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NETWORKS, FUR2 2018 VCAA 3

At the Zenith Post Office all computer systems are to be upgraded.

This project involves 10 activities, `A` to `J`.

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

  1. Determine the earliest starting time, in hours, for activity `I`.  (1 mark)
  2. The minimum completion time for the project is 15 hours.

     

    Write down the critical path.  (1 mark)

  3. Two of the activities have a float time of two hours.

     

    Write down these two activities.  (1 mark)

  4. For the next upgrade, the same project will be repeated but one extra activity will be added.
    This activity has a duration of one hour, an earliest starting time of five hours and a latest starting time of 12 hours.

     

    Complete the following sentence by filling in the boxes provided.  (1 mark)

     

    The extra activity could be represented on the network above by a directed edge from the

      end of activity   
 
 to the start of activity   
 
Show Answers Only
  1. `10\ text(hours)`
  2. `B – E – G – H – J`
  3.  `text(Activity)\ A\ text(and)\ C`
  4. `text(end of activity)\ E\ text(to the start of activity)\ J`
Show Worked Solution

a.  `text(Longest path to)\ I:`

`B -> E -> G`

`:.\ text(EST for)\ \ I` `= 2 + 3 + 5`
  `= 10\ text(hours)`

 
b.  `B – E – G – H – J`
 

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

♦ Mean mark 45%.

 


 

`:.\ text(Activity)\ A\ text(and)\ C\ text(have a 2 hour float time.)`
 

d.   `text(end of activity)\ E\ text(to the start of activity)\ J`

♦♦ Mean mark 25%.
 

`text(By inspection of forward and backward scanning:)`

`text(EST of 5 hours is possible after activity)\ E.`

`text(LST of 12 hours after activity)\ E -> text(edge has weight)`

`text(of 1 and connects to)\ J`

NETWORKS, FUR2 2018 VCAA 2

In one area of the town of Zenith, a postal worker delivers mail to 10 houses labelled as vertices `A` to `J` on the graph below.
 

  1. Which one of the vertices on the graph has degree 4?  (1 mark)

For this graph, an Eulerian trail does not currently exist.

  1. For an Eulerian trail to exist, what is the minimum number of extra edges that the graph would require.  (1 mark)
  2. The postal worker has delivered the mail at `F` and will continue her deliveries by following a Hamiltonian path from `F`.

     

    Draw in a possible Hamiltonian path for the postal worker on the diagram below.  (1 mark)

 

Show Answers Only
  1. `text(Vertex)\ F`
  2. `2`
  3.  `text(See Worked Solutions)`
Show Worked Solution

a.   `text(Vertex)\ F`
 

b. `text(Eulerian trail)\ =>\ text(all edges used exactly once.)`

`text(6 vertices are odd)`

`=> 2\ text(extra edges could create graph with only 2)`

`text(odd vertices)`

`:.\ text(Minimum of 2 extra edges.)`
 

c.  `text{One example (of a number) beginning at)\ F:}`
 


 

`text{(Note: path should not return to}\ F text{)}`

MATRICES, FUR2 2018 VCAA 2

The Westhorn Council must prepare roads for expected population changes in each of three locations: main town `(M)`, villages `(V)` and rural areas `(R)`.

The population of each of these locations in 2018 is shown in matrix  `P_2018`  below.
 

`P_2018 = [(2100),(1800),(1700)]{:(M),(V),(R):}`
 

The expected annual change in population in each location is shown in the table below.
 

     
 

  1. Write down matrix  `P_2019`, which shows the expected population in each location in 2019 (1 mark)
  2. The expected population in each of the three locations in 2019 can be determined from the matrix product.
     
    `qquad qquad P_2019 = F xx P_2018,`  where `F` is a diagonal matrix.
     
    Write down matrix  `F`.  (1 mark)

     

Show Answers Only
  1. `P_2019 = [(1.04 xx 2100),(0.99 xx 1800),(0.98 xx 1700)] = [(2184),(1782),(1666)]`
  2. `F = [(1.04, 0, 0),(0, 0.99, 0),(0, 0, 0.98)]`
Show Worked Solution

a.   `P_2019 = [(1.04 xx 2100),(0.99 xx 1800),(0.98 xx 1700)] = [(2184),(1782),(1666)]`

♦ Mean mark part (b) 40%.
COMMENT: Many students included 0.04, -0.01 and -0.02 in this matrix. Know why this is incorrect!

 

b.   `F = [(1.04, 0, 0),(0, 0.99, 0),(0, 0, 0.98)]`

MATRICES, FUR2 2018 VCAA 1

A toll road is divided into three sections, `E, F` and `G`.

The cost, in dollars, to drive one journey on each section is shown in matrix `C` below.
 

`C = [(3.58),(2.22),(2.87)]{:(E),(F),(G):}`
 

  1.  What is the cost of one journey on section `G`? (1 mark)
  2.  Write down the order of matrix `C`.  (1 mark)
  3.  One day Kim travels once on section `E` and twice on section `G`.

     

    His total toll cost for this day can be found by the matrix product  `M xx C`.

     

    Write down the matrix  `M`.  (1 mark)

Show Answers Only
  1. `$2.87`
  2. `text(Order:)\ 3 xx 1`
  3. `M = [(1, 0, 2)]`
Show Worked Solution

a.   `$2.87`
 

b.   `text(Order:)\ 3 xx 1`
 

c.   `M = [(1, 0, 2)]`

CORE, FUR2 2018 VCAA 5

After three years, Julie withdraws $14 000 from her account to purchase a car for her business.

For tax purposes, she plans to depreciate the value of her car using the reducing balance method.

The value of Julie’s car, in dollars, after `n` years, `C_n`, can be modelled by the recurrence relation shown below

`C_0 = 14\ 000, qquad C_(n + 1) = R xx C_n`

  1. For each of the first three years of reducing balance depreciation, the value of `R` is 0.85

     

    What is the annual rate of depreciation in the value of the car during these three years? (1 mark)

  2. For the next five years of reducing balance depreciation, the annual rate of depreciation in the value of the car is changed to 8.6%.

     

    What is the value of the car eight years after it was purchased?
    Round your answer to the nearest cent. (2 marks)

Show Answers Only
  1. `15 text(%)`
  2. `$5484.23\ text{(nearest cent)}`
Show Worked Solution

a.  `text(S)text(ince)\ \ R = 0.85,`

COMMENT: Note almost half of students answered incorrectly here!

`=> 85 text(% of the car’s value remains at the end of each)`

      `text{year (vs the value at the start of the same year.)}`

`:.\ text(Annual rate of depreciation) = 15 text(%)`

 

b.   `text(Value after 3 years)`

♦ Mean mark 42%.

`C_3` `= (0.85)^3 xx 14\ 000`
  `= $8597.75`

 
`:.\ text(Value after 8 years)`

`C_8` `= (0.914)^5 xx C_3`
  `= (0.914)^5 xx 8597.75`
  `= $5484.23\ text{(nearest cent)}`

CORE, FUR2 2018 VCAA 4

Julie deposits some money into a savings account that will pay compound interest every month.

The balance of Julie’s account, in dollars, after `n` months, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 12\ 000, qquad V_(n + 1) = 1.0062 V_n` 

  1.  How many dollars does Julie initially invest?  (1 mark)
  2.  Recursion can be used to calculate the balance of the account after one month.

    1. Write down a calculation to show that the balance in the account after one month, `V_1`, is  $12 074.40. (1 mark)
    2. After how many months will the balance of Julie’s account first exceed $12 300  (1 mark)
  3.  A rule of the form  `V_n = a xx b^n`  can be used to determine the balance of Julie's account after `n` months.

