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MATRICES, FUR1 2006 VCAA 2 MC

Let  `A = [(-2), (0)], B = [(0, 9)]`  and  `C = [2]`

Using these matrices, the matrix product that is not defined is

A.   `AB`

B.   `AC`

C.   `BA`

D.   `BC`

E.   `CB`

Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ D,`

   `B` `xx`  `C`
`1 xx 2`    `1 xx 1`

 

`text(Columns in)\ B !=\ text(Rows in)\ C`

`:. BC\ text(is undefined.)`

`rArr D`

MATRICES, FUR1 2006 VCAA 1 MC

The matrix `[(12, 36), (0, 24)]` is equal to

A.    `12[(0,3),(0,2)]` B.    `12[(1,3),(0,2)]`
       
C.    `12[(0,24),(−12,12)]` D.    `12[(0,24),(0,12)]`
       
E.    `12[(1,3),(−12,2)]`    
Show Answers Only

`B`

Show Worked Solution
`12[(1,3),(0,2)]` `= [(12 xx 1, 12 xx 3),(12 xx 0,12 xx 2)]`
  `= [(12,36),(0,24)]`

`rArr B`

MATRICES, FUR2 2014 VCAA 1

A small city is divided into four regions: Northern (`N`), Eastern (`E`), Southern (`S`) and Western (`W`).

The number of adult males (`M`) and the number of adult females (`F`) living in each of the regions in 2013 is shown in matrix `V` below.

`{:(qquadqquadqquadqquadMqquadqquadF),(V = [(1360,1460), (1680,1920), (900,1060), (1850,1770)]):}{:(),(N), (E), (S), (W):}`

  1. Write down the order of matrix `V`.   (1 mark)

  2. How many adult males lived in the Western region in 2013?   (1 mark)

  3. In terms of the population of the city, what does the sum of the elements in the second column of matrix `V` represent?   (1 mark)

An election is to be held in the city.

All of the adults in each of the regions of the city will vote in the election.

One of the election candidates, Ms Aboud, estimates that she will receive 45% of the male votes and 55% of the female votes in the election.

This information is shown in matrix `P` below.

`P = [(0.45), (0.55)]{:(M), (F):}`

  1. Explain, in terms of rows and columns, why the matrix product `V × P` is defined.   (1 mark)

The product of matrices `V` and `P` is shown below.

`V xx P = [(1360     1460), (1680     1920), (    900     1060), (1850     1770)] xx [(0.45),(0.55)] = [(w), (1812), (988), (1806)]`

  1. Using appropriate elements from the matrix product `V × P`, write a calculation to show that the value of `w` is `1415`.   (1 mark)

  2. How many votes does Ms Aboud expect to receive in the election?   (1 mark) 
Show Answers Only
  1. `4 xx 2`
  2. `1850`
  3. `text(The total number of adult females living in the small city.)`
  4. `text(It is defined because the number)`
    `text(of columns in matrix)\ V\ text(equals the)`
    `text(number of rows in matrix)\ P.`
  5. `1415`
  6. `6021`
Show Worked Solution

a.   `4 xx 2`
 

b.   `1850`
 

c.   `text(The total number of adult females living in the small city.)`
 

d.   `text(It is defined because the number of columns in)`

`text(matrix)\ V\ text(equals the number of rows in matrix)\ P.`

 

e.    `w` `= 1360 xx 0.45 + 1460 xx 0.55`
    `= 1415`

 

f.   `text(Expected votes for Ms Aboud)`

`= 1415 + 1812 + 988 + 1806`

`= 6021`

GRAPHS, FUR2 2012 VCAA 1

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

GRAPHS, FUR2 2012 VCAA 1

    1. Calculate the gradient of the line, `C`, drawn above.  (1 mark)
    2. Write an equation for the cost, `C`, in dollars, of making `n` phones.  (1 mark)

  2. The revenue, `R`, in dollars, obtained from selling `n` phones is given by  `R = 150n`.

    1. Draw this line on the graph above.  (1 mark)
    2. How many phones would need to be sold to obtain $54 000 in revenue?  (1 mark)
  3. Determine the number of phones that would need to be made and sold to break even.  (1 mark) 
Show Answers Only
    1. `50`
    2. `C = 20\ 000 + 50n`
    1. `text(See Worked Solutions)`
    2. `360`
  3. `200`
Show Worked Solution

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

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

 

a.ii.   `C = 50n + 20\ 000`

 

b.i.    GRAPHS, FUR2 2012 VCAA 1 Answer

 

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

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

 

c.   `text(Breakeven occurs when)`

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

GEOMETRY, FUR2 2012 VCAA 4

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

Triangle `OAB` is a right-angled triangle. 

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

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

Geometry and trigonometry, FUR2 2012 VCAA 4
 

  1. Calculate the length, `OA`.

     

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

  2. Determine the area of triangle `OAB`.

     

    Write your answer, in m², correct to one decimal place.  (1 mark)

  3. Triangles `OBC` and `OCD` are similar.

     

    The area of triangle `OBC` is 35 m².

     

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

  4. Determine angle `CDO`.

     

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

Show Answers Only
  1. `8.66\ text{m  (2 d.p.)}`
  2. `21.7\ text{m²  (1 d.p.)}`
  3. `68.6\ text(m²)`
  4. `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 1

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

  1. Calculate the area of this block of land.

     

    Write your answer in m².  (1 mark)

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

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

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

GEOMETRY, FUR2 2012 VCAA 11

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

     

    Write your answer in m³.  (1 mark)

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

Geometry and trigonometry, FUR2 2012 VCAA 1_2

  1. Calculate the total length of fencing that will be required. 

     

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

Show Answers Only
  1. `4250\ text(m²)`
  2. `1000\ text(m³)`
  3. `90.6\ text(m)`
Show Worked Solution
a.    `text(Area)` `= 50 xx 85`
    `= 4250\ text(m²)`

 

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

 

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

 

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

GEOMETRY, FUR2 2012 VCAA 1

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

 

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

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

`= 2 xx 20 + 2 xx 25.317…`

`= 90.635…`

`= 90.6\ text(m)`

CORE*, FUR2 2013 VCAA 1

Hugo is a professional bike rider.

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

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

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

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

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

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

Show Answers Only

  1. `$8000`
  2. i.  `text(See Worked Solutions)`
    ii. `$500`
  3. `12\ 000\ text(km)`

Show Worked Solution

a.   `$8000`
  

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

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

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

`= 8000-1500n`

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

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

  
c.   `text(After 2 years,)`

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

  
`:.\ text(Distance travelled)`

`= 3000/0.25`

`= 12\ 000\ text(km)`

GRAPHS, FUR2 2013 VCAA 2

Students at the camp can participate in two different watersport activities: canoeing and surfing.

The cost of canoeing is $30 per hour and the cost of surfing is $20 per hour.

The budget allows each student to spend up to $200, in total, on watersport activities.

The way in which a student decides to spend the $200 is described by the following inequality.

30 × hours canoeing + 20 × hours surfing ≤ 200

  1. Hillary wants to spend exactly two hours canoeing during the camp.

     

    Calculate the maximum number of hours she could spend surfing.  (1 mark)

  2. Dennis would like to spend an equal amount of time canoeing and surfing.

     

    If he spent a total of $200 on these activities, determine the maximum number of hours he could spend on each activity.  (1 mark)

Show Answers Only
  1. `7`
  2. `4\ text(hours)`
Show Worked Solution

a.   `text(Hillary spends 2 hrs canoeing)`

`text(Let)\ x = text(hours spent surfing)`

`30 xx 2 + 20x` `= 200`
`20x` `= 140`
`x` `= 7`

`:.\ text(Maximum hours surfing = 7)`

 

b.   `text(Let)\ t = text(time spent canoeing and surfing)`

`30 xx t + 20 xx t` `= 200`
`50t` `= 200`
`t` `= 4\ text(hours)`

`:.\ text(Dennis could spend a maximum)`

`text(of 4 hours on each activity.)`

GRAPHS, FUR2 2013 VCAA 1

The distance-time graph below shows the first two stages of a bus journey from a school to a camp.

GRAPHS, FUR2 2013 VCAA 1

  1. At what constant speed, in kilometres per hour, did the bus travel during stage 1 of the journey?  (1 mark)
  2. For how many minutes did the bus stop during stage 2 of the journey?  (1 mark)

The third stage of the journey is missing from the graph.

During stage 3, the bus continued its journey to the camp and travelled at a constant speed of 60 km/h for one hour.

  1. Draw a line segment on the graph above to represent stage 3 of the journey.  (1 mark)
  2. Find the average speed of the bus over the three hours.

    Write your answer in kilometres per hour.  (1 mark)

The distance, `D` km, of the bus from the school, `t` hours after departure is given by

`D = {(100t                      0 ≤ t ≤ 1.5), (150                        1.5 ≤ t ≤ 2), (60t + k            2≤ t ≤ 3):}`

  1. Determine the value of `k`.  (1 mark)
Show Answers Only
  1. `text(100 km/hr)`
  2. `text(30 minutes)`
  3. `text(See Worked Solutions)`
  4. `70\ text(km/hr)`
  5. `30`
Show Worked Solution

a.   `text(100 km/hr)`

 

b.   `text(30 minutes)`

 

c.    GRAPHS, FUR2 2013 VCAA 1 Answer

 

d.   `text(Average speed over 3 hours)`

♦ Mean mark of parts (c) and (d) consumed was 46%.

`= (text(Total distance))/3`

`= 210/3`

`= 70\ text(km/hr)`

 

e.   `text(When)\ t = 2,`

`D = 150,\ text(and)`

`D = 60 xx 2 + k`

`:. 120 + k` `= 150`
`:. k` `= 30`

GEOMETRY, FUR2 2013 VCAA 1

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

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

Geometry and Trig, FUR2 2013 VCAA 1_1
 

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

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

The angle `XGY` is 113°.
 

Geometry and Trig, FUR2 2013 VCAA 1_2
 

    1. Calculate the distance, `XY`, correct to the nearest metre.  (1 mark)
    2. Find the area of triangle `XGY`, correct to the nearest square metre.  (1 mark)
  2. Determine the angle of elevation of spectator `S` from competitor `Y`, correct to the nearest degree.

    Note that `X`, `G` and `Y` are on the same horizontal level.  (1 mark)
     

Geometry and Trig, FUR2 2013 VCAA 1_3

Show Answers Only
  1. `42\ text{m  (nearest m)}`
    1. `77\ text{m  (nearest m)}`
    2. `957\ text{m²  (nearest m²)}`
  3. `14^@\ \ text{(nearest degree)}`
Show Worked Solution

a.   `text(Using Pythagoras:)`

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

 

b.i.   `text(Using the cosine rule:)`

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

 

b.ii.   `text(Using the sine rule:)`

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

 

c.    Geometry-FUR2-2013-VCAA-1 Answer
`theta` `= text(angle of elevation of)\ S\ text(from)\ Y`
`tantheta` `= 13/52`
`:. theta` `= tan^(−1)\ 1/4`
  `= 14.036…`
  `= 14^@\ \ text{(nearest degree)}`

CORE*, FUR2 2014 VCAA 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.

  1. What is the annual rate of interest for this perpetuity investment?   (1 mark)

  2. After 10 years, how much money is still invested in the perpetuity?   (1 mark)

  1. 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.
    2. What is the expected price of this cricket equipment in 2015?   (1 mark)

    3. 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

  1. `3.75`
  2. `$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 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 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.

GRAPHS, FUR2 2014 VCAA 11

  1. How many kilograms of phosphorus are in 2 kg of Booster?  (1 mark)
  2. 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.

  1.           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.

  1. 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.

GRAPHS, FUR2 2014 VCAA 12 

    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
  1. `0.04\ text(kg)`
  2. `25\ text(kg)`
  3. `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)`
  6.  
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.   

