`

 

b.ii.   `R_2` `= [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(300),(520),(580)] + B_2`
    `= [(786),(288),(282),(644)] + [(-186),(0),(230),(-44)]= [(600),(288),(512),(600)]`

MATRICES, FUR2 2019 VCAA 2

The theme park has four locations, Air World `(A)`, Food World `(F)`, Ground World `(G)` and Water World `(W)`.

The number of visitors at each of the four locations is counted every hour.

By 10 am on Saturday the park had reached its capacity of 2000 visitors and could take no more visitors.

The park stayed at capacity until the end of the day

The state matrix, `S_0`, below, shows the number of visitors at each location at 10 am on Saturday.
 

`S_0 = [(600), (600), (400), (400)] {:(A),(F),(G),(W):}`
 

  1. What percentage of the park’s visitors were at Water World `(W)` at 10 am on Saturday?   (1 mark)

Let `S_n` be the state matrix that shows the number of visitors expected at each location `n` hours after 10 am on Saturday.

The number of visitors expected at each location `n` hours after 10 am on Saturday can be determined by the matrix recurrence relation below.
 

`{:(qquad qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad text(  this hour)),(qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad qquad qquad quad A qquad quad F qquad \  G \ quad quad W),({:S_0 = [(600), (600), (400), (400)], qquad S_(n+1) = T xx S_n quad quad qquad text(where):}\ T = [(0.1,0.2,0.1,0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`
 

  1. Complete the state matrix, `S_1`, below to show the number of visitors expected at each location at 11 am on Saturday.   (1 mark)

 
`S_1 = [(\ text{______}\ ), (\ text{______}\ ), (300),(\ text{______}\ )]{:(A),(F),(G),(W):}`
 

  1. Of the 300 visitors expected at Ground World `(G)` at 11 am, what percentage was at either Air World `(A)` or Food World `(F)` at 10 am?   (1 mark)

  2. The proportion of visitors moving from one location to another each hour on Sunday is different from Saturday.

     

    Matrix `V`, below, shows the proportion of visitors moving from one location to another each hour after 10 am on Sunday.

     

    `qquad qquad {:(qquadqquadqquadqquadqquadtext(this hour)),(qquad qquad qquad \ A qquad quad F qquad \  G \ quad quad W),(V = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]{:(A),(F),(G),(W):}\ text(next hour)):}`

     

     
    Matrix `V` is similar to matrix `T` but has the first two rows of matrix `T` interchanged.

     

  3. The matrix product that will generate matrix `V` from matrix `T` is
  4. `qquad qquad V = M xx T`
  5. where matrix `M` is a binary matrix.
  6. Write down matrix `M`.   (1 mark)
Show Answers Only
  1. `20%`
  2. `S_1 = [(300 ), (780), (300),(620)]{:(A),(F),(G),(W):}`
  3. `60%`
  4. `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
Show Worked Solution

a.   `text(Total visitors)\ =2000`

`text{Percentage}\  (W)` `= 400/2000= 20%`

 

b.   `S_1` `= TS_0`
    `= [(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)][(600),(600),(400),(400)]=[(300),(780),(300),(620)]`

 

c.  `text(Visitors from)\ A to G= 0.1 xx 600 = 60`

`text(Visitors from)\ F to G= 0.2 xx 600 = 120`

`:.\ text(Percentage of)\ G\ text(at 11 am)= (60 + 120)/300= 60%`
 

d.   `M = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)]`
 

 
`text(Check:)`

`M xx T = [(0,1,0,0), (1,0,0,0), (0,0,1,0), (0,0,0,1)][(0.1, 0.2, 0.1, 0.2),(0.3,0.4,0.6,0.3),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)] = [(0.3,0.4,0.6,0.3),(0.1,0.2,0.1,0.2),(0.1,0.2,0.2,0.1),(0.5,0.2,0.1,0.4)]`

Calculus, EXT2 C1 2002 HSC 2b

For  `n = 0, 1, 2,`...

let  `I_n = int_0 ^{(pi)/(4)} tan^(n) theta  d theta`.

  1.  Show that  `I _1 = (1)/(2) ln2`.   (1 mark)

  2.  Show that, for  `n >= 2`,
     
         `I_n + I_(n - 2) = (1)/(n-1)`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.         `I_n = int_0 ^{(pi)/(4)} tan^n theta \ d theta`

`I_1` `= int_0 ^{(pi)/(4)} tan theta \ d theta`
  `= [ -ln cos x ]_0 ^{(pi)/(4)}`
  `= [- ln cos\ (pi)/(4) + ln cos 0]`
  `= -ln 2^{-(1)/(2)}`
  `= (1)/(2) ln 2`

 

ii. `I_n` `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(n-2) theta ⋅ (sec^2 theta – 1) d theta`
    `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ sec^2 theta \ d theta – int_0 ^{(pi)/(4)} tan^(n – 2) theta \ d theta`

 
`text(Let) \ \ u = tan theta`

`(du)/(d theta) = sec^2 theta \ => \ du = sec^2 theta \ d theta`

`text(When)` `theta` `= (pi)/(4),`   `\ \ \ \u` `= 1`
  `theta` `=0,`   `u` `= 0`

 

`I_n` `= int_0 ^1 u^(n-2) \ du – I_(n – 2)`
  `= [(1)/(n-1) ⋅ u^(n-1)]_0 ^1 – I_(n – 2)`
  `= (1)/(n – 1) – I_(n-2)`

 
`:. \ I_n + I_(n – 2)=(1)/(n – 1)\ \ \ \ (n>=2)`

Calculus, EXT2 C1 2008 HSC 3c

For  `n >= 0`, let  `I_n = int_0 ^{(pi)/(4)} tan^(2n) theta  d theta`.   

  1. Show that for `n >= 1`,
     
         `I _n = (1)/(2n - 1) - I_(n-1)`.  (2 marks)

  2. Hence, or otherwise, calculate  `I_3`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `(52 – 15 pi)/(60)`
Show Worked Solution
i. `I_n` `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * (sec^2 theta – 1) \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * sec^2 theta  d theta – int_0 ^{(pi)/(4)} tan^(2n-2) theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2)theta * sec^2 theta \ d theta – I_(n-1)`

 

`text(Let) \ \ u = tan theta`

`(du)/(d theta) = sec^2 theta \ => \ du = \ sec^2 theta \ d theta`

`text(When)`    `theta` `= (pi)/(4)` `,` `\ \ \ \u` `= 1`
  `theta` `= 0` `,` `u` `= 0`

 

`I_n` `= int_0 ^1 u^(2n – 2) \ du – I_(n – 1)`
  `= [1/(2n -1) ⋅ u^(2n – 1)]_0 ^1 – I_(n – 1)`
  `= 1/(2n-1)(1^(2n-1)-0)  – I_(n – 1)`
  `= 1/(2n-1)  – I_(n – 1)`

 

ii.        `I_0 = int_0 ^{(pi)/(4)} d theta = (pi)/(4)`

`I_3` `= (1)/(5) – I_2`
  `= (1)/(5) – ((1)/(3) – I_1)`
  `= (1)/(5) – (1)/(3) + (1 – I_0)`
  `= (13)/(15) – (pi)/(4)`
  `= (52 – 15 pi)/(60)`

Calculus, EXT2 C1 2005 HSC 1c

Use integration by parts to evaluate  `int_1^e x^7 log_e x  dx`.   (3 marks)

