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Algebra, NAPX-p169224v02

A company ships crates overseas and calculates the cost of shipping per crate.

This company uses a formula for calculating the size and cost of shipping.

The formula is shown below:

Size = Length + Width + Height

The maximum size of crates to be shipped overseas is 350 cm.

Which of the following crates is oversized?

  Length Width Height
 
 200 60 80
 
 150 130 90
 
 160 100 70
 
 130 120 100
Show Answers Only

`text(Oversized: Length = 150, width = 130, height = 90)`

Show Worked Solution

`text(Check each option:)`

`text(Option 1 – 200 + 60 + 80 = 340)`

`text{Option 2 – 150 + 130 + 90 = 370 (Oversized)}`

`text(Option 3 – 160 + 100 + 70 = 330)`

`text(Option 4 – 130 + 120 + 100 = 350)`

Algebra, NAPX-p169224v01

A delivery company uses a formula to determine the cost of shipping different sizes of boxes.

The formula they use is as follows:

Size of box = length + width + height

The maximum size that can be shipped is 240 cm.

Which box is oversized?

  Length Width Height
 
 100 80 60
 
 70 60 90
 
 90 90 50
 
 90 110 50

Show Answers Only

`text(Oversized: Length = 90, width = 110, height = 50)`

Show Worked Solution

`text(Check each option:)`

`text(Option 1 – 100 + 80 + 60 = 240)`

`text(Option 2 – 70 + 60 + 90 = 220)`

`text(Option 3 – 90 + 90 + 50 = 230)`

`text{Option 4 – 90 + 110 + 50 = 250 (Oversized)}`

Number, NAPX-p109670v02

Kelly wants to give away some of the apples that came from her family’s farm.

The two small boxes shown below fit either 5 apples or 6 apples.

Kelly has 9 BOX A's and 10 BOX B's.

She shares the apples equally among 15 of her friends.

How many apples will each of her friends receive?

 
      5
 
      7
 
      8
 
   104
 
   105
Show Answers Only

`7`

Show Worked Solution
`text{Total apples}` `= (5 xx 9) + (6 xx 10) `
  `= 45 + 60 `
  `= 105`

 

`text{Apples per friend}` `= frac{105}{15}`
  `= 7`

Number, NAPX-p109670v01

John bought two different bags of bread rolls.
 

   
 

He bought 5 Bags A's and 10 Bag B's.

John then divided the bread rolls equally among 20 families.

How many bread rolls did each family receive?

 
      4
 
      5
 
      6
 
    70
 
    80
Show Answers Only

`4`

Show Worked Solution
`text{Total bread rolls}` `= (4 xx 5) + (6 xx 10) `
  `= 20 + 60 `
  `= 80`

 

`:.\ text{Bread rolls per family}` `= frac{80}{20}`
  `= 4`

Number, NAPX-p168435v02

A student needs 12 folder dividers for each subject.

This student is enrolled in 5 subjects.

A school supply store sells the folder dividers in packets of 8.

How many packets should the student buy?

 
     4
 
     6
 
     7
 
     8
Show Answers Only

`8`

Show Worked Solution

`text{Dividers required } \ = 12 xx 5 = 60`

`:.\ text{Packets required}` `= frac{60}{8}`  
  `=7.5`  
  `=8\ text{packets (round up)}`  

Number, NAPX-p168435v01

William needs 4 eggs for each cake he will bake.

He wants to make 12 cakes.

A certain store sells eggs in bags of 5.

How many bags must he buy in order to make 12 cakes?

 
     8
 
     9
 
   10
 
   12
Show Answers Only

`10`

Show Worked Solution

`text{Eggs required} \ = 4 xx 12 = 48`

`:.\ text{Bags required}` `= frac{48}{5}`  
  `=9.6`  
  `=10\ text{bags (round up)}`  

Statistics, NAPX-p168209v02

The time spent by Mark playing video games on his computer is recorded in a table.
 

   
 

What was the average time per day that Mark spent playing video games over this period?

 
     41 minutess
 
     57 minutes
 
     63 minutes
 
   342 minutes
Show Answers Only

`57 \ text{minutes}`

Show Worked Solution

`text(1 hour = 60 minutes.)`

`text{Average time}` `= frac{120+35+40+55+63+29}{6}`
  `= frac{342}{6}`
  `= 57 \ text{minutes}`

Statistics, NAPX-p168209v01

The table shown below records Emily's jogging time over six days.

 

What was the average time Emily jogged each day?

 
      48 minutes
 
      61 minutes
 
    288 minutes
 
    368 minutes
Show Answers Only

`48 \ text{minutes}`

Show Worked Solution

`text(1 hour = 60 minutes.)`

`text{Average time}` `= frac{29+36+70+40+53+60}{6}`
  `= frac{288}{6}`
  `= 48 \ text{minutes}`

Geometry, NAPX-p109101v02

Troy built a solid figure using cubes.
 


 

He paints all the outer sides red, including the base, and then separates the cubes.

How many faces are painted red?

 
    24
 
    30
 
    34
 
    36
Show Answers Only

`30`

Show Worked Solution

`text{Number of faces painted blue (top down, back to front)}`

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

`=30`

Geometry, NAPX-p109101v01

Sarah creates a solid figure using five cubes.
 


 

She paints all the outer sides blue, including the base, and then separates the cubes.

How many faces are painted blue?

 
    18
 
    22
 
    24
 
    26
Show Answers Only

`22`

Show Worked Solution

`text{Number of faces painted blue (top down, back to front)}`

`=5+4+4+4+5`

`=22`

Algebra, NAPX-p168204v02

`12.5 xx Z = 2.5`

Find the value of  `Z`  in order to make this number sentence correct

 
    1.35
 
    0.40
 
    5.0
 
    0.20
Show Answers Only

`0.20`

Show Worked Solution

`text{Check each option:}`

`12.5 xx 1.35 = 16.875 \ \ text{(Incorrect)}`

`12.5 xx 0.40 = 5 \ \ text{(Incorrect)}`

`12.5 xx 5.0 = 62.5 \ \ text{(Incorrect)}`

`12.5 xx 0.20 = 2.5 \ \ text{(Correct)}`

`therefore \ Z=0.2`

Algebra, NAPX-p168204v01

`1.36 xx B = 0.68`

Find the value of  `B`  that makes this number sentence correct.