GRAPHS, FUR2 2014 VCAA 1 Answer1

 

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).}`

GRAPHS, FUR2 2014 VCAA 1 Answer2 

CORE*, FUR2 2015 VCAA 3

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

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

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

  1. Determine the minimum amount they must invest in the perpetuity to fund the scholarship.   (1 mark)

  2. For how many years will they be able to provide the scholarship?   (1 mark)

Show Answers Only
  1. `12\ 500`
  2. `text(It will last forever.)`
Show Worked Solution

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

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

 

b.   `text(It will last forever.)`

CORE*, FUR2 2015 VCAA 1

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

The business charges customers $88 per hour.

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

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

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

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

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

  1. Calculate the new hourly rate (including GST).  (1 mark)
Show Answers Only
  1. `$8`
  2. `$352`
  3. `$99`
Show Worked Solution

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

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

 

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

 

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

 

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

GRAPHS, FUR2 2015 VCAA 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. 

Graphs, FUR2 2015 VCAA 4

  1. 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)):} :}`

 

  1. Find the online cost for 30 kg of luggage.  (1 mark)
  2. 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.)

  1. 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
  1. `$250`
  2. `$300`

  3.  

    Graphs-FUR2-2015-VCAA-4i

  4. `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)`

Graphs-FUR2-2015-VCAA-4i

 

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

`text(occurs when)`

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

GRAPHS, FUR2 2015 VCAA 3

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

Graphs, FUR2 2015 VCAA 3

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

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

yen = 95 × dollars – k

  1. Use the point `(100, 8075)` to show that the value of `k` is 1425.  (1 mark)
    1. Determine the intercept on the horizontal axis.  (1 mark)
    2. Interpret the intercept on the horizontal axis in the context of converting dollars to yen.  (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
    1. `$15`
    2. `text(There is a $15 fixed charge for)`

       

      `text(any conversion of dollars into yen.)`

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

 

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

 

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

 

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

GRAPHS, FUR2 2015 VCAA 2

Ben will use a currency exchange agency to buy some Japanese yen (the Japanese currency unit).

The graph below shows the relationship between Japanese yen and Australian dollars on a particular day.

This graph can be used to calculate a conversion between dollars and yen on that day.

Graphs, FUR2 2015 VCAA 2

  1. Ben converts his dollars into yen using this graph.

     

    How many yen does he receive for $200?  (1 mark)

  2. The slope of this graph is the exchange rate for converting dollars into yen on that particular day.

     

    How many yen will Ben receive for each dollar?  (1 mark)

Show Answers Only
  1. `18\ 000\ text(yen)`
  2. `90\ text(yen)`
Show Worked Solution

a.   `18\ 000\ text(yen)`

 

b.   `text(Using)\ \ (0, 0) and (200, 18\ 000)`

`text(Slope)` `= (y_2 – y_1)/(x^2 – x^1)`
  `= (18\ 000 – 0)/(200 – 0)`
  `= 90`

 

`:.\ text(Ben will receive 90 yen for)`

`text(each dollar.)`

GRAPHS, FUR2 2015 VCAA 1

Ben is flying to Japan for a school cultural exchange program.

The graph below shows the cost of a particular flight to Japan, in dollars, on each day in February.

Graphs, FUR2 2015 VCAA 1

  1. What is the cost, in dollars, of this flight to Japan on 19 February?  (1 mark)
  2. On how many days in February is the cost of this flight to Japan more than $1000?  (1 mark) 
Show Answers Only
  1. `$1200`
  2. `8`
Show Worked Solution

a.   `$1200`

b.   `8`

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.
 

VCAA 2014 3a

  1. 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)

    1. 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)

    2. 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)

    1. 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]`?
    2. Express the answer in milligrams per litre per hour, correct to two decimal places.   (2 marks)
    3. 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]`.
    4. 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^+`.

  1. 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
  1. `0.61\ text(mg/L)`
  2.  i. `0.33\ text(hours)`
  3. ii. `0.86\ text(hours)`
  4.  i. `−0.23\ text(mg/L/h)`
  5. ii. `t = 0.90\ text(or)\ t = 2.12`
  6. `4`
Show Worked Solution

a.  `text(Solve:)\ \ c^{′}(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.

VCAA 2014 2a

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.

  1. Write an expression for `h` in terms of `d`.   (2 marks)

  2. 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.

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

  2. Find the value of `h` when `S` is a minimum.   (1 mark)

Show Answers Only
  1. `h = 864/(pi d^2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `12/(pi^(1/3))\ text(m);quad108pi^(1/3)\ text(square metres)`
  4. `6/(root(3)(pi))m`
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 (d) 42%.
MARKER’S COMMENT: An exact answer required here!

`h= 864/(pi(12/(pi^(1/3)))^2)`

   `= 6/(root(3)(pi))\ \ text(m)`

Algebra, MET2 2014 VCAA 1

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

  1. Find the period and amplitude of the function `n`.   (2 marks)

  2. Find the maximum and minimum populations of wombats in this location.   (2 marks)

  3. Find  `n(10)`.   (1 mark)

  4. Over the 12 months from 1 March 2013, find the fraction of time when the population of wombats in this location was less than  `n(10)`.   (2 marks)

Show Answers Only
  1. `text(Period) = text(6 months);\ text(Amplitude) = 400`
  2. `text(Max) = 1600;\ text(Min) = 800`
  3. `1000`
  4. `1/3`
Show Worked Solution

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

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

`text(A)text(mplitude) = 400`
  

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

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

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

d.    `text(Solve)\ n(t)` `= 1000\ text(for)\ t ∈ [0,12]`

`t= 2,4,8,10`

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

♦ Mean mark 48%.

`=> n(t) < 1000\ \ text(for)`

`t ∈ (2,4)\ text(or)\ t ∈ (8,10)`

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

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.
    3. Find the probability that exactly one of the oranges has a diameter greater than 7 cm.
    4. Express the answer in the form `a/b`, where `a` and `b` are positive integers.   (2 marks)

  2. 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.

  1. 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.
    3. 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`
  2. `36/5\ text(cm)`
  3. `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 1

Harry operates a mobile pet care service. The call-out fee charged depends on the distance he has to travel to tend to a pet. The call-out fees for distances up to 30 km are shown on the graph below.

GRAPHS, FUR2 2006 VCAA 1

  1. According to this graph

     

    1. what is the call-out fee to travel a distance of 20 km?  (1 mark)
    2. what is the maximum distance travelled for a call-out fee of $10?  (1 mark)

A call-out fee of $50 is charged to travel distances of more than 30 km but less than or equal to 40 km.

  1. Draw this information on the graph above.  (1 mark) 
Show Answers Only
    1. `$30`
    2. `5\ text(km)`
  2.  
Show Worked Solution

a.i.   `$30`

 

a.ii.   `text(5 km)`

 

b.   

Calculus, MET2 2015 VCAA 1

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

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

VCAA 2015 1ai

  1. Write down the derivative  `f^{prime} (x)` of `f (x)`.   (1 mark)

  2.  i. Find the equation of the tangent to the graph of  `f` at the point  `P(1, 4/5)`.   (1 mark)

    ii. Find the coordinates of points `Q` and `S`.   (2 marks)

  3. Find the distance `PS` and express it in the form  `sqrt b/c`, where `b` and `c` are positive integers.  (2 marks)

VCAA 2015 1di

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

Show Answers Only
  1. `-3/5(x-4)(x-2)`
  2.  i.`y = -9/5x + 13/5`
    ii. `S (0, 13/5), \ \ Q (13/9, 0)`
  3. `sqrt 106/5`
  4. `108/5\ text(units²)`
Show Worked Solution
a.    `f(x)` `= 1/5 (x-2)^2 (5-x)`
  `f^{prime}(x)` `=1/5 xx 2(x-2)(5-x)-1/5 (x-2)^2`
    `= -3/5(x-4)(x-2)`

 

b.i.   `text(Solution 1)`

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

`text(Solution 2)`

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

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

  
b.ii.   `text(At)\ S,\ \ x=0`

`y =-9/5 xx 0 + 13/5 = 13/5`

`:. S(0,13/5)`
  

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

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

 
`:. Q(13/9,0)`
  

c.   `P(1, 4/5),\ \ S(0,13/5)`

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

 

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

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

`text{Solve (using technology):}`

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

`x = 1,7`

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

GEOMETRY, FUR2 2006 VCAA 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°.

GEOMETRY, FUR2 2006 VCAA 11 
 

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

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

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

     

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

  3. The bearing of `C` from `A` is 078°.

     

    Determine the bearing of `B` from `A`.  (1 mark)

  4. Determine the shortest distance from `X` to `C`.

     

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

  5. 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.

GEOMETRY, FUR2 2006 VCAA 12

  1. Determine the length of the fence `MN`.  (2 marks)
Show Answers Only
  1. `45^@`
  2. `77.8\ text(m)`
  3. `033^@`
  4. `102.9\ text(m)`
  5. `12\ 601\ text(m²)`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `33.7^@`
  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.)}`

 

c.   VCAA GEO FUR2 2006 1ci

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

`= 78 – 45`

`= 033^@`

 

d.   VCAA GEO FUR2 2006 1di

`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.

GEOMETRY, FUR2 2007 VCAA 11 

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.

GEOMETRY, FUR2 2007 VCAA 12

  1. Determine the length, in cm, of `QW`.  (1 mark)
  2. Calculate the angle `WAQ`. Write your answer in degrees, correct to one decimal place.  (1 mark)
  3. Calculate the angle `AWB` correct to one decimal place.  (1 mark)
  4. 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.

GEOMETRY, FUR2 2007 VCAA 2

  1. Show that, correct to the nearest centimetre, length `AW` is 34 cm.  (1 mark)
  2. 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.

GEOMETRY, FUR2 2007 VCAA 3

  1. Calculate the volume, in cm³, of the pyramid `ABCDY` in Figure 4.  (1 mark)
  2. Show that, correct to the nearest cm, length `AY` is 37 cm.  (2 marks)
  3. 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

  1. `12\ text(cm)`
  2. `20.6^@`
  3. `41.2^@\ text{(1 d.p.)}`
  4. `1/2`

Question 2

  1. `34\ text(cm)`
  2. `3344\ text(cm²)`

Question 3

  1. `7168\ text(cm³)`
  2. `37\ text(cm)`
  3. `420\ text(cm²)`
Show Worked Solution

`text(Question 1)`

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

 

b.   VCAA GEO FUR2 2007 1bi
`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)`

a.   VCAA GEO FUR2 2007 2ai

`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)`

 

b.   VCAA GEO FUR2 2007 3bi

`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…`

VCAA GEO FUR2 2007 3bii

`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 1

Two subjects, Biology and Chemistry, are offered in the first year of a university science course. 

The matrix `N` lists the number of students enrolled in each subject.

`N = [(460),(360)]{:(text(Biology)),(text(Chemistry)):}`

The matrix `P` lists the proportion of these students expected to be awarded an `A`, `B`, `C`, `D` or `E` grade in each subject.