Show Answers Only

`(7e^8 – 1)/(64)`

Show Worked Solution
  `u` `= log_e x,` `\ \ \ \ u^(′)` `= (1)/(x)`
  `v^(′)` `= x^7,` `v` `= (1)/(8) x^8`

 

`int_1^e x^7\ log_e x  dx` `= uv-int u^(′) v \ dx`
  `= [(x^8)/(8) ⋅ log_e x]_1 ^e-(1)/(8) int_1 ^e (1)/(x) ⋅ x^8 dx`
  `= ((e^8)/(8) log_e e-(1)/(8) log_e 1)-(1)/(8)[(1)/(8) x^8]_1 ^e`
  `= (e^8)/(8)-(1)/(8) ((e^8)/(8)-(1)/(8))`
  `= (e^8)/(8)-((e^8-1)/(64))`
  `= (7e^8 +1)/(64)`

Proof, EXT2 P1 SM-Bank 12

If  `ab`  is divisible by 3, prove by contrapositive that  `a`  or  `b`  is divisible by 3.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Statement)`

`ab \ text(is divisible by 3)\  => a \ text(or) \ b \ text(is divisible by) \ 3`

`text(Contrapositive)`

`a \ text(or) \ b not \ text(divisible by 3)\ => ab not \ text(divisible by 3)`

`text(Let) \ \ a` `= 3x + p, \ text(where) \ \ x ∈ Ζ \ \ text(and) \ \ p= 1 \ text(or) \ \ 2`
`b` `= 3y + q, \ text(where) \ \ y ∈ Ζ \ \ text(and) \ \ q= 1 \ text(or) \ \ 2`

 

`ab` `= (3x + p)(3y + q)`
  `= 9xy + 3qx + 3py + pq`
  `= 3(3xy + qx + py) + pq`

 
`text(Possible values of) \ \ pq = 1, 2, 4`

`=> \ pq \ text(is not divisible by 3)`

`:. \ ab \ text(is not divisible by 3)`

`:. \ text(By contrapositive, statement is true.)`

MATRICES, FUR2 2019 VCAA 1

The car park at a theme park has three areas, `A, B` and `C`.

The number of empty `(E)` and full `(F)` parking spaces in each of the three areas at 1 pm on Friday are shown in matrix `Q`  below.
 

`{:(qquad qquad qquad \ E qquad F),(Q = [(70, 50),(30, 20),(40, 40)]{:(A),(B),(C):}quad text(area)):}`
 

  1. What is the order of matrix `Q`?   (1 mark)

  2. Write down a calculation to show that 110 parking spaces are full at 1 pm.   (1 mark)

Drivers must pay a parking fee for each hour of parking.

Matrix `P`, below, shows the hourly fee, in dollars, for a car parked in each of the three areas.
 

`{:(qquad qquad qquad qquad qquad text{area}), (qquad qquad qquad A qquad quad quad B qquad qquad C), (P = [(1.30, 3.50, 1.80)]):}`
 

  1. The total parking fee, in dollars, collected from these 110 parked cars if they were parked for one hour is calculated as follows.  

     

     

    `qquad qquad qquad P xx L = [207.00]`

     

    where matrix  `L`  is a  `3 xx 1`  matrix.

     

    Write down matrix  `L`.   (1 mark)

The number of whole hours that each of the 110 cars had been parked was recorded at 1 pm. Matrix `R`, below, shows the number of cars parked for one, two, three or four hours in each of the areas `A, B` and `C`.

`{:(qquadqquadqquadqquadquadtext(area)),(quad qquadqquadquad \ A qquad B qquad C),(R = [(3, 1, 1),(6, 10, 3),(22, 7,10),(19, 2, 26)]{:(1),(2),(3),(4):}\ text(hours)):}`
 

  1. Matrix  `R^T`  is the transpose of matrix  `R`.

      

    Complete the matrix  `R^T`  below.   (1 mark)

      

    `qquad R^T = [( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , ), ( , , , , , , , , )]`
     

  2. Explain what the element in row 3, column 2 of matrix  `R^T`  represents.   (1 mark)

Show Answers Only
  1. `3 xx 2`
  2. `50 + 20 + 40 = 110`
  3. `L = [(50), (20), (40)]`
  4. `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
  5. `text(Number of cars parked in area)\ C\ text(for 2 hours).`
Show Worked Solution

a.  `text(Order) : 3 xx 2`
 

b.  `text(Add 2nd column): \ 50 + 20 + 40 = 110`
 

c.  `L = [(50), (20), (40)]`
 

d.  `R^T = [(3 ,6 , 22, 19), (1, 10, 7, 2), (1, 3, 10, 26)]`
 

e.   `e_32\ text(in)\ R^T =>` `text(number of cars parked in area)\ C`
    `text(for 2 hours.)`

Proof, EXT2 P1 SM-Bank 11

If  `a^2-4a + 3`  is even, `a ∈ Ζ`,

prove by contrapositive that  `a`  is odd. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contrapositive)`

`a not \ text(odd)` `\ =>\ \ a^2-4x + 3 not \ text(even)`
`text{(i.e)  If} \ \ a \ text(is even) ` `\ => \ a^2-4x + 3 \ \ text(is odd)`

 

`text(If) \ \ a \ \ text(is even) , ∃ \ k , k ∈ Ζ , text(such that) \ \ a = 2k`

`text(Substitute) \ \ 2k \ \ text(into) \ \ a^2-4x + 3`

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

 
`:. \ text(By contrapositive, if) \ \ a^2-4x + 3 \ \ text(is even) \ =>  \ a \ \ text(is odd.)`

Proof, EXT2 P1 SM-Bank 7

  1. Given  `a + b = 6`  and  `a, b > 0,` show
     
        `(1)/(a) + (1)/(b) >= (2)/(3)`   (2 marks)
      
  2. If  `a + b = c,` show
     
        `(1)/(a^2) + (1)/(b^2) >= (8)/(c^2)`   (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `text(Prove) \ \ (1)/(a) + (1)/(b) >= 2/3 :`

`(a – b)^2` `>= 0`
`(a + b)^2 – 4 ab` `>= 0`
`(a + b)^2` `>= 4ab`
`6 (a + b)` `>= 4 ab\ \ \ (a+b=6)`
`3 (a + b)` `>= 2 ab`
`(a + b)/(ab)` `>= (2)/(3)\ \ \ (a, b > 0\ => \ ab>0)`

 

ii.   `a + b = c`

`:. \ text(Statement true)`

Proof, EXT2 P1 SM-Bank 6

If  `x,  y,  z  ∈ R`  and  `x  ≠ y ≠ z`, then   

  1.  Prove  `x^2 + y^2 + z^2 > yz + zx + xy`   (2 marks)

  2.  If  `x + y + z = 1`, show  `yz+zx+xy<1/3`   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `x^2 + y^2 + z^2 – yz – zx – xy > 0`

`text(Multiply) × 2`

`2x^2 + 2y^2 + 2z^2 – 2yz – 2zx – 2xy > 0`

`(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 -2xz + x^2) > 0`

`(x – y)^2 + (y – z)^2 + (z – x)^2 > 0`

`text(Square of any rational number) > 0`

`:.\ text(Statement is true.)`

 

ii.    `(x + y + z)^2` `= x^2 + xy + xz + yx + y^2 + yz + zx + zy + z^2`
  `1` `= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz`

 
`text{Consider statement in part (i):}`

`x^2 + y^2 + z^2 – yz – zx – xy > 0`

`=> (x + y + z)^2 – (x^2 + y^2 + z^2 – yz – zx – xy)<1`

`3xy + 3yz + 3zx` `< 1`
`:. \ yz + zx + xy` `< (1)/(3)`

CORE, FUR2 2019 VCAA 8

Phil invests $200 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after `n` months, `A_n`, can be modelled by the recurrence relation

`A_0 = 200\ 000, qquad A_(n + 1) = 1.0035\ A_n - 3700`

  1. What monthly payment does Phil receive?   (1 mark)

  2. Show that the annual percentage compound interest rate for this annuity is 4.2%.   (1 mark)

At some point in the future, the annuity will have a balance that is lower than the monthly payment amount.

