SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Trigonometry, 2ADV T2 2016 HSC 8 MC

How many solutions does the equation  `|\ cos (2x)\ | = 1`  have for  `0 <= x <= 2 pi?`

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

`D`

Show Worked Solution

`|\ cos (2x)\ | = 1`

♦♦♦ Mean mark 23%.

`cos (2x) = +- 1`

`text(When)\ \ cos (2x)` `= 1`
`2x` `= 0, 2pi, 4 pi, …`
`:. x` `= 0, pi, 2 pi, …`

 

`text(When)\ \ cos (2x)` `= – 1`
`2x` `= pi, 3 pi, 5 pi, …`
`:. x` `= pi/2, (3 pi)/2, (5 pi)/2, …`

 

`:. x = 0, pi/2, pi, (3 pi)/2, 2 pi\ \ \ text(for)\ \ \ 0 <= x <= 2pi`

`=>  D`

Probability, MET1 2007 VCAA 6

Two events, `A` and `B`, from a given event space, are such that  `text(Pr)(A) = 1/5`  and  `text(Pr)(B) = 1/3`.

  1. Calculate  `text(Pr)(A^{′} ∩ B)`  when  `text(Pr)(A ∩ B) = 1/8`.  (1 mark)

  2. Calculate  `text(Pr)(A^{′} ∩ B)`  when `A` and `B` are mutually exclusive events.  (1 mark)

Show Answers Only

  1. `5/24`
  2. `1/3`

Show Worked Solution

a.   `text(Sketch Venn diagram:)`

♦♦ Mean mark (a) 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

met1-2007-vcaa-q6-answer3

`:. text(Pr)(A^{′} ∩ B)` `=text(Pr)(B)-text(Pr)(A ∩B)`
  `=1/3-1/8`
  `=5/24`

 

♦♦ Mean mark (b) 23%.

b.    met1-2007-vcaa-q6-answer4

`text(Mutually exclusive means)\ \ text(Pr)(A ∩ B)=0,`

`:. text(Pr)(A^{′} ∩ B) = 1/3`

Functions, MET1 2008 VCAA 10

Let  `f: R -> R,\ \ f(x) = e^(2x)-1`.

  1. Find the rule and domain of the inverse function  `f^(-1)`.   (2 marks)

  2. On the axes provided, sketch the graph of  `y = f(f^(-1)(x))`  for its maximal domain.   (1 mark)


     

    met1-2008-vcaa-q2
     

  3. Find  `f(-f^(-1)(2x))`  in the form  `(ax)/(bx + c)`  where `a`, `b` and `c` are real constants.   (2 marks)

Show Answers Only
  1. `f^(-1)(x) = 1/2log_e(x + 1), x ∈ (-1,∞)`
  2.  
    met1-2008-vcaa-q10-answer
  3. `f(-f^(-1)(2x)) = (-2x)/(2x + 1)`
Show Worked Solution

a.   `text(Let)\ \ y = f(x)`

`text(For Inverse, swap)\ x ↔ y`

`x` `= e^(2y)-1`
`x + 1` `= e^(2y)`
`2y` `= log_e(x + 1)`
`y` `= 1/2 log_e(x + 1)`
`text(Domain)(f^(-1))` `=\ text(Range)\ (f)`
  `= (-1,∞)`

 

`:. f^(-1)(x) = 1/2log_e(x + 1),quadx ∈ (-1,∞)`
  

b.   `f(f^(-1)(x)) = x`

♦ Mean mark part (b) 19%.
MARKER’S COMMENT: Few students were aware that a function of its own inverse function is the line  `y=x`  over the appropriate domain.

`text(Domain is)\ \ (-1, oo)`

met1-2008-vcaa-q10-answer

 

♦ Mean mark 34%.
c.    `-f^(-1)(2x)` `= -1/2 ln(2x + 1)`
  `:. f(-f^(-1)(2x))` `= e^(-log_e(2x + 1))-1`
    `=(2x+1)^-1-1`
    `= 1/(2x + 1)-1`
    `= (-2x)/(2x + 1)`

Calculus, MET1 2011 VCAA 10

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with  `AB = ED = 2\ text(cm)` and  `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let  `/_ CBD = theta`  where  `0 < theta < pi/2.`

 vcaa-2011-meth-10a

  1. Find `BD` and `CD` in terms of `a` and `theta`.   (2 marks)

  2. Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`.   (1 mark)

  3. Find  `(dL)/(d theta)`, and hence show that  `(dL)/(d theta) = 0`  when  `BD = 2CD`.   (2 marks)

  4. Find the maximum value of `L` if  `a = 3 sqrt 5`.   (1 mark)

Show Answers Only
  1. `BD = a cos theta,\ \ \ CD = a sin theta`
  2. `L = 4 + a + 2 a cos theta + a sin theta`
  3. `text(Proof)\ text{(See Worked Solutions)}`
  4. `L_max = 19 + 3 sqrt 5`
Show Worked Solution

a.   `text(In)\ \ Delta BCD,`

`cos theta` `= (BD)/a`
`:. BD` `= a cos theta`
`sin theta` `= (CD)/a`
`:. CD` `= a sin theta`

 

b.   `L` `= 4 + 2 BD + CD + a`
    `= 4 + 2a cos theta + a sin theta + a`
    `= 4 + a + 2a cos theta + a sin theta`

 

c.   `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta` `= 0`
`a cos theta` `= 2 a sin theta`
`:.  BD` `= 2CD\ \ text{(using part (a))}`

 

d.   `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark 5%.
`- 2 a sin theta+ a cos theta` `= 0`
`sin theta` `=1/2 cos theta`
`tan theta` `=1/2`

vcaa-2011-meth-10ai

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)` `= 4 + a + 2a cos theta + a sin theta`
  `= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
  `= 4 + 3 sqrt 5 + 12 + 3`
  `= 19 + 3 sqrt 5\ text(cm)`

Functions, MET1 2011 VCAA 4

If the function `f` has the rule  `f(x) = sqrt (x^2-9)`  and the function `g` has the rule  `g(x) = x + 5`

  1.  find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`  (2 marks)

  2.  state the maximal domain for which `f(g(x))` is defined.   (2 marks)

Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2-9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

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

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of `[-3,3)` that ignored the information from part (a).

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`

`:.\ text(Maximal domain is:)`

`x in (– oo, – 8] uu [– 2, oo)`

Calculus, MET1 2012 VCAA 10

Let  `f: R -> R,\ f(x) = e^(-mx) + 3x`,  where `m` is a positive rational number.

  1.  i. Find, in terms of `m`, the `x`-coordinate of the stationary point of the graph of  `y = f(x).`   (2 marks)

  2. ii. State the values of `m` such that the `x`-coordinate of this stationary point is a positive number.   (1 mark)

  3. For a particular value of `m`, the tangent to the graph of  `y = f(x)`  at  `x =-6`  passes through the origin.
  4. Find this value of `m`.   (3 marks)

Show Answers Only
  1.  i. `1/m log_e(m/3)`
  2. ii. `m > 3`
  3. `1/6`
Show Worked Solution

a.i.   `text(SP’s occur when)\ \ f^{prime}(x)=0`

`-me^(-mx) + 3` `= 0`
`me^(-mx)` `3`
`-mx` `= log_e (3/m)`
`:. x` `=-1/m log_e (3/m)`
  `= 1/m log_e (m/3), \ m>0`

 

♦♦♦ Part (a.ii.) mean mark 18%.
  ii.    `1/m log_e (m/3)` `> 0`
  `log_e (m/3)` `> 0`
  `m/3` `> 1`
  `:. m` `> 3`

 

b.   `text(Point of tangency:)\ \ (-6, e^(-6m)-18)`

♦♦ Mean mark (b) 33%.
MARKER’S COMMENT: Many confused `m` with the more common use of `m` for gradient in  `y=mx+c`.

`text(At)\ \ x=-6,`

`m_tan= f^{prime} (-6)= -me^(-6m) + 3`

`:.\ text(Equation of tangent:)`

`y-y_1` `= m (x-x_1)`
`y-(e^(-6m)-18)` `= (-me^(-6m) + 3) (x-(-6))`

 

`text(S)text{ince tangent passes through (0, 0):}`

`-e^(-6m) + 18` `= (-me^(-6m) + 3)(6)`
`-e^(-6m) + 18` `=-6 me^(-6m) + 18`
`e^(-6m)-6me^(-6m)` `= 0`
`e^(-6m) (1-6m)` `= 0`
`1-6m` `=0`
`:.m` `=1/6`

Calculus, MET1 2014 VCAA 10

A line intersects the coordinate axes at the points `U` and `V` with coordinates `(u, 0)` and `(0, v)`, respectively, where `u` and `v` are positive real numbers and  `5/2 <= u <= 6`.