`{:((qquadqquadqquadA,qquad\ B,qquadquadC,qquad\ D,qquadE)),(P = [(0.05,0.125,0.175,0.45,0.20)]):}`

  1. Write down the order of matrix `P`.   (1 mark)

  2. Let the matrix  `R = NP`.
  3.  i. Evaluate the matrix `R`.   (1 mark)

  4. ii. Explain what the matrix element `R_24`  represents.   (1 mark)

  5. Students enrolled in Biology have to pay a laboratory fee of $110, while students enrolled in Chemistry pay a laboratory fee of $95.
  6.  i. Write down a clearly labelled row matrix, called `F`, that lists these fees.   (1 mark)

  7. ii. Show a matrix calculation that will give the total laboratory fees, `L`, paid in dollars by the students enrolled in Biology and Chemistry. Find this amount.   (1 mark)

Show Answers Only
  1. `1 xx 5`
    1. `[(23,57.5,80.5,207,92),(18,45,63,162,72)]`
    2. `text(See Worked Solutions)`
    1.  `text(See Worked Solutions)`
    2. `text(See Worked Solutions)`
Show Worked Solution

a.   `1 xx 5`
 

b.i.   `R` `= NP`
    `= [(460), (360)] [0.05 quad 0.125 quad 0.175 quad 0.45 quad 0.20]`
    `= [(23,57.5,80.5,207,92),(18,45,63,162,72)]`

 
b.ii. `R_24\ text(represents the number of chemistry students)`

  `text(expected to get a)\ D.`
 

c.i.   `{:(qquad qquad qquad B quad qquad C),(F = [(110, 95)]):}`

 
c.ii.  `text(Total laboratory fees)`

`L = [(110, 95)] [(460), (360)] = [84\ 800]`

 

GEOMETRY, FUR2 2008 VCAA 3

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

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

GEOMETRY, FUR2 2008 VCAA 31
 

  1. 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)
     

GEOMETRY, FUR2 2008 VCAA 32

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

The angle, `TCN`, is 65°.

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

     

    Explain your answer showing appropriate calculations.  (2 marks)

Show Answers Only
  1. `42.7^@`
  2. `text(See Worked Solution)`
  3. `69^@`
  4. `111^@`
  5. `text(See Worked Solutions)`
Show Worked Solution
a.   VCAA GEO FUR2 2008 3ai
`text(Let)\ \ theta` `=\ text(angle of elevation)`
  `= 12/13`
`:. theta` `= 42.709…`
  `= 42.7^@\ text{(1 d.p.)}`

 

b.   VCAA GEO FUR2 2008 3bi

`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)}`

 

d.   VCAA GEO FUR2 2008 3di

`text(Bearing of)\ T\ text(from)\ N`

`= 180 – 69`

`= 111^@`

 

e.   VCAA GEO FUR2 2008 3ei

`ST\ text(is the shortest distance between the)`

MARKER’S COMMENT: Many students had significant difficulty in understanding the 3-dimensional diagram.

`text(tree and the shed.)`

`text(In)\ \ Delta NST,`

`sin 69^@` `= (ST)/12.6`
`:. ST` `= 12.6 xx sin 69^@`
  `= 11.76…\ text(m)`

 

`:.\ text(S) text(ince the tree is 12 m, and 12 m) > 11.76\ text(m),`

`text(it could hit the shed if it falls.)`

GEOMETRY, FUR2 2014 VCAA 2

The chicken coop has two spaces, one for nesting and one for eating.

The nesting and eating spaces are separated by a wall along the line `AX`, as shown in the diagrams below.
 

GEOMETRY, FUR2 2014 VCAA 21
 

`DX = 3.16\ text(m), ∠ADX = 45^@ and ∠AXD = 60^@.`

  1. Write down a calculation to show that the value of  `theta`  is 75°.  (1 mark)
  2. The sine rule can be used to calculate the length of the wall  `AX`.

     

    Fill in the missing numbers below.  (1 mark)
     

     

          GEOMETRY, FUR2 2014 VCAA 22
     

  3. What is the length of  `AX`?

     

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

  4. Calculate the area of the floor of the nesting space, `ADX`.

     

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

The height of the chicken coop is 1.8 m.

 

Wire mesh will cover the roof of the eating space.

 

The area of the walls along the lines  `AB, BC and CX`  will also be covered with wire mesh.

  1. What total area, in square metres, will be covered by wire mesh?

     

    Write your answer, correct to the nearest square metre.  (2 mark)

Show Answers Only
  1. `75^@`
  2. `(AX)/(sin 45^@) = (3.16)/(sin 75^@)`
  3. `2.31`
  4. `3.2\ text(m)^2`
  5. `17\ text(m)^2`
Show Worked Solution
a.    `theta` `= 180^@ − (45^@ + 60^@)`
    `= 75^@`

 

b.   `(AX)/(sin 45^@) = (3.16)/(sin 75^@)`

 

c.    `(AX)/(sin 45^@)` `= (3.16)/(sin 75^@)`
     
  `:. AX` `= (3.16 xx sin45^@)/(sin 75^@)`
    `= 2.313…`
    `=2.31\ text{m  (2 d.p.)}`

 

d.    `text(Area of)\ \ ΔADX` `= 1/2 xx b xx h`
    `= 1/2 xx 3.16 xx 2`
    `= 3.2\ text{m²  (1 d.p.)}`

 

e.   VCAA 2014 fur2 Q2ei
MARKER’S COMMENT: An excellent example where many students do not read the question carefully and give away easy marks.
`text(Roof area)` `=1/2 xx BC xx (AB + XC)`
  `= 1/2 xx 2 xx (3 + 1.84)`
  `= 4.84`

 

`text(Wall area)` `= (1.8 xx 3) + (1.8 xx 2) + (1.8 xx 1.84)`
  `= 12.312`

 

`:.\ text(Total area)` `= 4.84 + 12.312`
  `=17.152`
  `=17\ text{m²  (nearest m²)}`

GEOMETRY, FUR2 2014 VCAA 1

The floor of a chicken coop is in the shape of a trapezium.

The floor, `ABCD`, and the chicken coop are shown below.

 GEOMETRY, FUR2 2014 VCAA 1

`AB = 3\ text(m), BC = 2\ text(m  and)\ \ CD = 5\ text(m.)`

  1. What is the area of the floor of the chicken coop?

     

    Write your answer in square metres.  (1 mark)

  2. What is the perimeter of the floor of the chicken coop?

     

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

Show Answers Only
  1. `8\ text(m²)`
  2. `12.8\ text(m)`
Show Worked Solution
a.    `A` `=1/2 h (a+b)`
    `= 1/2 xx 2 xx (3 + 5)`
    `= 8\ text(m²)`

 

b.  

`text(Using Pythagoras,)`

`AD^2` `=sqrt(2^2+2^2)`
  `=2.82…\ text(m)`

 

`:.\ text(Perimeter)` `= 3 + 2 + 5 + 2.82…`
  `= 12.8\ text{m  (1 d.p.)}`

GEOMETRY, FUR2 2008 VCAA 2

A shed has the shape of a prism. Its front face, `AOBCD`, is shaded in the diagram below. `ABCD` is a rectangle and `M` is the mid point of `AB`.
 

GEOMETRY, FUR2 2008 VCAA 2

  1. Show that the length of `OM` is 1.6 m.  (1 mark)
  2. Show that the area of the front face of the shed, `AOBCD`, is 18 m².  (1 mark)
  3. Find the volume of the shed in m³.  (1 mark)
  4. All inside surfaces of the shed, including the floor, will be painted.

     

    1. Find the total area that will be painted in m².  (2 marks)

       

      One litre of paint will cover an area of 16 m².

    2. Determine the number of litres of paint that is required.  (1 mark)
Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `180`
    1. `208`
    2. `13\ text(litres)`
Show Worked Solution
a.  

`text(In)\ Delta AOM,\ text(using Pythagoras:)`

`OM` `= sqrt (3.4^2 – 3^2)`
  `= sqrt 2.56`
  `= 1.6\ text(m … as required)`

 

b.   `text(Area of front face of shed)`

MARKER’S COMMENT: “The “Show that” directive requires students to include all mathematical steps, as shown in the solution. Short cuts will not gain full marks!

`= text(Area)\ Delta AOB + text(Area)\ ABCD`

`= 1/2 xx 1.6 xx 6 + 2.2 xx 6`

`= 18\ text(m² … as required.)`

 

c.    `V` `= Ah`
    `= 18 xx 10`
    `= 180\ text(m³)`

 

d.i.    `text(Roof areas)` `= 2(1/2 xx 1.6 xx 6 + 3.4 xx 10)`
    `= 2 (4.8 + 34)`
    `= 77.6\ text(m²)`
  `text(Wall areas)` `= 2 (2.2 xx 6 + 2.2 xx 10)`
    `= 2 (13.2 + 22)`
    `= 70.4\ text(m²)`
  `text(Floor area)` `= 60\ text(m²)`

 

`:.\ text(Total Area to be painted)`

`= 77.6 + 70.4 + 60`

`= 208\ text(m²)`

 

  ii.   `text(Litres of paint required)`

`= 208/16`

`= 13\ text(litres)`

GEOMETRY, FUR2 2008 VCAA 1

A shed is built on a concrete slab. The concrete slab is a rectangular prism 6 m wide, 10 m long and 0.2 m deep.

 

Geometry and Trig, FUR2 2008 VCAA 1

  1. Determine the volume of the concrete slab in m³.  (1 mark)
  2. 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.

  1. What is the area of this surface on the plan?  (1 mark)

 

Show Answers Only
  1. `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 2009 VCAA 2

A yacht, `Y`, is 7 km from a lighthouse, `L`, on a bearing of 210° as shown in the diagram below.

 

GEOMETRY, FUR2 2009 VCAA 2

  1. A ferry can also be seen from the lighthouse. The ferry is 3 km from `L` on a bearing of 135°. On the diagram above, label the position of the ferry, `F`, and show an angle to indicate its bearing.  (1 mark)
  2. Determine the angle between `LY` and `LF`.  (1 mark)
  3. Calculate the distance, in km, between the ferry and the yacht correct to two decimal places.  (1 mark)
  4. Determine the bearing of the lighthouse from the ferry.  (1 mark)
Show Answers Only
  1.  
  2. `75^@`
  3. `6.87\ text(km)`
  4. `315^@`
Show Worked Solution
a.   

 

b.   `text(Angle between)\ LY\ text(and)\ LF\ \ text{(from graph)}`

`= 30 + 45`

`= 75^@`

 

c.   `text(Using the cosine rule,)`

`FY^2` `= 7^2 + 3^2 – 2 xx 7 xx 3 xx cos 75^@`
  `= 47.129…`
`:. FY` `= 6.865…`
  `= 6.87\ text{km  (2 d.p.)}`

 

d.   `text(Bearing of)\ L\ text(from)\ F\ text{(from graph)}`

`= 360 – 45`

`= 315^@`

GEOMETRY, FUR2 2009 VCAA 1

A ferry, `F`, is 400 metres from point `O` at the base of a 50 metre high cliff, `OC`.
 

GEOMETRY, FUR2 2009 VCAA 1 
 

  1. Show that the gradient of the line `FC` in the diagram is 0.125.  (1 mark)
  2. Calculate the angle of elevation of point `C` from `F`.

     

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

  3. Calculate the distance `FC`, in metres, correct to one decimal place.  (1 mark)
Show Answers Only
  1. `text(See Worked Solution.)`
  2. `7.1^@`
  3. `403.1\ text(m)`
Show Worked Solution
a.    `text(Gradient)` `= text(rise)/text(run)`
    `= 50/400`
    `= 0.125\ \ text(…   as required)`

 

b.    `tan\ /_ CFO` `= 50/400`
  `:. /_ CFO` `= tan^-1 0.125`
    `= 7.12…`
    `= 7.1^@\ text{(1 d.p.)}`

 

c.    `text(Using Pythagoras:)`
  `FC` `= sqrt(400^2 + 50^2)`
    `= 403.11…`
    `= 403.1\ text(m)`

Calculus, MET2 2013 VCAA 3

Tasmania Jones is in Switzerland. He is working as a construction engineer and he is developing a thrilling train ride in the mountains. He chooses a region of a mountain landscape, the cross-section of which is shown in the diagram below.

VCAA 2013 2a

The cross-section of the mountain and the valley shown in the diagram (including a lake bed) is modelled by the function with rule

`f(x) = (3x^3)/64-(7x^2)/32 + 1/2.`

Tasmania knows that  `A (0, 1/2)`  is the highest point on the mountain and that `C(2, 0)` and `B(4, 0)` are the points at the edge of the lake, situated in the valley. All distances are measured in kilometres.

