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Trigonometry, 2ADV T1 2013 HSC 14c

2013 14c

 
The right-angled triangle  `ABC`  has hypotenuse  `AB = 13`. The point  `D`  is on  `AC`  such that  `DC = 4`,  `/_DBC = pi/6` and  `/_ABD = x`.

Using the sine rule, or otherwise, find the exact value of  `sin x`.   (3 marks)

Show Answers Only

 `(7sqrt3)/26 text(.)`

Show Worked Solution

`text(Find)\ \ /_ADB`

`/_ADB` `= pi/6 + pi/2 \ \ \ text{(exterior angle of}\ Delta BDC text{)}`
  `= (2pi)/3\ text(radians)`

 
`text(Find)\ \ AD`

♦ Mean mark 36%.
STRATEGY TIP: The hint to use the sine rule should flag to students that they will be dealing in non-right angled trig (i.e. `Delta ABD`) and to direct their energies at initially finding `/_ADB` and `AD`.
`tan (pi/6)` `= 4/(BC)`
`1/sqrt3` `=4/(BC)`
`BC` `=4 sqrt3`

 

`text(Using Pythagoras:)`

`AC^2 + BC^2` `= AB^2`
`AC^2 + (4sqrt3)^2` `= 13^2`
`AC^2` `= 169\-48`
  `= 121`
`=>AC` `= 11`
`:.AD` `=AC\-DC`
  `= 11 -4`
  `=7`

 

`text(Using sine rule:)`

`(AB)/(sin /_BDA)` `= (AD)/(sinx)`
`13/(sin ((2pi)/3))` `=7/(sinx)`
`13 xx sinx` `= 7 xx sin ((2pi)/3)`
`sinx` `= 7/13 xx sin((2pi)/3)`
  `= 7/13 xx sqrt3/2`
  `= (7 sqrt3)/26`

 
`:.\ text(The exact value of)\ sinx = (7sqrt3)/26 text(.)`

Trigonometry, 2ADV T3 2013 HSC 13a

The population of a herd of wild horses is given by

`P(t) = 400 + 50 cos (pi/6 t)`

where  `t`  is time in months. 

  1. Find all times during the first 12 months when the population equals 375 horses.    (2 marks)

  2. Sketch the graph of  `P(t)`  for  `0 <= t <= 12`.    (2 marks)

Show Answers Only
  1. `t=4\ text(months and 8 months)`
  2. `text(See Worked Solutions.)`
Show Worked Solution

i.  `P(t) = 400 + 50 cos (pi/6 t)`

`text(Need to find)\ t\ text(when)\ P(t) = 375`

`375` `= 400 + 50 cos (pi/6 t)`
`50 cos (pi/6 t)` `=-25`
`cos (pi/6 t)` `= – 1/2`
   
`text(S)text(ince)\ \ cos(pi/3)=1/2, text(and cos is)`
`text(negative in)\ 2^text(nd) // 3^text(rd)\ text(quadrants:)`
`=>pi/6 t` `= (pi\ – pi/3),\ (pi + pi/3),\ (3pi\ – pi/3)`
  `= (2pi)/3,\ (4pi)/3,\ (8pi)/3,\ …`
`:.t` `= 4,\ 8,\ 16,\ …`

 
`:.\ text(In the 1st 12 months,)\ P(t) = 375\ text(when)`

`t=4\ text(months and 8 months.)`

 

♦ Mean mark 39%
ii. 2UA 2013 13a ans

Calculus, 2ADV C3 2013 HSC 8 MC

The diagram shows points  `A`,  `B`,  `C`  and  `D`  on the graph  `y = f(x)`.
 

2013 8 mc

 
 At which point is  `f^{′}(x) > 0`  and  ` f^{″}(x)= 0`? 

  1. `A`  
  2. `B`  
  3. `C`  
  4. `D`  
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 48%

`text(At)\ A,\ \ \ f^{′}(x) <0`

`text(At)\ C,\ \ \ f^{′}(x) =0`

`:.\ text(It cannot be)\ A\ text(or)\ C.`

`text(At)\ D,\ \ \ f^{″}(x) >0\ \ \ text{(concave up)}`

`text(At)\ B,\ \ \ f^{″}(x) =0\ \ \ text{(concavity changes)}`

`=> B`

Trigonometry, 2ADV T3 2013 HSC 6 MC

Which diagram shows the graph  `y=sin(2x + pi/3)`?
 

2013 6 mc1

2013 6 mc2

Show Answers Only

`D`

Show Worked Solution
♦♦ Mean mark 34%

`text(At)\ x = 0 text(,)\ \ y = sin (pi/3)  = sqrt3/2`

`=>\ text(It cannot be A or C)`
 

`text(Find)\ x\ text(when)\ y = 0,`

`sin (2x + pi/3)` `= 0`
`:.\ 2x + pi/3` `= 0\ \ \ \ \ text{(sin 0 = 0)}`
`2x` `=-pi/3`
`x` `= -pi/6`

`=>  D`

Financial Maths, STD2 F4 2011 HSC 28b

Norman and Pat each bought the same type of tractor for $60 000 at the same time. The value of their tractors depreciated over time.

The salvage value `S`, in dollars, of each tractor, is its depreciated value after `n` years.

Norman drew a graph to represent the salvage value of his tractor.
 

 2011 28b

  1. Find the gradient of the line shown in the graph.   (1 mark)

  2. What does the value of the gradient represent in this situation?   (1 mark)

  3. Write down the equation of the line shown in the graph.   (1 mark)

  4. Find all the values of `n` that are not suitable for Norman to use when calculating the salvage value of his tractor. Explain why these values are not suitable.   (2 marks)

Pat used the declining balance formula for calculating the salvage value of her tractor. The depreciation rate that she used was 20% per annum.

  1. What did Pat calculate the salvage value of her tractor to be after 14 years?   (2 marks)

  2. Using Pat’s method for depreciation, describe what happens to the salvage value of her tractor for all values of `n` greater than 15.   (1 mark)

Show Answers Only
  1. `text(Gradient) =-4000`
  2. `text(The amount the tractor depreciates each year.)`
  3. `S = 60\ 000\-4000n`
  4. `text(It is unsuitable to use)`
    `n<0\ text(because time must be positive)`
    `n>15\ text(because the tractor has no more value after 15 years and)`
    `text(therefore can’t depreciate further.)`
  5. `text(After 14 years, the tractor is worth $2638.83)`
  6. `text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases but remains)>0.`
  7.  
Show Worked Solution
♦♦♦ Mean mark 14%
COMMENT: The intercepts of both axes provide points where the gradient can be quickly found.
i.    `text(Gradient)` `= text(rise)/text(run)`
    `= (- 60\ 000)/15`
    `=-4000`

 

♦ Mean mark 37%

ii.   `text(The amount the tractor depreciates each year)`

 

♦♦ Mean mark 28%
COMMENT: Using the general form `y=mx+b` is quick here because you have the gradient (from part (i)) and the `y`-intercept is obviously `60\ 000`.
iii.   `text(S)text(ince)\ \ S = V_0\-Dn`
  `:.\ text(Equation of graph:)`
  `S = 60\ 000-4000n`

 

iv.   `text(It is unsuitable to use)` 

♦♦♦ Mean mark 20%
`n<0,\ text(because time must be positive:)`
`n>15,\ text(because it has no more value after 15)`
`text(years and therefore can’t depreciate further.)`

 

v.    `text(Using)\ S = V_0 (1-r)^n\ \ text(where)\ r = text(20%,)\ n = 14`
`S` `= 60\ 000 (1\-0.2)^14`
  `= 60\ 000 (0.8)^14`
  `= 2\ 638.8279…`

 

`:.\ text(After 14 years, the tractor is worth $2638.83`

 

♦ Mean mark 37%
vi.   `text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases)`
  `text(but remains > 0.)`

Algebra, STD2 A4 2011 HSC 28a

The air pressure, `P`, in a bubble varies inversely with the volume, `V`, of the bubble. 

  1. Write an equation relating `P`, `V` and `a`, where `a` is a constant.    (1 mark)

  2. It is known that `P = 3` when `V = 2`.

     

    By finding the value of the constant, `a`, find the value of `P` when `V = 4`.    (2 marks)

  3. Sketch a graph to show how `P` varies for different values of `V`.

     

    Use the horizontal axis to represent volume and the vertical axis to represent air pressure.   (2 marks)


Show Answers Only
  1. `P = a/V`
  2. `P = 1 1/2`
  3.  
     
Show Worked Solution
♦ Mean mark (i) 39%
COMMENT: Expressing the proportional relationship `P prop 1/V` as the equation `P=k/V` is a core skill here.
i. `P` `prop 1/V`
    `= a/V`

 

ii. `text(When)\ P=3,\ V = 2`
`3` `= a/2`
`a` `=6`

 

`text(Need to find)\ P\ text(when)\ V = 4`  

♦ Mean mark (ii) 47%
`P` `=6/4`
  `= 1 1/2`

  

♦♦ Mean mark (iii) 26%
COMMENT: An inverse relationship is reflected by a hyperbola on the graph.
iii.

Financial Maths, STD2 F5 2011 HSC 27d

Josephine invests $50 000 for 15 years, at an interest rate of 6% per annum, compounded annually.

Emma invests $500 at the end of each month for 15 years, at an interest rate of 6% per annum, compounded monthly. 

Financial gain is defined as the difference between the final value of an investment and the total contributions.

Who will have the better financial gain after 15 years? Using the Table below* and appropriate formulas, justify your answer with suitable calculations.   (4 marks)
  2UG-2011-27d1

Show Answers Only

`text(Josephine – see Worked Solutions)`

Show Worked Solution
♦ Mean mark 42%
COMMENT: Note that compound interest vs annuity comparisons are commonly tested.

`text(Josephine)`

`text(Investment)` `= 50\ 000 (1 + 0.06)^15`
  `= 50\ 000 (1.06)^15`
  `= $119\ 827.91`

 

`text(Financial gain)` `= 119\ 827.91-50\ 000`
  `= $69\ 827.91`

 

`text(Emma)`

`text{Monthly interest rate} = text(6%)-:12=text(0.5%)`

`text{# Monthly Payments}=12 xx 15=180`

`=>\ text{Annuity Factor = 290.8187    (from Table)}`

 

`text(Investment)` `= 500 xx 290.8187`
  `= $145\ 409.35`

 

`text(Financial gain)` `= 145\ 409.35\-text(total contributions)`
  `= 145\ 409.35\-(500 xx 12 xx 15)`
  `= 145\ 409.35\-90\ 000`
  `= $55\ 409.35`

 

`:.\ text(Josephine will have the better financial gain.)`

Statistics, STD2 S5 2011 HSC 27c

Two brands of light bulbs are being compared. For each brand, the life of the light bulbs is normally distributed.

