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CORE, FUR2 2016 VCAA 7

 

Ken has borrowed $70 000 to buy a new caravan.

He will be charged interest at the rate of 6.9% per annum, compounding monthly.

  1. For the first year (12 months), Ken will make monthly repayments of $800.

    1. Find the amount that Ken will owe on his loan after he has made 12 repayments.   (1 mark)
    2. What is the total interest that Ken will have paid after 12 repayments?   (1 mark)
  2. After three years, Ken will make a lump sum payment of $L in order to reduce the balance of his loan.
  3. This lump sum payment will ensure that Ken’s loan is fully repaid in a further three years.
  4. Ken’s repayment amount remains at $800 per month and the interest rate remains at 6.9% per annum, compounding monthly.
  5. What is the value of Ken’s lump sum payment, $L?
  6. Round your answer to the nearest dollar.   (2 marks)

Show Answers Only

a.i.`$65\ 076.22`

a.ii.`$4676.22`

b.  `$28\ 204.02`

Show Worked Solution

a.i.  `text(By TVM Solver,)`

♦ Mean mark 46%.

`N` `= 12`
`I(%)` `= 6.9`
`PV` `= 70\ 000`
`PMT` `= -800`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> FV = -65\ 076.219…`

`:.\ text(Ken will owe $65 076.22)`

♦♦ Mean mark 26%.

a.ii.    `text(Total interest paid)` `= text(Total repayments) – text(reduction in principal)`
    `= (12 xx 800) – (70\ 000 – 65\ 076.22)`
    `= $4676.22`

  
b.   `text(Find the loan balance after 3 years)`

♦♦♦ Mean mark 20%.
MARKER’S COMMENT: Correct input tables allowed for a method mark if answers were calculated incorrectly.

`N` `= 12 xx 3 = 36`
`I(%)` `= 6.9`
`PV` `= 70\ 000`
`PMT` `= -800`
`FV` `= ?`
`text(P/Y)` `=text(C/Y)=12`
   

`=> FV = 54\ 151.60`

  
`text(Find the loan amount that can be)`

`text(fully repaid by monthly payments)`

`text(of $800 over 3 years.)`

`N` `= 36`
`I(%)` `= 6.9`
`PV` `= ?`
`PMT` `= -800`
`FV` `= 0`
`text(P/Y)` `= text(C/Y)= 0`

 
`=> PV = 25\ 947.58`
  

`:.\ $L` `= 54\ 151.60 -25\ 947.58`
  `= $28\ 204.02`

CORE*, FUR2 2016 VCAA 6

Ken’s first caravan had a purchase price of $38 000.

After eight years, the value of the caravan was $16 000.

  1. Show that the average depreciation in the value of the caravan per year was $2750.   (1 mark)

  2. Let `C_n` be the value of the caravan `n` years after it was purchased.

     

    Assume that the value of the caravan has been depreciated using the flat rate method of depreciation.

     

    Write down a recurrence relation, in terms of `C_(n +1)` and `C_n`, that models the value of the caravan.   (1 mark)

  3. The caravan has travelled an average of 5000 km in each of the eight years since it was purchased.

     

    Assume that the value of the caravan has been depreciated using the unit cost method of depreciation.

     

    By how much is the value of the caravan reduced per kilometre travelled?   (1 mark)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `$0.55`
Show Worked Solution
a.    `text(Average depreciation)`

`= {(38\ 000-16\ 000)}/8`

`= $2750`


♦♦ Mean mark (a) 39%.
MARKER’S COMMENT: A “show that” question should include an equation
  

b.    `C_0` `= 38\ 000,`
  `C_(n+1)` `= C_n-2750`

♦♦ Mean mark (b) 33%.
MARKER’S COMMENT: A lack of attention to detail and careless errors were common!
  

c.    `text(Total kms travelled)` `= 8 xx 5000`
    `= 40\ 000`

 
`:.\ text(Depreciation per km)`

`= {(38\ 000-16\ 000)}/(40\ 000)`

`= $0.55`


♦♦ Mean mark (c) 29%.
  

CORE, FUR2 2016 VCAA 5

Ken has opened a savings account to save money to buy a new caravan.

