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Probability, MET1 2014 VCAA 9

Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.

Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.

  1. In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning.
  2. Find the probability that Sally walked Mack on at least one of these two mornings.  (2 marks)

  3. In the month of April, the probability of pleasant weather in the morning was `5/8`.
  4.  i. Find the probability that on a particular morning in April, Sally walked Mack.  (2 marks)

  5. ii. Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning.  (2 marks)

Show Answers Only

  1. `5/6`
    1. `19/32`
    2. `15/19`

Show Worked Solution

a.    `text{Pr(at least 1 walk)}` `= 1 – text{Pr(no walk)}`
    `= 1 – 1/4 xx 2/3`
    `= 5/6`

 

b.i.   `text(Construct tree diagram:)`
 

met1-2014-vcaa-q9-answer1 
 

`text(Pr)(PW) + text(Pr)(P′W)` `= 5/8 xx 3/4 + 3/8 xx 1/3`
  `= 19/32`

 

♦ Part (b)(ii) mean mark 38%.

b.ii.    `text(Pr)(P | W)` `= (text(Pr)(P ∩ W))/(text(Pr)(W))`
    `= (5/8 xx 3/4)/(19/32)`
    `= 15/32 xx 32/19`
    `= 15/19`

Probability, MET1 2014 VCAA 8

A continuous random variable, `X`, has a probability density function given by
 

`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`
 

The median of `X` is  `m`.

  1. Determine the value of  `m`.  (2 marks)
  2. The value of `m` is a number greater than 1.

     

    Find `text(Pr)(X < 1 | X <= m)`.  (2 marks)

Show Answers Only
  1. `−5log_e(1/2)\ \ text(or)\ \ 5log_e(2)\ \ text(or)\ \ log_e 32`
  2. `2(1 – e^(−1/5))`
Show Worked Solution
a.    `1/5 int_0^m e^(−x/5)dx` `= 1/2`
  `1/5 xx (−5)[e^(−x/5)]_0^m` `= 1/2`
  `[-e^(- x/5)]_0^m` `= 1/2`
  `-e^(−m/5) + 1` `= 1/2`
  `e^(−m/5)` `= 1/2`
  `- m/5` `= log_e(1/2)`

 

`:. m = −5log_e(1/2)\ \ \ (text(or)\ \ 5log_e(2),\ text(or)\ \ log_e 32)`

 

b.   `text(Using Conditional Probability:)`

♦ Part (b) mean mark 38%.
MARKER’S COMMENT: A common error was assuming `m` obtained in part (a) was equivalent to `text(Pr)(X<=m)`.
`text(Pr)(X < 1 | X <= m)` `= (text(Pr)(X < 1))/(text(Pr)(X <= m))`
  `= (1/5 int_0^1 e^(−x/5)dx)/(1/2)`
  `= (1/5(−5)[e^(−x/5)]_0^1)/(1/2)`
  `= −2[(e^(−1/5)) – e^0]`
  `= 2(1 – e^(−1/5))`

Calculus, MET1 2014 VCAA 7

If  `f^{prime}(x) = 2cos(x)-sin(2x)`  and  `f(pi/2) = 1/2`,  find  `f(x)`.   (3 marks)

Show Answers Only

`2sinx + 1/2cos(2x)-1`

Show Worked Solution
`f(x)` `= int(2cosx-sin2x)dx`
  `= 2sinx + 1/2cos(2x) + c`

  
`text(Substitute)\ \ f(pi/2) = 1/2:`

`1/2` `= 2sin(pi/2) + 1/2cos(pi) + c`
`1/2` `= 2-1/2 + c`
`c` `=-1`
`:. f(x)` `= 2sinx + 1/2cos(2x)-1`

Calculus, MET1 2014 VCAA 5

Consider the function  `f:[−1,3] -> R`,  `f(x) = 3x^2-x^3`.

  1. Find the coordinates of the stationary points of the function.   (2 marks)

  2. On the axes  below, sketch the graph of `f`.

     

    Label any end points with their coordinates.   (2 marks)

     

     
        met1-2014-vcaa-q5

     

  3. Find the area enclosed by the graph of the function and the horizontal line given by  `y = 4`.   (3 marks)

Show Answers Only
  1. `(0, 0) and (2, 4)`
  2.  
    met1-2014-vcaa-q5-answer3
  3. `27/4\ text(u²)`
Show Worked Solution
a.    `text(SP’s occur when)\ \ f^{′}(x)` `= 0`
`6x-3x^2`  `= 0` 
 `3x(2-x)` `=0`

`x = 0,\ \ text(or)\ \ 2`
 

`:.\ text{Coordinates are (0, 0) and (2, 4)}`

 

b.    met1-2014-vcaa-q5-answer3

 

♦ Mean mark (c) 48%.
c.    met1-2014-vcaa-q5-answer4

`text(Solution 1)`

`text(Area)` `= int_(−1)^2 4-(3x^2-x^3)dx`
  `= int_(−1)^2 4-3x^2 + x^3dx`
  `= [4x-x^3 + 1/4x^4]_(−1)^2`
  `= (8-8 + 4)-(−4-(−1) + 1/4)`
   
`:.\ text(Area)` `= 27/4 text(units²)`

 

`text(Solution 2)`

`text(Area)` `= 12-int_(−1)^2(3x^2-x^3)dx`
  `= 12-[x^3-1/4 x^4]_(−1)^2`
  `= 12-[(8-4)-(−1-1/4)]`
  `= 27/4\ text(units²)`

Functions, MET1 2014 VCAA 3

Solve  `2cos(2x) = -sqrt3`  for  `x`, where  `0 <= x <= pi`.   (2 marks)

Show Answers Only

`x = (5pi)/12, (7pi)/12`

Show Worked Solution
`cos(2x)` `= -sqrt3/2`
`2x` `= (5pi)/6, 2pi-(5pi)/6, 2pi+(5pi)/6`
  `=(5pi)/6, (7pi)/6, (17pi)/6,\ …`
`:. x` `=(5pi)/12, (7pi)/12\ \ \ text(for)\ \ 0 <= x <= pi`.

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`

Probability, MET1 2015 VCAA 8

For events `A` and `B` from a sample space, `text(Pr)(A | B) = 3/4`  and  `text(Pr)(B) = 1/3`.

  1. Calculate  `text(Pr)(A ∩ B)`.   (1 mark)

  2. Calculate  `text(Pr)(A^{′} ∩ B)`, where `A^{′}` denotes the complement of `A`.   (1 mark)

  3. If events `A` and `B` are independent, calculate  `text(Pr)(A ∪ B)`.   (1 mark)

Show Answers Only

  1. `1/4`
  2. `1/12`
  3. `5/6`

Show Worked Solution

a.   `text(Using Conditional Probability:)`

`text(Pr)(A | B)` `= (text(Pr)(A ∩ B))/(text(Pr)(B))`
`3/4` `= (text(Pr)(A ∩ B))/(1/3)`
`:. text(Pr)(A ∩ B)` `= 1/4`

 

b.    met1-2015-vcaa-q8-answer
`text(Pr)(A^{′} ∩ B)` `= text(Pr)(B)-text(Pr)(A ∩B)`
  `= 1/3-1/4`
  `= 1/12`

 

c.   `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.