    1. Complete this rule for Julie’s investment after `n` months by writing the appropriate numbers in the boxes provided below. (1 mark)
       
balance = 
 
  × 
 
  `n`

 

    1. What would be the value of  `n`  if Julie wanted to determine the value of her investment after three years?  (1 mark)
Show Answers Only
  1. `$12\ 000`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `4\ text(months)`
    1. `text(balance) = 12\ 000 xx 1.0062^n`
    2. `36`
Show Worked Solution

a.   `$12\ 000`
 

b.i.   `V_1` `= 1.0062 xx V_0`
    `= 1.0062 xx 12000`
    `= $12\ 074.40\ text(… as required.)`

 

b.ii.   `V_2` `= 1.0062 xx 12\ 074.40 = 12\ 149.26`
  `V_3` `= 1.0062 xx 12\ 149.26 = 12\ 224.59`
  `V_4` `= 1.0062 xx 12\ 224.59 = 12\ 300.38`

 
`:.\ text(After 4 months)`

 
c.i.  `text(balance) = 12\ 000 xx 1.0062^n`

 
c.ii.  `n = 12 xx 3 = 36`

CORE, FUR2 2018 VCAA 3

Table 3 shows the yearly average traffic congestion levels in two cities, Melbourne and Sydney, during the period 2008 to 2016. Also shown is a time series plot of the same data.

The time series plot for Melbourne is incomplete.
 

  1. Use the data in Table 3 to complete the time series plot above for Melbourne.  (1 mark)

     

                (Answer on the time series plot above.)

  2. A least squares line is used to model the trend in the time series plot for Sydney. The equation is
     
         `text(congestion level = −2280 + 1.15 × year)`
     

    1. Draw this least squares line on the time series plot.  (1 mark)

       

            (Answer on the time series plot above.)

    2. Use the equation of the least squares line to determine the average rate of increase in percentage congestion level for the period 2008 to 2016 in Sydney.

       

      Write your answer in the box provided below.  (1 mark)

 
  % per year
    1. Use the least squares line to predict when the percentage congestion level in Sydney will be 43%.  (1 mark)

 
The yearly average traffic congestion level data for Melbourne is repeated in Table 4 below.
 


 

  1. When a least squares line is used to model the trend in the data for Melbourne, the intercept of this line is approximately –1514.75556

     

    Round this value to four significant figures.  (1 mark)

  2. Use the data in Table 4 to determine the equation of the least squares line that can be used to model the trend in the data for Melbourne. The variable year is the explanatory variable.

     

    Write the values of the intercept and the slope of this least squares line in the appropriate boxes provided below.

     

    Round both values to four significant figures.  (2 marks)
     

congestion level = 
 
  + 
 
  × year

 

  1. Since 2008, the equations of the least squares lines for Sydney and Melbourne have predicted that future traffic congestion levels in Sydney will always exceed future traffic congestion levels in Melbourne.

     

    Explain why, quoting the values of appropriate statistics.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `1.15 text(%)`
    3. `2020`
  2. `-1515`
  3. `text(congestion level) = -1515 + 0.7667 xx text(year)`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.   

 

b.i.   

 

b.ii.  `text(The least squares line is 1.15% higher each year.)`

♦ Mean mark (b)(ii) 36%.
COMMENT: Major problems caused by part (b)(ii). Review!

  ` :.\ text(Average rate of increase) = 1.15 text(%)`

 

b.iii.    `text(Find year when:)`
  `43` `= -2280 + 1.15 xx text(year)`
  `text(year)` `= 2323/1.15`
    `= 2020`

 

c.  `-1515`

 

d.   `text(congestion level) = -1515 + 0.7667 xx text(year)`

 

e.   `text(Melbourne congestion level in 2008)`

♦♦♦ Mean mark 18%.

`= -1515 + 0.7667 xx 2008`

`= 24.5 text(%)`

 
`text{In 2008 Sydney has higher congestion (29.2 > 24.5)}`

`text(After 2008, Sydney congestion grows at 1.15% per)`

`text(year and Melbourne grows at 0.7667% per year.)`

`:.\ text(Sydney predicted to always exceed Melbourne.)`

CORE, FUR2 2018 VCAA 2

The congestion level in a city can be recorded as the percentage increase in travel time due to traffic congestion in peak periods (compared to non-peak periods).

This is called the percentage congestion level.

The percentage congestion levels for the morning and evening peak periods for 19 large cities are plotted on the scatterplot below.
 


 

  1. Determine the median percentage congestion level for the morning peak period and the evening peak period.

     

    Write your answers in the appropriate boxes provided below.  (2 marks)

Median percentage congestion level for morning peak period
%
Median percentage congestion level for evening peak period
%

 
A least squares line is to be fitted to the data with the aim of predicting evening congestion level from morning congestion level.

The equation of this line is.

evening congestion level = 8.48 + 0.922 × morning congestion level
 

  1. Name the response variable in this equation.  (1 mark)
  2. Use the equation of the least squares line to predict the evening congestion level when the morning congestion level is 60%.  (1 mark)
  3. Determine the residual value when the equation of the least squares line is used to predict the evening congestion level when the morning congestion level is 47%.

     

    Round your answer to one decimal place?  (2 marks)

  4. The value of the correlation coefficient `r` is 0.92

     

    What percentage of the variation in the evening congestion level can be explained by the variation in the morning congestion level?

     

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

Show Answers Only
  1. `52 text(%);\ 56 text(%)`
  2. `text(evening congestion level)`
  3. `63.8 text(%)`
  4. `-1.8 text{% (to 1 d.p.)}`
  5. `85 text(%)`
Show Worked Solution

a.   `19\ text(data points of morning peak)`

Mean mark 56%.
COMMENT: This question was surprisingly poorly answered. Review carefully!

`text(Median is 10th data point moving left to right.)`

`:.\ text{Median (morning peak) = 52%}`

 

`text(Median of evening peak is 10th data point)`

`text(moving bottom to top.)`

`:.\ text{Median (afternoon peak) = 56%}`
 

b.   `text(Response variable is evening congestion level.)`
 

c.    `text(evening congestion level)` `= 8.48 + 0.922 xx 60`
    `= 63.8 text(%)`

 
d.  `text(When morning level = 47%, Actual = 50%)`

Mean mark part (d) 53%.
COMMENT: Many students had problems at a number of stages in this part.

`text(Residual)` `=\ text(Actual evening congestion − predicted)`
  `= 50 – (8.48 + 0.922 xx 47)`
  `= -1.814`
  `= -1.8 text{% (to 1 d.p.)}`

 

e.    `r` `= 0.92`
  `r^2` `= 0.8464`
    `= 85 text{% (nearest whole)}`

 
`:. 85 text(% of the variations is explained.)`

CORE, FUR2 2018 VCAA 1

The data in Table 1 relates to the impact of traffic congestion in 2016 on travel times in 23 cities in the United Kingdom (UK).

The four variables in this data set are:

  • city — name of city
  • congestion level — traffic congestion level (high, medium, low)
  • size — size of city (large, small)
  • increase in travel time — increase in travel time due to traffic congestion (minutes per day).
  1. How many variables in this data set are categorical variables?  (1 mark)
  2. How many variables in this data set are ordinal variables  (1 mark)
  3. Name the large UK cities with a medium level of traffic congestion.  (1 mark)
  4. Use the data in Table 1 to complete the following two-way frequency table, Table 2.  (2 marks)
     

     


     

  5. What percentage of the small cities have a high level of traffic congestion?  (1 mark)

Traffic congestion can lead to an increase in travel times in cities. The dot plot and boxplot below both show the increase in travel time due to traffic congestion, in minutes per day, for the 23 UK cities.
 