  1. When `u = 6`, the line is a tangent to the graph of  `y = ax^2 + bx`  at the point `Q` with coordinates `(2, 4)`, as shown.

     


    met1-2014-vcaa-q10
     

     

    If `a` and `b` are non-zero real numbers, find the values of `a` and `b`.   (3 marks)

  2. The rectangle `OPQR` has a vertex at `Q` on the line. The coordinates of `Q` are `(2, 4)`, as shown.

     
    met1-2014-vcaa-q10_1
     

    1. Find an expression for `v` in terms of `u`.   (1 mark)

    2. Find the minimum total shaded area and the value of `u` for which the area is a minimum.   (2 marks)

    3. Find the maximum total shaded area and the value of `u` for which the area is a maximum.   (1 mark)

Show Answers Only
  1. `a = −3/2, b = 5`
    1. `v=(4u)/(u-2)`
    2. `8\ text{u²   (when}\ u=4)`
    3. `17\ text{u²}\ \  (text{when}\ u=5/2)`
Show Worked Solution

a.   `text(S)text{ince (2, 4) lies on parabola:}`

♦ Mean mark 48%.
MARKER’S COMMENT: The most common error incorrectly stated that (6,0) lay on the parabola.
`4` `= a(2)^2 + b(2)`
`4` `= 4a + 2b`
`2` `= 2a + b\ …\ (1)`

 

`y` `=ax^2+bx`
`dy/dx` `=2ax+b`

 

`text(At)\ \ x=2,`

`m_(QU)` `=m_(text(tang))`
`(4-0)/(2-6)` `= 2a(2) + b`
`−1` `= 4a + b\ …\ (2)`

 
`text(Subtract)\ \ (2)-(1)`

`−3 = 2a`

`:. a = −3/2, b = 5`

 

b.i.   `text(Solution 1)`

`text(Using similar triangles:)`

met1-2014-vcaa-q10-answer

`Delta VOU\ text(|||)\ Delta QPU`

♦♦ Mean mark (b.i.) 32%.
`v/4` `= u/(u -2)`
`:. v` `= (4u)/(u -2)`

 

`text(Solution 2)`

`m_(VQ)` `=m_(UQ)`
`(v-4)/(0-2)` `=(0-4)/(u-2)`
`v` `=8/(u-2) + 4`
  `=(8+4(u-2))/(u-2)`
  `=(4u)/(u-2)`

 

♦♦♦ Mean mark (b.ii.) 18%.
b.ii.    `text(Area)` `= 1/2uv-2 xx 4`
    `= 1/2u((4u)/(u -2))-8`
    `= (2u^2)/(u -2)-8`

 
`text(SP occurs when)\ \ (dA)/(du)=0,`

`(4u(u-2)-2u^2)/((u-2)^2)` `= 0`
`2u^2-8u` `= 0`
`2u(u-4)` `= 0`

 

`u = 0quadtext(or)quadu = 4`

`:. u = 4, \ \ u ∈ [5/2,6]`

`text(When)\ u=4,\ \ A` `=(2 xx 4^2)/(4-2)-8`
  `= 8\ text(u²)`

 
`text(Test areas at end points,)`

`text(When)\ u=5/2,\ A=17`

`text(When)\ u=6,\ A=10`

`:. A_(text(min))=8\ text{u²   (when}\ u=4)`
 

b.iii.   `text{As only one (local minimum) stationary point}`

♦♦♦ Mean mark (b.iii.) 8%.

`text(exists over)\ 5/2 <= u <= 6, text(the maximum area)`

`text(must occur at an endpoint.)`

`:. A_(text(max)) = 17\ text{u²}\ \  (text{when}\ u=5/2)`

Calculus, MET1 2015 VCAA 10

The diagram below shows a point, `T`, on a circle. The circle has radius 2 and centre at the point `C` with coordinates `(2, 0)`. The angle `ECT` is `theta`, where  `0 < theta <= pi/2`.
  

met1-2015-vcaa-q10
  

The diagram also shows the tangent to the circle at `T`. This tangent is perpendicular to `CT` and intersects the `x`-axis at point `X` and the `y`-axis at point `Y`.

  1. Find the coordinates of `T` in terms of `theta`.   (1 mark)

  2. Find the gradient of the tangent to the circle at `T` in terms of `theta`.   (1 mark)

  3. The equation of the tangent to the circle at `T` can be expressed as
  4. `qquad cos(theta)x + sin(theta)y = 2 + 2cos(theta)`
  5.  i. Point `B`, with coordinates `(2, b)`, is on the line segment `XY`.
  6.     Find `b` in terms of `theta`.   (1 mark)

  7. ii. Point `D`, with coordinates `(4, d)`, is on the line segment `XY`.
  8.     Find `d` in terms of `theta`.   (1 mark)

  9. Consider the trapezium `CEDB` with parallel sides of length `b` and `d`.
  10. Find the value of `theta` for which the area of the trapezium `CEDB` is a minimum. Also find the minimum value of the area.   (3 marks)

Show Answers Only
  1. `T(2 + 2costheta, 2 sintheta)`
  2. `(-1)/(tan(theta))`
  3.  i. `2/(sintheta)`
  4. ii. `(2-2 costheta)/(sintheta)`
  5. `theta = pi/3`
  6. `A_text(min) = 2sqrt3\ text(u²)`
Show Worked Solution
a.    `cos theta` `= (CM)/(CT)`
    `=(CM)/2`
  `CM` `= 2costheta`
♦♦♦ Part (a) mean mark 20%.
MARKER’S COMMENT: Many students did not include the “+2” in the `x`-coordinate.
`sintheta` `= (TM)/2`
`TM` `= 2sintheta`

 

`:. T\ text(has coordinates)\ \ (2 + 2costheta, 2 sintheta)`

 

♦♦♦ Part (b) mean mark 16%.
b.    `m_(CT)` `=(TM)/(CM)`
    `=(2 sin theta)/(2 cos theta)`
    `=tan theta`
     
  `:.m_(XY)` `=-1/tan theta,\ \ \ (CT ⊥ XY)`

 

c.i.   `text(Substitute)\ \ (2,b)\ \ text(into equation:)`

`2costheta + bsintheta` `= 2 + 2costheta`
`:. b` `= 2/(sintheta)`

 

c.ii.   `text(Substitute)\ \ (4,d)\ \ text(into equation:)`

♦ Part (c)(ii) mean mark 47%.
`4costheta + dsintheta` `= 2 + 2costheta`
`d sin theta` `=2-cos theta`
`:.d`  `= (2-2 costheta)/(sintheta)`

 

♦♦♦ Part (d) mean mark 19%.
d.    `text(A)_text(trap)` `= 1/2 xx 2 xx (b + d)`
    `= 2/(sintheta) + (2-2costheta)/(sintheta)`
    `= (4-2costheta)/(sintheta)`

 

`text(Stationary point when)\ \ (dA)/(d theta)=0`

`(2sin^2theta-costheta(4-2costheta))/(sin^2theta)` `= 0`
`2sin^2theta-4costheta + 2cos^2theta` `= 0`
`2[sin^2theta + cos^2theta]-4costheta` `= 0`
`2-4costheta` `= 0`
`costheta` `= 1/2`
`theta` `= pi/3,\ \ \ \ theta ∈ (0, pi/2)`

 

`A(pi/3)` `= (4-2(1/2))/(sqrt3/2)`
  `=3 xx 2/sqrt3`
  `= 2sqrt3`

 

`:. A_text(min) = 2sqrt3\ text(u²)`

Calculus, MET2 2010 VCAA 4

Consider the function  `f: R -> R,\ f(x) = 1/27 (2x-1)^3 (6-3x) + 1.`

  1. Find the `x`-coordinate of each of the stationary points of  `f` and state the nature of each of these stationary points.   (4 marks)

In the following, `f` is the function  `f: R -> R,\ f(x) = 1/27 (ax-1)^3 (b-3x) + 1` where `a` and `b` are real constants.