`text(Pr)(A ∩ B)` `= text(Pr)(A) xx Pr(B)`
`1/4` `= text(Pr)(A) xx 1/3`
`:. text(Pr)(A)` `= 3/4`

 

`text(Pr)(A ∪ B)` `= text(Pr)(A) + text(Pr)(B)-text(Pr)(A ∩ B)`
  `= 3/4 + 1/3-1/4`
`:. text(Pr)(A ∪ B)` `= 5/6`

Functions, MET1 2015 VCAA 5

On any given day, the depth of water in a river is modelled by the function

`h(t) = 14 + 8sin((pit)/12),\ \ 0 <= t <= 24`

where `h` is the depth of water, in metres, and  `t`  is the time, in hours, after 6 am. 

  1. Find the minimum depth of the water in the river.   (1 mark)

  2. Find the values of  `t`  for which  `h(t) = 10`.   (2 marks)

Show Answers Only
  1. `6\ text(m)`
  2. `14quadtext(or)quad22`
Show Worked Solution

a.   `h_(text(min))\ text(occurs when)\ \ sin((pit)/12)=-1`

MARKER’S COMMENT: Students who used calculus to find the minimum were less successful.
`:. h_(text(min))` `= 14-8`
  `= 6\ text(m)`

 

b.    `14 + 8sin(pi/12t)` `= 10`
  `sin(pi/12t)` `=-1/2`

 

`text(Solve in general:)`

`pi/12t` `=(7pi)/6 + 2pi n\ \ \ \ text(or)\ \ \ `  `pi/12t` `= (11t)/6 + 2pi n,`
`t` `= 14 + 24n` `t` `=22 + 24n`

 

`text(Substitute integer values for)\ n,`

`:. t = 14quadtext(or)quad22,\ \ \ (t ∈ [0,24])`

Calculus, MET1 2015 VCAA 4

Consider the function  `f:[-3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2-4)`.

  1. Find the coordinates of the stationary points of the function.   (2 marks)

The rule for  `f` can also be expressed as  `f(x) = 1/2(x-1)(x + 2)^2`.

  1. On the axes below, sketch the graph of  `f`, clearly indicating axis intercepts and turning points.

     

    Label the end points with their coordinates.   (2 marks)


      

    met1-2015-vcaa-q4
     

  2. Find the average value of  `f` over the interval  `0<=x<=2.`   (2 marks)

Show Answers Only
  1. `(0,-2), (-2,0)`
  2.  

     

    met1-2015-vcaa-q4-answer

  3. `1`
Show Worked Solution

a.   `text(Stationary points when)\ \ f^{prime}(x)=0,`

`1/2(3x^2 + 6x)` `= 0`
`3x(x + 2)` `= 0`

 

`:. x = 0, -2`

 

`:.\ text(Coordinates of stationary points:)`

`(0, -2), (-2,0)`

 

b.    met1-2015-vcaa-q4-answer
♦ Part (c) mean mark 50%.
MARKER’S COMMENT: Most students recalled the average value definition but then did not integrate correctly.

 

c.    `text(Avg value)` `= 1/(2-0) int_0^2 f(x) dx`
    `= 1/2 int_0^2 1/2(x^3 + 3x^2-4)dx`
    `= 1/4[1/4x^4 + x^3-4x]_0^2`
    `= 1/4[(16/4 + 2^3-4(2))-0]`
    `= 1`

Calculus, MET1 2015 VCAA 3

Evaluate  `int_1^4 (1/sqrtx)\ dx`.  (2 marks)

Show Answers Only

`2`

Show Worked Solution
MARKER’S COMMENT: This basic integral caused problems with many students answering with a log function.
`int_1^4 x^(−1/2)\ dx` `= 2[x^(1/2)]_1^4`
  `= 2[4^(1/2) – 1^(1/2)]`
  `= 2(2 – 1)`
  `= 2`

Calculus, MET1 2015 VCAA 2

Let  `f^{prime}(x) = 1-3/x`, where  `x != 0`.

Given that  `f(e) =-2`, find  `f(x)`.   (3 marks)

Show Answers Only

`x-3log_e(x) + 1-e`

Show Worked Solution
`f(x)` `= int 1-3x^(-1) dx`
`f(x)` `= x-3log_e |\ x\ | + c`

 

`text(Substitute)\ \ f(e) =-2,`

`-2` `= e-3log_e(e) + c`
`c` `= 1-e`

 

`:. f(x) = x-3log_e|\ x\ | + 1-e`

CORE*, FUR2 2006 VCAA 3

The company prepares for this expenditure by establishing three different investments.

  1. $7000 is invested at a simple interest rate of 6.25% per annum for eight years.
  2. Determine the total value of this investment at the end of eight years.   (2 marks)
  3. $10 000 is invested at an interest rate of 6% per annum compounding quarterly for eight years.
  4. Determine the total value of this investment at the end of eight years.
  5. Write your answer correct to the nearest dollar.   (1 mark)
  6. $500 is deposited into an account with an interest rate of 6.5% per annum compounding monthly.
  7. Deposits of $200 are made to this account on the last day of each month after interest has been paid.
  8. Determine the total value of this investment at the end of eight years.
  9. Write your answer correct to the nearest dollar.   (1 mark) 

Show Answers Only

  1. `$10\ 500`
  2. `$16\ 103`
  3. `$25\ 935`

Show Worked Solution

a.    `I` `= (PrT)/100`
    `= (7000 xx 6.25 xx 8)/100`
    `= $3500`

  
`:.\ text(Total value of investment)`

`= 7000 + 3500`

`= $10\ 500`
    

b.   `text(Compounding periods) = 8 xx 4 = 32`

`text(Interest rate)` `= (text(6%))/4`
  `= 1.5text(%  per quarter)`

  
`:.\ text(Total value of investment)`

`= PR^n`

`= 10\ 000(1.015)^32`

`= 16\ 103.24…`

`= $16\ 103\ \ text{(nearest $)}`
  

c.   `text(By TVM Solver,)`

`N` `= 8 xx 12 = 96`
`I(text(%))` `= 6.5`
`PV` `= 500`
`PMT` `= 200`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

   
`=> FV = −25\ 935.30…`

`:.\ text(Total value of investment is $25 935.)`

CORE*, FUR2 2006 VCAA 2

It is estimated that inflation will average 2% per annum over the next eight years.

If a new machine costs $60 000 now, calculate the cost of a similar new machine in eight years time, adjusted for inflation. Assume no other cost change.