 

  1. Describe the shape of the distribution of the increase in travel time for the 23 cities.  (1 mark)
  2. The data value 52 is below the upper fence and is not an outlier.

     

    Determine the value of the upper fence.  (1 mark)

Show Answers Only
  1. `3\ text(city, congestion level, size)`
  2. `2\ text(congestion level, size)`
  3. `text(Newcastle-Sunderland and Liverpool)`
  4. `text(See Worked Solutions)`
  5. `25 text(%)`
  6. `text(Positively skewed)`
  7. `52.5`
Show Worked Solution

a.   `3\ – text(city, congestion level, size)`
 

b.   `2\ – text(congestion level, size)`
 

c.   `text(Newcastle-Sunderland and Liverpool)`
 

d.   

 

e.    `text(Percentage)` `= text(Number of small cities high congestion)/text(Number of small cities) xx 100`
    `= 4/16 xx 100`
    `= 25 text(%)`

 
f.   `text(Positively skewed)`
 

g.   `IQR = 39 – 30 = 9`
 

`text(Calculate the Upper Fence:)`

`Q_3 + 1.5 xx IQR` `= 39 + 1.5 xx 9`
  `= 52.5`

MATRICES, FUR1 2018 VCAA 4 MC

Matrix `P` is a 4 × 4 permutation matrix.

Matrix `W` is another matrix such that the matrix product `P × W` is defined.

This matrix product results in the entire first and third rows of matrix `W` being swapped.

The permutation matrix `P` is

A.  `[(0,0,0,1),(0,1,0,0),(0,0,1,0),(0,0,0,1)]` B.  `[(0,0,1,0),(0,1,0,0),(1,0,0,0),(0,0,0,1)]` C.  `[(1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)]`
           
D.  `[(1,0,0,0),(0,0,0,0),(0,0,1,0),(0,0,0,0)]` E.  `[(1,0,0,0),(0,1,0,0),(1,0,0,0),(0,0,0,1)]`    
Show Answers Only

`B`

Show Worked Solution

`text(By trial and error:)`

`text(S)text(ince)\ P xx W\ \ text(is defined,)`

`=>\ W\ text(can be a 4 × 1 matrix)`
 

`text(Consider option)\ B,`

`[(0,0,1,0),(0,1,0,0),(1,0,0,0),(0,0,0,1)][(a),(b),(c),(d)] = [(c),(b),(a),(d)]`
 

`=> B`

MATRICES, FUR1 2018 VCAA 2 MC

The matrix product `[(4,2,0)] xx [(4),(12),(8)]` is equal to

  1. `[144]`
  2. `[(16),(24),(0)]`
  3. `4 xx [(1,2,0)] xx [(1),(12),(8)]`
  4. `2 xx [(2,1,0)] xx [(2),(6),(4)]`
  5. `4 xx [(2,1,0)] xx [(2),(6),(4)]`
Show Answers Only

`E`

Show Worked Solution
`[(4,2,0)] xx [(4),(12),(8)]` `= 2[(2,1,0)] xx 2[(2),(6),(4)]`
  `= 4 xx [(2,1,0)] xx [(2),(6),(4)]`

`=> E`

CORE, FUR1 2018 VCAA 21 MC

Which one of the following recurrence relations could be used to model the value of a perpetuity investment, `P_n`, after `n` months?

  1. `P_0 = 120\ 000,qquadP_(n + 1) = 1.0029 × P_n - 356`
  2. `P_0 = 180\ 000,qquadP_(n + 1) = 1.0047 × P_n - 846`
  3. `P_0 = 210\ 000,qquadP_(n + 1) = 1.0071 × P_n - 1534`
  4. `P_0 = 240\ 000,qquadP_(n + 1) = 0.0047 × P_n - 2232`
  5. `P_0 = 250\ 000,qquadP_(n + 1) = 0.0085 × P_n - 2125`
Show Answers Only

`B`

Show Worked Solution

`text(Perpetuity) => P_0 = P_1 = P_2 = … = P_n`

`text(By trial and error, consider option)\ B:`

`P_1` `= 1.0047 xx 180\ 000 – 846`
  `= 180\ 000`
  `= P_0`

 
`=> B`

CORE, FUR1 2018 VCAA 19 MC

Daniel borrows $5000, which he intends to repay fully in a lump sum after one year.

The annual interest rate and compounding period for five different compound interest loans are given below:
 
• Loan I – 12.6% per annum, compounding weekly
• Loan II – 12.8% per annum, compounding weekly
• Loan III – 12.9% per annum, compounding weekly
• Loan IV – 12.7% per annum, compounding quarterly
• Loan V – 13.2% per annum, compounding quarterly

When fully repaid, the loan that will cost Daniel the least amount of money is

  1. Loan I.
  2. Loan II.
  3. Loan III.
  4. Loan IV.
  5. Loan V.
Show Answers Only

`D`

Show Worked Solution

`text(Weekly compounding loans ⇒ Loan I is cheaper)`

`text(Repayment of Loan I)` `= 5000 xx (1 + 12.6/(52 xx 100))^52`
  `= $5670.55`

 
`text(Quarterly compounding loans ⇒ Loan IV is cheaper)`

`text(Repayment of Loan IV)` `= 5000 xx (1 + 2.7/(12 xx 100))^12`
  `= $5136.68`

 
`:.\ text(Loan IV is cheapest.)`

`=> D`

CORE, FUR1 2018 VCAA 17-18 MC

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

`V_0 = 46\ 000, quadqquadV_(n + 1) = 1.0034V_n + 500`

 
Part 1

What is the value of the regular payment added to the principal of this annuity investment?

  1.   $34.00
  2. $156.40
  3. $466.00
  4. $500.00
  5. $656.40

 
Part 2

Between the second and third years, the increase in the value of this investment is closest to

  1.   $656
  2.   $658
  3.   $661
  4. $1315
  5. $1975
Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Regular payment = $500)`

`=> D`
 

`text(Part 2)`

`V_1` `= 1.0034 xx 46\ 000 + 500 = $46\ 656.40`
`V_2` `= 1.0034 xx 46\ 656.40 + 500 = $47\ 315.03`
`V_3` `= 1.0034 xx 47\ 315.03 + 500 = $47\ 975.90`

 

`:.\ text(Increase)` `= 47\ 975.90 – 47\ 315.03`
  `= $660.87`

`=> C`

CORE, FUR1 2018 VCAA 16 MC

The quarterly sales figures for a large suburban garden centre, in millions of dollars, for 2016 and 2017 are displayed in the table below.

Using these sales figures, the seasonal index for Quarter 3 is closest to

  1.  1.28
  2.  1.30
  3.  1.38
  4.  1.46
  5.  1.48
Show Answers Only

`C`

Show Worked Solution

`text{Quarterly average (2016)}`

Mean mark 51%.

`= (1.73 + 2.87 + 3.34 + 1.23) ÷ 4`

`= 2.2925`

`text{Quarterly average (2017)}`

`= (1.03 + 2.45 + 2.05 + 0.78) ÷ 4`

`= 1.5775`

`:.\ text{Seasonal Index (Q3)}` `= (3.34/2.2925 + 2.05/1.5775)-: 2`
  `= 1.3782…`

`=> C`

CORE, FUR1 2018 VCAA 12 MC

A  `log_10(y)`  transformation was used to linearise a set of non-linear bivariate data.

A least squares line was then fitted to the transformed data.

