Band 5 – Page 58

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.

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

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On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data. (1 mark)
Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer. (1 mark)
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)

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Show Answers Only

`15`
`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)`

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

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

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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

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

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Show Answers Only

`6\ 760\ 000`
`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.

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)

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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)

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Show Answers Only

`text(0.125 L)`
`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.

Complete the tree diagram below. (2 marks)

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

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Show Answers Only

`6/11`

Show Worked Solution

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.

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.

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

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

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

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

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)

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

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

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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.

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)

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`7\ 780\ 340\ text(watts)`
`8000w^3`
`8.8\ text(million watts, or 8 748 000 watts)`
`w = 8.2\ text(m/s)\ \ \ text{(1 d.p.)}`
`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)

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`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.

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}

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

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

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

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$10$
$\dfrac{1}{9}$
$5$

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}

Draw Shape 4 of this pattern. (1 mark)

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

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

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

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Show Answers Only

`text(See Worked Solutions.)`
`302`
`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.

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

$6364.32
$6366.11
$6369.67
$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`

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

`$15\ 000`
`$20\ 000`
`$30\ 000`
`$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.

What was Simon’s total pay for that week?

$1189.28
$1296.30
$1334.55
$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°.

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

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

Show Answers Only

`055^@`
`13\ text(km)`
`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)

`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.

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)

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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)

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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)

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Show Answers Only

`$32\ 342`
`$50\ 000`
`$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)

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`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.

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

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

Find the equation of this line. (2 marks)

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Show Answers Only

`text(The correlation between height and)`

`text(mass is positive and strong.)`
`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.

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)
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`54\ text(metres)`

Show Worked Solution

(i)

(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^@`.

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

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

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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)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`312^@`
`2.8\ text(km)\ \ \ text{(1 d.p.)}`
`2.3\ text(km²)\ \ \ text{(1 d.p.)}`

Show Worked Solution

`/_ 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.

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)

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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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`1\ text(pm Tuesday)`
`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.

Find the interquartile range for boys. (1 mark)

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

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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)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`4`
`text(75% of girls spend 5 hours or less)`
`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(as median to upper quartile is 25%.)`
`=>\ text(This will be the same number only if the number of)`
`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.

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

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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
• 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)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`21.6^@`
`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.

Use Simpson’s Rule to find the approximate area of the lake’s surface. (3 marks)
The lake is 60 cm deep. Bozo the clown thinks he can empty the lake using a four-litre bucket.
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

`508\ text(m²)`
`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.

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

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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)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`5.2\ \ \ text{(to 1 d.p.)}`
`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)

--- 5 WORK AREA LINES (style=lined) ---

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.

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

--- 2 WORK AREA LINES (style=lined) ---
Explain the meaning of the gradient in the context of the data. (1 mark)

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

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

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

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Gradient = 6)`
`text(Males should grow 6 cm per)`

`text(year between the ages 11-16.)`
`h = 6a + 80`
`text(182 cm)`
`text(People slow and eventually stop growing)`
`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`.

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

--- 4 WORK AREA LINES (style=lined) ---

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.

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

--- 2 WORK AREA LINES (style=lined) ---
Between what lower and upper limits does the actual area of the top of the chopping board lie? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(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.

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)

--- 8 WORK AREA LINES (style=lined) ---

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.

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

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

--- 1 WORK AREA LINES (style=lined) ---
Describe how the two data sets differ in terms of the spread and skewness of their distributions. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`52\ text(minutes)`
`50.5\ text(minutes)`
`text(Spread)`
`text{Times without tolls have a tighter spread (range = 22)}`
`text{than times with tolls (range = 55).}`
`text(Skewness)`
`text(Times without tolls shows virtually no skewness while`
`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.

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

--- 4 WORK AREA LINES (style=lined) ---

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.

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

--- 1 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`1084.4`
`$ 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`.

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 WORK AREA LINES (style=lined) ---
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)

--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{Incorrect. Less than “or equal to 100” is correct.}`
`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.

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

--- 6 WORK AREA LINES (style=lined) ---
How far did the boat travel from `C` to `B`, correct to the nearest metre? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`53^circ`
`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.

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

--- 1 WORK AREA LINES (style=lined) ---
Mary has a `z`-score of 0. What mark did she achieve in the test? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`1.5`
`100`
`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)

--- 4 WORK AREA LINES (style=lined) ---

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?