2011 27c

  1. One of the Brand B light bulbs has a life of 400 hours. 

     

    What is the  `z`-score of the life of this light bulb?  (1 mark)

  2. A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:

     

    ‘Brand A light bulbs are more likely to be defective than Brand B light bulbs.’

     

    Is this claim correct? Justify your answer, with reference to  `z`-scores or standard deviations or the normal distribution.   (2 marks)

Show Answers Only
  1. `-2`
  2. `text(The claim is incorrect.)`
Show Worked Solution

i.    `z text{-score of Brand B bulb (400 hrs)}`

`= (x-mu)/sigma`
`= (400\-500)/50`
`= –2`

 

ii.   `z text{-score of Brand A bulb (400 hours)}`

♦ Mean mark 42%
MARKER’S COMMENT: A number of students found drawing normal curves in their solution advantageous.
`=(400-450)/25`
`=–2`

 
`text(S)text(ince the)\ z text(-score for both brands is –2,)`

`text(they are equally likely to be defective.)`

`:.\ text(The claim is incorrect.)`

Measurement, 2UG 2011 HSC 27b

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

Both places lie on the Equator. 

  1. Find the shortest distance between these two places, to the nearest kilometre. You may assume that the radius of the Earth is 6400 km.   (2 marks)
  2.  
  3. The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul?    (2 marks)
  4.  
Show Answers Only
  1. `10\ 165\ text(km)\ \ \ text{(nearest km)}`
  2. `152^@ text(E)`
Show Worked Solution
♦♦ Mean mark 29%
(i) `text(Longitude difference)` `= 109 + 160`
    `= 269^@`
`=> text(Shortest distance)\ text{(by degree)}` `= 360\-269`
  `= 91^@`
`:.\ text(Shortest distance)` `= 91/360 xx 2 pi r`
  `= 91/360 xx 2 xx pi xx 6400`
  `= 10\ 164.79…`
  `=10\ 165\ text(km)\ text{(nearest km)}`

 

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

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

Statistics, STD2 S1 2011 HSC 27a

A company sells handbags in Paris, New York and Florence. 

Use the data in the table to complete the area chart below.   (2 marks)

 2011 27a 

2UG 2011 27a

Show Answers Only

 2011 27a

Show Worked Solution
♦♦ Mean mark 27%
COMMENT: Most common error was to not realise Area charts show cumulative data.

2011 27a

Financial Maths, 2UG 2011 HSC 26c

Furniture priced at  $20 000  is purchased. A deposit of 15% is paid. 

The balance is borrowed using a flat-rate loan at 19% per annum interest, to be repaid in equal monthly instalments over five years.

What will be the amount of each monthly instalment? Justify your answer with suitable calculations.   (4 marks) 

Show Answers Only

 `$552.50`

Show Worked Solution

`text(Purchase price) = $20\ 000`

♦ Mean mark 44%
`text(Deposit)` `=\ text(15%)\ xx 20\ 000`
  `= $3000`
`:.\ text(Loan)` `= 20\ 000\-3000`
  `= $17\ 000`
`text(Using)\ I` `= Prn`
`I` `= 17\ 000 xx\ text(19%)\ xx 5`
  `= 16\ 150`
`=>\ text(Total amount to repay)` `= 17\ 000 + 16\ 150`
  `= $33\ 150`

 

`:.\ text(Monthly instalment)` `= (33\ 150)/(12 xx 5)`
  `= (33\ 150)/60`
  `= $552.50`

Probability, 2UG 2011 HSC 26a

The two spinners shown are used in a game.

2UG 2011 26a1

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

2UG 2011 26a2

  1. What is the value of `X` in the table?   (1 mark)
  2. What is the probability of obtaining a score less than 4?   (1 mark)
  3. On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3?   (1 mark)
  4. Elise plays a game using the spinners with the following financial outcomes.  

⇒ Win `$12` for a score of `4`

⇒ Win nothing for a score of less than `4`

⇒ Lose `$3` for a score of more than `4`

It costs `$5` to play this game. Will Elise expect a gain or a loss and how much will it be?

Justify your answer with suitable calculations.   (3 marks)

Show Answers Only
  1. `5`
  2. `1/2`
  3.  
  4. `2/3`
  5.  
  6. `text(Loss of)\ $1.50`
  7.  
Show Worked Solution

(i)   `X=3+2=5`

 

(ii)   `P(text{score}<4)=6/12=1/2`

 

(iii)   `P(3)=2/3`

 

(iv)   `P(4)=4/12=1/3`

♦ Mean mark 34%
MARKER’S COMMENT: Better responses remembered to deduct the $5 cost to play and recognised the negative result as a loss.
`P(text{score}<4)` `=6/12=1/2`
`P(text{score}>4)` `=2/12=1/6`

 

`text(Financial Expectation)`

`=(1/3xx12)+(1/2xx0)-(1/6xx3)-5`
`=4-0.5-5`
`=-1.50`

 

`:.\ text(Elise should expect a loss of $1.50) `

Financial Maths, STD2 F4 2010 HSC 25b

William wants to buy a car. He takes out a loan for  $28 000  at 7% per annum interest for four years. 

Monthly repayments for loans at different interest rates are shown in the spreadsheet.

2010 25b

How much interest does William pay over the term of this loan?   (2 marks)

Show Answers Only

`$4183.52`

Show Worked Solution
♦ Mean mark 42%
MARKER’S COMMENT: An incorrect table value used correctly in the following calculations received half-marks here. Show your working!

`text(Loan) = $28\ 000,\ \ \ \ r =\ text(7% p.a.)`

`text(Monthly repayment = $670.49`

`text(# Repayments) = 4 xx 12 = 48`

`text(Total repaid)` `= 48 xx 670.49`
  `= $32\ 183.52`

 

`:.\ text(Interest paid)` `=32\ 183.52\-28\ 000`
  `=$4183.52`

Statistics, STD2 S1 2011 HSC 25d

Data was collected from 30 students on the number of text messages they had sent in the previous 24 hours. The set of data collected is displayed.
 

2UG 2011 25d

  1. What is the outlier for this set of data? (1 mark)

  2. What is the interquartile range of the data collected from the female students? (1 mark)

Show Answers Only
  1. `71`
  2. `9`
Show Worked Solution

i.   `text(Outlier is 71)`

♦♦ Mean mark 34%
COMMENT: Ensure you can quickly and accurately find quartile values using stem and leaf graphs!

ii.   `text{Lower quartile = 9   (4th female data point)}`

`text{Upper quartile = 20   (11th female data point)}`

`:.\ text{Interquartile range (female)}=20-11=9`

Statistics, STD2 S1 2011 HSC 25a

A study on the mobile phone usage of NSW high school students is to be conducted.

Data is to be gathered using a questionnaire.

The questionnaire begins with the three questions shown.

2UG 2011 25a

  1. Classify the type of data that will be collected in Q2 of the questionnaire.  (1 mark)

  2. Write a suitable question for this questionnaire that would provide discrete ordinal data.   (1 mark)

  3. An initial study is to be conducted using a stratified sample.

     

    Describe a method that could be used to obtain a representative stratified sample.  (1 mark)

  4. Who should be surveyed if it is decided to use a census for the study?  (1 mark)

Show Answers Only
  1. `text(Categorical)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(A census would involve all high school students in NSW.)`
Show Worked Solution

i.  `text(Categorical)`

 

ii.  `text(How many outgoing calls do you make per day?)`

`text{(Ensure it can be answered with a numerical score.)}`

 

iii.  `text(The method could be to work out how many)`

♦♦♦ Mean mark 7%. Toughest mark to get in the 2011 exam!
COMMENT: Know and be able to describe random, systematic and stratified sampling!

`text{students are in each year and ask 10% of the}`

`text{students in each year. (Note the sample of}`

`text{students in each year must be  proportional to}`

`text{their percentage in the population).}`

 

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: A specific population needed (i.e. high school students).

iv.  `text(A census would involve all high school)`

`text(students in NSW.)`

Financial Maths, STD2 F4 2011 HSC 23c

An amount of $5000 is invested at 10% per annum, compounded six-monthly.

2UG 2011 23c

Use the table to find the value of this investment at the end of three years.   (2 marks)

Show Answers Only

`$6700`

Show Worked Solution
♦♦ Mean mark 28%
MARKER’S COMMENT: Remember that the number of periods is the number of “compounding periods” and when asked to use the table, use the table!

`text(Interest rate)= text(10% pa)= text(5% per 6 months)`

`text(Period)= 6\ \ \ \ text{(6 x 6 months in 3 years)}`

`=> text(Table value)=1.340`

`:.\ text(Value of investment)` `=5000xx1.34`
  `=$6700`

Measurement, STD2 M1 2010 HSC 28b

Moivre’s manufacturing company produces cans of Magic Beans. The can has a diameter of  10 cm and a height of  10 cm.

2010 28b1

  1. Cans are packed in boxes that are rectangular prisms with dimensions  30 cm × 40 cm × 60 cm.

     

    What is the maximum number of cans that can be packed into one of these boxes?   (1 mark)

  2. The shaded label on the can shown wraps all the way around the can with no overlap. What area of paper is needed to make the labels for all the cans in this box when the box is full?   (2 marks)

  3. The company is considering producing larger cans. Monica says if you double the diameter of the can this will double the volume.

     

    Is Monica correct? Justify your answer with suitable calculations.   (2 marks)

      
    The company wants to produce a can with a volume of 1570 cm³, using the least amount of metal. Monica is given the job of determining the dimensions of the can to be produced. She considers the following graphs.
     