 
    0.75
 
    0.60
 
    0.50
 
    0.20
Show Answers Only

`0.50`

Show Worked Solution

`text{Check each option:}`

`1.36 xx 0.75 = 1.02 \ \ text{(Incorrect)}`

`1.36 xx 0.60 = 0.816 \ \ text{(Incorrect)}`

`1.36 xx 0.50 = 0.68 \ \ text{(Correct)}`

`1.36 xx 0.20 = 0.272 \ \ text{(Incorrect)}`

`therefore \ B = 0.50`

Number, NAPX-p168203v02

The results of a men's 100 metre swimming race is recorded in the table below.

   

What could be the finishing time of the 2nd placed swimmer?

 
   46.28 seconds
 
   46.61 seconds
 
   46.48 seconds
 
   46.80 seconds
Show Answers Only

`text{46.48 seconds}`

Show Worked Solution

`text{The time of the 2nd swimmer must be between 46.37 and 46.52 seconds.}`

`therefore \ text{Time for 2nd could have been 46.48 seconds.}`

Number, NAPX-p168203v01

The result of a 100-metre dash was recorded in the table shown below.
 

   
 

What could be the time of the runner in 3rd place?

 
    13.85 seconds
 
    14.26 seconds
 
    14.58 seconds
 
    14.92 seconds
Show Answers Only

`14.58 \ text{seconds}`

Show Worked Solution

`text{The time of the 3rd runner must be between 14.29 and 14.84 seconds.}`

`therefore \ text{the time of the 3rd runner to finish was 14.58 seconds.}`

Probability, NAPX-p167447v02

Lester schedules a company meeting twice every 5 working days.

Today is a working day.

What is the probability that there is a meeting scheduled?

 
 `2/7`
 
 `0.40`
 
 `3/5`
 
 `text(25%)`

Show Answers Only

`0.40`

Show Worked Solution
`P` `= text(Favorable Events)/text(Total Possible Events)`
  `= 2/5`
  `= 0.40`

Probability, NAPX-p167447v01

Laura's country hut is visited by a possum twice every week.

What is the probability that the possum visits her hut today?

 
 `2/7`
 
 `2/5`
 
 `0.70`
 
 `text(25%)`

Show Answers Only

`2/7`

Show Worked Solution

`text{There are 7 days in a week.}`

`P` `= text(Favorable Events)/text(Total Possible Events)`
  `= 2/7`

Geometry, NAPX-p167443v02

Henry got lost on his way to visit his uncle’s house and made 3 U-turns before arriving.

In total, how many degrees does Henry turn through when making U-turns on his trip?

 
   150°
 
   270°
 
   540°
 
 1080°

Show Answers Only

`540^@`

Show Worked Solution

`text{One U-turn rotates the car by 180}^@`

`:. 3\ text(U-turns)` `= 3 xx 180`
  `= 540^@`

Number, NAPX-p167439v02

A pack of sugar weighs `1/4` of a kilogram.

Josh bought 6 packs for baking.

How many kilograms of sugar did he buy?

 
 `2/3`
 
 `1 1/2`
 
 `2 1/4`
 
 `3`

Show Answers Only

`1 1/2`

Show Worked Solution
`text{Weight of six packs}` `= 6 xx 1/4`
  `= 6/4`
  `= 1 1/2\ text(kg)`

Number, NAPX-p167439v01

A box of apples weighs `2/3` of a kilogram

Lou bought 3 boxes.

How many kilograms of apples did he bought?

 
    `1 frac{4}{9} \ text{kg}`
 
   `2 \ text{kg}`
 
   `frac{8}{9} \ text{kg}`
 
   `2 frac{2}{3} \ text{kg}`

Show Answers Only

`2\ text{kg}`

Show Worked Solution
`text(Total kilograms)` `=3 xx 2/3`  
  `=6/3`  
  `=2`  

Probability, NAPX-p116940v05

The arrow pictured below is spun once.
 


 

Which number is the spinner least likely to land on?

  1   2   3   1 or 2
 
 
 
 

Show Answers Only

`2`

Show Worked Solution

` text{Least likely to land on 2.}`

`text{It takes up the smallest area on the spinner}`

Probability, NAPX-p116940v04

A disk is thrown onto the table pictured below.
 


 

It has an equal chance of landing in any square.

Which numbered square is the disk least likely to land in?

  3   4   5 same chance for each number
 
 
 
 

Show Answers Only

`4`

Show Worked Solution

` text{Only 1 square is numbered 4 (all other numbers have 2 squares).}`

`therefore \ text{Least likely to land in square 4}`

Number, NAPX-p116689v02

The tallest living giraffe is measured at five thousand, seven hundred and eight millimetres tall.

Write this as a number in the box below

  millimetres

Show Answers Only

`5708 \ text{millimetres}`

Show Worked Solution

`5708 \ text{millimetres}`

Number, NAPX-p116689v05

The exact length of great white shark is measured as five thousand and ninety six millimetres.

Write this as a number in the box below

  millimetres

Show Answers Only

`5096 \ text{millimetres}`

Show Worked Solution

`5096 \ text{millimetres}`

Calculus, SPEC2 2020 VCAA 3

Let  `f(x) = x^2e^(−x)`.

  1. Find an expression for  `f′(x)`  and state the coordinates of the stationary points of  `f(x)`.  (2 marks)
  2. State the equation(s) of any asymptotes of  `f(x)`.  (1 mark)
  3. Sketch the graph of  `y = f(x)`  on the axes provided below, labelling the local maximum stationary point and all points of inflection with their coordinates, correct to two decimal places.  (3 marks)
     
         

Let  `g(x) = x^n e^(−x)`, where  `n ∈ Z`.

  1. Write down an expression for  `g″(x)`.  (1 mark)
  2.  i. Find the non-zero values of `x` for which  `g″(x) = 0`.  (1 mark)
  3. ii. Complete the following table by stating the value(s) of `n` for which the graph of  `g(x)`  has the given number of points of inflection.  (2 marks)
     
       
         
Show Answers Only
  1. `f(0) = 0; f(2) = 4e^(−2) text(and SP’s at)\ (0, 0), (2, 4e^(−2))`
  2. `y = 0`
  3.   
  4. `g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`
  5.  i. `x = x ± sqrtn`
  6. ii.
       
Show Worked Solution

a.   `f′(x) = 2xe^(−x) – x^2e^(−x)`

`text(SP’s when)\ \ f′(x) = 0:`

`x^2e^(−x)` `= 2xe^(−x)`
`x` `= 2\ \ text(or)\ \ 0`

 
`f(0) = 0; \ f(2) = 4e^(−2)`

`:. text(SP’s at)\ \ (0, 0) and  (2, 4e^(−2))`

 

b.   `text(As)\ \ x -> ∞, \ f(x) -> 0^+`

`:. text(Horizontal asymptote at)\ \ y = 0`

 

c.   

