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Probability, MET2 2010 VCAA 2

Shoddy Ltd produces statues that are classified as Superior or Regular and are entirely made by machines, on a construction line. The quality of any one of Shoddy’s statues is independent of the quality of any of the others on its construction line. The probability that any one of Shoddy’s statues is Regular is 0.8.

Shoddy Ltd wants to ensure that the probability that it produces at least two Superior statues in a day’s production run is at least 0.9.

Calculate the minimum number of statues that Shoddy would need to produce in a day to achieve this aim.  (3 marks)

Show Answers Only

`18`

Show Worked Solution

`text(Solution 1)`

`text(Let)\ \ X = text(Number of superior statues),`

♦♦ Mean mark 27%.

`X∼\ text(Bi) (n, 0.2)`

`n=18`

 

`text(Solution 2)`

`text(Pr) (X >= 2)` `>= 0.9`
`1 – text(Pr) (X = 0) – text(Pr) (X = 1)` `>= 0.9`
`1 – 0.8^n – ((n),(1))(0.2)^1 (0.8)^(n – 1)` `>= 0.9`
`n` `>= 17.9`
`:. n_min` `= 18`

Calculus, MET2 2010 VCAA 1

  1. Part of the graph of the function  `g: (-4, oo) -> R,\ g(x) = 2 log_e (x + 4) + 1`  is shown on the axes below

     

         

    1. Find the rule and domain of  `g^-1`, the inverse function of  `g`.  (3 marks)
    2. On the set of axes above sketch the graph of  `g^-1`. Label the axes intercepts with their exact values.  (3 marks)
    3. Find the values of `x`, correct to three decimal places, for which  `g^-1(x) = g(x)`.  (2 marks)
    4. Calculate the area enclosed by the graphs of  `g`  and  `g^-1`. Give your answer correct to two decimal places.  (2 marks)
  2. The diagram below shows part of the graph of the function with rule
     
    `qquad qquad qquad f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.

    • The graph has a vertical asymptote with equation  `x = –1`.
    • The graph has a y-axis intercept at 1.
    • The point `P` on the graph has coordinates  `(p, 10)`, where `p` is another real constant.
       

      VCAA 2010 1b

    1. State the value of `a`.  (1 mark)
    2. Find the value of `c`.  (1 mark)
    3. Show that  `k = 9/(log_e (p + 1)`.  (2 marks)
    4. Show that the gradient of the tangent to the graph of `f` at the point `P` is  `9/((p + 1) log_e (p + 1))`.  (1 mark)
    5. If the point  `(– 1, 0)`  lies on the tangent referred to in part b.iv., find the exact value of `p`.  (2 marks)
Show Answers Only
  1.   i. `g^-1(x) = e^((x-1)/2)-4,\ \ x in R`
  2.  ii. 
     
  3. iii. `– 3.914 or 5.503`
  4. iv. `52.63\ text(units²)`
  5.   i. `1`
  6.  ii. `1`
  7. iii. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.  iv. `text(Proof)\ \ text{(See Worked Solutions)}`
  9.   v. `e^(9/10)-1`
Show Worked Solution

a.i.   `text(Let)\ \ y = g(x)`

`text(Inverse:  swap)\ \ x harr y,\ \ text(Domain)\ (g^-1) = text(Range)\ (g)`

`x = 2 log_e (y + 4) + 1`

`:. g^{-1} (x) = e^((x-1)/2)-4,\ \ x in R`

 

ii.  

 

 iii.  `text(Intercepts of a function and its inverse occur)`

  `text(on the line)\ \ y=x.`

`text(Solve:)\ \ g(x) = g^{-1} (x)\ \ text(for)\ \ x`

`:. x dot = – 3.914 or x = 5.503\ \ text{(3 d.p.)}`

 

  iv.   `text(Area)` `= int_(-3.91432…)^(5.50327…) (g(x)-g^-1 (x))\ dx`
    `= 52.63\ text{u²   (2 d.p.)}`

 

b.i.   `text(Vertical Asymptote:)`

`x = – 1`

`:. a = 1`

 

  ii.   `text(Solve)\ \ f(0) = 1\ \ text(for)\ \ c,`

`c = 1`

 

iii.  `f(x)= k log_e (x + 1) + 1`

`text(S)text(ince)\ \ f(p)=10,`

`k log_e (p + 1) + 1` `= 10`
`k log_e (p + 1)` `= 9`
`:. k` `= 9/(log_e (p + 1))\ text(… as required)`

 

  iv.   `f^{′}(x)` `= k/(x + 1)`
  `f^{′}(p)` `= k/(p + 1)`
    `= (9/(log_e(p + 1))) xx 1/(p + 1)\ \ \ text{(using part (iii))}`
    `= 9/((p + 1)log_e(p + 1))\ text(… as required)`

 

  v.   `text(Two points on tangent line:)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students worked out the equation of the tangent which was unnecessary and time consuming.

`(p, 10),\ \ (– 1, 0)`

`f^{′} (p)` `= (10-0)/(p-(– 1))`
  `=10/(p+1)`
   

`text(Solve:)\ \ 9/((p + 1)log_e(p + 1))=10/(p+1)\ \ \ text(for)\ p,`

`:.p= e^(9/10)-1`

CORE*, FUR2 2008 VCAA 5

Michelle took a reducing balance loan for $15 000 to purchase her car. Interest is calculated monthly at a rate of 9.4% per annum.

In order to repay the loan Michelle will make a number of equal monthly payments of $350.

The final repayment will be less than $350.

  1. How many equal monthly payments of $350 will Michelle need to make?  (1 mark)
  2. How much of the principal does Michelle have left to pay immediately after she makes her final $350 payment? Find this amount correct to the nearest dollar.  (1 mark)

Exactly one year after Michelle established her loan the interest rate increased to 9.7% per annum. Michelle decided to increase her monthly payment so that the loan would be fully paid in three years (exactly four years from the date the loan was established).

  1. What is the new monthly payment Michelle will make? Write your answer correct to the nearest cent.  (2 marks)
Show Answers Only
  1. `52`
  2. `$147`
  3. `$388.30`
Show Worked Solution

a.   `text(By TVM Solver,)`

♦♦ Mean mark of all parts (combined) was 23%.
MARKER’S COMMENT: This part was not well done. An area where students consistently struggle each year.
`N` `= ?`
`I(text(%))` `= 9.4`
`PV` `= 15\ 000`
`PMT` `= −350`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>N = 52.422…`

`:. 52\ text(payments of $350 will be required.)`

 

b.   `text(Find principal left after 52 repayments.)`

`text(By TVM Solver,)`

`N` `= 52`
`I(text(%))` `= 9.4`
`PV` `= 15\ 000`
`PMT` `= −350`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 

`FV = −147.056…`

`:.\ text(Principal left to pay) = $147\ \ text{(nearest dollar)}`

 

c.   `text(Find principal left after 12 months.)`

`text(By TVM Solver,)`

`N` `= 12`
`I(text(%))` `= 9.4`
`PV` `= 15\ 000`
`PMT` `= −350`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=>FV = -12\ 086.602…`
 

`text(If loan repaid over the next 3 years,)`

`text(By TVM Solver,)`

`N` `= 3 xx 12 = 46`
`I(text(%))` `= 9.7`
`PV` `= 12\ 086.602…`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>PMT = 388.300…`

`:.\ text(New monthly payments is $388.30.)`

CORE*, FUR2 2008 VCAA 4

Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.

  1. By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation?  (1 mark)
  2. Determine the annual flat rate of depreciation correct to one decimal place.  (1 mark) 
Show Answers Only
  1. `$900`
  2. `5.3text{%  (1 d.p.)}`
Show Worked Solution
a.    `text(Depreciation)` `= (17\ 000 – 3500)/15`
    `= $13\ 500`

 
`:.\ text(Annual depreciation)`

`= (13\ 500)/15`

`= $900`

 

b.   `:.\ text(Flat rate of depreciation )`

`= 900/(17\ 000) xx 100text(%)`

`= 5.29…`

`= 5.3text{%  (1 d.p.)}`

CORE*, FUR2 2008 VCAA 3

Michelle purchased a $17 000 car. The car’s value depreciates at the rate of 10% per annum using the reducing balance method.

  1. By what amount, in dollars, does the car’s value depreciate during Michelle’s third year of ownership?  (2 marks)
  2. After how many years of ownership will the car’s value first be below $7000?  (1 mark) 
Show Answers Only
  1. `$1377`
  2. `text(After 9 years.)`
Show Worked Solution

a.   `text(Value after 2 years)`

`= 17\ 000 (1 – 0.1)^2`

`= 17\ 000(0.9)^2`

`= $13\ 770`

 

`:.\ text(Depreciation in 3rd year)`

`= 10text(%) xx 13\ 770`

`= $1377`

 

b.   `text(Find)\ n\ text(when)`

`17\ 000(0.9)^n` `=7000`
`n` `=8.42…`

 

`:.\ text(After 9 years, the car’s value)`

`text(will first drop below $7000.)`

CORE*, FUR2 2008 VCAA 2

Michelle decided to invest some of her money at a higher interest rate. She deposited $3000 in an account paying 8.2% per annum, compounding half yearly.

  1. Write down an expression involving the compound interest formula that can be used to find the value of Michelle’s $3000 investment at the end of two years. Find this value correct to the nearest cent.  (2 marks)
  2. How much interest will the $3000 investment earn over a four-year period?

     

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

Show Answers Only
  1. `$3523.09`
  2. `$1137.40`
Show Worked Solution

a.   `text(Compounding periods)\ (n) = 2 xx 2 = 4`

`text(Interest per half year) = 8.2/2 = 4.1text(%)`

`:. A` `= PR^n`
  `= 3000(1.041)^4`
  `= 3523.093…`
  `= $3523.09\ \ text{(nearest cent)}`

 

b.   `text(After 4 years)\ (n = 8),`

MARKER’S COMMENT: A TVM calculator could also be used to solve this question.
`A` `= 3000(1.041)^8`
  `= 4137.396…`

 

`:.\ text(Interest)` `= 4137.396-3000`
  `= 1137.396…`
  `= $1137.40\ \ text{(nearest cent)}`

GRAPHS, FUR2 2008 VCAA 3

An event involves running for 10 km and cycling for 30 km.

Let  `x`  be the time taken (in minutes) to run 10 km

 `y`  be the time taken (in minutes) to cycle 30 km

Event organisers set constraints on the time taken, in minutes, to run and cycle during the event.

Inequalities 1 to 6 below represent all time constraints on the event.

Inequality 1:   `x ≥ 0` Inequality 4:   `y <= 150`
Inequality 2:   `y ≥ 0` Inequality 5:   `y <= 1.5x`
Inequality 3:   `x ≤ 120` Inequality 6:   `y >= 0.8x`

 

  1. Explain the meaning of Inequality 3 in terms of the context of this problem.  (1 mark)

 

The lines  `y = 150`  and  `y = 0.8x`  are drawn on the graph below.

GRAPHS, FUR2 2008 VCAA 3

  1. On the graph above

     

    1. draw and label the lines  `x = 120`  and  `y = 1.5x`  (2 marks)
    2. clearly shade the feasible region represented by Inequalities 1 to 6.  (1 mark)

One competitor, Jenny, took 100 minutes to complete the run.

  1. Between what times, in minutes, can she complete the cycling and remain within the constraints set for the event?  (1 mark)
  2. Competitors who complete the event in 90 minutes or less qualify for a prize. 

     

    Tiffany qualified for a prize.

     

    1. Determine the maximum number of minutes for which Tiffany could have cycled.  (1 mark)
    2. Determine the maximum number of minutes for which Tiffany could have run.  (1 mark)
Show Answers Only
  1. `text(Inequality 3 means that the run must take)`
    `text(120 minutes or less for any competitor.)`
  2.  i. & ii.
    GRAPHS, FUR2 2008 VCAA 3 Answer
  3. `text(80 – 150 minutes)`
    1. `54\ text(minutes)`
    2. `50\ text(minutes)`
Show Worked Solution

a.   `text(Inequality 3 means that the run must take)`

`text(120 minutes or less for any competitor.)`

 

b.i. & ii.

GRAPHS, FUR2 2008 VCAA 3 Answer

 

c.   `text(From the graph, the possible cycling)`

♦♦ Mean mark of parts (c)-(d) (combined) was 19%.

`text(time range is between:)`

`text(80 – 150 minutes)`

 

d.i.   `text(Constraint to win a prize is)`

`x + y <= 90`

`text(Maximum cycling time occurs)`

`text(when)\ y = 1.5x`

`:. x + 1.5x` `<= 90`
`2.5x` `<= 90`
`x` `<= 36`

 

`:. y_(text(max))` `= 1.5 xx 36`
  `= 54\ text(minutes)`

 

d.ii.   `text(Maximum run time occurs)`

`text(when)\ \ y = 0.8x`

`:. x + 0.8x` `<= 90`
`1.8x` `<= 90`
`x` `<= 50`

 

`:. x_(text(max)) = 50\ text(minutes)`

CORE*, FUR2 2009 VCAA 5

In order to drought-proof the course, the golf club will borrow $200 000 to develop a water treatment facility. 

The club will establish a reducing balance loan and pay interest monthly at the rate of 4.65% per annum.

  1. $1500 per month will be paid on this loan.

     

    How much of the principal will be left to pay after five years?

     

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

  2. Determine the total interest paid on the loan over the five-year period.

     

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

  3. When the amount outstanding on the loan has reduced to $95 200, the interest rate increases to 5.65% per annum.

     

    Calculate the new monthly repayment that will fully repay this amount in 60 equal instalments.

     

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

Show Answers Only
  1. `$151\ 133.38`
  2. `$41\ 133.38`
  3. `$1825.03`
Show Worked Solution

a.   `text(Find principal after 5 years.)`

♦♦ Mean mark of all parts (combined) was 28%.

`text(By TVM Solver,)`

`N` `= 12 xx 5 = 60`
`I(text(%))` `= 4.65`
`PV` `= 200\ 000`
`PMT` `= −1500`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>FV = −151\ 133.38…`

`:. text(Principal left) = $151\ 133.38`

 

b.   `text(Total repayments)`

`= 60 xx 1500`

`= $90\ 000`

`text(Principal paid off after 5 years)`

`= 200\ 000 – 151\ 133.38`

`= $48\ 866.62`
 

`:.\ text(Total interest paid)`

`= 90\ 000 – 48\ 866.62`

`= $41\ 133.38`

 

c.   `text(By TVM Solver,)`

`N` `= 60`
`Itext(%)` `= 5.65`
`PV` `= 95\ 200`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>PMT = −1825.029…`

`:. text(New monthly repayment) = $1825.03.`

CORE*, FUR2 2009 VCAA 4

The golf club management purchased new lawn mowers for $22 000.