  1. Write down, in terms of `a` and `b`, the possible values of `x` for which `(x, f (x))` is a stationary point of `f`.   (3 marks)

  2. For what value of `a` does `f` have no stationary points?   (1 mark)

  3. Find `a` in terms of `b` if `f` has one stationary point.   (2 marks)

  4. What is the maximum number of stationary points that `f` can have?  (1 mark)

  5. Assume that there is a stationary point at `(1, 1)` and another stationary point `(p, p)` where  `p != 1`.
  6. Find the value of `p`.   (3 marks)

Show Answers Only
  1. `text(Point of inflection at)\ \ x = 1/2`
    `text(Local max at)\ \ x = 13/8`
  2. `x = (ab + 1)/(4a) or x = 1/a`
  3. `0`
  4. `3/b`
  5. `2`
  6. `4`
Show Worked Solution

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

`f(x)` `=1/27 (2x-1)^3 (6-3x) + 1`
`f^{′}(x)` `=- 1/9 (2x-1)^2 (8x-13) `

 
`text(Solve:)\ \ f^{′}(x)=0\ \ text(for)\ x,`

`:. x = 1/2 or x = 13/8`
 

 `text(Sketch the graph:)`

vcaa-graphs-fur2-2010-4ai

`=>\ text(Point of inflection at)\ \ x = 1/2`

`=>\ text(Local max at)\ \ x = 13/8`
 

b.   `text(S.P. occurs when)\ \ f prime (x) = 0`

♦ Mean mark (b) 46%.
MARKER’S COMMENT: Issues with use of CAS caused significant difficulties in this question.
`f(x)` `=1/27 (ax-1)^3 (b-3x) + 1`
`f^{′}(x)` `=1/9 (ax-1)^2 (ab+1-4ax)`

 
`text(Solve:)\ \ f^{′}(x) = 0\ \ text(for)\ \ x,`

`:. x = (ab + 1)/(4a)\ \ \text{or}\  \ x = 1/a`
 

c.   `text(For)\ \ x = (ab + 1)/(4a) or x = 1/a\ \ text(to exist,)`

♦ Mean mark part (c) 47%.

`a != 0`

`:.\ text(No stationary points when)\ \ a = 0`
 

d.   `text(If there is 1 S.P.,)`

♦♦♦ Mean mark (d) 16%.
`(ab + 1)/(4a)` `= 1/a`
`:. a` `= 3/b`

 

♦ Mean mark (e) 37%.

e.   `text(The maximum number of SP’s for a quartic)`

`text(polynomial is 3. In the function given, one of)`

`text(the SP’s is a point of inflection.)`

`:.f(x)\ \ text(has a maximum of 2 SP’s.)`

 

f.   `text{Solution 1 [by CAS]}` 

`text(Define)\ \ f(x) = 1/27 (x-1)^3 (b-3x) + 1`

`text(Solve:)\ \ f(p) = p, f^{′}(1) = 1  and f^{′}(p) = p\ \ text(for)\ \ p,`

`:. p = 4,\  \ \ (p != 1)`
 

`text(Solution 2)`

`text(SP’s occur at)\ \ (1,1) and (p,p),\ \ text(where,)`

♦♦♦ Mean mark (f) 9%.

`x = (ab + 1)/(4a) or x = 1/a`

`text(Consider)\ \ p=1/a,`

`f(p)` `=f(1/a)`
 

`=1/27 (a*1/a-1)^3(b-3*1/a)+1=1`
   

`f(p)=1,\ \ text(SP at)\ (1,1) and p!=1`

`=> p!=1/a`
 

`text(Consider)\ \ 1=1/a,`

`=> a=1 and  b=4p-1`

`f(1)=1`

`f(p)=p`

`1/27 (p-1)^3(4p-1-3p)+1` `=p`
`1/27(p-1)^4-(p-1)` `=0`
`(p-1)(1/27(p-1)^3-1)` `=0`
`(p-1)^3` `=27`
`p` `=4`

 

Calculus, MET2 2010 VCAA 3

An ancient civilisation buried its kings and queens in tombs in the shape of a square-based pyramid, `WABCD.`

The kings and queens were each buried in a pyramid with  `WA = WB = WC = WD = 10\ text(m).`

Each of the isosceles triangle faces is congruent to each of the other triangular faces.

The base angle of each of these triangles is `x`, where  `pi/4 < x < pi/2.`

Pyramid `WABCD` and a face of the pyramid, `WAB`, are shown here.
 

VCAA 2010 3a

`Z` is the midpoint of `AB.`

  1. i. Find `AB` in terms of `x`.   (1 mark)

  2. ii. Find `WZ` in terms of `x`.   (1 mark)

  3. Show that the total surface area (including the base), `S\ text(m)^2`, of the pyramid, `WABCD`, is given by 
  4.      `S = 400(cos^2 (x) + cos (x) sin (x))`.   (2 marks)

  5. Find `WY`, the height of the pyramid `WABCD`, in terms of `x`.    (2 marks)

  6. The volume of any pyramid is given by the formula  `text(Volume) = 1/3 xx text(area of base) xx text(vertical height)`.
  7. Show that the volume, `T\ text(m³)`, of the pyramid `WABCD`  is  `4000/3 sqrt(cos^4 x-2 cos^6 x)`.   (1 mark)

Queen Hepzabah’s pyramid was designed so that it had the maximum possible volume.

  1. Find  `(dT)/(dx)`  and hence find the exact volume of Queen Hepzabah’s pyramid and the corresponding value of `x`.   (4 marks)

Queen Hepzabah’s daughter, Queen Jepzibah, was also buried in a pyramid. It also had

`WA = WB = WC = WD = 10\ text(m.)`

The volume of Jepzibah’s pyramid is exactly one half of the volume of Queen Hepzabah’s pyramid. The volume of Queen Jepzibah’s pyramid is also given by the formula for `T` obtained in part d.

  1. Find the possible values of `x`, for Jepzibah’s pyramid, correct to two decimal places.   (2 marks)

Show Answers Only

a.i. `20 cos (x)`

a.ii.`10 sin (x)`

b.   `text(Proof)\ \ text{(See Worked Solutions)}`

c.   `10 sqrt (sin^2(x)-cos^2 (x))`

d.   `text(Proof)\ \ text{(See Worked Solutions)}`

e.   `x = cos^-1 (sqrt 3/3) -> T_max = (4000 sqrt 3\ m^3)/27`

f.   `x dot = 0.81 or x dot = 1.23`

Show Worked Solution
a.i.   `cos x` `= (1/2 AB)/10`
  `:. AB` `= 20 cos(x)`

 

  ii.   `sin (x)` `= (wz)/10`
  `:. wz` `= 10 sin (x)`

 

b.   `text(Area base)` `= (20 cos (x))^2`
    `= 400 cos^2(x)`
  `4 xx text(Area)_Delta` `= 4 xx (1/2 xx 20 cos (x) xx 10 sin (x))`
    `= 400 cos (x) sin (x)`
♦ Mean mark 47%.

 

`:. S` `= 400 cos^2 (x) + 400 cos (x) sin (x)`
  `= 400 (cos^2 (x) + cos (x) sin (x))\ \ text(… as required.)`

 

c.   `text(Using)\ \ Delta WYZ,`

 vcaa-graphs-fur2-2010-ci

`text(Using Pythagoras,)`

`WY` `= sqrt (10^2 sin^2 (x)-10^2 cos^2 (x))`
  `= 10 sqrt (sin^2(x)-cos^2 (x))`
♦♦♦ Mean mark part (d) 22%.

 

d.   `T` `= 1/3 xx text(base) xx text(height)`
    `= 1/3 xx (400 cos^2 (x)) xx (10 sqrt(sin^2 (x)-cos^2 (x)))`
    `= 4000/3 sqrt (cos^4 (x) (sin^2 (x)-cos^2 (x))`

 

`text(Using)\ \ sin^2 (x) = 1-cos^2 (x),`

`T` `= 4000/3 sqrt (cos^4 (x) (1-cos^2 (x)-cos^2 (x))`
  `= 4000/3 sqrt (cos^4 (x)-2 cos^6 (x))`

 

e.   `(dT)/(dx) = (8000 cos (x) sin (x) (3 cos^2 (x)-1))/(3 sqrt(1-2 cos^2 (x)))`

♦ Mean mark part (e) 45%.

 

`text(Stationary point when,)`

`(dT)/(dx) = 0\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = cos^-1 (sqrt 3/3)`

`:.T_max` `=T(cos^-1 (sqrt 3/3))`
  `= (4000 sqrt 3)/27\ \ text(m³)`

♦♦♦ Mean mark part (f) 16%.

 

f.   `text(Solve)\ \ T(x) = (2000 sqrt 3)/27\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = 0.81  or  x = 1.23\ \ text{(2 d.p.)}`

Probability, MET1 2015 VCAA 9

An egg marketing company buys its eggs from farm A and farm B. Let `p` be the proportion of eggs that the company buys from farm A. The rest of the company’s eggs come from farm B. Each day, the eggs from both farms are taken to the company’s warehouse.

Assume that `3/5` of all eggs from farm A have white eggshells and `1/5` of all eggs from farm B have white eggshells.