  1. What is the balance of the annuity when it first falls below the monthly payment amount?

     

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

  2. If the payment received each month by Phil had been a different amount, the investment would act as a simple perpetuity.

     

    What monthly payment could Phil have received from this perpetuity?   (1 mark)

Show Answers Only
  1. `$3700`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `$92.15`
  4. `$700`
Show Worked Solution

a.  `$3700`

b.   `text(Monthly rate)` `= 0.0035 = 0.35%`
  `text(Annual rate)` `= 12 xx 0.35 = 4.2%`

  
c.  `text(Find)\ N\ text(when)\ FV = 0\ \ text{(by TVM solver)}:`

`N` `= ?`
`I(%)` `= 4.2`
`PV` `= 200\ 000`
`PMT` `= 3700`
`FV` `= 0`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 
`=> N = 60.024951`

 
`text(Find)\ \ FV\ \ text(when)\ \ N = 60.024951\ \ text{(by TVM solver):}`

`=>FV = $92.15`  

d.  `text(Perpetuity) => text(monthly payment) = text(monthly interest)`

`:.\ text(Perpetuity payment)` `= 200\ 000 xx 4.2/(12 xx 100)`
  `= $700`

Proof, EXT2 P1 SM-Bank 5 MC

Four cards are placed on a table with a letter on one face and a shape on the other.
 

     

 
You are given the rule: "if N is on a card then a circle is on the other side."

Which cards need to be turned over to check if this rule holds?

  1. N and G
  2. G and triangle
  3. circle and N
  4. N and triangle
Show Answers Only

`=> \ D`

Show Worked Solution

`text(Solution 1)`

`text(L)text(ogically equivalent statements are:)`

`N` `=> \ text(circle)`  
`not\ text(circle)` `=> not N`  

 
`text(To confirm rule is not broken,)`

`N \ text(must be turned)`

`text{Triangle (not circle) must be turned – only other shape not a circle.}`

`=> D`
 

`text(Solution 2)`

`text(Consider the flip side of each card.)`

`text(If circle has) \ N \ text(on the other side (or not)) – text(tells us nothing.)`

`text(If) \ G \ text(has a circle on the other side (or not)) – text(tells us nothing.)`

`text(If) \ N \ text(doesn’t have a circle on other side) – text(rule broken.)`

`text(If triangle has an) \ N \ text(on other side) – text(rule broken.)`

`:. \ text(Need to turn) \ N \ text(and triangle)`

Financial Maths, 2ADV M1 SM-Bank 15

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

The value of the tools, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.
 

`V_0 = 60\ 000, qquad qquad  V_(n + 1) = 0.9 V_n`
 

  1. Use recursion to show that the value of the tools after two years.   (1 mark)

  2. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?  (1 mark)

  3. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.  (1 mark)

Show Answers Only
  1. `$48\ 600`
  2. `11\ text(years)`
  3. `V_(n + 1) = V_n – 0.08 V_0`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

 

b.  `text(Find)\ \ n\ \ text(such that:)`

COMMENT: Dividing by  `log_e 0.9`  is dividing by a negative number!

`60\ 000 xx 0.9^n` `< 20\ 000`  
`log_e 0.9^n` `<log_e (1/3)`  
`n` `> log_e (1/3) / log_e 0.9`  
  `> 10.427`  

 
`:.\ text(Phil will replace in the 11th year.)`

 

c.  `text(Annual depreciation) = 0.08 xx V_0`

`:. V_(n + 1) = V_n – 0.08V_0`

CORE, FUR2 2019 VCAA 7

Phil is a builder who has purchased a large set of tools.

The value of Phil’s tools is depreciated using the reducing balance method.

The value of the tools, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 60\ 000, qquad V_(n + 1) = 0.9 V_n` 

  1. Use recursion to show that the value of the tools after two years, `V_2` , is $48 600.   (1 mark)

  2. What is the annual percentage rate of depreciation used by Phil?   (1 mark)

  3. Phil plans to replace these tools when their value first falls below $20 000.

     

    After how many years will Phil replace these tools?   (1 mark)

  4. Phil has another option for depreciation. He depreciates the value of the tools by a flat rate of 8% of the purchase price per annum.

     

    Let `V_n` be the value of the tools after `n` years, in dollars.

     

    Write down a recurrence relation, in terms of `V_0, V_(n + 1)` and `V_n`, that could be used to model the value of the tools using this flat rate depreciation.   (1 mark)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(10%)`
  3. `11\ text(years)`
  4. `V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`
Show Worked Solution
a.   `V_0` `= 60\ 000`
  `V_1` `= 0.9 xx 60\ 000 = 54\ 000`
  `V_2` `= 0.9 xx 54\ 000 = $48\ 600`

  
b.  `text(Depreciation rate) = 0.1 = 10%`

 
c.  `text(Find)\ \ n\ \ text(such that)`

`60\ 000 xx 0.9^n = 20\ 000`

`=> n = 10.427\ \ \ text{(by CAS)}`

`:.\ text(Phil will replace in the 11th year.)`

 
d.  `text(Annual depreciation) = 0.08 xx V_0 = 4800`

`:.\ text(Recurrence relation is:)`

`V_0=60\ 000,\ \ \ V_(n + 1) = V_n-4800`

CORE, FUR2 2019 VCAA 6

The total rainfall, in millimetres, for each of the four seasons in 2015 and 2016 is shown in Table 5 below.

  1. The seasonal index for winter is shown in Table 6 below.

     

    Use the values in Table 5 to find the seasonal indices for summer, autumn and spring.

  2. Write your answers in Table 6, rounded to two decimal places.   (2 marks)

     


  3. The total rainfall for each of the four seasons in 2017 is shown in Table 7 below.

     


    Use the appropriate seasonal index from Table 6 to deseasonalise the total rainfall for winter in 2017.

     

    Round your answer to the nearest whole number.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `186\ text(mm)`
Show Worked Solution
a.  

`text{Average rainfall (2015)}\ = (142 + 156 + 222 + 120)/4 = 160`

`text{Average rainfall (2016)}\ = (135 + 153 + 216 + 96)/4= 150`

`text{SI (Summer)}` `= 1/2 (142/160 + 135/150)=0.89`
`text{SI (Autumn)}` `= 1/2(156/160 + 153/150)=1.00`
`text{SI (Spring)}` `= 1/2 (120/160 + 96/150)=0.70`

 

b.   `text{Winter (deseasonalised)}` `= 262/1.41`
    `~~ 186\ text(mm)`

CORE, FUR2 2019 VCAA 5

The scatterplot below shows the atmospheric pressure, in hectopascals (hPa), at 3 pm (pressure 3 pm) plotted against the atmospheric pressure, in hectopascals, at 9 am (pressure 9 am) for 23 days in November 2017 at a particular weather station.
 