  1. Find the coordinates of `G`, the deepest point in the lake.   (3 marks)

Tasmania’s train ride is made by constructing a straight railway line `AB` from the top of the mountain, `A`, to the edge of the lake, `B`. The section of the railway line from `A` to `D` passes through a tunnel in the mountain.

  1. Write down the equation of the line that passes through `A` and `B.`   (2 marks)

  2.  i. Show that the `x`-coordinate of `D`, the end point of the tunnel, is `2/3.`   (1 mark)

  3. ii. Find the length of the tunnel `AD.`   (2 marks)

In order to ensure that the section of the railway line from `D` to `B` remains stable, Tasmania constructs vertical columns from the lake bed to the railway line. The column `EF` is the longest of all possible columns. (Refer to the diagram above.)

  1.  i. Find the `x`-coordinate of `E.`   (2 marks)

  2. ii. Find the length of the column `EF` in metres, correct to the nearest metre.   (2 marks)

Tasmania’s train travels down the railway line from `A` to `B`. The speed, in km/h, of the train as it moves down the railway line is described by the function.

`V: [0, 4] -> R, \ V(x) = k sqrt x-mx^2,`

where `x` is the `x`-coordinate of a point on the front of the train as it moves down the railway line, and `k` and `m` are positive real constants.

The train begins its journey at `A (0, 1/2)`. It increases its speed as it travels down the railway line.

The train then slows to a stop at `B(4, 0)`, that is  `V(4) = 0.`

  1. Find `k` in terms of `m.`   (1 mark)

  2. Find the value of `x` for which the speed, `V`, is a maximum.   (2 marks)

Tasmania is able to change the value of `m` on any particular day. As `m` changes, the relationship between `k` and `m` remains the same.

  1. If, on one particular day, `m = 10`, find the maximum speed of the train, correct to one decimal place.   (2 marks)

  2. If, on another day, the maximum value of `V` is 120, find the value of `m.`   (2 marks)

Show Answers Only
  1. `G(28/9,−50/243)`
  2. `y = −1/8x + 1/2`
  3.  i. `text(See Worked Solutions)`
  4. ii. `sqrt65/12`
  5.  i. `(2(sqrt31 + 7))/9`
  6. ii. `336\ text(m)`
  7. `8\ text(m)`
  8. `2^(2/3)`
  9. `75.6\ text(km/h)`
  10. `10 xx 2^(2/3)`
Show Worked Solution

a.   `G\ text(occurs when)\ \ f^{′}(x)=0.`

`text(Solve:)\ \ (3x^3)/64-(7x^2)/32 + 1/2=0\ \ text(for)\ x`

`x= 28/9quadtext(for)quadx ∈ (2,4)`

`f(28/9) = −50/243`

`:. G(28/9,−50/243)`
 

b.    `m_(AB)` `= (1/2-0)/(0-4)`
    `= −1/8`

 
`text(Equation using point gradient formula,)`

`y-y_1` `= m(x-x_1)`
`y-1/2` `= −1/8(x-0)`
`:. y` `= −1/8x + 1/2`

 
c.i. `text{Solution 2 (using technology)}`

`text(Solve:)\ \ ` `3/64x^3-7/32x^2 + 1/2` `= −1/8x + 1/2\ \ text(for)\ x`

`x=0, 2/3 or 4`

`:.x=2/3,\ \ (0<x<4)`
 

`text(Solution 1)`

`text(Intersection occurs when:)`

`3/64x^3-7/32x^2 + 1/2` `= −1/8x + 1/2`
`3x^3-14x^2 + 8x` `= 0`
`x(3x^2-14x + 8)` `= 0`
`x(x-4)(3x-2)` `= 0`

 
`x = 2/3,\ \ (0<x<4)`

`:. x text(-coordinate of)\ D = 2/3`
 

c.ii.   `D (2/3, 5/12)qquadA(0,1/2)`

  `text(By Pythagoras,)`

`AD` `= sqrt((2/3-0)^2 + (5/12-1/2)^2)`
  `= sqrt65/12`

 

d.i.  `text(Let)\ \ z =\ EF`

♦♦♦ Mean mark part (d)(i) 24%.
MARKER’S COMMENT: Many students unfamiliar with this type of question.
 `z` `= (−1/8x + 1/2)-((3x^2)/64-7/32 x^2 + 1/2)`
  `=-1/8x-(3x^2)/64 + 7/32 x^2`

 
`text(Solve:)\ \ d/dx (-1/8x-(3x^2)/64 + 7/32 x^2) =0\ \ text(for)\ x`

`x= (2(sqrt31 + 7))/9,\ \ \ x > 0`
 

♦♦♦ Mean mark part (d)(ii) 22%.
d.ii.    `z((2sqrt31 + 14)/9)` `= 0.3360…\ text(km)`
    `= 336\ text{m  (nearest m)}`

 

e.    `V(4)` `= 0quadtext{(given)}`
  ` 0` `= k sqrt4-m xx 4^2`
  `:.k` `= 8m`

 

f.  `V(x) = 8msqrtx-mx^2`

`text(Solve:)\ \ V^{′}(x)` `= 0quadtext(for)quadx`

`x= 2^(2/3)`
 

g.  `text(When)\ \ m=10,\ \ k=8 xx 10=80`

♦ Mean mark 43%.

`:.V(x) = 80sqrtx-10x^2`

`text(Solve:)\ \ V^{′}(x)` `= 0quadtext(for)quadx`

`x= 2^(2/3)`

`V(2^(2/3))` `=75.59…`
  `= 75.6\ text(km/h)`

 

h.   `text(Maximum occurs at)\ x = 2^(2/3)`

♦ Mean mark 38%.

`V(x) = 8msqrtx-mx^2`

`text(Solve:)\ \ V(2^(2/3))` `= 120quadtext(for)quadm`
`:. m` `= 10 xx 2^(2/3)`

Graphs, MET2 2013 VCAA 1

Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature  `(Ttext{°C})` is given by  `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.

  1. State the maximum temperature in the greenhouse and the values of `t` when this occurs.   (2 marks)

  2. State the period of the function `T.`   (1 mark)

  3. Find the smallest value of `t` for which  `T = 26.`   (2 marks)

  4. For how many hours during the 24-hour time interval is  `T >= 26?`   (2 marks)

Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of  `y = sin(x)`  for  `0 <= x <= 2 pi`  and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`

  1. The line through points  `P((2 pi)/3, sqrt 3/2)`  and  `C (c, 0)`  is a tangent to the graph of  `y = sin (x)`  at point `P.`

    1. Find  `(dy)/(dx)`  when  `x = (2 pi)/3.`   (1 mark)

    2. Show that the value of `c` is  `sqrt 3 + (2 pi)/3.`   (1 mark)

In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.

  1. Let `X^{′}, P^{′}` and `C^{′}` be the image, under this transformation, of the points `X, P` and `C` respectively. 

     

    1. Find the values of `k` and `m`  if  `X^{′}P^{′} = 10`  and  `X^{′} C^{′} = 30.`   (2 marks)

    2. Find the coordinates of the point `P^{′}.`   (1 mark)

Show Answers Only
  1. `t = 0, or 16\ text(h)`
  2. `16\ text(hours)`
  3. `8/3`
  4. `8\ text(hours)`
  5.  i.  `-1/2`
    ii.  `text(See worked solution)`
  6.  i.  `k=(20sqrt3)/3, m=10sqrt3`
    ii.  `P^{′}((20pisqrt3)/3,10)`
Show Worked Solution

a.   `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`

`T_text(max)= 25 + 2 = 27^@C`

`text(Max occurs when)\ \ t = 0, or 16\ text(h)`

 

b.    `text(Period)` `= (2pi)/(pi/8)`
    `= 16\ text(hours)`

 

c.   `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`

`t`  `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]`
`t_text(min)` `= 8/3`

 

d.   `text(Consider the graph:)`

met2-2013-vcaa-sec1-answer1

`text(Time above)\ 26 text(°C)` `= 8/3 + (56/3-40/3)`
  `= 8\ text(hours)`

 

e.i.   `(dy)/(dx) = cos(x)`

`text(At)\ x = (2pi)/3,`

`(dy)/(dx)` `= cos((2pi)/3)=-1/2`

 

e.ii.  `text(Solution 1)`

`text(Equation of)\ \ PC,`

`y-sqrt3/2` `=-1/2(x-(2pi)/3)`
`y` `=-1/2 x +pi/3 +sqrt3/2`

 

`PC\ \ text(passes through)\ \ (c,0),`

`0` `=-1/2 c +pi/3 + sqrt3/2`
`c` `=sqrt3 + (2 pi)/3\ …\ text(as required)`

 

`text(Solution 2)`

`text(Equating gradients:)`

`- 1/2` `= (sqrt3/2-0)/((2pi)/3-c)`
`-1` `= sqrt3/((2pi-3c)/3)`
`3c-2pi` `= 3sqrt3`
`3c` `= 3 sqrt3 + 2pi`
`:. c` `= sqrt3 + (2pi)/3\ …\ text(as required)`

 

f.i.   `X^{′} ((2pi)/3 m,0)qquadP^{′}((2pi)/3 m, sqrt3/2 k)qquadC^{′} ((sqrt3 + (2pi)/3)m, 0)`

`X^{′}P^{′}` `= 10`
`sqrt3/2 k` `= 10`
`:. k` `= 20/sqrt3`
  `=(20sqrt3)/3`
♦♦♦ Mean mark part (f)(i) 14%.

 

`X^{′}C^{′}=30`

`((sqrt3 + (2pi)/3)m)-(2pi)/3 m` `= 30`
`:. m` `= 30/sqrt3`
  `=10sqrt3`
♦♦♦ Mean mark part (f)(ii) 12%.

 

f.ii.    `P^{′}((2pi)/3 m, sqrt3/2 k)` `= P^{′}((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)`
    `= P^{′}((20pisqrt3)/3,10)`

Calculus, MET2 2012 VCAA 5

The shaded region in the diagram below is the plan of a mine site for the Black Possum mining company.

All distances are in kilometres.

Two of the boundaries of the mine site are in the shape of the graphs of the functions

`f: R -> R,\ f(x) = e^x and g: R^+ -> R,\ g(x) = log_e (x).`

VCAA 2012 5a

    1. Evaluate `int_(−2)^0 f(x)\ dx`.   (1 mark)

    2. Hence, or otherwise, find the area of the region bounded by the graph of `g`, the `x` and `y` axes, and the line `y = –2`.   (1 mark)

    3. Find the total area of the shaded region.   (1 mark)

  2. The mining engineer, Victoria, decides that a better site for the mine is the region bounded by the graph of `g` and that of a new function  `k: (– oo, a) -> R,\ k(x) = – log_e(a-x)`, where `a` is a positive real number.
    1. Find, in terms of `a`, the `x`-coordinates of the points of intersection of the graphs of `g` and `k`.   (2 marks)

    2. Hence, find the set of values of `a`, for which the graphs of `g` and `k` have two distinct points of intersection.   (1 mark)

  3. For the new mine site, the graphs of `g` and `k` intersect at two distinct points, `A` and `B`. It is proposed to start mining operations along the line segment `AB`, which joins the two points of intersection.
  4. Victoria decides that the graph of `k` will be such that the `x`-coordinate of the midpoint of `AB` is `sqrt 2`.
  5. Find the value of `a` in this case.   (2 marks)

Show Answers Only
    1. `1-1/(e^2)`
    2. `1-1/(e^2)\ text(units²)`
    3. `e-1/(e^2)\ text(units²)`
    1. `(a ± sqrt(a^2-4))/2`
    2. `a > 2`
  3. `2sqrt2`
Show Worked Solution
ai.    `int_(−2)^0 f(x)\ dx` `=[e^x]_(-2)^0`
    `= e^0-e^(-2)`
    `=1-1/e^2`

 

a.ii.   `text(S)text(ince)\ \ g(x) = f^(−1)(x),`

♦ Mean mark part (a)(ii) 45%.

