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Algebra, NAP-C3-CA03

At an apple orchard, apples are picked and put in a basket.

The table below shows the total number of apples in the basket after each minute.

\begin{array} {|c|c|c|}
\hline \textbf{Minutes} & \textbf{Total number of apples} \\
\hline 1 & 4 \\
\hline 2 & 8 \\
\hline 3 & 12 \\
\hline 4 & 16 \\
\hline \end{array}

How many apples are in the basket after 10 minutes? 

`20` `30` `35` `40`
 
 
 
 
Show Answers Only

 `40\ text(apples)`

Show Worked Solution

`text(4 apples are put into the basket each minute.)`

`:.\ text(Apples in basket after 10 minutes)`

`=4 xx 10`

`= 40\ text(apples)`

Statistics, NAP-C3-CA02

For three days, Terry counted the colour of cars he saw drive past his house.
 

   

 
Which column on the graph below shows the total number of Red cars? 

 
 
 
 
Show Answers Only

 `text(The second bar from the left.)`

Show Worked Solution

`text(Option 2: 4 red cars passed.)`

 `text{(The second bar from the left)}`

Number, NAP-E4-CA05

Zilda has 4 children.

Each child needs 6 pens when they start school for the new year.

Zilda has no pens when she goes to the shop.

The shop sells its pens in packets of 5.

How many packets does Zilda need to buy?

`2`    `4`   `5`  `6`
 
 
 
 
Show Answers Only

`5\ text(packets)`

Show Worked Solution
`text(Pens required)` `= 4 xx 6`
  `= 24`

 

`:.\ text(Packets required)` `= 24/5`
  `= 4.8`
  `= 5`

Statistics, NAP-E4-CA03

Anne taught a kindergarten class and asked her students what their favorite fruit is.

Their answers were used to draw the pie chart below.
 

 

 
Around a quarter of her students preferred which fruit?

`text(Banana)` `text(Apple)` `text(Mango)` `text(Grapes)`
 
 
 
 
Show Answers Only

`text(Apple)`

Show Worked Solution

`text(A quarter of students will represent a sector that)`

`text(is the size of a quarter of the circle.)`

`:.\ text(Apple)`

Number, NAP-E4-CA01

The airforce buys 8 new fighter jets for $882.0 million.

Each jet costs the same amount.

What is the cost of 1 fighter jet?

`$110.25\ text(million)` `$441.0\ text(million)` `$874.0\ text(million)` `$7056\ text(million)`
 
 
 
 
Show Answers Only

`$110.25\ text(million)`

Show Worked Solution

`{$882.0\ text(million)}/ 8 = $110.25\ text(million)`

Number, NAP-F4-CA07

David buys a surfboard cover online that costs him $128.

The postage costs involved are listed in the table below.
 

 
The surfboard cover weighs 1.1 kg

What does David have to pay, in total, to purchase the surfboard and have it delivered?

`$140` `$145.50` `$149.75` `$153`
 
 
 
 
Show Answers Only

`$149.75`

Show Worked Solution

`text(Surfboard cover cost)\ +\ text(postage)`

`= $128 + $21.75`

`= $149.75`

Number, NAP-F4-NC04

Nigella placed a leg of lamb into her oven in the morning.

The 1st measurement of the lamb's temperature was  −6°C.

Fifteen minutes later, a second temperature reading was taken, which measured  16°C.

What was the change in temperature between the two measurements?

`text(decrease of)\ 22^@text(C)` `text(decrease of)\ 10^@text(C)` `text(increase of)\ 10^@text(C)` `text(increase of)\ 22^@text(C)`
 
 
 
 
Show Answers Only

`text(increase of)\  22^@text(C)`

Show Worked Solution
`text(Change in temperature)` `=16-(-6)`
  `=22^@ text(C)`

 

`:.\ text(An increase of  22°C.)`

Number, NAP-F4-NC03

A circle is divided into 5 equal areas and labelled, as shown in the diagram below.
 

 
What percentage of the circle's area has been labelled with an odd number?

`text(3%)` `text(15%)` `text(30%)` `text(40%)` `text(60%)`
 
 
 
 
 
Show Answers Only

`text(60%)`

Show Worked Solution

`text(S)text(ince all areas are equal,)`

`text(Percentage labelled with an odd number)`

`=3/5`

`=60text(%)`

Number, NAP-F4-CA04

The first three days of the Brisbane cricket test had the following attendances:

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Day 1}\rule[-1ex]{0pt}{0pt} & 20\ 156\\
\hline
\rule{0pt}{2.5ex}\text{Day 2}\rule[-1ex]{0pt}{0pt} & 18\ 397\\
\hline
\rule{0pt}{2.5ex}\text{Day 3}\rule[-1ex]{0pt}{0pt} & 29\ 981\\
\hline
\end{array}

What was the total crowd over the first 3 days, to the nearest `1000?`

`68\ 000` `69\ 000` `70\ 000` `71\ 000`
 
 
 
 
Show Answers Only

`69\ 000\ text{(nearest 1000)}`

Show Worked Solution

`20\ 156 + 18\ 397 + 29\ 981`

`= 68\ 534`

`= 69\ 000\ text{(nearest 1000)}`

Number, NAP-G4-NC02

Ben spent twice as much money as Mark.

If they spent a total of $90, how much did Ben spend?