`0.18`
`0.28`
`0.42`
`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%.

Statistics, STD2 S1 2010 HSC 16 MC

This back-to-back stem-and-leaf plot displays the test results for a class of 26 students.

What is the median test result for the class?

`44`
`46`
`48`
`49`

Show Answers Only

`B`

Show Worked Solution

♦♦ Mean mark 35%

`text(26 results given in the data)`

`=>text(Median is average of)\ 13^text(th)\ text(and)\ 14^text(th)`

`:.\ text(Median)`	`=(45+47)/2`
	`=46`

`=>B`

Financial Maths, STD2 F4 2009 HSC 24e

Jay bought a computer for $3600. His friend Julie said that all computers are worth nothing (i.e. the value is $0) after 3 years.

Find the amount that the computer would depreciate each year to be worth nothing after 3 years, if the straight line method of depreciation is used. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Explain why the computer would never be worth nothing if the declining balance method of depreciation is used, with 30% per annum rate of depreciation. Use suitable calculations to support your answer. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`$1200`
`text(See Worked Solutions.)`

Show Worked Solution

i.	`S`	`= V_0-Dn`
	`0`	`= 3600-D xx 3`
	`3D`	`= 3600`
	`D`	`= 3600/3`
		`= 1200`

`:.\ text(Annual depreciation = $1200`

♦ Mean mark 45%

ii	`text(Using)\ \ S = V_0 (1-r)^n`
	`text(where)\ r = text(30%)\ \ text(and)\ \ V_0 = 3600`

`S`	`=3600 (1-30/100)^n`
	`= 3600 (0.7)^n`

`(0.7)^n > 0\ text(for all)\ n`

`:.\ text(Salvage value is always)\ >0`

Algebra, STD2 A2 2009 HSC 24d

A factory makes boots and sandals. In any week

• the total number of pairs of boots and sandals that are made is 200
• the maximum number of pairs of boots made is 120
• the maximum number of pairs of sandals made is 150.

The factory manager has drawn a graph to show the numbers of pairs of boots (`x`) and sandals (`y`) that can be made.

Find the equation of the line `AD`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Explain why this line is only relevant between `B` and `C` for this factory. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
The profit per week, `$P`, can be found by using the equation `P = 24x + 15y`.

Compare the profits at `B` and `C`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`x + y = 200`
`text(S)text(ince the max amount of boots = 120)`

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`
`text(The profits at)\ C\ text(are $630 more than at)\ B.`

Show Worked Solution

i. `text{We are told the number of boots}\ (x),`

♦♦♦ Mean mark part (i) 14%.
Using `y=mx+b` is a less efficient but equally valid method, using `m=–1` and `b=200` (`y`-intercept).

`text{and shoes}\ (y),\ text(made in any week = 200)`

`=>text(Equation of)\ AD\ text(is)\ \ x + y = 200`

ii. `text(S)text(ince the max amount of boots = 120)`

♦ Mean mark 49%

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`

iii. `text(At)\ B,\ \ x = 50,\ y = 150`

♦ Mean mark 40%.

`=>$P (text(at)\ B)`	`= 24 xx 50 + 15 xx 150`
	`= 1200 + 2250`
	` = $3450`

`text(At)\ C,\ \ x = 120 text(,)\ y = 80`

`=> $P (text(at)\ C)`	`= 24 xx 120 + 15 xx 80`
	`= 2880 + 1200`
	`= $4080`

`:.\ text(The profits at)\ C\ text(are $630 more than at)\ B.`

Probability, STD2 S2 2009 HSC 23b

A personal identification number (PIN) is made up of four digits. An example of a PIN is

When all ten digits are available for use, how many different PINs are possible? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Rhys has forgotten his four-digit PIN, but knows that the first digit is either 5 or 6.
What is the probability that Rhys will correctly guess his PIN in one attempt? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`10\ 000`
`1/2000`

Show Worked Solution

♦ Mean mark 43%

i.	`#\ text(Combinations)`	`= 10 xx 10 xx 10 xx 10`
		`= 10\ 000`

♦♦♦ Mean mark 18%
MARKER’S COMMENT: A common error is finding the number of possible combinations but not then calculating the probability.

ii.	`#\ text(Combinations)`	`= 2 xx 10 xx 10 xx 10`
		`= 2000`

	`P text{(Correct PIN)}`	`= text{# Correct PINS}/text(# Combinations)`
		`=1/2000`

Measurement, STD2 M6 2009 HSC 23a

The point `A` is 25 m from the base of a building. The angle of elevation from `A` to the top of the building is 38°.