`text(POI when)\ \ f″(x) = 0`

`:. text(POI’s:)\ (0.59, 0.19), \ (3.41, 0.38)`

 

d.   `g′(x) = x^(n – 1) e^(−x)(n – x)`

`g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`

 

e.i.   `text(Solve:)\ \ x^2 – 2xn + n^2 – n = 0`

`x = n ± sqrtn`

♦♦♦ Mean mark (e)(ii) 9%.

 

e.ii.   

Probability, NAPX-p116940v03

A spinning wheel has sections labelled with different numbers.
 


 

Which of the numbers in the wheel is the spinner most likely to land on?

 
   1 or 3
 
   1
 
   2 or 4
 
   All of the colours are equally likely
Show Answers Only

`text{All of the colours are likely to land on}`

Show Worked Solution

`text{S}text{ince the spinner is divided into 8 equal parts and each colour has}`

`text{2 parts → all colours are equally likely.}`

Probability, NAPX-p116940v02

A spinning wheel has 3 different colours.
 


 

Which colour in the wheel is most likely to land on?

 
   White
 
   Black
 
   Grey
 
   All of the colours are equally likely
Show Answers Only

`text{Grey}`

Show Worked Solution

`text{By inspection, grey is the most likely as it it shades the}`

`text{largest area of the wheel.}`

Number, NAPX-p167229v02

A circle is divided into 8 equal parts, as shown in the image below.
 


 

What percentage of the circle’s area has been labelled with letters?

 
 30%
 
 37.5%
 
 42.5%
 
 45%
 
 47.5%

Show Answers Only

`text(37.5%)`

Show Worked Solution

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

`text(Percentage)` `= text(Number of letters)/text(Total number of sections) xx 100`
  `= 3/8 xx 100`
  `= 37.5text(%)`

Geometry, NAPX-p167209v02

Yohan was driving from the hospital to his house.
 


 

What directions best describe Yohan’s travel from the hospital to his house?

 
 East, north-east, north
 
 West, north-west, north
 
 East, north-east, south
 
 West, north-west, south

Show Answers Only

`text(East, north-east, north)`

Show Worked Solution

`text(The directions travelled by Yohan:)`
 

 
 

`text(The direction was East, North-East, and North)`

Geometry, NAPX-p167209v01

A man drives from his house to his office.
 


 

What directions best describe his way to the office?

 
 North, north-west, west, south-west
 
 North, north-east, east, south-east
 
 North, north-east, east, south
 
 North, north-west, east, south-east

Show Answers Only

`text(North, north-east, east, south-east)`

Show Worked Solution

`text(The directions travelled by the man:)`
 


 

`:.\ text(The direction was north, north-east, east, south-east.)`

Measurement, NAPX-p167202v02

Four triangular shaped playgrounds are shown below.

Which of these play grounds has the least surface area?

 
 
 
 

Show Answers Only

Show Worked Solution

`text(Check each option:)`

`text(Option 1 -)\ \ 1/2 xx 34 xx 15 = 255\ text(m)^2`

`text(Option 2 -)\ \ 1/2 xx 20 xx 26 = 260\ text(m)^2`

`text(Option 3 -)\ \ 1/2 xx40 xx 18 = 360\ text(m)^2`

`text(Option 4 -)\ \ 1/2 xx28 xx 25 = 350\ text(m)^2`
 

`:.\ text(The backyard with the least area is:`

Measurement, NAPX-p167202v01

In a suburb, four families measured the dimensions of their rectangular backyards.

Which backyard has the largest area?

 
 
 
 

Show Answers Only

   

Show Worked Solution

`text(Checking each option:)`

`text(Option 1:)\ 11 xx 18 = 198 text(m)^2`

`text(Option 2:)\ 16 xx 6 = 96 text(m)^2`

`text(Option 3:)\ 15 xx 10 = 150 text(m)^2`

`text(Option 4:)\ 14 xx 12 = 168 text(m)^2`
 

`:. text(The backyard with the largest area is the)\ 11\ text(m) xx 18\ text(m)`

`text(with a total area of 198 square metres.)`

   

Statistics, NAPX-p167236v02

A store sells second hand mobile phones.

The graph below shows the price of 2 similar second-hand phones.
 


 

Which of the following is true based on the graph shown?

 
 Phone A is older and less expensive than Phone B
 
 Phone B is older and more expensive than phone A
 
 Phone A is newer and more expensive than Phone B
 
 Phone A is older and more expensive than Phone B

Show Answers Only

`text(Phone A is newer and more expensive than Phone B)`

Show Worked Solution

`text(Phone A is left of Phone B → it is newer.)`

`text(Phone A is higher than Phone B → it is more expensive.)`

`:.\ text(Phone A is newer and more expensive than Phone B.)`

Statistics, NAPX-p167236v01

A man bought a plot of land in the past and now he is selling it.

The graph marks the price of the land when the man bought it and the price of the land now.
 


 

Which of the following is true based on the graph shown?

 
 The land is less expensive now than when it was purchased.
 
 The land is more expensive years ago than now.
 
 The land is more expensive now than years ago.
 
 The price of the land does not change with time.

Show Answers Only

`text(The land is more expensive now than years ago.)`

Show Worked Solution

`text(Z is further right on the x-axis → most recent price.)`

`text(Z is higher on the y-axis → more expensive.)`

`:.\ text(The land is more expensive now than years ago.)`

Number, NAPX-p167229v01

A circle is divided into 8 equal parts, as shown in the diagram below.
 


 

What percentage of the circle’s area has been labelled with even numbers?

 
 37.5%
 
 50%
 
 57.5%
 
 62.5%
 
 70%

Show Answers Only

`text(62.5%)`

Show Worked Solution

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

`text(Percentage)` `= text(Number of even numbers)/text(Total number of sections) xx 100`
  `= 5/8 xx 100`
  `= 62.5text(%)`

Algebra, SPEC2 2020 VCAA 7 MC

For non-zero constants `a` and `b`, where  `b < 0`, the expression  `1/(ax(x^2 + b))`  in partial fraction form with linear denominators, where  `A, B`  and  `C`  are real constants, is

  1. `A/(ax) + (Bx + C)/(x^2 + b)`
  2. `A/(ax) + B/(x + sqrtb) + C/(x - sqrtb)`
  3. `A/x + B/(ax + sqrt|b|) + C/(ax - sqrt|b|)`
  4. `A/x + B/(x + sqrt|b|) + C/(x - sqrt|b|)`
  5. `A/(ax) + B/((x + sqrtb)^2) + C/(x + sqrtb)`
Show Answers Only

`D`

Show Worked Solution

♦♦♦ Mean mark 26%.
MARKER’S COMMENT: Option A results from not considering `b<0`.
`1/(ax(x^2 + b))` `= 1/a (A_1/x + B_1/(x + sqrt|b|) + C_1/(x – sqrt|b|))`
  `= A/x + B/(x + sqrt|b|) + C/(x – sqrt|b|)`

 
`=>D`

 

Trigonometry, SPEC2 2020 VCAA 4 MC

Let  `f(x) = sqrt(x - 1)/x`  over its implied domain and  `g(x) = text(cosec)^2 x`  for  `0 < x < pi/2`.