  1. An egg is selected at random from the set of all eggs at the warehouse.
  2. Find, in terms of `p`, the probability that the egg has a white eggshell.  (1 mark)

  3. Another egg is selected at random from the set of all eggs at the warehouse.

     

    1. Given that the egg has a white eggshell, find, in terms of `p`, the probability that it came from farm B.  (2 marks)

    2. If the probability that this egg came from farm B is 0.3, find the value of `p`.  (1 mark)

Show Answers Only

  1. `(2p+1)/5`
    1. `(1 – p)/(2p + 1)`
    2. `7/16`

Show Worked Solution

a.  

met1-2015-vcaa-q9-answer1

 

`text(Pr)(AW) + text(Pr)(BW)` `= p xx 3/5 + (1-p) xx 1/5`
  `=(3p)/5+1/5-p/5`
  `= (2p+1)/5`

 

♦ Part (b) mean mark 41%.
MARKER’S COMMENT: Algebraic fractions “were not handled well”!

b.i.    `text(Pr)(B | W)` `= (text(Pr)(B ∩ W))/(text(Pr)(W))`
    `= ((1 – p)/5)/((2p + 1)/5)`
    `=(1-p)/5 xx 5/(2p+1)`
    `= (1 – p)/(2p + 1)`

 

♦♦♦ Part (c) mean mark 19%.
STRATEGY: Previous parts of a question are gold dust for directing your strategy in many harder questions.

b.ii.    `text(Pr)(B | W)` `= 3/10`
  `(1 – p)/(2p + 1)` `= 3/10`
  `10 – 10p` `= 6p + 3`
  `7` `= 16p`
  `:. p` `= 7/16`

CORE*, FUR2 2006 VCAA 4

A company anticipates that it will need to borrow $20 000 to pay for a new machine.

It expects to take out a reducing balance loan with interest calculated monthly at a rate of 10% per annum.

The loan will be fully repaid with 24 equal monthly instalments.

Determine the total amount of interest that will be paid on this loan.

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

Show Answers Only

`$2150\ \ text{(nearest $)}`

Show Worked Solution

`text(Find monthly payment by TVM Solver,)`

♦♦ Mean mark 21%.
`N` `= 2 xx 12 = 24`
`I(text(%))` `= 10`
`PV` `= 20\ 000`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=>PMT = -922.898…`

 

`text(Total repayments)`

`= 24 xx 922.898…`

`= $22\ 149.65…`
 

`:.\ text(Total interest paid)`

`= 22\ 149.56…-20\ 000`

`= 2149.56…`

`= $2150\ \ text{(nearest $)}`

CORE*, FUR2 2007 VCAA 2

Khan decides to extend his home office and borrows $30 000 for building costs. Interest is charged on the loan at a rate of 9% per annum compounding monthly.

Assume Khan will pay only the interest on the loan at the end of each month.

  1. Calculate the amount of interest he will pay each month.   (1 mark)

Suppose the interest rate remains at 9% per annum compounding monthly and Khan pays $400 each month for five years.

  1. Determine the amount of the loan that is outstanding at the end of five years.

     

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

Khan decides to repay the $30 000 loan fully in equal monthly instalments over five years.

The interest rate is 9% per annum compounding monthly.

  1. Determine the amount of each monthly instalment. Write your answer correct to the nearest cent.   (1 mark) 
Show Answers Only
  1. `$225`
  2. `$16\ 801`
  3. `$622.75`
Show Worked Solution

a.   `text(Interest paid each month)`

♦ Mean mark of all parts (combined) was 37%.
MARKER’S COMMENT: Many students were confused by the fact that part (a) did not have a given loan period.

`= 1/12 xx 0.09 xx 30\ 000`

`= $225`
  

b.   `text(Find principal left after 5 years.)`

`text(By TVM Solver:)`

`N` `= 5 xx 12 = 60`
`I(text(%))` `= 9`
`PV` `= 30\ 000`
`PMT` `= −400`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

`=> FV = −16\ 800.77…`

`:. $16\ 801\ text(is outstanding after 5 years.)`
  

c.   `text(By TVM Solver,)`

`N` `= 5 xx 12 = 60`
`I(text(%))` `= 9`
`PV` `= 30\ 000`
`PMT` `= ?`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

`=>PMT = −622.750…`

`:.\ text(Monthly installment is $622.75)`

GRAPHS, FUR2 2007 VCAA 3

Gas is generally cheaper than petrol. 

A car must run on petrol for some of the driving time.

Let  `x`  be the number of hours driving using gas

 `y`  be the number of hours driving using petrol

Inequalities 1 to 5 below represent the constraints on driving a car over a 24-hour period.

Explanations are given for Inequalities 3 and 4.

Inequality 1:   `x ≥ 0`

Inequality 2:   `y ≥ 0`

Inequality 3:   `y ≤ 1/2x`     The number of hours driving using petrol must not exceed half the number of hours driving using gas.
Inequality 4:   `y ≥ 1/3x`     The number of hours driving using petrol must be at least one third the number of hours driving using gas.

Inequality 5:   `x + y ≤ 24`

 

  1. Explain the meaning of Inequality 5 in terms of the context of this problem.  (1 mark)

The lines  `x + y = 24`  and  `y = 1/2x`  are drawn on the graph below.

GRAPHS, FUR2 2007 VCAA 3

  1. On the graph above

     

    1. draw the line  `y = 1/3x`  (1 mark)
    2. clearly shade the feasible region represented by Inequalities 1 to 5.  (1 mark)

On a particular day, the Goldsmiths plan to drive for 15 hours. They will use gas for 10 of these hours.

  1. Will the Goldsmiths comply with all constraints? Justify your answer.  (1 mark)

On another day, the Goldsmiths plan to drive for 24 hours.

Their car carries enough fuel to drive for 20 hours using gas and 7 hours using petrol.

  1. Determine the maximum and minimum number of hours they can drive using gas while satisfying all constraints.  (2 marks)

     

     

    Maximum = ___________ hours

     

     

    Minimum = ___________ hours

Show Answers Only
  1. `text(Inequality 5 means that the total hours driving)`
    `text(with gas PLUS the total hours driving with petrol)`
    `text(must be less than or equal to 24 hours.)`
  2. i. & ii.
    GRAPHS, FUR2 2007 VCAA 3 Answer
  3. `text(If they drive for 10 hours on gas,)`
    `text(5 hours is driven on petrol,)`
    `text{(10, 5) is in the feasible region.}`
    `:.\ text(They comply with all constraints.)`
  4. `text(Maximum = 18 hours)`
    `text(Minimum = 17 hours)`
Show Worked Solution

a.   `text(Inequality 5 means that the total hours driving)`

`text(with gas PLUS the total hours driving with petrol)`

`text(must be less than or equal to 24 hours.)`

♦♦ Mean mark of parts (b)-(d) (combined) was 33%.

 

b.i. & ii.

GRAPHS, FUR2 2007 VCAA 3 Answer

 

c.   `text(If they drive for 10 hours on gas, 5 hours)`

MARKER’S COMMENT: A mark was only awarded if a reference was made to the 5 hours driving.

`text(is driven on petrol.)`

`=>\ text{(10, 5) is in the feasible region.}`

`:.\ text(They comply with all constraints.)`

 

d.   `text(Maximum = 18 hours)`

♦♦♦ “Very few” obtained both answers here.
MARKER’S COMMENT: Many ignored that the Goldsmiths planned to drive for 24 hours.

`text{(6 hours of petrol available)}`

`text(Minimum = 17 hours)`

`text{(7 hours of petrol is the highest available)}`

 

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.)`

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.`

GRAPHS, FUR2 2009 VCAA 4

Cheapstar Airlines wishes to find the optimum number of flights per day on two of its most popular routes: Alberton to Bisley and Alberton to Crofton.

Let  `x`  be the number of flights per day from Alberton to Bisley

 `y`  be the number of flights per day from Alberton to Crofton

Table 4 shows the constraints on the number of flights per day and the number of crew per flight.

GRAPHS, FUR2 2009 VCAA 41

The lines  `x + y = 10`  and  `3x + 5y = 41`  are graphed below.

GRAPHS, FUR2 2009 VCAA 42

A profit of $1300 is made on each flight from Alberton to Bisley and a profit of $2100 is made on each flight from Alberton to Crofton.

Determine the maximum total profit that Cheapstar Airlines can make per day from these flights.  (2 marks)

Show Answers Only

`$17\ 300`

Show Worked Solution

GRAPHS, FUR2 2009 VCAA 4 Answer

`P = 1300x + 2100y`

♦♦♦ Mean mark 9%.
MARKER’S COMMENT: Many students incorrectly substituted the point of intersection (4.5,5.5) into the equation. Integer values within the feasible region were required.

`text(Within the feasible area, profit is)`

`text(maximised at)\ (2,7).`

`:.\ text(Maximum profit)`

`= 1300 xx 2 + 2100 xx 7`

`= $17\ 300` 

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.
    3. 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?
  3. 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.)}`

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 4

A flying fox suspension wire begins at `V`, 12.5 metres above `T` as shown in the diagram below. It ends at `S`, 4.5 metres above `K`.
 