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

Show Answers Only

`$70\ 300\ \ text{(nearest $)}`

Show Worked Solution

`text(Find value)\ (A)\ text(in 8 years.)`

`A` `= PR^n`
  `= 60\ 000(1.02)^8`
  `= 70\ 299.562…`
  `= $70\ 300\ \ text{(nearest $)}`

CORE*, FUR2 2006 VCAA 1

A company purchased a machine for $60 000.

For taxation purposes the machine is depreciated over time.

Two methods of depreciation are considered.

  1. Flat rate depreciation

    The machine is depreciated at a flat rate of 10% of the purchase price each year.

    i.      By how many dollars will the machine depreciate annually?   (1 mark)

    ii.  Calculate the value of the machine after three years.   (1 mark)

  2. iii. After how many years will the machine be $12 000 in value?   (1 mark)

  3. Reducing balance depreciation

    The value, `V`, of the machine after `n` years is given by the formula `V=60\ 000 xx(0.85)^n`.

    i.      By what percentage will the machine depreciate annually?   (1 mark)

    ii.  Calculate the value of the machine after three years.   (1 mark)

  4. iii. At the end of which year will the machine's value first fall below $12 000?   (1 mark)

  1. At the end of which year will the value of the machine first be less using flat rate depreciation than it will be using reducing balance depreciation?  (2 marks)

Show Answers Only
  1. i.   `$6000`
    ii.  `$42\ 000`
    iii. `8\ text(years)`
  2. i.   `text(15%)` 
    ii.      `$36\ 847.50`
    iii. `10\ text(years)`
  3. `text(7th year)`
Show Worked Solution
a.i.    `text(Annual depreciation)` `= 10text(%) xx 60\ 000`
    `= $6000`

a.ii.   `text(After 3 years,)`

`text(Value)` `= 60\ 000-(3 xx 6000)`
  `= $42\ 000`

a.iii.   `text(Find)\ n\ text(when value = $12 000)`

`12\ 000` `= 60\ 000-6000 xx n`
`6000n` `= 48\ 000`
`:.n` `=(48\ 000)/6000`
  `= 8\ text(years)`

 

b.i.    `1-r` `= 0.85`
   `r` `= 0.15`

`:.\ text(Annual depreciation is 15%.)`
  

b.ii.   `text(After 3 years,)`

`text(Value)` `= 60\ 000 xx (0.85)^3`
  `= $36\ 847.50`

 

b.iii.   `text(Find)\ n\ text(when)\ \ V = $12\ 000`

`12\ 000` `= 60\ 000 xx (0.85)^n`
`(0.85)^n` `= 0.2`
`:. n` `= 9.90…\ \ text(years)`

  
`:.\ text(Machine value falls below $12 000)`

`text(after 10 years.)`
  

c.   `text(Sketching both graphs,)`

BUSINESS, FUR2 2006 VCAA 1 Answer

`text(From the graph, at the end of the 7th year the)`

`text(value using flat rate drops below reducing)`

`text(balance for the 1st time.)`

CORE*, FUR2 2007 VCAA 3

Khan paid $900 for a fax machine.

This price includes 10% GST (goods and services tax).

  1. Determine the price of the fax machine before GST was added. Write your answer correct to the nearest cent.  (1 mark)

  2. Khan will depreciate his $900 fax machine for taxation purposes.
  3. He considers two methods of depreciation.
  4. Flat rate depreciation
  5. Under flat rate depreciation the fax machine will be valued at $300 after five years.
    1. Calculate the annual depreciation in dollars.   (1 mark)
    2. ii. Determine the value of the fax machine after five years.   (1 mark)

Show Answers Only

  1. `$818.18`
  2. i. `$120`
    ii. `$325` 

Show Worked Solution

a.   `text(Let $)P = text(price ex-GST)`

MARKER’S COMMENT: Reverse GST questions continue to cause problems for many students.

`:. P + 10text(%)P` `= 900`
`1.1P` `= 900`
`P` `= 900/1.1`
  `= 818.181…`
  `= $818.18\ \ text(nearest cent)`

  
b.i.   `text(Annual depreciation)`

`= ((900-300))/5`

`= $120`

 

b.ii.   `text(Value after 5 years)`

`= 900-(250 xx 0.46 xx 5)`

`= $325`

CORE*, FUR2 2007 VCAA 1

Khan wants to buy some office furniture that is valued at $7000.

    1. A store requires 25% deposit. Calculate the deposit.   (1 mark)
    2. The balance is to be paid in 24 equal monthly instalments. No interest is charged.
    3. Determine the amount of each instalment. Write your answer in dollars and cents.   (1 mark)

Another store offers the same $7000 office furniture for $500 deposit and 36 monthly instalments of $220.

 

    1. Determine the total amount paid for the furniture at this store.   (1 mark)
    2. Calculate the annual flat rate of interest charged by this store.
    3. Write your answer as a percentage correct to one decimal place.   (2 marks)

A third store has the office furniture marked at $7000 but will give 15% discount if payment is made in cash at the time of sale.

  1. Calculate the cash price paid for the furniture after the discount is applied.   (1 mark) 

Show Answers Only

  1. i.  `$1750`
  2. ii.`$218.75`
  3. i.  `$8420`
  4. ii.  `7.3text{%}`
  5. `$5950`

Show Worked Solution

a.i.    `text(Deposit)` `= 25text(%) xx 7000`
    `= $1750`

 
a.ii.   `text(Installment amount)`

`= ((7000-1750))/24`

`= $218.75`
 

b.i.    `text(Total paid)` `= 500 + 36 xx 220`
    `= $8420`

 

b.ii.   `text(Total interest paid)`

`= 8420-7000`

`= $1420`

 

`I` `= (PrT)/100`
`1420` `= (6500 xx r xx 3)/100`
`:. r` `= (1420 xx 100)/(6500 xx 3)`
  `= 7.282…`
  `= 7.3text{%  (1 d.p.)}`

 

c.    `text(Cash price)` `= 7000-15text(%) xx 7000`
    `= 7000-1050`
    `= $5950`

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)}`

 

GRAPHS, FUR2 2007 VCAA 2

The Goldsmiths’ car can use either petrol or gas. 

The following equation models the fuel usage of petrol, `P`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.

`P = 12 - 0.02s`

The line  `P = 12 - 0.02s`  is drawn on the graph below for average speeds up to 110 km/h.

GRAPHS, FUR2 2007 VCAA 2

  1. Determine how many litres of petrol the car will use to travel 100 km at an average speed of 60 km/h.

     

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

The following equation models the fuel usage of gas, `G`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.

`G = 15 - 0.06s`

  1. On the axes above, draw the line  `G = 15 - 0.06s`  for average speeds up to 110 km/h.  (1 mark)
  2. Determine the average speeds for which fuel usage of gas will be less than fuel usage of petrol.  (1 mark)

The Goldsmiths'’ car travels at an average speed of 85 km/h. It is using gas.