The equation of this least squares line is

`log_10(y) = 3.1 - 2.3x`

This equation is used to predict the value of `y` when  `x = 1.1`

The value of `y` is closest to

  1. –0.24
  2.   0.57
  3.   0.91
  4.   1.6
  5.   3.7
Show Answers Only

`E`

Show Worked Solution
`log_10y` `= 3.1 – 2.3(1.1)`
  `= 0.57`
`:.y` `= 10^(0.57)`
  `= 3.71…`

 
`=> E`

CORE, FUR1 2018 VCAA 11 MC

Freya uses the following data to generate the scatterplot below.

The scatterplot shows that the data is non-linear.

To linearise the data, Freya applies a reciprocal transformation to the variable `y`.

She then fits a least squares line to the transformed data.

With `x` as the explanatory variable, the equation of this least squares line is closest to

  1. `1/y = −0.0039 + 0.012x`
  2. `1/y = −0.025 + 1.1x`
  3. `1/y = 7.8 - 0.082x`
  4. `y = 45.3 + 59.7 xx 1/x`
  5. `y = 59.7 + 45.3 xx 1/x`
Show Answers Only

`A`

Show Worked Solution

`text(By trial and error:)`

COMMENT: Carefully choosing a data point to test all options can sometimes eliminate all incorrect answers.

`text(Consider Option)\ A,`

`text{Using (9,9) to test .. }`

`1/9` `~~ -0.0039 + 0.012x`
`0.111…` `~~0.1041\ \ text{(a close approximation)}`
   

`text(All other options are not close.)`

`=> A`

CORE, FUR1 2018 VCAA 10 MC

In a study of the association between a person’s height, in centimetres, and body surface area, in square metres, the following least squares line was obtained.

body surface area = –1.1 + 0.019 × height

Which one of the following is a conclusion that can be made from this least squares line?

  1. An increase of 1 m² in body surface area is associated with an increase of 0.019 cm in height.
  2. An increase of 1 cm in height is associated with an increase of 0.019 m² in body surface area.
  3. The correlation coefficient is 0.019
  4. A person’s body surface area, in square metres, can be determined by adding 1.1 cm to their height.
  5. A person’s height, in centimetres, can be determined by subtracting 1.1 from their body surface area, in square metres.
Show Answers Only

`B`

Show Worked Solution

`text(By definition of the equation, option)\ B\ text(is a)`

♦ Mean mark 51%.

`text(correct conclusion.)`

`=> B`

CORE, FUR1 2018 VCAA 7-9 MC

The scatterplot below displays the resting pulse rate, in beats per minute, and the time spent exercising, in hours per week, of 16 students. A least squares line has been fitted to the data.
 

 
Part 1

Using this least squares line to model the association between resting pulse rate and time spent exercising, the residual for the student who spent four hours per week exercising is closest to

  1. –2.0 beats per minute.
  2. –1.0 beats per minute.
  3. –0.3 beats per minute.
  4.   1.0 beats per minute.
  5.   2.0 beats per minute.

 
Part 2

The equation of this least squares line is closest to

  1. resting pulse rate = 67.2 – 0.91 × time spent exercising
  2. resting pulse rate = 67.2 – 1.10 × time spent exercising
  3. resting pulse rate = 68.3 – 0.91 × time spent exercising
  4. resting pulse rate = 68.3 – 1.10 × time spent exercising
  5. resting pulse rate = 67.2 + 1.10 × time spent exercising

 
Part 3

The coefficient of determination is 0.8339

The correlation coefficient `r` is closest to

  1. –0.913
  2. –0.834
  3. –0.695
  4.   0.834
  5.   0.913
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`text(Predicted rate = 64 bpm)`

`:.\ text(Residual)` `=\ text(actual − predicted)`
  `= 63 – 64`
  `= −1.0\ text(bpm)`

`=> B`

 

`text(Part 2)`

♦ Mean mark 46%.

`text(By elimination:)`

`ytext(-intecept) != 67.2\ text(since the)\ xtext(-axis starts)`

`text(at)\ \ x = 1\ \ (text(not)\ \ x = 0)`

`:.\ text(Eliminate)\ A, B\ text(and)\ E.`
 

`text(Consider the gradient.)`

`text(For each horizontal run of 1 unit, the line)`

`text(decreases vertically more than 1 unit.)`

`:.\ text(Eliminate)\ C.`

`=> D`

 

`text(Part 3)`

♦ Mean mark 45%.

`text(The correlation is clearly negative.)`

`:. r` `= −sqrt(0.8339)`
  `= −0.913`

`=>A`

CORE, FUR1 2018 VCAA 6 MC

Data was collected to investigate the association between the following two variables:

    • age (29 and under, 30–59, 60 and over)
    • uses public transport (yes, no)

Which one of the following is appropriate to use in the statistical analysis of this association?

  1. a scatterplot
  2. parallel box plots
  3. a least squares line
  4. a segmented bar chart
  5. the correlation coefficient r
Show Answers Only

`=> D`

Show Worked Solution

`text(A segmented bar chart is the best way to display the association)`

`text(between these two categorical variables.)`

`=> D`

CORE, FUR1 2018 VCAA 3-5 MC

The pulse rates of a population of Year 12 students are approximately normally distributed with a mean of 69 beats per minute and a standard deviation of 4 beats per minute.
 

Part 1

A student selected at random from this population has a standardised pulse rate of  `z = –2.5`

This student’s actual pulse rate is

  1.  59 beats per minute.
  2.  63 beats per minute.
  3.  65 beats per minute.
  4.  73 beats per minute.
  5.  79 beats per minute.
     

Part 2

Another student selected at random from this population has a standardised pulse rate of  `z=–1.`

The percentage of students in this population with a pulse rate greater than this student is closest to

  1.   2.5%
  2.   5%
  3.  16%
  4.  68%
  5.  84%
     

Part 3

A sample of 200 students was selected at random from this population.

The number of these students with a pulse rate of less than 61 beats per minute or greater than 73 beats per minute is closest to

  1.  19
  2.  37
  3.  64
  4.  95
  5. 190
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`ztext(-score)` `= (x – barx)/s`
`−2.5` `= (x – 69)/4`
`x – 69` `= −10`
`:. x` `= 59`

 
`=> A`
 

`text(Part 2)`

`ztext(-score) = −1`
 

`text(% above)` `= 34 + 50`
  `= 84 text(%)`

`=> E`
 

`text(Part 3)`

`z-text(score)\ (61)` `= (61 – 69)/4 = −2`
`z-text(score)\ (73)` `= (73 – 69)/4 = 1`

 


 

`text(Percentage)` `= 2.5 + 16`
  `= 18.5text(%)`

 

`:.\ text(Number of students)` `= 18.5 text(%) xx 200`
  `= 37`

`=> B`

Networks, STD2 N3 SM-Bank 31

Murray is building a new garage. The project involves activities `A` to `L`.
 

 

The network diagram shows these activities and their completion times in days.

  1. Which TWO activities immediately precede activity `G`?  (1 mark)

  2. By completing the diagram shown, calculate the minimum time required to build the new garage.  (2 marks)

  3. Hence, what is the float time for activity `E`?  (1 mark)

Show Answers Only
  1. `text(Activity)\ C\ text(and)\ D.`
  2. `30\ text(days)`
  3. `6\ text(days)`
Show Worked Solution

a.   `text(Activity)\ C\ text(and)\ D.`

 

b.   

`text(Critical path)\ \ A – D – G – L`

`= 1 + 10 + 13 + 6`

`= 30\ text(days)`

 

c.   `text(Float time of Activity)\ E`

`= 14 – 8`

`= 6\ text(days)`

Networks, STD2 N2 SM-Bank 30

An amusement park has five toilet areas, `A`, `B`, `C`, `D` and `E`, which are connected by pathways.