`=>\ text(Area is the same as)\ int_(−2)^0f(x)\ dx,`

`:. text(Area) = 1-1/(e^2)\ text(u²)`

 

♦ Mean mark part (a)(iii) 36%.
a.iii.    `text(Area)` `=int_0^1 (e^x)\ dx + text{Area from part (a)(ii)}`
    `= [e^x]_0^1 + (1-1/(e^2))`
    `= e-1/(e^2)\ text(u²)`

 

b.i.    `g(x)` `= k(x)`
  `log_e (x)` `=- log_e(a-x)`
  `log_e (x)+log_e(a-x)` `=0`
  `log_e(x(a-x))` `=0`
  `ax-x^2` `=1`
  `x^2-ax+1` `=0`

 
`:.x= (a ± sqrt(a^2-4))/2`

 

♦♦♦ Mean mark (b.ii.) 11%.

b.ii.   `text(For 2 solutions:)`

`b^2-4ac` `>0`
`a^2-4` `>0`
`:. a` `>2,\  \ (a>0)`

 

c.   `xtext(-coordinate of Midpoint)= sqrt2`

♦♦♦ Mean mark (c) 15%.
`((a + sqrt(a^2-4))/2 +(a-sqrt(a^2-4))/2)/2` `= sqrt2`
`a + sqrt(a^2-4) +a-sqrt(a^2-4)`  `=4 sqrt2`
`a` `=2sqrt2quadtext(for)quada > 2`

Calculus, MET2 2012 VCAA 4

Tasmania Jones is in the jungle, searching for the Quetzalotl tribe’s valuable emerald that has been stolen and hidden by a neighbouring tribe. Tasmania has heard that the emerald has been hidden in a tank shaped like an inverted cone, with a height of 10 metres and a diameter of 4 metres (as shown below).

The emerald is on a shelf. The tank has a poisonous liquid in it.

VCAA 2012 4a

  1. If the depth of the liquid in the tank is `h` metres.
      
     i.     find the radius, `r` metres, of the surface of the liquid in terms of `h`.   (1 mark)

    ii. show that the volume of the liquid in the tank is  `(pi h^3)/75\ text(m³)`.   (1 mark)

The tank has a tap at its base that allows the liquid to run out of it. The tank is initially full. When the tap is turned on, the liquid flows out of the tank at such a rate that the depth, `h` metres, of the liquid in the tank is given by

`h = 10 + 1/1600 (t^3 - 1200t)`,

where `t` minutes is the length of time after the tap is turned on until the tank is empty.

  1. Show that the tank is empty when  `t = 20`.   (1 mark)

  2. When  `t = 5` minutes, find
      
    i.     the depth of the liquid in the tank.   (1 mark)

    ii. No longer in syllabus

  1. The shelf on which the emerald is placed is 2 metres above the vertex of the cone.

    From the moment the liquid starts to flow from the tank, find how long, in minutes, it takes until  `h = 2`.

    (Give your answer correct to one decimal place.)   (2 marks)

  2. As soon as the tank is empty, the tap turns itself off and poisonous liquid starts to flow into the tank at a rate of 0.2 m³/minute.
      
    How long, in minutes, after the tank is first empty will the liquid once again reach a depth of 2 metres?   (2 marks)

  3. In order to obtain the emerald, Tasmania Jones enters the tank using a vine to climb down the wall of the tank as soon as the depth of the liquid is first 2 metres. He must leave the tank before the depth is again greater than 2 metres.
      
    Find the length of time, in minutes, correct to one decimal place, that Tasmania Jones has from the time he enters the tank to the time he leaves the tank.   (1 mark)

Show Answers Only
  1.  i. `h/5`
    ii. `(pih^3)/75`
  2. `text(See Worked Solutions)`
  3.  i. `405/64\ text(m)`
    ii. `text(No longer in course)`
  4. `12.2\ text(min)`
  5. `(8pi)/15\ text(min)`
  6. `9.5\ text(min)`
Show Worked Solution

a.i.   `text(Using similar triangles:)`

`r/h` `= 2/10`
`:. r` `= h/5`

 

a.ii.    `V` `= 1/3 pir^2h`
    `= 1/3 pi(h/5)^2h`
  `:. V` `= (pih^3)/75\ \ \ text(… as required)`

 

b.    `h(20)` `= 10 + 1/1600(20^3-1200 xx 20)`
    `= 10 + 1/1600(-16\ 000)`
    `= 10-10`
    `= 0`

 

c.i.   `text(When)\ \ t=5,`

`h(5)` `=10 + 1/1600 (5^3-1200xx5)`
  `= 405/64\ text(m)`

 

c.ii.   `text(No longer in syllabus.)`

 

d.    `text(Solve:)\ \ 10 + 1/1600 (t^3-1200t)` `= 2`

 

`t= 12.2\ text(min)quadtext(for)quadt ∈ (0,20)`

 

e.   `text(When)\ \ h=2,`

♦♦♦ Mean mark part (e) 17%.
`text(Volume)` `=(pi 2^3)/75=(8pi)/75\ text(m³)`
   

`text(Time to fill back up to)\ \ h=2`

`=(8pi)/75 -: 0.2`

`=(8pi)/15\ text(minutes)`

 

f.   `text(Length of time)`

♦♦♦ Mean mark part (f) 11%.

`=\ text(Time for 2m to empty + Time to Fill to 2m)`

`= (20-12.16799…) + (8pi)/15`

`= 9.5\ text(min)\ \ text{(1 d.p.)}`

Calculus, MET2 2012 VCAA 2

Let  `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`

  1. Sketch the graph of  `y = f(x)` on the set of axes below. Label the axes intercepts with their coordinates and label each of the asymptotes with its equation.   (3 marks)

           VCAA 2012 2a
     

  2.   i. Find `f^{′}(x)`.   (1 mark)

  3.  ii. State the range of  `f ^{′}`.   (1 mark)

  4. iii. Using the result of part ii. explain why `f` has no stationary points.   (1 mark)

  5. If  `(p, q)`  is any point on the graph of  `y = f(x)`, show that the equation of the tangent to  `y = f(x)`  at this point can be written as  `(2p-4)^2 (y-3) = -2x + 4p-4.`   (2 marks)

  6. Find the coordinates of the points on the graph of  `y = f(x)`  such that the tangents to the graph at these points intersect at  `(-1, 7/2).`   (4 marks)

  7. A transformation  `T: R^2 -> R^2`  that maps the graph of  `f` to the graph of the function  `g: R text(\{0}) -> R,\ g(x) = 1/x`  has rule
  8.      `T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
  9. Find the values of `a, c` and `d`.   (2 marks)

Show Answers Only
  1. met2-2012-vcaa-sec2-answer
  2.   i. `f^{′}(x) = (−2)/((2x-4)^3)`
     ii. `text(Range) = (−∞,0)`
    iii. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
  5. `a = 2, c = −4, d = −3`
Show Worked Solution

a.   `text(Asymptotes:)`

`x = 2`

`y = 3`

met2-2012-vcaa-sec2-answer

 

b.i.   `f^{′}(x) = (−2)/((2x-4)^2)`

 

b.ii.   `text(Range) = (−∞,0), or  R^-`

MARKER’S COMMENT: Incorrect notation in part (b)(ii) was common, including `{-oo,0}, -R, (0,-oo)`.

 

b.iii.    `text(As)\ \ ` `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})`
    `f^{′}(x) != 0`

 
`:. f\ text(has no stationary points.)`
 

c.   `text(Point of tangency) = P(p,1/(2p-4) + 3)`

♦♦ Mean mark 29%.
`m_text(tang)` `= f^{′}(p)`
  `= (-2)/((2p-4)^2)`

 
`text(Equation of tangent using:)`

`y-y_1` `= m(x-x_1)`
`y-(1/(2p-4) + 3)` `= (-2)/((2p-4)^2)(x-p)`
`y-3` `= (-2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)`
`(2p-4)^2(y-3)` `=-2x + 2p + 2p-4`
`:. (2p-4)^2(y-3)` `=-2x + 4p-4\ \ text(… as required)`

 

d.   `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`

♦♦♦ Mean mark 19%.

`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(-1) + 4p-4\ \ text(for)\ p:`

`:. p = 1,\ text(or)\ 5`

`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`

`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`
 

e.   `text(Determine transformations that that take)\ f -> g:`

`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`

`text(factor of 2 from the)\ \ ytext(-axis).`

`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`

`text(Translate the graph 4 units to the left and 3)`

`text(units down to obtain)\ \ g(x).`
 

`text(Using the transformation matrix,)`

`x^{′}` `=ax+c`
`y^{′}` `=y+d`

 
`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`

`x^{′}=2x-4`

`=> a=2,\ \ c=-4`
 

`f -> g:\ \ y -> y^{′} + 3`

`y^{′}=y -3`

`=>\ \ d=-3`

GRAPHS, FUR2 2011 VCAA 1

Michael is preparing to hike through a national park.

He decides to make some trail mix to eat on the hike.

The trail mix consists of almonds and raisins.

The table below shows some information about the amout of carbohydrate and protein contained in each gram of almonds and raisins.

GRAPHS, FUR2 2011 VCAA 11 

  1. If Michael mixed 180 g of almonds and 250 g of raisins to make some trail mix, calculate the weight, in grams, of carbohydrate in the trail mix.  (1 mark)

Michael wants to make some trail mix that contains 72 g of protein. He already has 320 g of almonds.

  1. How many grams of raisins does he need to add?  (2 marks)

The trail mix Michael takes on his hike must satisfy his dietary requirements.

Let `x` be the weight, in grams, of almonds Michael puts into the trail mix.

Let `y` be the weight, in grams, of raisins Michael puts into the trail mix.

Inequalities 1 to 4 represents Michael's dietary requirements for the weight of carbohydrate and protein in the trail mix.

Inequality 1 `x >= 0`
Inequality 2 `y >= 0`
Inequality 3 (carbohydrate)   `0.2x + 0.8y >= 192`
Inequality 4 (protein) `0.2x + 0.04y <= 40`    

 

Michael also requires a minimum of 16 g of fibre in the trail mix.

Each gram of almonds contains 0.1 g of fibre.

Each gram of raisins contains 0.04 g of fibre.

  1. Write down an inequality, in terms of `x` and `y`, that represents this dietary requirement.

     

     

    Inequality 5 (fibre) _________________________  (1 mark)

The graphs of  `0.2x + 0.8y = 192`  and  `0.2x + 0.04y = 40`  are shown below.

GRAPHS, FUR2 2011 VCAA 12

  1. On the graph above

     

    1. draw the straight line that relates to Inequality 5  (1 mark)
    2. shade the region that satisfies Inequalities 1 to 5.  (1 mark)
  2. What is the maximum weight, in grams, of trail mix that satisfies Michael's dietary requirements?  (1 mark)

Michael plans to carry at least 500 g of trail mix on his hike.

He would also like this trail mix to cantain the greatest possible weight of almonds.

The trail mix must satisfy all of Michael's dietary requirements.

  1. What is the weight of the almonds, in grams, in this trail mix?  (2 marks) 
Show Answers Only
  1. `236\ text(g)`
  2. `200\ text(g)`
  3. `0.1x + 0.04y >= 16`
  4. i. & ii.
  5. `1000`
  6. `125\ text(g)`
Show Worked Solution
a.    `text(Carbohydrates)` `= 180 xx 0.2 + 250 xx 0.8`
    `= 236\ text(grams)`

 

b.   `text(Almond protein)`

`= 320 xx 0.2`

`= 64\ text(g)`

`=>\ text(8 grams of protein still required.)`

 

`:.\ text(Amount of raisins needed)`

`= 8/0.04`

`= 200\ text(g)`

 

c.   `text(Inequality 5)`

`0.1x + 0.04y >= 16`

 

d.i. & ii.   