`$15` `$30` `$60` `$180`
 
 
 
 
Show Answers Only

`$60` 

Show Worked Solution

`text(Strategy 1)`

`text(Ratio)`

`text{Ben : Mark = 2 : 1  (3 parts)}`

`text(1 part)\ = $90-:3=$30`

`:.\ text(Ben spent  2 × $30 = $60)`

 

`text(Strategy 2)`

`text(Let)\ \ $x = text(Amount Mark spent)`

`2x + x` `= 90`
`x` `= $30`

 

`:.\ text(Ben spent $60.)`

Measurement and Geometry, NAP-G4-CA04

Petunia walks along the path from `A` to `B.`

How far does she walk in kilometres?

`0.14` `1.4` `14` `1400`
 
 
 
 
Show Answers Only

`1.4\ text(km)`

Show Worked Solution
`text(Distance)` `= 14 xx 100\ text(m)`
  `= 1.4\ text (km)`

Measurement, NAP-I4-CA06

Peter left home at 9:15 in the morning and did not return until 5:25 in the afternoon.

How long was Peter away from his house?

 
 3 hours 50 minutes
 
 4 hours 10 minutes
 
 7 hours 50 minutes
 
 8 hours 10 minutes
 
 13 hours 20 minutes
Show Answers Only

`8\ text(hours)\ 10\ text(minutes)`

Show Worked Solution
`text(Time away)` `= 2 text(h)\ 45\ text(min) + 5 text(h)\ 25\ text(min)`
  `= 8\ text(hours)\ 10\ text(minutes)`

Quadratic, EXT1 2016 HSC 14c

The point  `T(2at,at^2)` lies on the parabola `P_1` with the equation  `x^2=4ay`.

The tangent to the parabola `P_1` at `T` meets the directrix at `D`.

The normal to the parabola `P_1` at `T` meets the vertical line through `D` at the point `R`, as shown in the diagram.

 

ext1-2016-hsc-q14

  1. Show that the point `D` has coordinates `(at - a/t, −a)`.  (1 mark)
  2. Show that the locus of `R` lies on another parabola `P_2`.  (3 marks)
  3. State the focal length of the parabola `P_2`.  (1 mark)

It can be shown that the minimum distance between `R` and `T` occurs when the normal to `P_1` at `T` is also the normal to `P_2` at `R`.  (Do NOT prove this.)

  1. Find the values of `t` so that the distance between `R` and `T` is a minimum.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `a/4`
  4. `+- sqrt 2`
Show Worked Solution

i.     `text(Show)\ \ D (text{at}\ -a/t, -a)`

`text(T)text(angent equation at)\ \ T:`

`y = tx – at^2`

`D\ \ text(occurs when)\ \ y = -a,`

`tx – at^2` `= -a`
`tx` `= at^2 – a`
`x` `= at – a/t`

`:. D\ text(has coordinates)\ \ (text{at}\ -a/t, -a)`

 

ii.    `text(Normal equation at)\ \ T:`

♦♦ Mean mark 32%.
MARKER’S COMMENT: Normal equation is found in the Reference Sheet. Save time by using it!

`x + ty = 2at + at^3`

`R\ text(occurs when)\ \ x = at – a/t`

`at – a/t + ty` `= 2at + at^3`
`ty` `= at + a/t + at^3`
`y` `= a + a/t^2 + at^2`
  `= a(1 + 1/t^2 + t^2)\ \ …\ text{(*)}`

 

`:. R (a (t – 1/t), a (1 + 1/t^2 + t^2))`

`x^2` `= a^2 (t – 1/t)^2`
  `= a^2 (t^2 – 2 + 1/t^2)`
  `= a + (1+1/t^2 + t^2 – 3)`
  `= a^2 (y/a – 3)\ \ text{(see (*) above)}`
  `= ay – 3a^2`

 

`:.\ text(Locus of)\ R\ text(is)\ \ x^2 = ay – 3a^2`

 

iii.   `text(In the form)\ \ x^2 = 4ay,`

♦♦♦ Mean mark 15%.
`x^2` `= a(y – 3a)`
  `= 4 · a/4 (y – 3a)`

 

`:.\ text(Focal length) = a/4`

 

iv.   `text(Equation of)\ \ P_2`

♦♦♦ Mean mark 11%.
`x^2` `= ay – 3a^2`
`y` `= x^2/a + 3a`
`y prime` `= (2x)/a`

`text(At)\ \ R,\ \ x = a (t – 1/t)`

`:.\ text(Gradient of normal at)\ \ R`

`=(-a)/(2x)`

`= (-a)/(2a(t – 1/t)) xx t/t`

`= (-t)/(2(t^2 – 1))`

 

`text(Gradient of normal at)\ \ T:`

`x + ty` `= 2at + at^3`
`y` `= -1/t x + 2a + at^2`
`:. m` `= -1/t`

 

`text(Distance)\ \ RT\ \ text(is a minimum when)`

`(-t)/(2(t^2 – 1))` `= -1/t`
`t^2` `= 2t^2 – 2`
`t^2` `= 2`
`:. t` `= +- sqrt 2`

Binomial, EXT1 2016 HSC 14b

Consider the expansion of  `(1 + x)^n`, where `n` is a positive integer.