The table shows the length of some of the pathways, in metres.
 

 
The following network diagram is drawn to represent this information and a correct minimum spanning tree is shown by the solid lines.

Complete the network diagram including a possible value for each of the two edges `AC` and `BE`, and justify why `AC` and `BE` were not included as part of the minimum spanning tree.  (3 marks)

Show Answers Only

 
`BE > 500, AC > 450`

Show Worked Solution

`text(Consider Kruskal’s algorithm for the given minimum spanning tree:)`

`CB ->\ text{1st edge (200 – minimum weight)}`

`BD ->\ text{2nd edge (300)}`

`AD ->\ text{3rd edge (450)}`

`=> AC=460\ \ text{(choose any value such that}\ \ AC>450)`
 

`DE ->\ text{4th edge (500)}`

`=> BE=510\ \ text{(choose any value such that}\ \ BE>500)`
 

`:. BE > 500\ (BE\ text(must be greater than)\ DE)`

`:. AC > 450\ (AC\ text(must be greater than)\ AD)`

Networks, STD2 N2 SM-Bank 28

In central Queensland, there are four petrol stations `A`, `B`, `C` and `D`. The table shows the length, in kilometres, of roads connecting these petrol stations.
 


 

  1. Construct a network diagram to represent the information in the table.  (2 marks)

  2. A petrol tanker needs to refill each station. It starts at Station `A` and visits each station.

     

    Calculate the shortest distance that can be travelled by the petrol tanker. In your answer, include the order the petrol stations are refilled.  (2 marks)

Show Answers Only
  1.  
  2. `380\ text(km)`
Show Worked Solution
a.   

 

b.   `text(Shortest Path from)\ A\ (text(visiting all stations))`

`A – B – D – C`

`text(Distance)` `= 170 + 90 + 120`
  `= 380\ text(km)`

Networks, STD2 N2 SM-Bank 25 MC

In a town, there are four cafes `W`, `X`, `Y` and `Z`. The table shows the distances, in metres, of paved footpath connecting the cafes.

A coffee supplier needs to visit each cafe.

What is the shortest distance she needs to walk along the paved footpath if she starts at cafe `W`?

  1. 260 m
  2. 320 m
  3. 330 m
  4. 360 m
Show Answers Only

`text(B)`

Show Worked Solution

`text(Possible paths:)`

`W – Z – X – Y = 120 + 70 + 140 = 330\ text(m)`

`W – Z – Y – X = 120 + 100 + 140 = 360\ text(m)`

`W – X – Z – Y = 150 + 70 + 100 = 320\ text(m)`

`W – X – Y – Z = 150 + 140 + 100 = 390\ text(m)`

`=>\ text(B)`

Measurement, STD2 M7 SM-Bank 7

The scale on a given map is `1:80\ 000`.

If the actual distance between two points is 3.4 kilometres, how far apart on the map would be the two points be, in centimetres?  (2 marks)

Show Answers Only

`4.25\ text(cm)`

Show Worked Solution

`text(Actual distance on map)`

`= text(real distance)/text(scale)`

`= text(3.4 km)/(80\ 000)`

`= text(3400 m)/(80\ 000)`

`= 0.0425\ text(m)`

`= 4.25\ text(cm)`

Measurement, STD2 M2 SM-Bank 2

Bert is in Moscow, which is three hours behind of Coordinated Universal Time (UTC).

Karen is in Sydney, which is eleven hours ahead of UTC.

  1. Bert is going to ring Karen at 9 pm on Tuesday, Moscow time. What day and time will it be in Sydney when he rings?  (1 mark)

  2. Karen is going to fly from Sydney to Moscow. Her flight will leave on Wednesday at 8 am, Sydney time, and will take 15 hours. What day and time will it be in Moscow when she arrives?  (2 marks)

Show Answers Only
  1. `11\ text(am Wednesday)`
  2. `9\ text(am Wednesday)`
Show Worked Solution

a.   `text(Bert is 14 hours behind Karen.)`

`text(At 9 pm in Moscow:)`

`text(Time in Sydney)` `= 9\ text(pm + 14 hours)`
  `= 11\ text(am Wednesday)`

 

b.    `text(Arrival time)` `= 8\ text(am + 15 hours)`
    `= 11\ text{pm Wednesday (Sydney time)}`

 

`:.\ text(Time in Moscow)` `= 11\ text(pm less 14 hours)`
  `= 9\ text(am Wednesday)`

Measurement, STD2 M7 SM-Bank 6

In a raffle, the total prize money is shared among the first two tickets drawn in the ratio  6 : 4.

The prize for the second ticket drawn is $360.

What is the total prize money?  (2 marks)

Show Answers Only

`$900`

Show Worked Solution

`4/10 xx text(Total prize money = $360)`

`=> 1/10 xx text(Total prize money = $90)`

`:.\ text(Total prize money)` `= 90 xx 10`
  `= $900`

Measurement, STD2 M7 SM-Bank 4

Blood pressure is measured using two numbers: systolic pressure and diastolic pressure. If the measurement shows 120 systolic and 80 diastolic, it is written as 120/80.

The bars on the graph show the normal range of blood pressure for people of various ages.

  1. What is the normal range of blood pressure for a 53-year-old?  (2 marks)

  2. Ralph, aged 53, had a blood pressure reading of 173 over 120. A doctor prescribed Ralph a medication to reduce his blood pressure to be within the normal range. To check that the medication was being effective, the doctor measured Ralph's blood pressure for 10 weeks and recorded the following results.  
     

     
    With reference to the data provided, comment on the effectiveness of the medication during the 10-week period in returning Ralph’s blood pressure to the normal range.  (3 marks)

Show Answers Only

a.   `text{Normal range (systolic): 117 – 145}`

       `text{Normal range (diastolic): 81 – 89}`

b.   `text(Effectiveness of medication)`

`text(• systolic and diastolic pressure reduced below original)`

   `text(readings from week 3 onwards.)`

`text(• systolic pressure in normal range from week 5 onwards.)`

`text(• diastolic pressure in normal range from week 7 onwards.)`

`text(• blood pressure only in the normal range from week 7.)`

Show Worked Solution

a.   `text{Normal range (systolic): 117 – 145}`

`text{Normal range (diastolic): 81 – 89}`

 

b.   `text(Effectiveness of medication)`

`text(• systolic and diastolic pressure reduced below original)`

     `text(readings from week 3 onwards.)`

`text(• systolic pressure in normal range from week 5 onwards.)`

`text(• diastolic pressure in normal range from week 7 onwards.)`

`text(• blood pressure only in the normal range from week 7.)`

Measurement, STD2 M1 SM-Bank 26 MC

During a flood, 12.5 hectares of land was covered by water to a depth of 30 cm.

How many kilolitres of water covered the land?

(1 hectare = 10 000 m² and 1 m³ = 1000 L)

  1. 3.75 kL
  2. 37.5 kL
  3. 37 500 kL
  4. 37 500 000 kL
Show Answers Only

`=>\ text(C)`

Show Worked Solution

`text(Volume of water in m³:)`

`text(Vol)` `= Ah`
  `= 12.5 xx 10\ 000 xx 30\ text(cm)`
  `= 12.5 xx 10\ 000 xx 0.3`
  `= 37\ 500 \ text(m³)`
  `= 37\ 500\ text(kL)`

 
`=>\ text(C)`

Algebra, SPEC2 2018 VCAA 2 MC

Consider the function  `f` with rule  `f(x) = 1/sqrt(sin^(-1)(cx + d))`, where  `c, d in R`  and  `c > 0`.