Gas costs 80 cents per litre.

  1. Determine the cost of the gas used to travel 100 km.

     

    Write your answer in dollars and cents.  (2 marks)

Show Answers Only
  1. `10.8\ text(litres)`
  2.  
    GRAPHS, FUR2 2007 VCAA 2 Answer
  3. `text(75 km/hr) < text(speed) <= 110\ text(km/hr)`
  4. `$7.92`
Show Worked Solution

a.   `text(When)\ S = 60\ text(km/hr)`

`P` `= 12 – 0.02 xx 60`
  `= 10.8\ text(litres)`

 

b.   `text(When)\ s = 110,`

`G = 15 – 0.06 xx 100 = 8.4`

`:.\ text{(0, 15) and (110, 8.4) are on the line}`

GRAPHS, FUR2 2007 VCAA 2 Answer

 

c.   `text(Intersection of graphs occur when)`

MARKER’S COMMENT: Since fuel usage is less for gas, a speed of 75 km/hr was incorrect, as it was equal.
`12 – 0.02s` `= 15 – 0.06s`
`0.04s` `= 3`
`s` `= 75`

 

`:.\ text(Gas usage is less than fuel for)`

`text(average speeds over 75 km/hr.)`

 

d.   `text(When)\ x = 85,`

`text(Gas usage)` `= 15 – 0.06 xx 85`
  `= 9.9\ text(L/100 km.)`

 

`:.\ text(C)text(ost of gas for 100km journey)`

`= 9.9 xx 0.80`

`= $7.92`

GRAPHS, FUR2 2007 VCAA 1

The Goldsmith family are going on a driving holiday in Western Australia.

On the first day, they leave home at 8 am and drive to Watheroo then Geraldton.

The distance––time graph below shows their journey to Geraldton.

GRAPHS, FUR2 2007 VCAA 1

At 9.30 am the Goldsmiths arrive at Watheroo.

They stop for a period of time.

  1. For how many minutes did they stop at Watheroo?  (1 mark)

After leaving Watheroo, the Goldsmiths continue their journey and arrive in Geraldton at 12 pm.

  1. What distance (in kilometres) do they travel between Watheroo and Geraldton?  (1 mark)
  2. Calculate the Goldsmiths'’ average speed (in km/h) when travelling between Watheroo and Geraldton.  (1 mark)

The Goldsmiths leave Geraldton at 1 pm and drive to Hamelin. They travel at a constant speed of 80 km/h for three hours. They do not make any stops.

  1. On the graph above, draw a line segment representing their journey from Geraldton to Hamelin.  (1 mark)

 

Show Answers Only
  1. `text(30 minutes)`
  2. `190 \ text(km)`
  3. `95\ text(km/hr)`
  4.  
    GRAPHS, FUR2 2007 VCAA 1 Answer
Show Worked Solution

a.   `text(30 minutes)`

 

b.   `text(Distance travelled)`

`= 310 – 120`

`= 190\ text(km)`

 

c.   `text(Time taken = 2 hours)`

`:.\ text(Average speed)` `= 190/2`
  `= 95\ text(km/hr)`

 

d.    GRAPHS, FUR2 2007 VCAA 1 Answer

 

Calculus, MET2 2010 VCAA 1

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

     

         

    1. Find the rule and domain of  `g^-1`, the inverse function of  `g`.   (3 marks)

    2. On the set of axes above sketch the graph of  `g^-1`. Label the axes intercepts with their exact values.   (3 marks)

    3. Find the values of `x`, correct to three decimal places, for which  `g^-1(x) = g(x)`.   (2 marks)

    4. Calculate the area enclosed by the graphs of  `g`  and  `g^-1`. Give your answer correct to two decimal places.   (2 marks)

  2. The diagram below shows part of the graph of the function with rule
  3.         `f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.
     

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

      VCAA 2010 1b

    1. State the value of `a`.   (1 mark)

    2. Find the value of `c`.   (1 mark)

    3. Show that  `k = 9/(log_e (p + 1)`.   (2 marks)

    4. Show that the gradient of the tangent to the graph of `f` at the point `P` is  `9/((p + 1) log_e (p + 1))`.   (1 mark)

    5. If the point  `(-1, 0)`  lies on the tangent referred to in part b.iv., find the exact value of `p`.   (2 marks)

Show Answers Only
  1.   i. `g^-1(x) = e^((x-1)/2)-4,\ \ x in R`
  2.  ii. 
     
  3. iii. `3.914 or 5.503`
  4. iv. `52.63\ text(units²)`
  5.   i. `1`
  6.  ii. `1`
  7. iii. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.  iv. `text(Proof)\ \ text{(See Worked Solutions)}`
  9.   v. `e^(9/10)-1`
Show Worked Solution

a.i.   `text(Let)\ \ y = g(x)`

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

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

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

 

ii.  

 

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

  `text(on the line)\ \ y=x.`

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

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

 

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

 

b.i.   `text(Vertical Asymptote:)`

`x =-1`

`:. a = 1`

 

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

`c = 1`

 

iii.  `f(x)= k log_e (x + 1) + 1`

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

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

 

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

 

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

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

`(p, 10),\ \ (-1, 0)`

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

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

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

CORE*, FUR2 2008 VCAA 4

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

  1. By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation?   (1 mark)

  2. Determine the annual flat rate of depreciation correct to one decimal place.   (1 mark) 

Show Answers Only
  1. `$900`
  2. `5.3text{%  (1 d.p.)}`
Show Worked Solution
a.    `text(Depreciation)` `= (17\ 000 – 3500)/15`
    `= $13\ 500`

 
`:.\ text(Annual depreciation)`

`= (13\ 500)/15`

`= $900`

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

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

`= 5.29…`

`= 5.3text{%  (1 d.p.)}`

CORE*, FUR2 2008 VCAA 2

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

  1. Write down an expression involving the compound interest formula that can be used to find the value of Michelle’s $3000 investment at the end of two years. Find this value correct to the nearest cent.   (2 marks)

  2. How much interest will the $3000 investment earn over a four-year period?

     

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

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

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

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

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

 

b.   `text(After 4 years)\ (n = 8),`

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

 

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

CORE*, FUR2 2008 VCAA 1

Michelle has a bank account that pays her simple interest.

The bank statement below shows the transactions on Michelle’s account for the month of July.
  


 

  1. What amount, in dollars, was deposited in cash on 11 July?   (1 mark)

Interest for this account is calculated on the minimum monthly balance at a rate of 3% per annum.

