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Vectors, SPEC2 2018 VCAA 14 MC

The scalar resolute of  `underset ~a = 3 underset ~i - 2 underset ~k`  in the direction of  `underset ~b = - underset ~i + 2 underset ~j + 3 underset ~k`  is

A.   `-(9 sqrt 13)/13`

B.   `-9/14(-underset ~i + 2 underset ~j + 3 underset ~k)`

C.   `-(9 sqrt 14)/14`

D.   `-9/13 (3 underset ~i - 2 underset ~k)`

E.   `- sqrt 14/2` 

Show Answers Only

`C`

Show Worked Solution
`underset ~a ⋅ underset ~hat b`

 `= (3 xx (-1) + 0 xx 2 + (-2) xx 3)/sqrt((-1)^2 + 2^2 + 3^2)`
  `= (-3-6)/sqrt 14`
  `= -9/sqrt 14 xx sqrt 14/sqrt 14`
  `= -(9 sqrt 14)/14`

 
`=>  C`

Vectors, SPEC2 2018 VCAA 13 MC

The position vector of a particle that is moving along a curve at time `t` is given by  `underset ~r(t) = 3 cos (t) underset ~i + 4 sin (t) underset ~j, \ t >= 0`.

The first time when the speed of the particle is a minimum is

  1. `3`
  2. `pi/2`
  3. `tan^(-1) (4/3)`
  4. `(3 pi)/2`
  5. `9` 
Show Answers Only

`B`

Show Worked Solution
`underset ~dot r(t)` `= -3 sin(t) underset ~i + 4 cos (t) underset ~j`
`|underset ~dot r(t)|` `= sqrt (9 sin^2(t) + 16 cos^2(t))`
  `= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
  `= sqrt (9 + 7 cos^2(t))`

 
`text(Minimised for)\ cos(t) = 0`

`:.  t_1 = pi/2`

`=>  B`

Calculus, SPEC2 2018 VCAA 10 MC

The differential equation that best represents the direction field above is

A.  `(dy)/(dx) = (2x + y)/(y - 2x)`

B.  `(dy)/(dx) = (x + 2y)/(2x - y)`

C.  `(dy)/(dx) = (2x - y)/(x + 2y)`

D.  `(dy)/(dx) = (x - 2y)/(y - 2x)`

E.  `(dy)/(dx) = (2x + y)/(2y - x)` 

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=>  A`

Calculus, SPEC2 2018 VCAA 9 MC

A solution to the differential equation  `(dy)/(dx) = 2/{sin(x + y)-sin(x-y)}`  can be obtained from

  1. `int 1\ dx = int 2 sin(y)\ dy`
  2. `int cos(y)\ dy = int text{cosec}(x)\ dx`
  3. `int cos(x)\ dx = int text{cosec}(y)\ dy`
  4. `int sec(x)\ dx = int sin(y)\ dy`
  5. `int sec(x)\ dx = int text{cosec}(y)\ dy` 
Show Answers Only

`D`

Show Worked Solution
`(dy)/(dx)` `= 2/{sin(x) cos(y) + sin(y) cos(x)-(sin(x) cos(y)-sin(y) cos(x))}`
  `= 2/{2 sin(y) cos(x)}`
  `= 1/{sin(y) cos(x)}`

 
`sin(y) *(dy)/(dx)= sec(x)`

`int sin (y)\ dy= int sec(x)\ dx`

`=>  D`

Calculus, SPEC2 2018 VCAA 8 MC

Using a suitable substitution, `int_0^(pi/6) tan^2 (x) sec^2(x)\ dx`  can be expressed as

A.  `int_0^(1/sqrt 3) (u^4 + u^2) du`

B.  `int_1^(2/sqrt 3) (u^4 + u^2) du`

C. `int_0^(1/(sqrt 3)) u\ du`

D. `int_0^(pi/6) u^2\ du`

E. `int_0^(1/sqrt 3) u^2\ du` 

Show Answers Only

`E`

Show Worked Solution
`text(Let)\ \ u` `= tan(x),`
`(du)/(dx)` `= sec^2 (x)\ \ =>\ du = sec^2(x)\ dx`
`u(0)` `= tan(0) = 0`
`u (pi/6)` `= tan (pi/6) = 1/sqrt 3`

 
`int_0^(pi/6) tan^2 (x)\ sec^2 (x)\ dx = int_0^(1/sqrt 3) u^2 du`

`=>  E`

Complex Numbers, SPEC2 2018 VCAA 6 MC

The complex numbers  `z, iz `  and  `z + iz`, where  `z in C\ text(\{0})`, are plotted in the Argand plane, forming the vertices of a triangle.

The area of this triangle is given by

  1. `|z|`
  2. `|z| + |z|^2`
  3. `(|z|^2)/2`
  4. `|z|^2`
  5. `(sqrt 3 |z|^2)/2` 
Show Answers Only

`C`

Show Worked Solution

`text(Area)` `= 1/2 b h`
  `= 1/2 |z + iz – z| |z + iz – iz|`
  `= 1/2 |iz| |z|`
  `= 1/2 |i| |z|^2`
  `= (|z|^2)/2`

 
`=>  C`

Calculus, SPEC1 2018 VCAA 9

A curve is specified parametrically by  `underset ~r(t) = sec(t) underset ~i + sqrt 2/2 tan(t) underset ~j, \ t in R`.

  1.  Show that the cartesian equation of the curve is  `x^2 - 2y^2 = 1`.  (2 marks)
  2.  Find the `x`-coordinates of the points of intersection of the curve  `x^2 - 2y^2 = 1`  and the line  `y = x - 1`.  (1 mark)
  3.  Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the `x`-axis.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 1 or x = 3`
  3. `(2 pi)/3`
Show Worked Solution

a.     `x = sec(t), qquad y = sqrt 2/2 tan(t)`

`x^2 = sec^2(t), qquad y^2 = 1/2 tan^2(t)`

`x^2 = sec^2(t), qquad 2y^2 = tan^2(t)`

`1 + tan^2(t)` `= sec^2(t)`
`1 + 2y^2` `= x^2`
`:.x^2 – 2y^2` `=1\ \ text(.. as required)`

 

b.    `x^2 – 2(x – 1)^2` `= 1`
  `x^2 – 2(x^2 – 2x + 1)` `=1`
  `x^2 – 2x^2 + 4x – 2` `=1`
  `-x^2 + 4x – 2 – 1` `=0`
  `x^2 – 4x + 3` `=0`
  `(x – 3) (x – 1)` `=0`

 
`:. x = 1 or x = 3`

♦♦ Mean mark 30%.

 

c.   `x^2 – {:2y_1:}^2` `=1`
  `{:2y_1:}^2` `=x^2 – 1`
  `{:y_1:}^2` `= (x^2 – 1)/2`
  `{:y_2:}^2` `= (x – 1)^2`

 

`V` `=pi int_1^3 {:y_1:}^2 – {:y_2:}^2 \ dx`
  `= pi int_1^3 (x^2 – 1)/2 – (x-1)^2\ dx`
  `= pi [x^3/6 – x/2 – (x – 1)^3/3]_1^3`
  `= pi [(3^3/6 – 3/2 – 2^3/3) – (1^3/6 – 1/2 – 0)]`
  `= pi (9/2 – 3/2 – 8/3 – 1/6 + 1/2)`
  `= pi (7/2 – 8/3 – 1/6)`
  `= pi ((21 – 16 – 1)/6)`
  `= (2 pi)/3`

Calculus, SPEC1 2018 VCAA 8

A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute.