    2010 28b2

  4. What radius and height should Monica recommend that the company use to minimise the amount of metal required to produce these cans? Justify your choice of dimensions with reference to the graphs and/or suitable calculations.   (2 marks)

Show Answers Only
  1. `72\ text(cans)`
  2. `20\ 358\ text{cm²  (nearest cm²)}`
  3. `text(Monica is incorrect because the volume)`

     

    `text(doesn’t double.  It increases by a factor of 4.)`

  4. `text(Radius 6.3 cm and height 12.6 cm.)`
Show Worked Solution
♦♦ Mean mark 27%
i.    `text(Maximum # Cans)` `= 4 xx 3 xx 6`
    `= 72\ text(cans)`

 

♦ Mean mark 38%
MARKER’S COMMENT: Many students didn’t account for the clearance of 0.5 cm at the top and bottom of each can.
ii.    `text(Label Area)\ text{(1 can)}` `= 2 pi rh`
    `= 2 xx pi xx 5 xx 9`
    `= 90 pi`
    `= 282.7433…\ text(cm²)`

 

`:.\ text(Label Area)\ text{(72 cans)}`

`= 72 xx 282.7433…`

`= 20\ 357.52…`

`= 20\ 358\ text(cm²)`  `text{(nearest cm²)}`

 

♦ Mean mark 44%
MARKER’S COMMENT: Many students performed calculations in this part without concluding if Monica is correct or not. Read the question carefully.
iii.   `text(Original volume)` `= pi r^2 h`
    `= pi xx 5^2 xx 10`
    `= 785.398…\ text(cm³)`
`text(If the diameter doubles, radius) = 10\ text(cm)`
`text(New volume)` `= pi xx 10^2 xx 10`
  `= 3141.592…\ text(cm³)`

 

`:.\ text(Monica is incorrect because the volume)`
`text(doesn’t double.  It increases by a factor of 4.)`

 

♦♦ Mean mark 26%
iv.
`text(Minimum metal used when surface area is a minimum.)`
`text(From graph, minimum surface area when)\ r = 6.3\  text(cm)`
`text(When)\ r = 6.3\  text(cm,)\ h = 12.6\  text(cm)\ \ \ text{(from graph)}`
`:.\ text(She should recommend radius 6.3 cm and height 12.6 cm)`

Financial Maths, STD2 F4 2010 HSC 28a

The table shows monthly home loan repayments with interest rate changes from February to October 2009.

 2010 28a

  1. What is the change in monthly repayments on a  $250 000  loan from February 2009 to April 2009?  (1 mark)

  2. Xiang wants to borrow  $307 000  to buy a house.

     

    Xiang’s bank approves loans for customers if their loan repayments are no more than 30% of their monthly gross salary.

     

    Xiang’s monthly gross salary is $6500.

     

    If she had applied for the loan in October 2009, would her bank have approved her loan?

     

    Justify your answer with suitable calculations.    (3 marks)

  3. Jack took out a loan at the same time and for the same amount as Xiang.

     

    Graphs of their loan balances are shown.
     
          2010 28a2

    Identify TWO differences between the graphs and provide a possible explanation for each difference, making reference to interest rates and/or loan repayments.   (2 marks)

Show Answers Only
  1. `text(Monthly repayments decrease by $15)`
  2. `text(S)text(ince repayments of 1987.29 > 1950, the loan)`
    `text(would not have been approved.)`
  3. `text(Differences)`
  4. `text(Jack’s loan balance falls more sharply for first 12 years)`
  5. `text(Jack’s loan balance falls less sharply between years 12-30.)`
  6.  
  7. `text(Explanation)`
  8.  
  9. `text(Jack made larger repayments for first 12 years, or)`
  10. `text(Jack made the same repayments but had a lower interest rate)`
  11. `text(for the first 12 years.)`
  12. `text(Jack made smaller repayments in years 12-30.)`
Show Worked Solution
i.    `text(Repayment)\ text{(Feb 09)}` `= 1588`
  `text(Repayment)\ text{(Apr 09)}` `= 1573`
`text(Difference) = 1588\-1573 = 15`
`:.\ text(Monthly repayments decrease by $15)`

 

 

♦ Mean mark 39%
MARKER’S COMMENT: Borrowing $307,000 can be achieved by borrowing $300,000, and then `7` times the table repayment value for borrowing $1000. 
ii.    `text(Loan) = $307\ 000`
`text{Repayments (Oct 09)}` `= 1942 + (7 xx 6.47)`
  `= 1942 + 45.29`
  `= $1987.29\ text(per month)`

 

`text(30% Gross salary)` `= 6500 xx\ text(30%)`
  `= $1950\ text(per month)`

 

`:.\ text(S)text(ince repayments of $1987.29 > $1950, the loan)`
`text(would not have been approved.)`

 

 

 

iii.  `text(Differences)`

`text(Jack’s loan balance falls more sharply for first 12 years)`

`text(Jack’s loan balance falls less sharply between years 12-30.)`

`text{Explanation(s)}`

♦ Mean mark 36%
MARKER’S COMMENT: Explanations were generally poor and many failed to refer directly to the graphs shown, or reference Xiang or Jack directly.

`text(Jack made larger repayments for 1st 12 years, or)`

`text(Jack made the same repayments but had a)`

`text(lower interest rate for the first 12 years.)`

`text(Jack made smaller repayments in years 12-30.)`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

  1. Use the graph to find the tax payable on a taxable income of $21 000.  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  `A`  is  `1/3`.    (1 mark)

  3. How much of each dollar earned between  $21 000  and  $39 000  is payable in tax?   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between  $21 000  and  $39 000.   (2 marks)

Show Answers Only
  1. `$3000\ \ \ text{(from graph)}`
  2. `1/3`
  3. `33 1/3\ text(cents per dollar earned)`
  4. `text(Tax payable on)\ I = 1/3 I\-4000`
Show Worked Solution
i.

 `text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

 

ii.  `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%
`text(Gradient at)\ A` `= (y_2\-y_1)/(x_2\ -x_1)`
  `= (9000-3000)/(39\ 000 -21\ 000)`
  `= 6000/(18\ 000)`
  `= 1/3\ \ \ \ \ text(… as required)`

 

iii.  `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.
`text(Tax)` ` = 1/3\ text(of each dollar earned)`
  ` = 33 1/3\ text(cents per dollar earned)`

 

iv.  `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I` `= 3000 + 1/3 (I\-21\ 000)`
  `= 3000 + 1/3 I\-7000`
  `= 1/3 I\-4000`

Statistics, STD2 S1 2010 HSC 27b

The graphs show the distribution of the ages of children in Numbertown in 2000 and 2010.
  

  1. In 2000 there were 1750 children aged 0–18 years.

     

    How many children were aged 12–18 years in 2000?   (1 mark)

  2. The number of children aged 12–18 years is the same in both 2000 and 2010.

     

    How many children aged 0–18 years are there in 2010?    (1 mark)

  3. Identify TWO changes in the distribution of ages between 2000 and 2010. In your answer, refer to measures of location or spread or the shape of the distributions.   (2 marks)

  4. What would be ONE possible implication for government planning, as a consequence of this change in the distribution of ages?   (1 mark)

Show Answers Only

i.    `875`

ii.    `3500`

iii.  `text{Changes in distribution (include 2 of the following):}`

  • `text(the lower quartile age is lower in 2010)`
  • `text(the median is lower in 2010)`
  • `text(the upper quartile age is lower in 2010)`
  • `text(the interquartile range is greater in 2010)`
  • `text(2010 is positively skewed while 2000 is negatively)`

iv.  `text(Implication for government planning:)`

`text(Since the children are getting younger in 2010,)`

  • `text(Approve and build more childcare facilities)`
  • `text(Build more school and public playgrounds)`
Show Worked Solution

i.    `text{Since the median = 12 years}`

♦ Mean mark (i) 45%

`=>\ text{50% of children are aged 12–18 years}`

`:.\ text{Children aged 12–18}\ = 50\text{%}\ xx 1750 = 875`

 

♦♦ Mean mark (ii) 25%

ii.   `text{Upper quartile (2010) = 12 years}`

`text{Children in upper quartile = 875 (from part (i))}`

`:.\ text{Children aged 0–18}\ =4 xx 875= 3500`
 

iii.  `text{Changes in distribution (include 2 of the following):}`

♦ Mean mark (iii) 35%
MARKER’S COMMENT: A number of students incorrectly identified “positive” skew as “negative” skew here.
  • `text(the lower quartile age is lower in 2010)`
  • `text(the median is lower in 2010)`
  • `text(the upper quartile age is lower in 2010)`
  • `text(the interquartile range is greater in 2010)`
  • `text(2010 is positively skewed while 2000 is negatively)`

iv.  `text(Implication for government planning:)`

♦ Mean mark (iv) 46%
MARKER’S COMMENT: Answers should reflect the 1 mark allocation.

`text(Since the children are getting younger in 2010,)`

  • `text(Approve and build more childcare facilities)`
  • `text(Build more school and public playgrounds)`

Algebra, 2UG 2010 HSC 27a

Fully simplify  `(4x^2)/(3y) -: (xy)/5`.   (3 marks) 

Show Answers Only

 `(20x)/(3y^2)`

Show Worked Solution
♦♦ Mean mark 33%.
MARKER’S COMMENT: Remember to “invert and multiply” when dividing a fraction by a fraction (see Worked Solution).
`(4x^2)/(3y) -: (xy)/5` `= (4x^2)/(3y) xx 5/(xy)`
  `= (20x^2)/(3xy^2)`
  `= (20x)/(3y^2)`

Probability, 2UG 2010 HSC 26c

Tai plays a game of chance with the following outcomes. 

• `1/5`  chance of winning  `$10` 

• `1/2`  chance of winning  `$3` 

• `3/10`  chance of losing  `$8` 

The game has a  `$2`  entry fee. 

What is his financial expectation from this game?    (2 marks)

Show Answers Only

 `text(A loss of $0.90.)`

Show Worked Solution
♦♦ Mean mark 31%
MARKER’S COMMENT: A common error was not to deduct the chance of losing or the entry fee. BE CAREFUL!

`text(Financial Expectation)`

`= (1/5 xx 10) + (1/2 xx 3)\-(3/10 xx 8)\-2`

`= 2 + 1.5\-2.4\-2`

`=-0.9`
 

`:.\ text(The financial expectation is a loss of $0.90.)`

Statistics, STD2 S1 2010 HSC 26b

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
 

2010 26b

  1. Find the value of  Χ  in the table.   (1 mark)

  2. On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data.   (1 mark)
  3. Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer.   (1 mark)
  4. Prior to the opening of the new shopping centre, the median number of motor vehicles passing the school between  2.30 pm  and  4.00 pm  was 57 vehicles per day.

     

    What problem could arise from the change in the median number of motor vehicles passing the school before and after the opening of the new shopping centre?