The amount of money in the savings account after `n` years, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 15000, qquad qquad qquad V_(n + 1) = 1.04 xx V_n`

  1. How much money did Ken initially deposit into the savings account?  (1 mark)

  2. Use recursion to write down calculations that show that the amount of money in Ken’s savings account after two years, `V_2`, will be $16 224.  (1 mark)

  3. What is the annual percentage compound interest rate for this savings account?  (1 mark)

  4. The amount of money in the account after `n` years, `Vn` , can also be determined using a rule.
    i.     Complete the rule below by writing the appropriate numbers in the boxes provided.   (1 mark)

    `V_n =` 
     
    		 `­^n xx`
     
  5. ii. How much money will be in Ken’s savings account after 10 years?   (1 mark)

Show Answers Only
  1. `$15000`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `4 text(%)`
  4. i. `text(See Worked Solutions)`
    ii. `$22\ 203.66` 
Show Worked Solution
a.    `text(Initial deposit)` `= V_0`
    `= $15\ 000`
b.    `V_0` `= $15\ 000`
  `V_1` `= 1.04 xx 15\ 000`
    `= $15\ 600`
  `V_2` `= 1.04 xx 15\ 600`
    `= $16\ 224\ text(… as required.)`

MARKER’S COMMENT: (b) Stating `V_2 =1.04^2 xx 15\ 600` `=16\ 224` is not using recursion as required here and did not gain a mark.
c.    `text(Annual compound interest)` `= 0.04 xx 100`
    `= 4 text(%)`

 

d.i.   `V_n` `= 1.04^n xx V_0`
d.ii.    `V_10` `= 1.04^10 xx 15\ 000`
    `= $22\ 203.664…`
    `= $22\ 203.66\ text{(nearest cent)}`

♦ Mean mark (d)(ii) 47%.
MARKER’S COMMENT: Rounding to $22 203.70 lost a mark!

CORE, FUR2 2016 VCAA 4

The time series plot below shows the minimum rainfall recorded at the weather station each month plotted against the month number (1 = January, 2 = February, and so on).

Rainfall is recorded in millimetres.

The data was collected over a period of one year.
 

  1. Five-median smoothing has been used to smooth the time series plot above.

     

    The first four smoothed points are shown as crosses (×).

     

    Complete the five-median smoothing by marking smoothed values with crosses (×) on the time series plot above.   (2 marks)

The maximum daily rainfall each month was also recorded at the weather station.

The table below shows the maximum daily rainfall each month for a period of one year.

The data in the table has been used to plot maximum daily rainfall against month number in the time series plot below.
 

  1. Two-mean smoothing with centring has been used to smooth the time series plot above.

     

    The smoothed values are marked with crosses (×).

     

    Using the data given in the table, show that the two-mean smoothed rainfall centred on October is 157.25 mm.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution
a.   
♦ Mean mark of both Parts (a) and (b) was 49%.
MARKER’S COMMENT: Use the accurate table data when available. Reading values from the graph will cause inaccuracies.
b.    `text(Mean)\ _text(Sep-Oct)` `= (124 + 140)/2`
    `= 132\ text(mm)`
  `text(Mean)\ _text(Oct-Nov)` `= (140 + 225)/2`
    `= 182.5\ text(mm)`

 
`:.\ text{Two mean (smoothed) for October}`

`= (132 +182.5)/2`

`= 157.25\ text(mm … as required)`

CORE, FUR2 2016 VCAA 3

The data in the table below shows a sample of actual temperatures and apparent temperatures recorded at a weather station. A scatterplot of the data is also shown.

The data will be used to investigate the association between the variables apparent temperature and actual temperature.
 

  1. Use the scatterplot to describe the association between apparent temperature and actual temperature in terms of strength, direction and form.   (1 mark)

  2.  i. Determine the equation of the least squares line that can be used to predict the apparent temperature from the actual temperature.
  3. Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
  4. Round your answers to two significant figures.   (3 marks)
     apparent temperature `=`    
 
 `+`  
 
 `xx`   actual temperature
  1. ii. Interpret the intercept of the least squares line in terms of the variables apparent temperature and actual temperature.   (1 mark)

  2. The coefficient of determination for the association between the variables apparent temperature and actual temperature is 0.97
  3. Interpret the coefficient of determination in terms of these variables.   (1 mark)

  4. The residual plot obtained when the least squares line was fitted to the data is shown below.
     
     
  5.  i. A residual plot can be used to test an assumption about the nature of the association between two numerical variables.
  6.     What is this assumption?   (1 mark)
  7. ii. Does the residual plot above support this assumption? Explain your answer.   (1 mark)
Show Answers Only