The rule for  `f(g(x))`  and the range, respectively, are given by

  1. `f(g(x)) = text(cosec)^2(sqrt(x - 1)/x), [1, ∞)`
  2. `f(g(x)) = text(cosec)^2(sqrt(x - 1)/x), [2, ∞)`
  3. `f(g(x)) = sin(x)cos(x), [−0.5, 0.5]\\ {0}`
  4. `f(g(x)) = sin(x)cos(x), (0, 1/2)`
  5. `f(g(x)) = 1/2 sin(2x), (0, 1/2]`
Show Answers Only

`E`

Show Worked Solution

♦♦♦ Mean mark 28%.
`f(g(x))` `= sqrt(text(cosec)^2(x) – 1)/(text(cosec)^2(x))`
  `= sqrt(cot^2(x))/(text(cosec)^2(x))`
  `= (cos(x))/(sin(x)) · sin^2(x)`
  `= cos(x)sin(x)`
  `= 1/2 sin(2x)`

 
`text(Given)\ \ 0 < x < pi/2`,

`text(Range)\ \ f(g(x)) :\ \ (0, 1/2]`

`=>E`

Number, NAPX-Z3-CA06 v1

What number is exactly halfway between `4 frac{1}{4}` and `6 frac{3}{4}`

 
 
   `4 frac{3}{4}`
 
 
   `5`
 
 
   `5 frac{1}{4}`
 
 
   `5 frac{1}{2}`
Show Answers Only

`5 frac{1}{2}`

Show Worked Solution
  `text{Halfway}` `= (4 frac{1}{4} + 6 frac{3}{4}) \ div 2`
    `= 11/2`
    `= 5 frac{1}{2}`

NETWORKS, FUR1 2020 VCAA 10 MC

The directed network below shows the sequence of activities, `A` to `I`, that is required to complete an office renovation.

The time taken to complete each activity, in weeks, is also shown.
 


 

The project manager would like to complete the office renovation in less time.

The project manager asks all the workers assigned to activity `H` to also work on activity `F`.

This will reduce the completion time of activity `F` to three weeks.

The workers assigned to activity `H` cannot work on both activity `H` and activity `F` at the same time.

No other activity times will be changed.

This change to the network will result in a change to the completion time of the office renovation.

Which one of the following is correct?

  1. The completion time will be reduced by one week if activity `F` is completed before activity `H` is started.
  2. The completion time will be reduced by three weeks if activity `F` is completed before activity `H` is started.
  3. The completion time will be reduced by one week if activity `H` is completed before activity `F` is started.
  4. The completion time will be reduced by three weeks if activity `H` is completed before activity `F` is started.
  5. The completion time will be increased by three weeks if activity `H` is completed before activity `F` is started.
Show Answers Only

`A`

Show Worked Solution

`text{Original forward scan (note}\ F\ text{is 6 origionally but}`

♦♦♦ Mean mark 24%.

`text{is reduced to 3 for the adjusted critical path):}`
 


 

`text(Original critical path is:)\ ACEFGI = 2+5+3+6+4+5=25`

`text(If activity)\ F\ text(is completed in 3 weeks, and then)`

`text(activity)\ H\ text(starts, new critical path is:)`

`ACEF\ text{(dummy)}\ HI = 24\ text(weeks)`

`=>  A`

NETWORKS, FUR1 2020 VCAA 9 MC

The flow of liquid through a series of pipelines, in litres per minute, is shown in the directed network below.
 


 

Five cuts labelled A to E are shown on the network.

The number of these cuts with a capacity equal to the maximum flow of liquid from the source to the sink, in litres per minute, is

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

`C`

Show Worked Solution

`text(Minimum cut) = 33`

`text(Cut)\ A = 8 + 15 + 15 = 38`

`text(Cut)\ B = 8 + 15 + 10 = 33`

`text(Cut)\ C = 15 + 8 + 10 = 33`

`text(Cut)\ D = 15 + 8 + 10 = 33`

`text(Cut)\ E = 15 + 8 + 5 + 10 = 38`

`text{(Note that it is arguable that the flow of 5 should not be counted}`

`text{in Cut}\ E,\ text(making the correct answer)\ D.\ text(This is because this)`

`text{edge cannot hold any flow in the network as given.)}`

`=>  C`

CORE, FUR1 2020 VCAA 26 MC

Ray deposited $5000 in an investment account earning interest at the rate of 3% per annum, compounding quarterly.

A rule for the balance, `R_n` , in dollars, after `n` years is given by

  1. `R_n = 5000 xx 0.03^n`
  2. `R_n = 5000 xx 1.03^n`
  3. `R_n = 5000 xx 0.03^(4n)`
  4. `R_n = 5000 xx 1.0075^n`
  5. `R_n = 5000 xx 1.0075^(4n)`
Show Answers Only

`E`

Show Worked Solution

`text(Quarterly rate) = 0.03/4 = 0.0075`

♦♦♦ Mean mark 20%.

`text(Compounding periods) = 4n`

`:.\ text(Balance after)\ n\ text(years)`

`= 5000 xx 1.0075^(4n)`
 

`=>  E`

Calculus, MET1 2020 VCAA 8

Part of the graph of  `y = f(x)`, where  `f:(0, ∞) -> R, \ f(x) = xlog_e(x)`, is shown below.
 


 

The graph of `f` has a minimum at the point `Q(a, f(a))`, as shown above.

  1. Find the coordinates of the point `Q`.  (2 marks)
  2. Using  `(d(x^2log_e(x)))/(dx) = 2x log_e(x) + x`, show that  `xlog_e(x)`  has an antiderivative  `(x^2log_e(x))/2 - (x^2)/4`.  (1 mark)
  3. Find the area of the region that is bounded by `f`, the lines  `x = a`  and the horizontal axis for  `x ∈ [a, b]`, where `b` is the `x`-intercept of `f`.  (2 marks)
  4. Let  `g: (a, ∞) -> R, \ g(x) = f(x) + k`  for  `k ∈ R`.