A least squares line has been fitted to the scatterplot as shown.

The equation of this line is

pressure 3 pm = 111.4 + 0.8894 × pressure 9 am

  1. Interpret the slope of this least squares line in terms of the atmospheric pressure at this weather station at 9 am and at 3 pm.   (1 mark)

  2. Use the equation of the least squares line to predict the atmospheric pressure at 3 pm when the atmospheric pressure at 9 am is 1025 hPa.
  3. Round your answer to the nearest whole number.   (1 mark)

  4. Is the prediction made in part b. an example of extrapolation or interpolation?   (1 mark)

  5. Determine the residual when the atmospheric pressure at 9 am is 1013 hPa.
  6. Round your answer to the nearest whole number.   (1 mark)

  7. The mean and the standard deviation of pressure 9 am and pressure 3 pm for these 23 days are shown in Table 4 below.

    1. Use the equation of the least squares line and the information in Table 4 to show that the correlation coefficient for this data, rounded to three decimal places, is  `r` = 0.966   (1 mark)

    2. What percentage of the variation in pressure 3 pm is explained by the variation in pressure 9 am?
    3. Round your answer to one decimal place.   (1 mark)

  1. The residual plot associated with the least squares line is shown below.
     

    1. The residual plot above can be used to test one of the assumptions about the nature of the association between the atmospheric pressure at 3 pm and the atmospheric pressure at 9 am.
    2. What is this assumption?   (1 mark)

    3. The residual plot above does not support this assumption.
    4. Explain why.   (1 mark)

Show Answers Only
  1. `text(An increase in 1hPa of pressure at 9 am is associated)`
    `text(with an increase of 0.8894 hPa of pressure at 3 pm.)`
  2. `1023\ text(hPa)`
  3. `text(Interpolation)`
  4. `3\ text(hPa)`
    1. `0.966`
    2. `93.3%`
    1. `text(The assumption is that a linear relationship)`
      `text(exists between the pressure at 9 am and the)`
      `text(pressure at 3 pm.)`
    2. `text(The residual plot does not appear to be random.)`
Show Worked Solution

a.    `text(An increase in 1hPa of pressure at 9 am is associated)`

`text(with an increase of 0.8894 hPa of pressure at 3 pm.)`

 

b.   `text(pressure 3 pm)` `= 111.4 + 0.8894 xx 1025`
    `= 1023\ text(hPa)`

 

c.  `text{Interpolation (1025 is within the given data range)}`

 

d.   `text(Residual)` `= text(actual) – text(predicted)`
    `= 1015 – (111.4 + 0.8894 xx 1013)`
    `= 1015 – 1012.36`
    `= 2.63…`
    `~~ 3\ text(hPa)`

 

e.i.   `r= b (s_x)/(s_y)`

    `= 0.8894 xx 4.5477/4.1884`
    `= 0.96569…`
    `= 0.966`

 

e.ii.   `r` `= 0.966`
  `r^2` `= 0.9331`
    `= 93.3%`

 

f.i.   `text(The assumption is that a linear relationship)`
 

`text(exists between the pressure at 9 am and the)`

`text(pressure at 3 pm.)`

 

f.ii.  `text(The residual plot does not appear to be random.)`

CORE, FUR2 2019 VCAA 4

The relative humidity (%) at 9 am and 3 pm on 14 days in November 2017 is shown in Table 3 below.

A least squares line is to be fitted to the data with the aim of predicting the relative humidity at 3 pm (humidity 3 pm) from the relative humidity at 9 am (humidity 9 am).

  1. Name the explanatory variable.   (1 mark)

  2. Determine the values of the intercept and the slope of this least squares line.
  3. Round both values to three significant figures and write them in the appropriate boxes provided.   (1 mark)

humidity 3 pm = 
 
   +  
 
   × humidity 9 am  (1 mark)
  1. Determine the value of the correlation coefficient for this data set.
  2. Round your answer to three decimal places.   (1 mark)

Show Answers Only
  1. `text(humidity 9 am)`
  2. `text(humidity 3 pm) = -1.26 + 0.765 xx text(humidity 9 am)`
  3. `r = 0.871`
Show Worked Solution

a.  `text(humidity 9 am)`
 

b.  `text(Input all data points into CAS:)`

`text(humidity 3 pm) = -1.26 + 0.765 xx text(humidity 9 am)`
 

c.    `r = 0.871\ \ text{(3 d.p.)}`

Proof, EXT2 P1 SM-Bank 8

Prove that  `log_2 11`  is irrational.   (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ log_2 11\ \ text(is rational.)`

`:. log_2 11 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`qlog_2 11` `=p`  
`log_2 11^q` `=p`  
`11^q` `=2^p`  

 
`text(LHS is odd.)`

`text(RHS is even.)`

`:.\ text(Contradiction: integer values)\ \ p, q\ \ text(do not exist.)`

`:.\ log_2 11\ \ text(is irrational.)`

Proof, EXT2 P1 SM-Bank 9

If  `n`  is a positive integer,

prove  `sqrt(10n+2)`  is always irrational.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt(10n+2)\ \ text(is rational.)`

`:. sqrt(10n+2) = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`10n+2` `=p^2/q^2`  
`2q^2(5n+1)` `=p^2\ …\ (1)`  

 
`p^2\ \ text(is even) \ =>\ p\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ p=2k`
 

`text{Substitute}\ \ p=2k\ \ text{into (1)}`

`2q^2(5n+1)` `=(2k)^2`  
`q^2(5n+1)` `=2k^2`  

 

`q^2(5n+1)\ \ text(is even since) \ =>\ 2k^2\ text(is even)`

`=> q^2\ \ text(is even since)\ \ 5n + 1\ \ text(can be odd or even).`

`=> q\ \ text(is even)`

`=> \ text(Contradiction:)\  p and q\ \ text(have a common factor of 2)`

`:.\ sqrt(10n+2)\ \ \ text(is irrational.)`

Proof, EXT2 P1 SM-Bank 4

Prove that  `root3 2`  is irrational.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ root3 2\ \ text(is rational.)`

`:. root3 2 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`2` `=p^3/q^3`  
`2q^3` `=p^3\ …\ (1)`  

 
`p^3\ \ text(is even) \ =>\ p\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ p=2k`
 

`text{Substitute}\ \ q=2k\ \ text{into (1)}`

`2q^3` `=(2k)^3`  
`2q^3` `=8k^3`  
`q^3` `=4k^3`  

 

`q^3\ \ text(is even) \ =>\ q\ text(is even)`

`:. p and q\ \ text(have a common factor of 2)`

`=> \ text(Contradiction)`

`:.\ root3 2\ \ \ text(is irrational.)`

Proof, EXT2 P1 SM-Bank 2

Prove that  `sqrt3`  is irrational.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

 

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt3\ \ text(is rational.)`

`:. sqrt3 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`3` `=p^2/q^2`  
`3q^2` `=p^2\ …\ (1)`  

 
`text(If)\ \ q^2\ \ text(is even):`

COMMENT: `(2//q)`  can be used for “2 divides `q`”  or  `q`  is divisible by 2.