  1. Use the flat rate depreciation method with a depreciation rate of 12% per annum to find the depreciated value of the lawn mowers after four years.  (2 marks)
  2. Use the reducing balance depreciation method with a depreciation rate of 16% per annum to calculate the depreciated value of the lawn mowers after four years. Write your answer in dollars correct to the nearest cent.  (1 mark)
  3. After 4 years, which method, flat rate depreciation or reducing balance depreciation, will give the greater depreciation? Write down the greater depreciation amount in dollars correct to the nearest cent.  (1 mark)
Show Answers Only
  1. `$11\ 440`
  2. `$10\ 953.17`
  3. `$11\ 046.83`
Show Worked Solution

a.   `text(When)\ n = 4,`

`text(Depreciation)` `= 22\ 000 xx 0.12 xx 4`
  `= $10\ 560`

 
`:.\ text(Depreciated Value)`

`= 22\ 000 – 10\ 560`

`= $11\ 440`
 

b.   `r = 16text(%)`

`text(Value)` `= 22\ 000(1 – r)^n`
  `= 22\ 000(0.84)^4`
  `= 10\ 953.169…`
  `= $10\ 953.17`

 

c.   `text(Reducing balance gives a greater depreciated)`

`text(amount after 4 years.)`

`text(Greater depreciation amount)`

`= 22\ 000 – 10\ 953.17`

`= $11\ 046.83`

CORE*, FUR2 2009 VCAA 3

The golf club’s social committee has $3400 invested in an account which pays interest at the rate of 4.4% per annum compounding quarterly.

  1. Show that the interest rate per quarter is 1.1%.  (1 mark)
  2. Determine the value of the $3400 investment after three years.

     

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

  3. Calculate the interest the $3400 investment will earn over six years.

     

    Write your answer in dollars correct to the nearest cent.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `$3876.97`
  3. `$1020.86`
Show Worked Solution

a.   `text(Interest rate per quarter)`

`= 4.4/4`

`= 1.1text(%  …as required.)`
 

b.   `text(Compounding periods = 12)`

`A` `= PR^n`
  `= 3400(1.011)^12`
  `= 3876.973…`
  `= $3876.97`

 

c.   `text(Compounding periods) = 6 xx 4 = 24`

`A` `= 3400(1.011)^24`
  `= 4420.858…`

 
`text(Interest earned over 6 years)`

`= 4420.86- 3400`

`= $1020.86\ \ text{(nearest cent)}`

CORE*, FUR2 2010 VCAA 4

A home buyer takes out a reducing balance loan of $250 000 to purchase an apartment.

Interest on the loan will be calculated and paid monthly at the rate of 6.25% per annum.

  1. The loan will be fully repaid in equal monthly instalments over 20 years.

     

    1. Find the monthly repayment, in dollars, correct to the nearest cent.  (1 mark)
    2. Calculate the total interest that will be paid over the 20 year term of the loan.
      Write your answer correct to the nearest dollar.  (2 marks)
  2. After 60 monthly repayments have been made, what will be the outstanding principal on the loan?

     

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

By making a lump sum payment after nine years, the home buyer is able to reduce the principal on his loan to $100 000. At this time, his monthly repayment changes to $1250. The interest rate remains at 6.25% per annum, compounding monthly.

  1. With these changes, how many months, in total, will it take the home buyer to fully repay the $250 000 loan?  (1 mark)
Show Answers Only
  1. i.  `$1827.32`
  2. ii.  `$188\ 557\ \ text{(nearest $)}`
  3. `$213\ 118`
  4. `212\ text(months)`
Show Worked Solution

a.i.   `text(By TVM Solver,)`

♦♦ This question (all parts) were generally poorly answered, although exact data unavailable.
`N` `= 20 xx 12 = 240`
`I(text(%))` `= 6.25`
`PV` `= 250\ 000`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>PMT = −1827.320…`

`:. text(Monthly repayment) = $1827.32`

 

a.ii.   `text(Total of all repayments)`

`= 240 xx 1827.32`

`= $438\ 556.80`

`:.\ text(Total interest paid)`

`= 438\ 556.80 – 250\ 000`

`= 188\ 556.80`

`= $188\ 557\ \ text{(nearest $)}`

 

b.   `text(By TVM Solver:)`

`N` `= 60`
`I(text(%))` `= 6.25`
`PV` `= 250\ 000`
`PMT` `= −1827.32`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>FV = −213\ 117.807…`

`:.\ text(After 60 repayments, the loan balance is $213 118.)`

 

c.   `text(By TVM Solver:)`

`N` `= ?`
`I(text(%))` `= 6.25`
`PV` `= 100\ 000`
`PMT` `= −1250`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> N = 103.75…`

MARKER’S COMMENT: A common error was to forget to add the initial 108 months.

`:.\ text(Total time to repay loan)`

`= 104 + 9 xx 12`

`= 212\ text(months)`

CORE*, FUR2 2010 VCAA 3

Simple Saver is a simple interest investment in which interest is paid annually.

Growth Plus is a compound interest investment in which interest is paid annually.

Initially, $8000 is invested with both Simple Saver and Growth Plus.

The graph below shows the total value (principal and all interest earned) of each of these investments over a 15 year period.

The increase in the value of each investment over time is due to interest
 

BUSINESS, FUR2 2010 VCAA 3
 

  1. Which investment pays the highest annual interest rate, Growth Plus or Simple Saver?

     

    Give a reason to justify your answer.  (1 mark)

  2. After 15 years, the total value (principal and all interest earned) of the Simple Saver investment is $21 800.

     

    Find the amount of interest paid annually.  (1 mark)

  3. After 15 years, the total value (principal and all interest earned) of the Growth Plus investment is $24 000.

     

    1. Write down an equation that can be used to find the annual compound interest rate, `r`.  (1 mark)
    2. Determine the annual compound interest rate.

       

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

Show Answers Only
  1. `text(Simple Saver has the highest annual)`
    `text(interest rate because after 1 year,)`
    `text(the value of investment is higher.)`
  2. `$920`
    1. `24\ 000 = 8000 (1 + r/100)^15`
    2. `7.6text{%  (1 d.p.)}`
Show Worked Solution

a.   `text(Simple Saver has the highest annual)`

♦♦♦ Part (a) was “very” poorly answered although exact data unavailable.
MARKER’S COMMENT: Most students ignored the word “rate” and instead referred to the eventual return of each investment.

`text(interest rate because after 1 year,)`

`text(the value of investment is higher.)`

 

b.   `text(Total interest earned)`

`= 21\ 800 – 8000`

`= $13\ 800`

`:.\ text(Interest paid annually)`

`= (13\ 800)/15`

`= $920`

 

c.i.   `text(Using)\ A = PR^n,`

`24\ 000 = 8000 (1 + r/100)^15`

 

c.ii.    `(1 + r/100)^15` `= 3`
  `1 + r/100` `= 1.0759…`
  `:. r` `= 0.0759…`
    `= 7.6text{%  (1 d.p.)}`

CORE*, FUR2 2010 VCAA 2

$360 000 is invested in a perpetuity at an interest rate of 5.2% per annum.

  1. Find the monthly payment that the perpetuity provides.  (1 mark)
  2. After six years of monthly payments, how much money remains invested in the perpetuity?  (1 mark)
Show Answers Only
  1. `$1560`
  2. `$360\ 000`
Show Worked Solution

a.   `text(Monthly repayment)`

`= 1/12 xx (360\ 000 xx 5.2/100)`

`= 1/12 xx 18\ 720`

`= $1560`

♦♦ Part (b) was poorly answered.
MARKER’S COMMENT: Many students did not understand the concept of a perpetuity.

 

b.   `$360\ 000`

 

GRAPHS, FUR2 2010 VCAA 3

Let `x` be the number of Softsleep pillows that are sold each week and `y` be the number of Resteasy pillows that are sold each week.

A constraint on the number of pillows that can be sold each week is given by

Inequality 1:   `x + y ≤ 150`

  1. Explain the meaning of Inequality 1 in terms of the context of this problem.  (1 mark)

Each week, Anne sells at least 30 Softsleep pillows and at least `k` Resteasy pillows.

These constraints may be written as

Inequality 2:   `x ≥ 30`

 

Inequality 3:   `y ≥ k`

The graphs of  `x + y = 150`  and  `y = k`  are shown below.

GRAPHS, FUR2 2010 VCAA 3

  1. State the value of `k`.  (1 mark)
  2. On the axes above

     

    1. draw the graph of  `x = 30`  (1 mark)
    2. shade the region that satisfies Inequalities 1, 2 and 3.  (1 mark)
  3. Softsleep pillows sell for $65 each and Resteasy pillows sell for $50 each.

     

    What is the maximum possible weekly revenue that Anne can obtain?  (2 marks)

Anne decides to sell a third type of pillow, the Snorestop.

She sells two Snorestop pillows for each Softsleep pillow sold. She cannot sell more than 150 pillows in total each week.

  1. Show that a new inequality for the number of pillows sold each week is given by

     

          Inequality 4:   `3x + y ≤ 150`

     

     

    where     `x`  is the number of Softsleep pillows that are sold each week

     

        and     `y`  is the number of Resteasy pillows that are sold each week.  (1 mark)

Softsleep pillows sell for $65 each.

Resteasy pillows sell for $50 each.

Snorestop pillows sell for $55 each.

  1. Write an equation for the revenue, `R` dollars, from the sale of all three types of pillows, in terms of the variables `x` and `y`.  (1 mark)
  2. Use Inequalities 2, 3 and 4 to calculate the maximum possible weekly revenue from the sale of all three types of pillow.  (2 marks) 
Show Answers Only
  1. `text(Inequality 1 means that the combined number of Softsleep)`
    `text(and Resteasy pillows must be less than 150.)`
  2. `45`
  3. i. & ii.
    GRAPHS, FUR2 2010 VCAA 3 Answer
  4. `$9075`
  5. `text(See Worked Solutions)`
  6. `R = 175x + 50y`
  7. `$8375`
Show Worked Solution

a.   `text(Inequality 1 means that the combined number of Softsleep)`

`text(and Resteasy pillows must be less than 150.)`

 

b.   `k = 45`

 

c.i. & ii.   

GRAPHS, FUR2 2010 VCAA 3 Answer

 

d.   `text(Checking revenue at boundary)`

`text(At)\ (30,120),`

`R = 65 xx 30 + 50 xx 120 = $7950`

`text(At)\ (105,45),`

`R = 65 xx 105 + 50 xx 45 = $9075`

`:. text(Maximum weekly revenue) = $9075`

 

e.   `text(Let)\ z = text(number of SnoreStop pillows)`

♦♦ Mean mark of parts (e)-(g) (combined) was 24%.

`:. x + y + z <= 150,\ text(and)`

`z = 2x\ \ text{(given)}`

`:. x + y + 2x` `<= 150`
`3x + y` `<= 150\ \ …text(as required)`

 

f.    `R` `= 65x + 50y + 55(2x)`
    `= 65x + 50y + 110x`
    `= 175x + 50y`

 

g.   

GRAPHS, FUR2 2010 VCAA 3 Answer1

`text(New intersection occurs at)\ (35,45)`

`:.\ text(Maximum weekly revenue)`

`= 175 xx 35 + 50 xx 45`

`= $8375`

GEOMETRY, FUR2 2010 VCAA 3

A concrete square pyramid with volume 1.8 m³ sits on the flat top of the hill.

The length of the square base of the pyramid is `x` metres. The height of the pyramid, `VT`, is 2.5 metres.

GEOMETRY, FUR2 2010 VCAA 3

Find the value of `x`, correct to two decimal places.  (2 marks)

 

Show Answers Only

`1.47\ text(m)`

Show Worked Solution
♦ Mean mark 41%.
`text(Volume)` `= 1/3  Ah`
`1.8` `= 1/3 xx x^2 xx 2.5`
`x^2` `= (3 xx 1.8)/2.5`
  `= 2.16`
`:. x` `= 1.469…`
  `= 1.47\ text(m)`

 

CORE*, FUR2 2011 VCAA 4

Tania takes out a reducing balance loan of $265 000 to pay for her house.

Her monthly repayments will be $1980.

Interest on the loan will be calculated and paid monthly at the rate of 7.62% per annum.

  1. i.  How many monthly repayments are required to repay the loan? Write your answer to the nearest month.  (1 mark)
  2. ii.  Determine the amount that is paid off the principal of this loan in the first year. Write your answer to the nearest cent.  (1 mark)

Immediately after Tania made her twelfth payment, the interest rate on her loan increased to 8.2% per annum, compounding monthly.

Tania decided to increase her monthly repayment so that the loan would be repaid in a further nineteen years.

  1. Determine the new monthly repayment.

     

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

Show Answers Only
  1. i.  `300`
  2. ii.  `$3694.25\ \ text{(nearest cent)}`
  3. `$2265.04`
Show Worked Solution

a.i.   `text(By TVM Solver:)`

`N` `= ?`
`I(text(%))` `= 7.62`
`PV` `= 265\ 000`
`PMT` `= −1980`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> N = 299.573…`

`:.\ text(After 300 months, the loan will be repaid.)`

 

a.ii.   `text(After 12 months, by TVM Solver:)`

`N` `= 12`
`I(text(%))` `= 7.62`
`PV` `= 265\ 000`
`PMT` `= −1980`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> FV = −261\ 305.74…`

`:.\ text(Amount paid off after 1 year)`

`= 265\ 000 – 261\ 305.747…`

`= 3694.252…`

`= $3694.25\ \ text{(nearest cent)}`

 

b.   `text(By TVM Solver:)`

`N` `= 19 xx 12 = 228`
`Itext(%)` `= 8.2`
`PV` `= 261\ 305.75`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> PMT = −2265.043`

`:.\ text(New monthly repayment is $2265.04)`

CORE*, FUR2 2011 VCAA 3

Tania purchased a house for $300 000.

She will have to pay stamp duty based on this purchase price.

Stamp duty rates are listed in the table below.
 

     BUSINESS, FUR2 2011 VCAA 3
 

  1. Calculate the amount of stamp duty that Tania will have to pay.  (1 mark)
  2. Assuming that her house will increase in value at a rate of 3.17% per annum, what will the value of Tania's house be after 5 years?