  1. Calculate the interest for July, correct to the nearest cent.   (2 marks)

Show Answers Only
  1. `$620`
  2. `$15.30`
Show Worked Solution
a.    `text(Deposit)` `= 6870.67-6250.67`
    `= $620`

 

b.   `text(Minimum Balance) = $6120.86`

`:.\ text(Interest)` `=(PrT)/100`
  `= 6120.86 xx 3/100 xx 1/12`
  `= 15.302…`
  `= $15.30`

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)`

GRAPHS, FUR2 2008 VCAA 2

Tiffany decides to enter a charity event involving running and cycling. 

There is a $35 fee to enter.

  1. Write an equation that gives the total amount, `R` dollars, collected from entry fees when there are `x` competitors in the event.  (1 mark)

The event costs the organisers $50 625 plus $12.50 per competitor.

  1. Write an equation that gives the total cost, `C`, in dollars, of the event when there are `x` competitors.  (1 mark)
    1. Determine the number of competitors required for the organisers to break even.  (1 mark)

The number of competitors who entered the event was 8670.

  1. Determine the profit made by the organisers.  (1 mark)
Show Answers Only
  1. `R = 35x`
  2. `C = 50\ 625 + 12.5x`
    1. `2250`
    2. `$144\ 450`
Show Worked Solution

a.   `R = 35x`

 

b.   `C = 50\ 625 + 12.50x`

 

c.i.   `text(Break even when)\ R = C`

`35x` `= 50\ 625 + 12.5x`
`22.5x` `= 50\ 625`
`x` `= (50\ 625)/22.5`
  `= 2250`

 

`:. 2250\ text(competitors required to break even.)`

 

c.ii.   `text(When)\ \ x = 8670,`

`text(Profit)` `= R – C`
  `= 35 xx 8670 – (50\ 625 + 12.5 xx 8670)`
  `= 303\ 450 – 159\ 000`
  `= $144\ 450`

CORE*, FUR2 2009 VCAA 4

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

  1. Use the flat rate depreciation method with a depreciation rate of 12% per annum to find the depreciated value of the lawn mowers after four years.   (2 marks)

  2. Use the reducing balance depreciation method with a depreciation rate of 16% per annum to calculate the depreciated value of the lawn mowers after four years. Write your answer in dollars correct to the nearest cent.   (1 mark)

  3. After 4 years, which method, flat rate depreciation or reducing balance depreciation, will give the greater depreciation? Write down the greater depreciation amount in dollars correct to the nearest cent.   (1 mark)

Show Answers Only
  1. `$11\ 440`
  2. `$10\ 953.17`
  3. `$11\ 046.83`
Show Worked Solution

a.   `text(When)\ n = 4,`

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

 
`:.\ text(Depreciated Value)`

`= 22\ 000 – 10\ 560`

`= $11\ 440`
 

b.   `r = 16text(%)`

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

 

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

`text(amount after 4 years.)`

`text(Greater depreciation amount)`

`= 22\ 000 – 10\ 953.17`

`= $11\ 046.83`

CORE*, FUR2 2009 VCAA 3

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

  1. Show that the interest rate per quarter is 1.1%.   (1 mark)

  2. Determine the value of the $3400 investment after three years.

     

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

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

     

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

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

a.   `text(Interest rate per quarter)`

`= 4.4/4`

`= 1.1text(%  …as required.)`
 

b.   `text(Compounding periods = 12)`

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

 

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

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

 
`text(Interest earned over 6 years)`

`= 4420.86- 3400`

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

CORE*, FUR2 2009 VCAA 2

Rebecca will need to borrow $250 to buy a golf bag.

  1. If she borrows the $250 on her credit card, she will pay interest at the rate of 1.5% per month.

     

    Calculate the interest Rebecca will pay in the first month.

     

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

  2. If Rebecca borrows the $250 from the store’s finance company she will pay $6 interest per month.

     

    Calculate the annual flat interest rate charged. Write your answer as a percentage correct to one decimal place.   (1 mark)

Show Answers Only
  1. `$3.75`
  2. ` text(28.8%)`
Show Worked Solution

a.   `text(Interest in the 1st month)`

`= 1.5text(%) xx 250`

`= $3.75`
 

b.   `text(Annual interest) = 12 xx 6 = $72`

`:.\ text(Annual flat interest rate)`

`= 72/250 xx 100text(%)`

`= 28.8text(%)`

GRAPHS, FUR2 2009 VCAA 3

Another company, Cheapstar Airlines, uses the two equations below to calculate the total cost of a flight. 

The passenger fare, in dollars, for a given distance, in km, is calculated using the equation

fare = `20` + `0.47` × distance.

The charge, in dollars, for a particular excess luggage weight, in kg, is calculated using the equation

charge = `m` × (excess luggage weight)².

Suzie will fly 450 km with 15 kg of excess luggage on Cheapstar Airlines. 

She will pay $299 for this flight. 

Determine the value of `m`.  (2 marks)

Show Answers Only

`m = 0.3`

Show Worked Solution

`text{Total cost = fare + charge (luggage)}`

`text(fare)` `= 20 + 0.47 xx 450`
  `= $231.50`

 

`:.\ text(Amount left for luggage)`

`= 299 – 231.50`

`= $67.50`

`67.50` `= m xx 15^2`
`:. m` `= (67.50)/(15^2)`
  `= 0.3`

GRAPHS, FUR2 2009 VCAA 2

Luggage over 20 kg in weight is called excess luggage.

Fair Go Airlines charges for transporting excess luggage.

The charges for some excess luggage weights are shown in Table 2.

GRAPHS, FUR2 2009 VCAA 21

  1. Complete this graph by plotting the charge for excess luggage weight of 10 kg. Mark this point with a cross (×).  (1 mark)

    GRAPHS, FUR2 2009 VCAA 22

  2. A graph of the charge against (excess luggage weight)² is to be constructed.

     

    Fill in the missing (excess luggage weight)² value in Table 3 and plot this point with a cross (×) on the graph below.  (1 mark)

    GRAPHS, FUR2 2009 VCAA 23

GRAPHS, FUR2 2009 VCAA 24

  1. The graph above can be used to find the value of `k` in the equation below.

     

          charge = `k` × (excess luggage weight)²

     

    Find `k`.  (1 mark)

  2. Calculate the charge for transporting 12 kg of excess luggage.

     

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

Show Answers Only
  1.  
    GRAPHS, FUR2 2009 VCAA 2 Answer
  2.  
    GRAPHS, FUR2 2009 VCAA 2 Answer1
    GRAPHS, FUR2 2009 VCAA 2 Anwer2
  3. `0.45`
  4. `$64.80`
Show Worked Solution
a.   

GRAPHS, FUR2 2009 VCAA 2 Answer

 

b.    GRAPHS, FUR2 2009 VCAA 2 Answer1

GRAPHS, FUR2 2009 VCAA 2 Anwer2

 

c.   `text(Using the point)\ (100,45),`

`45` `= k xx 100`
`:. k` `= 0.45`

 

d.   `text(Charge for 12 kg excess)`

`= 0.45 xx 12^2`

`= $64.80`

GRAPHS, FUR2 2009 VCAA 1

Fair Go Airlines offers air travel between destinations in regional Victoria. 