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

b.i.  `text(apparent temperature) = -1.7 xx 0.94 xx text(actual temperature)`

b.ii.  `text(When actual temperature is 0°C, on average,)`

`text(the apparent temperature is)\ − 1.7^@\text(C.)`

c.  `text(97% of the variation in the apparent temperature can be explained)`

`text(by the variation in the actual temperature.)`

d.i.  `text(There is a linear relationship between the two variables.)`

d.ii.  `text(The random pattern supports the assumption.)`

`text{(Students should refer to randomness or a lack of pattern}`

`text{explicitly here).}`

Show Worked Solution

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

b.i.   `text(By calculator:)`

`text(apparent temperature) = -1.7 xx 0.94 xx text(actual temperature)`
 

♦♦ Mean mark of part (b)(ii) – 28%.
MARKER’S COMMENT: “the predicted apparent temp is -1.7°C” also gained a mark.
b.ii.    `text(When actual temperature is 0°C, on average,)`
 

`text(the apparent temperature is)\ − 1.7^@\text(C.)`

 

♦ Mean mark 49%.
IMPORTANT: Any mention of causality loses a mark!

c.  `text(97% of the variation in the apparent temperature can be explained)`

`text(by the variation in the actual temperature.)`
  

d.i.  `text(There is a linear relationship between the two variables.)`

♦ Mean mark of both parts of (d) was 46%.

d.ii.  `text(The random pattern supports the assumption.)`

`text{(Students should refer to randomness or a lack of pattern}`

`text{explicitly here).}`

CORE, FUR2 2016 VCAA 2

A weather station records daily maximum temperatures.

  1. The five-number summary for the distribution of maximum temperatures for the month of February is displayed in the table below.

 

  1. There are no outliers in this distribution.
  2.  i. Use the five-number summary above to construct a boxplot on the grid below.   (1 mark)

 

  1. ii. What percentage of days had a maximum temperature of 21°C, or greater, in this particular February?   (1 mark)

  2. The boxplots below display the distribution of maximum daily temperature for the months of May and July.
     

  3.   i. Describe the shapes of the distributions of daily temperature (including outliers) for July and for May.   (1 mark)

  4.  ii. Determine the value of the upper fence for the July boxplot.   (1 mark)

  5. iii. Using the information from the boxplots, explain why the maximum daily temperature is associated with the month of the year. Quote the values of appropriate statistics in your response.   (1 mark)

Show Answers Only
a.i.   

a.ii.   `text(75%)`

b.i.    `text(July – Positively skewed with an outlier.)`
  `text(May – Symmetrical with no outliers.)`

b.ii.  `15.5^@\text(C)`

b.iii. `text{The median temperature in May (14.5°C)}`

`text(differs from the median temperature in July)`

`text{(just over 9°C). This difference is why the}`

`text(maximum daily temperature is associated)`

`text(with the month.)`

Show Worked Solution
a.i.   

a.ii.   `text(75%)`

MARKER’S COMMENT: Incorrect May descriptors included “evenly or normally distributed”, “bell shaped” and “symmetrically skewed.”
b.i.    `text(July – Positively skewed with an outlier.)`
  `text(May – Symmetrical with no outliers.)`

 

b.ii.    `text(Upper fence)` `= Q_3 + 1.5 xx IQR`
    `= 11 + 1.5 xx (11 – 8)`
    `= 11 + 4.5`
    `= 15.5^@\text(C)`
♦♦ Mean mark (b)(iii) – 30%.
COMMENT: Refer to the difference in medians. Just quoting the numbers was not enough to gain a mark here.

b.iii. `text{The median temperature in May (14.5°C)}`

`text(differs from the median temperature in July)`

`text{(just over 9°C). This difference is why the}`

`text(maximum daily temperature is associated)`

`text(with the month.)`

NETWORKS, FUR1 2016 VCAA 6-7 MC

The directed graph below shows the sequence of activities required to complete a project.

All times are in hours.
 

 
Part 1

The number of activities that have exactly two immediate predecessors is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4

 
Part 2

There is one critical path for this project.

Three critical paths would exist if the duration of activity

  1. I were reduced by two hours.
  2. E were reduced by one hour.
  3. G were increased by six hours.
  4. K were increased by two hours.
  5. F were increased by two hours. 
Show Answers Only

`text(Part 1:)\ C`

`text(Part 1:)\ B`

Show Worked Solution

`text(Part 1)`

`I\ text(and)\ J`

`=> C`

 

`text(Part 2)`

♦ Mean mark of Part 2: 45%.