     

    Write your answer to the nearest thousand dollars.  (1 mark)

Tania bought her house at the start of 2011.

  1. If the rate of increase in value remains at 3.17% per annum, at the start of which year will the value of Tania's house first exceed $450 000?  (2 marks)
Show Answers Only
  1. `$13\ 070`
  2. `$351\ 000`
  3. `2024`
Show Worked Solution
a.    `text(Stamp duty)` `= 2870 + 6text(%) xx (300\ 000 – 130\ 000)`
    `= $13\ 070`
♦♦ Part (a) was “poorly answered” although exact data is unavailable.
MARKER’S COMMENT: A majority of students did not understand how to use the table.

 

b.   `text(Using)\ \ \ A = PR^n`

`text(Value)` `= 300\ 000(1.0317)^5`
  `= 350\ 661.7…`
  `= $351\ 000\ \ text{(nearest $1000)}`

 

c.   `text(Find)\ n\ text(when)\ \ \ A > $450\ 000`

`300\ 000 xx 1.0317^n` `= 450\ 000`
`n` `~~12.99…`

 

MARKER’S COMMENT: Many students who correctly found `n` lost a mark by failing to identify the exact year. 

`:.\ text(After 13 years, in 2024, the house value)`

`text(will be over)\ $450\ 000.`

CORE*, FUR2 2011 VCAA 2

Tom and Patty both decided to invest some money from their savings.

Each chose a different investment strategy.

Tom's investment strategy

•  Deposit $5600 into an account with an interest rate of 7.2% per annum, compounding monthly.

•  Immediately after interest is paid into his investment account on the last day of each month, deposit a further $200 into the account.

  1. Determine the total amount in Tom's investment account at the end of the first month.  (1 mark)

Patty's investment strategy

•  Invest $8000 at the start of the year at an interest rate of 7.2% per annum, compounding annually.

  1. The following expression can be used to determine the value of Patty's investment at the end of the first year. Complete the expression by filling in the box.  (1 mark)

BUSINESS, FUR2 2011 VCAA 2

At the end of twelve months, Patty has more money in her investment account than Tom.

  1. How much more does she have?

     

    Write your answer to the nearest cent.  (2 marks)

  2. What annual compounding rate of interest would Patty need in order to earn $1000 interest in one year on her $8000 investment?

     

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

Show Answers Only
  1. `$5833.60`
  2. `text(Value of investment)\ = 8000 xx (1 + 7.2/100)`
  3. `$78.42\ \ text{(nearest cent)}`
  4. `text(12.5%)`
Show Worked Solution

a.   `text(Tom’s investment after 1 month)`

`= 5600 xx (1 + 7.2/(12 xx 100)) + 200`

`= $5833.60`

 

b.   `text(After 1 year,)`

`text(Value of investment) = 8000 xx (1 + 7.2/100)`

 

c.   `text(Tom’s investment after 12 months,)`

`text(by TVM Solver,)`

`N` `= 12`
`I(text(%))` `= 7.2`
`PV` `= 5600`
`PMT` `= 200`
`text(P/Y)` `= 12`
`text(C/Y)` `= 12`

 

`=> FV = −8497.58…`

 

`text(Patty’s investment after 12 months)`

MARKER’S COMMENT: Many students valued Patty’s investment correctly but then deducted Tom’s after 1 month (from part a), instead of 12.

`= 8000 xx (1 + 7.2/100)`

`= $8576`
 

`:.\ text(Extra value of Patty’s investment)`

`= 8576 – 8497.580…`

`= 78.419…`

`= $78.42\ \ text{(nearest cent)}`

 

d.    `text(Using)\ \ \ I` `= (PrT)/100`
  `1000` `= (8000 xx r xx 1)/100`
  `r` `= (1000 xx 100)/8000`
    `= 12.5text(%)`

CORE*, FUR2 2012 VCAA 4

Arthur invested $80 000 in a perpetuity that returns $1260 per quarter. Interest is calculated quarterly.

  1. Calculate the annual interest rate of Arthur’s investment.  (1 mark)
  2. After Arthur has received 20 quarterly payments, how much money remains invested in the perpetuity?  (1 mark)
  3. Arthur’s wife, Martha, invested a sum of money at an interest rate of 9.4% per annum, compounding quarterly.

     

     

    She will be paid $1260 per quarter from her investment.

     

     

    After ten years, the balance of Martha’s investment will have reduced to $7000.

     

     

    Determine the initial sum of money Martha invested.

     

     

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

Show Answers Only
  1. `6.3text(%)`
  2. `$80\ 000`
  3. `$35\ 208`
Show Worked Solution

a.   `text(Let)\ \ r =\ text(annual interest rate)`

♦♦ Mean mark of all parts (combined) was 26%.
`80\ 000 xx r/(4 xx 100)` `= 1260`
`:. r`  `= (1260 xx 400)/(80\ 000)`
   `= 6.3text(%)`

 

b.   `$80\ 000`

`text{(The principal invested in a perpetuity}`

`text{remains unchanged.)}`

 

c.   `text(Find)\ PV\ text(using TVM Solver:)`

`N` `= 4 xx 10 = 40`
`I(text(%))` `= 9.4`
`PV` `= ?`
`PMT` `= 1260`
`FV` `= 7000`
`text(P/Y)` `= text(C/Y) = 4`

 

`=> PV = −35\ 208.002…`

`:.\ text{Martha initially invested $35 208 (nearest $)}`

CORE*, FUR2 2012 VCAA 3

An area of a club needs to be refurbished.

$40 000 is borrowed at an interest rate of 7.8% per annum. 

Interest on the unpaid balance is charged to the loan account monthly. 

Suppose the $40 000 loan is to be fully repaid in equal monthly instalments over five years.

  1. Determine the monthly payment, correct to the nearest cent.  (1 mark)
  2. If, instead, the monthly payment was $1000, how many months will it take to fully repay the $40 000?  (1 mark)
  3. Suppose no payments are made on the loan in the first 12 months.
  4. i.  Write down a calculation that shows that the balance of the loan account after the first 12 months will be $43 234 correct to the nearest dollar.  (1 mark)
  5. ii. After the first 12 months, only the interest on the loan is paid each month. 

     

        Determine the monthly interest payment, correct to the nearest cent.   (1 mark)  

Show Answers Only
  1. `$807.23`
  2. `text(47 months)`
  3. i.  `text(See Worked Solutions)`
  4. ii.  `$281.02\ \ text{(nearest cent)}`
Show Worked Solution

a.   `text(Find monthly payment by TVM solver:)`

♦♦ Mean mark of all parts (combined) 28%.
`N` `= 5 xx 12 = 60`
`I(text(%))` `= 7.8`
`PV` `= 40\ 000`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> PMT = −807.232…`

`:.\ text{Monthly payment is $807.23 (nearest cent)}`

 

b.   `text(Find)\ n\ text(when loan fully paid:)`

`N` `= ?`
`I(text(%))` `= 7.8`
`PV` `= 40\ 000`
`PMT` `= −1000`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> N = 46.47…`

`:.\ text(Loan will be fully repaid after 47 months.)`

 

c.i.   `text(Loan balance after 12 months)`

`= 40\ 000 xx (1 + (7.8)/(12 xx 100))^12`

`= 43\ 233.99…`

`= $43\ 234\ \ text{(nearest $) … as required}`

 

c.ii.   `text(Interest paid each month)`

`= 43\ 234 xx (7.8)/(12 xx 100)`

`= 281.021`

`= $281.02\ \ text{(nearest cent)}`

CORE*, FUR2 2012 VCAA 2

The value of the equipment will be depreciated using the unit cost method. 

The initial value of the equipment is $8360. It will depreciate by 22 cents per hour of use. 

On average, the equipment will be used for 3800 hours each year.

  1. Calculate the depreciated value of the equipment after three years.  (1 mark)
  2. Show that, in any one year, the flat rate method of depreciation with a depreciation rate of 10% per annum will give the same annual depreciation as the unit cost method.  (1 mark)
  3. After how many years will equipment be written off with a depreciated value of $0?  (1 mark)
  4. Suppose the reducing balance method is used to depreciate the equipment instead of the unit cost method.

     

    The initial value of the equipment is $8360. It will depreciate at a rate of 14% per annum of the reducing balance.

     

    Find, correct to the nearest dollar, the depreciated value of the equipment after ten years.  (1 mark)

Show Answers Only
  1. `$5852`
  2. `text(See Worked Solution)`
  3. `text(10 years)`
  4. `$1850\ \ text{(nearest $)}`
Show Worked Solution

a.   `text(Unit cost depreciation per year)`

`= 3800 xx 0.22`

`= $836`

`:.\ text(After 3 years, depreciated value)`

`= 8360 – (3 xx 836)`

`= $5852`

 

b.   `text(10% Depreciation per year)`

`= 10text(%) xx 8360`

`= $836\ \ …text(as required.)`

 

c.   `text(Depreciated value of $0 occurs when)`

`8360 – 836n` `= 0`
`836n` `= 8360`
`n` `= 10\ text(years)`

 

d.   `text(After 1 year,)`

`V_1` `= 8360(1 – 0.14)`
  `= 8360(0.86)`

 

`text(After 10 years,)`

`V_10` `= 8360(0.86)^10`
  `= 1850.08…`
  `= $1850\ \ text{(nearest $)}`

NETWORKS, FUR1 2008 VCAA 8-9 MC

The network below shows the activities that are needed to finish a particular project and their completion times (in days).

 

networks-fur1-2008-vcaa-8-mc

Part 1

The earliest start time for Activity `K`, in days, is

A.     7

B.   15

C.   16

D.   19

E.   20

 

Part 2

This project currently has one critical path.

A second critical path, in addition to the first, would be created by

A.   increasing the completion time of `D` by 7 days.

B.   increasing the completion time of `G` by 1 day.

C.   increasing the completion time of `I` by 2 days.

D.   decreasing the completion time of `C` by 1 day.

E.   decreasing the completion time of `H` by 2 days.

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(EST for Activity)\ K`

`=\ text(Duration)\ ACFI`

`= 2 + 5 + 6 + 3`

`= 16`

`=> C`

 

`text(Part 2)`

♦♦ Mean mark of Part 2 was 35%.

`text(Original critical path is)`

`ACFHJL\ text{(22 days)}`

`text(Consider option)\ A,`

`text(New critical path is created)`

`ABDJL\ text{(22 days)}`

`=> A`

NETWORKS, FUR1 2008 VCAA 6 MC

networks-fur1-2008-vcaa-6-mc

 
For the graph above, the capacity of the cut shown is

A.   `33`

B.   `36`

C.   `40`

D.   `42`

E.   `46`

Show Answers Only

`=> D`

Show Worked Solution

`text(Capacity of the cut)`

♦♦ Mean mark 35%.
MARKER’S COMMENT: For an individual flow to contribute to the “cut”, it must flow from the source to the sink.

`= 4 + 2 + 7 + 9 + 8 + 6 + 6`

`= 42`

`=> D`

 

NETWORKS, FUR1 2008 VCAA 5 MC

A connected planar graph has five vertices, `A`, `B`, `C`, `D` and `E`.

The degree of each vertex is given in the following table.
 

networks-fur1-2008-vcaa-5-mc
 

Which one of the following statements regarding this planar graph is true?

A.   The sum of degrees of the vertices equals 15.

B.   It contains more than one Eulerian path.

C.   It contains an Eulerian circuit.

D.   Euler’s formula  `v + f = e + 2`  could not be used.

E.   The addition of one further edge could create an Eulerian path.

Show Answers Only

`=> E`

Show Worked Solution

`text(Consider)\ E,`

♦ Mean mark 41%.

`text(If one edge added, the planar graph would)`

`text(have exactly 2 vertices that are odd, and an)`

`text(Eulerian path could exist.)`

`=> E`

NETWORKS, FUR1 2010 VCAA 6-7 MC

In the network below, the values on the edges give the maximum flow possible between each pair of vertices. The arrows show the direction of flow. A cut that separates the source from the sink in the network is also shown.
 

vcaa-networks-fur1-2010-6-7

 
Part 1

The capacity of this cut is

A.   `14`

B.   `18`

C.   `23`

D.   `31`

E.   `40`

 

Part 2

The maximum flow between source and sink through the network is

A.    `7`

B.   `10`

C.   `11`

D.   `12`

E.   `20`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ B`

Show Worked Solution

Part 1

`text(Capacity of the cut)`

♦ Mean mark 50%.
COMMENT: A quarter of students incorrectly included the “8” which is flowing in the opposite direction.

`= 11 + 5 + 7`

`= 23`

`=>  C`

 

Part 2

vcaa-networks-fur1-2010-6-7i

`text(The maximum flow)`

♦♦♦ Mean mark 24%.

`=\ text{minimum cut (see above)}`

`= 4 + 2 + 3 + 1`

`= 10`

`=>  B`

NETWORKS, FUR1 2006 VCAA 3 MC

networks-fur1-2006-vcaa-3-mc
 

Which one of the following statements is true regarding the network above?

A.   `ABCDEFG` is a Hamiltonian circuit.

B.   Only one Hamiltonian path exists.

C.   `CBAGFEDC` is an Eulerian circuit.

D.   At least two Eulerian paths exist.

E.   There are no circuits.

Show Answers Only

`D`

Show Worked Solution

`text(The network has exactly 2 vertices with)`

♦ Mean mark 43%.

`text(odd-degrees.)`

`:.\ text(At least two Eulerian paths exist.)`

`rArr D`

NETWORKS, FUR1 2007 VCAA 5-6 MC

The following network shows the activities that are needed to complete a project and their completion times (in hours).
 


 

Part 1

Which one of the following statements regarding this project is false?

A.   Activities `A, B` and `C` all have the same earliest start time.

B.   There is only one critical path for this project.

C.   Activity `J` may start later than activity `H.`

D.   The shortest path gives the minimum time for project completion.

E.   Activity `L` must be on the critical path.

 

Part 2

The earliest start time for activity `L`, in hours, is

A.   11

B.   12

C.   14

D.   15

E.   16

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ E`

Show Worked Solution

`text (Part 1)`

♦ Mean mark 43%.

`A, B, C,\ text(and)\ E\ text(can be shown to be true.)`

`=>  D`

 

`text (Part 2)`

`text(Critical path is)\ CDFKL`

`:.\ text(EST for)\ L` `= 5 + 0 + 4 + 7`
  `= 16\ text(hours)`

  
`=>  E`

NETWORKS, FUR1 2009 VCAA 5-6 MC

The network shows the activities that are needed to complete a particular project.

networks-fur1-2009-vcaa-5-6-mc

Part 1

The total number of activities that need to be completed before activity `L` may begin is

A.   `2`

B.   `4`

C.   `6`

D.   `7`

E.   `8`

 

Part 2

The duration of every activity is initially 5 hours. For an extra cost, the completion times of both activity `F` and activity `K` can be reduced to 3 hours each.