Table 1 shows the fares for some distances travelled.

GRAPHS, FUR2 2009 VCAA 11

  1. What is the maximum distance a passenger could travel for $160?  (1 mark)

The fares for the distances travelled in Table 1 are graphed below.

GRAPHS, FUR2 2009 VCAA 13

  1. The fare for a distance longer than 400 km, but not longer than 550 km, is $280.

     

    Draw this information on the graph above.  (1 mark)

Fair Go Airlines is planning to change its fares.

A new fare will include a service fee of $40, plus 50 cents per kilometre travelled.

An equation used to determine this new fare is given by

fare = `40 + 0.5` × distance.

  1. A passenger travels 300 km.

     

    How much will this passenger save on the fare calculated using the equation above compared to the fare shown in Table 1?  (1 mark)

  2. At a certain distance between 250 km and 400 km, the fare, when calculated using either the new equation or Table 1, is the same.

     

    What is this distance?  (2 marks)

  3. An equation connecting the maximum distance that may be travelled for each fare in Table 1 on page 16 can be written as

     

             fare = `a` + `b` × maximum distance.

     

    Determine `a` and `b`.  (2 marks)

Show Answers Only
  1. `text(250 km)`
  2.  
    GRAPHS, FUR2 2009 VCAA 1 Answer
  3. `$30`
  4. `360\ text(km)`
  5. `b = 2/5\ text(and)\ a = 60`
Show Worked Solution

a.   `text(250 km)`

 

b.   

GRAPHS, FUR2 2009 VCAA 1 Answer

 

c.    `text(New fare)` `= 40 + 0.5 xx 300`
    `= $190`

 

`text(Fare from the table = $220`

`:.\ text(Passenger will save $30.)`

 

d.   `text(In table 1, a fare of $220 applies for travel)`

`text(between 250 – 400 km.)`

`:. 220` `= 40 + 0.5d`
`0.5d` `= 180`
`d` `= 360\ text(km)`

 

e.   `text(Equations in required form are:)`

`100` `= a + b xx 100\ \ …(1)`
`160` `= a + b xx 250\ \ …(2)`

 

`text(Subtract)\ (2) – (1)`

`60` `= 150b`
`:. b` `= 60/150 = 2/5`

 

`text(Substitute)\ b = 2/5\ text{into (1)}`

`100` `= a + 2/5 xx 100`
`:. a` `= 60`

 

CORE*, FUR2 2010 VCAA 3

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

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

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

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

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

BUSINESS, FUR2 2010 VCAA 3
 

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

     

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

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

     

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

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

     

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

    2. Determine the annual compound interest rate.

       

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

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

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

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

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

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

b.   `text(Total interest earned)`

`= 21\ 800-8000`

`= $13\ 800`

`:.\ text(Interest paid annually)`

`= (13\ 800)/15`

`= $920`
  

c.i.   `text(Using)\ A = PR^n,`

`24\ 000 = 8000 (1 + r/100)^15`
  

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

CORE*, FUR2 2010 VCAA 2

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

  1. Find the monthly payment that the perpetuity provides.   (1 mark)

  2. After six years of monthly payments, how much money remains invested in the perpetuity?   (1 mark)

Show Answers Only
  1. `$1560`
  2. `$360\ 000`
Show Worked Solution

a.   `text(Monthly repayment)`

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

`= 1/12 xx 18\ 720`

`= $1560`

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

b.   `$360\ 000`

CORE*, FUR2 2010 VCAA 1

The cash price of a large refrigerator is $2000.

  1. A customer buys the refrigerator under a hire-purchase agreement.
  2. She does not pay a deposit and will pay $55 per month for four years.
    1. Calculate the total amount, in dollars, the customer will pay.   (1 mark)

    2. Find the total interest the customer will pay over four years.   (1 mark)

    3. Determine the annual flat interest rate that is applied to this hire-purchase agreement.
    4. Write your answer as a percentage.   (1 mark)

  3. Next year the cash price of the refrigerator will rise by 2.5%.
  4. The following year it will rise by a further 2.0%.
  5. Calculate the cash price of the refrigerator after these two price rises.   (1 mark)

Show Answers Only

  1. i.  `$2640`
  2. ii.  `$640`
  3. iii. `text(8%)`
  4. `$2091`

Show Worked Solution

a.i.   `text(Total amount paid)`

`= 55 xx 4 xx 12`

`= $2640`
  

a.ii.    `text(Total interest)` `= 2640-2000`
    `= $640`

 

a.iii.    `I` `= (PrT)/100`
  `640` `= (2000 xx r xx 4)/100`
  `:. r` `= (640 xx 100)/(2000 xx 4)`
    `= 8text(%)`

  
b.   `text(After 1 year,)`

`P = 2000(1.025) = $2050`

`text(After 2 years,)`

`P = 2050(1.02) = $2091`

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`

GRAPHS, FUR2 2010 VCAA 2

Anne sells Resteasy pillows.

Last week she sold 35 Softsleep and `m` Resteasy pillows.

The selling price per pillow is shown in Table 1 below.

 

GRAPHS, FUR2 2010 VCAA 2

The total revenue from pillow sales last week was $4275.

Find `m`, the number of Resteasy pillows sold.  (1 mark)

Show Answers Only

`40`

Show Worked Solution
`text(Total Revenue)` `= 65 xx 35 + 50 xx m`
`4275` `= 2275 + 50m`
`50m` `= 2000`
`m` `= 40`

GRAPHS, FUR2 2010 VCAA 1

Anne sells Softsleep pillows for $65 each.

  1. Write an equation for the revenue, `R` dollars, that Anne receives from the sale of `x` Softsleep pillows.  (1 mark)
  2. The cost, `C` dollars, of making `x` Softsleep pillows is given by

`C = 500 + 40x`

Find the cost of making 30 Softsleep pillows.  (1 mark)

The revenue, `R`, from the sale of `x` Softsleep pillows is graphed below.

 

GRAPHS, FUR2 2010 VCAA 1

  1. Draw the graph of  `C = 500 + 40x`  on the axes above.  (1 mark)
  2. How many Softsleep pillows will Anne need to sell in order to break even?  (1 mark)
Show Answers Only
  1. `R = 65x`
  2. `$1700`
  3.  
    GRAPHS, FUR2 2010 VCAA 1 Answer
  4. `text(20 pillows)`
Show Worked Solution

a.   `R = 65x`

 

b.   `text(When)\ x = 30,`

`C` `= 500 + 40 xx 30`
  `= $1700`

 

c.   