     

    Briefly recommend a solution to this problem.   (2 marks)

Show Answers Only
  1. `15`
  2.  
  3.  
  4. `text(Problems)`
  5. `text(- increased traffic delays)`
  6. `text(- increased danger to students leaving school)`

  7.  

    `text(Solutions)`

  8. `text(- signpost alternative routes around school)`
  9. `text(- decrease the speed limit in the area)`
Show Worked Solution
i. `X` `= 25\ -10`
    `= 15`

 

♦♦♦ Mean mark 18%
MARKER’S COMMENT: The ogive was poorly drawn with many students incorrectly joining the middle of each column rather than from corner to corner.
ii.
♦♦ Mean mark 25%
MARKER’S COMMENT: Many students did not “show by drawing lines on the graph” as the question asked.

iii.  `text(Median)\ ~~155`

♦ Mean mark 47%
MARKER’S COMMENT: Short answers were often the best. Be concise when you can.
iv. `text(Problems)`
  `text(- increased traffic delays)`
  `text(- increased danger to students leaving school)`
   
  `text(Solutions)`
  `text(- signpost alternative routes around school)`
  `text(- decrease the speed limit in the area)`

Probability, STD2 S2 2010 HSC 26a

A design of numberplates has a two-digit number, two letters and then another two-digit number, for example

2010 26a1

  1. How many different numberplates are possible using this design?   (1 mark)

Jo can order a numberplate with ‘JO’ in the middle but will have to have randomly selected numbers on either side.

Jo’s birthday is 30 December 1992, so she would like a numberplate with either 

2010 26a2

  1. What is the probability that Jo is issued with one of the numberplates she would like?   (2 marks)

Show Answers Only
  1. `6\ 760\ 000`
  2. `P = 1/5000`
Show Worked Solution
♦ Mean mark 41%
i.    `text(# Combinations)` `=10 xx 10 xx 26 xx 26 xx 10 xx 10`
    `=6\ 760\ 000`

 

♦♦ Mean mark 30%
IMPORTANT: Since the middle letters of “JO” can be guaranteed, the focus becomes purely on the 4 surrounding digits.

ii.   `text(# Possible numberplates)`

`=10 xx 10 xx 10 xx 10`

`=10\ 000`
 

`:.P (30\ text(JO)\ 12) + P (19\ text(JO)\ 92)`
`= 1/(10\ 000) + 1/(10\ 000)`
`= 1/5000`

Measurement, STD2 M7 2013 HSC 30c

Joel mixes petrol and oil in the ratio  40 : 1  to make fuel for his leaf blower. 

  1. Joel pours 5 litres of petrol into an empty container to make fuel for his leaf blower.

     

    How much oil should he add to the petrol to ensure that the fuel is in the correct ratio?   (1 mark)

  2. Joel has 4.1 litres of fuel left in his container after filling his leaf blower.

     

    He wishes to use this fuel in his lawnmower. However, his lawnmower requires the petrol and oil to be mixed in the ratio  25 : 1.

     

    How much oil should he add to the container so that the fuel is in the correct ratio for his lawnmower?   (3 marks)

Show Answers Only
  1. `text(0.125 L)`
  2. `60\ text(mL)`
Show Worked Solution
♦ Mean mark 50% 
i.    `text(Petrol : Oil) = 40\ :\ 1`
  `5\ text(Litres :)\ X = 40\ :\ 1`
`:. 5/X` `= 40/1`
`40X` `=5`
`X` `= 5/40 = 0.125\ text(L)`

 

`:.\ text(Joel should add 0.125 L of oil)`

 

ii.  `text(4.1 L)\ = 4100\ text(mL)`

`text(In ratio  40:1,  there is)`

♦♦♦ Mean mark 8% 
STRATEGY: The critical information required is how much petrol is in the container. Once known, the oil required to satisfy a 25:1 ration can be easily found.

`=>\ text(4000 mL Petrol and 100 mL Oil)`

 

`text(Lawnmower ratio) = 25:1`

`text(Let Oil required for 4000 mL Petrol)\ = X\ text(mL)`

`X/4000` `=1/25`
`25X` `=4000`
`X` `=4000/25`
  `=160\ text(mL)`

 
`text(4 L of petrol requires 160 mL of oil)`

`text(Container already has 100 mL of oil)`

`:.\ text(Oil to add)\ ` `=160\ -100`
  `=60\ text(mL)`

Probability, STD2 S2 2013 HSC 30b

In a class there are 15 girls (G) and 7 boys (B). Two students are chosen at random to be class representatives.

  1. Complete the tree diagram below.    (2 marks)
     
     

  2. What is the probability that the two students chosen are of the same gender?    (2 marks)

Show Answers Only
  1.  
  2. `6/11`
Show Worked Solution

i.  

ii.    `Ptext{(same gender)}` `=P(G,G) + P(B,B)`
    `=(15/22 xx 14/21) + (7/22 xx 6/21)`
    `=210/462 + 42/462`
    `=252/462`
    `=6/11`
♦ Mean mark (ii) 40%.

Algebra, STD2 A4 2013 HSC 30a

Wind turbines, such as those shown, are used to generate power.

2013 30a

In theory, the power that could be generated by a wind turbine is modelled using the equation

`T = 20\ 000w^3`

where `T` is the theoretical power generated, in watts 
  `w` is the speed of the wind, in metres per second.

 

  1. Using this equation, what is the theoretical power generated by a wind turbine if the wind speed is 7.3 m/s ?  (1 mark)

In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.

  1. If `A` is the actual power generated, in watts, write an equation for `A` in terms of `w`.    (1 mark)

The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.
 
        2013 30a2

  1. Using the graph, or otherwise, find the difference between the theoretical power and the actual power generated when the wind speed is 9 m/s.   (1 mark)

A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.

  1. What is the minimum wind speed required for the farm to be self-sufficient?    (1 mark)

A more accurate formula to calculate the power (`P`) generated by a wind turbine is

`P = 0.61 xx pi xx r^2 × w^3` 

where     `r` is the length of each blade, in metres
  `w` is the speed of the wind, in metres per second. 

 
Each blade of a particular wind turbine has a length of 43 metres.The turbine operates at a wind speed of 8 m/s.

  1. Using the formula above, if the wind speed increased by 10%, what would be the percentage increase in the power generated by this wind turbine?   (3 marks)

Show Answers Only
  1. `7\ 780\ 340\ text(watts)`
  2. `8000w^3`
  3. `8.8\ text(million watts, or 8 748 000 watts)`
  4. `w = 8.2\ text(m/s)\ \ \ text{(1 d.p.)}`
  5. `text(33%)`
Show Worked Solution
i.    `T=20\ 000w^3`
  `text(If)\ \ w = 7.3`
`T` `=20\ 000 xx (7.3)^3`
  `= 7\ 780\ 340\ text(watts)`

  

♦ Mean mark 34%
ii.    `text(We know)\ A = 40% xx T`
`=> A` `=0.4 xx 20\ 000 xx w^3`
  `=8000w^3`

  

♦ Mean mark 38%
iii.    `text(Solution 1)`
  `text(At)\ w=9`
  `A = text(5.8 million watts)\ \ \ text{(from graph)}`
  `T = text(14.6 million watts)\ \ \ text{(from graph)}`
`text(Difference)` `= text(14.6 million)\-text(5.8 million)`
  `= text(8.8 million watts)`

 

`text(Alternative Solution)`
`text(At)\ w=9`
`T` `= 20\ 000 xx 9^3`
  `= 14\ 580\ 000\ text(watts)`
`A` `= 8000 xx 9^3`
  `= 5\ 832\ 000\ text(watts)`
`text(Difference)` `=14\ 580\ 000\-5\ 832\ 000`
  `=8\ 748\ 000\ \ text(watts)`

  

♦♦ Mean mark 25%
COMMENT: Students need to be comfortable in finding the cube roots of values – a calculation that can be required in a number of topic areas and is regularly examined.
iv.    `text(Find)\ w\ text(if)\ A=4.4\ text(million)`
`8000w^3` `= 4\ 400\ 000`
`w^3` `= (4\ 400\ 000)/8000`
  `= 550`
`:. w` `= root(3)(550)`
  `=8.1932…`
  `=8.2\ text(m/s)\ \ \ text{(1 d.p.)}`

 

`:.\ text(The minimum wind speed required is 8.2 m/s)`

 

♦ Mean mark 41%
MARKER’S COMMENT: Students are reminded that a % increase requires them to find the difference in power generated at different wind speeds and divide this result by the original power output, as shown in the Worked Solution.
v.    `text(Find)\ P\ text(when)\ w=8\ text(and)\ r=43`
`P` `= 0.61 xx pi xx r^2 xx w^3`
  `= 0.61 xx pi xx 43^2 xx 8^3`
  `= 1\ 814\ 205.92\ text(watts)`

 

`text(When speed of wind)\ uarr10%`
`w′ = 8 xx 110text(%) = 8.8\ text(m/s)`
 

`text(Find)\ P\ text(when)\ w′ = 8.8`

`P` `=0.61 xx pi xx 43^2 xx 8.8^3`
  `= 2\ 414\ 708.08\ text(watts)`

 

`text(Increase in Power)` `=2\ 414\ 708.08\-1\ 814\ 205.92`
  `= 600\ 502.16`

 

`:.\ text(% Power increase)` `= (600\ 502.16)/(1\ 814\ 205.92)`
  `= 0.331`
  `= text(33%)\ \ \ text{(nearest %)}`

 

Measurement, STD2 M6 2010 HSC 26d

Find the area of triangle `ABC`, correct to the nearest square metre.   (3 marks)

Show Answers Only

`717\ text(m²)`    `text{(nearest m²)}`

Show Worked Solution
♦♦ Mean mark 32%.
TIP: The allocation of 3 marks to this question should flag the need for more than 1 step.
`cos/_C` `=(AC^2 + CB^2-AB^2)/(2 xx AC xx CB)`
  `=(50^2 + 40^2-83^2)/(2 xx 50 xx 40)`
  `= -0.69725…`
`/_C` `=134.2067…^@`

 

`text(Using Area) = 1/2 ab\ sinC :`
`text(Area)\ Delta ABC` `=1/2 xx 50 xx 40 xx sin134.2067…^@`
  `=716.828…`
  `=717\ text(m²)\ \ \ \ text{(nearest m²)}`

Probability, 2UG 2013 HSC 29d

Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown in the table.