     

    1. Find the value of `k` for which  `y = 2x`  is a tangent to the graph of `g`.  (1 mark)
    2. Find all values of `k` for which the graphs of `g` and `g^(−1)` do not intersect.  (2 marks)
Show Answers Only
  1. `Q(1/e, −1/e)`
  2. `text(See Worked Solutions)`
  3. `1/4 – 3/(4e^2)\ text(u)^2`
  4. i.  `e`
  5. ii.  `k ∈ (1, ∞)`
Show Worked Solution

a.   `y = xlog_e x`

`(dy)/(dx)` `= x · 1/x + log_e x`
  `= 1 + log_e x`

 
`text(Find)\ x\ text(when)\ (dy)/(dx) = 0:`

`1 + log_e x` `= 0`
`log_e x` `= −1`
`x` `= 1/e`
`y` `= 1/e log_e (e^(−1))`
  `= −1/e`

 
`:. Q(1/e, −1/e)`

 

b.    `int 2x log_e(x) + x\ dx` `= x^2 log_e (x) + c`
  `2 int x log_e(x)\ dx` `= x^2 log_e (x) – intx\ dx + c`
  `:. int x log_e(x)\ dx` `= (x^2 log_e (x))/2 – (x^2)/4 \ \ (c = 0)`

 

c.   

`text(When)\ \ x log_e x = 0 \ => \ x = 1`

`=> b = 1`

`:.\ text(Area)` `= −int_(1/e)^1 x log_e(x)\ dx`
  `= [(x^2)/4 – (x^2 log_e(x))/2]_(1/e)^1`
  `= (1/4 – 0) – (1/(4e^2) – (log_e(e^(−1)))/(2e^2))`
  `= 1/4 – (1/(4e^2) + 1/(2e^2))`
  `= 1/4 – 3/(4e^2) \ text(u)^2`

 

d.i.   `text(When)\ \ f′(x) = m_text(tang) = 2,`

`1 + log_e(x)` `= 2`
`x` `= e`

 
`text(T)text(angent meets)\ \ g(x)\ \ text(at)\ \ (e, 2e)`

`g(e)` `= f(e) + k`
`2e` `= e log_e e + k`
`:.k` `= e`

 

d.ii. `text(Find the value of)\ k\ text(when)\ \ y = x\ \ text(is a tangent to)\ g(x):`

`text(When)\ \ f′(x) = 1,`

`1 + log_e(x)` `= 1`
`x` `= 1`

 
`text(T)text(angent occurs at)\ (1, 1)`

`g(1) = f(1) + k \ => \ k = 1`
 

`:.\ text(Graphs don’t intersect for)\ k ∈ (1, ∞)`

Calculus, EXT2 C1 2020 HSC 16b

Let  `I_n = int_0^(frac{pi}{2}) sin^(2n + 1)(2theta)\ d theta, \ n = 0, 1, ...`

  1. Prove that  `I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`.  (3 marks)

  2. Deduce that  `I_n = frac{2^(2n)(n!)^2}{(2n +1)!}`.  (3 marks)

Let  `J_n = int_0^1 x^n (1 - x)^n\ dx , \ n = 0, 1, 2,...`

  1. Using the result of part (ii), or otherwise, show that  `J_n = frac{(n!)^2}{(2n + 1)!}`.  (3 marks)

  2. Prove that  `(2^n n!)^2 ≤ (2n + 1)!`.  (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}`
Show Worked Solution

i.    `text{Prove} \ \ I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`

♦ Mean mark (i) 48%.

`I_n = int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`

`text{Integrating by parts:}`

`u = sin^(2n) (2 theta)`   `u^(′) = 2n sin^(2n -1) (2 theta)  xx -frac(1)(2) cos (2 theta)`
`v = -frac{1}{2} cos (2 theta)`   `v^(′) = sin 2 theta`

 

`I_n` `= [ sin^(2n) (2 theta) * -frac{1}{2} cos (2 theta)]_0^(frac{pi}{2}) -2n int_0^(frac{pi}{2}) sin^(2n -1) (2 theta) * 2 cos (2 theta) * -frac{1}{2} cos (2 theta)\ d theta`
`I_n` `= 0 + 2n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) * cos^2 (2 theta)\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) (1 – sin^2 (2 theta))\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) ( 2 theta) – sin^(2n+1) (2 theta)\ d theta`
`I_n` `= 2n (I_(n-1) – I_n)`
`I_n + 2 n  I_n` `= 2 n I_(n-1)`
`I_n (2n + 1)` `= 2 n I_(n-1)`
`therefore I_n` `= frac{2n}{2n +1} I_(n-1)`
♦ Mean mark (ii) 36%.

 

ii.     `I_0` `= int_0^(frac{pi}{2}) sin (2 theta)\ d theta`
    `= [ -frac(1)(2) cos (2 theta) ]_0^(frac{pi}{2}`
    `=( -frac{1}{2} cos pi + frac{1}{2} cos 0 )`
    `= 1`
     
`I_n` `= frac{2n}{2n + 1} I_(n-1)`
`I_(n-1)` `= frac{2(n -1)}{2n -1} I_(n-2)`
  `vdots`
`I_1` `= frac{2}{3} I_0`

 

`I_n` `= frac{2n}{2n + 1} xx frac{2(n-1)}{2n-1} xx frac{2(n-2)}{2n-3} xx … xx frac{2}{3} xx 1`
  `= frac{2n}{2n+1} xx frac{2n}{2n} xx frac{2(n-1)}{2n-1} xx frac{2(n-1)}{2n-2} xx … xx frac{2}{3} xx frac{2}{2} xx 1`
  `= frac{2^n (n xx (n-1) xx .. xx 1) xx 2^n (n xx (n – 1) xx … xx 1)}{(2n + 1)!}`
  `= frac{2^(2n) (n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iii) 16%.

 

iii.   `J_n = int_0^1 x^n (1-x)^n\ dx ,  \ n = 0, 1, 2, …`

`text{Let} \ \ x` `= sin^2 theta`
`frac{dx}{d theta}` `= 2 sin theta \ cos theta \ => \ dx = 2 sin theta \ cos theta \ d theta`

 

`text{When}`    `x = 0 \ ,` ` \ theta = 0`
  `x = 1  \ ,` ` \ theta = frac{pi}{2}`

 

`J_n` `= int_0^(frac{pi}{2}) (sin^2 theta)^n (1 – sin^2 theta)^n * 2 sin theta \ cos theta \ d theta`
  `= int_0^(frac{pi}{2}) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) 2^(2n) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n+1) (2 theta)\ d theta`
  `= frac{1}{2^(2n)} * frac{2^(2n) (n!)^2}{(2n+1)!}\ \ \ text{(using part (ii))}`
  `= frac{(n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iv) 10%.