  1. Show that  `2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`.  (1 mark)
  2. Show that  `n2^(n - 1) = ((n),(1)) + 2((n),(2)) + 3((n),(3)) + … + n((n),(n))`.  (1 mark)
  3. Hence, or otherwise, show that  `sum_(r = 1)^n ((n),(r))(2r - n) = n`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text[(See Worked Solutions)}`
  2. `text(Proof)\ \ text[(See Worked Solutions)}`
  3. `text(Proof)\ \ text[(See Worked Solutions)}`
Show Worked Solution

i.     `text(Show)`

`2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`

`text(Using binomial expansion)`

`(1 + x)^n = ((n), (0)) + ((n), (1))x + ((n),(2)) x^2 + … + ((n), (n)) x^n`

`text(Let)\ \ x = 1,`

`2^n = ((n), (0)) + ((n), (1)) + ((n), (2)) + … + ((n), (n))`

`text(… as required.)`

 

ii.   `text(Differentiate both sides of expansion,)`

`n (1 + x)^(n – 1) = ((n), (1)) + 2 ((n), (2))x + 3 ((n), (3)) x^2 + … + n ((n), (n)) x^(n – 1)`

`text(Let)\ \ x = 1,`

`n2^(n – 1) = ((n), (1)) + 2 ((n), (2)) + 3 ((n), (3)) + … + n ((n), (n))`

`text(… as required.)`

 

iii.  `text{Multiply part (i)} xx n`

♦♦♦ Mean mark 14%.
`n2^n` `= n[((n), (0)) + ((n), (1)) + … + ((n), (n))]`
  `= sum_(r = 0)^n ((n), (r)) n\ \ text{…  (1)}`

 

`text{Multiply part (ii)} xx 2`

`2 xx n2^(n – 1)` `= 2[((n), (1)) + 2 ((n), (2)) + … + n ((n), (n))]`
`n2^n` `= sum_(r = 1)^n ((n), (r)) 2r\ \ text{…  (2)}`

 

 `text(Subtract) qquad (2) – (1)`

`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 0)^n ((n), (r))n` `= n2^n – n2^n`
`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 1)^n ((n), (r)) n – n` `= 0`
`sum_(r = 1)^n ((n), (r)) (2r – n)` `= n\ \ text(…  as required)`

Plane Geometry, EXT1 2016 HSC 13c

The circle centred at `O` has a diameter `AB`. From the point `M` outside the circle the line segments `MA` and `MB` are drawn meeting the circle at `C` and `D` respectively, as shown in the diagram. The chords `AD` and `BC` meet at `E`. The line segment `ME` produced meets the diameter `AB` at `F`.

ext1-2016-hsc-q13_1

Copy or trace the diagram into your writing booklet.

  1. Show that `CMDE` is a cyclic quadrilateral.  (2 marks)
  2. Hence, or otherwise, prove that `MF` is perpendicular to `AB`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.  
`/_ ACB` `= /_ ADB = 90^@ qquad text{(angle in semi-circle)}`
`/_ MCE` `= 90^@ qquad (/_ MCA\ text{is a straight angle)}`
`/_ MDA` `= 90^@ qquad (/_ MDB\ text{is a straight angle)}`

 
`:. CMDE\ text(is a cyclic quad)`

`qquad text{(opposite angles are supplementary)}`
 

ii.   `text(Consider)\ \ Delta MAB,`

♦♦♦ Mean mark 14%.

`CB\ \ text(is an altitude.)`

`AD\ \ text(is an altitude.)`

`text(S) text(ince the altitudes of a triangle are)`

`text(concurrent)\ \ text{(in}\ Delta MAB, text{at}\ Etext{),}`

`=> MF\ \ text(must be an altitude.)`

`:. MF _|_ AB`

Calculus, 2ADV C3 2016 HSC 16b

Some yabbies are introduced into a small dam. The size of the population, `y`, of yabbies can be modelled by the function

`y = 200/(1 + 19e^(-0.5t)),`

where `t` is the time in months after the yabbies are introduced into the dam.

  1. Show that the rate of growth of the size of the population is
  2. `qquad qquad (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`.  (2 marks)

  3. Find the range of the function `y`, justifying your answer.  (2 marks)

  4. Show that the rate of growth of the size of the population can be written as
  5. `qquad qquad y/400 (200-y)`.  (1 mark)

  6. Hence, find the size of the population when it is growing at its fastest rate.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `10 <= y < 200`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `100`
Show Worked Solution
i.   `y` `= 200/(1 + 19 e^(-0.5t))`
  `(dy)/(dt)` `= 200/(1 + 19 e^(-0.5t))^2 xx d/(dt) (1 + 19 e^(-0.5t))`
    `= (-200)/(1 + 19 e^(-0.5t))^2 xx -0.5 xx 19 e^(-0.5t)`
    `= (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2\ \ text(… as required)`

 

ii.   `text(When)\ \ t = 0,`

♦♦♦ Mean mark (ii) 21%.

`y = 200/(1 + 19) = 10`

`text(As)\ \ t -> oo,\ \ (1 + 19^(-0.5t)) -> 1`

`:. y -> 200`

`:.\ text(Range)\ \ \ 10 <= y < 200`

 

iii.  `(dy)/(dt) = (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`

♦♦♦ Mean mark (iii) 18%.

`text(S) text(ince)\ \ y = 200/(1 + 19 e^(-0.5t))`

`=> (1 + 19 e^(-0.5t)) = 200/y`

`=> 19 e^(-0.5t) = 200/y-1 = (200-y)/y`

`text(Substituting into)\ \ (dy)/(dt):`

`(dy)/(dt)` `= (100 ((200-y)/y))/(200/y)^2`
  `= 100 ((200-y)/y) xx y^2/200^2`
  `= y/400 (200-y)\ \ text(… as required)`

 

iv.   `(dy)/(dt) = -y^2/400 + y/2`

♦♦♦ Mean mark (iv) 14%.