GRAPHS, FUR2 2010 VCAA 1 Answer

 

d.   `text(Breakeven occurs when:)`

`text(Revenue)` `=\ text(C)text(osts)`
`65x` `= 500 + 40x`
`25x` `= 500`
`x` `= 20`

 

`:.\ text(Anne needs to sell 20 pillows to breakeven.)`

GEOMETRY, FUR2 2010 VCAA 1

In the plan below, the entry gate of an adventure park is located at point `G`.

A canoeing activity is located at point `C`.

The straight path `GC` is 40 metres long.

The bearing of `C` from `G` is 060°.

 

Geometry anad Trig, FUR2 2011 VCAA 1_1
 

  1. Write down the size of the angle that is marked `x^@` in the plan above.  (1 mark)
  2. What is the bearing of the entry gate from the canoeing activity?  (1 mark)
  3. How many metres north of the entry gate is the canoeing activity?  (1 mark)

`CW` is a 90 metre straight path between the canoeing activity and a water slide located at point `W`.

`GW` is a straight path between the entry gate and the water slide.

The angle `GCW` is 120°.

 

GEOMETRY, FUR2 2010 VCAA 12
 

    1. Find the area that is enclosed by the three paths, `GC`, `CW` and `GW`.

       

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

    2. Show that the length of path `GW` is 115.3 metres, correct to one decimal place.  (1 mark)

Straight paths `CK` and `WK` lead to the kiosk located at point `K`.

These two paths are of equal length.

The angle `KCW` is 10°.

 

GEOMETRY, FUR2 2010 VCAA 13

    1. Find the size of the angle `CKW`.  (1 mark)
    2. Find the length of path `CK`, in metres, correct to one decimal place.  (1 mark)
Show Answers Only
  1. `120^@`
  2. `240^@`
  3. `20\ text(m)`
    1. `1558.8\ text{m²  (1 d.p.)}`
    2. `text(See Worked Solutions.)`
    1. `160^@`
    2. `45.7\ text{m  (1 d.p.)}`
Show Worked Solution
a.    `x^@ + 60^@` `= 180^@`
   `:. x^@` `= 120^@`

 

b.   `text(Bearing of)\ G\ text(from)\ C`

MARKER’S COMMENT: True bearings (3 figure) are preferred to quadrant bearings although S60°W was accepted.

`= 360 – 120`

`= 240^@`

 

c.   

GEOMETRY, FUR2 2010 VCAA 1 Answer1

`text(Let)\ d = text(distance north of)\ C\ text(from)\ G`

`cos60^@` `= d/40`
`:. d` `= 40 xx cos60`
  `= 20\ text(m)`

 

d.i.    `A` `= 1/2ab sinC`
    `= 1/2 xx 40 xx 90 xx sin120^@`
    `= 1558.84…`
    `= 1558.8\ text{m²  (1 d.p.)}`

 

d.ii.   `text(Using the cosine rule,)`

`GW` `= sqrt(40^2 + 90^2 + 2 xx 40 xx 90 xx cos120^@)`
  `= sqrt(13\ 300)`
  `= 115.32…`
  `= 115.3\ text{(1 d.p.)  …as required.}`

 

e.i.   `DeltaCKW\ text(is isosceles)`

`:. angleCKW` `= 180 – (2 xx 10)`
  `= 160^@`

 

e.ii.   `text(Using the sine rule,)`

`(CK)/(sin10^@)` `= 90/(sin160^@)`
`:. CK` `= (90 xx sin10^@)/(sin160^@)`
  `= 45.69…`
  `= 45.7\ text{m  (1 d.p.)}`

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 2011 VCAA 3

Tania purchased a house for $300 000.

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

Stamp duty rates are listed in the table below. 

     BUSINESS, FUR2 2011 VCAA 3
 

  1. Calculate the amount of stamp duty that Tania will have to pay.   (1 mark)

  2. Assuming that her house will increase in value at a rate of 3.17% per annum, what will the value of Tania's house be after 5 years?

     

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

Tania bought her house at the start of 2011.

  1. If the rate of increase in value remains at 3.17% per annum, at the start of which year will the value of Tania's house first exceed $450 000?   (2 marks)

Show Answers Only
  1. `$13\ 070`
  2. `$351\ 000`
  3. `2024`
Show Worked Solution
a.    `text(Stamp duty)` `= 2870 + 6text(%) xx (300\ 000-130\ 000)`
    `= $13\ 070`
♦♦ Part (a) was “poorly answered” although exact data is unavailable.
MARKER’S COMMENT: A majority of students did not understand how to use the table.

 

b.   `text(Using)\ \ \ A = PR^n`

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

 

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

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

 

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

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

`text(will be over)\ $450\ 000.`

CORE*, FUR2 2011 VCAA 2

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

Each chose a different investment strategy.

Tom's investment strategy

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

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

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

Patty's investment strategy

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

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

BUSINESS, FUR2 2011 VCAA 2

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

  1. How much more does she have?
  2. Write your answer to the nearest cent.  (2 marks)

  3. What annual compounding rate of interest would Patty need in order to earn $1000 interest in one year on her $8000 investment?
  4. Write your answer correct to one decimal place.  (1 mark)

Show Answers Only

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

Show Worked Solution

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

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

`= $5833.60`
  

b.   `text(After 1 year,)`

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

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

`text(by TVM Solver,)`

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

 

`=> FV = −8497.58…`

 

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

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

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

`= $8576`
 

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

`= 8576-8497.580…`

`= 78.419…`

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

 

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

CORE*, FUR2 2011 VCAA 1

Tony plans to take his family on a holiday.

The total cost of $3630 includes a 10% Goods and Services Tax (GST).

  1. Determine the amount of GST that is included in the total cost.  (1 mark)

During the holiday, the family plans to visit some theme parks.

The prices of family tickets for three theme parks are shown in the table below.

BUSINESS, FUR2 2011 VCAA 1

  1. What is the total cost for the family if it visits all three theme parks?  (1 mark)

If Tony purchases the Movie Journey family ticket online, the cost is discounted to $202.40

  1. Determine the percentage discount.  (1 mark)
Show Answers Only
  1. `$330`
  2. `$462`
  3. `text(8%)`
Show Worked Solution

a.   `text(Let $)P\ text(be the cost ex-GST)`

`P + 10text(%)P` `= 3630`
`1.1P` `= 3630`
`P` `= 3630/1.1`
  `= $3300`
   
`:.\ text(GST)` `= 10text(%) xx $3300`
  `= $330`

 

b.   `text(C)text(ost to visit all 3 parks)`

`= 82 + 220 + 160`

`= $462`

 

c.    `text(Savings)` `= 220 – 202.40`
    `= 17.60`
`:.\ text(Discount)` `= (17.60)/220`
  `= 0.08`
  `= 8text(%)`

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)}`

CORE*, FUR2 2012 VCAA 2

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

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

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

  1. Calculate the depreciated value of the equipment after three years.   (1 mark)

  2. Show that, in any one year, the flat rate method of depreciation with a depreciation rate of 10% per annum will give the same annual depreciation as the unit cost method.   (1 mark)

  3. After how many years will equipment be written off with a depreciated value of $0?   (1 mark)

  4. Suppose the reducing balance method is used to depreciate the equipment instead of the unit cost method.

     

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

     

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

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

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

`= 3800 xx 0.22`

`= $836`

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

`= 8360-(3 xx 836)`

`= $5852`
  

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

`= 10text(%) xx 8360`

`= $836\ \ …text(as required.)`
  

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

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

  
d.   `text(After 1 year,)`

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

  
`text(After 10 years,)`

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

CORE*, FUR2 2012 VCAA 1

A club purchased new equipment priced at $8360. A 15% deposit was paid.