  1.  Show that the differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by
    `qquad (dQ)/(dt) = -(3Q)/(16 + 2t)`  (1 mark)
      
  2.  Solve the differential equation given in part a. to find `Q` as a function of `t`.
      
    Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)
Show Answers Only
  1.  `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  `Q = 32/(16 + 2t)^(3/2)`
Show Worked Solution

a. `Q_0 = 0.5, \ V_0 = 16`

♦ Net mean mark of both parts 44%.

`V(t)` `= 16 + (5 – 3) t`
  `= 16 + 2t`

 

`text(Concentration)\ (C)` `= Q/V= Q/(16 + 2t)\ text(kg/L)`

 

`(dQ)/(dt)` `= 0 xx 5 – 3C`
  `= -(3Q)/(16 + 2t)`

MARKER’S COMMENT: Many students took the common factor of 2 from  `16+2t`. This wasn’t necessary and complicated the arithmetic in part b.

 

b.   `-1/(3Q) * (dQ)/(dt) = 1/(16 + 2t)`

`int -1/(3Q)\ dQ` `= int 1/(16 + 2t) dt`
`-1/3 int 1/Q\ dQ` `= 1/2 int 2/(16 + 2t)\ dt`
`-1/3 ln Q` ` = [1/2 ln(16 + 2t)] + c`

 
`text(When)\ \ t=0,\ \ Q=0.5,`

`-1/3 ln (1/2)` `= 1/2 ln (16) +c`
`c` `= -1/2 ln (16) -1/3 ln (1/2)`

 

`-1/3 ln Q` `= 1/2 ln(16 + 2t) -1/2 ln(16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= 1/2 ln ((16 + 2t)/16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= ln (((16 + 2t)^(1/2))/4) – ln (2^(-1/3))`
`ln (Q^(-1/3))` `= ln (((16 + 2t)^(1/2))/(2^2 ⋅ 2^(-1/3)))`
`Q^(-1/3)` `= ((16 + 2t)^(1/2))/(2^(5/3))`
`Q` `= (((16 + 2t)^(1/2))/(2^(5/3)))^-3`
  `= (16 + 2t)^(- 3/2)/(2^(-5))`
`:. Q` `= 32/((16 + 2t)^(3/2))`

Vectors, SPEC1 2018 VCAA 6

A particle of mass 2 kg moves under a force  `underset ~ F`  so that its position vector  `underset ~ r`  at anytime `t` is given by  `underset ~r = sin(t) underset ~i + cos(t) underset ~j + t^2 underset ~k`.  Distances are measured in metres and time is measured in seconds.

Find the change in momentum, in kg ms¯², from  `t = pi/2`  to  `t = pi`.  (3 marks)

Show Answers Only

`2 (- underset ~i + underset ~j +  pi underset ~k)`

Show Worked Solution
`underset ~ dot r (t)` `= cos(t) underset ~i – sin(t) underset ~j + 2t underset~k`
`underset ~ dot r (pi)` `= cos(pi) underset ~i – sin(pi) underset ~j + 2 pi underset ~k`
  `= – underset ~i + 2 pi underset ~k`
   
`underset ~ dot r (pi/2)` `= cos(pi/2) underset ~i – sin (pi/2) underset ~j + pi underset ~k`
  `= – underset ~j + pi underset ~k`
   
`:. m Delta r` `= 2(underset ~ dot r (pi) – underset ~ dot r (pi/2))`
  `= 2(- underset ~i + 2 pi underset ~k – (- underset ~j + pi underset ~k))`
  `= 2 (- underset ~i + underset ~j +  pi underset ~k)`

Graphs, SPEC1 2018 VCAA 5

Sketch the graph of  `f(x) = (x + 1)/(x^2 - 4)`  on the axes provided below, labelling any asymptotes with their equations and any intercepts with their coordinates.  (4 marks)

Show Answers Only

Show Worked Solution

`text(Using partial fractions:)`

`(x + 1)/(x^2 – 4)= A/(x – 2) + B/(x + 2)`

`A(x + 2) + B (x – 2) = x + 1`

 
`text(When)\ \ x = 2,`

`4A` `= 3 => \ A=3/4`

 
`text(When)\ \ x = -2`

`-4B` `= -1 =>\ B=1/4`

 
`f(x)= 1/(4(x + 2)) + 3/(4(x – 2))`
 

`text(As)\ \ x->oo, \ f(x)->0^+`

`text(As)\ \ x->- oo, \ f(x)->0^-`

`f(0)= – 1/4`
 

`text(Find)\ x\ text(when)\ \ f(x)=0:`

`x + 1` `= 0`
`x` `= -1`

Complex Numbers, SPEC1 2018 VCAA 2

  1. Show that  `1 + i = sqrt 2\ text(cis)(pi/4)`.  (1 mark)
  2. Evaluate  `(sqrt 3 - i)^10/(1 + i)^12`, giving your answer in the form  `a + bi`, where  `a, b ∈ R`.  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-8 – 8 sqrt 3 i`
Show Worked Solution
a.   `r` `= sqrt(1^2 + 1^2)`
    `= sqrt 2`
  `theta` `= tan^(-1) (1/1) = pi/4`
     
  `:. 1 + i` `= sqrt 2\ text(cis)(pi/4)`

 

b.   
 `r_2` `= sqrt((sqrt 3)^2 + (-1)^2)`
    `= sqrt (3 + 1)`
    `= 2`

 
`theta_2 = -tan^(-1) (1/sqrt 3)=-pi/6`

`sqrt 3 – i` `= 2\ text(cis)(- pi/6)`
`(sqrt 3 – i)^10` `= 2^10\ text(cis) ((-10 pi)/6)`
   
`=>(1 + i)^12` `= (sqrt 2)^12text(cis)((12pi)/4)`
  `=2^6 text(cis)(3pi)`

 

`:. (sqrt 3 – i)^10/(1 + i)^12` `= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
  `= 2^4 text(cis)((-5pi)/3 – (9pi)/3)`
  `= 16 text(cis)((-14pi)/3)`
  `= 16 text(cis) ((-2pi)/3)`
  `= 16(-1/2 + (-sqrt3)/2 i)`
  `= -8 – 8 sqrt 3 i`

Mechanics, SPEC1 2018 VCAA 1

Two objects of masses 5 kg and 8 kg are attached by a light inextensible string that passes over a smooth pulley. The 8 kg mass is on a smooth plane inclined at 30° to the horizontal. The 5 kg mass is hanging vertically, as shown in the diagram below.
 


 

  1. On the diagram above, show all forces acting on both masses.  (1 mark)
  2. Find the magnitude, in `text(ms)^(-2)`, and state the direction of the acceleration of the 8 kg mass.  (3 marks)
Show Answers Only
a.  

b.   `a = 9/13\ text(ms)^(-2)\ text(up the incline)`

Show Worked Solution
a.  

 

b.   
 `sum F` `= 5 text(g) – 8 text(g)\ sin (30^@) = (5 + 8)a\ \ text{(up slope → +)}`
  `5 text(g) – 4 text(g)` `= 13a`
  `:. a` `= 9/13\ text(ms)^(-2)\ text(up the incline)`

Calculus, MET1 2018 VCAA 8

Let  `f: R -> R, \ f(x) = x^2e^(kx)`, where `k` is a positive real constant.

  1.  Show that  `fprime(x) = xe^(kx)(kx + 2)`.  (1 mark)
  2. Find the value of `k` for which the graphs of  `y = f(x)`  and  `y = fprime(x)`  have exactly one point of intersection.  (3 marks)

Let  `g(x) = −(2xe^(kx))/k`. The diagram below shows sections of the graphs of  `f` and `g` for  `x >= 0`.
 