`text(Sketching the parabola:)`

`(-y^2)/400 + y/2` `= 0`
`-y^2 + 200y` `= 0`
`y (200-y)` `= 0`

 

hsc-2016-16bi

`:.\ text(Maximum)\ \ (dy)/(dt)\ \ text(occurs when)\ \ y = 100.`

Proof, EXT2 P2 2016 HSC 16c

In a group of `n` people, each has one hat, giving a total of `n` different hats. They place their hats on a table. Later, each person picks up a hat, not necessarily their own.

A situation in which none of the `n` people picks up their own hat is called a derangement.

Let  `D(n)`  be the number of possible derangements.

  1. Tom is one of the `n` people. In some derangements Tom finds that he and one other person have each other's hat.

     

    Show that, for  `n > 2`, the number of such derangements is  `(n - 1) D (n - 2).`  (1 mark)

  2. By also considering the remaining possible derangements, show that, for  `n > 2,`

    `qquad qquad D(n) = (n - 1) [D(n - 1) + D(n - 2)].`  (2 marks)

  3. Hence, show that  `D(n) - nD(n - 1) = -[D(n - 1) - (n - 1) D(n - 2)]`,  for  `n > 2.`  (1 mark)

  4. Given  `D(1) = 0`  and  `D(2) = 1`,  deduce that  `D(n) - n D(n - 1) = (-1)^n`,  for  `n > 1.`  (1 mark)

  5. Prove by mathematical induction, or otherwise, that for all integers  `n >= 1,\ D(n) = n! sum_(r = 0)^n (-1)^r/(r!).`  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(Tom and one other person can have each)`

♦♦♦ Mean mark 22%.

`text(other’s hat in)\ (n – 1)\ text(combinations.)`

`text(There are)\ \ D(n – 2)\ \ text(possibilities for the)`

`text(rest of the people selecting the wrong hats.)`

`:. text(Number of derangements) = (n – 1)D(n – 2)`

 

ii.   `text(Consider all possible derangements:)`

♦♦♦ Mean mark 6%.

`text(Any of)\ n\ text(people choose the wrong hat.)`

`text(Remainder)\ (n – 1)\ text(people can select the)`

`text(wrong hat in)\ D(n – 1)\ text(ways.)`

`:. nD(n – 1)\ text(derangements.)`

`text{This includes part (i) combinations.}`

 

`:.\ text(Remaining possible derangements)`

`= nD(n – 1) – (n – 1)D(n – 2)`

`= nD(n – 1) – D(n – 1)`

`= (n – 1)D(n – 1)`

`:. D(n)` `= (n – 1)D(n – 1) + (n – 1)D(n – 2)`
  `= (n – 1)[D(n – 1) + D(n – 2)]`

 

♦♦ Mean mark part (iii) 37%.
iii.    `D(n)` `= (n – 1)[D(n – 1) + D(n – 2)]`
    `= nD(n – 1) – D(n – 1) + (n – 1)D(n – 2)`

 

`:. Dn – nD(n – 1) = −[D(n – 1) – (n – 1)D(n – 2)]`

 

iv.   `D(1) = 0, D(2) = 1`

♦♦♦ Mean mark 1%!

`D(2) – 2D(1) = 1`

`D(3) – 3D(2) = −[D(2) – 2D(1)] = −1`

`D(4) – 4D(3) = −[D(3) – 3D(2)] = −(−1) = 1`

`D(5) – 5D(4) = −[D(4) – 4D(3)] = −1`

 

`:. D(n) – nD(n – 1) = (−1)^n\ text(for)\ n > 1`

 

v.   `text(Prove)\ D(n) = n! sum_(r = 0)^n ((−1)^r)/(r!)\ text(for)\ n >= 1`

`text(When)\ n = 1,`

`D(1) = 1! sum_(r = 0)^1 ((−1)^r)/(r!) = 1 – 1 = 0`

`text(S)text(ince)\ D(1) = 0\ (text(given)),`

`:.\ text(True for)\ n = 1`

♦♦♦ Mean mark 13%.

 

`text(Assume true for)\ \ n = k,`

`text(i.e.)\ \ D(k) = k! sum_(r = 0)^k ((−1)^r)/(r!)`

`text(Prove true for)\ \ n=k+1,`

`text(i.e.)\ \ D(k+1) = (k+1)!sum_(r = 0)^(k+1) ((−1)^r)/(r!)`

`D(k+1)`

`= (k + 1)D(k) + (−1)^(k + 1)qquad(text{from part (iv)})`
`= (k + 1) · k! sum_(r = 0)^k ((−1)^r)/(r!) + (−1)^(k + 1)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!)) + (−1)^(k + 1) · ((k + 1)!)/((k + 1)!)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!) + ((−1)^(k+1))/((k+1)!))`
`= (k + 1)! sum_(r = 0)^(k + 1) ((−1)^r)/(r!)`

 

`=> text(True for)\ n = k + 1.`

`:. text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Complex Numbers, EXT2 N2 2016 HSC 16b

  1. The complex numbers  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.

     

    Show that  `u^2 + v^2 = uv.`  (2 marks)

  2. Give an example of non-zero complex numbers  `u`  and  `v`, so that  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.  (1 mark)

Show Answers Only
  1. `text(Show Worked Solutions)`
  2. `u = 1/2 + sqrt3/2 i`
Show Worked Solution

i.   `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark 23%.