 

e.   `text(Maximum weight to satisfy)`

`= 1000\ text(grams)`

 

f.   `text(New constraint)`

`x + y >= 500`

VCAA GRAPHS FUR2 2011 1dii

`text(In the new feasible region, the maximum amount)`

`text(of almonds occurs at the intersection of)`

`x + y = 500\ text{  …  (1)}`

♦♦ Part f was “Poorly answered” although exact data unavailable.
MARKER’S COMMENT: Ensure you incorporate the new constraint!

`0.2x + 0.04y = 40\ text{  …  (2)}`

 

`text(Substitute)\ \ y = 500 – x\ \ text{from (1) into (2)}`

`0.2x + 0.04 (500 – x)` `= 40`
`0.2x + 20 – 0.04x` `= 40`
`0.16x` `= 20`
`x` `= 125`

 

`:.\ text(Weight of almonds) = 125\ text(g)`

MATRICES, FUR2 2012 VCAA 2

Rosa uses the following six-digit pin number for her bank account: 216342. 

With her knowledge of matrices, she decides to use matrix multiplication to disguise this pin number.

First she writes the six digits in the 2 × 3 matrix `A`.

`A = [(2,6,4),(1,3,2)]`

Next she creates a new matrix by forming the matrix product, `C = BA`,

where `B = [(1,-1),(2,-1)]`

    1. Determine the matrix `C = BA`.   (1 mark)

    2. From the matrix `C`, Rosa is able to write down a six-digit number that disguises her original pin number. She uses the same pattern that she used to create matrix `A` from the digits 216342.

       

      Write down the new six-digit number that Rosa uses to disguise her pin number.   (1 mark)

  2. Show how the original matrix `A` can be regenerated from matrix `C`.   (1 mark) 
Show Answers Only
    1. `C = [(1,3,2),(3,9,6)]`
    2. `133926`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.i.   `C` `= BA`
    `= [(1,-1),(2,-1)][(2,6,4),(1,3,2)]`
    `= [(1,3,2),(3,9,6)]`

 

a.ii. `133926`

`text(The pattern in matrix)\ C\ text(is)`

 

b.   `C` `= BA`
  `B^-1 C` `= B^-1 B A`
  `:. A` `= B^-1 C`

MATRICES, FUR2 2012 VCAA 1

Matrix `F` below shows the flight connections for an airline that serves four cities, Anvil (`A`), Berga (`B`), Cantor (`C`) and Dantel (`D`).

`{:(qquad qquad qquad qquad quad text(from)),((qquad qquad quad\ A,B,C,D)),(F = [(0\ ,1\ ,0\ ,0),(1,0,1,0),(0,0,0,1),(0,1,0,0)]):}{:(),(),(A),(B),(C),(D):}{:(),(qquad text(to)):}`

In this matrix, the `1` in column `C` row `B`, for example, indicates that, using this airline, you can fly directly from Cantor to Berga. The `0` in column `C` row `D`, for example, indicates that you cannot fly directly from Cantor to Dantel.

  1. Complete the following sentence.
  2. On this airline, you can fly directly from Berga to ________ and ________.   (1 mark)

  3. List the route that you must follow to fly from Anvil to Cantor.   (1 mark)

  4. Evaluate the matrix product `G = KF`, where `K = [1,1,1,1]`.   (1 mark)

     

  5. In the context of the problem, what information does matrix `G` contain?   (1 mark)

Show Answers Only
  1. `text(Anvil and Dantel)`
  2. `text(Anvil – Berga – Dantel – Cantor)`
  3. `[1,2,1,1]`
  4. `text(Matrix)\ G\ text(contains the information of the total)`
    `text(amount of direct flights out of each of 4 cities,)`
    `text(that go to another city in the same group.)`
Show Worked Solution

a.   `text(Anvil and Dantel)`
 

b.   `text(Anvil – Berga – Dantel – Cantor)`
 

c.   `G` `= KF`
    `= [1,1,1,1][(0,1,0,0),(1,0,1,0),(0,0,0,1),(0,1,0,0)]`
    `= [1,2,1,1]`

 
d.    `text(Matrix)\ G\ text(contains the information of the total)`

`text(amount of direct flights out of each of 4 cities,)`

`text(that go to another city in the same group.)`

MATRICES, FUR2 2014 VCAA 2

There are three candidates in the election: Ms Aboud (`A`), Mr Broad (`B`) and Mr Choi (`C`).

The election campaign will run for six months, from the start of January until the election at the end of June.

A survey of voters found that voting preference can change from month to month leading up to the election.

The transition diagram below shows the percentage of voters who are expected to change their preferred candidate from month to month.
 

MATRICES, FUR2 2014 VCAA 2

    1. Of the voters who prefer Mr Choi this month, what percentage are expected to prefer Ms Aboud next month?   (1 mark)

    2. Of the voters who prefer Ms Aboud this month, what percentage are expected to change their preferred candidate next month?   (1 mark)

In January, 12 000 voters are expected in the city. The number of voters in the city is expected to remain constant until the election is held in June.

The state matrix that indicates the number of voters who are expected to have a preference for each candidate in January, `S_1`, is given below.

`S_1 = [(6000), (3840), (2160)]{:(A), (B), (C):}`

  1. How many voters are expected to change their preference to Mr Broad in February?   (1 mark)

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

`{:(qquadqquadqquadqquadtext(this month)),(qquadqquadqquad{:(\ A,qquadB,qquadC):}),(T = [(0.75,0.10,0.20),(0.05,0.80,0.40),(0.20,0.10,0.40)]{:(A),(B),(C):}qquadtext(next month)):}`

    1. Evaluate the matrix `S_3 = T^2S_1` and write it down in the space below.
    2. Write the elements, correct to the nearest whole number.   (1 mark)

    3. `S_3 = [(qquadqquadqquad),(),()]`
    4. What information does matrix `S_3` contain?   (1 mark)

  2. Using matrix `T`, how many votes would the winner of the election in June be expected to receive?
  3. Write your answer, correct to the nearest whole number.   (1 mark)

Show Answers Only
    1. `text(20%)`
    2. `text(25%)`
  2. `1164`
    1.  
      `S_3 = [(4900), (4634), (2466)]{:(A), (B), (C):}`
    2. `S_3\ text(contains the predicted number of)`
      `text(preferences for each candidate for)`
      `text(March.)`
  4. `5303`
Show Worked Solution

a.i.   `text(20%)`
 

a.ii.   `text(5%) + text(20%) = 25text(%)`
 

b.   `text(Voters expected to change to)\ B`

`= 5text(%) xx 6000 + 40text(%) xx 2160`

`= 1164`

 

c.i.    `S_3` `= T^2S_1`
    `= [(4900), (4634), (2466)]{:(A), (B), (C):}`

 

c.ii.   `S_3\ text(contains the predicted number of preferences)`

  `text(for each candidate for March.)`

 

MARKER’S COMMENT: A common error used `S_text(June)“=S_6=T^6 xx S_1`. Know why this is incorrect.
d.    `S_6` `= T^5S_1`
    `= [(4334), (5303), (2363)]{:(A), (B), (C):}`

 

`:. B\ text(is the expected winner with 5303 votes.)`

MATRICES, FUR2 2015 VCAA 2

The ability level of students is assessed regularly and classified as beginner (`B`), intermediate (`I`) or advanced (`A`).

After each assessment, students either stay at their current level or progress to a higher level.

Students cannot be assessed at a level that is lower than their current level.

The expected number of students at each level after each assessment can be determined using the transition matrix, `T_1`, shown below.

`{:(qquad qquad text(before assessment)), (qquad qquad qquad quad {:(B, qquad quad I, quad A):}), (T_1 = [(0.50, 0, 0), (0.48, 0.80, 0), (0.02, 0.20, 1)] {:(B), (I), (A):} qquad text(after assessment)):}`

  1. The element in the third row and third column of matrix `T_1` is the number 1.
  2. Explain what this tells you about the advanced-level students.   (1 mark)

Let matrix `S_n` be a state matrix that lists the number of students at beginner, intermediate and advanced levels after `n` assessments.

The number of students in the school, immediately before the first assessment of the year, is shown in matrix `S_0` below.

`S_0 = [(20), (60), (40)] {:(B), (I), (A):}`

    1. Write down the matrix `S_1` that contains the expected number of students at each level after one assessment.
    2. Write the elements of this matrix correct to the nearest whole number.   (1 mark)

    3. How many intermediate-level students have become advanced-level students after one assessment?   (1 mark)

Show Answers Only
  1. `text(All advanced students before assessment)`
    `text(remain advanced students after)`
    `text(assessment.)`
    1. `[(10),(58),(52)]`
    2. `12`
Show Worked Solution

a.    `text(All advanced students before assessment)`

`text(remain advanced students after)`

`text(assessment.)`

 

b.i.    `S_1` `= T_1 S_0`
    `= [(0.5, 0, 0), (0.48, 0.80, 0), (0.02, 0.20, 1)] [(20), (60), (40)]`
    `= [(10), (58), (52)]`

 

b.ii.   `text(Intermediate)\  ->\  text(Advanced)`

`= 0.20 xx 60`

`= 12\ text(students)`

CORE, FUR2 2006 VCAA 1

Table 1 shows the heights (in cm) of three groups of randomly chosen boys aged 18 months, 27 months and 36 months respectively.

Core, FUR2 2006 VCAA 11

  1. Complete Table 2 by calculating the standard deviation of the heights of the 18-month-old boys.

     

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

     

    Core, FUR2 2006 VCAA 12

A 27-month-old boy has a height of 83.1 cm.

  1. Calculate his standardised height (`z` score) relative to this sample of 27-month-old boys.
  2. Write your answer correct to one decimal place.   (1 mark)

The heights of the 36-month-old boys are normally distributed.

A 36-month-old boy has a standardised height of 2.

  1. Approximately what percentage of 36-month-old boys will be shorter than this child?   (1 mark)

Using the data from Table 1, boxplots have been constructed to display the distributions of heights of 36-month-old and 27-month-old boys as shown below. 

     Core, FUR2 2006 VCAA 13

  1. Complete the display by constructing and drawing a boxplot that shows the distribution of heights for the 18-month-old boys.   (2 marks)

  2. Use the appropriate boxplot to determine the median height (in centimetres) of the 27-month-old boys.   (1 mark)

The three parallel boxplots suggest that height and age (18 months, 27 months, 36 months) are positively related.

  1. Explain why, giving reference to an appropriate statistic.   (1 mark)

Show Answers Only
  1. `3.8`
  2. `−1.4`
  3.  `2.5 text(%)`
  4.  
    Core, FUR2 2006 VCAA 13 Asnwer
  5. `89.5`
  6. `text(Median height increases as age increases.)`
Show Worked Solution

a.   `text(By calculator,)`

`text(standard deviation) = 3.8`

 

b.    `z` `= (x – barx)/s`
    `= (83.1 – 89.3)/(4.5)`
    `=-1.377…`
    `= -1.4\ \ text{(1 d.p.)}`

 

♦♦ MARKER’S COMMENT: Attention required here as this standard question was “very poorly answered”.

c.  `text{2.5%  (see graph below)}`

CORE, FUR2 2006 VCAA Answer 111

 

d.   `text(Range = 76 – 89.8,)\ Q_1 = 80,\ Q_3 = 85.8,\ text(Median = 83,)`

Core, FUR2 2006 VCAA 13 Asnwer

e.   `89.5`

MARKER’S COMMENT: A boxplot statistic was required, so mean values were not relevant.

 

f.   `text(The median height increases with age.)`

CORE, FUR2 2007 VCAA 3

The table below displays the mean surface temperature (in °C) and the mean duration of warm spell (in days) in Australia for 13 years selected at random from the period 1960 to 2005.
 