 

Let `A` be the area of the region bounded by the curves  `y = f(x), \ y = g(x)` and the line  `x = 2`.

  1. Write down a definite integral that gives the value of  `A`.  (1 mark)
  2. Using your result from part a., or otherwise, find the value of `k` such that  `A = 16/k`.  (3 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(One point of intersection)\ x = 0\ text(when)\ k = 1.`
  3. `int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`
  4. `1/2 ln 4\ \ text(or)\ \ ln2`
Show Worked Solution

a.   `f(x) = x^2e^(kx)`

`fprime(x)` `= 2x · e^(kx) + x^2 · k · e^(kx)`
  `= xe^(kx)(kx + 2)\ \ …text(as required)`

 

b.  `text(Intersection occurs when:)`

♦♦♦ Mean mark 7%.
MARKER’S COMMENT: Most students found the correct quadratic equation but solved for `k`.

`x^2e^(kx)` `= xe^(kx)(kx + 2)`
`x^2 – x(kx + 2)` `= 0,\ \ e^(kx) != 0`
`x^2 – kx^2 – 2x` `= 0`
`x^2(1 – k) – 2x` `= 0`
`x[x(1 – k) – 2]` `= 0`

 

`:.x = 0\ \ text(or)\ \ x(1 – k) – 2` `= 0`
`x` `= 2/(1 – k)`

 
`text(S)text(ince)\ \ x = 2/(1 – k)\ \ text(is undefined when)\ \ k = 1,`

`=>\ text(One point of intersection only at)\ \ x = 0\ \ text(when)\ \ k = 1.`

 

c.    `A` `= int_0^2 f(x)\ dx – int_0^2 g(x)\ dx`
    `= int_0^2 x^2e^(kx)\ dx – int_0^2 −(2xe^(kx))/k \ dx`
    `= int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`

♦♦ Mean mark 29%.

 

d.    `int_0^2(x^2e^(kx) + (2xe^(kx))/k)\ dx` `= 16/k`
  `1/k int_0^2(kx^2e^(kx) + 2xe^(kx))\ dx` `= 16/k`
  `1/k int_0^2(xe^(kx)(kx + 2))\ dx` `= 16/k`
  `1/k[x^2e^(kx)]_0^2` `= 16/k\ \ \ text{(using part a)}`
  `[2^2 · e^(2k) – 0]` `= 16`
  `e^(2k)` `= 4`
  `2k` `= ln 4`
  `k` `= 1/2 ln 4\ \ text(or)\ \ ln2`

Probability, MET1 2018 VCAA 6

Two boxes each contain four stones that differ only in colour.

Box 1 contains four black stones.

Box 2 contains two black stones and two white stones.

A box is chosen randomly and one stone is drawn randomly from it.

Each box is equally likely to be chosen, as is each stone.

  1.  What is the probability that the randomly drawn stone is black?  (2 marks)
  2.  It is not known from which box the stone has been drawn.
  3. Given that the stone that is drawn is black, what is the probability that it was drawn from Box 1?  (2 marks)
Show Answers Only
  1. `3/4`
  2. `2/3`
Show Worked Solution
a.    `text(Pr)(text(Black))` `= text{Pr(Box 1)} · text{Pr(B)} + text{Pr(Box 2)} · text{Pr(B)}`
    `= 0.5 xx 4/4 + 0.5 xx 2/4`
    `= 3/4`

 

b.    `text(Pr)(text(Box 1 | black))` `= (text{Pr(Box 1)}\ ∩\ text{Pr(Black)})/text{Pr(Black)}`
    `= (1/2)/(3/4)`
    `= 2/3`

Algebra, STD2 A2 2007 HSC 27b*

A clubhouse uses four long-life light globes for five hours every night of the year. The purchase price of each light globe is $6.00 and they each cost  `$d`  per hour to run.

  1. Write an equation for the total cost (`$c`) of purchasing and running these four light globes for one year in terms of  `d`.    (2 marks)

  2. Find the value of  `d`  (correct to three decimal places) if the total cost of running these four light globes for one year is $250.   (1 mark)

  3. If the use of the light globes increases to ten hours per night every night of the year, does the total cost double? Justify your answer with appropriate calculations.   (1 mark)

Show Answers Only
  1. `$c = 24 + 7300d`
  2. `0.031\ $ text(/hr)\ text{(3 d.p.)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Purchase price) = 4 xx 6 = $24`

`text(Running cost)` `= text(# Hours) xx text(Cost per hour)`
  `= 4 xx 5 xx 365 xx d`
  `= 7300d`
 
`:.\ $c = 24 + 7300d`

 

ii.  `text(Given)\ $c = $250`

`=>250` `= 24 + 7300d`
`7300d` `= 226`
`d` `= 226/7300`
  `= 0.03095…`
  `= 0.031\ $ text(/hr)\ text{(3 d.p.)}`

 

iii.  `text(If)\ d\ text(doubles to 0.062)\ \ $text(/hr)`

`$c` `= 24 + 7300 xx 0.062`
  `= $476.60`
   
`text(S) text(ince $476.60 is less than)\ 2 xx $250\ ($500),`
`text(the total cost increases to less than double)`
`text(the original cost.)`

 

Measurement, STD2 M6 SM-Bank 4

The diagram shows three checkpoints A, B and C. Checkpoint C is due east of Checkpoint A. The bearing of Checkpoint B from Checkpoint A is N22°E and the bearing of Checkpoint C from Checkpoint B is S68°E. The distance between Checkpoint A and Checkpoint B is 42 kilometres.
 


 

  1. Mark the given information on the diagram and explain why `angleABC` is 90°.  (2 marks)

  2. Find the distance, to the nearest kilometre, between Checkpoint A and Checkpoint C.  (2 marks)

  3. If a runner is travelling 12.6 km/h, how long does it take her to travel between Checkpoint A and Checkpoint B, in hours and minutes?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `112\ text{km}`
  3. `3text(h 20 mins)`
Show Worked Solution
i.   
`angle ABC` `= 22 + 68`
  `= 90°`

 

ii.  `text(In)\ \ DeltaABC,`

`cosangleBAC` `= (AB)/(AC)`
`cos68°` `= 42/(AC)`
`AC` `= 42/(cos68°)`
  `= 112.11…`
  `= 112\ text{km  (nearest km)}`

 

iii.    `text(Travel time)` `= text(dist)/text(speed)`
    `= 42/12.6`
    `= 3.333…`
    `= 3text(h 20 mins)`

Statistics, STD2 S1 SM-Bank 1

Write down the five-number summary for the dataset 

`3, \ 7, \ 8, \ 11, \ 13, \ 18.`  (2 marks)

Show Answers Only
`text(Minimum value:)`   `3`
`text(First quartile:)`   `7`
`text(Median:)`   `(11 + 8)/2 = 9.5`
`text(Third quartile:)`   `13`
`text(Maximum value:)`   `18`
Show Worked Solution
`text(Minimum value:)`   `3`
`text(First quartile:)`   `7`
`text(Median:)`   `(11 + 8)/2 = 9.5`
`text(Third quartile:)`   `13`
`text(Maximum value:)`   `18`

Statistics, STD2 S5 SM-Bank 5

The diastolic measurement for blood pressure in 30-year-old people is normally distributed, with a mean of 82 and standard deviation of 16.