`text(The possible paths are:)`

`ADIL – 19\ text(mins)`

`BEIL – 20\ text(mins)`

`BFJL – 17\ text(mins)`

`CGJL – 13\ text(mins)`

`CHKL – 19\ text(mins)`

`text(If)\ E\ text(were reduced by 1 hour,)\ BEIL\ text(will)`

`text{take 19 minutes (i.e. 3 critical paths of 19}`

`text{minutes would exist).}`

`=> B`

NETWORKS, FUR1 2016 VCAA 4 MC

The minimum spanning tree for the network below includes the edge with weight labelled `k`.
 

 
The total weight of all edges for the minimum spanning tree is 33.

The value of `k` is

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

`E`

Show Worked Solution

`text(Minimum spanning tree)`

`text(Total weight)` `= k + 5 + 5 + 3 + 2 + 4 + 2 + 1 + 6`
`33` `= k + 28`
`:. k` `= 5`

`=> E`

GRAPHS, FUR1 2016 VCAA 7 MC

Simon grows cucumbers and zucchinis.

Let `x` be the number of cucumbers that are grown.

Let `y` be the number of zucchinis that are grown.

For every two cucumbers that are grown, Simon grows at least three zucchinis.

An inequality that represents this situation is

  1. `y >= x/2 + 3`
  2. `y >= x/3 + 2`
  3. `y <= x/3 + 2`
  4. `y >= (2x)/3`
  5. `y >= (3x)/2`
Show Answers Only

`E`

Show Worked Solution

`text(Inequality can be restated:)`

`text(For every one cucumber grown, Simon)`

`text(grows at least)\ 3/2\ text(zucchinis.)`

`:. y >= (3x)/2`

`=> E`

GEOMETRY, FUR1 2016 VCAA 7 MC

The diagram below shows a rectangular-based right pyramid, ABCDE.

In this pyramid, AB = DC = 24 cm, AD = BC = 10 cm and AE = BE = CE = DE = 28 cm.

The height, OE, of the pyramid, in centimetres, is closest to

  1. `10.4`
  2. `13.0`
  3. `24.8`
  4. `25.3`
  5. `30.9`
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ AC,\ text(using Pythagoras,)`

`AC^2` `= 24^2 + 10^2`
  `= 676`
`AC` `= 26`

 

`text(Consider)\ DeltaAEO,`

`AO` `= 1/2AC`
  `= 13`

 

`text(Using Pythagoras,)`

`OE^2 + AO^2` `= AE^2`
`OE^2` `= 28^2 – 13^2`
  `= 615`
`:. OE` `= 24.79…`

`=> C`

MATRICES, FUR1 2016 VCAA 8 MC

The matrix below shows the result of each match between four teams, A, B, C and D, in a bowling tournament. Each team played each other team once and there were no draws.
 

`{:(qquadqquadqquadqquadqquadqquadquadtext(loser)),(qquadqquadqquadqquadqquadquadAquadBquadCquadD),(text(winner)quad{:(A),(B),(C),(D):}[(0,0,1,0),(1,0,0,1),(0,1,0,1),(1,0,0,0)]):}`
 

In this tournament, each team was given a ranking that was determined by calculating the sum of its one-step and two-step dominances. The team with the highest sum was ranked number one (1). The team with the second-highest sum was ranked number two (2), and so on.

Using this method, team C was ranked number one (1).

Team A would have been ranked number one (1) if the winner of one match had lost instead.

That match was between teams

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

`A`

Show Worked Solution

`text(Test one and two step dominances after)`

♦ Mean mark 42%.

`text(swapping one result until)\ A\ text(is ranked)`

`text(number one.)`

`text(If we reverse the result and swap)\ A\ text(vs)\ B\ (text(now)\ A\ text(wins,))`

`text(Total dominances:)`
  

`{:(qquadqquadqquadqquadqquad\ text(lose)),(qquadqquadqquadqquadAquadBquadCquadDqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadtext(total)),(text(win)quad{:(A),(B),(C),(D):}[(0,1,1,0),(0,0,0,1),(0,1,0,1),(1,0,0,0)] + [(0,1,1,0),(0,0,0,1),(0,1,0,1),(1,0,0,0)]^2 = [(0,2,1,2),(1,0,0,1),(1,1,0,2),(1,1,1,0)]{:(5),(2),(4),(3):}):}`

 

`:.\ text(If the result in)\ A\ text(vs)\ B\ text(was reversed,)`

`A\ text(would have ranked number one.)`

`=> A`

MATRICES, FUR1 2016 VCAA 7 MC

Each week, the 300 students at a primary school choose art (A), music (M) or sport (S) as an afternoon activity.