If this is done, the completion time for the project will be

A.   decreased by 2 hours.

B.   decreased by 3 hours.

C.   decreased by 4 hours.

D.   decreased by 6 hours.

E.   unchanged. 

Show Answers Only

`text(Part 1:)\ D`

`text(Part 1:)\ E`

Show Worked Solution

`text(Part 1)`

`A, B, C, D, E, H\ text(and)\ I\ text(must be)`

`text(completed before)\ L.`

`=>  D`

 

`text(Part 2)`

♦ Mean mark 45%.

`F\ text(and)\ K\ text(are not on any critical path and a)`

`text(reduction of 3 hours on either activity will not change)`

`text(the completion time for the project.)`

`=>E`

NETWORKS, FUR1 2009 VCAA 3 MC

networks-fur1-2009-vcaa-3-mc1

 
The maximum flow from source to sink through the network shown above is

A.     `6`

B.     `7`

C.     `8`

D.   `11`

E.   `16`

Show Answers Only

`B`

Show Worked Solution

networks-fur1-2009-vcaa-3-mc-answer 
 

♦ Mean mark 44%.
`text(Maximum flow)` `=\ text(minimum cut)`
  `= 1 + 4 + 2`
  `= 7`

`=>  B`

NETWORKS, FUR1 2013 VCAA 7 MC

A connected graph consists of five vertices and four edges.

Consider the following five statements.

•  The graph is planar.

•  The graph has more than one face.

•  All vertices are of even degree.

•  The sum of the degrees of the vertices is eight.

•  The graph cannot have a loop.

How many of these statements are always true for such a graph?

A.   `1`

B.   `2`

C.   `3`

D.   `4`

E.   `5`

Show Answers Only

`C`

Show Worked Solution

`text(The possible graphs with 5 vertices)`

`text(and 4 edges are:)`

vcaa-networks-fur1-2013-7

`text(By inspection, the true statements are:)`

♦ Mean mark 39%.

`•\ text(graph is planar)`

`•\ text(sum of degrees is eight)`

`•\ text(graph can’t have a loop)`

 `=>  C`

NETWORKS, FUR2 2006 VCAA 3

The five musicians are to record an album. This will involve nine activities.

The activities and their immediate predecessors are shown in the following table.

The duration of each activity is not yet known.
 

NETWORKS, FUR2 2006 VCAA 31
 

  1. Use the information in the table above to complete the network below by including activities `G`, `H` and `I`.  (2 marks)
     

NETWORKS, FUR2 2006 VCAA 32

There is only one critical path for this project.

  1. How many non-critical activities are there?  (1 mark)

The following table gives the earliest start times (EST) and latest start times (LST) for three of the activities only. All times are in hours.


Networks, FUR2 2006 VCAA 3_3

  1. Write down the critical path for this project.  (1 mark)

The minimum time required for this project to be completed is 19 hours.

  1. What is the duration of activity `I`?  (1 mark)

The duration of activity `C` is 3 hours.

  1. Determine the maximum combined duration of activities `F` and `H`.  (1 mark) 
Show Answers Only
  1.  
    networks-fur2-2006-vcaa-3-answer
  2. `5`
  3. `B − E − G − I`
  4. `text(7 hours)`
  5. `text(8 hours)`
Show Worked Solution
a.    networks-fur2-2006-vcaa-3-answer

 

b.   `text(Possible critical paths are,)`

`ADGI, BEGI\ text(or)\ CFHI`

`:.\ text(Non-critical activities)`

`= 9 – 4 = 5`

 

c.   `text(Critical activities have zero slack time.)`

♦ Mean mark of parts (c)-(e) (combined) was 36%.

`:. A\ text(and)\ C\ text(are non-critical.)`

`:. B − E − G − I\ \ text(is the critical path.)`

 

d.    `text(Duration of)\ \ I` `= 19 – 12`
    `= 7\ text(hours)`

 

e.   `text(Maximum time for)\ F\ text(and)\ H`

`=\ text(LST of)\ I – text(duration)\ C – text(slack time of)\ C`

`= 12 – 3 – 1`

`= 8\ text(hours)`

MATRICES*, FUR2 2006 VCAA 2

The five musicians, George, Harriet, Ian, Josie and Keith, compete in a music trivia game.

Each musician competes once against every other musician.

In each game there is a winner and a loser.

The results are represented in the dominance matrix, Matrix 1, and also in the incomplete directed graph below.

On the directed graph an arrow from Harriet to George shows that Harriet won against George.

 

NETWORKS, FUR2 2006 VCAA 2
 

  1. Explain why the figures in bold in Matrix 1 are all zero.  (1 mark)

One of the edges on the directed graph is missing.

  1. Using the information in Matrix 1, draw in the missing edge on the directed graph above and clearly show its direction.  (1 mark)

The results of each trivia contest (one-step dominances) are summarised as follows.


networks-fur2-2006-vcaa-2_2 

In order to rank the musicians from first to last in the trivia contest, two-step (two-edge) dominances will be considered.

The following incomplete matrix, Matrix 2, shows two-step dominances.

`{:(qquadqquadqquadtext(Matrix 2)),(qquadqquad{:GquadHquadI\ quadJquad\ K:}),({:(G),(H),(I),(J),(K):}[(0,1,1,2,0),(1,0,1,1,1),(1,0,0,0,0),(0,0,1,0,1),(2,0,1,x,0)]):}`

 

  1. Explain the two-step dominance that George has over Ian.  (1 mark)
  2. Determine the value of the entry `x` in Matrix 2.  (1 mark)
  3. Taking into consideration both the one-step and two-step dominances, determine which musician was ranked first and which was ranked last in the trivia contest.  (2 marks)
Show Answers Only
  1. `text(A musicians does not compete against him/herself.)`
  2.  
    networks-fur2-2006-vcaa-2-anwer
  3. `text(Two step dominance occurs because George is dominant)`

     

    `text(over Keith who is in turn dominant over Ian.)`

  4. `2`
  5. `text{First is Keith (8), last is Ian (2)}`
Show Worked Solution

a.   `text(A musicians does not compete against him/herself.)`

 

b.   `text(Josie won against George.)`

 

networks-fur2-2006-vcaa-2-anwer

 

c.   `text(Two step dominance occurs because George is dominant)`

`text(over Keith who is in turn dominant over Ian.)`

 

d.   `text(Following the edges on network diagram:)`

`text(Keith over Harriet who beats Josie.)`

`text(Keith over Ian who beats Ian.)`

`:. x = 2`

 

e.    `D_1 + D_2 =` `[(0,1,2,2,1),(2,0,2,2,1),(1,0,0,1,0),(1,0,1,0,1),(2,1,2,3,0)]{:(G – 6),(H – 7),(I – 2),(J – 3),(K – 8):}`

 

`text{Summing the rows (above),}`

`:.\ text{First is Keith (8), last is Ian (2).}`

NETWORKS, FUR2 2007 VCAA 4

A community centre is to be built on the new housing estate.

Nine activities have been identified for this building project.

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

NETWORKS, FUR2 2007 VCAA 4
 

  1. Determine the minimum time, in weeks, to complete this project.  (1 mark)
  2. Determine the slack time, in weeks, for activity `D`.  (2 marks)

The builders of the community centre are able to speed up the project.

Some of the activities can be reduced in time at an additional cost.

The activities that can be reduced in time are `A`, `C`, `E`, `F` and `G`.

  1. Which of these activities, if reduced in time individually, would not result in an earlier completion of the project?  (1 mark)

The owner of the estate is prepared to pay the additional cost to achieve early completion.

The cost of reducing the time of each activity is $5000 per week.

The maximum reduction in time for each one of the five activities, `A`, `C`, `E`, `F`, `G`, is 2 weeks.

  1. Determine the minimum time, in weeks, for the project to be completed now that certain activities can be reduced in time.  (1 mark)
  2. Determine the minimum additional cost of completing the project in this reduced time.  (1 mark)
Show Answers Only
  1. `19\ text(weeks)`
  2. `5\ text(weeks)`
  3. `A, E, G`
  4. `text(15 weeks)`
  5. `$25\ 000`
Show Worked Solution

a.   `B-C-F-H-I\ \ text(is the critical path.)`

♦ Mean mark of all parts (combined) 40%.
`:.\ text(Minimum time)` `= 4 + 3 + 4 + 2 + 6`
  `= 19\ text(weeks)`

 

b.    `text(EST of)\ D` `= 4`
  `text(LST of)\ D` `= 9`
`:.\ text(Slack time of)\ D` `= 9 – 4`
  `= 5\ text(weeks)`

 

c.   `A, E,\ text(and)\ G\ text(are not currently on)`

`text(the critical path, therefore crashing)`

`text(them will not affect the completion)`

`text(time.)`

 

d.   `text(Reduce)\ C\ text(and)\ F\ text(by 2 weeks.)`

`text(However, a new critical path)`

`B − E − H − I\ text(takes 16 weeks.)`

`:.\ text(Also reduce)\ E\ text(by 1 week.)`

`:.\ text(Minimum time = 5 weeks)`

 

e.    `text(Additional cost)` `= 5 xx $5000`
    `= $25\ 000`

NETWORKS, FUR2 2007 VCAA 3

As an attraction for young children, a miniature railway runs throughout the new housing estate.

The trains travel through stations that are represented by nodes on the directed network diagram below.

The number of seats available for children, between each pair of stations, is indicated beside the corresponding edge.
 

NETWORKS, FUR2 2007 VCAA 3

 
Cut 1, through the network, is shown in the diagram above.

  1. Determine the capacity of Cut 1.  (1 mark)
  2. Determine the maximum number of seats available for children for a journey that begins at the West Terminal and ends at the East Terminal.  (1 mark)

On one particular train, 10 children set out from the West Terminal.

No new passengers board the train on the journey to the East Terminal.

  1. Determine the maximum number of children who can arrive at the East Terminal on this train.  (1 mark)
Show Answers Only
  1. `43`
  2. `22`
  3. `7`
Show Worked Solution

a.   `text(The capacity of Cut 1)`

♦♦ Mean mark for all parts (combined) was 33%.
MARKER’S COMMENT: A common error was counting the edge with “10” in the reverse direction (it should be ignored).

`=14 + 8 + 13 + 8`

`= 43`

 

b.    networks-fur2-2007-vcaa-3-answer
`text(Maximum seats)` `=\ text(minimum cut)`
  `= 6 + 7 + 9`
  `= 22`

 

c.  `text{The path (edge weights) of the train setting out with}`

`text(10 children starts with: 11 → 13.)`

`text(At the next station, a maximum of 7 seats are available)`

`text(which remain until the East Terminal.)`
  

`:.\ text(Maximum number of children arriving is 7.)`

NETWORKS, FUR2 2007 VCAA 2

The estate has large open parklands that contain seven large trees.

The trees are denoted as vertices `A` to `G` on the network diagram below.

Walking paths link the trees as shown.

The numbers on the edges represent the lengths of the paths in metres.
 

NETWORKS, FUR2 2007 VCAA 2

  1. Determine the sum of the degrees of the vertices of this network.  (1 mark)
  2. One day Jamie decides to go for a walk that will take him along each of the paths between the trees.

     

    He wishes to walk the minimum possible distance.

     

    1. State a vertex at which Jamie could begin his walk.  (1 mark)
    2. Determine the total distance, in metres, that Jamie will walk.  (1 mark)

Michelle is currently at `F`.

She wishes to follow a route that can be described as the shortest Hamiltonian circuit.

  1. Write down a route that Michelle can take.  (1 mark) 
Show Answers Only
  1. `24`
    1. `C\ text(or)\ G`
    2. `2800\ text(m)`

  3. `F − G − A − B − C − D − E − F,\ text(or)`
    `F − E − D − C − B − A − G − F`

Show Worked Solution

a.   `text(Sum of degrees of vertices)`

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

`= 4 + 2 + 5 + 2 + 4 + 4 + 3`

`= 24`

 

b.i.   `C\ text(or)\ G`

`text(An Euler path is required and)`

`text(therefore the starting point is at)`

`text(a vertex with an odd degree.)`

 

b.ii.   `2800\ text(m)`

MARKER’S COMMENT: Many students incorrectly found the shortest Hamiltonian path.

 

c.    `F − G − A − B − C − D − E − F,\ text(or)`

`F − E − D − C − B − A − G − F`

NETWORKS, FUR1 2014 VCAA 8 MC

Which one of the following statements about critical paths is true?

  1. There can be only one critical path in a project.
  2. A critical path always includes at least two activities.
  3. A critical path will always include the activity that takes the longest time to complete.
  4. Reducing the time of any activity on a critical path for a project will always reduce the minimum completion time for the project.
  5. If there are no other changes, increasing the time of any activity on a critical path will always increase the completion time of a project.
Show Answers Only

`E`

Show Worked Solution

`=>  E` 

♦ Mean mark 40%.

 

MATRICES*, FUR1 2015 VCAA 8 MC

There are five teams in a table tennis competition.

Every team played one match against every other team, and each match had a winner and a loser.

The results of the matches are summarised in the directed graph below. For example, an arrow from Lions to Eagles indicates that Lions defeated Eagles.
 

NETWORKS, FUR1 2015 VCAA 8 MC
 

In determining the ranking of these teams, the total of each team’s one-step dominances and two-step dominances will be calculated.

The team with the highest total will be ranked first.

The team with the next highest total will be ranked second, and so on.