  1. Calculate the deposit.   (1 mark)

    1. Determine the amount of money that the club still owes on the equipment after the deposit is paid.   (1 mark)
    2. The amount owing will be fully repaid in 12 installments of $650.
    3. Determine the total interest paid.   (1 mark)

Show Answers Only

  1. `$1254`
  2. i.  `$7106`
  3. ii. `$684`

Show Worked Solution

a.    `text(Deposit)` `= 15text(%) xx 8360`
    `= $1254`

 

b.i.    `text(Amount still owed)` `= 8360-1254`
    `= $7106`

 

b.ii.    `text(Total repayments)` `= 12 xx 650`
    `= $7800`

 
`:.\ text(Total interest paid)`

`= 7800-7106`

`= $694`

NETWORKS, FUR1 2008 VCAA 8-9 MC

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

 

networks-fur1-2008-vcaa-8-mc

Part 1

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

A.     7

B.   15

C.   16

D.   19

E.   20

 

Part 2

This project currently has one critical path.

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

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

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

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

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

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

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(EST for Activity)\ K`

`=\ text(Duration)\ ACFI`

`= 2 + 5 + 6 + 3`

`= 16`

`=> C`

 

`text(Part 2)`

♦♦ Mean mark of Part 2 was 35%.

`text(Original critical path is)`

`ACFHJL\ text{(22 days)}`

`text(Consider option)\ A,`

`text(New critical path is created)`

`ABDJL\ text{(22 days)}`

`=> A`

NETWORKS, FUR1 2008 VCAA 4 MC

A simple connected graph with 3 edges has 4 vertices.

This graph must be

A.   a complete graph.

B.   a tree.

C.   a non-planar graph.

D.   a graph that contains a loop.

E.   a graph that contains a circuit.

Show Answers Only

`=> B`

Show Worked Solution

`text(Consider the graph below,)`
 

networks-fur1-2008-vcaa-4-mc-answer

`=> B`

NETWORKS, FUR1 2010 VCAA 6-7 MC

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

vcaa-networks-fur1-2010-6-7

 
Part 1

The capacity of this cut is

A.   `14`

B.   `18`

C.   `23`

D.   `31`

E.   `40`

 

Part 2

The maximum flow between source and sink through the network is

A.    `7`

B.   `10`

C.   `11`

D.   `12`

E.   `20`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ B`

Show Worked Solution

Part 1

`text(Capacity of the cut)`

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

`= 11 + 5 + 7`

`= 23`

`=>  C`

 

Part 2

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

`text(The maximum flow)`

♦♦♦ Mean mark 24%.

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

`= 4 + 2 + 3 + 1`

`= 10`

`=>  B`

NETWORKS, FUR1 2010 VCAA 4 MC

A board game consists of nine labelled squares as shown.

A player must start at square `J` and, moving one square at a time, aim to finish at square `R`.

Each move may only be to the right one square or down one square.

A player who lands on square `N` must stay there and cannot move again.

A player can only stop moving when they reach `N` or `R`.
 

vcaa-networks-fur1-2010-4
 

A digraph that shows all the possible moves that a player could make to reach `N` or `R` from `J` is

vcaa-networks-fur1-2010-4ai

vcaa-networks-fur1-2010-4aii

vcaa-networks-fur1-2010-4aiii

Show Answers Only

`E`

Show Worked Solution

`=>  E`

NETWORKS, FUR1 2006 VCAA 7 MC

A complete graph with six vertices is drawn.

This network would best represent

  1. the journey of a paper boy who delivers to six homes covering the minimum distance.
  2. the cables required to connect six houses to pay television that minimises the length of cables needed.
  3. a six-team basketball competition where all teams play each other once.
  4. a project where six tasks must be performed between the start and finish.
  5. the allocation of different assignments to a group of six students.  
Show Answers Only

`C`

Show Worked Solution

`text(A complete graph has all vertices connected)`

`text(directly to all other vertices without any)`

`text(parallel edges or loops.)`

`rArr C`

NETWORKS, FUR1 2006 VCAA 6 MC

networks-fur1-2006-vcaa-6-mc

 
In the directed graph above the weight of each edge is non-zero.

The capacity of the cut shown is

A.   a + b + c + d + e

B.   a + c + d + e

C.   a + b + c + e

D.   a + b + c – d + e

E.   a – b + c – d + e 

Show Answers Only

`C`

Show Worked Solution

`text(Capacity of the cut)`

`= a + b + c + e`

`rArr C`

MATRICES*, FUR1 2007 VCAA 8 MC

There are five teams, `A, B, C, D` and `E`, in a volleyball competition. Each team played each other team once in 2007.

The results are summarised in the directed graph below. An arrow from `A` to `E` signifies that `A` defeated `E.`
 


 

In 2007, the team that had the highest number of two-step dominances was

A.   team `A`

B.   team `B`

C.   team `C`

D.   team `D`

E.   team `E`

Show Answers Only

`B`

Show Worked Solution

`text(2-step dominance matrix)` 

`{: (\ quad A quad B quad C quad D quad E), ([(0, 0, 1, 2, 1), (3, 0, 0, 2, 1), (0, 3, 0, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 0, 0)]{:(A – 4), (B – 6), (C – 3), (D – 2), (E – 1):}\ \ \ \ {:text(“2-step” wins):}):}`

 
`=>  B`

NETWORKS, FUR1 2007 VCAA 7 MC

The minimal spanning tree for the network below includes two edges with weightings `x` and `y.`
 

 
The length of the minimal spanning tree is 19.

The values of `x` and `y` could be

A.   `x = 1 and y = 7`

B.   `x = 2 and y = 5`

C.   `x = 3 and y = 5`

D.   `x = 4 and y = 5`

E.   `x = 5 and y = 6`

Show Answers Only

`C`

Show Worked Solution

`text(Minimal spanning tree is:)`
 

vcaa-networks-fur1-2007-7i

 
`:.\ text(Minimal spanning tree)`

`19` `= y + 3 + x + 2 + 1 + 5`
`8` `= x + y`

 
`=>  C`