GEOMETRY, FUR2 2010 VCAA 4
 

At `P`, the flying fox wire passes over `H`.

The horizontal distances `TH` and `HK` are 95 metres and 65 metres respectively.

Calculate the vertical distance, `PH`, in metres.  (2 marks)

Show Answers Only

`7.75\ text(m)`

Show Worked Solution

GEOMETRY, FUR2 2010 VCAA 4 Answer 
 

`Delta VAS\ text(|||)\ Delta PBS\ \ (A A A)`

♦♦♦ Mean mark 11%.
`(PB)/(BS)` `= (VA)/(AS)`
`(PB)/(65)` `= 8/160`
`:. PB` `= (8 xx 65)/160`
  `= 3.25\ text(m)`

 

`:. PH` `= PB + BH`
  `= 3.25 + 4.5`
  `= 7.75\ text(m)`

 

`text(Alternative solution)`

`text(In)\ Delta VAS,`

`tantheta` `= 8/160`
`:. theta` `= 2.862…^@`

 

`text(In)\ DeltaPBS,`

`tantheta` `= (PB)/65`
`PB` `= 65 xx tan2.862…^@`
  `= 3.25\ text(m)`

 

`:. PH = 3.25 + 4.5 = 7.75\ 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. How many monthly repayments are required to repay the loan? Write your answer to the nearest month.   (1 mark)
    2. 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.
  2. 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 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.
    1. 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)
    2. After the first 12 months, only the interest on the loan is paid each month.
    3. 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)}`

NETWORKS, FUR1 2006 VCAA 8 MC

Euler’s formula, relating vertices, faces and edges, does not apply to which one of the following graphs?

networks-fur1-2006-vcaa-8-mc-ab

networks-fur1-2006-vcaa-8-mc-cd

networks-fur1-2006-vcaa-8-mc-e

Show Answers Only

`D`

Show Worked Solution

`text(Euler’s formula only applies to)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: Over a third of students incorrectly chose option `A`, unaware that any graph with 4 or less vertices must be planar.

`text(planar graphs.)`

`text(Consider option)\ D,`

networks-fur1-2006-vcaa-8-mc-answer

`text(Only option)\ D\ text(cannot be redrawn)`

`text(as a planar graph.)`

`rArr D`

NETWORKS, FUR1 2010 VCAA 9 MC

The table below shows the time (in minutes) that each of four people, Aiden, Bing, Callum and Dee, would take to complete each of the tasks `U, V, W` and `X.`
 

     vcaa-networks-fur1-2010-9
 

If each person is allocated one task only, the minimum total time for this group of people to complete all four tasks is

A.   `22\ text(minutes)`

B.   `28\ text(minutes)`

C.   `29\ text(minutes)`

D.   `30\ text(minutes)`

E.   `32\ text(minutes)`

Show Answers Only

`B`

Show Worked Solution

vcaa-networks-fur1-2010-9i

 
`:.\ text(Minimum time)`

`= 9 + 5 + 6 + 8`

`= 28\ text(minutes)`

`=>  B`

NETWORKS, FUR1 2006 VCAA 9 MC

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


 

The project is to be crashed by reducing the completion time of one activity only.

This will reduce the completion time of the project by a maximum of

A.   1 hour 

B.   2 hours

C.   3 hours

D.   4 hours

E.   5 hours

Show Answers Only

`D`

Show Worked Solution

`text(The critical path is)\ BDCEHJ\ text{(19 hours)}`

♦♦♦ Mean mark 17%.
MARKER’S COMMENT: When choosing an activity to crash, take care that a new critical path is not created.

 
`text(Also,)\ ACEHJ\ text{(15 hours)}`

`:.\ text(Activity)\ B\ text(could be crashed by 4 hours)`

`text(without a new critical path emerging.)`

`rArr D`

MATRICES*, FUR1 2009 VCAA 9 MC

Five soccer teams played each other once in a tournament. In each game there was a winner and a loser.

A table of one-step and two-step dominances was prepared to summarise the results.
 

     networks-fur1-2009-vcaa-9-mc
 

One result in the tournament that must have occurred is that

A.   Elephants defeated Bears.

B.   Elephants defeated Aardvarks.

C.   Aardvarks defeated Donkeys.

D.   Donkeys defeated Bears.

E.   Bears defeated Chimps. 

Show Answers Only

`B`

Show Worked Solution

`text(Consider option)\ A,`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: The critical step here is to realise that if Team A beats Team B, then Team B’s one-step dominances become Team A’s two-step dominances.

`text(If elephants defeat bears, elephants)`

`text(will have a 2-step dominance over)`

`text(3 other teams.)`

`:.\ text(Incorrect.)`

 

`text(Consider option)\ B,`

`text(If elephants defeat aardvarks, elephants)`

`text(will have a 2-step dominance over)`

`text(1 other team.)`

`:.\ text(Correct.)`

`=>  B`

NETWORKS, FUR1 2009 VCAA 8 MC

An undirected connected graph has five vertices.

Three of these vertices are of even degree and two of these vertices are of odd degree.

One extra edge is added. It joins two of the existing vertices.

In the resulting graph, it is not possible to have five vertices that are

A.   all of even degree.

B.   all of equal degree.

C.   one of even degree and four of odd degree.

D.   three of even degree and two of odd degree.

E.   four of even degree and one of odd degree. 

Show Answers Only

`E`

Show Worked Solution

`text(Consider an example of the graph)`

♦♦♦ Mean mark 25%.

`text{described (below):}`
 

matrices-fur1-2009-vcaa-8-mc-answer
 

`A\ text(is possible – join)\ V\ text(and)\ Z`

`B\ text(is possible – join)\ V\ text(and)\ Z`

`C\ text(is possible – join)\ W\ text(and)\ Y`

`D\ text(is possible – join)\ V\ text(and)\ X`

`E\ text(is NOT possible)`

`=>  E`

NETWORKS, FUR1 2011 VCAA 9 MC

An Euler path through a network commences at vertex `P` and ends at vertex `Q`.

Consider the following five statements about this Euler path and network.

•  In the network, there could be three vertices with degree equal to one.

•  The path could have passed through an isolated vertex.

•  The path could have included vertex `Q` more than once.

•  The sum of the degrees of vertices `P` and `Q` could equal seven.

•  The sum of the degrees of all vertices in the network could equal seven.

How many of these statements are true?

A.   `0`

B.   `1`

C.   `2`

D.   `3`

E.   `4`

Show Answers Only

`B`

Show Worked Solution

`text(“The path could have included vertex)\ Q`

♦♦ Mean mark 30%.

`text(more than once” is the only true statement.)`

`=>  B`

NETWORKS, FUR1 2011 VCAA 7 MC

Andy, Brian and Caleb must complete three activities in total (K, L and M)

The table shows the person selected to complete each activity, the time it will take to complete the activity in minutes and the immediate predecessor for each activity.
 

   

 
All three activities must be completed in a total of 40 minutes.

The instant that Andy starts his activity, Caleb gets a telephone call.

The maximum time, in minutes, that Caleb can speak on the telephone before he must start his allocated activity is

A.    5

B.   13

C.   18

D.   24

E.   34

Show Answers Only

`D`

Show Worked Solution

`text(Maximum speaking time)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students incorrectly answered the earliest starting time, C.

`= 40 – text(duration of)\ M`

`= 40 – 16`

`= 24\ text(minutes)`

`=>  D`

NETWORKS, FUR1 2012 VCAA 9 MC

John, Ken and Lisa must work together to complete eight activities, `A, B, C, D, E, F, G` and `H`, in minimum time.

The directed network below shows the activities, their completion times in days, and the order in which they must be completed.

net9

Several activities need special skills. Each of these activities may be completed only by a specified person.

  • Activities `A` and `F` may only be completed by John.
  • Activities `B` and `C` may only be completed by Ken.
  • Activities `D` and `E` may only be completed by Lisa.
  • Activities `G` and `H` may be completed by any one of John, Ken or Lisa.

With these conditions, the minimum number of days required to complete these eight activities is

A.  14

B.  17

C.  20

D.  21

E.  24

Show Answers Only

`D`

Show Worked Solution

`text(The minimum days require the following path:)`

♦♦♦ Mean mark 18%.

`text(John completes)\ A.`

`text(Lisa completes)\ D\ text(then)\ E\ text{(restriction)}.`

`H\ text(is completed.)`

`text(All other activities can be completed)`

`text(during this time.)`

`text(Minimum days)` `= 3 + 7 + 6 + 5`
  `= 21`

`rArr D`

NETWORKS, FUR1 2012 VCAA 8 MC

networks-fur1-2012-vcaa-8-mc 
 

Eight activities, `A, B, C, D, E, F, G\ text(and)\ H`, must be completed for a project.