The domain of  `f` is

A.  `x > -d/c`

B.  `-d/c < x <= (1 - d)/c`

C.  `(-1 - d)/c <= x <= (1 - d)/c`

D. `x in R\ text(\) {-d/c}`

E.  `x in R` 

Show Answers Only

`B`

Show Worked Solution
`cx + d` `in [-1, 1]`
`cx` `in [-1-d, 1-d]`
`x` `in [(-1-d)/c, (1 – d)/c]\ \ …\ (1)`

 

`sin^(-1) (cx + d)` `> 0`
`cx + d` `>0`
`cx` `> -d`
`x` `> -d/c\ \ …\ (2)`

 

`(1) nn (2):`

`x in (-d/c, (1 – d)/c]`

 
`=>  B`

Algebra, STD2 A4 EQ-Bank 8

Two friends, Sequoia and Raven, sold organic chapsticks at the the local market.

Sequoia sold her chapsticks for $4 and Raven sold hers for $3 each. In the first hour, their total combined sales were $20.

If Sequoia sold `x` chapsticks and Raven sold `y` chapsticks, then the following equation can be formed:

`4x + 3y = 20`

In the first hour, the friends sold a total of 6 chapsticks between them.

Find the number of chapsticks each of the friends sold during this time by forming a second equation and solving the simultaneous equations graphically.  (5 marks)

Show Answers Only

`text(Sequoia sold 2 and Raven sold 4.)`

Show Worked Solution
`text(Graphing)\ \ 4x + 3y` `= 20`
`y` `= −4/3x + 20/3`

 
`ytext(-intercept) = (0, 20/3)`

`xtext(-intercept)= (5, 0)`

`text(Gradient)\ = -4/3`
 

`text(S)text(econd equation:)`

`x + y = 6`

`text(From the graph,)`

`text(Sequoia sold 2 and Raven sold 4.)`

Algebra, STD2 A4 SM-Bank 7

The graph of the line  `y = 4 - x`  is shown.
 


 

By graphing  `y = 2x + 1`  on the grid provided, find the point of intersection of  `y = 4 - x`  and  `y = 2x + 1`.  (3 marks)

Show Answers Only

`(1, 3)`

Show Worked Solution

`text(Graphing)\ y = 2x + 1 :`

`ytext(-intercept = 1)`

`text(Gradient = 2)`
 

 
`:.\ text{Point of intersection is (1, 3).}`

Algebra, STD2 A4 SM-Bank 6

A student was asked to solve the following simultaneous equations.

`y = 2x - 8`

`x - 4y + 3 = 0`

After graphing the equations, the student found the point of intersection to be `(5,2)`?

Is the student correct? Support your answer with calculations.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text{Substitute (5, 2) into}\ \ y = 2x – 8`

`text(LHS) = y = 2`

`text(RHS) = 2(5) – 8 = 2`

`:.\ text(LHS = RHS)`
 

`text{Substitute (5, 2) into}\ \ x – 4y + 3 = 0`

`text(LHS)` `= 5 – 4(2) + 3`
  `= 0`
  `=\ text(RHS)`

 
`=> (5,2)\ text(satisfies both equations.)`

`:.\ text(Student is correct.)`

Probability, STD2 S2 SM-Bank 1

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 red tokens and 3 black tokens. The player keeps the two tokens drawn.

  1.  Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)
     
     
       
     
  2.  What is the probability that a player draws at least one red token. Give your answer in exact form.  (2 marks)

Show Answers Only
  1.  
  2. `187/190`
Show Worked Solution
i.   

 

ii.   `P(text(at least one red))`

`= 1 – P(BB)`

`= 1 – 3/20 · 2/19`

`= 187/190`

Financial Maths, STD2 F4 SM-Bank 1

An investment fund purchases 4500 shares of Bank ABC for a total cost of $274 500 (ignore any transaction costs).

The investment fund is paid a divided of $3.66 per share in the first year.

  1. What was the purchase price of 1 share?  (1 mark)
  2. Calculate the divided yield.  (1 mark)
Show Answers Only
  1. `$61`
  2. `text(6%)`
Show Worked Solution

i.   `text(Price per share)`

`= (274\ 500)/4500`

`= $61`
 

ii.   `text(Dividend Yield)`

`= text(Dividend)/text(Share Price) xx 100`

`= 3.66/61 xx 100`

`= 6text(%)`

Functions, 2ADV F2 SM-Bank 1

  1.  Draw the graph  `y = ln x`.  (1 mark)

  2.  Explain how the above graph can be transformed to produce the graph
     
             `y = 3ln(x + 2)`
     
    and sketch the graph, clearly identifying all intercepts.  (3 marks)

Show Answers Only

  1.  

  2.  
Show Worked Solution

i.

 

ii.   `text(Transforming)\ \ y = ln x => \ y = ln(x + 2)`

`y = ln x\ \ =>\ text(shift 2 units to left.)`
 

`text(Transforming)\ \ y = ln(x + 2)\ \ text(to)\ \ y = 3ln(x + 2)`

`=>\ text(increase each)\ y\ text(value by a factor of 3)`
 

Calculus, 2ADV C1 SM-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3 - t^2 - 3t + 11`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 2`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only
  1.  `17\ text(ms)^(−1)`
  2.  `(1 + sqrt19)/6`
Show Worked Solution

i.   `x =2t^3 – t^2 – 3t + 11` 

`v = (dx)/(dt) = 6t^2 – 2t – 3`

 
`text(When)\ t = 2,`

`v` `= 6 xx 2^2 – 2 · 2 – 3`
  `= 17\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`6t^2 – 2t – 3` `= 0`
`:. t` `= (2 ±sqrt((−2)^2 – 4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`

Calculus, 2ADV C1 SM-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2 - 3x`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −3x`
    `y = 3x – 9`
  2. `(3/2, −9/4)`
Show Worked Solution
i.   `y` `= x^2 – 3x`
  `= x(x – 3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 0\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x – 3`

 
`text(At)\ \ x = 0 \ => \ (dy)/(dx) = -3`

`T_1\ text(has)\ \ m = −3,\ text{through (0, 0)}`

`y – 0` `= -3(x – 0)`
`y` `= -3x`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 3`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y – 0` `= 3(x – 3)`
`y` `= 3x – 9`

 

ii.   `text(Intersection occurs when:)`

`3x – 9` `= -3x`
`6x` `= 9`
`x` `= 3/2`

  

`y = -3 xx 3/2 = −9/2`

`:.\ text(Intersection at)\ \ (3/2, −9/2)`

Trigonometry, 2ADV T2 SM-Bank 33

Given  `sectheta = −37/12`  for  `0 < theta < pi`,

find the exact value of  `text(cosec)\ theta`.  (2 marks)

Show Answers Only

`37/35`

Show Worked Solution

`sec\ theta = −37/12 => cos\ theta = −12/37`
 

`text(Graphically:)`

`x=sqrt(37^2 – 12^2)=35`
 

`:.\ text(cosec)\ theta` `= 1/(sintheta)`
  `= 1/(35/37)`
  `= 37/35`

Trigonometry, 2ADV T2 SM-Bank 32

Express  `5cot^2 x - 2text(cosec)\ x + 2`  in terms of  `text(cosec)\ x`  and hence solve

`5cot^2 x - 2text(cosec)\ x + 2 = 0`  for  `0 < x < 2pi`.  (3 marks)

Show Answers Only

`x = pi/2`

Show Worked Solution
`cot^2 x` `= (cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x)/(sin^2 x)`
  `= text(cosec)^2 x – 1`

 