`=> q\ \ text(is even)\ \ (2//q)`

`=>p^2\ \ text(is even)\ \ =>\ p\ \ text(is even)\ \ (2//p)`

`p,q\ \ text(have a common factor of 2`

`text(Contradiction)\ =>\ sqrt3\ \ text(is irrational for)\ \ p, q\ \ text(even.)`
 

`text(If)\ \ q^2\ \ text(is odd):`

`=> 3q^2\ \ text(is odd)\ =>\ q\ \ text(is odd)`

`=>p^2\ \ text(is odd)\ \ =>\ p\ \ text(is odd)`

`text(Let)\ \ p=2x+1, \ \ q=2y+1,\ \ x,y inZZ`
 

`text{Substitute into (1)}`

`3(2y+1)^2` `=(2x+1)^2`  
`3(4y^2+4y+1)` `=4x^2+4x+1`  
`12y^2+12y+3` `=4x^2+4x`  
`6y^2+6y+1` `=2x^2+2x`  

 

`text(LHS) = 6(y^2+y)+1\ \ text(which is odd)`

`text(RHS) = 2(x^2+x)\ \ text(which is even)`

`text(Contradiction)\ =>\ sqrt3\ \ text(is irrational for)\ \ p, q\ \ text(odd.)`

`:.\ sqrt3\ \ \ text(is irrational.)`

CORE, FUR2 2019 VCAA 3

The five-number summary for the distribution of minimum daily temperature for the months of February, May and July in 2017 is shown in Table 2.

The associated boxplots are shown below the table.

Explain why the information given above supports the contention that minimum daily temperature is associated with the month. Refer to the values of an appropriate statistic in your response.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(The appropriate statistic to explore a possible association)`

`text(is the median.)`

`text(The median of the minimum daily temperature exhibits a clear)`

`text(down trend as the months of the year increase. This supports the)`

`text(contention of an association between the variables.)`

CORE, FUR2 2019 VCAA 2

 

The parallel boxplots below show the maximum daily temperature and minimum daily temperature, in degrees Celsius, for 30 days in November 2017.
 

  1. Use the information in the boxplots to complete the following sentences.
  2. For November 2017
  3.    i. the interquartile range for the minimum daily temperature was _____ °C   (1 mark)

  4.  ii. the median value for maximum daily temperature was _____ °C higher than the median value for minimum daily temperature   (1 mark)

  5. iii. the number of days on which the maximum daily temperature was less than the median value for minimum daily temperature was _____    (1 mark)

  1. The temperature difference between the minimum daily temperature and the maximum daily temperature in November 2017 at this location is approximately normally distributed with a mean of 9.4 °C and a standard deviation of 3.2 °C.
  2. Determine the number of days in November 2017 for which this temperature difference is expected to be greater than 9.4 °C.  (1 mark)

Show Answers Only

    1. `5text(°C)`
    2. `10text(°C)`
    3. `1\ text(day)`
  2. `15\ text(days)`

Show Worked Solution

a.i.  `text(IQR)\ = 17-12= 5text(°C)`
 

a.ii.    `text{Median (maximum temperature)}` `= 25`
  `text{Median (minimum temperature)}` `= 15`

 
`:.\ text(Maximum is 10°C higher)`
 

a.iii.  `text{Median (minimum temperature)} = 15text(°C)`

   `text(1 day) => text(maximum temperature is below)\ 15text(°C)`
 

b.    `text(Number of days)` `= 0.50 xx 30`
    `= 15\ text(days)`

CORE, FUR2 2019 VCAA 1

Table 1 shows the day number and the minimum temperature, in degrees Celsius, for 15 consecutive days in May 2017.
 

  1. Which of the two variables in this data set is an ordinal variable?   (1 mark)

The incomplete ordered stem plot below has been constructed using the data values for days 1 to 10.

  1. Complete the stem plot above by adding the data values for days 11 to 15.   (1 mark)

  2. The ordered stem plot below shows the maximum temperature, in degrees Celsius, for the same 15 days.

     

     

    Use this stem plot to determine

  3.  i. the value of the first quartile `(Q_1)`   (1 mark)

  4. ii. the percentage of days with a maximum temperature higher than 15.3 °C.   (1 mark)

Show Answers Only
  1. `text(Day number)`
  2. `text(See Worked Solutions)`
    1. `12.2`
    2. `text(20%)`
Show Worked Solution

a.   `text(Day number)`

b.  

 
c.i.   `15\ text(data points)`

 `Q_1 = 12.2\ text{(4th data point)}`
 

c.ii.    `text(% Days)` `= 3/15 xx 100`
    `= 20%`

GRAPHS, FUR1 2019 VCAA 7 MC

The feasible region for a linear programming problem is shaded in the diagram below.
 


 

The equation of the objective function for this problem is of the form
 

`P = mx + ny`,    where  `m > 0`  and  `n > 0`
 

The dotted line passing through the points `V` and `W` in the diagram has the same slope as the objective function for this problem.

The minimum value of the objective function can be determined by calculating its value at

  1. point `S` only.
  2. point `T` only.
  3. point `U` only.
  4. any point along the line segment  `RU`.
  5. any point along the line segment  `ST`.
Show Answers Only

`E`

Show Worked Solution

`VW\ text(||)\ ST`

`:.\ text(Minimum value is any point along)\ ST`

`=>  E`

GRAPHS, FUR1 2019 VCAA 5 MC

Giovanni makes and sells toy train sets.

Each toy train set costs $40.00 to make and sells for $65.00

Giovanni also has a fixed cost for making a batch of toy train sets.

Last Sunday, Giovanni sold a batch of 35 toy train sets for a profit of $250.

Giovanni’s fixed cost for this batch of train sets is

  1.    $250
  2.    $625
  3.    $875
  4.  $1400
  5.  $2275
Show Answers Only

`B`

Show Worked Solution

`text(Revenue) = 35 xx 65 = $2275`

`text(C)text(ost)` `= F + 35 xx 40`
  `= F + 1400`

 

`text(Profit)` `=\ text(Revenue – costs)`
`250` `= 2275 – (F + 1400)`
`F` `= 2275 – 1400 – 250`
  `= $625`

 
`=>  B`

GRAPHS, FUR1 2019 VCAA 4 MC

The graph below shows a straight line that crosses the `x`-axis at  `(-10, 0)`, passes through the point  `(-6, 8)`  and crosses the `y`-axis at  `(0, q)`.
 


 

What is the value of `q`?

  1.  14
  2.  16
  3.  18
  4.  20
  5.  22
Show Answers Only

`D`

Show Worked Solution

`M = (8 – 0)/(-6 + 10) = 2`

`text(Equation of line)\ \ m = 2,\ text(through)\ (-10, 0):`

`y – 0` `= 2(x + 10)`
`y` `= 2x + 20`

 
`text(When)\ \ x = 0, quad y = 20,`

`:. q = 20`

`=>  D`

GRAPHS, FUR1 2019 VCAA 3 MC

A shop sells two types of discs: compact discs (CDs) and digital video discs (DVDs).

CDs are sold for $7.00 each and DVDs are sold for $13.00 each.

Bonnie bought a total of 16 discs for $178.00

How many DVDs did Bonnie buy?