Show that the height of the building is approximately 19.5 m. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
A car is parked 62 m from the base of the building.

What is the angle of depression from the top of the building to the car?

Give your answer to the nearest minute. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

i. `text{Proof (See Worked Solutions)}`

ii. `17°28^{′}`

Show Worked Solution

i. `text(Need to prove height (h) ) ~~ 19.5\ text(m)`

`tan 38^@`	`= h/25`
`h`	`= 25 xx tan38^@`
	`= 19.5321…`
	`~~ 19.5\ text(m)\ \ text(… as required.)`

ii.

`text(Let)\ \ /_ \ text(Elevation (from car) ) = theta`

♦♦ Mean mark 33%
MARKER’S COMMENT: If >30 “seconds”, round to the higher “minute”.

`tan theta`	`= h/62`
	`= 19.5/62`
	`= 0.3145…`
`:. theta`	`= 17.459…`
	`= 17°27^{′}33^{″}..`
	`=17°28^{′}\ \ text{(nearest minute)}`

`:./_ \ text(Depression to car) =17°28^{′}\ \ text{(alternate to}\ theta text{)}`

Measurement, STD2 M6 2009 HSC 22 MC

In the diagram, `AD` and `DC` are equal to 30 cm.

What is the length of `AB` to the nearest centimetre?

`28\ text(cm)`
`31\ text(cm)`
`34\ text(cm)`
`39\ text(cm)`

Show Answers Only

`A`

Show Worked Solution

♦ Mean mark of 35%

`Delta ADC\ text(is isosceles)`

`/_DAB = /_DCA`	`= x^@`
`2x + 80^@`	`= 180^@\ \ \ (text{Angle sum of}\ DeltaADC)`
`2x`	`= 100^@`
`x`	`= 50^@`

`/_ DBA`	`= 180\-(50 + 60)\ \ \ (text{Angle sum of}\ Delta ADB)`
	`= 70^@`

`text(Using sine rule:)`

`(AB)/sin60`	`= 30/sin70`
`AB`	`= (30 xx sin60)/sin70`
	`= 27.648…\ text(cm)`

`=> A`

Financial Maths, STD2 F1 2009 HSC 20 MC

Lou bought a plasma TV which was priced at $3499. He paid $1000 deposit and got a loan for the balance that was paid off by 24 monthly instalments of $135.36.

What simple interest rate per annum, to the nearest percent, was charged on his loan?

Show Answers Only

`B`

Show Worked Solution

♦♦ Mean mark 34%
COMMENT: A multi-step question targeting higher bands that can be a time-trap for many students.

`text(Loan)`	`=\ text(Price – deposit)`
	`= 3499-1000`
	`= $2499`

`text(Total repaid)`	`= 24 xx 135.36`
	`= $3248.64`

`:.\ text(Interest paid)`	`= 3248.64\-2499`
	`= $749.64`

`text(Simple Interest)`	`= Prn`
`749.64`	`= 2499 xx r xx 2`
`:. r`	`= 749.64/(2 xx 2499)`
	`= 0.1499…`
	`~~15 text(%)`

`=> B`

Algebra, STD2 A1 2009 HSC 16 MC

The time for a car to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?

Show Answers Only

`A`

Show Worked Solution

`T`	`prop 1/S`
`T`	`= k/S`

`text{By elimination:}`

`text(As S) uarr text(, T) darr => text(cannot be B or D)`

♦ Mean mark 38%

`text(C is incorrect because it graphs a linear relationship)`

`=> A`

Algebra, STD2 A2 2009 HSC 14 MC

If `A = 6x + 10`, and `x` is increased by 2, what will be the corresponding increase in `A` ?

`2x`
`6x`
`2`
`12`

Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 50%.
STRATEGY: Substituting real numbers into the equation can work well in these type of questions. eg. If `x=0,\ A=10` and when `x=2,\ A=22`.

`A = 6x + 10`

`text(If)\ x\ text(increases by 2)`

`A\ text(increases by)\ 6 xx 2 = 12`

`=> D`

Algebra, STD2 A2 2009 HSC 13 MC

The volume of water in a tank changes over six months, as shown in the graph.