2013 29d

It costs `$4` to play one game. Will Jane expect a gain or a loss, and how much will it be? Justify your answer with suitable calculations.    (3 marks)

Show Answers Only

 `text(Loss of $1.50.)`

Show Worked Solution

`P (H,H) = 1/2 xx 1/2 = 1/4`

♦♦ Mean mark 31%
MARKER’S COMMENT: Students must be comfortable with how to set out and calculate such financial expectation examples, as well as interpreting their result.
`P (H\ text{and}\ T)` `= P (H,T) + P (T,H)`
  `= (1/2 xx 1/2) + (1/2 xx 1/2)`
  `= 1/2`

`P (T,T)  = 1/2 xx 1/2 = 1/4`

`text(Financial Expectation)`

`=(1/4 xx 6) + (1/2 xx 1) + (1/4 xx 2)-4`
`=1.50 + 0.5 + 0.5 -4`
`=-1.50`

 

`:.\ text(Jane should expect a loss of $1.50.)`

Probability, STD2 S2 2011 HSC 24b

A die was rolled 72 times. The results for this experiment are shown in the table.
  

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number obtained} \rule[-1ex]{0pt}{0pt} & \textit{Frequency} \\
\hline
\rule{0pt}{2.5ex} \ 1 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \ 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} \ 3 \rule[-1ex]{0pt}{0pt} & \textbf{A} \\
\hline
\rule{0pt}{2.5ex} \ 4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} \ 5 \rule[-1ex]{0pt}{0pt} & 12 \\
\hline
\rule{0pt}{2.5ex} \ 6 \rule[-1ex]{0pt}{0pt} & 15 \\
\hline
\end{array}

  1. Find the value of  `A`.   (1 mark)

  2. What was the relative frequency of obtaining a 4.   (1 mark)

  3. If the die was unbiased, which number was obtained the expected number of times?   (1 mark)

Show Answers Only
  1. \(10\)
  2. \(\dfrac{1}{9}\)
  3. \(5\)
Show Worked Solution
i.     \(\text{Since die rolled 72 times}\)
\(\therefore\ A\) \(=72-(16+11+8+12+15)\)
  \(=72-62\)
  \(=10\)
♦ Mean mark 38%
IMPORTANT: Many students confused ‘relative frequency’ with ‘frequency’ and incorrectly answered 8.
ii.     \(\text{Relative frequency of 4}\) \(=\dfrac{8}{72}\)
  \(=\dfrac{1}{9}\)

 

iii.  \(\text{Expected frequency of any number}\)
\(=\dfrac{1}{6}\times 72\)
\(=12\)
 
\(\therefore\ \text{5 was obtained the expected number of times.}\)

Algebra, STD2 A2 2011 HSC 23b

Sticks were used to create the following pattern. 

The number of sticks used is recorded in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape $(S)$} \rule[-1ex]{0pt}{0pt} & \;\;\; 1 \;\;\; & \;\;\; 2 \;\;\; & \;\;\; 3 \;\;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of sticks $(N)$}\; \rule[-1ex]{0pt}{0pt} & \;\;\; 5 \;\;\; & \;\;\; 8 \;\;\; & \;\;\; 11 \;\;\; \\
\hline
\end{array}

  1. Draw Shape 4 of this pattern.  (1 mark)

  2. How many sticks would be required for Shape 100?    (1 mark)

  3. Is it possible to create a shape in this pattern using exactly 543 sticks?

     

    Show suitable calculations to support your answer.    (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `302`
  3. `text(No)`
Show Worked Solution

  i.     `text(Shape 4 is shown below:)`

ii.     `text(S)text(ince)\ \ N=2+3S`

♦ Mean mark 48%.
MARKER’S COMMENT: Students should attempt to find a “rule” in such questions, and use this formula to solve the question, as per the Worked Solution.  
`text(If)\ \ S` `=100`,
`N` `=2+(3xx100)`
  `=302`

 

iii.    `543` `=2+3S`
  `3S` `=541`
  `S` `=180 1/3`

 

`text(S)text(ince S is not a whole number, 543 sticks)`

`text(will not create a figure.)`

Financial Maths, STD2 F4 2011 HSC 22 MC

Ying borrowed $250 000 to buy a house. The interest rate and monthly repayment for her loan are shown in the spreadsheet.

2UG 2011 22

What is the total interest charged for the first four months of this loan?

  1.   $6364.32
  2.   $6366.11
  3.   $6369.67
  4.   $6376.25
Show Answers Only

`A`

Show Worked Solution

`text(Month 3)`

`P+I-R` `=251\ 032.04-1871.94`
  `=249\ 160.10`

`text(Month 4)`

`P` `=249\ 160.10`
`:.I` `=249\ 160.10xx0.0765/12`
  `=1588.40`

 

`:.\ text(Total interest)` `=1593.75+1591.98+1590.19+1588.40`
   `=6364.32`

`=> A`

Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost `C`, in dollars, for the centre
to host an event, where `x` people attend, is given by:

`C = 10\ 000 + 50x`

The centre charges $100 per person. Its income `I`, in dollars, is given by:

`I = 100x`
 

2UG 2011 20

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

  1.  `$15\ 000`
  2. `$20\ 000`
  3. `$30\ 000` 
  4. `$40\ 000`
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

`text(When)\ x=500,\ I=100xx500=$50\ 000`

`text(Breakeven when)\ \ x=200\ \ \ text{(from graph)}`

`text(When)\ \ x=200,\ I=100xx200=$20\ 000`

`text(Difference)` `=50\ 000-20\ 000`
  `=$30\ 000`

 
`=> C`

Financial Maths, STD2 F1 2011 HSC 19 MC

Simon is a mechanic who receives a normal rate of pay of $22.35 per hour for a 40-hour
week.

When he is needed for emergency call-outs he is paid a special allowance of $150 for that
week. Additionally, every time he is called out to an emergency he is paid for a minimum
of 4 hours at double time.

In the week beginning 2 February, 2011 Simon worked 40 hours normal time and was
needed for emergency call-outs. His emergency call-out log book for the week is shown.

2011 19 mc

What was Simon’s total pay for that week?

  1.   $1189.28
  2.   $1296.30
  3.   $1334.55
  4.  $1446.30
Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 41%
COMMENT: To reduce errors, calculate each element of pay separately in your working and then add up the total, as per the Worked Solution.
`text(Normal pay)` `=40xx22.35`
  `=894.00`

 
`text(Special Allowance)= 150.00`

`text(Emergency Calls)` `= (5xx2xx22.35)+(4xx2xx22.35)`
  `=223.50+178.80`
  `=402.30`

 

`text(Total Weekly Pay)` `=894.00+150.00+402.30`
  `=1446.30`

 
`=> D`

Measurement, 2UG 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.

2UG 2011 24c

Copy the diagram into your writing booklet and show all the information on it.

  1. What is the bearing of `C` from `B`?   (1 mark)
  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)
  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

 

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
♦♦ Mean mark 24% 
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier.
(i)     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD` `=180-121\ text{(cointerior with}\ \ /_A text{)}`
  `=59^@`
`/_DBC` `=114-59`
  `=55^@`

  

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

(ii)    `text(Using cosine rule:)`

♦ Mean mark 39%
`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

(iii)    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

♦♦♦ Mean mark 15%
MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Financial Maths, STD2 F5 2009 HSC 27a

The table shows the future value of a $1 annuity at different interest rates over different numbers of time periods. 
 

2UG-2009-27a

  1. What would be the future value of a $5000 per year annuity at 3% per annum for 6 years, with interest compounding yearly?   (1 mark)

  2. What is the value of an annuity that would provide a future value of  $407100  after 7 years at 5% per annum compound interest?   (1 mark)

  3. An annuity of $1000 per quarter is invested at 4% per annum, compounded quarterly for 2 years. What will be the amount of interest earned?    (3 marks)

Show Answers Only
  1. `$32\ 342`
  2. `$50\ 000`
  3. `$285.70`
Show Worked Solution

i.  `text(Table factor when)\ \ n = 6,\ \ \ r =\ 3text(%) \ => \ 6.4684`

`:.\ FV` `= 5000 xx 6.4684`
  `= $32\ 342`


ii.  `text(Table factor when)\ \ n = 7,\ \ \ r =\ text(5%)` 

♦ Mean mark 45%
MARKER’S COMMENT: A common error was to multiply $407 100 by 8.1420 rather than divide.

`=> 8.1420`

`text(Let)\ \ A = text(annuity)`

`FV` `= A xx 8.1420`
`A` `= (FV)/8.1420`
  `= (407\ 100)/8.1420`
  `= $50\ 000`

 

iii.  `n=8\ \ \ (text(8 quarters in 2 years) )`

♦♦ Mean mark 31%
MARKER’S COMMENT: When questions asked for the interest paid on annuities, remember to subtract the total principal amounts contributed.

`r = text(4%)/4 =\ text{1%  per quarter}`

`:.\ text(Table factor) => 8.2857`

`FV` `=1000 xx 8.2857`
  `=8285.70`

 

`text(Interest)` `= FV (text(annuity) )\ – text(Principal)`
  `= 8285.70\ – (8 xx 1000)`
  `= 285.70`

 

`:.\ text(Interest earned is $285.70)`

Algebra, STD2 A4 2009 HSC 28c

The height above the ground, in metres, of a person’s eyes varies directly with the square of the distance, in kilometres, that the person can see to the horizon.

A person whose eyes are 1.6 m above the ground can see 4.5 km out to sea.

How high above the ground, in metres, would a person’s eyes need to be to see an island that is 15 km out to sea? Give your answer correct to one decimal place.   (3 marks)

Show Answers Only

 `17.8\ text(m)\ \ text{(to 1 d.p.)}`

Show Worked Solution
♦♦ Mean mark 22%
CRITICAL STEP: Reading the first line of the question carefully and establishing the relationship `h=k d^2` is the key part of solving this question.

`h prop d^2`

`h=kd^2`

`text(When)\ h = 1.6,\ d = 4.5`

`1.6` `= k xx 4.5^2`
`:. k` `= 1.6/4.5^2`
  `= 0.07901` `…`

 

`text(Find)\ h\ text(when)\ d = 15`

`h` `= 0.07901… xx 15^2`
  `= 17.777…`
  `= 17.8\ text(m)\ \ \ text{(to 1 d.p.)}`

Statistics, STD2 S4 2009 HSC 28b

The height and mass of a child are measured and recorded over its first two years. 