  1.   3
  2.   5
  3.   7
  4.   9
  5. 11
Show Answers Only

`E`

Show Worked Solution
`text(Let)\ \ a` `=\ text(number of CDs sold)`
`b` `=\ text(number of DVDs sold)`
`a + b` `=\ text{16  … (1)}`
`7a + 13b` `=\ text{178  … (2)}`

 
`text(Solve simultaneously:)`

`a = 5`

`b = 11`

`=>  E`

Calculus, EXT2 C1 2008 HSC 1e

It can be shown that

`qquad (8(1-x))/((2-x^2)(2-2x+x^2)) = (4-2x)/(2-2x+x^2)-(2x)/(2-x^2)`    (do NOT prove this)
 

Use this result to evaluate  `int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`   (4 marks)

Show Answers Only

`pi/2`

Show Worked Solution

`int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`

`=int_0^1 (4-2x)/(2-2x+x^2)\ dx-int_0^1 (2x)/(2-x^2)\ dx`

`=int_0^1 (2-(2x-2))/(2-2x+x^2)\ dx + [log_e|2-x^2|]_0^1`

`=int_0^1 2/(1+(1-x)^2)-int_0^1 (2x-2)/(2-2x+x^2) + [log_e|2-x^2|]_0^1`

`=[-2tan^(-1)(1-x)]_0^1-[log_e(2-2x+x^2)]_0^1 + [log_e|2-x^2|]_0^1`

`=-2tan^(-1)0+2tan^(-1) 1-(log_e1-log_e2) + (log_e 1-log_e2)`

`=0+2 xx pi/4 + log_e2-log_e 2`

`=pi/2`

Calculus, EXT2 C1 2003 HSC 1d

  1.  Find the real numbers  `a`  and  `b`  such that
     
    `qquad (5x^2-3x+13)/((x-1)(x^2+4)) ≡ a/(x-1) + (bx-1)/(x^2+4)`   (2 marks)

  2.  Hence find  `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx`   (2 marks)

Show Answers Only
  1. `a = 3, b = 2`
  2. `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`
Show Worked Solution

i.   `(5x^2-3x+13)/((x-1)(x^2+4)) ≡ (a(x^2+4) + (bx-1)(x-1))/((x-1)(x^2+4))`

 
`text(Equating numerators:)`

`5x^2-3x+13` `=ax^2+4a+bx^2-bx-x+1`  
  `=(a+b)x^2+(-b-1)x+4a+1`  

 

`-b-1` `=-3\ \ =>\ \ b=2`  
`a` `=3`  

 

ii.    `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx` `=int 3/(x-1)\ dx-int (2x)/(x^2+4)\ dx-int 1/(4+x^2)\ dx`
    `=3 log_e|x-1|-log_e |x^2+4|-1/2 tan^(-1)(x/2)+C`

GEOMETRY, FUR1 2019 VCAA 6 MC

All cities in China are in the same time zone.

Wuhan (31° N, 114° E) and Chengdu (31° N, 104° E) are both cities in China.

On one day in January, the sun set in Wuhan at 5.50 pm.

Assuming that 15° of longitude equates to a one-hour time difference, the time the sun set in Chengdu on the same day is

  1. 4.20 pm
  2. 5.10 pm
  3. 5.50 pm
  4. 6.30 pm
  5. 7.20 pm
Show Answers Only

`D`

Show Worked Solution

`text(Longitudinal difference)`

`= 114 – 104`

`= 10^@`
 

`text(Time difference)` `= 10/15 xx 60`
  `= 40\ text(minutes)`

 
`text(Wuhan is further East)`

`=>\ text(Chengdu 40 minutes behind)`
  

`:.\ text(Sunset in Chengdu)`

`= 5:50\ text(pm) + 40\ text(mins)`

`= 6:30\ text(pm)`
 

`=>  D`

GEOMETRY, FUR1 2019 VCAA 5 MC

In triangle  `ABC`, the point  `D`  lies on the line  `AC`, as shown in the diagram below.
 


 

The length of the line  `BC`  is equal to the length of the line  `BD`.

Which one of the following statements is true?

  1. The angle  `BAD`  is equal to the angle  `ADB`.
  2. The angle  `ADB`  is equal to the angle  `BDC`.
  3. The sum of the angle  `ACB`  and the angle  `ADB`  is equal to 180°.
  4. The sum of the angle  `ABD`  and the angle  `CBD`  is equal to 180°.
  5. The sum of the angle  `ABC`  and the angle  `ACB`  is equal to the angle  `ADB`.
Show Answers Only

`C`

Show Worked Solution


 

`text(Let)\ \ angle ACB = x`

`angle BDC = x\ \ \ text{(angles opposite equal side in isosceles Δ)}`

`angle ACB + angle ADB` `=x+180-x`  
  `=180`  

`=>  C`

GEOMETRY, FUR1 2019 VCAA 4 MC

Triangle `M`, shown below, has side lengths of 3 cm, 4 cm and 5 cm.
 


 

Four other triangles have the following side lengths:

    • Triangle `N` has side lengths of 3 cm, 6 cm and 8 cm.
    • Triangle `O` has side lengths of 4 cm, 8 cm and 12 cm.
    • Triangle `P` has side lengths of 6 cm, 8 cm and 10 cm.
    • Triangle `Q` has side lengths of 9 cm, 12 cm and 15 cm.

The triangles that are similar to triangle `M` are

  1. triangle `N` and triangle `O`.
  2. triangle `N`, triangle `O` and triangle `P`.
  3. triangle `O` and triangle `P`.
  4. triangle `O` and triangle `Q`.
  5.  triangle `P` and triangle `Q`.
Show Answers Only

`E`

Show Worked Solution

`text(Similarity test: all side pairs are in the same ratio.)`

`text(Triangle)\ P:`

`3/6 = 4/8 = 5/10 =>\ text(similar)`
 

`text(Triangle)\ Q:`

`3/9 = 4/12 = 5/15 =>\ text(similar)`
 

`text(Triangles)\ N, O \ => \ text(ratios not equal)`

`=>  E`

GEOMETRY, FUR1 2019 VCAA 2 MC

Town B is located on a bearing of 060° from Town A.

The diagram that could illustrate this is

A.   B.  
C.   D.  
E.      
Show Answers Only

`A`

Show Worked Solution

`=>  A`

Calculus, EXT2 C1 2004 HSC 1d

  1. Find real numbers  `a`  and  `b`  such that
     
    `qquad (x^2-7x+4)/((x+1)(x-1)^2) ≡ a/(x+1) + b/(x-1) - 1/(x-1)^2`   (2 marks)
     
  2. Hence find  `int (x^2-7x+4)/((x+1)(x-1)^2)\ dx`   (2 marks)

 

Show Answers Only
  1. `a = 3, b = -2`
  2. `3 log_e|x+1| – 2log_e |x-1| + 1/(x-1)+C`
Show Worked Solution

i.   `(x^2-7x+4)/((x+1)(x-1)^2) ≡ (a(x-1)^2 +b(x^2-1) -(x+1))/((x+1)(x-1)^2)`

`text(Equating numerators:)`

`x^2-7x+4` `=ax^2-2ax+a+bx^2-b-x-1`  
  `=(a+b)x^2 + (-2a-1)x +a-b-1`  

 

`a+b` `=1\ \ …\ (1)`  
`a-b-1` `=4`  
`a-b` `=5\ \ …\ (2)`  

 
`(1) + (2)`

`2a` `=6`
`a` `= 3`
`b` `=-2`

 

ii.    `int (x^2-7x+4)/((x+1)(x-1)^2)\ dx` `=int 3/(x+1)\ dx – int 2/(x-1)\ dx – int 1/(x-1)^2\ dx`
    `=3 log_e|x+1| – 2log_e |x-1| + 1/(x-1)+C`

NETWORKS, FUR1 2019 VCAA 6 MC

The map below shows all the road connections between five towns, `P, Q, R, S` and `T`.
 