The graph above shows these activities and their usual duration in hours.

The duration of each activity can be reduced by one hour.

To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is

A.   1

B.   2

C.   3

D.   4

E.   5

Show Answers Only

`C`

Show Worked Solution

`text(Paths)\ AFH\ text(and)\ BCFH => 18\ text(hours)`

♦♦ Mean mark 30%.

`text(Paths)\ AEG\ text(and)\ BCEG => 17\ text(hours)`

`text(Reducing)\ A\ text(and)\ B\ text(by 1 hour each)`

`text(reduces each path above by 1 hour.)`

 

`:. text(Need to now reduce)\ AFH\ text(and)`

`BCFH\ text(by 1 hour to 16 hours.)`

`text(Reducing)\ F\ text(or)\ H\ text(by 1 hour brings)`

`text(the critical path down to 16 hours.)`

`rArr C`

NETWORKS, FUR1 2012 VCAA 7 MC

Vehicles from a town can drive onto a freeway along a network of one-way and two-way roads, as shown in the network diagram below.

The numbers indicate the maximum number of vehicles per hour that can travel along each road in this network. The arrows represent the permitted direction of travel.

One of the four dotted lines shown on the diagram is the minimum cut for this network.
  

networks-fur1-2012-vcaa-7-mc-1

 
The maximum number of vehicles per hour that can travel through this network from the town onto the freeway is

A.  `310`

B.  `330`

C.  `350`

D.  `370`

E.  `390`

Show Answers Only

`C`

Show Worked Solution

`text(Consider each “minimum cut” line,)`

♦♦♦ Mean mark 23%.

`text(Line 1: doesn’t seperate the town and freeway)`

`text(Line 2: 240 + 110 = 350)`

`text(Line 3: 240 + 60 + 90 = 390)`

`text(Line 4: 280 + 90 = 370)`

`:.\ text(Line 2 gives the maximum flow)`

`rArr C`

NETWORKS, FUR1 2013 VCAA 8 MC

   
 

The graph above shows five activities, `A, B, C, D\ text(and)\ E`, that must be completed to finish a project.

The number next to each letter shows the completion time, in hours, for the activity.

Each of the five activities can have its completion time reduced by a maximum of one hour at a cost of $100 per hour.

The least cost to achieve the greatest reduction in the time taken to finish the project is

A.  $100

B.  $200

C.  $300

D.  $400

E.  $500

Show Answers Only

`D`

Show Worked Solution

`text(Clearly, we reduce)\ E\ text(by 1 hour.)`

♦♦♦ Mean mark 23%.

`text(New critical path is)\ B – D – E\ \ text{(23 hours)}`

`text(Next, reduce)\ B and D\ text(by 1 hour each.)`

`text(A new critical path emerges:)`

`A – C – E\ \ text{(23 hours)}`

 

`:.\ text(Reduce)\ A\ text(or)\ C\ text(by 1 hour so)`

`text(that)\ A – C – E\ \ text{(22 hours) matches}`

`B – D – E\ \ text{(22 hours).}`

`:.\ text(Least cost for maximum time reduction)`

`= 4 xx 100`

`= $ 400`

`=>  D`

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, FUR1 2014 VCAA 9 MC

A network of tracks connects two car parks in a festival venue to the exit, as shown in the directed graph below.
 

 
The arrows show the direction that cars can travel along each of the tracks and the numbers show each track’s capacity in cars per minute.

Five cuts are drawn on the diagram.

The maximum number of cars per minute that will reach the exit is given by the capacity of

A.  Cut A

B.  Cut B

C.  Cut C

D.  Cut D

E.  Cut E

Show Answers Only

`D`

Show Worked Solution

`text(Cut)\ A and B\ text(don’t separate both)`

♦♦♦ Mean mark 24%.
COMMENT: Note that the “4” is not included in Cut D as it is flowing in the opposite direction.

`text(car parks from the exit.)`

`text(Cut)\ D\ text(has the minimum cut of)`

`text(the three cuts remaining.)`

`=>  D`

NETWORKS, FUR1 2014 VCAA 6-7 MC

Consider the following four graphs.
 


 

Part 1

How many of these four graphs have an Eulerian circuit?

A.  `0`

B.  `1`

C.  `2`

D.  `3`

E.  `4`

 

Part 2

How many of these four graphs are planar?

A.  `0`

B.  `1`

C.  `2`

D.  `3`

E.  `4`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(An Euler circuit cannot exist if any vertices)`

`text(have an odd degree.)`

`=>  B` 

♦♦♦ Mean mark part 2: 7%!
MARKER’S COMMENT: A majority of students chose option A, not understanding that a graph with intersecting edges can still be planar.

 

`text(Part 2)`

`=>  E`

 

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. i. 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. ii. 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 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 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.
  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 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`

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[(qquadqquadqquadqquadqquad),(),()][(qquadquad),(qquadquad),(qquadquad)] = [(qquadquad),(qquadquad),(qquadquad)]`
     

     

  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 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.
    5. When does this first occur?   (1 mark)

    6. 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.
  4. Assuming 30% of adults lay eggs each week, the population matrix after one week, `S_1`, is now given by
  5. `qquad S_1 = TS_0 + BS_0`
  6. 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)

    2. This pattern continues. The population matrix after `n` weeks, `S_n`, is given by
    3. `qquad qquad qquad S_n = TS_(n - 1) + BS_(n - 1)`
    4. Determine the number of live eggs in this insect population after two weeks.  (1 mark)

Show Answers Only
  1. `700`
  2. `50text(%)`
  3.  
    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 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.
    2. Plot the corresponding points on the graph.   (2 marks)

    3. Use matrices to determine the expected total number of female ducks in the colony in the long term.
    4. 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)]`

`\Rightarrow 61\ text(ducks at end of year 4.)`

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

`\Rightarrow 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 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.
  2. The matrix describing the population after one year, `S_1`, is now given by the new rule
  3. `S_1 = T S_0 + 500\ M\ S_0`
  4. 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)

  5. This pattern continues so that the matrix describing the population after `n` years, `S_n`, is given by the rule
  6.       `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

a.i.   `6000`

a.ii.  `text(See Worked Solutions)`

b.i.   `S_1= [(0),(4000),(650),(7150)]`

b.ii. `S_4= [(0),(0),(331.25),(11\ 468.75)]`

c.i.   `text{13 years}`

c.ii.  `1325`

d.    `text(Add 10 000 eggs, remove 3000 baby trout and add 150)`

`text(150 adult trout to keep the population constant.)`

e.i.   `S_1= [(200\ 000),(4000),(650),(7150)]`

e.ii.   `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 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 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 and add 150 adult)`

`text(trout to keep the 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)]`

MATRICES, FUR1 2008 VCAA 7-9 MC

A large population of mutton birds migrates each year to a remote island to nest and breed. There are four nesting sites on the island, A, B, C and D.

Researchers suggest that the following transition matrix can be used to predict the number of mutton birds nesting at each of the four sites in subsequent years. An equivalent transition diagram is also given.
 

`{:(qquad qquad qquad qquad {:text(this year):}), (qquad qquad quad quad \ {:(A,\ \ B,\ \ C,\ D):}), (T = [(0.4, 0, 0.2, 0),(0.35, 1, 0.15, 0), (0.15, 0, 0.55, 0), (0.1, 0, 0.1, 1)] {:(A), (B), (C), (D):} quad {:text(next year):}):}`      VCAA MATRICES FUR2 2008 7i

Part 1

Two thousand eight hundred mutton birds nest at site C in 2008.

Of these 2800 mutton birds, the number that nest at site A in 2009 is predicted to be

A.    `560`

B.    `980`

C.  `1680`

D.  `2800`

E.  `3360`

 

Part 2

This transition matrix predicts that, in the long term, the mutton birds will

A.  nest only at site A.

B.  nest only at site B.

C.  nest only at site A and C.

D.  nest only at site B and D.

E.  continue to nest at all four sites.

 

Part 3

Six thousand mutton birds nest at site B in 2008.

Assume that an equal number of mutton birds nested at each of the four sites in 2007. The same transition matrix applies.

The total number of mutton birds that nested on the island in 2007 was

A.  `6000`

B.  `8000`

C.  `12\ 000`

D.  `16\ 000`

E.  `24\ 000`

Show Answers Only

`text(Part 1:)    A`

`text(Part 2:)   D`

`text(Part 3:)   D`

Show Worked Solution

`text(Part 1:)`

`text(20% of birds at site)\ A\ text(in 2008 are predicted)`

`text(to move to site)\ C.`

`:.\ text(Number of birds)`

`= 20 text(%) xx 2800`

`= 560`

`=>   A`

 

`text(Part 2:)`

`text(Consider)\ n\ text{large  (say}\ n = 50 text{)},`

`T^50 = [(0, 0, 0, 0), (0.75, 1, 0.66, 0), (0, 0, 0, 0), (0.25, 0, 0.33, 0)]`

`=>   D`

 

`text(Part 3:)`

`text(Let)\ \ x = text(mutton birds at each site in 2007)`

`text(In 2008, 6000 birds nest at)\ B.`

♦♦♦ Mean mark 25%.