`5cot^2 x – 2text(cosec)\ x + 2` `= 0`
`5(text(cosec)^2 x – 1) – 2text(cosec)\ x + 2` `= 0`
`5text(cosec)^2 x – 2text(cosec)\ x – 3` `= 0`
`(5text(cosec)\ x + 3)(text(cosec)\ x – 1)` `= 0`
`text(cosec)\ x` `= −3/5` `text(cosec)\ x` `= 1`
`sinx` `= −5/3` `sinx` `= 1`
`(text(no solution))` `x` `= pi/2`

 
`:. x = pi/2`

Trigonometry, 2ADV T2 SM-Bank 31

Given  `cottheta = −24/7`  for  `−pi/2 < theta < pi/2`, find the exact value of

  1.  `sectheta`  (2 marks)

  2.  `sintheta`  (1 mark)

Show Answers Only
  1. `25/24`
  2. `-7/25`
Show Worked Solution

i.   `cot theta = −24/7\ \ =>\ \ tan theta=– 7/24`

`text(Graphically, given)\ −pi/2 < theta < pi/2:`

`x= sqrt(24^2 + 7^2)=25`
 

`sectheta` `= 1/(costheta)`
  `= 1/(24/25)`
  `= 25/24`

 

ii.   `sintheta = −7/25`

Functions, 2ADV F1 EQ-Bank 27

The stopping distance of a car on a certain road, once the brakes are applied, is directly proportional to the square of the speed of the car when the brakes are first applied.

A car travelling at 70 km/h takes 58.8 metres to stop.

How far does it take to stop if it is travelling at 105 km/h?  (3 marks)

Show Answers Only

`132.3\ text(metres)`

Show Worked Solution

`text(Let)\ \ d\ text(= stopping distance)`

`d ∝ s^2`

`d = ks^2`
 

`text(Find)\ k,`

`58.8` `= k xx 70^2`
`k` `= 58.8/(70^2)`
  `= 0.012`

 
`text(Find)\ \ d\ \ text(when)\ \ s = 105:`

`d` `= 0.012 xx 105^2`
  `= 132.3\ text(metres)`

Functions, 2ADV F1 EQ-Bank 26

Fuifui finds that for Giant moray eels, the mass of an eel is directly proportional to the cube of its length.

An eel of this species has a length of 25 cm and a mass of 4350 grams.

What is the expected length of a Giant moray eel with a mass of 6.2 kg? Give your answer to one decimal place.  (3 marks)

Show Answers Only

`28.1\ text{cm}`

Show Worked Solution

`text(Mass) prop text(length)^3`

`m = kl^3`
 

`text(Find)\ k:`

`4350` `= k xx 25^3`
`k` `= 4350/25^3`
  `= 0.2784`

 
`text(Find)\ \ l\ \ text(when)\ \ m = 6200:`

`6200` `= 0.2784 xx l^3`
`l^3` `= 6200/0.2784`
`:. l` `= 28.13…`
  `= 28.1\ text{cm  (to 1 d.p.)}`

Functions, 2ADV F1 SM-Bank 25

Damon owns a swim school and purchased a new pool pump for $3250.

He writes down the value of the pool pump by 8% of the original price each year.

  1.  Construct a function to represent the value of the pool pump after `t` years.   (1 mark)

  2.  Draw the graph of the function and state its domain and range.   (2 marks)

Show Answers Only
  1. `text(Value) = 3259 – 260t`
  2. `text(Domain)\ {t: 0 <= t <= 12.5}`
    `text(Range)\ {y: 0 <= y <= 3250}`
Show Worked Solution
i.   `text(Depreciation each year)` `= 8text(%) xx 3250`
  `= $260`

 
`:.\ text(Value) = 3250 – 260t`
 

ii.   

`text(Find)\ \ t\ \ text(when value = 0)`

`3250 – 260t` `= 0`
`t` `= 3250/260`
  `= 12.5\ text(years)`

 
`text(Domain)\ \ {t: 0 <= t <= 12.5}`

`text(Range)\ \ {y: 0 <= y <= 3250}`

Functions, 2ADV F1 SM-Bank 23

Find the values of `k` for which the expression  `x^2-3x + (4-2k)`  is always positive.  (3 marks)

Show Answers Only

`k < 7/8`

Show Worked Solution

`x^2-3x + (4-2k) > 0`

`x^2-3x + (4-2k) = 0\ \ text(is a concave up parabola)`

`=>\ text{Always positive (no roots) if}\ \ Delta < 0`
 

`b^2-4ac < 0`

`(−3)^2-4 · 1 · (4-2k)` `< 0`
`9-16 + 8k` `< 0`
`8k` `< 7`
`k` `< 7/8`

Functions, 2ADV F1 EQ-Bank 22

Worker A picks a bucket of blueberries in `a` hours. Worker B picks a bucket of blueberries in `b` hours.

  1.  Write an algebraic expression for the fraction of a bucket of blueberries that could be picked in one hour if A and B worked together.  (2 marks)

  2.  What does the reciprocal of this fraction represent?  (1 mark)

Show Answers Only
  1. `(a + b)/(ab)`
  2. `text(The reciprocal represents the number of hours it would)`
  3. `text(take to fill one bucket, with A and B working together.)`
Show Worked Solution

i.    `text(In one hour:)`

COMMENT: Note that the question asks for “a fraction”.

`text(Worker A picks)\ 1/a\ text(bucket.)`

`text(Worker B picks)\ 1/b\ text(bucket.)`
 

`:.\ text(Fraction picked in 1 hour working together)`

`= 1/a + 1/b`

`= (a + b)/(ab)`
 

ii.   `text(The reciprocal represents the number of hours it would)`

`text(take to fill one bucket, with A and B working together.)`

Calculus, 2ADV C1 SM-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = sqrt(20t - 2t^2)`  metres per second

Find the acceleration of the particle when  `t = 4`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

`sqrt3/6\ \ text(ms)^(−2)`

Show Worked Solution

`v = sqrt(20t – 2t^2)`

`alpha` `= (dv)/(dt)`
  `= 1/2 · (20t – 2t^2)^(−1/2) · (20 – 4t)`

 
`text(When)\ \ t = 4,`

`alpha` `= 1/2(20 · 4 – 2 · 4^2)^(−1/2)(20 – 16)`
  `= 2/(sqrt48)`
  `= 2/(4sqrt3) xx sqrt3/sqrt3`
  `= sqrt3/6\ \ text(ms)^(−2)`

Trigonometry, EXT1 T2 EQ-Bank 7

Find the exact value of  `cos((11pi)/12)`.  (2 marks)

Show Answers Only

`−((sqrt2 + sqrt6))/4`

Show Worked Solution
`cos((11pi)/12)` `= cos((2pi)/3 + pi/4)`
  `= cos((2pi)/3) · cos(pi/4) – sin((2pi)/3) · sin(pi/4)`
  `= cos(pi – pi/3) · 1/sqrt2 – sin (pi – pi/3) · 1/sqrt2`
  `= −1/2 · 1/sqrt2 – sqrt3/2 · 1/sqrt2`
  `= −((1 + sqrt3))/(2sqrt2)`
  `= −((sqrt2 + sqrt6))/4`

Trigonometry, EXT1 T2 EQ-Bank 5

If  `sintheta = −4/6`  and  `−pi/2 < theta < pi/2`,

determine the exact value of  `costheta`  in its simplest form.  (2 marks)

Show Answers Only

`sqrt5/3`

Show Worked Solution

`text(Consider the angle graphically:)`

COMMENT: Pay careful attention to the range of  `theta`.