 

ext2-2016-hsc-16b-answer

`=> u` `= v text(cis)(±pi/3)`
`u^3` `= v^3text(cis)(±pi)`
`u^3` `= − v^3`
`u^3 + v^3` `= 0`

 
`(u + v)(u^2 – uv + v^2) = 0`

`text(S)text(ince)\ u != v,`

`u^2 – uv + v^2` `= 0`
`:. u^2 + v^2` `= uv`

 

ii.   `text(Let)\ \ v = 1,`

♦♦ Mean mark 33%.
`:. u` `= text(cis)(pi/3)`
  `= cos\ pi/3 + isin\ pi/3`
  `= 1/2 + sqrt3/2 i`

Complex Numbers, EXT2 N2 2016 HSC 16a

  1. The complex numbers  `z = cos theta + i sin theta`  and  `w = cos alpha + i sin alpha`, where  `-pi < theta < pi`  and  `-pi < alpha <= pi`, satisfy

        `1 + z + w = 0.`

    By considering the real and imaginary parts of      `1 + z + w`, or otherwise, show that  `1, \ z`  and  `w` form the vertices of an equilateral triangle in the Argand diagram.  (3 marks)

  2. Hence, or otherwise, show that if the three non-zero complex numbers  `2i, \ z_1`  and  `z_2`  satisfy

     

     
        `|\ 2i\ | = |\ z_1\ | = |\ z_2\ |`  AND  `2i + z_1 + z_2 = 0.`

    then they form the vertices of an equilateral triangle in the Argand diagram.      (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.    `z` `= costheta + isintheta, \ |\ z\ | = 1`
  `w` `= cosalpha + isinalpha, \ |\ w\ | = 1`

 

`text(S)text(ince)\ 1 + z + w = 0`

♦♦♦ Mean mark 25%.
STRATEGY: A clear graphical image simplifies both parts of this question significantly.
`text(Im)(1 + z + w)` `= 0`
`sintheta + sinalpha` `= 0`
`sintheta` `= −sinalpha`
`:. theta` `= −alpha\ …\ (1)`

 

`text(Re)(1 + z + w)` `= 0`
`1 + costheta + cosalpha` `= 0`
`2costheta` `= −1qquad(text(S)text(ince)\ cosalpha  = cos(−theta) = costheta)`
`costheta` `= −1/2`
`:. theta` `= (2pi)/3`
`alpha` `= −(2pi)/3`

 
ext2-2016-hsc-16a-answer2 

 
`=>\ text(All points are on the unit circle)`

`text(separated by)\ (2pi)/3\ text(radians.)`

`:.\ text(They are vertices of an equilateral triangle.)`

 

ii.   `|\ 2i\ | = 2`

`text(Let)\ \ z_1` `= 2(costheta + isintheta), \  |\ z_1\ | = 2`
`z_2`  `= 2(cosalpha + isinalpha), \  |\ z_2\ | = 2`

 

♦♦♦ Mean mark 5%.
`text(Re)(2i + z_1 + z_2)` `= 0`
`2(costheta + cosalpha)` `= 0`
`:. costheta` `= −cosalpha`
`:. theta` `= pi – alpha`

 

`text(Im)(2i + z_1 + z_2)` `= 0`
`2(1 + sintheta + sinalpha)` `= 0`
`sintheta + sinalpha` `= −1`
`2sintheta` `= −1qquad(text(S)text(ince)\ sinalpha = sin(pi – theta) = sintheta)`
`sintheta` `= −1/2`
`:. theta` `= (7pi)/6`
`:. alpha` `= −pi/6`

 
ext2-2016-hsc-16a-answer3 
 

`=>\ text(All points are on the 2 unit)`

`text(circle separated by)\ (2pi)/3\ text(radians.)`

`:. 2i, z_1\ text(and)\ z_2\ text(are vertices of an)`

`text(equilateral triangle.)`

Polynomials, EXT2 2016 HSC 15c

  1. Use partial fractions to show that

    `qquad (3!)/(x(x + 1) (x + 2) (x + 3)) = 1/x - 3/(x + 1) + 3/(x + 2) - 1/(x + 3).`  (2 marks)

  2. Suppose that for `n` a positive integer

     

    `qquad qquad (n!)/(x(x + 1) … (x + n)) = a_0/x + a_1/(x + 1) + … + a_k/(x + k) + … + a_n/(x + n).`

    Show that  `a_k = (-1)^k ((n), (k)).`  (3 marks)

  3. Hence, or otherwise, find the limiting sum of

     

    `qquad qquad 1 - 1/2 ((n), (1)) + 1/3 ((n), (2)) - 1/4 ((n), (3)) + … + (-1)^n/(n + 1).`  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `1/(n + 1)`
Show Worked Solution

i.   `text(Show)`

`(3!)/(x(x + 1)(x + 2)(x + 3)) = 1/x – 3/(x + 1) + 3/(x + 2) – 1/(x + 3)`

`text(LHS)` `= (a_0)/x + (a_1)/(x + 1) + (a_2)/(x + 2) + (a_3)/(x + 3)`
  `= [(a_0(x + 1)(x + 2)(x + 3) + a_1x(x + 2)(x + 3) + a_2x(x + 1)(x + 3) +`
  ` a_3x(x + 1)(x + 2)]//(x(x + 1)(x + 2)(x + 3))`

 