`text(Using the diagram,)`

`x + 0.35x + 0.15x` `= 6000`
 `1.5x` `= 6000`
 `x` `= 4000`

 

`:.\ text(Total number nested in 2007)`

`= 4 xx 4000`

`= 16\ 000\ text(birds)`

`=>   D`

MATRICES, FUR1 2010 VCAA 9 MC

Robbie completed a test of four multiple-choice questions.

Each question had four alternatives, A, B, C or D.

Robbie randomly guessed the answer to the first question.

He then determined his answers to the remaining three questions by following the transition matrix
 

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

Which of the following statements is true?

A.  It is impossible for Robbie to give the same answer to all four questions.

B.  Robbie would always give the same answer to the first and fourth questions

C.  Robbie would always give the same answer to the second and third questions.

D.  If Robbie answered A for question one, he would have answered B for question two

E.  It is possible that Robbie gave the same answer to exactly three of the four questions.

Show Answers Only

`B`

Show Worked Solution

`text(Consider when the first guess is)\ \ C,`

♦♦ Mean mark 29%.

`text(i.e. the state matrix is)\ \ [(0), (0), (1), (0)]`

`text(Using the transition matrix,)`

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

`:.\ text(The 2nd answer is)\ \ B`

 

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

`:.\ text(The 3rd answer is)\ \ D`

 

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

`:.\ text(The 4th answer is)\ \ C`

 

`text(Other guesses for the first answer can)`

`text(be similarly followed to show)\ B\ text(is true.)`

`=>   B`

MATRICES, FUR1 2010 VCAA 6 MC

Vince, Nev and Rani all service office equipment.

The matrix `T` shows the time that it takes (in minutes) for each of Vince (V), Nev (N) and Rani (R) to service a photocopier (P) a fax machine (F) and a scanner (S).
 

`{:(qquad qquad qquad {:(V,\ N,\ R):}), (T = [(12, 15, 14),(8, 7, 8), (20, 19, 17)] {:(P), (F), (S):}):}`
 

The matrix `U` below displays the number of photocopiers, fax machines and scanners to be serviced in three schools, Alton (A), Borton (B) and Carlon (C).
 

`{:(qquad qquad qquad {:(P, F, S):}), (U = [(5,\ 3,\ 2),(4,\ 4,\ 3), (6,\ 1,\ 2)] {:(A), (B), (C):}):}`
 

A matrix that displays the time that it would take each of Vince, Nev and Rani, working alone, to service the photocopiers, fax machines and scanners in each of the three schools is

A.  `[(17, 18, 16), (12, 11, 11), (26, 20, 19)]` B.  `[(204, 110, 97), (116, 60, 53), (278, 153, 131)]`
       
C.  `[(124, 134, 128), (140, 145, 139), (120, 135, 126)]` D.  `[(7, 12, 12), (4, 3, 5), (14, 18, 15)]`
       
E.  `[(60, 15, 28), (32, 35, 24), (120, 19, 34)]`    

 

Show Answers Only

`C`

Show Worked Solution

`text(Matrix)\ \ UT\ \ text(gives the information.)`

♦♦♦ Mean mark 19%.
MARKER’S COMMENT: Most students incorrectly calculated matrix `TU`. Make sure you know why this is incorrect!
`UT` `= [(5, 3, 2), (4, 4, 3), (6, 1, 2)] [(12, 15, 14), (8, 7, 8), (20, 19, 17)]`
  `= [(124, 134, 128), (140, 145, 139), (120, 135, 126)]`

 
`=>   C`

MATRICES, FUR1 2010 VCAA 3 MC

The total cost of one ice cream and three soft drinks at Catherine’s shop is $9.

The total cost of two ice creams and five soft drinks is $16.

Let `x` be the cost of an ice cream and `y` be the cost of a soft drink

The matrix `[(x), (y)]` is equal to
 

A.  `[(1, 3), (2, 5)] [(x), (y)]`

B.  `[(1, 3), (2, 5)] [(9), (16)]`

C.  `[(1, 2), (3, 5)] [(9), (16)]`

D.  `[(– 5, 2), (3, – 1)] [(9), (16)]`

E.  `[(– 5, 3), (2, – 1)] [(9), (16)]`

Show Answers Only

`E`

Show Worked Solution

`[(1, 3), (2, 5)][(x), (y)] = [(9), (16)]`

♦♦ Mean mark 35%.
`:. [(x), (y)]` `= [(1, 3), (2, 5)]^-1 [(9), (16)]`
  `= [(– 5, 3), (2, – 1)] [(9), (16)]`

 
`=>   E`

MATRICES, FUR1 2013 VCAA 9 MC

`P, Q, R`  and  `S`  are matrices such that the matrix product  `P = QRS`  is defined.

Matrix  `Q`  and matrix  `S`  are square, non-zero matrices for which  `Q + S`  is not defined.

Which one of the following matrix expressions is defined?

A.   `R - S`

B.   `Q + R`

C.   `P^2`

D.   `R^-1`

E.   `P × S`

Show Answers Only

`E`

Show Worked Solution

`text(If)\ Q(n xx n)\ text(and)\ S(m xx m)\ text(are square)`

♦ Mean mark 39%.

`text(matrices and)\ Q + S\ text(is not defined,)`

`m != n.`

`text(S)text(ince)\ QRS\ text(is defined, we can deduce)`

`text(that)\ R\ text(is)\ n xx m,\ text(and)\ P\ text(is)\ n xx m.`

`text(Consider)\ E,`

`P xx S\ text(is defined because the number)`

`text(of columns in)\ P(m)\ text(equals the number)`

`text(of rows in)\ S(m).`

`text(All other options can be shown to not)`

`text(be defined.)`

`rArr E`

MATRICES, FUR1 2006 VCAA 6 MC

If  `A = [(1,3), (6,4), (0,0)]`  and the matrix product  `XA = [(4,1),(1,4), (3,5)],` then the order matrix `X` is

A.   `(2 xx 2)`

B.   `(2 xx 3)`

C.   `(3 xx 1)`

D.   `(3 xx 2)`

E.   `(3 xx 3)`

Show Answers Only

`E`

Show Worked Solution
   `X`  `xx`  `A` `=`    `XA`
`a xx b`    `3 xx 2`      `3 xx 2`
♦♦♦ Mean mark 23%.

 

`b = 3\ \ text{(columns in}\ X = text(rows in)\ A)`

`a = 3\ \ text{(rows in}\ X = text(rows in)\ XA)`

`:. XA\ text(is order)\ 3 xx 3.`

`rArr E`

MATRICES, FUR1 2011 VCAA 9 MC

Matrix `A` is a 3 x 3  matrix. Seven of the elements in matrix `A` are zero.

Matrix `B` contains six elements, none of which are zero.

Assuming the matrix product  `AB`  is defined, the minimum number of zero elements in the product matrix  `AB`  is

A.   `0`

B.   `1`

C.   `2`

D.   `4`

E.   `6`

Show Answers Only

`C`

Show Worked Solution

`text(Matrix)\ B\ text(is)\ 3 xx 2`

`text(If non-zero elements of matrix)\ A`

`text(are in the same row,)`
 

`[(a,0,b),(0,0,0),(0,0,0)][(c,d),(e,f),(g,h)] = [(x,y),(0,0),(0,0)]`
 

`text(where)\ a,b,c …, x, y != 0`

 

`text(If non-zero elements of matrix)\ B`

`text(are not in the same row,)`
 

`[(a,0,0),(b,0,0),(0,0,0)][(c,d),(e,f),(g,h)] = [(ac,ad),(bc,bd),(0,0)]`

 

`:.\ text(Minimum number of zero elements)`

`text(in matrix)\ AB\ text(is 2.)`

`=> C`

MATRICES, FUR2 2015 VCAA 3

A new model for the number of students in the school after each assessment takes into account the number of students who are expected to leave the school after each assessment.

After each assessment, students are classified as beginner (`B`), intermediate (`I`), advanced (`A`) or left the school (`L`).

Let matrix `T_2` be the transition matrix for this new model.

Matrix `T_2`, shown below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.
  