The ranking of these five teams from first to last is

A.   Lions, Rebels, Dingoes, Eagles, Heavies

B.   Lions, Rebels, Eagles, Dingoes, Heavies

C.   Rebels, Lions, Dingoes, Eagles, Heavies

D.   Rebels, Lions, Eagles, Dingoes, Heavies

E.   Eagles, Lions, Rebels, Dingoes, Heavies

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ D_1 =\ text(1-step dominance matrix)`

  `{:(quad qquad qquad qquad L quad E quad\ R quad\ D quad\ H), (D_1 = [(0,0,0,1,0),(1,0,1,0,0),(1,0,0,0,0),(0,1,1,0,0),(1,1,1,1,0)]{:(L),(E),(R),(D),(H):}\ \ \ \ \ \ text(loses)):}`

 

`text(Let)\ \ D_2 =\ text(2-step dominance matrix)`

`D_2 = [(0,1,1,0,0),(1,0,0,1,0),(0,0,0,1,0),(2,0,1,0,0),(2,1,2,1,0)]`

 

`{:(\ \ \ \ quad qquad qquad qquad qquad L quad E quad\ R quad\ D quad\ H), (D_1+D_2 = [(0,1,1,1,0),(2,0,1,1,0),(1,0,0,1,0),(2,1,2,0,0),(3,2,3,2,0)]), ({:qquad qquad qquad qquad qquad\ \ \ \ 8 quad\ \ 4 quad\ 7 quad\ 5 quad\ 0:}):}`

 

`:.\ text(Summing up each column, the ranking from)`

`text(highest to lowest is)\ LRDEH.`

 
`=> A`

NETWORKS, FUR2 2013 VCAA 3

The rangers at the wildlife park restrict access to the walking tracks through areas where the animals breed.

The edges on the directed network diagram below represent one-way tracks through the breeding areas. The direction of travel on each track is shown by an arrow. The numbers on the edges indicate the maximum number of people who are permitted to walk along each track each day.
 

NETWORKS, FUR2 2013 VCAA 31
 

  1. Starting at `A`, how many people, in total, are permitted to walk to `D` each day?  (1 mark)

One day, all the available walking tracks will be used by students on a school excursion.

The students will start at `A` and walk in four separate groups to `D`.

Students must remain in the same groups throughout the walk.

    1. Group 1 will have 17 students. This is the maximum group size that can walk together from `A` to `D`.

       

      Write down the path that group 1 will take.  (1 mark)

    2. Groups 2, 3 and 4 will each take different paths from `A` to `D`.

       

      Complete the six missing entries shaded in the table below.  (2 marks)

       


      NETWORKS, FUR2 2013 VCAA 32

Show Answers Only
  1. `37`
    1. `A − B − E − C − D`
    2.  `text{One possible solution is:}`
       
      Networks, FUR2 2013 VCAA 3_2 Answer1
Show Worked Solution
a.    `text(Maximum flow)` `=\ text(minimum cut through)\ CD and ED`
    `= 24 + 13`
    `= 37`
♦ Mean mark of all parts (combined) was 41%.

 

`:.\ text(A maximum of 37 people can walk)`

`text(to)\ D\ text(from)\ A.`

 

b.i.   `A − B − E − C − D`

 

b.ii.   `text(One solution using the second possible largest)`

  `text(group of 11 students and two groups from the)`

  `text(remaining 9 students is:)`

Networks, FUR2 2013 VCAA 3_2 Answer1

NETWORKS, FUR2 2013 VCAA 2

A project will be undertaken in the wildlife park. This project involves the 13 activities shown in the table below. The duration, in hours, and predecessor(s) of each activity are also included in the table.
 
NETWORKS, FUR2 2013 VCAA 21
 

Activity `G` is missing from the network diagram for this project, which is shown below.

 
NETWORKS, FUR2 2013 VCAA 22
 

  1. Complete the network diagram above by inserting activity `G`.  (1 mark)
  2. Determine the earliest starting time of activity `H`.  (1 mark)
  3. Given that activity `G` is not on the critical path
    1. write down the activities that are on the critical path in the order that they are completed  (1 mark)
    2. find the latest starting time for activity `D`.  (1 mark)
  4. Consider the following statement.

    ‘If just one of the activities in this project is crashed by one hour, then the minimum time to complete the entire project will be reduced by one hour.’

    Explain the circumstances under which this statement will be true for this project.  (1 mark)

  5. Assume activity `F` is crashed by two hours.

    What will be the minimum completion time for the project?  (1 mark) 

Show Answers Only
  1.  
    networks-fur2-2013-vcaa-2-answer
  2. `7\ text(hours)`
    1. `A − F − I − M`
    2. `14\ text(hours)`
  4. `text(The statement will only be true if the crashed activity)`
    `text(is on the critical path)\ \ A − F − I − M.`
  5. `text(36 hours)`
Show Worked Solution
a.    networks-fur2-2013-vcaa-2-answer

 

b.    `text(EST of)\ H` `= 4 + 3`
    `= 7\ text(hours)`

 

c.i.   `A − F − I − M`

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

 

c.ii.  networks-fur2-2013-vcaa-23-answer

`G\ text(precedes)\ I`

`:. text(LST of)\ G = 20 – 4 = 16\ text(hours)`

`:. text(LST of)\ D = 16 – 2 = 14\ text(hours)`

 

d.   `text(The statement will only be true if the crashed activity)`

MARKER’S COMMENT: Most students struggled with part (d).

`text(is on the critical path)\ \ A − F − I − M.`

 

e.   `A − F − I − M\ text(is 37 hours.)`

`text(If)\ F\ text(is crashed by 2 hours, the new)`

`text(new critical path is)`

`C − E − H − G − I − M\ text{(36 hours)}`

`:.\ text(Minimum completion time = 36 hours)`

MATRICES*, FUR2 2008 VCAA 4

The children are taken to the zoo where they observe the behaviour of five young male lion cubs. The lion cubs are named Arnold, Barnaby, Cedric, Darcy and Edgar. A dominance hierarchy has emerged within this group of lion cubs. In the directed graph below, the directions of the arrows show which lions are dominant over others.
 

NETWORKS, FUR2 2008 VCAA 41
 

  1. Name the two pairs of lion cubs who have equal totals of one-step dominances.  (2 marks)
  2. Over which lion does Cedric have both a one-step dominance and a two-step dominance?  (1 mark)

In determining the final order of dominance, the number of one-step dominances and two-step dominances are added together.

  1. Complete the table below for the final order of dominance.  (1 mark)

     


    NETWORKS, FUR2 2008 VCAA 42

Over time, the pattern of dominance changes until each lion cub has a one-step dominance over two other lion cubs.

  1. Determine the total number of two-step dominances for this group of five lion cubs.  (1 mark)
Show Answers Only
  1. `text{Arnold and Edgar (1 each)}`
    `text{Barnaby and Cedric (2 each)}`
  2. `text(Edgar)`
  3.  
    networks-fur2-2008-vcaa-4-answer1
  4. `20`
Show Worked Solution

a.   `text(One step dominances:)`

`text{Arnold and Edgar (1 each)}`

`text{Barnaby and Cedric (2 each)}`

 

b.   `text(Edgar)`

 

c.    networks-fur2-2008-vcaa-4-answer

networks-fur2-2008-vcaa-4-answer1

 

d.   `text(Each lion has a one step dominance over 2 others.)`

`=>\ text(Each lion must have a two step)`

`text(dominance over 2 × 2 = 4 lions)`

 

`:.\ text(Total 2 step dominances in group)`

`= 5 xx 4`

`= 20`

NETWORKS, FUR2 2008 VCAA 2

Four children, James, Dante, Tahlia and Chanel each live in a different town. 

The following is a map of the roads that link the four towns, `A`, `B`, `C` and `D`.
 

NETWORKS, FUR2 2008 VCAA 21
 

  1. How many different ways may a vehicle travel from town `A` to town `D` without travelling along any road more than once?  (1 mark)

James’ father has begun to draw a network diagram that represents all the routes between the four towns on the map. This is shown below.


NETWORKS, FUR2 2008 VCAA 22
 

In this network, vertices represent towns and edges represent routes between towns that do not pass through any other town.

    1. One more edge needs to be added to complete this network. Draw in this edge clearly on the diagram above.  (1 mark)
    2. With reference to the network diagram, explain why a motorist at `A` could not drive each of these routes once only and arrive back at `A`.  (1 mark)
Show Answers Only
  1. `7`
    1.  
      networks-fur2-2008-vcaa-2-answer
    2. `text(Driving each route once and arriving back at)\ A`

       

      `text(requires an Eulerian circuit where all vertices must)`

       

      `text(be an even degree.)`

       

      `text(The vertices at)\ C\ text(and)\ B\ text(are odd.)`

       

      `:.\ text(No Eulerian circuit exists.)`

Show Worked Solution

a.   `text(Let the two unnamed intersections be)\ T_1\ text{(top) and}\ T_2.`

`text(The possible paths are:)`

♦♦ Exact data unavailable but “few students” answered this question correctly.

`ACD, ACT_2BD, ACT_2T_1BD, AT_1T_2CD,`

`AT_1T_2BD, AT_1BD, AT_1,BT_2CD.`

`:. 7\ text(different ways from)\ A\ text(to)\ D.`

 

b.i.    networks-fur2-2008-vcaa-2-answer

 

b.ii.   `text(Driving each route once and arriving back at)`

MARKER’S COMMENT: Be specific! Note that “an Eulerian circuit requires all vertices of an even degree” did not gain a mark here.

   `A\ text(requires an Eulerian circuit where all)`

   `text(vertices must be an even degree.)`

   `text(The vertices at)\ C\ text(and)\ B\ text(are odd.)`

   `:.\ text(No Eulerian circuit exists.)`

NETWORKS, FUR2 2009 VCAA 4

A walkway is to be built across the lake.

Eleven activities must be completed for this building project.

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

NETWORKS, FUR2 2009 VCAA 4
 

  1. What is the earliest start time for activity `E`?  (1 mark)
  2. Write down the critical path for this project.  (1 mark)
  3. The project supervisor correctly writes down the float time for each activity that can be delayed and makes a list of these times.

     

    Determine the longest float time, in weeks, on the supervisor’s list.  (1 mark)

A twelfth activity, `L`, with duration three weeks, is to be added without altering the critical path.

Activity `L` has an earliest start time of four weeks and a latest start time of five weeks.

 

NETWORKS, FUR2 2009 VCAA 4

  1. Draw in activity `L` on the network diagram above.  (1 mark)
  2. Activity `L` starts, but then takes four weeks longer than originally planned.

     

    Determine the total overall time, in weeks, for the completion of this building project.  (1 mark)

Show Answers Only
  1. `7`
  2. `BDFGIK`
  3. `H\ text(or)\ J\ text(can be delayed for)`
    `text(a maximum of 3 weeks.)`
  4.  
    NETWORKS, FUR2 2009 VCAA 4 Answer
  5. `text(25 weeks)`
Show Worked Solution

a.   `7\ text(weeks)`

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

 

b.   `BDFGIK`

 

c.   `H\ text(or)\ J\ text(can be delayed for a maximum)`

`text(of 3 weeks.)`

 

d.    NETWORKS, FUR2 2009 VCAA 4 Answer

 

e.   `text(The new critical path is)\ BLEGIK.`

`L\ text(now takes 7 weeks.)`

`:.\ text(Time for completion)`

`= 4 + 7 + 1 + 5 + 2 + 6`

`= 25\ text(weeks)`

NETWORKS, FUR2 2010 VCAA 4

In the final challenge, each of four teams has to complete a construction project that involves activities `A` to `I`.
 

NETWORKS, FUR2 2010 VCAA 4
 

Table 1 shows the earliest start time (EST), latest start time (LST) and duration, in minutes, for each activity. 

The immediate predecessor is also shown. The earliest start time for activity `F` is missing.

  1. What is the least number of activities that must be completed before activity `F` can commence?  (1 mark)
  2. What is the earliest start time for activity `F`?  (1 mark)
  3. Write down all the activities that must be completed before activity `G` can commence.  (1 mark)
  4. What is the float time, in minutes, for activity `G`?  (1 mark)
  5. What is the shortest time, in minutes, in which this construction project can be completed?  (1 mark)
  6. Write down the critical path for this network.  (1 mark)
Show Answers Only
  1. `2`
  2. `9\ text(minutes)`
  3. `A\ text(and)\ C`
  4. `4\ text(minutes)`
  5. `16\ text(minutes)`
  6. `A-B-D-H`
Show Worked Solution

a.   `2`

 

b.    `text(EST for)\ F` `= 5 + 4`
    `= 9\ text(minutes)`

 

c.   `A\ text(and)\ C`

 

d.    `text(Float time for)\ G` `= 13 – 9`
    `= 4\ text(minutes)`

 

e.   `text(Shortest construction time)`

`= 5 + 6 + 2 + 3`

`= 16\ text(minutes)`

 

f.   `A-B-D-H`

NETWORKS, FUR2 2010 VCAA 3

The following network diagram shows the distances, in kilometres, along the roads that connect six intersections `A`, `B`, `C`, `D`, `E` and `F`.
 

  1. If a cyclist started at intersection `B` and cycled along every road in this network once only, at which intersection would she finish?  (1 mark)
  2. The next challenge involves cycling along every road in this network at least once.

     

    Teams have to start and finish at intersection `A`.

     

    The blue team does this and cycles the shortest possible total distance.

     

    1. Apart from intersection `A`, through which intersections does the blue team pass more than once?  (1 mark)
    2. How many kilometres does the blue team cycle?  (1 mark)
  3. The red team does not follow the rules and cycles along a bush path that connects two of the intersections.

     

    This route allows the red team to ride along every road only once.

     

    Which two intersections does the bush path connect?  (1 mark)

Show Answers Only
  1. `D`
    1. `B, D\ text(and)\ F`
    2. `32\ text(km)`
  3. `B\ text(and)\ D`
Show Worked Solution

a.   `D`

 

b.i.   `text(Consider the path)`

`AFDEFBCDFBA`

`:.\ text(Passes through)\ B, D,\ text(and)\ F\ text(more)`

`text(than once.)`

 

b.ii.   `text{Total distance (Blue)}`

`= 6 + 3 + 3 + 4 + 2 + 4 + 2 + 3 + 2 + 3`

`= 32\ text(km)`

 

c.   `text(An Euler circuit can only exist when the)`

`text(degree of all vertices is even.)`

`:.\ text(The bush track joins)\ B\ text(and)\ D.`

NETWORKS, FUR2 2011 VCAA 4

Stormwater enters a network of pipes at either Dunlop North (Source 1) or Dunlop South (Source 2) and flows into the ocean at either Outlet 1 or Outlet 2.

On the network diagram below, the pipes are represented by straight lines with arrows that indicate the direction of the flow of water. Water cannot flow through a pipe in the opposite direction.

The numbers next to the arrows represent the maximum rate, in kilolitres per minute, at which stormwater can flow through each pipe.

 

NETWORKS, FUR2 2011 VCAA 4_1
 

  1. Complete the following sentence for this network of pipes by writing either the number 1 or 2 in each box.  (1 mark)

NETWORKS, FUR2 2011 VCAA 4_2

  1. Determine the maximum rate, in kilolitres per minute, that water can flow from these pipes into the ocean at Outlet 1 and Outlet 2.  (2 marks)

 

A length of pipe, show in bold on the network diagram below, has been damaged and will be replaced with a larger pipe.