`a_0 = lim_(x -> 0) (3!)/((x + 1)(x + 2)(x + 3)) = 6/6 = 1`

`a_1 = lim_(x -> −1) (3!)/(x(x + 2)(x + 3)) = 6/((−1)(1)(2)) = −3`

`a_2 = lim_(x -> −2) (3!)/(x(x + 1)(x + 3)) = 6/((−2)(−1)(1)) = 3`

`a_3 = lim_(x -> − 3)(3!)/(x(x + 1)(x + 2)) = 6/((−3)(−2)(−1)) = −1`

 

ii.   `text(Given that)`

`(n!)/(x(x + 1) … (x + n)) = (a_0)/x + (a_1)/(x + 1) + … + (a_k)/(x + k) + … +  (a_n)/(x + n)`

 

`text(Show)\ a_k = (−1)^k((n),(k))`

`a_k` `= lim_(x -> −k) (n!(x + k))/(x(x + 1) … (x + k – 1)(x + k)(x + k + 1) …(x + n))`
  `= (n!)/((−k)(−k + 1) … (−k + k – 1)(−k + k + 1) … (−k + n))`
   
  `=>\ text(S)text(ince negative for)\ k\ text(odd,)`
 `a_k` `= ((−1)^kn!)/(k(k – 1) … (2)(1)(1)(2) … (n – k))`
  `= ((−1)^kn!)/(k!(n – k)!)`
  `= (−1)^k ((n),(k))`

 

iii.   `(n!)/(x(x + 1) … (x + n))`

`= (a_0)/x + (a_1)/(x + 1) + … + (a_n)/(x + n)`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(x + n)((n),(n))`

 

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

`(n!)/((1)(2)(3) … (n + 1))`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(n + 1)`

`text(LHS)` `= (n!)/((n + 1)!)`
  `= 1/(n + 1)`

 

`:.\ text(Limiting sum) = 1/(n + 1)`

Algebra, STD2 A4 2016 HSC 29b

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only
  1. `1.5\ text(kg)`
  2. `10text(%)`
Show Worked Solution

i.   `text(When)\ x = 0,`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)`

`= 0.1`

`= 10text(%)`

Probability, STD2 S2 2016 HSC 23 MC

A group of 485 people was surveyed. The people were asked whether or not they smoke. The results are recorded in the table.
 

A person is selected at random from the group.

What is the approximate probability that the person selected is a smoker OR is male?

  1. 33%
  2. 18%
  3. 68%
  4. 87%
Show Answers Only

`=> C`

Show Worked Solution

`P(text(Smoker or a male))`

`= (text(Total males + female smokers))/(text(Total surveyed))`

`= (264 + 68)/485`

`= 0.684…`
 

`=> C`

♦♦ Mean mark 34%.

Algebra, 2UG 2016 HSC 18 MC

The value of `E` varies directly with the square of `S`.

It is known that  `E = 20`  when  `S = 10`.

What is the value of `E` when  `S = 40`?

  1. `40`
  2. `80`
  3. `320`
  4. `400`
Show Answers Only

`=> C`

Show Worked Solution

`E ∝ S^2`

♦♦♦ Mean mark 22%.
STRATEGY: A well structured answer is important in solving this question type. See solution.

`E = kS^2`

`text(Find)\ k:`

`20` `= k(10)^2`
`:. k` `= 0.2`

 

`text(When)\ S = 40,`

`E` `= 0.2 xx 40^2`
  `= 320`

 
`=> C`

Statistics, STD2 S1 2016 HSC 7 MC

Which set of data is classified as categorical and nominal?

  1. blue, green, yellow
  2. small, medium, large
  3. 5.2 cm, 6 cm, 7.21 cm
  4. 4 people, 5 people, 9 people 
Show Answers Only

`A`

Show Worked Solution

`text(Categorical and nominal data is)`

♦♦ Mean mark 26%.

`text(qualitative and not ordered.)`

`=> A`

Trigonometry, 2ADV T2 2016 HSC 8 MC

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

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

`D`

Show Worked Solution

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

♦♦♦ Mean mark 23%.

`cos (2x) = +- 1`

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

 

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

 

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

`=>  D`

Probability, MET1 2007 VCAA 6

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

  1. Calculate  `text(Pr)(A′ ∩ B)`  when  `text(Pr)(A ∩ B) = 1/8`.  (1 mark)
  2. Calculate  `text(Pr)(A′ ∩ B)`  when `A` and `B` are mutually exclusive events.  (1 mark)
Show Answers Only
  1. `5/24`
  2. `1/3`
Show Worked Solution

a.   `text(Sketch Venn diagram:)`

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

met1-2007-vcaa-q6-answer3

`:. text(Pr)(A′ ∩ B)` `=text(Pr)(B) – text(Pr)(A ∩B)`
  `=1/3 – 1/8`
  `=5/24`

 

♦♦ Mean mark 23%.
b.    met1-2007-vcaa-q6-answer4

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

`:. text(Pr)(A′ ∩ B) = 1/3`

Functions, MET1 2008 VCAA 10

Let  `f: R -> R,\ \ f(x) = e^(2x)-1`.