  1.   A person is considered to have low blood pressure if the diastolic measurement is 66 or less.

     

      What percentage of 30-year-old people have low blood pressure?  (1 mark)

  2.   Calculate the `z`-score for a diastolic measurement of 70.  (1 mark)

Show Answers Only
  1. `16text(%)`
  2. `−0.75`
Show Worked Solution
a.    `ztext(-score)\ (66)` `= (66 – 82)/16`
    `= −1`

 

`:.\ text(Percentage)` `= 100 – (50+34)`
  `= 16text(%)`

 

b.    `ztext(-score)\ (70)` `= (70 – 82)/16`
    `= −0.75`

Measurement, STD2 M6 SM-Bank 7 MC

Jeet walks 5 km from his home on a bearing of 153°. He then walks due north until he arrives a point which is due east of his home.

How far east, to the nearest 0.1 km, is Jeet from home?

  1.  2.3 km
  2.  2.5 km
  3.  4.9 km
  4.  9.8 km
Show Answers Only

`A`

Show Worked Solution

`text(Jeet finishes at)\ P`

`text(Find)\ \ OP:`

`cos 63°` `= (OP)/5`
`:. OP` `= 5 xx cos 63°`
  `= 2.26…`

 
`=> A`

Algebra, STD2 A4 SM-Bank 7 MC

A computer application was used to draw the graphs of the equations

`x + y = 4`  and  `x - y = 4`

Part of the screen is shown.
 

Which row of the table correctly matches the equations with the lines drawn and identifies the solution when the equations are solved simultaneously?

Show Answers Only

`A`

Show Worked Solution

`text(Line 1:) \ \ x + y = 4`

`text(Line 2:) \ \ x – y = 4`

`text(Intersection at) \ (4, 0).`

`=> A`

Algebra, STD2 A4 2014 HSC 26d

Draw each graph on the grid below and hence solve the simultaneous equations.  (3 marks)

`y = 2x + 1`

`x - 2y - 4 = 0`   (3 marks)
 

Show Answers Only

`x=-2,\ y=-3`

Show Worked Solution

`text(Solution is at the intersection:)\ \ x=-2,\ y=-3`

Statistics, STD2 S1 2018 HSC 17 MC

The area chart shows the number of students involved in tennis or cricket at a school over a number of years.
 


 

In which year was the number of students involved in tennis equal to the number of students involved in cricket?

  1. 2013
  2. 2014
  3. 2015
  4. 2016
Show Answers Only

`C`

Show Worked Solution

`text(Area Charts are cumulative,)`

`text(Consider 2015,)`

`text(Cricket players = 30)`

`text(Tennis players) = 60 – 30 = 30`

`=> C`

Algebra, STD2 A4 2011 HSC 26b

Jack needs to find the number of years, `t`, it will take for a population of bats to first exceed 18 000.

He uses a ‘guess-and-check’ method to estimate `t` in the following equation

`5 xx 3^t = 18\ 000.`

Here is his working:

  1. Jack’s next guess is  `t = 6`. Show Jack’s correct working for this guess, including the calculation and conclusion.  (1 mark)

  2. Continue using the ‘guess-and-check’ method to find the number of years, `t`, it will take for the population to first exceed 18 000, if `t` is a whole number. Include the calculations and conclusions.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8`
Show Worked Solution

i.   `text(When)\ t = 6,`

`5 xx 3^6 = 3645`

`=>\ text(Too small)`

 

ii.   `text(When)\ \ t = 7,`

`5 xx 3^7 = 10\ 935`

`=>\ text(Too small)`
 

`text(When)\ \ t = 8,`

`5 xx 3^8 = 32\ 805`

`=>\ text(exceeds 18 000)`

`:. t = 8`

Probability, STD2 S2 2018 HSC 30d

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 tokens labelled 10 cents and 3 tokens labelled $2. The player wins the total value of the two tokens drawn.

Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)

 


Show Answers Only

Show Worked Solution

Financial Maths, STD2 2014 HSC 27a

Alex is buying a used car which has a sale price of  $13 380. In addition to the sale price there are the following costs:

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
  2. Calculate the Stamp Duty payable.   (1 mark)
  3.  
  4. Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following: 
    1.  
    2. 2014 27a2
  5. Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)
  6.  
  7. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
  8. What extra cover is provided by the comprehensive car insurance?   (1 mark)

 

Show Answers Only
  1. `$402`
  2. `$985.74`
  3. `text(Comprehensive insurance covers Alex)`
  4. `text(for damage done to his own car as well.)`
  5.  
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
(i)    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

(ii)   `text(Base rate) = $845`

`text(FSL) =\ text(1%) xx 845 = $8.45`
 

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`

 

`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

`:.\ text(Total cost)` `= 845 + 8.45 + 46.94 + 85.35`
  `= $985.74`

 

♦ Mean mark 34%.
(iii)   `text(Comprehensive insurance covers Alex)`
  `text(for damage done to his own car as well.)`

Measurement, STD2 M2 SM-Bank 4

Lee wants to call home to Sydney (34°S, 151°E) from Pateete, Tahiti (17°S, 149°W) when it is 7 pm on Friday in Sydney.

Sydney is eleven hours ahead of UTC. Papeete is ten hours behind UTC.

Give the time and day in Papeete when she should make the call. (Ignore time zones.)  (2 marks)

Show Answers Only

`10text{pm (Thurs)}`

Show Worked Solution

`text(Sydney is 21 hours ahead.)`

`:.\ text(Time in Papeete)`

`= 7text{pm (Fri) less 21 hours}`

`= 10text{pm (Thurs)}`

Measurement, STD2 M2 2011 HSC 27b

Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W.

Both places lie on the Equator. 

  1. Calculate the shortest distance between these two places (`d`), to the nearest kilometre, using

     

      `d=theta/360 xx  2pir`  where  `theta=91°`  and  `r=6400` km   (1 mark)

  2. The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul?    (2 marks)

Show Answers Only
  1. `10\ 165\ text(km)\ \ \ text{(nearest km)}`
  2. `152^@ text(E)`
Show Worked Solution
i.  `text(Shortest distance)` `= 91/360 xx 2 pi r`
  `= 91/360 xx 2 xx pi xx 6400`
  `= 10\ 164.79…`
  `=10\ 165\ text(km)\ text{(nearest km)}`

 

♦♦ Mean mark 33%
ii. `text(Latitude)`
  `4^@\ text(South of Jarvis Island)`
  `text(S)text(ince Jarvis Island is on equator)`
  `=> text(Latitude is)\ 4^@ text(S)`
   
  `text(Longitude)`
  `text(Jarvis Island is)\ 160^@ text(W)`
  `text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)`
  `text(which is)\ 28^@\ text{past meridian (180°)}`
`=>\ text(Longitude)` `= (180 -28)^@ text(E)`
  `= 152^@ text(E)`

 
`:.\ text(Position is)\ (4^@text{S}, 152^@text{E})`

Probability, STD2 S2 2011 HSC 26a

The two spinners shown are used in a game.

2UG 2011 26a1

Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.

2UG 2011 26a2

  1. What is the value of `X` in the table?   (1 mark)

  2. What is the probability of obtaining a score less than 4?   (1 mark)

  3. On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3?   (1 mark)

Show Answers Only
  1. `5`
  2. `1/2`
  3. `2/3`
Show Worked Solution

i.   `X=3+2=5`
 

ii.   `P(text{score}<4)=6/12=1/2`
 

iii.   `P(3)=2/3`

Financial Maths, STD2 F4 2008 HSC 24c

Heidi’s funds in a superannuation scheme have a future value of  $740 000  in 20 years time. The interest rate is 4% per annum and earnings are calculated six-monthly.