The transition matrix below shows how the students’ choices change from week to week.
 

`{:(qquadqquadqquadqquadtext(this week)),((qquadqquadqquadA,quadM,quadS)),(T=[(0.5,0.4,0.1),(0.3,0.4,0.4),(0.2,0.2,0.5)]{:(A),(M),(S):}quad{:text(next week):}):}`

 

Based on the information above, it can be concluded that, in the long term

  1. no student will choose sport.
  2. all students will choose to stay in the same activity each week.
  3. all students will have chosen to change their activity at least once.
  4. more students will choose to do music than sport.
  5. the number of students choosing to do art and music will be the same.
Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ n\ text(large)\ (n = 50),\ text(and begin with)`

`text(the students spread equally among the sports.)`

`[(0.5,0.4,0.1),(0.3,0.4,0.4),(0.2,0.2,0.5)]^(50)[(100),(100),(100)]~~[(105),(109),(86)]`

 

`:.\ text(There will be 109 students in music)`

`text(and 86 in sport.)`

`=> D`

MATRICES, FUR1 2016 VCAA 5 MC

Let `M = [(1,2,3,4),(3,4,5,6)]`.

The element in row `i` and column `j` of `M` is `m_(ij)`.

The elements of `M` are determined by the rule

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

`C`

Show Worked Solution

`text(Check)\ m_14 = 4,`

`text(Option)\ A:\  1 + 4 – 1 = 4\ \ text{(correct)}`

`text(Option)\ B:\  2 – 4 + 1 = −1\ \ text{(incorrect)}`

`text(Option)\ C:\  2 + 4 – 2 = 4\ \ text{(correct)}`

`text(Option)\ D:\  1 + 8 – 2 = 7\ \ text{(incorrect)}`

`text(Option)\ E:\  1 + 4 + 1 = 6\ \ text{(incorrect)}`

 

`text(Check)\ m_22 = 4,`

`text(Option)\ A:\  2 + 2 – 1 = 3\ \ text{(incorrect)}`

`text(Option)\ C:\  4 + 2 – 2 = 4\ \ text{(correct)}`

`=> C`

MATRICES, FUR1 2016 VCAA 4 MC

The table below shows the number of each type of coin saved in a moneybox.
 

 
The matrix product that displays the total number of coins and the total value of these coins is
 

A.    `[(5,10,20,50)][(15),(32),(48),(24)]` B.    `[(15,32,48,24)][(1,5),(1,10),(1,20),(1,50)]`
C.    `[(5,10,20,50)][(1,15),(1,32),(1,48),(1,24)]` D.    `[(15,32,48,24)][(5),(10),(20),(50)]`
E.    `[(5,10,20,50),(15,32,48,24)][(1),(1),(1),(1)]`    
Show Answers Only

`B`

Show Worked Solution
`[(15,32,48,24)]quad[(1,5),(1,10),(1,20),(1,50)]`
`qquadquad1xx4qquadqquadqquadqquadqquad\ 4xx2`
♦♦ Mean mark 34%.

 

`= [15 + 32 + 48 + 24qquadqquad15 xx 5 + 32 xx 10 + 48 xx 20 + 24 xx 50]` 

`=[119  qquad 2555]`

`= {:[text(total number)qquadtext(total value)]:}`

`=> B`

CORE, FUR1 2016 VCAA 23 MC

Sarah invests $5000 in a savings account that pays interest at the rate of 3.9% per annum compounding quarterly. At the end of each quarter, immediately after the interest has been paid, she adds $200 to her investment.

After two years, the value of her investment will be closest to

  1. $5805
  2. $6600
  3. $7004
  4. $7059
  5. $9285
Show Answers Only

`D`

Show Worked Solution

`text(By TVM Solver, after 2 years)`

`N` `= 2 xx 4 = 8`
`I(text(%))` `= 3.9`
`PV` `= −5000`
`PMT` `= −200`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 4`
   
`:. FV` `= 7059.25`

 
`=> D`

CORE, FUR1 2016 VCAA 22 MC

The first three lines of an amortisation table for a reducing balance home loan are shown below.