 

NETWORKS, FUR2 2011 VCAA 4_3
 

  1. The new pipe must enable the greatest possible rate of flow of stormwater into the ocean from Outlet 2.

     

    What minimum rate of flow through the pipe, in kilolitres per minute, will achieve this?  (1 mark)

Show Answers Only
  1. `text(Storm water from Source 2 cannot reach Outlet 1)`
  2. `text(Outlet 1: 700 kL/min)`
    `text(Outlet 2: 700 kL/min)`
  3. `text(300 kL per min)`
Show Worked Solution

a.   `text(Storm water from Source 2 cannot reach Outlet 1)`

♦ Mean mark of all parts (combined) was 35%.

 

b.    NETWORKS, FUR2 2011 VCAA 4 Answer

 
`text(The minimum cut includes the 200 kL/min pipe from Source 1.)`

`:.\ text(Maximum rates are)`

`text(Outlet 1: 700 kL/min)`

`text(Outlet 2: 700 kL/min)`

 

c.   `text(The next smallest cut in the lower pipe system is 800.)`

`:.\ text(The minimum flow through the new pipe that will achieve)`

`text(this is 300 kL/min.)`

NETWORKS, FUR2 2011 VCAA 3

A section of the Farnham showgrounds has flooded due to a broken water pipe. The public will be stopped from entering the flooded area until repairs are made and the area has been cleaned up.

The table below shows the nine activities that need to be completed in order to repair the water pipe. Also shown are some of the durations, Earliest Start Times (EST) and the immediate predecessors for the activities.
 

NETWORKS, FUR2 2011 VCAA 3 

  1. What is the duration of activity `B`?  (1 mark)
  2. What is the Earliest Start Time (EST) for activity `D`?  (1 mark)
  3. Once the water has been turned off (Activity `B`), which of the activities `C` to `I` could be delayed without affecting the shortest time to complete all activities?  (1 mark)

It is more complicated to replace the broken water pipe (Activity `E`) than expected. It will now take four hours to complete instead of two hours.

  1. Determine the shortest time in which activities `A` to `I` can now be completed.  (1 mark)

Turning on the water to the showgrounds (Activity `H`) will also take more time than originally expected. It will now take five hours to complete instead of one hour.

  1. With the increased duration of Activity `H` and Activity `E`, determine the shortest time in which activities `A` to `I` can be completed.  (1 mark)
Show Answers Only
  1. `text(2 hours)`
  2. `text(3 hours)`
  3. `text(Activities)\ F and H`
  4. `13\ text(hours)`
  5. `14\ text(hours)`
Show Worked Solution
a.    `text(Duration of)\ B` `= text(EST of)\ C`
    `= 2\ text(hours)`
♦ Mean mark of all parts (combined) was 42%.

 

b.   `text(EST of)\ C = 3\ text(hours)`

 

c.   `text(Activities)\ F and H`

MARKER’S COMMENT: Many students incorrectly included `G` in this answer (note that `G` is not on the critical path).

 

d.   `text(Shortest time)\ (A\ text(to)\ I)`

`= 2 + 1 + 1 + 4 + 4 + 1`

`= 13\ text(hours)`

 

e.   `text(New shortest time)`

`= 2 + 1 + 1 + 4 + 5 + 1`

`= 14\ text(hours)`

NETWORKS, FUR2 2014 VCAA 4

To restore a vintage train, 13 activities need to be completed.

The network below shows these 13 activities and their completion times in hours.
 

NETWORKS, FUR2 2014 VCAA 4
 

  1. Determine the earliest starting time of activity `F`.  (1 mark)

The minimum time in which all 13 activities can be completed is 21 hours.

  1. What is the latest starting time of activity `L`?  (1 mark)
  2. What is the float time of activity `J`?  (1 mark)

Just before they started restoring the train, the members of the club needed to add another activity, `X`, to the project.

Activity `X` will take seven hours to complete.

Activity `X` has no predecessors, but must be completed before activity `G` starts.

  1. What is the latest starting time of activity `X` if it is not to increase the minimum completion time of the project?  (1 mark)

Activity `A` can be crashed by up to four hours at an additional cost of $90 per our.

This may reduce the minimum completion time for the project, including activity `X`.

  1. Determine the least cost of crashing activity `A` to give the greatest reduction in the minimum completion time of the project.  (1 mark)
Show Answers Only
  1. `text(7 hours)`
  2. `text(18 hours)`
  3. `text(2 hours)`
  4. `text(4 hours)`
  5. `$270`
Show Worked Solution

a.   `5 + 2 = 7\ text(hours)`

♦ Mean mark for all parts (combined) was 42%.

 

b.   `text(Latest starting time of)\ L`

`= text(Length of critical path – duration of)\ L`

`= 21 – 3`

`= 18\ text(hours)`

 

c.   `text(Float time of)\ J`

`=\ text(LST – EST)`

`= 13 – 11`

`= 2\ text(hours)`

 

d.   `X\ text(precedes)\ G`

`text(EST of)\ G = 11`

`:. text(LST of)\ X = 11`

 

`text(EST)\ text(of)\ X`

`= text(LST of)\ X – text(duration of)\ X`

`= 11 – 7`

`= 4\ text(hours)`

 

e.   `text(Longer paths are)`

`A – C – G – K = 21\ text{hours (critical path)}`

`A – D – E – H – K = 20\ text(hours)`

`A – D – F – J – M = 19\ text(hours)`

`A – D – E – I – M = 18\ text(hours)`

`B – E – H – K = 18\ text(hours)`

`B – F – J – M = 17\ text(hours)`

 

`:.\ text(Reduce)\ \ A – C – G – K\ \ text(by 3 hours to get)`

`text{to 18 hours (equals}\ \ B – E – H – K)`
 

`:.\ text(Least cost)` `= 3 xx 90`
  `= $270`

NETWORKS, FUR2 2015 VCAA 3

Nine activities are needed to prepare a daily delivery of groceries from the factory to the towns.

The duration, in minutes, earliest starting time (EST) and immediate predecessors for these activities are shown in the table below.
 

   Networks, FUR2 2015 VCAA 31
 

The directed network that shows these activities is shown below.
 

 Networks, FUR2 2015 VCAA 32
 

All nine of these activities can be completed in a minimum time of 26 minutes.

  1. What is the EST of activity `D`?  (1 mark)
  2. What is the latest starting time (LST) of activity `D`?  (1 mark)
  3. Given that the EST of activity `I` is 22 minutes, what is the duration of activity `H`?  (1 mark)
  4. Write down, in order, the activities on the critical path.  (1 mark)
  5. Activities `C` and `D` can only be completed by either Cassie or Donna.

     

    One Monday, Donna is sick and both activities `C` and `D` must be completed by Cassie. Cassie must complete one of these activities before starting the other.

     

    What is the least effect of this on the usual minimum preparation time for the delivery of groceries from the factory to the five towns?  (1 mark)

  6. Every Friday, a special delivery to the five towns includes fresh seafood. This causes a slight change to activity `G`, which then cannot start until activity `F` has been completed.

    1. On the directed graph below, show this change without duplicating any activity.  (1 mark)
       

      Networks, FUR2 2015 VCAA 32
       

    2. What effect does the inclusion of seafood on Fridays have on the usual minimum preparation time for deliveries from the factory to the five towns?  (1 mark)

Show Answers Only
  1. `text(3 minutes)`
  2. `text(4 minutes)`
  3. `text(3 minutes)`
  4. `A – C – F – H – I`
  5. `text(The critical path is increased by)`
    `text(7 minutes to 33 minutes.)`
    1.  
      Networks, FUR2 2015 VCAA 3 Answer
    2. `text(The critical path is increased by)`
      `text(2 minutes to 28 minutes.)`

Show Worked Solution

a.   `text(3 minutes)`
 

b.   `text(4 minutes)`
 

c.   `text(3 minutes)`
 

d.   `A – C – F – H – I`
 

e.   `text(The critical path is increased by 7 minutes)`

`text(to 33 minutes.)`

 

f.i.    Networks, FUR2 2015 VCAA 3 Answer

 

f.ii.   `text(The critical path is increased by 2 minutes)`

`text(to 28 minutes.)`

MATRICES, FUR2 2006 VCAA 3

Market researchers claim that the ideal number of bookshops (`x`), sports shoe shops (`y`) and music stores (`z`) for a shopping centre can be determined by solving the equations

`2x + y + z = 12`

`x-y+z=1`

`2y-z=6`

  1. Write the equations in matrix form using the following template.  (1 mark)

     

     
    `qquad[(qquadqquadqquad),(),()][(quad),(quad),(quad)] = [(quad),(quad),(quad)]`
     

     

  2. Do the equations have a unique solution? Provide an explanation to justify your response.  (1 mark)
  3. Write down an inverse matrix that can be used to solve these equations.  (1 mark)
  4. Solve the equations and hence write down the estimated ideal number of bookshops, sports shoe shops and music stores for a shopping centre.  (1 mark)
Show Answers Only
  1.  
    `[(2,1,1),(1,-1,1),(0,2,-1)][(x),(y),(z)] = [(12),(1),(6)]`
  2.  `text(Yes. See worked solutions.)`
  3.  
    `[(2,1,1),(1,-1,1),(0,2,-1)]^(-1) = [(-1,3,2),(1,-2,-1),(2,-4,-3)]`
  4. `text(3 bookshops, 4 sports shoe shops, 2 music stores.)`
Show Worked Solution
a.    `[(2,1,1),(1,-1,1),(0,2,-1)][(x),(y),(z)] = [(12),(1),(6)]`
♦ Mean mark 35% for all parts (combined).

 

b.    `text(det)\ [(2,1,1),(1,-1,1),(0,2,-1)] = 1 != 0`

 
`:.\ text(A unique solution exists.)`

 

c.   `text(By CAS,)`

`[(2,1,1),(1,-1,1),(0,2,-1)]^(-1) = [(-1,3,2),(1,-2,-1),(2,-4,-3)]`

 

d.    `[(x),(y),(z)]` `= [(-1,3,2),(1,-2,-1),(2,-4,-3)][(12),(1),(6)]`
    `= [(3),(4),(2)]`

 

`:.\ text(Estimated ideal numbers are:)`

`text(3 bookshops)`

`text(4 shoe shops)`

`text(2 music stores)`

MATRICES, FUR2 2006 VCAA 2

A new shopping centre called Shopper Heaven (`S`) is about to open. It will compete for customers with Eastown (`E`) and Noxland (`N`).

Market research suggests that each shopping centre will have a regular customer base but attract and lose customers on a weekly basis as follows.

80% of Shopper Heaven customers will return to Shopper Heaven next week
12% of Shopper Heaven customers will shop at Eastown next week
8% of Shopper Heaven customers will shop at Noxland next week

76% of Eastown customers will return to Eastown next week
9% of Eastown customers will shop at Shopper Heaven next week
15% of Eastown customers will shop at Noxland next week

85% of Noxland customers will return to Noxland next week
10% of Noxland customers will shop at Shopper Heaven next week
5% of Noxland customers will shop at Eastown next week

  1. Enter this information into transition matrix `T` as indicated below (express percentages as proportions, for example write 76% as 0.76).  (2 marks)

     

     
    `qquad{:(qquadqquadqquadtext(this week)),((qquadqquadqquadS,E,N)),(T = [(qquadqquadqquadqquad),(),()]{:(S),(E),(N):}{:qquadtext(next week):}):}`
     

During the week that Shopper Heaven opened, it had 300 000 customers.

In the same week, Eastown had 120 000 customers and Noxland had 180 000 customers.

  1. Write this information in the form of a column matrix, `K_0`, as indicated below.  (1 mark)

     

     
    `qquadK_0 = [(qquadqquadqquad),(),()]{:(S),(E),(N):}`
     

  2. Use `T` and `K_0` to write and evaluate a matrix product that determines the number of customers expected at each of the shopping centres during the following week.  (2 marks)
  3. Show by calculating at least two appropriate state matrices that, in the long term, the number of customers expected at each centre each week is given by the matrix    (2 marks)

     

     
    `qquadK = [(194\ 983),(150\ 513),(254\ 504)]`

Show Answers Only
  1.  
    `{:((qquadqquadqquad\ S,qquadE,qquadN)),(T = [(0.8,0.09,0.10),(0.12,0.76,0.05),(0.08,0.15,0.85)]{:(S),(E),(N):}):}`
  2.  
    `K_0 = [(300\ 000),(120\ 000),(180\ 000)]{:(S),(E),(N):}`
  3.  
    `TK_0 = [(268\ 800),(136\ 200),(195\ 000)]`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.     `{:((qquadqquadqquad\ S,qquadE,qquadN)),(T = [(0.8,0.09,0.10),(0.12,0.76,0.05),(0.08,0.15,0.85)]{:(S),(E),(N):}):}`

 

b.     `K_0 = [(300\ 000),(120\ 000),(180\ 000)]{:(S),(E),(N):}`

 

c.   `text(Customers expected at each centre the next week,)`

`TK_0` `= [(0.80,0.09,0.10),(0.12,0.76,0.05),(0.08,0.15,0.85)][(300\ 000),(120\ 000),(180\ 000)]`
  `= [(268\ 800),(136\ 200),(195\ 000)]`

 

d.   `text(Consider)\ \ T^nK_0\ \ text(when)\ n\ text(large),`

`text(say)\ n=50, 51`

`T^50K_0` `= [(0.8,0.09,0.10),(0.12,0.76,0.05),(0.08,0.15,0.85)]^50[(300\ 000),(120\ 000),(180\ 000)]`
  `= [(194\ 983),(150\ 513),(254\ 504)]`

 

`T^51K_0` `= [(0.8,0.09,0.10),(0.12,0.76,0.05),(0.08,0.15,0.85)]^51[(300\ 000),(120\ 000),(180\ 000)]`
  `= [(194\ 983),(150\ 513),(254\ 504)]`
  ` = T^50K_0`

MATRICES, FUR2 2007 VCAA 2

To study the life-and-death cycle of an insect population, a number of insect eggs (`E`), juvenile insects (`J`) and adult insects (`A`) are placed in a closed environment.

The initial state of this population can be described by the column matrix
 

`S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
 

A row has been included in the state matrix to allow for insects and eggs that die (`D`).