What single amount could be invested now to produce the same result over the same period of time at the same interest rate?  (3 marks)

Show Answers Only

`$335\ 138.91`

Show Worked Solution
`FV` `= PV(1 + r)^n`
`740\ 000` `= PV(1 + 2/100)^40`
`:. PV` `= (740\ 000)/((1.02)^40)`
  `= 335\ 138.907…`
  `= $335\ 138.91`

Algebra, STD2 A2 SM-Bank 3

The average height, `C`, in centimetres, of a girl between the ages of 6 years and 11 years can be represented by a line with equation

`C = 6A + 79`

where `A` is the age in years. For this line, the gradient is 6.

  1. What does this indicate about the heights of girls aged 6 to 11?   (1 mark)

  2. Give ONE reason why this equation is not suitable for predicting heights of girls older than 12.   (1 mark)

Show Answers Only
  1. `text(It indicates that 6-11 year old girls, on average, grow 6 cm per year.)`
  2. `text(Girls eventually stop growing, and the equation doesn’t factor this in.)`
Show Worked Solution

i.  `text(It indicates that 6-11 year old girls, on average, grow)`

`text(6 cm per year.)`
 

ii. `text(Girls eventually stop growing, and the equation doesn’t)`

`text(factor this in.)`

Algebra, STD2 A1 SM-Bank 10

For adults (18 years and older), the Body Mass Index is given by

`B = m/h^2`  where  `m = text(mass)`  in kilograms and  `h = text(height)`  in metres.

The medically accepted healthy range for  `B`  is  `21 <= B <= 25`.

What is the minimum weight for a 163 cm adult female to be considered healthy? (2 marks)

Show Answers Only

`55.8\ text(kg)`

Show Worked Solution

`B = m/h^2`

`h = 163\ text(cm) = 1.63\ text(m)`
 

`text(Given)\ \ 21 <= B <= 25,`

`=>B = 21\ \ text(for minimum healthy weight.)`

`21` `= m/1.63^2`
`:. m` `= 21 xx 1.63^2`
  `= 55.794…`
  `= 55.8\ text(kg)\ text{(1 d.p.)}`

Measurement, STD2 M2 SM-Bank 3

An aircraft travels at an average speed of  913 km/h. It departs from a town in Kenya  (0°, 38°E)  on Tuesday at 10 pm and flies east to a town in Borneo  (0°, 113°E).

  1. Calculate the distance (`d`), to the nearest kilometre, between the two towns, using

     

    `d=theta/360 xx 2 pi r`  where  `theta = 75^@`  and  `r=6400` km  (2 marks)

  2. How long will the flight take? (Answer to the nearest hour.)   (1 mark)

  3. What will be the day and local time in Borneo when the aircraft arrives? (Ignore time zones.)   (2 marks)

Show Answers Only
  1. `8378\ text(km)`
  2. `9\ text(hours)`
  3. `text(12 midday on Wednesday)`
Show Worked Solution
i.   `text(Angular difference in longitude)`

`= 113 – 38`

`= 75^@`
 

`:.\ text(Distance)` `= 75/360 xx 2 xx pi xx 6400`
  `= 8377.58…`
  `= 8378\ text(km)\ text{(nearest km)}`

 

ii.    `text(Flight time)` `= text(Distance)/text(Speed)`
    `= 8378/913`
    `= 9.176…`
    `= 9\ text(hours)\ text{(nearest hr)}`

 

iii.   `text(Time Difference)` `= 75 xx 4`
    `= 300\ text(minutes)`
    `= 5\ text(hours)`

 
`text(Kenya is further East)`

COMMENT: NESA Nov-19 syllabus updates … use longitude to calculate time differences (1° = 4 mins).

`=>\ text(Kenya is +5 hours)`
 

`:.\ text(Arrival time in Kenya)`

`= text{10 pm (Tues) + 5 hrs + 9 hrs}\ text{(flight)}`

`= 12\ text(midday on Wednesday)`

Algebra, MET1 2018 VCAA 5

Let  `f: (2, ∞) -> R`, where  `f(x) = 1/((x - 2)^2)`.

State the rule and domain of  `f^(−1)`.  (3 marks)

Show Answers Only

`f^(−1)(x) = 1/sqrtx + 2`

`text(Domain of)\ \ f^(−1)(x):\ (0, ∞)`

Show Worked Solution

`y = 1/((x – 2)^2)`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= 1/((y – 2)^2)`
`(y – 2)^2` `= 1/x`
`y – 2` `= 1/sqrtx`
`y` `= 1/sqrtx + 2`

 
`:. f^(−1)(x) = 1/sqrtx + 2`

`text(Domain of)\ \ f^(−1)(x):\ (0, ∞)\ \ text(or)\ \ R^+`

Calculus, MET1 2018 VCAA 2

The derivative with respect to `x` of the function  `f:(1,∞) -> R`  has the rule  `f′(x) = 1/2 - 1/(2x - 2)`.

Given that  `f(2) = 0`, find  `f(x)`  in terms of `x`.  (3 marks)

Show Answers Only

`f(x) =1/2[x – ln(x – 1)] – 1`

Show Worked Solution

`fprime(x) = 1/2 – 1/(2x – 2)`

`f(x)` `= int(1/2 – 1/(2x -2))\ dx`
  `= 1/2 int(1 – 1/(x – 1))\ dx`
  `= 1/2[x – ln|x – 1|] + c`

 
`text(Given)\ \ f(2) = 0,`

`0 = 1/2(2 – ln 1) + c`

`c = −1`

`:. f(x) = 1/2[x – ln|x – 1|] – 1`

Probability, MET2 2018 VCAA 14 MC

Two events, `A` and `B`, are independent, where  `text(Pr) (B) = 2 text(Pr) (A)`  and  `text(Pr) (A uu B) = 0.52`

`text(Pr) (A)`  is equal to

  1. 0.1
  2. 0.2
  3. 0.3
  4. 0.4
  5. 0.5
Show Answers Only

`B`

Show Worked Solution

`text(If)\ \ p_A = text(Pr)(A) \ =>\  2p_A = text(Pr)(B)`

`text(Pr)(A uu B)` `= text(Pr)(A) + text(Pr)(B) – text(Pr)(A) xx text(Pr)(B)`
`0.52` `= p_A + 2p_A – p_A ⋅ 2p_A`
`2(p_A)^2 – 3 p_A + 0.52` `= 0`
`:. p_A` `= 0.2`

  
`=>   B`

Calculus, MET1 2018 VCAA 1b

Let  `f(x) = (e^x)/(cos(x))`.

Evaluate  `f′(pi)`.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`f′(x) = (e^x)/(cos(x))`

`u` `= e^x` `v` `= cos(x)`
`u′` `= e^x` `v′` `= −sin(x)`
`f′(x)` `= (u′v – uv′)/(v^2)`
  `= (e^x · cos(x) + e^x sin(x))/(cos^2(x))`

 

`f′(pi)` `= (e^pi · cospi + e^pi sinpi)/(cos^2 pi)`
  `= (e^pi(−1) + e^pi · 0)/((−1)^2)`
  `= −e^pi`

Networks, STD2 N3 SM-Bank 33

A project requires nine activities (A–I) to be completed. The duration, in hours, and the immediate predecessor(s) of each activity are shown in the table below.
 