Consider the overall decrease in the volume of water.

What is the average percentage decrease in the volume of water per month over this time, to the nearest percent?

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 48%
COMMENT: Remember that % decrease requires the decrease in volume to be divided by the original volume (50,000L).

`text(Initial Volume)`	`= 50\ 000\ text(L)`
`text(Final volume)`	`= 18\ 000\ text(L)`
`text(Decrease)`	`= 50\ 000-18\ 000`
	`= 32\ 000\ text{L (over 6 months)}`

`text(Loss per month)`	`= (32\ 000)/6`
	`= 5333.33…\ text(L per month)`
`text(% loss per month)`	`= (5333.33…)/(50\ 000)`
	`=10.666… %`

`=> B`

Probability, 2UG 2009 HSC 7 MC

Two people are to be selected from a group of four people to form a committee.

How many different committees can be formed?

(A) `6`

(B) `8`

(C) `12`

(D) `16`

Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 38%

`text(# Committees)`	`= (4 xx 3)/(2 xx 1)`
	`=6`

`=> A`

Algebra, STD2 A1 2011 HSC 21 MC

A train departs from Town A at 3.00 pm to travel to Town B. Its average speed for the
journey is 90 km/h, and it arrives at 5.00 pm. A second train departs from Town A at
3.10 pm and arrives at Town B at 4.30 pm.

What is the average speed of the second train?

135 km/h
150 km/h
216 km/h
240 km/h

Show Answers Only

`A`

Show Worked Solution

`text(1st train)`

♦ Mean mark 49%

`text(Travels 2hrs at 90km/h)`

`text(Distance)`	`=text(Speed)xxtext(Time)`
	`=90xx2`
	`=180\ text(km)`

`text(2nd train)`

`text(Travels 180 km in 1 hr 20 min)\ (4/3\ text(hrs))`

`text(Speed)`	`=text(Distance)/text(Time)`
	`=180-:4/3`
	`=180xx3/4`
	`=135\ text(km/h)`

`=> A`

Algebra, STD2 A1 2011 HSC 18 MC

Which of the following correctly expresses `a` as the subject of `s= ut+1/2at^2 `?

`a=(2(s-ut))/t^2`
`a=(2s-ut)/t^2`
`a=(1/2(s-ut))/t^2`
`a=(1/2s-ut)/t^2`

Show Answers Only

`A`

Show Worked Solution

`s`	`=ut+1/2at^2`
`1/2at^2`	`=s-ut`
`at^2`	`=2(s-ut)`
`a`	`=(2(s-ut))/t^2`

`=>A`

Measurement, STD2 M1 2010 HSC 17 MC

During a flood 1.5 hectares of land was covered by water to a depth of 17 cm.

How many kilolitres of water covered the land? (1 hectare = 10 000 m²)

2.55 kL
2550 kL
255 000 kL
2 550 000 kL

Show Answers Only

`B`

Show Worked Solution

♦♦ Mean mark 34%
NOTE: The unit conversion 1 m³ = 1000 L is contained in the Formulae and Data sheet given out in the exam.

`text(Area covered)`	`=1.5xx10\ 000`
	`=15\ 000\ text(m²)`
`text(Volume)`	`=Ah`
	`=15\ 000xx0.17`
	`=2550\ text(m³)`

`text{1 m³}`	`=1000\ text(L)=1\ text(kL)`
`:.\ text(Volume)`	`=2550\ text(kL)`

`=>B`

Algebra, STD2 A4 2010 HSC 13 MC

The number of hours that it takes for a block of ice to melt varies inversely with the temperature. At 30°C it takes 8 hours for a block of ice to melt.

How long will it take the same size block of ice to melt at 12°C?

3.2 hours
20 hours
26 hours
45 hours

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 50%

`text{Time to melt}\ (T) prop1/text(Temp) \ => \ T=k/text(Temp)`

`text(When) \ T=8, text(Temp = 30)`

`8`	`=k/30`
`k`	`=240`

`text{Find}\ T\ text{when Temp = 12:}`

`T`	`=240/12`
	`=20\ text(hours)`

`=> B`

Measurement, STD2 M6 2010 HSC 10 MC

A plane flies on a bearing of 150° from `A` to `B`.

What is the bearing of `A` from `B`?