  1. What is the total number of insects in the population (including eggs) at the beginning of the study?  (1 mark)

In this population

    • eggs may die, or they may live and grow into juveniles
    • juveniles may die, or they may live and grow into adults
    • adults will live a period of time but they will eventually die.

In this population, the adult insects have been sterilised so that no new eggs are produced. In these circumstances, the life-and-death cycle of the insects can be modelled by the transition matrix
 

`{:(qquadqquadqquadqquadquadtext(this week)),((qquadqquadqquadE,quad\ J,quadA,\ D)),(T = [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)]{:(E),(J),(A),(D):}):}`
 

  1. What proportion of eggs turn into juveniles each week?  (1 mark)
    1. Evaluate the matrix product  `S_1 = TS_0`  (1 mark)
    2. Write down the number of live juveniles in the population after one week.  (1 mark)
    3. Determine the number of live juveniles in the population after four weeks. Write your answer correct to the nearest whole number.  (1 mark)
    4. After a number of weeks there will be no live eggs (less than one) left in the population.
      When does this first occur?  (1 mark)
    5. Write down the exact steady-state matrix for this population.  (1 mark)
  3. If the study is repeated with unsterilised adult insects, eggs will be laid and potentially grow into adults.

     

    Assuming 30% of adults lay eggs each week, the population matrix after one week, `S_1`, is now given by
     
    `qquad S_1 = TS_0 + BS_0`
      
    where   `B = [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]`   and   `S_0 = [(400),(200),(100),(0)]{:(E),(J),(A),(D):}`
     

    1. Determine `S_1`  (1 mark)
       

       

      This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
       
      `qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
       

    2. Determine the number of live eggs in this insect population after two weeks.  (1 mark)
Show Answers Only
  1. `700`
  2. `50text(%)`
    1. `[(160),(280),(180),(80)]{:(E),(J),(A),(D):}`
    2. `280`
    3. `56`
    4. `text(7th week)`
    5. `[(0),(0),(0),(700)]`
    1. `[(190),(280),(180),(80)]`
    2. `130`
Show Worked Solution

a.   `400 + 200 + 100 + 0 = 700`

 

b.   `50text(%)`

 

c.i.    `S_1` ` = TS_0`
    `= [(0.4,0,0,0),(0.5,0.4,0,0),(0,0.5,0.8,0),(0.1,0.1,0.2,1)][(400),(200),(100),(0)]`
    `= [(160),(280),(180),(80)]{:(E),(J),(A),(D):}`

 

c.ii.   `280`

 

c.iii.    `S_4` ` = T^4S_0`
    `= [(10.24),(56.32),(312.96),(320.48)]{:(E),(J),(A),(D):}\ \ \ text{(by graphics calculator)}`

 

`:. 56\ text(juveniles still alive after 4 weeks.)`

 

c.iv.   `text(Each week, only 40% of eggs remain.)`
 

`text(Find)\ \ n\ \ text(such that)`
`400 xx 0.4^n` `< 1`
`0.4^n` `<1/400`
`n` `> 6.5`

 

`:.\ text(After 7 weeks, no live eggs remain.)`

 

c.v.   `text(Consider)\ \ n\ \ text{large (say}\ \ n = 100 text{)},`

`[(0.4, 0, 0, 0), (0.5, 0.4, 0, 0), (0, 0.5, 0.8, 0), (0.1, 0.1, 0.2, 1)]^100 [(400), (200), (100), (0)] ~~ [(0), (0), (0), (700)]`

 

d.i.   `S_1` `= TS_0 + BS_0`
    `= [(160),(280),(180),(80)] + [(0,0,0.3,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(400),(200),(100),(0)]`
    `= [(190),(280),(180),(80)]`

 

♦♦ Mean mark for part (d) was 30%.
d.ii.   `S_2` `= TS_1 + BS_1`
    `= [(130), (207), (284), (163)]`

 

`:.\ text(There are 130 live egss after 2 weeks.)`

MATRICES, FUR2 2009 VCAA 4

A series of extra rehearsals commenced in April. Each week participants could choose extra dancing rehearsals or extra singing rehearsals.

A matrix equation used to determine the number of students expected to attend these extra rehearsals is given by
 

`L_(n + 1) = [(0.85,0.25),(0.15,0.75)] xx L_n - [(5),(7)]`
 

where `L_n` is the column matrix that lists the number of students attending in week `n`.

The attendance matrix for the first week of extra rehearsals is given by
 

`L_1 = [(95),(97)]{:(text(dancing)),(text(singing)):}`
 

  1. Calculate the number of students who are expected to attend the extra singing rehearsals in week 3.  (1 mark)
  2. Of the students who attended extra rehearsals in week 3, how many are not expected to return for any extra rehearsals in week 4?  (1 mark)
Show Answers Only
  1. `text(68 students)`
  2. `12`
Show Worked Solution

a.   `text(Using matrix equation,)`

`L_2` `= T xx L_1 – [(5),(7)]`
  `= [(0.85,0.25),(0.15,0.75)][(95),(97)] – [(5),(7)]`
  `= [(100),(80)]`
`L_3` `= [(0.85,0.25),(0.15,0.75)][(100),(80)] – [(5),(7)]`
  `= [(100),(68)]`

 

`:.\ text(68 students are expected to attend singing)`

`text(in week 3.)`

 

b.    `L_4` `= [(0.85,0.25),(0.15,0.75)][(100),(68)] – [(5),(7)]`
    `= [(97),(59)]`

 

`:.\ text(Students expected not to return)`

`= (100 + 68) – (97 + 59)`

`= 12`

MATRICES, FUR2 2009 VCAA 3

In 2009, the school entered a Rock Eisteddfod competition.

When rehearsals commenced in February, all students were asked whether they thought the school would make the state finals. The students’ responses, ‘yes’, ‘no’ or ‘undecided’ are shown in the initial state matrix `S_0`.
 

`S_0 = [(160),(120),(220)]{:(text(yes)),(text(no)),(text(undecided)):}`
 

  1. How many students attend this school?  (1 mark)

Each week some students are expected to change their responses. The changes in their responses from one week to the next are modelled by the transition matrix `T` shown below.
 

`{:(qquadqquadqquadtext( response this week)),(qquadqquadquadtext( yes       no     undecided)),(T = [(0.85quad,0.35quad,0.60),(0.10quad,0.40quad,0.30),(0.05quad,0.25quad,0.10)]{:(text(yes)),(text(no)),(text(undecided)):}qquad{:(text(response)),(text(next week)):}):}`
 

The following diagram can also be used to display the information represented in the transition matrix `T`.

MATRICES, FUR2 2009 VCAA 3

    1. Complete the diagram above by writing the missing percentage in the shaded box.  (1 mark)
    2. Of the students who respond ‘yes’ one week, what percentage are expected to respond ‘undecided’ the next week when asked whether they think the school will make the state finals?  (1 mark)
    3. In total, how many students are not expected to have changed their response at the end of the first week?  (2 marks)
  2. Evaluate the product  `S_1 = TS_0`, where `S_1` is the state matrix at the end of the first week.  (1 mark)
  3. How many students are expected to respond ‘yes’ at the end of the third week when asked whether they think the school will make the state finals?  (1 mark)
Show Answers Only
  1. `500`
    1. `text(25%)`
    2. `text(5%)`
    3. `206`
  3. `S_1 = [(310),(130),(60)]`
  4. `361`
Show Worked Solution

a.   `text(Total students attending)`

`= 160 + 120 + 220`

`= 500`

 

b.i.   `text(25%)`

 

b.ii.   `text(5%)`

 

b.iii.   `text(Students not expected to change)`

`= 0.85 xx 160 + 0.4 xx 120 + 0.1 xx 220`

`= 206`

 

c.    `S_1` `=TS_0`
    `= [(0.85,0.35,0.60),(0.10,0.40,0.30),(0.05,0.25,0.10)][(160),(120),(220)]`
    `= [(310),(130),(60)]`

 

d.    `S_3` `= T^3 S_0` 
    `= [(0.85,0.35,0.60),(0.10,0.40,0.30),(0.05,0.25,0.10)]^3[(160),(120),(220)]`
    `= [(361),(91.1),(47.9)]` 

 

`:. 361\ text(students expected to respond “yes” at end of week 3.)`

MATRICES, FUR2 2010 VCAA 4

The Dinosaurs (`D`) and the Scorpions (`S`) are two basketball teams that play in different leagues in the same city.

The matrix `A_1` is the attendance matrix for the first game. This matrix shows the number of people who attended the first Dinosaur game and the number of people who attended the first Scorpion game.
 

`A_1 = [(2000),(1000)]{:(D),(S):}`
 

The number of people expected to attend the second game for each team can be determined using the matrix equation

`A_2 = GA_1`

where `G` is the matrix     `{:(qquadqquadqquadtext(this game)),((qquadqquadqquadD,qquad\ S)),(G = [(1.2,-0.3),(0.2,0.7)]{:(D),(S):}qquad{:text(next game):}):}`

 

    1. Determine `A_2`, the attendance matrix for the second game.  (1 mark)
    2. Every person who attends either the second Dinosaur game or the second Scorpion game will be given a free cap. How many caps, in total, are expected to be given away?  (1 mark)

Assume that the attendance matrices for successive games can be determined as follows.

`A_3 = GA_2`

`A_4 = GA_3`

and so on such that `A_(n + 1) = GA_n`

  1. Determine the attendance matrix (with the elements written correct to the nearest whole number) for game 10.  (1 mark)
  2. Describe the way in which the number of people attending the Dinosaurs’ games is expected to change over the next 80 or so games.  (1 mark)

The attendance at the first Dinosaur game was 2000 people and the attendance at the first Scorpion game was 1000 people.

Suppose, instead, that 2000 people attend the first Dinosaur game, and 1800 people attend the first Scorpion game.

  1. Describe the way in which the number of people attending the Dinosaurs’ games is expected to change over the next 80 or so games.  (1 mark)
Show Answers Only
    1.  
      `A_2 = [(2100),(1100)]`
    2. `3200\ text(people)`
  2.  
    `A_10 = [(2613),(1613)]`
  3. `text(Attendance at the Dinosaur’s games increases gradually)`

     

    `text(to 3000, at which level it remains steady.)`

  4. `text(Attendence at the Dinosaur’s games decreases)`

     

    `text(gradually to 600, where it remains steady.)`

Show Worked Solution
a.i.    `A_2` `= GA_1`
    `= [(1.2,-0.3),(0.2,0.7)][(2000),(1000)]`
    `= [(2100),(1100)]`

 

a.ii.   `text(Total attending second games)`

`= 2100 + 1100`

`= 3200\ text(people)`

 

b.    `A_10` `= GA_9`
    `= G^9A_1`
    `= [(1.2,-0.3),(0.2,0.7)]^9[(2000),(1000)]`
    `= [(2613),(1613)]`

 

c.    `A_80` `= G^79A_1`
    `= [(3000),(2000)]`
`A_81` `= G^80A_1`
  `= [(3000),(2000)]`

 

`:.\ text(Attendance at the Dinosaur’s games increases)`

`text(gradually to 3000, at which level it remains)`

`text(steady.)`

 

d.   `text(Using the new initial attendences,)`

`A_80 = [(1.2,-0.3),(0.2,0.7)]^79[(2000),(1800)] = [(600),(400)]`

`A_81 = [(1.2,-0.3),(0.2,0.7)]^80[(2000),(1800)] = [(600),(400)]`

 

`:.\ text(Attendence at the Dinosaur’s games decreases)`

`text(gradually to 600, where it remains steady.)`

NETWORKS, FUR2 2012 VCAA 3

Four tasks, `W`, `X`, `Y` and `Z`, must be completed. 

Four workers, Julia, Ken, Lana and Max, will each do one task. 

Table 1 shows the time, in minutes, that each person would take to complete each of the four tasks.
 

Networks, FUR2 2012 VCAA 3_1
 

The tasks will be allocated so that the total time of completing the four tasks is a minimum. 

The Hungarian method will be used to find the optimal allocation of tasks. Step 1 of the Hungarian method is to subtract the minimum entry in each row from each element in the row.
 

Networks, FUR2 2012 VCAA 3_2
 

  1. Complete step 1 for task `X` by writing down the number missing from the shaded cell in Table 2.  (1 mark)

The second step of the Hungarian method ensures that all columns have at least one zero.

The numbers that result from this step are shown in Table 3 below.
 

Networks, FUR2 2012 VCAA 3_3
 

  1. Following the Hungarian method, the smallest number of lines that can be drawn to cover the zeros is shown dashed in Table 3.

     

    These dashed lines indicate that an optimal allocation cannot be made yet.

     

    Give a reason why.  (1 mark)

  2. Complete the steps of the Hungarian method to produce a table tasks can be made.

     

    Two blank tables have been provided for working if needed.  (1 mark)

     

     
        Networks, FUR2 2012 VCAA 3_4

          Networks, FUR2 2012 VCAA 3_4
     

  3. Write the name of the task that each person should do for the optimal allocation of tasks.  (2 marks)
     

Networks, FUR2 2012 VCAA 3_5

Show Answers Only
  1. `17`
  2. `text(Allocating four tasks to four people requires)`

     

    `text(four lines and there are only three.)`

  3.  
    Networks, FUR2 2012 VCAA 3_1 Answer
  4.  
    Networks, FUR2 2012 VCAA 3_2 Answer
Show Worked Solution

a.   `17`

 

b.   `text(Allocating four tasks to four people)`

♦♦ Exact data unavailable although examiners highlighted part (b) as “poorly answered”.

`text(requires four lines and there are)`

`text(only three.)`

 

c.    Networks, FUR2 2012 VCAA 3_1 Answer

 

d.    Networks, FUR2 2012 VCAA 3_2 Answer

NETWORKS, FUR2 2012 VCAA 2

Thirteen activities must be completed before the produce grown on a farm can be harvested. 

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

 

NETWORKS, FUR2 2012 VCAA 2
  

  1. Determine the earliest starting time, in days, for activity `E`.  (1 mark)
  2. A dummy activity starts at the end of activity `B`.

     

    Explain why this dummy activity is used on the network diagram.  (1 mark)

  3. Determine the earliest starting time, in days, for activity `H`.  (1 mark)
  4. In order, list the activities on the critical path.  (1 mark)
  5. Determine the latest starting time, in days, for activity `J`.  (1 mark)
Show Answers Only
  1. `12\ text(days)`
  2. `F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`
    `text(have)\ B\ text(and)\ C\ text(as predecessors.)`
    `text(S)text(ince there cannot be 2 activities called)\ B,`
    `text{a dummy activity is drawn as an extension of}`
    `B\ text(to show that it is also a predecessor of)\ G\ text(and)`
    `H\ text{(with zero time).}`
  3. `15\ text(days)`
  4. `A-B-H-I-L-M`
  5. `25\ text(days)`
Show Worked Solution
a.    `text(EST of)\ E` `= 10 + 2`
    `= 12\ text(days)`
♦ Mean mark of all parts (combined) 47%.