 

  1. Sketch the network, identifying each activity and its duration.  (2 marks)

  2. Identify the critical path and the minimum completion time of the project.  (2 marks)

Show Answers Only
  1.   
  2. `text(Critical path is)\ ACFGI.`
    `text(20 hours)`
Show Worked Solution

i.  `text(Sketch the network:)`

 

 
ii.   `text(Completion time of possible paths:)`

`ABEGI = 4 + 3 + 5 + 4 + 3 = 19\ text(hours)`

`ACFGI = 4 + 7 + 2 + 4 + 3 = 20\ text(hours)`

`ADHI = 4 + 2 + 5 + 3 = 14\ text(hours)`
 

`:.\ text(Critical path is)\ ACFGI.`

`:.\ text{Minimum completion time  = 20 hours}`

Calculus, MET1 2018 VCAA 1a

If  `y = (−3x^3 + x^2 - 64)^3`, find  `(dy)/(dx)`.  (1 mark)

Show Answers Only

`3(−3x^3 + x^2 – 64)^2 · (−9x^2 + 2x)`

Show Worked Solution

Mean mark 58%!
MARKER’S COMMENT: Poor use use brackets was notable.

`y` `= (−3x^3 + x^2 – 64)^3`
`(dy)/(dx)` `= 3(−3x^3 + x^2 – 64)^2 · (−9x^2 + 2x)`

 

Probability, MET2 2018 VCAA 13 MC

In a particular scoring game, there are two boxes of marbles and a player must randomly select one marble from each box. The first box contains four white marbles and two red marbles. The second box contains two white marbles and three red marbles. Each white marble scores −2 points and each red marble scores +3 points. The points obtained from the two marbles randomly selected by a player are added together to obtain a final score.

What is the probability that the final score will equal +1?

  1. `2/3`
  2. `1/5`
  3. `2/5`
  4. `2/15`
  5. `8/15`
Show Answers Only

`E`

Show Worked Solution

`text(Final score requires a red and white.)`

`:. P(+1)` `= P(WR) + P(RW)`
  `= 4/6 xx 3/5 + 2/6 xx 2/5`
  `= 8/15`

 
`=>   E`

Probability, MET2 2018 VCAA 12 MC

The discrete random variable `X` has the following probability distribution.
 

Let `mu` be the mean of `X`.

`text(Pr) (X < mu)`  is

  1. `1/2`
  2. `1/4`
  3. `17/20`
  4. `4/5`
  5. `7/10`
Show Answers Only

`E`

Show Worked Solution
`mu` `= (1 xx 9/20) + (2 xx 1/10) + (3 xx 1/20) + (6 xx 3/20)`
  `= 17/10`

 

`text(Pr) (X < 17/10)` `= text(Pr) (X = 0) + text(Pr) (X = 1)`
  `= 7/10`

`=>   E`

Algebra, MET2 2018 VCAA 10 MC

The function `f` has the property  `f (x + f (x)) = f (2x)`  for all non-zero real numbers `x`.

Which one of the following is a possible rule for the function?

  1. `f(x) = 1 - x`
  2. `f(x) = x - 1`
  3. `f(x) = x`
  4. `f(x) = x/2`
  5. `f(x) = (1 - x)/2`
Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option C:)`

`x + f(x)` `= x + x = 2x`
`f(2x)` `=2x`
`:. f(x + f(x))` `= f(2x)`

 
`=>   C`

Calculus, MET2 2018 VCAA 9 MC

A tangent to the graph of  `y = log_e(2x)`  has a gradient of 2.

This tangent will cross the `y`-axis at

A.   `0`

B.   `-0.5`

C.   `-1`

D.   `-1 - log_e(2)`

E.   `-2 log_e(2)`

Show Answers Only

`C`

Show Worked Solution
`y` `= ln 2x`
`(dy)/(dx)` `= 1/x`

 
`text(When)\ \ (dy)/(dx) = 2:`

`1/x` `= 2`
`x` `= 1/2`

 
`text(When)\ \ x = 1/2,\ \ y = ln 1 = 0`

`text(Find equation)\ \ m = 2,\ \ text(through)\ \ (1/2, 0):`

`y – 0` `= 2 (x – 1/2)`
`y` `= 2x – 1`

 
`=>   C`

Algebra, MET2 2018 VCAA 6 MC

Let  `f` and `g` be two functions such that  `f(x) = 2x`  and  `g(x + 2) = 3x + 1`.

The function  `f (g(x))`  is

  1. `6x - 5`
  2. `6x + 1`
  3. `6x^2 + 1`
  4. `6x - 10`
  5. `6x + 2`
Show Answers Only

`D`

Show Worked Solution
`g(x + 2)` `= 3x + 1`
`g((x – 2) + 2)` `= 3(x – 2) + 1`
`g(x)` `= 3x – 5`

 

`f(x)` `= 2x`
`f(g(x))` `= 2(3x – 5)`
  `= 6x – 10`

 
`=>   D`

Calculus, MET2 2018 VCAA 5 MC

Consider  `f(x) = x^2 + p/x,\ x != 0,\ p in R`.

There is a stationary point on the graph of  `f` when  `x = -2`.

The value of  `p`  is

  1.  −16
  2.  –8
  3. 2
  4. 8
  5. 16
Show Answers Only

`A`

Show Worked Solution

`f(x) = x^2 + p/x`

`f^{′}(x) = 2x-p/x^2`

`text(S.P. occurs when)\ \ f^{′}(x) = 0 and x = -2\ text{(given)}`

`0` `= 2(-2)-p/(-2)^2`
`:. p` `= -16`

 
`=>   A`

GEOMETRY, FUR2 2018 VCAA 2

Frank travelled from Melbourne (38° S, 145° E) to a tennis tournament in Ho Chi Minh City, Vietnam, (11° N, 107° E).

Frank departed Melbourne at 10.30 pm on Monday, 5 February 2018.

Frank arrived in Ho Chi Minh City at 8.00 am on Tuesday, 6 February 2019.

The time difference between Melbourne and Ho Chi Minh City is four hours.

  1.  How long did it take Frank to travel from Melbourne to Ho Chi Minh City?

     

     Give your answer in hours and minutes.   (1 mark)

Ho Chi Minh City is located at latitude 11° N and longitude 107° E.

Assume that the radius of Earth is 6400 km.

  1. i. Write a calculation that shows that the radius of the small circle of Earth at latitude 11° N is 6282 km, rounded to the nearest kilometre.  (1 mark)
  2. ii. Iloilo City, in the Philippines, is located at latitude 11° N and longitude 123° E.

     

        Find the shortest small circle distance between Ho Chi Minh City and Iloilo City.

     

        Round your answer to the nearest kilometre.  (1 mark)

Show Answers Only
    1. `text(13 hours 30 min.)`
    2. i. `text(See Worked Solutions)`
    3. ii. `1754\ text(km)`
Show Worked Solution

a.   `text{8am Ho Chi Minh (Tues) = 12 midday Melbourne (Tues)}`

♦♦ Mean mark 29%.
COMMENT: Review carefully!

`:.\ text(Travel time)` `=\ text{10:30 pm (Mon) to 12 midday (Tues)}`
  `=\ text(13 hours 30 min.)`

 

b.i.  `text(Let)\ \ x=\ text(small circle radius)`

Mean mark part (b)(i) 52%.

 


 

`cos 11^@` `= x/6400`
`:. x` `= 6400 xx cos 11^@`

 

b.ii.  `text(Iloilo City)\ ->\ text(same latitude)`

♦♦ Mean mark part (b)(ii) 33%.
MARKER’S COMMENT: Many students incorrectly used 6400 as the radius in part (b)(ii). Understand why  `6400 xx cos 11°`  is the correct radius here!

 `text(Longitudinal difference) = 123 – 107 = 16^@`
 


 

`:.\ text{Small circle distance (arc)}`

`= 16/360 xx 2 xx pi xx (6400 xx cos 11^@)`

`= 1754.38…`

`= 1754\ text{km (nearest km)}`

GRAPHS, FUR1 2018 VCAA 06 MC

Amy makes and sells quilts.