The interest rate for this home loan is 4.8% per annum compounding monthly.

The loan is to be repaid with monthly payments of $1500.
 

 
The amount of payment number 2 that goes towards reducing the principal of the loan is

  1.  $486
  2.  $502
  3.  $504
  4.  $996
  5.  $998
Show Answers Only

`B`

Show Worked Solution
`text(Interest)` `= 249\ 500 xx (4.8text(%))/12`
  `= 249\ 500 xx 0.048/12`
  `= $998`

 

`:.\ text(Amount that reduces principal)`

`= 1500 – 998`

`= $502`

`=> B`

CORE, FUR1 2016 VCAA 21 MC

Juanita invests $80 000 in a perpetuity that will provide $4000 per year to fund a scholarship at a university.

The graph that shows the value of this perpetuity over a period of five years is
 

 

Show Answers Only

`B`

Show Worked Solution

`text(A perpetuity lasts indefinitely by only)`

`text(paying out interest.)`

`:.\ text(The graph should show a value that)`

`text(does not change over the years.)`

`=> B`

CORE, FUR1 2016 VCAA 14-16 MC

The table below shows the long-term average of the number of meals served each day at a restaurant. Also shown is the daily seasonal index for Monday through to Friday.

 

Part 1

The seasonal index for Wednesday is 0.84

This tells us that, on average, the number of meals served on a Wednesday is

  1. 16% less than the daily average.
  2. 84% less than the daily average.
  3. the same as the daily average.
  4. 16% more than the daily average.
  5. 84% more than the daily average.

 

Part 2

Last Tuesday, 108 meals were served in the restaurant.

The deseasonalised number of meals served last Tuesday was closest to

  1.   `93`
  2. `100`
  3. `110`
  4. `131`
  5. `152`

 

Part 3

The seasonal index for Saturday is closest to

  1. `1.22`
  2. `1.31`
  3. `1.38`
  4. `1.45`
  5. `1.49`
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

`1 – 0.84 = 0.16`

`:.\ text(A seasonal index of 0.84 tell us)`

`text(16% less meals are served.)`

`=> A`

 

`text(Part 2)`

`text{Deseasonalised number (Tues)}`

`= text(actual number)/text(seasonal index)`

`= 108/0.71`

`~~ 152`

`=> E`

 

`text(Part 3)`

`text(S)text(ince the same number of deseasonalised)`

`text(meals are served each day.)`

`text{S.I. (Sat)}/190` `= 1.10/145`
`text{S.I. (Sat)}` `= (1.10 xx 190)/145`
  `= 1.44…`

`=> D`

CORE, FUR1 2016 VCAA 11-12 MC

The table below gives the Human Development Index (HDI) and the mean number of children per woman (children) for 14 countries in 2007.

A scatterplot of the data is also shown. 
 

 

Part 1

The scatterplot is non-linear.

A log transformation applied to the variable children can be used to linearise the scatterplot.

With HDI as the explanatory variable, the equation of the least squares line fitted to the linearised data is closest to

  1. log(children) = 1.1 – 0.0095 × HDI
  2. children = 1.1 – 0.0095 × log(HDI)
  3. log(children) = 8.0 – 0.77 × HDI
  4. children = 8.0 – 0.77 × log(HDI)
  5. log(children) = 21 – 10 × HDI

 

Part 2

There is a strong positive association between a country’s Human Development Index and its carbon dioxide emissions.

From this information, it can be concluded that

  1. increasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
  2. decreasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
  3. this association must be a chance occurrence and can be safely ignored.
  4. countries that have higher human development indices tend to have higher levels of carbon dioxide emissions.
  5. countries that have higher human development indices tend to have lower levels of carbon dioxide emissions.
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`=>A`

 

`text(Part 2)`

`text(A strong positive association does not mean)`

`text(an increase in one variable causes an increase)`

`text(in the other.)`

`=> D`

CORE, FUR1 2016 VCAA 8 MC

Parallel boxplots would be an appropriate graphical tool to investigate the association between the monthly median rainfall, in millimetres, and the

  1. monthly median wind speed, in kilometres per hour.
  2. monthly median temperature, in degrees Celsius.
  3. month of the year (January, February, March, etc.).
  4. monthly sunshine time, in hours.
  5. annual rainfall, in millimetres.
Show Answers Only

`C`

Show Worked Solution

`text(Parallel boxplots can be used to investigate an)`

♦ Mean mark 45%.