  1. Find the rule and domain of the inverse function  `f^(−1)`.  (2 marks)
  2. On the axes provided, sketch the graph of  `y = f(f^(−1)(x))`  for its maximal domain.  (1 mark)
     

     

    met1-2008-vcaa-q2
     

  3. Find  `f(−f^(−1)(2x))`  in the form  `(ax)/(bx + c)`  where `a`, `b` and `c` are real constants.  (2 marks)
Show Answers Only
  1. `f^(−1)(x) = 1/2log_e(x + 1), x ∈ (−1,∞)`
  2.  
    met1-2008-vcaa-q10-answer
  3. `f(−f^(−1)(2x)) = (−2x)/(2x + 1)`
Show Worked Solution

a.   `text(Let)\ \ y = f(x)`

`text(For Inverse, swap)\ x ↔ y`

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

 

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

 

b.   `f(f^(−1)(x)) = x`

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

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

met1-2008-vcaa-q10-answer

 

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

Calculus, MET1 2011 VCAA 10

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

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

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

 vcaa-2011-meth-10a

  1. Find `BD` and `CD` in terms of `a` and `theta`.  (2 marks)
  2. Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`.  (1 mark)
  3. Find  `(dL)/(d theta)`,  and hence show that  `(dL)/(d theta) = 0`  when  `BD = 2CD`.  (2 marks)
  4. Find the maximum value of `L` if  `a = 3 sqrt 5`.  (1 mark)
Show Answers Only
  1. `BD = a cos theta,\ \ \ CD = a sin theta`
  2. `L = 4 + a + 2 a cos theta + a sin theta`
  3. `text(Proof)\ text{(See Worked Solutions)}`
  4. `L_max = 19 + 3 sqrt 5`
Show Worked Solution

a.   `text(In)\ \ Delta BCD,`

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

 

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

 

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

♦ Mean mark 35%.

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

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

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

 

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

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

 

 vcaa-2011-meth-10ai

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

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

 

Functions, MET1 2011 VCAA 4

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

  1.  find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`(2 marks)
  2.  state the maximal domain for which  `f(g(x))`  is defined.  (2 marks)
Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2 – 9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

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

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

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

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

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

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

`x in (– oo, – 8] uu [– 2, oo)`

Calculus, MET1 2012 VCAA 10

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

  1.  i. Find, in terms of `m`, the `x`-coordinate of the stationary point of the graph of  `y = f(x)`.  (2 marks)
  2. ii. State the values of `m` such that the `x`-coordinate of this stationary point is a positive number.  (1 mark)
  3. For a particular value of `m`, the tangent to the graph of  `y = f(x)`  at  `x = – 6`  passes through the origin.
  4. Find this value of `m`.  (3 marks)
Show Answers Only
  1.  i. `1/m log_e(m/3)`
  2. ii. `m > 3`
  3. `1/6`
Show Worked Solution

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

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

 

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

 

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

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

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

`m_tan` `= f prime (– 6)`
  `= – me^(– 6m) + 3`

 

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

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

 

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

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

Calculus, MET1 2014 VCAA 10

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

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

     


    met1-2014-vcaa-q10
     

     

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

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

     
    met1-2014-vcaa-q10_1
     

    1. Find an expression for `v` in terms of `u`.  (1 mark)
    2. Find the minimum total shaded area and the value of `u` for which the area is a minimum.  (2 marks)
    3. Find the maximum total shaded area and the value of `u` for which the area is a maximum.  (1 mark)
Show Answers Only
  1. `a = −3/2, b = 5`
    1. `v=(4u)/(u-2)`
    2. `8\ text{u²   (when}\ u=4)`
    3. `17\ text{u²}\ \  (text{when}\ u=5/2)`
Show Worked Solution

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

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

 

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

 

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

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

 

`text(Subtract)\ \ (2) – (1)`

`−3 = 2a`

`:. a = −3/2, b = 5`

 

b.i.   `text(Solution 1)`

`text(Using similar triangles:)`

met1-2014-vcaa-q10-answer

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

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

 

`text(Solution 2)`

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

 

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

 

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

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

 

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

`:. u = 4, \ \ u ∈ [5/2,6]`

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

 

`text(Test areas at end points,)`

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

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

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

 

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

♦♦♦ Mean mark 8%.

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

`text(must occur at an endpoint.)`

 

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

Calculus, MET1 2015 VCAA 10

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

met1-2015-vcaa-q10
  

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

  1. Find the coordinates of `T` in terms of `theta`.  (1 mark)
  2. Find the gradient of the tangent to the circle at `T` in terms of `theta`.  (1 mark)
  3. The equation of the tangent to the circle at `T` can be expressed as
  4. `qquad cos(theta)x + sin(theta)y = 2 + 2cos(theta)`
  5.  i. Point `B`, with coordinates `(2, b)`, is on the line segment `XY`.
  6.     Find `b` in terms of `theta`.  (1 mark)
  7. ii. Point `D`, with coordinates `(4, d)`, is on the line segment `XY`.
  8.     Find `d` in terms of `theta`.  (1 mark)
  9. Consider the trapezium `CEDB` with parallel sides of length `b` and `d`.
  10. Find the value of `theta` for which the area of the trapezium `CEDB` is a minimum. Also find the minimum value of the area.  (3 marks)
Show Answers Only
  1. `T(2 + 2costheta, 2 sintheta)`
  2. `(−1)/(tan(theta))`
  3.  i. `2/(sintheta)`
  4. ii. `(2 – 2 costheta)/(sintheta)`
  5. `theta = pi/3`
  6. `A_text(min) = 2sqrt3\ text(u²)`
Show Worked Solution
a.    `cos theta` `= (CM)/(CT)`
    `=(CM)/2`
  `CM` `= 2costheta`
♦♦♦ Part (a) mean mark 20%.
MARKER’S COMMENT: Many students did not include the “+2” in the `x`-coordinate.
`sintheta` `= (TM)/2`
`TM` `= 2sintheta`

 

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

 

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

 

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

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

 

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

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

     

    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 prime (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 part (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 prime (x) = 0\ \ text(for)\ \ x,`

`:. x = (ab + 1)/(4a) 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 part (d) 16%.
`(ab + 1)/(4a)` `= 1/a`
`:. a` `= 3/b`

 

♦ Mean mark part (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)`

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

♦♦♦ Mean mark part (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`

 

`text{Solution 2 [by CAS]}` 

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

`text(Solve:)\ \ f(p) = p, f prime (1) = 1  and f prime (p) = p\ \ text(for)\ \ p,`

`:. p = 4,\  \ \ (p != 1)`

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
    1. `20 cos (x)`
    2. `10 sin (x)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `10 sqrt (sin^2(x) – cos^2 (x))`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `x = cos^-1 (sqrt 3/3) -> T_max = (4000 sqrt 3\ m^3)/27`
  6. `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.