 

b.   `F\ text(has)\ B\ text(as a predecessor while)\ G\ text(and)\ H`

`text(have)\ B\ text(and)\ C\ text(as predecessors.)`

`text(S)text(ince there cannot be 2 activities called)\ B,`

`text{a dummy activity is drawn as an extension of}`

`B\ text(to show that it is also a predecessor of)\ G\ text(and)`

`H\ text{(with zero time).}`

 

♦♦ Exact data unavailable but “few students” were able to correctly deal with the dummy activity in this question.
c.    `text(EST of)\ H` `= 10 + 5`
    `= 15\ text(days)`

 

d.   `text(The critical path is)`

`A-B-H-I-L-M`

 

e.   `text(The shortest time to complete all the activities)`

MARKER’S COMMENT: A correct calculation based on an incorrect critical path in part (d) gained a consequential mark here. Show your working!

`= 10 + 5 + 4 + 3  + 4 + 2`

`= 28\ text(days)`

 

`:.\ text(LST of)\ J` `= 28 − 3`
  `= 25\ text(days)`

NETWORKS, FUR2 2012 VCAA 1

Water will be pumped from a dam to eight locations on a farm.

The pump and the eight locations (including the house) are shown as vertices in the network diagram below.

The numbers on the edges joining the vertices give the shortest distances, in metres, between locations.
 

NETWORKS, FUR2 2012 VCAA 1
 

    1. Determine the shortest distance between the house and the pump.  (1 mark)
    2. How many vertices on the network diagram have an odd degree?  (1 mark)
    3. The total length of all edges in the network is 1180 metres.

       

      A journey starts and finishes at the house and travels along every edge in the network.

       

      Determine the shortest distance travelled.  (1 mark)

The total length of pipe that supplies water from the pump to the eight locations on the farm is a minimum.

This minimum length of pipe is laid along some of the edges in the network.

    1. On the diagram below, draw the minimum length of pipe that is needed to supply water to all locations on the farm.  (1 mark)

       

         
        NETWORKS, FUR2 2012 VCAA 1
       

    2. What is the mathematical term that is used to describe this minimum length of pipe in part i.?  (1 mark)
Show Answers Only
    1. `160\ text(m)`
    2. `2`
    3. `1250\ text(m)`
    1.  
      NETWORKS, FUR2 2012 VCAA 1 Answer
    2. `text(Minimal spanning tree)`
Show Worked Solution

a.i.   `text(Shortest distance)`

`=70 + 90`

`= 160\ text(m)`

MARKER’S COMMENT: Many students, surprisingly, had problems with part (a)(ii).

 

a.ii.   `2\ text{(the house and the top right vertex)}`

 

a.iii.   `text{An Eulerian path is possible if it starts at}`

♦♦ “Very poorly answered”.
MARKER’S COMMENT: An Euler circuit is optimal but not possible here because of the two odd degree vertices.

   `text{the house (odd vertex) and ends at the top}`

   `text{right vertex (the other odd vertex). However,}`

   `text{70 metres must be added to return to the}`

   `text{house.}`

`:.\ text(Total distance)` `= 1180 + 70`
  `= 1250\ text(m)`

 

b.i.    NETWORKS, FUR2 2012 VCAA 1 Answer

 

b.ii.   `text(Minimal spanning tree)`

MATRICES, FUR2 2011 VCAA 3

A breeding program is started in the wetlands. It is aimed at establishing a colony of native ducks.

The matrix `W_0` displays the number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) that are introduced to the wetlands at the start of the breeding program.

`W_0 = [(32),(64)]{:(J),(A):}`

 

  1. In total, how many female ducks are introduced to the wetlands at the start of the breeding program?  (1 mark)

The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year 1 of the breeding program is determined using the matrix equation

`W_1 = BW_0`

In this equation, `B` is the breeding matrix

`{:((qquadqquadqquad\ J,qquadA)),(B = [(0,2),(0.25,0.5)]{:(J),(A):}):}`

 

  1. Determine `W_1`  (1 mark)

The number of juvenile female ducks (`J`) and the number of adult female ducks (`A`) in the colony at the end of Year `n` of the breeding program is determined using the matrix equation

`W_n = BW_(n - 1)`

The graph below is incomplete because the points for the end of Year 3 of the breeding program are missing.

 

MATRICES, FUR2 2011 VCAA 3

    1. Use the matrices to calculate the number of juvenile and the number of adult female ducks expected in the colony at the end of Year 3 of the breeding program.

       

      Plot the corresponding points on the graph.  (2 marks)

    2. Use matrices to determine the expected total number of female ducks in the colony in the long term.

       

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

The breeding matrix `B` assumes that, on average, each adult female duck lays and hatches two female eggs for each year of the breeding program.

If each adult female duck lays and hatches only one female egg each year, it is expected that the duck colony in the wetland will not be self-sustaining and will, in the long run, die out.

The matrix equation

`W_n = PW_(n - 1)`

with a different breeding matrix

`{:((qquadqquadqquad\ J,qquadA)),(P = [(0,1),(0.25,0.5)]{:(J),(A):}):}`

 

and the initial state matrix

`W_0 = [(32),(64)]{:(J),(A):}`

models this situation.

  1. During which year of the breeding program will the number of female ducks in the colony halve?  (1 mark)

Changing the number of juvenile and adult female ducks at the start of the breeding program will also change the expected size of the colony.

  1. Assuming the same breeding matrix, `P`, determine the number of juvenile ducks and the number of adult ducks that should be introduced into the program at the beginning so that, at the end of Year 2, there are 100 juvenile female ducks and 50 adult female ducks.  (2 marks)

 

Show Answers Only
  1. `96`
  2. `W_1 = [(128),(40)]`
  3.  
    1.  
      MATRICES, FUR2 2011 VCAA 3 Answer
    2. `144`
  4. `text(year 5)`
  5. `400\ text(juvenile ducks and 0 adult)`
    `text(ducks should be introduced.)`
Show Worked Solution

a.   `text(Total female ducks introduced)`

`= 32 + 64`

`= 96`

 

b.    `W_1` `= BW_0`
    `= [(0,2),(0.25,0.5)][(32),(64)]`
    `= [(128),(40)]`

 

c.i.    MATRICES, FUR2 2011 VCAA 3 Answer

`W_n = BW_(n – 1)`

`W_2 = [(0,2),(0.25,0.5)][(128),(40)] = [(80),(52)]`

`W_3 = [(0,2),(0.25,0.5)][(80),(52)] = [(104),(46)]`

 

`:.\ text{Plot (3, 104) and (3, 46)}`

 

 

c.ii.   `text(Consider)\ n = text(50 and 51,)`

`W_50 = [(0,2),(0.25,0.5)]^49[(32),(64)] = [(96),(48)]`

`W_51 = [(0,2),(0.25,0.5)]^50[(32),(64)] = [(96),(48)]`

 

`:.\ text(Total female ducks in the long term)`

`= 96 + 48`

`= 144`

 

d.   `text(Initial female ducks = 96)`

♦♦♦ Mean mark of parts (d)-(e) was 20%.

`W_n = PW_(n – 1)`

`W_4 = [(0,1),(0.25,0.5)]^3[(32),(64)] = [(28),(33)]`

`:. 51\ text(ducks at end of year 4.)`

`W_5 = [(0,1),(0.25,0.5)]^4[(32),(64)] = [(23),(18.5)]`

 

`:. 41.5\ text(ducks at end of year 5.)`

`:.\ text{Numbers halve (drop below 48) in year 5.}`

 

e.    `text(Let )` `a = text(initial juvenile ducks)`
    `b = text(initial adult ducks)`

 

`text(Find)\ a\ text(and)\ b\ text(such that,)`

`[(0,1),(0.25,0.5)]^2[(a),(b)]= [(100),(50)]`

`[(0.25,0.5),(0.125,0.5)][(a),(b)] = [(100),(50)]`

`:. [(a),(b)]` `= [(0.25,0.5),(0.125,0.5)]^(-1)[(100),(50)]`
  `= [(400),(0)]`

 

`:. 400\ text(juvenile ducks and 0 adult ducks)`

`text(should be introduced.)`

MATRICES, FUR2 2013 VCAA 2

10 000 trout eggs, 1000 baby trout and 800 adult trout are placed in a pond to establish a trout population.

In establishing this population

    • eggs (`E`) may die (`D`) or they may live and eventually become baby trout (`B`)
    • baby trout (`B`) may die (`D`) or they may live and eventually become adult trout (`A`)
    • adult trout (`A`) may die (`D`) or they may live for a period of time but will eventually die.

From year to year, this situation can be represented by the transition matrix `T`, where
 

`{:(qquadqquadqquadqquadqquadtext(this year)),((qquadqquadqquadE,quad\ B,quad\ A,\ D)),(T = [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]):}{:(),(),(E),(B),(A),(D):}{:(),(),(qquadtext(next year)):}`
 

  1. Use the information in the transition matrix `T` to
     

    1. determine the number of eggs in this population that die in the first year  (1 mark)
    2. complete the transition diagram below, showing the relevant percentages.  (2 marks)

       

       
          Matrices, FUR2 2013 VCAA 2_a

The initial state matrix for this trout population, `S_0`, can be written as
 

`S_0 = [(10\ 000),(1000),(800),(0)]{:(E),(B),(A),(D):}`
 

Let `S_n` represent the state matrix describing the trout population after `n` years.

  1. Using the rule  `S_n = T S_(n – 1)`, determine each of the following.

     

    1. `S_1`  (1 mark)
    2. the number of adult trout predicted to be in the population after four years  (1 mark)
  2. The transition matrix `T` predicts that, in the long term, all of the eggs, baby trout and adult trout will die.

     

    1. How many years will it take for all of the adult trout to die (that is, when the number of adult trout in the population is first predicted to be less than one)?  (1 mark)
    2. What is the largest number of adult trout that is predicted to be in the pond in any one year?  (1 mark)
  3. Determine the number of eggs, baby trout and adult trout that, if added to or removed from the pond at the end of each year, will ensure that the number of eggs, baby trout and adult trout in the population remains constant from year to year.  (2 marks)

The rule  `S_n = T S_(n – 1)`  that was used to describe the development of the trout in this pond does not take into account new eggs added to the population when the adult trout begin to breed.

  1. To take breeding into account, assume that 50% of the adult trout lay 500 eggs each year.

     

    The matrix describing the population after one year, `S_1`, is now given by the new rule
     
         `S_1 = T S_0 + 500\ M\ S_0` 
     

    where      `T=[(0,0,0,0),(0.40,0,0,0),(0,0.25,0.50,0),(0.60,0.75,0.50,1.0)], M=[(0,0,0.50,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]\ text(and)\ S_0=[(10\ 000),(1000),(800),(0)]`
     
     

    1. Use this new rule to determine `S_1`.  (1 mark)

This pattern continues so that the matrix describing the population after `n` years, `S_n`, is given by the rule

     `S_n = T\ S_(n – 1) + 500\ M\ S_(n – 1)`

  1. Use this rule to determine the number of eggs in the population after two years  (2 marks)
Show Answers Only
    1. `6000`
    2. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `text(See Worked Solutions)`
    1. `text(13 years)`
    2. `text(Largest number of adult trout is 1325.)`
  2. `text(Add 10 000 eggs, remove 3000 baby trout and add 150)`

     

    `text(150 adult trout to keep the population constant.)`

    1. `text(See Worked Solutions)`
    2. `text(See Worked Solutions)`
Show Worked Solution

a.i.   `text(60% of eggs die in 1st year,)`

`:.\ text(Eggs that die in year 1)`

`= 0.60 xx 10\ 000`

`= 6000`
 

MARKER’S COMMENT: A 100% cycle drawn at `D` was a common omission. Do not draw loops and edges of 0%!
a.ii.   

Matrices-FUR2-2013-VCAA-2_a Answer
b.i.    `S_1` `= TS_0`
    `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(10\ 000),(1000),(800),(0)]`
    `= [(0),(4000),(650),(7150)]`

 

b.ii.    `S_4` `= T^4S_0`
    `= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)]^4[(10\ 000),(1000),(800),(0)]`
    `= [(0),(0),(331.25),(11\ 468.75)]`

 

`:. 331\ text(trout is the predicted population)`

`text(after 4 years.)`

 

c.i.    `S_12 = T^12S_0 = [(0),(0),(1.29),(11\ 791)]`

`S_13 = T^13S_0 = [(0),(0),(0.65),(11\ 799)]`

 

`:.\ text{It will take 13 years (when the trout}`

`text{population drops below 1).}`

 

c.ii.    `S_1 = TS_0 = [(0),(4000),(650),(7150)]`

 

`text(After 1 year, 650 adult trout.)`

`text(Similarly,)`

`S_2 = T^2S_0 = [(0),(0),(1325),(10\ 475)]`

`S_3 = T^3S_0 = [(0),(0),(662.5),(11\ 137.5)]`

`S_4 = T^4S_0 = [(0),(0),(331),(11\ 469)]`

 

`:.\ text(Largest number of adult trout = 1325.)`

 

d.    `S_0 – S_1 = [(10\ 000),(1000),(800),(0)] – [(0),(4000),(650),(7150)] = [(10\ 000),(−3000),(150),(−7150)]`

 

`:.\ text(Add 10 000 eggs, remove 3000 baby trout)`

`text(and add 150 adult trout to keep the)`

`text(population constant.)`

 

e.i.    `S_1` `= TS_0 + 500MS_0`
    `= [(0),(4000),(650),(7150)] + 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(10\ 000),(1000),(800),(0)]`
    `= [(0),(4000),(650),(7150)] + 500[(400),(0),(0),(0)]`
    `= [(200\ 000),(4000),(650),(7150)]`

 

e.ii.    `S_2` `= TS_1 + 500MS_1`
   

`= [(0,0,0,0),(0.4,0,0,0),(0,0.25,0.5,0),(0.6,0.75,0.5,1)][(200\ 000),(4000),(650),(7150)]`

       `+ 500 xx [(0,0,0.5,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)][(200\ 000),(4000),(650),(7150)]`
    `= [(162\ 500),(80\ 000),(1325),(130\ 475)]`