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Height (cm), } H \rule[-1ex]{0pt}{0pt} & \text{45} & \text{50} & \text{55} & \text{60} & \text{65} & \text{70} & \text{75} & \text{80} \\
\hline \rule{0pt}{2.5ex} \text{Mass (kg), } M \rule[-1ex]{0pt}{0pt} & \text{2.3} & \text{3.8} & \text{4.7} & \text{6.2} & \text{7.1} & \text{7.8} & \text{8.8} & \text{10.2} \\
\hline
\end{array}

This information is displayed in a scatter graph. 
 

  1. Describe the correlation between the height and mass of this child, as shown in the graph.   (1 mark)

  2. A line of best fit has been drawn on the graph.

     

    Find the equation of this line.   (2 marks)

Show Answers Only
  1. `text(The correlation between height and)`

     

    `text(mass is positive and strong.)`

  2. `M = 0.23H-8`
Show Worked Solution

i.  `text(The correlation between height and)`

♦ Mean mark 48%. 

`text(mass is positive and strong.)`

 

ii.  `text(Using)\ \ P_1(40, 1.2)\ \ text(and)\ \ P_2(80, 10.4)`

♦♦♦ Mean mark 18%. 
MARKER’S COMMENT: Many students had difficulty due to the fact the horizontal axis started at `H= text(40cm)` and not the origin.
`text(Gradient)` `= (y_2-y_1)/(x_2-x_1)`
  `= (10.4-1.2)/(80-40)`
  `= 9.2/40`
  `= 0.23`

 

`text(Line passes through)\ \ P_1(40, 1.2)`

`text(Using)\ \ \ y-y_1` `= m(x-x_1)`
`y-1.2` `= 0.23(x-40)`
`y-1.2` `= 0.23x-9.2`
`y` `= 0.23x-8`

 
`:. text(Equation of the line is)\ \ M = 0.23H-8`

Algebra, STD2 A4 2009 HSC 28a

Anjali is investigating stopping distances for a car travelling at different speeds. To model this she uses the equation
 

`d = 0.01s^2+ 0.7s`,
 

where `d` is the stopping distance in metres and `s` is the car’s speed in km/h.

The graph of this equation is drawn below.

2009 28a

  1. Anjali knows that only part of this curve applies to her model for stopping distances.

     

    In your writing booklet, using a set of axes, sketch the part of this curve that applies for stopping distances.  (1 mark)

  2. What is the difference between the stopping distances in a school zone when travelling at a speed of 40 km/h and when travelling at a speed of 70 km/h?   (2 marks)

Show Answers Only
  1.  
    2UG-2009-28a-Answer
  2. `54\ text(metres)`
Show Worked Solution
(i)

2UG-2009-28a-Answer

(ii) `text(When)\ \ s = 40`

♦ Mean mark 41%
COMMENT: Students could easily have used the graph for calculating `d` at `text(40 km/h)`, although the formula was required when the speed increased to `text(70 km/h)`.
`d` `= 0.01(40^2)+ 0.7 (40)`
  `= 16+28`
  `= 44\ text(m)`

 

`text(When)\ \ s = 70`

`d` `= 0.01 (70^2) +0.7(70)`
  `= 49 +49`
  `= 98\ text(m)`

 

`:.\ text(Difference)` `= 98\ -44`
  `= 54\ text(metres)`

Probability, STD2 S2 2009 HSC 27c

In each of three raffles, 100 tickets are sold and one prize is awarded.

Mary buys two tickets in one raffle. Jane buys one ticket in each of the other two raffles.

Determine who has the better chance of winning at least one prize. Justify your response using probability calculations.   (4 marks)  

Show Answers Only
`P(text(Mary wins) )` `= 2/100`
  `= 1/50`

 

`P(text(Jane wins at least 1) )` `= 1-P (text(loses both) )`
  `= 1-99/100 xx 99/100`
  `= 1-9801/(10\ 000)`
  `= 199/(10\ 000)`

 
`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

Show Worked Solution
`P(text(Mary wins) )` `= 2/100`
  `= 1/50`

 

`P(text(Jane wins at least 1) )` `= 1-P (text(loses both) )`
  `= 1-99/100 xx 99/100`
  `= 1-9801/(10\ 000)`
  `= 199/(10\ 000)`

 
`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Very few students calculated Jane’s chance of winning correctly. Note the use of “at least” in the question. Finding `1-P`(complement) is the best strategy here.

Measurement, STD2 M6 2009 HSC 27b

A yacht race follows the triangular course shown in the diagram. The course from  `P`  to  `Q`  is 1.8 km on a true bearing of 058°.

At  `Q`  the course changes direction. The course from  `Q`  to  `R`  is 2.7 km and  `/_PQR = 74^@`.
 

 2009-2UG-27b
 

  1. What is the bearing of  `R`  from  `Q`?   (1 mark)

  2. What is the distance from  `R`  to  `P`?     (2 marks)

  3. The area inside this triangular course is set as a ‘no-go’ zone for other boats while the race is on.

     

    What is the area of this ‘no-go’ zone?   (1 mark)

Show Answers Only
  1. `312^@`
  2. `2.8\ text(km)\ \ \ text{(1 d.p.)}`
  3. `2.3\ text(km²)\ \ \ text{(1 d.p.)}`
Show Worked Solution
i.    2UG-2009-27b-Answer

`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`

♦♦♦ Mean mark 18%.
TIP: Draw North-South parallel lines through relevant points to help calculate angles as shown in the Worked Solutions.

`text(Bearing of)\ R\ text(from)\ Q`

`= 180^@ + 58^@ + 74^@`
`= 312^@`

 

ii.   `text(Using Cosine rule:)`

♦ Mean mark 36%
`RP^2` `=RQ^2` + `PQ^2` `- 2` `xx RQ` `xx PQ` `xx cos` `/_RQP`
  `= 2.7^2` + `1.8^2` `- 2` `xx 2.7` `xx 1.8` `xx cos74^@`
  `=7.29 + 3.24\ – 2.679…`
  `=7.851…`
`:.RP` `= sqrt(7.851…)`
  `=2.8019…`
  `~~ 2.8\ text(km)  (text(1 d.p.) )`

 

iii.   `text(Using)\ \ A = 1/2 ab sinC`

♦ Mean mark 44%
`A` `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@`
  `= 2.3358…`
  `= 2.3\ text(km²)`

 

`:.\ text(No-go zone is 2.3 km²)`

Measurement, STD2 M2 2009 HSC 26b

John lives in Denver and wants to ring a friend in Osaka.

  1. In Denver it is 9 pm Monday. Given Osaka has a UTC of +9 and Denver has a UTC of –7, what time and day is it in Osaka then?   (1 mark) 

  2. John’s friend in Osaka sent him a text message which happened to take 14 hours to reach him. It was sent at 10 am Thursday, Osaka time.

     

    What was the time and day in Denver when John received the text?    (2 marks)

Show Answers Only
  1. `1\ text(pm Tuesday)`
  2. `8\ text(am Thursday)`
Show Worked Solution
i.   `text{Denver is behind Osaka time by 16 hours.}`

 

`:.\ text(Time in Osaka)` `= 9\ text(pm Monday plus 16 hours)`
  `= 1\ text(pm Tuesday)`

 

♦ Mean mark 34%
ii.   `text(Denver is 16 hours behind Osaka)`
   `:.\ text(John will receive the text at 10 am Thursday)`
   `text(less 16 hours plus 14 hours.)`
   `text{(i.e. 8 am Thursday.)}`

Statistics, STD2 S1 2009 HSC 26a

In a school, boys and girls were surveyed about the time they usually spend on the internet over a weekend. These results were displayed in box-and-whisker plots, as shown below. 
 

2UG-2009-26a

  1. Find the interquartile range for boys.   (1 mark)

  2. What percentage of girls usually spend 5 or less hours on the internet over a weekend?  (1 mark)

  3. Jenny said that the graph shows that the same number of boys as girls usually spend between 5 and 6 hours on the internet over a weekend.

     

    Under what circumstances would this statement be true?    (1 mark)

Show Answers Only
  1. `4`
  2. `text(75% of girls spend 5 hours or less)`
  3. `text(5-6 hours for girls accounts for 25% of all girls.)`
  4. `text(5-6 hours for boys accounts for 25% of all boys,)`
  5. `text(as median to upper quartile is 25%.)`
     
  6. `=>\ text(This will be the same number only if the number of)`
  7. `text(all girls surveyed equals the number of boys surveyed.)`
Show Worked Solution
i.    `text(Interquartile range)` `= 6` `- 2`
    `= 4`

 

♦♦ Mean mark part ii: 31%
ii.    `text(Upper quartile = 5`
  `:.\ text(75% of girls spend 5 or less hours)`

 

♦♦♦ Mean mark part iii: 9%
iii.    `text(5-6 hours for girls accounts for 25% of all girls.)`
  `text(5-6 hours for boys accounts for 25% of all boys,)`
  `text{(median to the upper quartile represents 25%.)}`
  `=>\ text(This will only be the same number if the number of)`
  `text(all girls surveyed equals the number of boys surveyed.)`

Statistics, STD2 S5 2009 HSC 25d

In Broken Hill, the maximum temperature for each day has been recorded. The mean of these maximum temperatures during spring is 25.8°C, and their standard deviation is 4.2° C. 

  1. What temperature has a `z`-score of  –1?    (1 mark)

  2. What percentage of spring days in Broken Hill would have maximum temperatures between 21.6° C and 38.4°C?

     

    You may assume that these maximum temperatures are normally distributed and that

  3.  

    • 68% of maximum temperatures have `z`-scores between –1 and 1
    • 95% of maximum temperatures have `z`-scores between –2 and 2
    • 99.7% of maximum temperatures have `z`-scores between –3 and 3.   (3 marks)

Show Answers Only
  1. `21.6^@`
  2. `text(83.85%)`
Show Worked Solution

i.   `mu = 25.8\ \ \ sigma = 4.2`

`text(Using)\ \ \ \ z` `= (x-mu)/sigma`
`-1` `= (x-25.8)/4.2`
`x` `- 25.8` `= -4.2`
`x` `= 21.6°`

 

`:.\ 21.6^@\ text(has a)\ z text(-score of)  -1` 

 

ii.    `z text(-score of)\ 21.6 = -1`

♦ Mean mark 41%
MARKER’S COMMENT: Many students failed to use the answer to (d)(i), costing them valuable exam time.