`text(S)text(ince)\ sintheta\ text(is negative) => underbrace(text(4th quadrant))_(−pi/2 <\ theta\ < pi/2)`

`text(Using Pythagoras:)`

`x^2 = 6^2-4^2`

`x = sqrt20 = 2sqrt5`

`:. costheta= (2sqrt5)/6= sqrt5/3`

Trigonometry, EXT1 T2 SM-Bank 2

Find the exact value of  `cos\ pi/8`.  (2 marks)

Show Answers Only

`(sqrt(sqrt2 + 2))/2`

Show Worked Solution

`text(Using:)\ \ cos2A = 2cos^2A – 1`

`2cos^2\ pi/8 – 1` `= cos\ pi/4`
`2cos^2\ pi/8` `= 1/sqrt2 + 1`
`cos^2\ pi/8` `= (1 + sqrt2)/(2sqrt2) xx sqrt2/sqrt2`
  `= (sqrt2 + 2)/4`
`:. cos\ pi/8` `= sqrt((sqrt2 + 2)/4)`
  `= (sqrt(sqrt2 + 2))/2`

Functions, EXT1 F1 SM-Bank 3

A circle has centre `(5,3)` and radius 3.

  1.  Describe, with inequalities, the region that consists of the interior of the circle and more than 2 units above the `x`-axis.  (2 marks)

  2.  Sketch the region.  (1 mark)

Show Answers Only
  1. `(x-5)^2 + (y-3)^2 < 9\ ∩\ y > 2`

  2.  

Show Worked Solution

i.   `text(Equation of circle:)`

`(x-5)^2 + (y-3)^2 = 3^2`
 

`:.\ text(Region is:)`

`(x-5)^2 + (y- 3)^2 < 9\ ∩\ y > 2`

COMMENT: The broken line on the graph represents an excluded boundary.

 

ii.   

Combinatorics, EXT1 A1 EQ-Bank 3

Find the coefficient of  `x^4`  in the expansion of  `(x^2 - 3/x)^5`.  (2 marks)

Show Answers Only

`90`

Show Worked Solution

`text(General term:)`

`T_k` `= \ ^5C_k(x^2)^(5 – k) · (−3/x)^k`
  `= \ ^5C_k · x^(10 – 2k)(−3)^k · x^(−k)`
  `= \ ^5C_k · x^(10 – 3k) · (−3)^k`

 
`text(Coefficient of)\ \ x^4\ \ text(occurs when)`

`10 – 3k` `= 4`
`3k` `= 6`
`k` `= 2`

 
`:.\ text(Coefficient of)\ \ x^4`

`= \ ^5C_2·(−3)^2`

`= 90`

Combinatorics, EXT1 A1 SM-Bank 1

`(1-2sqrt2)^6 = x + ysqrt2`

Evaluate the value of `x` and `y`.  (2 marks)

Show Answers Only

`x = 1593, \ y =-1100`

Show Worked Solution

`text(Using the binomial expansion:)`

`(1-2sqrt2)^6`

`= \ ^6C_0 + \ ^6C_1(-2sqrt2) + \ ^6C_2(-2sqrt2)^2 + \ ^6C_3(-2sqrt2)^3 + \ ^6C_4(-2sqrt2)^4`

`+ \ ^6C_5(-2sqrt2)^5 + \ ^6C_6(-2sqrt2)^6`
 

`= 1-12sqrt2 + 120 -320sqrt2 + 960-768sqrt2 + 512`

`= 1593-1100sqrt2`
 

`:.x = 1593, \ y =-1100`

Trigonometry, 2ADV T1 EQ-Bank 2

Determine all possible dimensions for triangle  `ABC`  given  `AB = 6.2\ text(cm)`, `angleABC = 35°`  and  `AC = 4.1`.

Give all dimensions correct to one decimal place.  (3 marks)

Show Answers Only

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Show Worked Solution

`text(Using the sine rule:)`

`(sinangleACB)/6.2` `= (sin35^@)/4.1`
`sinangleACB` `= (6.2 xx sin35^@)/4.1`
  `= 0.8673…`
`angleACB` `= 60.15…^@\ text(or)\ 119.84…^@`

  
`text(If)\ \ angleACB = 60.15^@,`

`angleBAC = 180 – (35 + 60.15) = 84.85^@`
 

`(BC)/(sin84.85)` `= 4.1/(sin35^@)`
`BC` `= 7.11…`
  `= 7.1\ text(cm)`

 
`text(If)\ \ angleACB = 119.85^@,`

`angleBAC = 180 – (35 + 119.85) = 25.15^@`
 

`(BC)/(sin25.15)` `= 4.1/(sin35^@)`
`BC` `= 3.03…`
  `= 3.0\ text(cm)`

 
`:.\ text(Possible dimensions are:)`

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Calculus, EXT1 C3 SM-Bank 1

The region enclosed by the semicircle  `y = sqrt(1 - x^2)`  and the `x`-axis is to be divided into two pieces by the line  `x = h`, when  `0 <= h <1`.
 


 

The two pieces are rotated about the `x`-axis to form solids of revolution. The value of `h` is chosen so that the volumes of the solids are in the ratio `2 : 1`.

Show that `h` satisfies the equation  `3h^3 - 9h + 2 = 0`.  (3 marks)

Show Answers Only

`text(Show Worked Solution)`

Show Worked Solution

(i)   `text(Volume of smaller solid)`

`= pi int_h^1 (sqrt(1 – x^2))^2\ dx`

`= pi int_h^1 1 – x^2\ dx`

`= pi[x – (x^3)/3]_h^1`

`= pi[(1 – 1/3) – (h – (h^3)/3)]`

`= pi(2/3 – h + (h^3)/3)`

 
`text(S)text(ince smaller solid is)\ 1/3\ text(volume of sphere,)`

`pi(2/3 – h + (h^3)/3)` `= 1/3 xx 4/3 · pi · 1^3`
`(h^3)/3 – h + 2/3` `= 4/9`
`3h^3 – 9h + 6` `= 4`
`:. 3h^3 – 9h + 2` `= 0\ \ text(… as required)`

Calculus, SPEC2 2017 VCAA 7 MC

With a suitable substitution `int_1^2 x^2 sqrt(2 - x)\ dx` can be expressed as

  1. `−int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  2. `int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  3. `int_0^1(−4u^(1/2) + 4u^(3/2) - u^(5/2)) du`
  4. `−int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  5. `int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
Show Answers Only

`D`

Show Worked Solution
`u` `= 2 – x`
`x` `= 2 – u\ \ =>\ \ x^2 = (2 – u)^2`

 
`(du)/(dx) = −1\ \ =>\ \ du = – dx`

`u(2)` `= 0`
`u(1)` `= 2 – 1=1`

 
`int_1^2 – (2 – u)^2sqrtu\ du`

`= −int_1^0(2 – u)^2 u^(1/2)\ du`

`= −int_1^0(4 – 4u + u^2) u^(1/2)\ du`

`= −int_1^0 (4u^(1/2) – 4u^(3/2) + u^(5/2))\ du`

 
`=>D`

Complex Numbers, SPEC2 2017 VCAA 4 MC

The solutions to  `z^n = 1 + i, \ n ∈ Z^+`  are given by

  1. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ R`
  2. `2^(1/n)text(cis)(pi/(4n) + 2pik), k ∈ Z`
  3. `2^(1/(2n))text(cis)(pi/4 + (2pik)/n), k ∈ R`
  4. `2^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  5. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
Show Answers Only

`E`

Show Worked Solution
`z^n` `= sqrt(1^2 + 1^2)text(cis)(tan^(−1)(1/1))`
  `= sqrt2text(cis)(pi/4)`
`z_1` `= (2^(1/2))^(1/n)text(cis)(pi/(4n))`
`z_k` `= (2^(1/2))^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  `= 2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`

 
`=>E`