     

    Find, in terms of `p`, the probability that the egg has a white eggshell.  (1 mark)

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

     

    1. Given that the egg has a white eggshell, find, in terms of `p`, the probability that it came from farm B.  (2 marks)
    2. If the probability that this egg came from farm B is 0.3, find the value of `p`.  (1 mark)
Show Answers Only
  1. `(2p+1)/5`
    1. `(1 – p)/(2p + 1)`
    2. `7/16`
Show Worked Solution
a.  

met1-2015-vcaa-q9-answer1

 

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

 

♦ Part (b) mean mark 41%.
MARKER’S COMMENT: Algebraic fractions “were not handled well”!
b.i.    `text(Pr)(B | W)` `= (text(Pr)(B ∩ W))/(text(Pr)(W))`
    `= ((1 – p)/5)/((2p + 1)/5)`
    `=(1-p)/5 xx 5/(2p+1)`
    `= (1 – p)/(2p + 1)`

 

♦♦♦ Part (c) mean mark 19%.
STRATEGY: Previous parts of a question are gold dust for directing your strategy in many harder questions.
b.ii.    `text(Pr)(B | W)` `= 3/10`
  `(1 – p)/(2p + 1)` `= 3/10`
  `10 – 10p` `= 6p + 3`
  `7` `= 16p`
  `:. p` `= 7/16`

CORE*, FUR2 2006 VCAA 4

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

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

 

`=>PMT = −922.898…`

 

`text(Total repayments)`

`= 24 xx 922.898…`

`= $22\ 149.65…`
 

`:.\ text(Total interest paid)`

`= 22\ 149.56… – 20\ 000`

`= 2149.56…`

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

CORE*, FUR2 2007 VCAA 2

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

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

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

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

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

     

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

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

The interest rate is 9% per annum compounding monthly.

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

a.   `text(Interest paid each month)`

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

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

`= $225`

 

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

`text(By TVM Solver:)`

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

 

`=> FV = −16\ 800.77…`

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

 

c.   `text(By TVM Solver,)`

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

 

`=>PMT = −622.750…`

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

GRAPHS, FUR2 2007 VCAA 3

Gas is generally cheaper than petrol. 

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

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

 `y`  be the number of hours driving using petrol

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

Explanations are given for Inequalities 3 and 4.

Inequality 1:   `x ≥ 0`

Inequality 2:   `y ≥ 0`

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

Inequality 5:   `x + y ≤ 24`

 

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

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

GRAPHS, FUR2 2007 VCAA 3

  1. On the graph above

     

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

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

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

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

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

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

     

     

    Maximum = ___________ hours

     

     

    Minimum = ___________ hours

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

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

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

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

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

 

b.i. & ii.

GRAPHS, FUR2 2007 VCAA 3 Answer

 

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

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

`text(is driven on petrol.)`

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

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

 

d.   `text(Maximum = 18 hours)`

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

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

`text(Minimum = 17 hours)`

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

 

CORE*, FUR2 2008 VCAA 5

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

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

The final repayment will be less than $350.

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

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

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

a.   `text(By TVM Solver,)`

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

 

`=>N = 52.422…`

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

 

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

`text(By TVM Solver,)`

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

 

`FV = −147.056…`

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

 

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

`text(By TVM Solver,)`

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

 
`=>FV = -12\ 086.602…`
 

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

`text(By TVM Solver,)`

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

 

`=>PMT = 388.300…`

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

GRAPHS, FUR2 2008 VCAA 3

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

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

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

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

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

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

 

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

 

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

GRAPHS, FUR2 2008 VCAA 3

  1. On the graph above

     

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

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

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

     

    Tiffany qualified for a prize.

     

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

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

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

 

b.i. & ii.

GRAPHS, FUR2 2008 VCAA 3 Answer

 

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

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

`text(time range is between:)`

`text(80 – 150 minutes)`

 

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

`x + y <= 90`

`text(Maximum cycling time occurs)`

`text(when)\ y = 1.5x`

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

 

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

 

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

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

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

 

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

CORE*, FUR2 2009 VCAA 5

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

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

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

     

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

     

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

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

     

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

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

     

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

     

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

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

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

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

`text(By TVM Solver,)`

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

 

`=>FV = −151\ 133.38…`

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

 

b.   `text(Total repayments)`

`= 60 xx 1500`

`= $90\ 000`

`text(Principal paid off after 5 years)`

`= 200\ 000 – 151\ 133.38`

`= $48\ 866.62`
 

`:.\ text(Total interest paid)`

`= 90\ 000 – 48\ 866.62`

`= $41\ 133.38`

 

c.   `text(By TVM Solver,)`

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

 

`=>PMT = −1825.029…`

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