`text(association between categorical and numerical)`

`text(variables.)`

`text(S)text(ince rainfall is numerical, the other variable must)`

`text(be categorical. Only)\ C\ text(is categorical.)`

`=> C`

CORE, FUR1 2016 VCAA 6-7 MC

Part 1

The histogram below shows the distribution of the number of billionaires per million people for 53 countries.
 

Using this histogram, the percentage of these 53 countries with less than two billionaires per million people is closest to

  1. `text(49%)`
  2. `text(53%)`
  3. `text(89%)`
  4. `text(92%)`
  5. `text(98%)`

 

Part 2

The histogram below shows the distribution of the number of billionaires per million people for the same 53 countries as in Part 1, but this time plotted on a `log_10` scale.
 

Based on this histogram, the number of countries with one or more billionaires per million people is

  1.   `text(1)`
  2.   `text(3)`
  3.   `text(8)`
  4.   `text(9)`
  5. `text(10)`
Show Answers Only

`text(Part 1:)\ D`

`text(Part 1:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Percentage with less than 2)`

`= text(countries with less than 2)/text(total countries)`

`= 49/((49 + 2 + 1 + 1))`

`= 49/53`

`~~ 92.4text(%)`

`=> D`

 

`text(Part 2)`

`text(Let)\ \ x=\ text(number of billionaires per million,)`

`text(Number of countries where)\ \ x >= 1,`

♦ Mean mark 45%.
MARKER’S COMMENT: On a `log_10` scale, 1 is plotted as 0 because `log_10\ 1 = 0`, or `10^0=1`

`=> log_10 x >= 0`

`:.\ text(Number of countries)`

`= 9 + 1`

`= 10`

`=> E`

CORE, FUR1 2016 VCAA 1-2 MC

The blood pressure (low, normal, high) and the age (under 50 years, 50 years or over) of 110 adults were recorded. The results are displayed in the two-way frequency table below.
 

     

Part 1

The percentage of adults under 50 years of age who have high blood pressure is closest to

  1.  11%
  2.  19%
  3.  26%
  4.  44%
  5.  58%

Part 2

The variables blood pressure (low, normal, high) and age (under 50 years, 50 years or over) are

  1. both nominal variables.
  2. both ordinal variables.
  3. a nominal variable and an ordinal variable respectively.
  4. an ordinal variable and a nominal variable respectively.
  5. a continuous variable and an ordinal variable respectively.
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ B`

Show Worked Solution

`text(Part 1)`

`text(Percentage)` `= text(Under 50 with high BP)/text(Total under 50)`
  `= 11/58`
  `~~ 19text(%)`

 
`=> B`

 

`text(Part 2)`

♦♦ Mean mark of Part 2: 31%.
MARKER’S COMMENT: Many students incorrectly identified the age (under 50, 50 or over) as nominal.

`text(Blood pressure is an ordinal variable)`

`text(because it is categorical data that can)`

`text(have an order.)`

`text(Under 50 and over 50, likewise, is an)`

`text(ordinal variable.)`

`=> B`

Statistics, NAP-F4-NC02

This graph shows the number of bats in a fruit tree at 15 minute intervals over 4 hours.
 

 
At which time were the highest number of bats in the fruit tree?

`5:45` `6:00` `6:15` `6:30`
 
 
 
 
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`text(6:15 pm)`

Show Worked Solution

`text(The highest data point is one interval past 6:00 pm.)`

`:.\ text(The highest number were in the tree at 6:15 pm.)`

Number and Algebra, NAP-B2-19

Mandy has 46 jelly beans. Jenny has 34 Jelly beans.

How many jelly beans should Mandy give to Jenny if Mandy wanted them both to have the same number of jelly beans?

`6` `12` `40` `80`
 
 
 
 
Show Answers Only

`6\ text(jelly beans)`

Show Worked Solution

`text(If Mandy gives Jenny 6 jelly beans,)`

`text(Mandy’s jelly beans)` `=46-6`
  `=40`
`text(Jenny’s jelly beans)` `=34+6`
  `=40`

 

`:. 6\ text(jelly beans.)`

Measurement, NAP-B2-14 SA

James needs to catch a bus to the city from Greenville. Below is the bus timetable.

What is the latest time James can catch the bus to get to the city before 2 pm?
Show Answers Only

`text(12:40)`

Show Worked Solution

`text(The 12:40 pm bus leaving Greenville arrives at)`

`text(Town Hall at 1:55 pm and is the latest bus)`

`text(arriving before 2 pm.)`

Number and Algebra, NAP-C1-24

Cecil is packing away his club's lawn bowls after a practice session.