 

iv.   `text{If} \ \ I_n ≤ 1,`

`2^(2n) (n!)^2` ` ≤ (2n + 1)!`
`(2^n n!)^2` `≤ (2n + 1)!`

  
`text{Show} \ \ I_n ≤ 1 :`

`text{Consider the graphs}`

`y = sin(2 theta) \ \ text{and}\ \ y = sin^(2n + 1) (2 theta) \ \ text{for} \ \ 0 ≤ theta ≤ frac{pi}{2}`
 

`int_0^(frac{pi}{2}) sin(2 theta)` `= [ – frac{1}{2} cos (2 theta) ]_0^(frac{pi}{2})`
  `= – frac{1}{2} cos \ pi + frac{1}{2} cos \ 0`
  `= 1`

 
`y = sin(2 theta) \ => \ text{Range} \ [0, 1] \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`

`sin^(2n+1) (2 theta)` `≤ sin (2 theta) \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`
`sin^(2n+1) (2 theta)` `≤ 1`
`I_n` `≤ 1`
`therefore (2^n n!)^2` `≤ (2n + 1) !`

Measurement, STD1 M5 2020 HSC 28

Two similar right-angled triangles are shown.
 


 

The length of side `AB` is 8 cm and the length of side `EF` is 4 cm.

The area of triangle `ABC` is 20 cm2.

Calculate the length in centimetres of side `DF` in Triangle II, correct to two decimal places.   (4 marks)

Show Answers Only

`7.55\ \text{cm}`

Show Worked Solution

`text{Consider} \ Δ ABC :`

`text{Area}` `= frac{1}{2} xx AB xx BC`
`20` `= frac{1}{2} xx 8 xx BC`
`therefore \ BC` `= 5`

 

`text{Using Pythagoras in} \ Δ ABC :`

♦♦♦ Mean mark 11%.

`AC = sqrt(8^2 + 5^2) = sqrt89`

 

`text{S} text{ince} \ Δ ABC\ text{|||}\ Δ DEF,`

`frac{AC}{BC}` `= frac{DF}{EF}`
`frac{sqrt89}{5}` `= frac{DF}{4}`
`therefore \ DF` `= frac{4 sqrt89}{5}`
  `= 7.547 …`
  `= 7.55 \ text{cm (to 2 d.p.)}`

Trigonometry, EXT1 T3 2020 HSC 14b

  1. Show that  `sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`.  (2 marks)

  2. By letting  `x = 4sin theta`  in the cubic equation  `x^3-12x + 8 = 0`.

     

    Show that  `sin (3theta) = 1/2`. (2 marks)

  3. Prove that  `sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`.  (3 marks)

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

i.   `text(Prove:)\  \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`

`text(LHS)` `= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)`
  `= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1 – 2sin^2theta))`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)`
  `= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta`
  `= 0`

 

ii.   `text(Show)\ \ sin(3theta) = 1/2`

`text{Using part (i):}`

`(sin(3theta))/4` `= 3/4 sintheta-sin^3 theta`
`sin(3theta)` `= 3sintheta-4sin^3theta\ …\ (1)`

 
`x^3-12x + 8 = 0`

`text(Let)\ \ x = 4 sin theta`

`(4sintheta)^3-12(4sintheta) + 8` `= 0`
`64sin^3theta-48sintheta` `= 0`
`−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}` `= −8`
`−16 sin(3theta)` `= −8`
`sin(3theta)` `= 1/2`
♦♦♦ Mean mark (iii) 21%.

 

iii.   `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`

`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`

`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`

`text(When)\ \ sin3theta = 1/2,`

`3theta` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …`
`theta` `= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …`

 
`:.\ text(Solutions)`

`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`

`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`

`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`
 

`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`

`α + β + γ = −b/a = 0`

`αβ + βγ + αγ = c/a = −12`

`(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2` `= (α + β + γ)^2 – 2(αβ + βγ + αγ)`
`16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)` `= 0-2(−12)`
`:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18` `= 24/16=3/2`

Algebra, STD1 A3 2020 HSC 29

There are two tanks on a property, Tank A and Tank B. Initially, Tank A holds 1000 litres of water and Tank B is empty.

  1. Tank A begins to lose water at a constant rate of 20 litres per minute.

     

    The volume of water in Tank A is modelled by  `V = 1000 - 20t`  where `V` is the volume in litres and  `t`  is the time in minutes from when the tank begins to lose water.

     

    On the grid below, draw the graph of this model and label it as Tank A.   (1 mark)
     

     

  2. Tank B remains empty until  `t=15`  when water is added to it at a constant rate of 30 litres per minute.
     By drawing a line on the grid (above), or otherwise, find the value of  `t`  when the two tanks contain the same volume of water.  (2 marks)
  3. Using the graphs drawn, or otherwise, find the value of  `t`  (where  `t > 0`) when the total volume of water in the two tanks is 1000 litres.  (1 mark)

Show Answers Only
  1.  `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}` 
      

     
  2. `29 \ text{minutes}`
  3. `45 \ text{minutes}`
Show Worked Solution

a.     `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`

♦ Mean mark part (a) 45%.
 


 

b.   `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`

♦♦ Mean mark part (b) 24%.
 

   

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

 
c.   `text{Strategy 1}`

♦♦♦ Mean mark part (c) 5%.

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when  t = 45)`
  

`text{Strategy 2}`

`text{Total Volume}` `=text{T} text{ank A} + text{T} text{ank B}`
`1000` `= 1000 – 20t + (t – 15) xx 30`
`1000` `= 1000 – 20t + 30t – 450 `
`10t` `= 450`
`t` `= 45 \ text{minutes}`

Algebra, STD1 A2 2020 HSC 10 MC

A plumber charges a call-out fee of $90 as well as $2 per minute while working.

Suppose the plumber works for  `t`  hours.

Which equation expresses the amount the plumber charges ($`C`) as a function of time (`t`  hours)?

  1.  `C = 2 + 90t`
  2.  `C = 90 + 2t`
  3.  `C = 120 + 90t`
  4.  `C = 90 + 120t`
Show Answers Only

`D`

Show Worked Solution

♦♦♦ Mean mark 21%.
`text(Hourly rate)` `= 60 xx 2`
  `= $120`

 
`therefore  C = 90 + 120t`

Financial Maths, STD2 F5 2020 HSC 37

Wilma deposited a lump sum into a new bank account which earns 2% per annum compound interest.