CORE, FUR2 2007 VCAA 31
 

This data set has been used to construct the scatterplot below. The scatterplot is incomplete.

  1. Complete the scatterplot below by plotting the bold data values given in the table above. Mark the point with a cross (×).  (1 mark)

          

     

  2. Mean surface temperature is the explanatory variable.

     

    1. Determine the equation of the least squares regression line for this set of data. Write the equation in terms of the variables mean duration of warm spell and mean surface temperature. Write the value of the coefficients correct to one decimal place.  (2 marks)

    2. Plot the least squares regression line on Scatterplot 1.  (1 mark)

The residual plot below was constructed to test the assumption of linearity for the relationship between the variables mean duration of warm spell and the mean surface temperature.

CORE, FUR2 2007 VCAA 33

  1. Explain why this residual plot supports the assumption of linearity for this relationship.  (1 mark)

  2. Write down the percentage of variation in the mean duration of a warm spell that is explained by the variation in mean surface temperature. Write your answer correct to the nearest per cent.  (1 mark)

  3. Describe the relationship between the mean duration of a warm spell and the mean surface temperature in terms of strength, direction and form.  (2 marks)

Show Answers Only
  1.  
    CORE, FUR2 2007 VCAA 3 Answer

    1. `text{See part b.ii. below.}`
    2. `text(Mean duration of warm spell)`
      `= -776.9 + 60.3 xx text(mean surface temperature)`
  2. `text(Linearity is supported because there is no)`
    `text(pattern to the residual data.)`
  3. `83text(%)`
  4. `text(Strong, positive, and linear.)`
Show Worked Solution
a.    CORE, FUR2 2007 VCAA 3 Answer

 

b.i.   `text(Mean duration of warm spell)`

`= -776.9 + 60.3 xx text(mean surface temperature)`

 

MARKER’S COMMENT: A consistent error in these type of questions is taking 2 points that are too close together!

b.ii.   `text(Taking extreme points on the above graph,)`

  `text(When)\ \ x = 13.2, \ y = -776.9 + 60.3 xx 13.2 = 19.06`

`:.\ text(Passes through)\ \ A (13.2, 19.06)`

  `text(When)\ x = 13.8, y = -776.9 + 60.3 xx 13.8 = 55.24`

`:.\ text(Passes through)\ \ B (13.8, 55.24)`

`text(*See the regression line plotted above.)`

 

c.   `text(Linearity is supported because there is no)`

`text(pattern to the residual data.)`
 

d.   `text(By Calculator,)`

`r^2 = 0.828… = 83text{%  (nearest %)}`
 

e.   `text(Strong, positive, and linear.)`

CORE, FUR2 2007 VCAA 2

The mean surface temperature (in °C) of Australia for the period 1960 to 2005 is displayed in the time series plot below.

2007 2-1

  1. In what year was the lowest mean surface temperature recorded?   (1 mark)

The least squares method is used to fit a trend line to the time series plot.

  1.   i. The equation of this trend line is found to be
  2.          mean surface temperature = – 12.361 + 0.013 × year
  3.      Use the trend line to predict the mean surface temperature (in °C) for 2010.
  4.      Write your answer correct to two decimal places.   (1 mark)

  5.  ii. The actual mean surface temperature in the year 2000 was 13.55°C.
  6.      Determine the residual value (in °C) when the trend line is used to predict the mean surface temperature for this year.
  7.      Write your answer correct to two decimal places.   (1 mark)

  8. iii. By how many degrees does the trend line predict Australia's mean surface temperature will rise each year?
  9.      Write your answer correct to three decimal places.   (1 mark)

Show Answers Only
  1. `1964`
    1. `13.77 text(°C)`
    2. `-0.09`
    3. `0.013 text(°C)`
Show Worked Solution

a.   `1964`
 

b.i.   `text{Mean surface temperature (2010)}`

`= -12.361 + 0.013 xx 2010`

`=13.769`

`= 13.77 text{°C  (2 d.p.)}`
 

b.ii.   `text{Predicted mean surface temp (2010)}`

MARKER’S COMMENT: A common error was to omit the negative sign.

`= -12.361 + 0.013 xx 2000`

`= 13.639 text(°C)`

 

`:.\ text(Residual)` `= 13.55-13.639`
  `= -0.089`
  `=-0.09\ text{(2 d.p.)}`

 

b.iii.   `text(S)text(ince the gradient of the equation = 0.013,)`

  `text(the temperature is predicted to rise 0.013°C)`

  `text(each year.)`

CORE, FUR2 2007 VCAA 1

The histogram below shows the distribution of mean yearly rainfall (in mm) for Australia over 103 years.
 

CORE, FUR2 2007 VCAA 1

  1. Describe the shape of the histogram.   (1 mark)

  2. Use the histogram to determine
  3.  i. the number of years in which the mean yearly rainfall was 500 mm or more.   (1 mark)

  4. ii. the percentage of years in which the mean yearly rainfall was between 500 mm and 600 mm.
  5.     Write your answer correct to one decimal place.  (1 mark)

Show Answers Only
  1. `text(Positively skewed)`
    1. `26`
    2. `text(19.4%)`
Show Worked Solution

a.   `text(Positively skewed.)`
 

b.i.   `text{Years with rainfall over 500 mm}`

`=13+7+3+1+1+1 =26`
 

b.ii.   `text{Years with rainfall between 500 and 600 mm (%)}`

`=20/103 xx 100text(%)`

`= 19.41…`

`=19.4 text{%  (1 d.p.)}`

CORE, FUR2 2008 VCAA 3

The arm spans (in cm) were also recorded for each of the Years 6, 8 and 10 girls in the larger survey. The results are summarised in the three parallel box plots displayed below.
 

CORE, FUR2 2008 VCAA 3

  1. Complete the following sentence.
  2. The middle 50% of Year 6 students have an arm span between _______ and _______ cm.   (1 mark)
  3. The three parallel box plots suggest that arm span and year level are associated.
  4. Explain why.   (1 mark)

  5. The arm span of 110 cm of a Year 10 girl is shown as an outlier on the box plot. This value is an error. Her real arm span is 140 cm. If the error is corrected, would this girl’s arm span still show as an outlier on the box plot? Give reasons for your answer showing an appropriate calculation.   (2 marks)

Show Answers Only
  1. `text(124 and 148)`
  2. `text(The median arm span increases with the year)`
    `text(level, or the range/IQR decreases as the year.)`
    `text(level increases.)`
  3. `text(See Worked Solution)`
Show Worked Solution

a.   `text(124 and 148 cm)`
 

b.   `text(The median arm span increases with the year)`

♦ Sub 50% mean.
MARKER’S COMMENT: Use specific metrics! Stating “arm span increases” did not receive a mark.

`text(level, or the range/IQR decreases as the year.)`

`text(level increases.)`

 

c.   `text(Consider the Year 10 boxplot,)`

MARKER’S COMMENT: The final comparison here, “Since 140 < 145” is worth a full mark.

`Q_1=160, \ Q_3=170,`

`=> IQR=170-160=10`

`Q_1-1.5 xx text(IQR)= 160-1.5 xx 10= 145`

`text(S)text(ince 140 < 145,)`

`:. 140\ text(will remain an outlier.)`

CORE, FUR2 2008 VCAA 2

In a larger survey, Years 6, 8 and 10 girls were asked what they did (walked, sat, stood, ran) for most of the time during a typical school lunch time. The results are displayed in the percentage segmented bar chart below.
 

CORE, FUR2 2008 VCAA 2 

Does the percentage segmented bar chart support the opinion that, for these girls, the lunch time activity (walked, sat or stood, ran) undertaken is associated with year level? Justify your answer by quoting appropriate percentages.   (2 marks)

Show Answers Only

`text(Yes. The table above shows that from year 6 to)`

`text(to year 8 to year 10, the percentage that ran)`

`text(changed from 78% to 40% to 10%.)`

Show Worked Solution

CORE, FUR2 2008 VCAA 2 Answer

`text(Yes. The table above shows that from year 6)`

MARKER’S COMMENT: An association does not have to be a consistent increase or decrease over the years.

`text(to year 8 to year 10, the percentage that ran)`

`text(changed from 78% to 40% to 10%.)`
 

CORE, FUR2 2009 VCAA 4

  1. Table 2 shows the seasonal indices for rainfall in summer, autumn and winter. Complete the table by calculating the seasonal index for spring.   (1 mark)

     

    CORE, FUR2 2009 VCAA 4

  2. In 2008, a total of 188 mm of rain fell during summer.

     

    Using the appropriate seasonal index in Table 2, determine the deseasonalised value for the summer rainfall in 2008. Write your answer correct to the nearest millimetre.   (1 mark)

  3. What does a seasonal index of 1.05 tell us about the rainfall in autumn?   (1 mark)

Show Answers Only
  1. `1.10`
  2. `241\ text(mm)`
  3. `text(The Autumn rainfall is 5% above the average)`
    `text(for the four seasons of the year.)`
Show Worked Solution

a.   `text(Seasonal index for Spring)`

`=4-(0.78 + 1.05 + 1.07)`

`= 1.10`

 

b.   `text{Deseasonalised value (Summer)}`

`= 188/0.78`

`=241.02…`

`=241\ text{mm (nearest mm)}`

♦ Part (c) was “poorly answered” (no exact data).
MARKER’S COMMENT: A common error was to say rainfall was above average monthly rainfall.

 

c.   `text(The Autumn rainfall is 5% above the average)`

`text(for the four seasons of the year.)`

 

CORE, FUR2 2009 VCAA 2

The time series plot below shows the rainfall (in mm) for each month during 2008.
 

CORE, FUR2 2009 VCAA 2
 

  1. Which month had the highest rainfall?   (1 mark)

  2. Use three-median smoothing to smooth the time series. Plot the smoothed time series on the plot above.
  3. Mark each smoothed data point with a cross (×).   (2 marks)

  4. Describe the general pattern in rainfall that is revealed by the smoothed time series plot.   (1 mark)

Show Answers Only
  1. `text(November)`
  2.  
    CORE, FUR2 2009 VCAA 2 Answer
     
  3. `text(Until April, the rainfall increases each month)`
    `text(and then it remains relatively constant until)`
    `text(November).`
  4.  
Show Worked Solution

a.   `text(November)`

 

b.    CORE, FUR2 2009 VCAA 2 Answer
MARKER’S COMMENT: Locate medians graphically by inspection. Explaining a general pattern with more than one trend proved challenging.

c.   `text(Until April, there is an increase in rainfall)`

`text(and then it remains relatively constant until)`

`text(November.)`

 

MATRICES, FUR2 2015 VCAA 1

Students in a music school are classified according to three ability levels: beginner (`B`), intermediate (`I`) or advanced (`A`).

Matrix `S_0`, shown below, lists the number of students at each level in the school for a particular week.

`S_0 = [(20), (60), (40)] {:(B), (I), (A):}`

  1. How many students in total are in the music school that week?   (1 mark)

The music school has four teachers, David (`D`), Edith (`E`), Flavio (`F`) and Geoff (`G`).

Each teacher will teach a proportion of the students from each level, as shown in matrix `P` below.

`{:(qquadqquadqquad{:(Dquad,Eqquad,Fqquad,G):}),(P = [(0.25,0.5,0.15,0.1)]):}`

The matrix product, `Q = S_0P`, can be used to find the number of students from each level taught by each teacher.

  1.  i. Complete matrix `Q`, shown below, by writing the missing elements in the shaded boxes.   (1 mark) 

 Matrices, FUR2 2015 VCAA 1

  1. ii. How many intermediate students does Edith teach?  (1 mark)

The music school pays the teachers $15 per week for each beginner student, $25 per week for each intermediate student and $40 per week for each advanced student.

These amounts are shown in matrix `C` below.