 

The road connections could be represented by the adjacency matrix

A.   `{:(qquad qquad P\ \ Q\ \ R\ \ \ S\ \ \ T), ({:(P),(Q),(R),(S),(T):}[(1, 3, 0, 2, 2),(3,0,1,1,1),(0,1,0,1,0),(2,1,1,0,2),(2,1,0,2,0)]):}` B.   `{:(qquad qquad P\ \ Q\ \ R\ \ \ S\ \ \ T), ({:(P),(Q),(R),(S),(T):}[(1, 2, 0, 2, 2),(2,0,1,1,1),(0,1,0,1,0),(2,1,1,0,2),(2,1,0,2,0)]):}`
       
C.   `{:(qquad qquad P\ \ Q\ \ R\ \ \ S\ \ \ T), ({:(P),(Q),(R),(S),(T):}[(0, 3, 0, 2, 2),(3,0,1,1,1),(0,1,0,1,0),(2,1,1,0,2),(2,1,0,2,0)]):}` D.   `{:(qquad qquad P\ \ Q\ \ R\ \ \ S\ \ \ T), ({:(P),(Q),(R),(S),(T):}[(0, 2, 0, 2, 2),(2,0,1,1,1),(0,1,0,1,0),(2,1,1,1,2),(2,1,0,2,0)]):}`
       
E.   `{:(qquad qquad P\ \ Q\ \ R\ \ \ S\ \ \ T), ({:(P),(Q),(R),(S),(T):}[(1, 2, 0, 2, 2),(2,0,1,1,1),(0,1,0,1,0),(2,1,1,1,1),(2,1,0,1,0)]):}`    

 

Show Answers Only

`A`

Show Worked Solution

`text(Consider town)\ P:`

`text(By Elimination,)`

`P\ \ text(to)\ \ P = 1 quad text{(a loop exists)} \ -> \ text(Eliminate)\ C, D`

`P\ \ text(to)\ \ Q = 3 \ -> \ text(Eliminate)\ B, E`

`=>  A`

Networks, STD2 N2 SM-Bank 34

The following diagram shows the distances, in metres, along a series of cables connecting a main server to seven points, `A` to `G`, in a computer network.
 


 

Calculate the minimum length of cable, in metres, required to ensure that each of the seven points is connected to the main server directly or via another point.   (2 marks)

Show Answers Only

`203\ text(m)`

Show Worked Solution

`text(Using Prim’s Algorithm:)`

`text{Edge 1: main server to D (15)}`

`text{Edge 2: DE (28)}`

`text{Edge 3: EG (30)}`

`text{Edge 4: GF (32)}`

`text{Edge 5: EA (35)}`

`text{Edge 6: AB (28)}`

`text{Edge 7: GC (35)}`
 

`:.\ text(Minimum length)`

`= 15 + 28 + 30 + 32 + 35 + 28 + 35`

`= 203\ text(m)`

NETWORKS, FUR1 2019 VCAA 5 MC

The following diagram shows the distances, in metres, along a series of cables connecting a main server to seven points, `A` to `G`, in a computer network.
 


 

The minimum length of cable, in metres, required to ensure that each of the seven points is connected to the main server directly or via another point is

  1. 175
  2. 203
  3. 208
  4. 221
  5. 236
Show Answers Only

`B`

Show Worked Solution

`text(Using Prim’s Algorithm:)`

`text{Edge 1: main server to D (15)}`

`text{Edge 2: DE (28)}`

`text{Edge 3: EG (30)}`

`text{Edge 4: GF (32)}`

`text{Edge 5: EA (35)}`

`text{Edge 6: AB (28)}`

`text{Edge 7: GC (35)}`
 

`:.\ text(Minimum length)`

`= 15 + 28 + 30 + 32 + 35 + 28 + 35`

`= 203\ text(m)`

`=>  B`

NETWORKS, FUR1 2019 VCAA 4 MC

Two graphs, labelled Graph 1 and Graph 2, are shown below.
 


 

Which one of the following statements is not true?

  1. Graph 1 and Graph 2 are isomorphic.
  2. Graph 1 has five edges and Graph 2 has six edges.
  3. Both Graph 1 and Graph 2 are connected graphs.
  4. Both Graph 1 and Graph 2 have three faces each.
  5. Neither Graph 1 nor Graph 2 are complete graphs.
Show Answers Only

`A`

Show Worked Solution

`text(Graph 1 → 5 edges)`

`text(Graph 2 → 6 edges)`

`:.\ text(Cannot be isomorphic.)`

`=>  A`

NETWORKS, FUR1 2019 VCAA 3 MC

The flow of water through a series of pipes is shown in the network below

The numbers on the edges show the maximum flow through each pipe in litres per minute.
 

 
The capacity of Cut `Q`, in litres per minute, is

  1. 11
  2. 13
  3. 14
  4. 16
  5. 17
Show Answers Only

`C`

Show Worked Solution

`text(Capacity of cut)\ Q`

`= 5 + 6 + 3`

`= 14`

`=>  C`

MATRICES, FUR1 2019 VCAA 7 MC

The communication matrix below shows the direct paths by which messages can be sent between two people in a group of six people, `U` to `Z`.
 

`{:(qquad qquad qquad qquad qquad qquad qquad qquad text(receiver)),(qquad qquad qquad qquad qquad quad U quad V quad W \ \ \ X \ \ \ Y quad \ Z), (text(sender) qquad {:(U),(V),(W),(X),(Y), (Z):}[(0, 1, 1, 0, 1, 1), (1, 0, 1, 0, 1, 0), (1, 1, 0, 1, 0, 1), (0, 1, 0, 0, 1, 1), (0, 0, 1, 1, 0, 1), (1, 1, 0, 1, 1, 0)]):}`
 

A 1 in the matrix shows that the person named in that row can send a message directly to the person named in that column.

For example, the 1 in row 4, column 2 shows that `X` can send a message directly to `V`.

In how many ways can `Y` get a message to `W` by sending it directly to one other person?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`A`

Show Worked Solution

`Y\ text(can send directly to)\ W, X and Z.`

`text(Consider the next message after)\ Y\ text(sends to)`

`W:\ text(can’t send directly to)\ W`

`X:\ text(can’t send directly to)\ W`

`Z:\ text(can’t send directly to W)`

`=>  A`

Combinatorics, EXT1 A1 SM-Bank 5

  1. In how many ways can the letters of COOKBOOK be arranged in a line?  (1 mark)

  2. What is the probability that a random rearrangement of the letters has four O's together?  (2 marks)

Show Answers Only
  1. `840`
  2. `1/14`
Show Worked Solution

i.   `text(Combinations) = (8!)/(4!2!) = 840`

 

ii.   `text(Treat four O’s as one letter)`

`text(Combinations) = 5! = 120`

`text(Adjusting for 2 Ks:)`

`text(Combinations) = (5!)/(2!) = 60`

`:. P(text(4 O’s together))`

`= 60/840`

`= 1/14`

Combinatorics, EXT1 A1 EQ-Bank 4

How many numbers greater than 6000 can be formed with the digits 1, 4, 5, 7, 8 if no digit is repeated.  (2 marks)

Show Answers Only

`168`

Show Worked Solution

`text(4 digit numbers:)`

`text(Must start with 7 or 8)`

`text(Combinations (4 digits)) = 2 xx\ ^4P_3 = 2 xx 4 xx 3 xx 2 xx 1= 48`
 

`text(5 digit numbers:)`

`text(Combinations) = 5! = 120`

`:.\ text(Total combinations = 168)`

MATRICES, FUR1 2019 VCAA 6 MC

A water park is open from 9 am until 5 pm.