Present value interest factors for an annuity of $1 for various interest rates (`r`) and numbers of periods (`N`) are given in the table.
 


 

Wilma was able to make the following withdrawals from this account.

  • $1000 at the end of each year for twenty years (starting one year after the account is opened)
  • $3000 each year for ten years starting 21 years after the account is opened.

Calculate the minimum lump sum Wilma must have deposited when she opened the new account.   (3 marks)

Show Answers Only

`$ 34 \ 486`

Show Worked Solution

`text{Annuity 1:}\ PV\ text{of $1000 annuity for 20 years at} \ \ r = 0.02`

♦♦♦ Mean mark 23%.

`PV\ text{factor} = 16.351`

`therefore \ PV\ text{Annuity 1}` `= 16.351 xx 1000`
  `= $16 \ 351`

 
`text{Annuity 2:}\ PV\ text{of $3000 annuity for years 21-30 at} \ \ r = 0.02`

`PV\ text{Annuity 2}` `= PVtext{(30 years)} – PVtext{(20 years)}`
  `= 3000 xx 22.396 – 3000 xx 16.351`
  `= $ 18 \ 135`

 

`:.\ text{Lump sum required}` `= 16 \ 351 + 18\ 135`
  `= $ 34 \ 486`

Calculus, EXT1 C2 2020 HSC 13c

Suppose  `f(x) = tan(cos^(-1)(x))`  and  `g(x) = (sqrt(1-x^2))/x`.

The graph of  `y = g(x)`  is given.
 

  1. Show that  `f^(′)(x) = g^(′)(x)`.  (4 marks)

  2. Using part (i), or otherwise, show that  `f(x) = g(x)`.  (3 marks) 

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

i.   `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.
`f^(′)(x)` `= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
  `= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
  `= -1/(x^2sqrt(1-x^2))`

 
`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)` `= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
  `= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
  `= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
  `= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
  `= -1/(x^2sqrt(1-x^2))`
  `=f^(′)(x)`

 

ii.   `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`
 

`text(Find)\ c:`

`f(1)` `= tan(cos^(-1) 1)`
  `= tan 0`
  `= 0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c`

`:. c = 0`

`:. f(x) = g(x)`

Statistics, STD2 S5 2020 HSC 35

The intelligence Quotient (IQ) scores for adults in City A are normally distributed with a mean of 108 and a standard deviation of 10.

The IQ score for adults in City B are normally distributed with a mean of 112 and a standard deviation of 16.

  1. Yin is an adult who lives in City A and has an IQ score of 128.
    What percentage of the adults in this city have an IQ score higher than Yin's?   (2 marks) 

  2. There are 1 000 000 adults living in City B.
    Calculate the number of adults in City B that would be expected to have an IQ score lower than Yin's.  (2 marks)

  3. Simon, an adult who lives in City A, moves to City B. The  `z` -score corresponding to his IQ score in City A is the same as the `z`-score corresponding to his IQ score in City B.
    By first forming an equation, calculate Simon's IQ score. Give your answer correct to one decimal place.   (3 marks)

Show Answers Only
  1. `text(2.5%)`
  2. `840 \ 000`
  3. `101.3`
Show Worked Solution

a.    `text{In City A:}`

♦ Mean mark 50%.

`z text{-score}\ (128) = frac {x -mu}{sigma} = frac{128 – 108}{10} = 2`
 


 

`therefore\ text{2.5% have a higher IQ in City} \ A.`
 

b.    `text{In City B:}`

♦ Mean mark 44%.

`z text{-score}\ (128) = frac{128 – 112}{16} = 1`
 


 

`therefore \ text{Adults in City} \ B \ text{with an IQ}\ <  128`

`= 84text(%) xx 1 \ 000 \ 000`

`= 840 \ 000`
 

c.    `z text{-score in City A}\ = z text{-score in City B}`

♦♦♦ Mean mark 17%.
`frac{x – 108}{10} ` `= frac{x -112}{16} \ \ text{(multiply b.s.} \ xx 160 text{)}`
`16 (x – 108)` `= 10 (x – 112)`
`16 x – 1728` `= 10 x – 1120`
`6x` `= 608`
`x` `= 101.3`

  
`therefore \ text{Simon’s IQ} = 101.3 \ text{(to 1 d.p.)}`

Statistics, STD2 S4 2020 HSC 36

A cricket is an insect. The male cricket produces a chirping sound.

A scientist wants to explore the relationship between the temperature in degrees Celsius and the number of cricket chirps heard in a 15-second time interval.

Once a day for 20 days, the scientist collects data. Based on the 20 data points, the scientist provides the information below.

  • A box-plot of the temperature data is shown.
     
       
  • The mean temperature in the dataset is 0.525°C below the median temperature in the dataset.
  • A total of 684 chirps was counted when collecting the 20 data points.

The scientist fits a least-squares regression line using the data `(x, y)`, where `x` is the temperature in degrees Celsius and `y` is the number of chirps heard in a 15-second time interval. The equation of this line is

`y = −10.6063 + bx`,

where `b` is the slope of the regression line.

The least-squares regression line passes through the point  `(barx, bary)`, where  `barx`  is the sample mean of the temperature data and  `bary`  is the sample mean of the chirp data.

Calculate the number of chirps expected in a 15-second interval when the temperature is 19° Celsius. Give your answer correct to the nearest whole number.  (5 marks)

Show Answers Only

`29\ text(chirps)`

Show Worked Solution

`y = −10.6063 + bx`

♦♦♦ Mean mark 18%.

`text(Find)\ b:`

`text(Line passes through)\ \ (barx, bary)`

`barx` `= 22 – 0.525`
  `= 21.475`
`bary` `= text(total chirps)/text(number of data points)`
  `= 684/20`
  `= 34.2`

 

`34.2` `= −10.6063 + b(21.475)`
`:.b` `= 44.8063/21.475`
  `~~ 2.0864`

 
`text(If)\ \ x = 19,`

`y` `= −10.6063 + 2.0864 xx 19`
  `= 29.03`
  `= 29\ text(chirps)`

Algebra, STD2 A4 2020 HSC 33

The graph shows the number of bacteria, `y`, at time `n` minutes. Initially (when `n = 0`) the number of bacteria is 1000.
 