One bag carries 4 lawn bowls, as shown below.

Cecil needs to pack away 16 bowls.

Which of these shows how Cecil could work out the number of bags he needs?

`16 ÷ 4` `16 xx 4` `16 - 4` `16 + 4`
 
 
 
 
Show Answers Only

`16 ÷ 4`

Show Worked Solution

`text(S)text(ince each bag can carry 4 bowls, and there)`

`text(are 16 bowls in total,)`

`text(The number of bags = 16 ÷ 4)`

Number and Algebra, NAP-B1-21

What is the difference between the most sick days and the least sick days?

`11` `12` `13` `14`
 
 
 
 
Show Answers Only

`13`

Show Worked Solution

`text{Most sick days (Bella) = 15}`

`text{Least sick days (Kim) = 2}`

`:.\ text(Difference)` `= 15 – 2`
  `= 13`

Number and Algebra, NAP-B1-18

4 groups of 6 plums is the same number of plums as 3 groups of

`8` `6` `4` `3`
 
 
 
 
Show Answers Only

`8`

Show Worked Solution
`text(Number of plums)` `= 4 xx 6`
  `= 24`

`text(If 3 groups,)`

`text(Number in each group)`

`= 24 ÷ 3`

`= 8`

Probability, NAP-C1-22

Rudolph takes 4 marbles out of the bag at the same time.

Which of these is impossible?

 
 3 striped marbles and 1 white marble.
 
 3 white marbles and 1 grey marble.
 
 2 star marbles and 2 white marbles.
 
 2 striped marbles and 2 grey marbles.
Show Answers Only

`text(2 striped marbles and 2 grey marbles.)`

Show Worked Solution

`text{S}text{ince there is only 1 grey marble in the bag,}`

`text(2 striped marbles and 2 grey marbles is impossible.)`

Measurement, NAP-C1-19

Simon makes a calendar for March but forgets to label the days of the week.

He knows the 2nd of March is a Thursday.

What day of the week is 20 March?

 
 `text(Wednesday)`
 
 `text(Tuesday)`
 
 `text(Sunday)`
 
 `text(Monday)`
Show Answers Only

`text(Monday)`

Show Worked Solution

`text(Monday)`

Statistics, NAP-C1-16

Some students at a country school were asked to name their favourite animal.

The four most popular choices were put on the graph below.

  • Dogs were the most popular.
  • Sheep were more popular than cats.
  • Horses were less popular than sheep.

Which column shows sheep on the graph?
 

 

`text(A)` `text(B)` `text(C)` `text(D)`
 
 
 
 
Show Answers Only

`text(B)`

Show Worked Solution

`text(Column D is dogs.)`

`text(S)text(ince horses are less popular than sheep)`

`text(and sheep are more popular than cats.)`

`=>\ text(Sheep are more popular than horses and cats.)`

`=>\ text(Column B is sheep.)`

Number and Algebra, NAP-C1-13

Bobby has 76 football cards.

Shania has 117 football cards.

How many football cards do Bobby and Shania have altogether?

`183` `184` `193` `194`
 
 
 
 
Show Answers Only

`193`

Show Worked Solution
`text(Total cards)` `=117+76`
  `=187+6`
  `=193`

Number and Algebra, NAP-D1-23 SA

Kelly is buying tennis balls from the sporting goods shop.

It costs $12 to buy 4 balls.

If each ball costs the same, how many tennis balls can Kelly buy for $60?
Show Answers Only

`20`

Show Worked Solution
`text(C)text(ost of 1 ball)` `= 12 ÷ 4`
  `=$3`

 

`:.\ text(Number of balls she can buy)`

`= 60 ÷ 3`

`= 20`

Number and Algebra, NAP-D1-16

A chocolate bar has a mass of 100 grams.

The dotted line shows where Sharon cuts the chocolate bar.

Sharon takes the larger piece.

About what mass is Sharon's piece of chocolate?

`text(30 grams)` `text(50 grams)` `text(70 grams)` `text(90 grams)`
 
 
 
 
Show Answers Only

`text(70 grams)`

Show Worked Solution

`text(The cut is on a line where the smaller piece is)`

`text{just over one quarter (25 g) of the chocolate bar.}`

`:.\ text(The larger piece is about 70 grams.)`