`text(Find)\ z text(-score of 38.4)`

`z\ (38.4)` `= (38.4\ – 25.8)/4.2=3`

 

`text(68% of scores are between)\ z= –1\ text(and 1)`

`=>\ text(34%)\ text(are between)\ z=–1\ text(and 0)`

`text(99.7% of scores are between)\ z= –3\ text(and 3)`

`=>\ text(49.85%)\ text(are between)\ z=0\ text(and 3)`

 

`:.\ text(% Temps between 21.6° and 38.4°)`

`=\ text(34% + 49.85%)`
`=\ text(83.85%)`

 

 

Measurement, 2UG 2009 HSC 25c

There is a lake inside the rectangular grass picnic area `ABCD`, as shown in the diagram.
 

2UG-2009-25c
 

  1. Use Simpson’s Rule to find the approximate area of the lake’s surface.   (3 marks)
  2. The lake is 60 cm deep. Bozo the clown thinks he can empty the lake using a four-litre bucket.
  3. How many times would he have to fill his bucket from the lake in order to empty the lake?  (Note that 1 m³ = 1000 L)`.    (2 marks)

 

 

Show Answers Only
  1. `508\ text(m²)`
  2. `text(76 200 times)`
Show Worked Solution
(i)     `text(Area of lake = Area of rectangle)\ – text(Area of grass)`
`text(Area of rectangle)` `= 24 xx 55`
  `= 1320\ text(m²)`

 

`text(Area of grass)` `~~ h/3 [y_0 + 4y_1 + y_2]\ \ \ text(… applied twice)`
  `~~ 12/3 [20 + 4 xx 5 + 10] + 12/3 [35 + 4 xx 22 + 30]`
  `~~ 4[50] + 4[153]`
  `~~ 200 + 612`
  `~~ 812\ text(m²)`

 

`:.\ text(Area of lake)` `~~ 1320\ – 812`
  `~~ 508\ text(m²)`

 

♦ Mean mark 44%
STRATEGY: Most students who did calculations in cm² and cm³ made errors. Keeping calculations in metres is much easier here.
(ii)    `V` `= Ah`
    `= 508 xx 0.6`
    `= 304.8\ text(m³)`
    `= 304\ 800\ text(L)\ \ \ text{(since 1 m³ = 1000  L)}`

 

`text(# Times to fill bucket)` `= (304\ 800)/4`
  `= 76\ 200`

 

`:.\ text(Bozo would have to fill his bucket)`

`text(76 200 times to empty the lake.)`

Statistics, STD2 S5 2013 HSC 29b

Ali’s class sits two Geography tests. The results of her class on the first Geography test are shown.

`58,\ \ 74,\ \ 65,\ \ 66,\ \ 73,\ \ 71,\ \ 72,\ \ 74,\ \ 62,\ \ 70`

The mean was 68.5 for the first test. 

  1. Calculate the standard deviation for the first test. Give your answer correct to one decimal place.    (1 mark)

  2. On the second Geography test, the mean for the class was 74.4 and the standard deviation was 12.4.

     

    Ali scored 62 on the first test. Calculate the mark that she needed to obtain in the second test to ensure that her performance relative to the class was maintained.   (3 marks)

Show Answers Only
  1. `5.2\ \ \ text{(to 1 d.p.)}`
  2. `text(Ali needs to score 58.9)`
Show Worked Solution
♦ Mean mark 39%
COMMENT: Make sure you are confident with this function on your calculator!
i. `sigma` `=5.2201…`
    `=5.2`  `text{(to 1 d.p.)}`

 

ii. `z =(x-mu)/sigma`
`z text{-score (1st test)}` `= (62-68.5)/5.2`
  `=-1.25`

 
`text(2nd test has)\ z text(-score of)\-1.25 :`

♦♦ Mean mark 21%.
MARKER’S COMMENT: When “performance relative to the class is maintained”, `z text(-scores)` are the same in each test.
`-1.25` `= (x-74.4)/12.4`
`x-74.4` `=-15.5`
`x` `=58.9`

 

`:.\ text(Ali needs to score 58.9)`

Algebra, STD2 A1 2013 HSC 29a

Sarah tried to solve this equation and made a mistake in Line 2. 

`(W+4)/3-(2W-1)/5` `=1` `text(... Line 1)`
`5W+ 20-6W-3` `=15` `text(... Line 2)`
`17-W` `=15` `text(... Line 3)`
`W` `=2` `text(... Line 4)`

 
Copy the equation in Line 1 and continue your solution to solve this equation for `W`.

Show all lines of working.   (2 marks)

Show Answers Only
`(W+4)/3-(2W-1)/5` `=1` `text(… Line 1)`
`5W+ 20-6W+ 3` `=15` `text(… Line 2)`
`23-W` `=15` `text(… Line 3)`
`W` `=8` `text(… Line 4)`
Show Worked Solution
♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 15 (from Line 1 to Line 2), indicating both sides must have been multiplied by 15.
`(W+4)/3-(2W-1)/5` `=1` `text(… Line 1)`
`5W+ 20-6W+3` `=15` `text(… Line 2)`
`23-W` `=15` `text(… Line 2)`
`W` `=8` `text(… Line 4)`

Measurement, 2UG 2013 HSC 28c

A ship sails due South from Channel-Port-aux-Basques, Canada,  `47^@ text(N)\ 59^@ text(W)`  to Barbados, `13^@ text(N)\ 59^@ text(W)`.

How far did the ship sail, to the nearest kilometre? Assume that the radius of Earth is 6400 km.  (2 marks)

Show Answers Only

 `3798\ text{km (nearest km)}`

Show Worked Solution
♦ Mean mark 50%
`text(Angular difference in latitude)` `=47-13`
  `=34^@`
`text{(No difference in longitude)}`
`text(Distance)` `=34/360` `xx 2 pi r`
  `=34/360` `xx 2` `xx pi` `xx 6400`
  `= 3797.836…`
  `= 3798` `text{km   (nearest km)}`

Statistics, STD2 S4 2013 HSC 28b

Ahmed collected data on the age (`a`) and height (`h`) of males aged 11 to 16 years.

He created a scatterplot of the data and constructed a line of best fit to model the relationship between the age and height of males.
 

  1. Determine the gradient of the line of best fit shown on the graph.   (1 mark)

  2. Explain the meaning of the gradient in the context of the data.   (1 mark)

  3. Determine the equation of the line of best fit shown on the graph.  (2 marks)

  4. Use the line of best fit to predict the height of a typical 17-year-old male.   (1 mark)

  5. Why would this model not be useful for predicting the height of a typical 45-year-old male?   (1 mark)

Show Answers Only
  1. `text(Gradient = 6)`
  2. `text(Males should grow 6 cm per)`

     

    `text(year between the ages 11-16.)`

  3. `h = 6a + 80`
  4. `text(182 cm)`
  5. `text(People slow and eventually stop growing)`

  6.  

    `text(after they become adults.)`

Show Worked Solution

i.    `text{Gradient}\ =(176-146)/(16-11)=30/5=6`
 

ii.   `text{Males should grow 6cm per year between the}`

`text{ages 11–16.}`
 

♦♦ Mean marks of 38%, 26% and 25% respectively for parts (i)-(iii).
MARKER’S COMMENT: Interpreting gradients has been consistently examined in recent history and almost always poorly answered. 

iii.   `text{Gradient = 6,  Passes through (11, 146)}`

`y-y_1` `=m(x-x_1)`
`h-146` `=6(a-11)`
`:. h` `=6a-66+146`
  `=6a + 80`

 

iv.   `text{Substitue}\ \ a=17\ \ \text{into equation from part (iii):}`

`h=(6 xx 17) +80=182`

`:.\ text{A typical 17 year old is expected to be 182cm.}`
  

v.    `text(People slow and eventually stop growing)`
  `text(after they become adults.)`

Measurement, STD2 M6 2013 HSC 28a

A compass radial survey of the field `ABCD` has been conducted from `O`.
 

2013 28a
 

Find the area of the section `ABO`, to the nearest square metre.   (2 marks)

 

Show Answers Only

 `2127\ text(m²)`

Show Worked Solution
♦ Mean mark 50%
`/_` `AOB` `=21^@ + (360` `-310)`
  `=71^@`

`text(Using Area) = 1/2 ab sinC`

`text(Area)` `=1/2` `xx 60` `xx 75` `xx sin71^@`
  `=2127.4167…`
  `=2127` `text(m²)\ \ \ text{(nearest m²)}`

Measurement, STD2 M1 2013 HSC 27d

A rectangular wooden chopping board is advertised as being 17 cm by 25 cm, with each side measured to the nearest centimetre.

  1. Calculate the percentage error in the measurement of the longer side.   (1 mark)

  2. Between what lower and upper limits does the actual area of the top of the chopping board lie?     (2 marks)

Show Answers Only
  1. `text(2%)`
  2. `404.25\ text{cm}^2 and 446.25\ text{cm}^2`
Show Worked Solution

i.    `text(Longer side) = 25\ text(cm)`

♦♦ Mean mark 23%
MARKER’S COMMENT: Be aware that measurements accurate to the nearest cm have an absolute error for calculation purposes of 0.5 cm.

`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 1 = 0.5\ text{cm}`

`text{% error}` `=\ frac{text{absolute error}}{text{measurement}} xx 100%`  
  `=0.5/25 xx 100%`  
  `=2%`  

 

ii.   `text(Area) = l xx b`

♦ Mean mark 35%
`text{Area (upper)}` `=25.5 xx 17.5`
  `=446.25\ text{cm}^2`

 

`text{Area (lower)}` `=24.5 xx 16.5`
  `=404.25\ text{cm}^2`

 
`:.\ text{Area is between 404.25 cm}^2\ text{and 446.25 cm}^2.`

Financial Maths, STD2 F1 2013 HSC 27b

The table shows the tax payable to the Australian Taxation Office for different taxable incomes.

2013 27b

Peta has a gross annual salary of  $84 000. She has tax deductions of $1000 for work-related travel and $500 for stationery. The Medicare levy that she pays is calculated at 1.5% of her taxable income.

Peta has already paid $18 500 in tax.