 

  1. Find the number of bacteria at 40 minutes.   (1 mark)

  2. The number of bacteria can be modelled by the equation  `y = A xx b^n`, where `A` and `b` are constants.
    Use the guess and check method to find, to two decimal places, an upper and lower estimate for the value of `b`. The upper and lower estimates must differ by 0.01. (2 marks)

Show Answers Only
  1. `4000`
  2. `text{Show Worked Solutions}`
Show Worked Solution

a.    `text{When} \ \ n = 40,`

`text{Number of Bacteria} \ (y) = 4000`
 

b.   `A = 1000 \ => \ y = 1000  b^n`

♦♦♦ Mean mark 14%.

`text{By inspection, graph passes through (40, 4000)}`

`=> \ 4000 = 1000  b^40`
 

`text(Guess and check possible values of)\ b:`

`text{If} \ \ b = 1.03 \ , \ \ y = 1000 xx 1.03^40 = 3262 \ text{(too low)}`

`text{If} \ \ b = 1.04 \ , \ \ y = 1000 xx 1.04^40 = 4801 \ text{(too high)}`

`therefore \ 1.03 < b < 1.04`

Combinatorics, EXT1 A1 2020 HSC 14a

  1. Use the identity `(1 + x)^(2n) = (1 + x)^n(1 + x)^n`

     

    to show that
     
        `((2n),(n)) = ((n),(0))^2 + ((n),(1))^2 + … + ((n),(n))^2`,
     
    where `n` is a positive integer.  (2 marks)

  2. A club has `2n` members, with `n` women and `n` men.

     

    A group consisting of an even number `(0, 2, 4, …, 2n)` of members is chosen, with the number of men equal to the number of women.
     
    Show, giving reasons, that the number of ways to do this is `((2n),(n))`.  (2 marks)

  3. From the group chosen in part (ii), one of the men and one of the women are selected as leaders.

     

    Show, giving reasons, that the number of ways to choose the even number of people and then the leaders is
     

     

        `1^2 ((n),(1))^2 + 2^2((n),(2))^2 + … + n^2((n),(n))^2`.  (2 marks)

  4. The process is now reversed so that the leaders, one man and one woman, are chosen first. The rest of the group is then selected, still made up of an equal number of women and men.

     

    By considering this reversed process and using part (ii), find a simple expression for the sum in part (iii).  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `n^2 xx \ ^(2n – 2)C_(n – 1)`
Show Worked Solution

i.   `text(Expand)\ \ (1 + x)^(2n):`

♦♦ Mean mark part (i) 26%.

`\ ^(2n)C_0 + \ ^(2n)C_1 x^2 + … + \ ^(2n)C_n x^n + … \ ^(2n)C_(2n) x^(2n)`

`=> text(Coefficient of)\ \ x^n = \ ^(2n)C_n`
 

`text(Expand)\ \ (1 + x)^n (1 + x)^n:`

`[\ ^nC_0 + \ ^nC_1 x + … + \ ^nC_n x^n][\ ^nC_0 + \ ^nC_1 x + … + \ ^nC_n x^n]`

`=> \ text(Coefficient of)\ \ x^n`

`= \ ^nC_0 · \ ^nC_n + \ ^nC_1 · \ ^nC_(n – 1) + … + \ ^nC_(n – 1) · \ ^nC_1 + \ ^nC_n · \ ^nC_0`

`= (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_(n – 1))^2 + (\ ^nC_n)^2\ \ \ (\ ^nC_k = \ ^nC_(n – k))`
 

`text(Equating coefficients:)`

`\ ^(2n)C_n = (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_n)^2`

♦♦ Mean mark part (ii) 23%.

 

ii.   `text(Number of men = Number of women)\ \ (M = W)`

`text(If)\ \ M = W = 0:`  `text(Ways) = \ ^nC_0 · \ ^nC_0 = (\ ^nC_0)^2`
`text(If)\ \ M = W = 1:`  `text(Ways) = \ ^nC_1 · \ ^nC_1 = (\ ^nC_1)^2`

`vdots`

`text(If)\ \ M = W = n:  text(Ways) = \ ^nC_n · \ ^nC_n = (\ ^nC_n)^2`
 

`:.\ text(Total combinations)`

`= (\ ^nC_0)^2 + (\ ^nC_1)^2 + … + (\ ^nC_n)^2`

`= \ ^(2n)C_n\ \ \ text{(from part (i))}`

 

iii.   `text(Let)\ \ M_L = text(possible male leaders)`

♦♦ Mean mark part (iii) 26%.

`text(Let)\ \ W_L = text(possible female leaders)`

`text(If)\ \ M = W = 0 => text(no leaders)`

`text(If)\ \ M = W = 1:  text(Ways) = \ ^nC_1 xx M_L xx \ ^nC_1 xx W_L = 1^2 (\ ^nC_1)^2`

`text(If)\ \ M = W = 2:  text(Ways) = \ ^nC_2 xx 2 xx \ ^nC_2 xx 2 = 2^2 (\ ^nC_2)^2`

`vdots`

`text(If)\ \ M = W = n:  text(Ways) = \ ^nC_n xx n xx \ ^nC_2 xx n = n^2 (\ ^nC_n)^2`
 

`:.\ text(Total combinations)`

`= 1^2(\ ^nC_1)^2 + 2^2(\ ^nC_2)^2 + … + n^2(\ ^nC_n)^2`

♦♦♦ Mean mark part (iv) 16%.

 

iv.  `text(If)\ \ M = W = 1:  text(Ways)` `= M_L xx \ ^(n – 1)C_0 xx W_L xx \ ^(n – 1)C_0`
    `= n xx \ ^(n – 1)C_0 xx n xx \ ^(n – 1)C_0`
    `= n^2(\ ^(n – 1)C_0)^2`
`text(If)\ \ M = W = 2:  text(Ways)` `= n xx \ ^(n – 1)C_1 xx n xx \ ^(n – 1)C_1`
  `= n^2(\ ^(n – 1)C_1)^2`

`vdots`

`text(If)\ \ M = W = n:\ text(Ways)` `= n xx \ ^(n – 1)C_(n – 1) xx n xx \ ^(n – 1)C_(n – 1)`
  `= n^2(\ ^(n – 1)C_(n – 1))^2`

 
`:.\ text(Total combinations)`

`= n^2(\ ^(n – 1)C_0)^2 + n^2(\ ^(n – 1)C_1)^2 + … + n^2(\ ^(n – 1)C_(n – 1))^2`

`= n^2 xx \ ^(2(n – 1))C_(n – 1)\ \ \ text{(using part (i))}`

`= n^2 xx \ ^(2n – 2)C_(n – 1)`