`{:(qquad qquad quad{:(B quad, I quad,A):}),(C = [(15, 25, 40)]):}`

The amount paid to each teacher each week can be found using a matrix calculation.

  1.  i. Write down a matrix calculation in terms of `Q` and `C` that results in a matrix that lists the amount paid to each teacher each week.   (1 mark)

  2. ii. How much is paid to Geoff each week?   (1 mark)

Show Answers Only
  1. `120`
    1. `[(5, 10, 3, 2), (15, 30, 9, 6), (10, 20, 6, 4)]`
    2. `30`
    1. `C xx Q`
    2. `$340`
Show Worked Solution

a.   `20 + 60 + 40 = 120`
 

b.i.   `Q` `= S_0 P`
    `= [(20), (60), (40)] [(0.25, 0.5, 0.15, 0.1)]`
    `= [(5, 10, 3, 2), (15, 30, 9, 6), (10, 20, 6, 4)]`

 
b.ii.   `text(Edith teaches 30 intermediate students.)`

 
c.i.   `C xx Q`

 
c.ii.   `CQ = [(15, 25, 40)] [(5, 10, 3, 2), (15, 30, 9, 6), (10, 20, 6, 4)]`

`:.\ text(Geoff’s pay)`

`= 15 xx 2 + 25 xx 6 + 40 xx 4`

`= $340`

CORE, FUR2 2009 VCAA 1

Table 1 shows the number of rainy days recorded in a high rainfall area for each month during 2008.
 

CORE, FUR2 2009 VCAA 11

The dot plot below displays the distribution of the number of rainy days for the 12 months of 2008.
 

CORE, FUR2 2009 VCAA 12

  1. Circle the dot on the dot plot that represents the number of rainy days in April 2008.   (1 mark)

  2. For the year 2008, determine

     

  3.  i. the median number of rainy days per month.   (1 mark)
  4. ii. the percentage of months that have more than 10 rainy days. Write your answer correct to the nearest per cent.   (1 mark) 
Show Answers Only
  1.  
    CORE, FUR2 2009 VCAA 12 Answer
    1. `15.5`
    2. `text(92%)`
Show Worked Solution
a.    CORE, FUR2 2009 VCAA 12 Answer

 

b.i.    `text(Median)` `= text{(6th + 7th)}/2`
    `=(15+16)/2`
    `=15.5`

 

b.ii.   `text(Months with more than 10 rainy days)`

`=11/12 xx text(100%)`

`=91.66…`

`=92text(%)\ \ text{(nearest %)}`

CORE, FUR2 2010 VCAA 3

Table 2 shows the Australian gross domestic product (GDP) per person, in dollars, at five yearly intervals for the period 1980 to 2005.
 

CORE, FUR2 2010 VCAA 31 

  1. Complete the time series plot above by plotting the GDP for the years 2000 and 2005.   (1 mark)

  2. Briefly describe the general trend in the data.   (1 mark)

In Table 3, the variable year has been rescaled using 1980 = 0, 1985 = 5 and so on. The new variable is time.

CORE, FUR2 2010 VCAA 32

  1. Use the variables time and GDP to write down the equation of the least squares regression line that can be used to predict GDP from time. Take time as the independent variable.   (2 marks)

  2. In the year 2007, the GDP was $34 900. Find the error in the prediction if the least squares regression line calculated in part c. is used to predict GDP in 2007.   (2 marks)

Show Answers Only
  1.  
    CORE, FUR2 2010 VCAA 31 Answer
  2. `text(An increasing trend.)`
  3. `text(GDP = 20 000 + 524 × time)`
  4. `752\ text(below the actual GDP)`
Show Worked Solution
a.    CORE, FUR2 2010 VCAA 31 Answer

 

b.   `text(An increasing trend.)`

 

c.   `text(By calculator,)`

MARKER’S COMMENT: Students are expected to use the variables in the diagram rather than just `x` and `y`.

`text(GDP) = 20\ 000 + 524 × text(time)`
 

d.   `text(In 2007, time = 27,)`

`text{GDP (Predicted)}\ = 20\ 000 + 524 xx 27= 34\ 148`

`:.\ text(Error)\ = 34\ 900-34\ 148= 752\ \ text{(less than real GDP)}`

CORE, FUR2 2010 VCAA 2

In the scatterplot below, average annual female income, in dollars, is plotted against average annual male income, in dollars, for 16 countries. A least squares regression line is fitted to the data.
 


 

The equation of the least squares regression line for predicting female income from male income is

female income = 13 000 + 0.35 × male income

  1. What is the explanatory variable?  (1 mark)

  2. Complete the following statement by filling in the missing information.

     

    From the least squares regression line equation it can be concluded that, for these countries, on average, female income increases by `text($________)` for each $1000 increase in male income.  (1 mark)

    1. Use the least squares regression line equation to predict the average annual female income (in dollars) in a country where the average annual male income is $15 000.  (1 mark)

    2. The prediction made in part c.i. is not likely to be reliable.

       

      Explain why.  (1 mark)


Show Answers Only

  1. `text(Male income)`
  2. `$350`
    1. `$18\ 250`
    2. `text(The model established by the regression)`
      `text(equation cannot be relied upon outside the)`
      `text(range of the given data set.)`
  4.  

Show Worked Solution

a.   `text(Male income)`
 

b.   `text(Increase in female income)`

`= 0.35 xx 1000`

`= $350`
 

c.i.   `text(Average annual female income)`

`= 13\ 000 + 0.35 xx 15\ 000`

`= $18\ 250`

♦♦ This part was poorly answered (exact data unavailable).
MARKER’S COMMENT: Many students offered “real world” explanations which did not gain a mark here.

 
c.ii.   `text(The model established by the regression)`

   `text(equation cannot be relied upon outside the)`

   `text(range of the given data set.)`

CORE, FUR2 2010 VCAA 1

Table 1 shows the percentage of women ministers in the parliaments of 22 countries in 2008.
 

CORE, FUR2 2010 VCAA 11
 

  1. What proportion of these 22 countries have a higher percentage of women ministers in their parliament than Australia?  (1 mark)

  2. Determine the median, range and interquartile range of this data.  (2 marks)

The ordered stemplot below displays the distribution of the percentage of women ministers in parliament for 21 of these countries. The value of Canada is missing.
 

    CORE, FUR2 2010 VCAA 12
 

  1. Complete the stemplot above by adding the value for Canada.  (1 mark)

  2. Both the median and the mean appropriate measures of centre for this distribution.
  3. Explain why.  (1 mark)

Show Answers Only

  1. `0.5`
  2. `text(Median= 28, Range = 56, IQR = 17)`
  3. `1 | 246`
  4. `text(S)text(ince the distribution is approximately)`

     

    `text(symmetric, the median and mean will be)`

     

    `text(appropriate measures of the centre.)`

Show Worked Solution

a.   `11/22 = 0.5`

b.   `text(22 data points,)`

`text(Median)` `=\ text{(11th + 12th)}/2`
  `= (32 + 24)/2`
  `= 28`

 

`text(Range)` `= 56 – 0`
  `= 56`

 
`Q_1=21 and Q_3=38`

`text(IQR)` `= 38 – 21`
  `= 17`

 

c.   `1 | 2 quad 4 quad 6`

♦♦ Part (d) was “poorly answered”.
MARKER’S COMMENT: The use of “symmetric” gained a mark while “evenly distributed” was deemed too vague.

 

d.   `text(S)text(ince the distribution is approximately)`

`text(symmetric, the median and mean will be)`

`text(appropriate measures of the centre.)`

CORE, FUR2 2011 VCAA 2

Table 1 shows information about a particular country. It shows the percentage of women by age at first marriage, for the years 1986, 1996 and 2006.

CORE, FUR2 2011 VCAA 2

  1. Of the women who first married in 1986, what percentage were aged 20 to 29 years inclusive?   (1 mark)

  2. Does the information in Table 1 support the opinion that, for the years 1986, 1996 and 2006, the age of women at first marriage was associated with years of marriage?
  3. Justify your answer by quoting appropriate percentages. It is sufficient to consider one age group only when justifying your answer.   (2 marks) 
Show Answers Only
  1. `text(65.5%)`
  2. `text(Yes. See Worked solutions.)`
Show Worked Solution

a.   `text(Percentage aged 20 – 29 who married in 1986)`

`=42.1 + 23.4`

`= 65.5 text(%)`
 

b.   `text(Yes.)`

`text(Considering the 25 – 29 age group, the)`

`text{percentage of women increased from 23.4%}`

`text{(1986), to 31.7% (1996), to 34.5% (2006).}`

CORE, FUR2 2011 VCAA 1

The stemplot in Figure 1 shows the distribution of the average age, in years, at which women first marry in 17 countries.
 

CORE, FUR2 2011 VCAA 11
 

  1. For these countries, determine
    1. the lowest average age of women at first marriage  (1 mark)

    2. the median average age of women at first marriage  (1 mark)

The stemplot in Figure 2 shows the distribution of the average age, in years, at which men first marry in 17 countries.
 

CORE, FUR2 2011 VCAA 12

  1. For these countries, determine the interquartile range (IQR) for the average age of men at first marriage.  (1 mark)

  2. If the data values displayed in Figure 2 were used to construct a boxplot with outliers, then the country for which the average age of men at first marriage is 26.0 years would be shown as an outlier.
  3. Explain why is this so. Show an appropriate calculation to support your explanation.  (2 marks)

Show Answers Only

    1. `text(25 years)`
    2. `text(28.2 years)`
  2. `text(1.1 years)`
  3. `text(See Worked Solutions)`

Show Worked Solution

a.i.   `text(Lowest age = 25 years)`

 

a.ii.   `text(Median age of 17 data point is the)`

   `text(9th point = 28.2 years)`

 

b.   `Q_L = 29.9, Q_U = 31.0,`

`:. IQR` `= Q_U − Q_L`
  `= 31.0 − 29.9`
  `= 1.1\ text(years)`

 

c.   `1.5 xx IQR = 1.5 xx 1.1 = 1.65`

MARKER’S COMMENT: Many students correctly calculated 28.25 but then failed to discuss how 26.0 related to it.

`Q_1 – IQR` `=29.9-1.65`
  `=28.25`

 

`text(S)text(ince  26.0 < 28.25,)`

`:. 26.0\ text(is an outlier.)`

CORE, FUR2 2012 VCAA 1

The dot plot below displays the maximum daily temperature (in °C) recorded at a weather station on each of the 30 days in November 2011. 
  

CORE, FUR2 2012 VCAA 1

  1. From this dot plot, determine

     

  2.  i. the median maximum daily temperature, correct to the nearest degree  (1 mark)
  3. ii. the percentage of days on which the maximum temperature was less than 16 °C.
  4.     Write your answer, correct to one decimal place.  (1 mark)

Records show that the minimum daily temperature for November at this weather station is approximately normally distributed with a mean of 9.5 °C and a standard deviation of 2.25 °C.

  1. Determine the percentage of days in November that are expected to have a minimum daily temperature less than 14 °C at this weather station.
  2. Write your answer, correct to one decimal place.  (1 mark)

Show Answers Only

    1. `20 text(°C)`
    2. `23.3text(%)`
  2. `97.5text(%)`

Show Worked Solution

a.i.   `text(The median is the average of the 15th and 16th)`

`\ \ text{data points (30 data points in total).}`.

`:.\ text(Median = 20 °C)`
 

a.ii.   `text{Percentage of days less than 16 °C}`

`= 7/30 xx 100text(%)`

`=23.333…`

`=23.3text{%  (to 1 d.p.)}`
 

b.    `z text{-score (14)}` `=(x-barx)/s`
    `=(14-9.5)/2.25`
    `=2`

 

CORE, FUR2 2012 VCAA 1 Answer 
 

`text(Percentage of days with a minimum below 14 °C)`

`= 95text(%) + 2.5text(%)`

`= 97.5text(%)`