Will Peta receive a tax refund or will she owe money to the Australian Taxation Office? Justify your answer by calculating the refund or amount owed.   (4 marks)

Show Answers Only

 `text(Peta owes the tax office $1209.50.)`

Show Worked Solution
`text(Total Deductions)` `=1000+500`
  `=$1500`
`text(Taxable Income)` `= text(Gross Income)-text(Total Deductions)`
  `= 84\ 000-1500`
  `= $82\ 500`

 

`text(Using the tax table:)`

♦ Mean mark 44%
IMPORTANT: Note that ‘Tax’ and the ‘Medicare Levy’ are calculated separately using the ‘Taxable Income’ figure and added together to find the amount owed to the ATO.
`text(Tax)` `= 17\ 547+ 0.37 xx (82\ 500- 80\ 000)`
  `= 17\ 547 +925`
  `= $18\ 472`

 

`text(Medicare owing)` `=\ text(1.5%) xx 82\ 500`
  `= $1237.50`

 

`text(Owed to ATO)` `= 18\ 472+1237.50`
  `=$19\ 709.50`
   
`text(Tax paid)` `= $18\ 500`

 

`text(Difference owing)` `=19\ 709.50-18\ 500`
  `= $1209.50`

 
`:.\ text(Peta owes the tax office $1209.50.`

Statistics, STD2 S1 2013 HSC 26f

Jason travels to work by car on all five days of his working week, leaving home at 7 am each day. He compares his travel times using roads without tolls and roads with tolls over a period of 12 working weeks.

He records his travel times (in minutes) in a back-to-back stem-and-leaf plot.
 

2013 26f
 

  1. What is the modal travel time when he uses roads without tolls?  (1 mark)

  2. What is the median travel time when he uses roads without tolls?   (1 mark)

  3. Describe how the two data sets differ in terms of the spread and skewness of their distributions.   (2 marks)

Show Answers Only
  1. `52\ text(minutes)`
  2. `50.5\ text(minutes)`
  3. `text(Spread)`
  4. `text{Times without tolls have a tighter spread (range = 22)}`
  5. `text{than times with tolls (range = 55).}`

  6.  

    `text(Skewness)`

  7. `text(Times without tolls shows virtually no skewness while`
  8. `text(times with tolls are positively skewed.)`
Show Worked Solution

i.  `text(Modal time) = 52\ text(minutes)`

♦ Mean mark 36%
MARKER’S COMMENT: Finding a median proved challenging for many students. Take note!

 

ii.  `text(30 times with no tolls)`

`text(Median)` `=\ text(Average of 15th and 16th)`
  `=(50 + 51)/2`
  `= 50.5\ text(minutes)`

 

♦ Mean mark 39%

 

 iii.  `text(Spread)`

`text{Times without tolls have a much tighter}`

`text{spread (range = 22) than times with tolls}`

`text{(range = 55).}`

`text(Skewness)`

`text(Times without tolls shows virtually no skewness)`

`text(while times with tolls are positively skewed.)`

Financial Maths, STD2 F4 2013 HSC 26e

Kimberley has invested $3500.  

Interest is compounded half-yearly at a rate of 2% per half-year.

2013 26e

Use the table to calculate the value of her investment at the end of 4 years.  (2 marks)

Show Answers Only

 `$4102`

Show Worked Solution
♦ Mean mark 44%
COMMENT: Structure your answer: 1-Find the interest rate per compounding period (same in this case). 2-Find the number of compounding periods.

`r =\ text(2% per half-year)`

`n = 8 \ \ \ \  text{(8 half-years in 4 years)}`

`=>\ text(Table Factor = 1.172)`

`text(Investment)` `= 3500` ` xx 1.172`
  `= $4102`

 

`:.\ text(After 4 years, investment value is $4102)`

Measurement, STD2 M1 2013 HSC 26d

A section of Jim’s electricity bill is shown.

2013 26d

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

  2. How much will Jim save if he uses 154 kWh of energy at the Off-peak rate rather than at the Peak rate?    (2 marks)

Show Answers Only
  1. `1084.4`
  2. `$ 58.78\ \ \ (text(nearest cent) )`
Show Worked Solution
i.    `A` `=\ text(Last reading + Energy used)`
    `= 560.9 + 523.5`
    `= 1084.4`
♦ Mean mark part (i) 43% and part  (ii) 46%.

 

ii.    `text(C)text(ost at off-peak)` `= 154 xx 9.6`
    `= 1478.4\ text(cents)`
`text(C)text(ost at peak)` `=154 xx 47.77`
  `= 7356.58\ text(cents)`

 

`:.\ text(Saving)` `=7356.58` `-1478.4`
  `=5878.18`
  `= $58.78`    `text{(nearest cent)}`

Probability, STD2 S2 2013 HSC 26c

The probability that Michael will score more than 100 points in a game of bowling is `31/40`. 

  1. A commentator states that the probability that Michael will score less than 100 points in a game of bowling is  `9/40`.

     

    Is the commentator correct? Give a reason for your answer.   (1 mark)

  2. Michael plays two games of bowling. What is the probability that he scores more than 100 points in the first game and then again in the second game?   (1 mark)

Show Answers Only
  1. `text{Incorrect. Less than “or equal to 100” is correct.}`
  2. `961/1600`
Show Worked Solution
♦♦♦ Mean mark 11%

i.   `text(The commentator is incorrect. The correct)`

`text(statement is)\ Ptext{(score} <=100 text{)} =9/40`

`text{(i.e. less than “or equal to 100” is the correct statement)}`

 

♦ Mean mark 34%
ii. `\ \ \ P(text{score >100 in both})` `= 31/40 xx 31/40` 
    `= 961/1600`

Measurement, STD2 M6 2010 HSC 24d

The base of a lighthouse, `D`, is at the top of a cliff 168 metres above sea level. The angle of depression from `D` to a boat at `C` is 28°. The boat heads towards the base of the cliff, `A`, and stops at `B`. The distance `AB` is 126 metres.
 

  1. What is the angle of depression from `D` to `B`, correct to the nearest degree?   (3 marks)

  2. How far did the boat travel from `C` to `B`, correct to the nearest metre?   (2 marks)

Show Answers Only
  1. `53^circ`
  2. `190\ text(m)`
Show Worked Solution
♦♦ Mean mark 31%
i.    `tan/_ADB` `=126/168`
  ` /_ADB` `=36.8698…`
    `=36.9^circ\ \ \ \ text{(to 1 d.p)}` 

 

`/_text(Depression)\ D\ text(to)\ B` `=90-36.9`
  `=53.1`
  `=53^circ\ text{(nearest degree)}`

 

ii.     `text(Find)\ CB:`

♦♦ Mean mark 31%
MARKER’S COMMENT: Solve efficiently by using right-angled trigonometry. Many students used non-right angled trig, adding to the calculations and the difficulty.
`/_ADC+28` `=90`
 `/_ADC` `=62^circ`
`tan 62^circ` `=(AC)/168`
`AC` `=168xxtan 62^circ`
  `=315.962…`

 

`CB` `=AC-AB`
  `=315.962…-126`
  `=189.962…`
  `=190\ text(m (nearest m))`

Statistics, STD2 S5 2010 HSC 24c

The marks in a class test are normally distributed. The mean is 100 and the standard deviation is 10.

  1. Jason's mark is 115. What is his  `z`-score?  (1 mark)

  2. Mary has a `z`-score of 0. What mark did she achieve in the test?   (1 mark)

  3. What percentage of marks lie between 80 and 110?

     

    You may assume the following:

     

    • 68% of marks have a `z`-score between –1 and 1

     

    • 95% of marks have a `z`-score between  –2 and 2

     

    • 99.7% of marks have a `z`-score between –3 and 3.   (2 marks) 

Show Answers Only
  1. `1.5`
  2. `100`
  3. `81.5%`
Show Worked Solution

i.     ` text(Given) \ \ mu=100,\ \ sigma=10`

MARKER’S COMMENT: Too may students had calculator errors in this question, giving away easy marks. BE CAREFUL!

`text(If mark is 115,`

`ztext(-score)` `=(115-mu)/sigma`
  `=(115-100)/100`
  `=1.5`

 

ii.     `z text(-score = 0 when mark equals the mean)`

`:.\ text(Mary’s score was)\ 100`

 

♦ Mean mark 42%
iii.   `ztext(-score of)\ 110` `=(110-100)/10=1`
`ztext(-score of)\ 80` `=(80-100)/10=–2`

 

`text(68% of marks lie between)\ z= –1 \ text(and)\  1`

 `=>text(34%  lie between)\ z= 0\ text(and)\ 1`

`text(95%  of marks lie between)\ z= –2 \ text(and)\  2`

 `=> text(47.5%  lie between)\ z= –2\ text(and)\ 0`

 

`:.\ text(% marks between 80 and 110`

`=\ text(34% + 47.5%)`

`=\ text(81.5%)`

Algebra, STD2 A4 2010 HSC 24b

Ashley makes picture frames as part of her business. To calculate the cost,  `C`, in dollars, of making  `x`  frames, she uses the equation  `C=40+10x`.

She sells the frames for $20 each and determines her income,  `I`, in dollars, using the equation  `I=20x`.
 

Use the graph to solve the two equations simultaneously for  `x`  and explain the significance of this solution for Ashley's business.   (2 marks)

Show Answers Only

`x=4`

Show Worked Solution

`text(From the graph, intersection occurs at)\ x=4`

♦ Mean mark 36%.
MARKER’S COMMENT: The intersection on the graph is the same point at which the two simultaneous equations are solved for the given value of `x`.

`=>\ text(Break-even point occurs at)\ x=4`

`text(i.e. when 4 frames sold)`

`text(Income)` `=20xx4=$80\ \ \ text(is equal to)`
`text(C)text(osts)` `=40+(10xx4)=$80`

 

`text(If)\ <4\ text(frames sold)=>\ text(LOSS for business)`

`text(If)\ >4\ text(frames sold)=>\ text(PROFIT)`

Probability, STD2 S2 2010 HSC 20 MC

Lou and Ali are on a fitness program for one month. The probability that Lou will finish the program successfully is 0.7 while the probability that Ali will finish successfully is 0.6. The probability tree shows this information

 

What is the probability that only one of them will be successful ?

  1. `0.18`
  2. `0.28`
  3. `0.42`
  4. `0.46`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ Ptext{(Lou successful)}=P(L) = 0.7, \ P(\text{not}\ L) = 0.3`

`text(Let)\ \ Ptext{(Ali successful)}=P(A) = 0.6, \ P(\text{not}\ A) = 0.4`

`P text{(only 1 successful)}` `=P(L)xxP(text(not)\ A)+P(text(not)\ L)xxP(A)`
  `=(0.7xx0.4)+(0.3xx0.6)`
  `=0.28+0.18`
  `=0.46`

 
`=>  D`

♦ Mean mark 48%.