The fixed cost to produce the quilts is $800.

Each quilt costs an additional $35 to make.

Amy made and sold a batch of 80 quilts for a profit of $1200.

The selling price of each quilt was

  1.  $15.00
  2.  $22.50
  3.  $30.00
  4.  $45.00
  5.  $60.00
Show Answers Only

`E`

Show Worked Solution
`text(C)text(ost of 80 quilts)` `= 800 + 80 xx 35`
  `= $3600`

Mean mark 51%.

`text(Profit)` `=\ text(Revenue) – text(C)text(ost)`
`1200` `= 80 xx text(sale price) – 3600`
`:.\ text(sale price)` `= (1200 + 3600)/80`
  `= $60`

`=> E`

GRAPHS, FUR1 2018 VCAA 04 MC

A team of four students competes in a 4 × 100 m relay race.

Each student in the race runs 100 m.

The order in which each student runs in the race is shown in the table below.
 


 

The following line-segment graph represents the race for Joanne, Sam and Kristen, where `d` is the distance, in metres, from the starting point and `t` is the recorded time, in seconds.

Elle’s line segment is missing.
 


 

The equation representing Elle’s line segment is

  1. `d = 4t`
  2. `d = 4t + 20`
  3. `d = 5t`
  4. `d = 5t - 5`
  5. `d = 6.25t - 81.25`
Show Answers Only

`B`

Show Worked Solution
`m_d` `= (200 – 100)/(45 – 20)`
  `= 4`

 

`d_text(Elle)` `= 100 + 4(t – 20)`
  `= 100 + 4t – 80`
  `= 4t + 20`

 
`=> B`

GRAPHS, FUR1 2018 VCAA 03 MC

The three inequalities below were used to construct the feasible region for a linear programming problem.

`x` `<3`
`y` `<6`
`x + y` `> 6`

 
A point that lies within this feasible region is

  1. `(0, 6)`
  2. `(1, 5)`
  3. `(2, 4)`
  4. `(2, 5)`
  5. `(3, 6)`
Show Answers Only

`D`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option)\ D:`

`(2,5)`

`x < 3 -> 2 < 3\ \ text(satisfies)`

`y < 6 -> 5 < 6\ \ text(satisfies)`

`x + y > 6 -> 2 + 5 > 6\ \ text(satisfies)`
 

`:. (2,5)\ text(lies within the feasible region.)`

`=> D`

GRAPHS, FUR2 2018 VCAA 3

Robert wants to hire a geologist to help him find potential gold locations.

One geologist, Jennifer, charges a flat fee of $600 plus 25% commission on the value of gold found.

The following graph displays Jennifer’s total fee in dollars.
 


 

Another geologist, Kevin, charges a total fee of $3400 for the same task.

  1. Draw a graph of the line representing Kevin’s fee on the axes above.  (1 mark)

     

    `qquad qquad`(answer on the axes above.)

  2. For what value of gold found will Kevin and Jennifer charge the same amount for their work?  (1 mark)
  3. A third geologist, Bella, has offered to assist Robert.
  4. Below is the relation that describes Bella’s fee, in dollars, for the value of gold found.

  5. `qquad  text{fee (dollars)} = {(quad 500),(1000),(2600),(4000):}qquad qquad quad{:(qquad quad 0 <),(2000 <=),(6000 <=),(quad):}{:(text(value of gold found) < 2000),(text(value of gold found) < 6000),(text(value of gold found) < 10\ 000),(text(value of gold found) >= 10\ 000):}`


    The step graph below representing this relation is incomplete.

     

    Complete the step graph by sketching the missing information.  (2 marks)
     

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `$11\ 200`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.  

 

b.    `text(Let)\ \ x = text(value of gold)`

♦ Mean mark 41%.

`text(Jennifer’s total fee) = 600 + 0.25x`

`text(Equating fees:)`

`600 + 0.25x` `= 3400`
`0.25x` `= 2800`
`:.x` `= 2800/0.25`
  `= $11\ 200`

 

c.  

GEOMETRY, FUR1 2018 VCAA 8 MC

A cone with a radius of 2.5 cm is shown in the diagram below.

The slant edge, `x`, of this cone is also shown.

The volume of this cone is 36 cm³.

The surface area of this cone, including the base, can be found using the rule 

`text(surface area) = πr(r + x)`.

The total surface area of this cone, including the base, in square centimetres, is closest to

  1.  20
  2.  42
  3.  63
  4.  67
  5.  90
Show Answers Only

`D`

Show Worked Solution
`text(Volume of cone)` `= 1/3pir^2h`
`36` `= 1/3pi xx 2.5^2 xx h`
`h` `= 108/(pi xx 2.5^2)`
  `= 5.50…`

 
`text(Using Pythagoras,)`

Mean mark 51%.

`x` `= sqrt(2.5^2 + 5.50…^2)`
  `= 6.04…\ \ text(cm)`

 

`:.\ text(Surface Area)` `= pi xx 2.5(2.5 + 6.04)`
  `= 67.07…\ \ text(cm²)`

`=> D`

Trigonometry, 2ADV T1 SM-Bank 1 MC

A windscreen wiper blade can clean a large area of windscreen glass, as shown by the shaded area in the diagram below.
 


 

The windscreen wiper blade is 30 cm long and it is attached to a 9 cm long arm.

The arm and blade move back and forth in a circular arc with an angle of 110° at the centre.

The area cleaned by this blade, in square centimetres, is closest to

  1.    786
  2.  1382
  3.  2573
  4.  4524
Show Answers Only

`B`

Show Worked Solution
`text(Area cleaned)` `=\ text(large sector − small sector)`
  `= 110/360 xx pi xx 39^2 – 110/360 xx pi xx 9^2`
  `= 1382.3…\ \ text(cm²)`

`=>B`

GEOMETRY, FUR1 2018 VCAA 07 MC

A windscreen wiper blade can clean a large area of windscreen glass, as shown by the shaded area in the diagram below.
 

 

The windscreen wiper blade is 30 cm long and it is attached to a 9 cm long arm.

The arm and blade move back and forth in a circular arc with an angle of 110° at the centre.

The area cleaned by this blade, in square centimetres, is closest to

  1.    786
  2.  1382
  3.  2573
  4.  2765
  5.  4524
Show Answers Only

`B`

Show Worked Solution
`text(Area cleaned)` `=\ text(large sector − small sector)`
  `= 110/360 xx pi xx 39^2 – 110/360 xx pi xx 9^2`
  `= 1382.3…\ \ text(cm²)`

`=>B`

GEOMETRY, FUR1 2018 VCAA 5 MC

On the first day of June 2019 in the city of St Petersburg, Russia, (60° N, 30° E) the sun is expected to rise at 3.48 am.

On this same day, the sun is expected to rise in Helsinki, Finland, (60° N, 25° E) approximately

  1. at the same time.
  2. 20 minutes earlier.
  3. 20 minutes later.
  4. one hour earlier.
  5. one hour later.
Show Answers Only

`C`

Show Worked Solution

`text(Difference in longitude) = 30 – 25 = 5°`

Mean mark 51%.

`text(S)text(ince 15° = 1 hour difference)`

`text(Time difference)` `= 5/15 xx 60`
  `= 20\ text(minutes)`

  
`text(St Petersburg is further east.)`

`:.\ text(Sunrise in Helsinki is 20 mins later.)`

`=> C`

GRAPHS, FUR2 2018 VCAA 2

The weight of gold can be recorded in either grams or ounces.