There are three activities, the pool `(P)`, the slide `(S)` and the water jets `(W)`, at the water park.

Children have been found to change their activity at the water park each half hour, as shown in the transition matrix, `T`, below.
 

`{:(qquad qquad qquad quadtext(this half year)),(qquad qquad qquad quad P qquad qquad S qquad quad W), (T = [(0.80,0.20,0.40),(0.05,0.60,0.10),(0.15,0.20,0.50)] {:(P),(S),(W):} text( next half year)):}`
 

A group of children has come to the water park for the whole day.

The percentage of these children who are expected to be at the slide `(S)` at closing time is closest to

  1.  14%
  2.  20%
  3.  24%
  4.  25%
  5.  62%
Show Answers Only

`A`

Show Worked Solution

`text(Park is open 8 hours → 15 × 30-minute transitions.)`

`text(After 15 transitions, matrix approaches a steady state.)`
 

`[(0.80,0.20,0.40),(0.05,0.60,0.10),(0.15,0.20,0.50)]^15 ~~ [(0.62,0.62,0.62),(0.14,0.14,0.14),(0.24,0.24,0.24)]`
 

`=>  A`

MATRICES, FUR1 2019 VCAA 5 MC

`[(2, 2, 0, -2), (2, 0, -2, 0), (0, 2, 0, -2), (2, 2, -2, 0)][(v), (w), (x), (y)]=[(2), (6), (-8), (4)]`
 

Which one of the following systems of simultaneous linear equations best represents the matrix equation above?

A.   `2v + 2w - 2y` `= 2` B.   `2v + 2w - 2y` `= 2`
  `2v - 2x` `= 6`   `2v - 2x` `= 6`
  `2v - 2y` `= -8`   `2w - 2y` `= -8`
  `2v + 2w` `= 4`   `2v + 2w - 2x` `= 4`
           
C.   `2v + 2w + 2y` `= 2` D.   `2v + 2w - 2y` `= 2`
  `2v - 2x` `= 6`   `2v - 2x` `= 6`
  `2w - 2y` `= -8`   `2w - 2y` `= -8`
  `2v + 2w - 2x` `= 4`   `2v + 2w + 2x` `= 4`
           
E.   `2v + 2w - 2y` `= 2`      
  `2v - 2w` `= 6`      
  `2w - 2y` `= -8`      
  `2v + 2w - 2x` `= 4`      
Show Answers Only

`B`

Show Worked Solution

`text(Creating equations from the matrix:)`

`2v + 2w – 2y` `= 2`
`2v – 2x` `= 6`
`2w – 2y` `= -8`
`2v + 2w – 2x` `= 4`

 

`=>  B`

MATRICES, FUR1 2019 VCAA 4 MC

Stella completed a multiple-choice test that had 10 questions.

Each question had five possible answers, `A, B, C, D` and `E`.

For question number one, Stella chose the answer `E`.

Stella chose each of the nine remaining answers, in order, by following the transition matrix, `T`, below
 

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

What answer did Stella choose for question number six?

  1. `A`
  2. `B`
  3. `C`
  4. `D`
  5. `E`
Show Answers Only

`E`

Show Worked Solution

`text(1st question) -> E`

`text(2nd question) -> C`

`text(3rd question) -> A`

`text(then)\ \ D -> B -> E`
 

`=>  E`

MATRICES, FUR1 2019 VCAA 3 MC

Consider the matrix `P`, where  `P = [(3, 2, 1), (5, 4, 3)]`.

The element in row `i` and column `j` of matrix `P` is  `p_(ij)`.

The elements in matrix `P` are determined by the rule

  1. `p_(ij) = 4 - j`
  2. `p_(ij) = 2i + 1`
  3. `p_(ij) = i + j + 1`
  4. `p_(ij) = i + 2j`
  5. `p_(ij) = 2i - j + 2`
Show Answers Only

`E`

Show Worked Solution

`text(Consider option E):`

`P_11 = 2 xx 1 – 1 + 2 = 3`

`P_12 = 2 xx 1 – 2 + 2 = 2`

`P_13 = 2 xx 1 – 3 + 2 = 1`

`P_21 = 2 xx 2 – 1 + 2 = 5`

`text(etc…)`

`=>  E`

CORE, FUR1 2019 VCAA 22 MC

A machine is purchased for $30 000

It produces 24 000 items each year.

The value of the machine is depreciated using a unit cost method of depreciation.

After three years, the value of the machine is $18 480.

A rule for the value of the machine after `n` units are produced, `V_n`, is

  1. `V_n = 0.872 n`
  2. `V_n = 24\ 000 n - 3840`
  3. `V_n = 30\ 000 - 24\ 000 n`
  4. `V_n = 30\ 000 - 0.872 n`
  5. `V_n = 30\ 000 - 0.16 n`
Show Answers Only

`E`

Show Worked Solution
`text(Unit cost)` `= (30\ 000 – 18\ 480)/(3 xx 24\ 000)`
  `= 0.16`

 
`:. V_n = 30\ 000 – 0.16 n`

`=>  E`

CORE, FUR1 2019 VCAA 20 MC

Consider the following amortisation table for a reducing balance loan.
 


 

The annual interest rate for this loan is 3.6%.

Interest is calculated immediately before each payment.

For this loan, the repayments are made

  1. weekly.
  2. fortnightly
  3. monthly.
  4. quarterly.
  5. yearly
Show Answers Only

`C`

Show Worked Solution

`text(Consider payment #1:)`

`text(Interest rate)` `= 900/(300\ 000)`
  `= 0.3%`

 
`text(S) text(ince annual rate = 3.6%, repayments are monthly.)`

`=>  C`

CORE, FUR1 2019 VCAA 18 MC

The value of a compound interest investment, in dollars, after `n` years, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 100\ 000, qquad V_(n + 1) = 1.01 V_n`

The interest rate, per annum, for this investment is

  1.     `0.01 text(%)`
  2.     `0.101 text(%)`
  3.     `1 text(%)`
  4.     `1.01 text(%)`
  5. `101 text(%)`
Show Answers Only

`C`

Show Worked Solution

`V_(n + 1) = 1.01 V_n`

`text(Interest) = 1%`

`=>  C`

CORE, FUR1 2019 VCAA 15-16 MC

The time series plot below shows the monthly rainfall at a weather station, in millimetres, for each month in 2017.
 


 

Part 1

The median monthly rainfall for 2017 was closest to

  1.  53 mm
  2.  82 mm
  3.  96 mm
  4. 103 mm
  5. 111 mm

 
Part 2

If seven-mean smoothing is used to smooth this time series plot, the number of smoothed data points would be

  1.  3
  2.  5
  3.  6
  4.  8
  5. 10
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(12 data points)`

`text(Median)` `= text{(6th + 7th)}/2`
  `~~ (97 + 109)/2`
  `~~ 103\ text(mm)`

`=>  D`
 

`text(Part 2)`

`text(January – July)\ =>\ text(First data point)`

`text(February – August)\ =>\ text(S)text(econd data point)`

`qquad vdots`

`text(June – December)\ =>\ text(Sixth data point)`

`=>  C`