`{:(qquadqquadqquadquadtext(before assessment)),(qquadqquadqquadquad{:(B,\ qquadI,qquadA,quadL):}),(T_2 = [(0.30,0,0,0),(0.40,0.70,0,0),(0.05,0.20,0.75,0),(0.25,0.10,0.25,1)]{:(B),(I),(A),(L):}qquadtext(after assessment)):}`

  1. An incomplete transition diagram for matrix `T_2` is shown below.
  2. Complete the transition diagram by adding the missing information.   (2 marks)


     

    Matrices, FUR2 2015 VCAA 3

The number of students at each level, immediately before the first assessment of the year, is shown in matrix `R_0` below.

`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`

Matrix `T_2`, repeated below, contains the percentages of students who are expected to change their ability level or leave the school after each assessment.

`{:(qquadqquadqquadquadtext(before assessment)),(qquadqquadqquadquad{:(B,\ qquadI,qquadA,quadL):}),(T_2 = [(0.30,0,0,0),(0.40,0.70,0,0),(0.05,0.20,0.75,0),(0.25,0.10,0.25,1)]{:(B),(I),(A),(L):}qquadtext(after assessment)):}`

  1. What percentage of students is expected to leave the school after the first assessment?   (1 mark)

  2. How many advanced-level students are expected to be in the school after two assessments.
  3. Write your answer correct to the nearest whole number.   (1 mark)

  4. After how many assessments is the number of students in the school, correct to the nearest whole number, first expected to drop below 50?   (1 mark)

Another model for the number of students in the school after each assessment takes into account the number of students who are expected to join the school after each assessment.

Let `R_n` be the state matrix that contains the number of students in the school immediately after `n` assessments.

Let `V` be the matrix that contains the number of students who join the school after each assessment. 

Matrix `V` is shown below.

`V = [(4),(2),(3),(0)]{:(B),(I),(A),(L):}`

The expected number of students in the school after `n` assessments can be determined using the matrix equation

`R_(n + 1) = T_2 xx R_n + V`

where

`R_0 = [(20),(60),(40),(0)]{:(B),(I),(A),(L):}`
 

  1. Consider the intermediate-level students expected to be in the school after three assessments.
  2. How many are expected to become advanced-level students after the next assessment?
  3. Write your answer correct to the nearest whole number.   (2 marks)

Show Answers Only
  1.  
  2. `17.5 text(%)`
  3. `43`
  4. `5`
  5. `7`
Show Worked Solution
a.   

 

b.   `T_2 R_0` `= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)] [(20), (60), (40), (0)]= [(6), (50), (43), (21)]`

 
`:.\ text(Percentage that leave school)`

♦♦ Mean mark of parts (b)-(e) was 32%.

`= 21/120`

`= 17.5 text(%)`
 

c.   `text(After 2 assessments,)`

`(T_2)^2 R_0` `= [(0.3, 0, 0, 0), (0.4, 0.7, 0, 0), (0.05, 0.2, 0.75, 0), (0.25, 0.1, 0.25, 1)]^2 [(20), (60), (40), (0)]=[(1.8), (37.4), (42.55), (38.25)]`

 
`:.\ text(43 advanced-level students expected to remain.)`
 

d.   `text(After 4 assessments,)`

`(T_2)^4 R_0\ text(shows 54 students left)`

`text(After 5 assessments,)`

`(T_2)^5 R_0\ text(shows 43 students left.)`

`:.\ text(Numbers drop below 50 after 5 assessments.)`
 

MARKER’S COMMENT: Understand why the common error `R_3 =` `(T_2)^3R_0 + V` is incorrect!
e.   `R_1` `= T_2 R_0 + V`
    `= [(6), (50), (43), (21)] + [(4), (2), (3), (0)] = [(10), (52), (46), (21)]`
  `R_2` `= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(10), (52), (46), (21)] + [(4), (2), (3), (0)] = [(7), (42.4), (48.4), (40.2)]`
   `R_3` `= [(0.30, 0, 0, 0), (0.40, 0.70, 0, 0), (0.05, 0.20, 0.75, 0), (0.25, 0.10, 0.25, 1)] [(7), (42.4), (48.4), (40.2)] + [(4), (2), (3), (0)] = [(6.1), (34.48), (48.13), (58.29)]`

 

`:.\ text(After 3 assessments, the number expected to)`

`text(move from Intermediate to Advanced)`

`= 20text(%) xx 34.48`

`= 6.896`

`= 7\ text{students  (nearest whole)}`

MATRICES, FUR2 2012 VCAA 3

When a new industrial site was established at the beginning of 2011, there were 350 staff at the site. 

The staff comprised 100 apprentices (`A`), 200 operators (`O`) and 50 professionals (`P`). 

At the beginning of each year, staff can choose to stay in the same job, move to a different job at the site or leave the site (`L`).

The number of staff in each category at the beginning of 2011 is given in the matrix

`S_2011 = [(100), (200), (50), (0)]{:(A), (O), (P), (L):}`

The transition diagram below shows the way in which staff are expected to change their jobs at the site each year.

Matrices, FUR2 2012 VCAA 3

  1. How many staff at the site are expected to be working in their same jobs after one year?   (1 mark)

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

`{:(qquad qquad qquad qquad qquad qquad text(this year)),((qquad qquad qquad\ A, qquad O, qquad P, qquad L)),(T = [(0.70, 0, 0, 0),(0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)]):} {:(), (), (A), (O), (P), (L):} {:(), (), (qquad text(next year)):}`

  1. Explain the meaning of the entry in the fourth row and fourth column of transition matrix `T`.   (1 mark)

If staff at the site continue to change their jobs in this way, the matrix `S_n` will contain the number of apprentices (`A`), operators (`O`), professionals (`P`) and staff who leave the site (`L`) at the beginning of the `n`th year.

  1. Use the rule  `S_(n + 1) = TS_n`  to find

     

    1. `S_2012`   (1 mark)

    2. the expected number of operators at the site at the beginning of 2013   (1 mark)

    3. the beginning of which year the number of operators at the site first drops below 30   (1 mark)

    4. the total number of staff at the site in the longer term.   (1 mark)

Suppose the manager decides to bring 30 new apprentices, 20 new operators and 10 new professionals to the site at the beginning of each year.

The matrix `S_(n + 1)` will then be given by

`S_(n + 1) = T S_n + A`   where   `S_2011 = [(100), (200), (50), (0)] {:(A), (O), (P), (L):}`   and   `A = [(30), (20), (10), (0)] {:(A), (O), (P), (L):}`

  1. Find the expected number of operators at the site at the beginning of 2013.   (2 marks) 

Show Answers Only
  1. `275`
  2. `text(Any worker who has left the)`
    `text(site will not return.)`
    1. `[(70), (170), (65), (45)]`
    2. `143`
    3. `2021`
    4. `0`
  4. `182`
Show Worked Solution

a.  `text(Same jobs after 1 year)`

♦ Mean mark for parts (a)-(d) (combined) was 37%.

`= 0.7 xx 100 + 0.8 xx 200 + 0.9 xx 50`

`= 275`
 

b.    `text(Any worker who has left the site will not return.)`

MARKER’S COMMENT: Most students couldn’t explain the meaning of the 1.0 figure in this context.

 

c.i.   `S_2012` `= T S_2011`
    `= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (200), (50), (0)]`
    `= [(70),(150),(65),(45)]`

 

c.ii.   `S_2013` `= T S_2012`
    `= [(49), (143), (75.5), (82.5)]`

 
`:.\ text(143 operators are expected at the site at the start of 2013.)`
 

c.iii.   `S_2020` `= T^9 S_2011`
    `= [(0.70, 0, 0, 0), (0.10, 0.8, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)]^9 [(100), (200), (50), (0)] = [(4), (36.2), (78), (231.8)]`
  `S_2021` `= T^10 S_2011`
    `= [(2.8), (29.4), (73.8), (244)]`

 

`:.\ text(At the start of 2021, the number of operators drops below 30.)`

MARKER’S COMMENT: “In the 10th year” recieved no marks – make sure you answer with specific years when actual years are used.

 
c.iv.   `text(Consider)\ n\ text(large:)`

`T^100 S_2011 = [(0), (0), (0), (350)]`

`:.\ text(NO staff remain at the site in the long term.)`
 

d.   `S_2012` `= T S_2011 + A`
    `= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (200), (50), (0)] + [(30), (20), (10), (0)]= [(100), (190), (75), (45)]`
     
  `S_2013` `= T S_2012 + A`
    `= [(0.70, 0, 0, 0), (0.10, 0.80, 0, 0), (0, 0.10, 0.90, 0), (0.20, 0.10, 0.10, 1.00)] [(100), (190), (75), (45)] + [(30), (20), (10), (0)]= [(100), (182), (96.5), (91.5)]`
MARKER’S COMMENT: A common error was  `S_2013“=T^2S_2011“+A`. Know why this is not correct!

 

`:.\ text(182 operators are expected on site at the )`

`text(start of 2013.)`