The following graph shows the relationship between weight in grams and weight in ounces.
 


 

The relationship between weight measured in grams and weight measured in ounces is shown in the equation

     weight in grams = `M` × weight in ounces
 

  1. Show that  `M = 28.35`  (1 mark)
  2. Robert found a gold nugget weighing 0.2 ounces.

     

    Using the equation above, calculate the weight, in grams, of this gold nugget.  (1 mark)

  3. Last year Robert sold gold to a buyer at $55 per gram.
    The buyer paid Robert a total of $12 474.

     

    Using the equation above, calculate the weight, in ounces, of this gold.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `5.67\ text(g)`
  3. `8\ text(ounces)`
Show Worked Solution

a.    `text{Substitute (20, 567) into equation:}`

Mean mark 53%.
MARKER’S COMMENT: Giving the equation and writing “solve” did not receive a mark in part (a).

`567` `= M xx 20`
`:. M` `= 567/20`
  `= 28.35\ text(… as required.)`

 

b.   `text{Weight (grams)}` `= 28.35 xx 0.2`
    `= 5.67\ text(g)`

 

c.   `text(Total grams sold)` `= 12474/55`
    `= 226.8\ text(g)`

 

`226.8` `= 28.35 xx W_text(ounces)`
`W_text(ounces)` `= 226.8/28.35`
  `= 8\ text(ounces)`

GEOMETRY, FUR2 2018 VCAA 3

Frank owns a tennis court.

A diagram of his tennis court is shown below

Assume that all intersecting lines meet at right angles.

Frank stands at point `A`. Another point on the court is labelled point `B`.
 

  1. What is the straight-line distance, in metres, between point `A` and point `B`?

     

    Round your answer to one decimal place.   (1 mark)

  2. Frank hits a ball when it is at a height of 2.5 m directly above point `A`.

     

    Assume that the ball travels in a straight line to the ground at point `B`.

     

    What is the straight-line distance, in metres, that the ball travels?

     

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

Frank hits two balls from point `A`.

For Frank’s first hit, the ball strikes the ground at point `P`, 20.7 m from point `A`.

For Frank’s second hit, the ball strikes the ground at point `Q`.

Point `Q` is `x` metres from point `A`.

Point `Q` is 10.4 m from point `P`.

The angle, `PAQ`, formed is 23.5°.
 


 

    1. Determine two possible values for angle `AQP`.

       

      Round your answers to one decimal place.   (1 mark)

    2. If point `Q` is within the boundary of the court, what is the value of `x`?

       

      Round your answer to the nearest metre.  (1 mark)

Show Answers Only
  1. `18.8\ text{m (to 1 d.p.)}`
  2. `19\ text(m)\ text{(nearest m)}`
    1. `52.5^@,\ 127.5^@`
    2. `13\ text(m)\ text{(nearest m)}`
Show Worked Solution
a.    `text(Using Pythagoras,)`
  `AB` `= sqrt(4.1^2 + (6.4 + 6.4 + 5.5)^2`
    `= 18.75…`
    `= 18.8\ text{m (to 1 d.p.)}`

 

b. `text(Let)\ \ d = text(distance travelled)`
  


 

`d` `= sqrt(2.5^2 + 18.8^2)`
  `= 18.96…`
  `= 19\ text{m (nearest m)}`

 

c.i.  

`/_AQP ->\ text(2 possibilities)`

♦♦ Mean mark 25%.

`text(Using Sine rule,)`

`(sin/_AQP)/20.7` `= (sin 23.5^@)/10.4`
`sin /_AQP` `= (20.7 xx sin 23.5^@)/10.4`
  `= 0.7936…`
`:. /_AQP` `= 52.5^@ or 127.5^@\ \ \ text{(to 1 d.p.)}`

 

♦♦♦ Mean mark 18%.

c.ii.    `Q\ text(is within court when)\ \ /_AQP = 127.5^@`
  `/_APQ = 180 – (127.5 + 23.5) = 29^@`
   
  `text(Using sine rule,)`
  `x/(sin 29^@)` `= 10.4/(sin 23.5^@)`
  `:. x` `= (10.4 xx sin 29^@)/(sin 23.5^@)`
    `= 12.64…`
    `= 13\ text{m (nearest m)}`

NETWORKS, FUR1 2018 VCAA 2 MC

Niko drives from his home to university.

The network below shows the distances, in kilometres, along a series of streets connecting Niko’s home to the university.

The vertices `A`, `B`, `C`, `D` and `E` represent the intersection of these streets.
 


 

The shortest path for Niko from his home to the university could be found using

  1. a minimum cut.
  2. Prim’s algorithm.
  3. Dijkstra’s algorithm.
  4. critical path analysis.
  5. the Hungarian algorithm.
Show Answers Only

`C`

Show Worked Solution

`text(Djikstra’s algorithm can be used to find the)`

`text(shortest path.)`

`text(Note that Prim’s algorithm can be used to)`

`text(find the minimum spanning tree but doesn’t)`

`text(necessary provide the shortest path from 2)`

`text(nominated vertices.)`

`=> C`

GEOMETRY, FUR2 2018 VCAA 1

Tennis balls are packaged in cylindrical containers.

Frank purchases a container of tennis balls that holds three standard tennis balls, stacked one on top of the other.

This container has a radius of 3.4 cm and a height of 20.4 cm, as shown in the diagram below.
 

  1. What is the diameter, in centimetres, of this container?   (1 mark)
  2. What is the total outside surface area of this container, including both ends?

     

    Write your answer in square centimetres, rounded to one decimal place.  (1 mark)

A standard tennis ball is spherical in shape with a radius of 3.4 cm.

    1. Write a calculation that shows that the volume, rounded to one decimal place, of one standard tennis ball is 164.6 cm³.  (1 mark)
    2. Write a calculation that shows that the volume, rounded to one decimal place, of the cylindrical container that can hold three standard tennis balls is 740.9 cm³. (1 mark)
    3. How much unused volume, in cubic centimetres, surrounds the tennis balls in this container?

       

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

Show Answers Only
  1. `6.8\ text(cm)`
  2. `508.4\ text(cm)^2\ text{(to 1 d.p.)}`
    1. `164.6\ text(cm)^3\ text{(to 1 d.p.)}`
    2. `740.9\ text(cm)^3\ text{(to 1 d.p.)}`
    3. `247\ text(cm)^3\ text{(nearest cm}^3 text{)}`
Show Worked Solution
a.    `text(Diameter)` `= 2 xx text(radius)`
    `= 2 xx 3.4`
    `= 6.8\ text(cm)`

 

b.    `text(S.A.)` `= 2 pi r^2 + 2 pi rh`
    `= 2 xx pi xx 3.4^2 + 2 xx pi xx 3.4 xx 20.4`
    `= 508.43…`
    `= 508.4\ text(cm)^2\ text{(to 1 d.p.)}`

 

c.i.    `text(Volume)` `= 4/3 pi r^3`
    `= 4/3 xx pi xx 3.4^3`
    `= 164.6\ text(cm)^3\ text{(to 1 d.p.)}`

 

c.ii.    `text(Volume)` `= Ah`
    `= pi xx 3.4^2 xx 20.4`
    `= 740.86…`
    `= 740.9\ text(cm)^3\ text{(to 1 d.p.)}`

 

c.iii.    `text(Unused volume)` `= text(cylinder volume) – text(volume of balls)`
    `= 740.9 – 3 xx 164.6`
    `= 247.1`
    `= 247\ text(cm)^3\ text{(nearest cm}^3 text{)}`