`30^@`
`150^@`
`210^@`
`330^@`

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 34%

`/_TBA=30^@\ \ \ text{(angle sum of triangle)}`

`:.\ text(Bearing of)\ A\ text{from}\ B`

`=360-30`

`=330^@`

`=> D`

Algebra, STD2 A4 2012 HSC 30c

In 2010, the city of Thagoras modelled the predicted population of the city using the equation

`P = A(1.04)^n`.

That year, the city introduced a policy to slow its population growth. The new predicted population was modelled using the equation

`P = A(b)^n`.

In both equations, `P` is the predicted population and `n` is the number of years after 2010.

The graph shows the two predicted populations.

Use the graph to find the predicted population of Thagoras in 2030 if the population policy had NOT been introduced. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
In each of the two equations given, the value of `A` is 3 000 000.

What does `A` represent? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
The guess-and-check method is to be used to find the value of `b`, in `P = A(b)^n`.

(1) Explain, with or without calculations, why 1.05 is not a suitable first estimate for `b`. (1 mark)

(2) With `n = 20` and `P = 4\ 460\ 000`, use the guess-and-check method and the equation `P = A(b)^n` to estimate the value of `b` to two decimal places. Show at least TWO estimate values for `b`, including calculations and conclusions. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
The city of Thagoras was aiming to have a population under 7 000 000 in 2050. Does the model indicate that the city will achieve this aim?

Justify your answer with suitable calculations. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`6\ 600\ 000`
`text(The population in 2010.)`
`text{(1) See Worked Solution}`

`(2)\ \ b = 1.03, 1.02`
`text(See Worked Solution)`

Show Worked Solution

i. `text(2030 occurs at)\ \ n = 20\ \ text(on the)\ x text(-axis.)`

`text(Expected population (no policy) ) = 6\ 600\ 000`

COMMENT: Common ADV/STD2 content in new syllabus.

ii. `A\ text(represents the population when)\ \ n=0`

`text(which is the population in 2010.)`

♦ Mean mark part (ii) 48%

iii. (1) `P = A(1.05)^n\ text(would be steeper and lie above)`

`P = A(1.04)^n\ text(since)\ 1.05 > 1.04`

iii. (2) `text(Let)\ \ b = 1.03`

`P`	`= 3\ 000\ 000 xx 1.03^20`
	`= 5\ 418\ 000`

`text(Let)\ \ b = 1.02`

`P`	`= 3\ 000\ 000 xx 1.02^20`
	`= 4\ 457\ 800`

`:. b = 1.02`

iv. `text(In 2050,)\ n = 40`

`P`	`= 3\ 000\ 000 xx 1.02^40`
	`= 6\ 624\ 119\ \ (text(nearest whole))`

`text(S)text(ince the population is below 7 million,)`

`text(the model will achieve the aim.)`

Measurement, STD2 M6 2012 HSC 29c

Raj cycles around a course. The course starts at `E`, passes through `F`, `G` and `H` and finishes at `E`. The distances `EH` and `GH` are equal.

What is the length of `EF`, to the nearest kilometre? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
What is the total distance that Raj cycles, to the nearest kilometre? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`22\ text{km (nearest km)}`
`202\ text{km (nearest km)}`

Show Worked Solution

i. `text(Find)\ EF:`

♦ Mean mark 48%.

`/_ FGE`	`= 180\-(139 + 31) \ \ text{(angle sum of}\ Delta EFGtext{)}`
	`= 180-170`
	`= 10^@`

`text(Using Sine rule:)`

`(EF)/sin10^@`	`= 82/sin139^@`
`EF`	`= (82 xx sin10^@)/sin139^@`
	`= 21.70406…`
	`= 22\ text{km (nearest km)}`

ii. `text(Let)\ \ d = text(total distance cycled)`

`text(Find)\ EH`

`text(S)text(ince)\ Delta EGH\ text(is isosceles, and)\ /_EHG = 90^@`

`/_GEH = /_HGE = 45^@`

`text{(angles opposite equal sides in}\ Delta EGHtext{)}`

♦ Mean mark 37%.
MARKER’S COMMENT: Students could also have used Pythagoras or the Sine rule to calculate `GH`.

`sin45^@`	`= (GH)/82`
`GH`	`= 82 xx sin45^@`
	`= 57.983…`

`:. d`	`= EF + FG + GH + EH`
	`= 21.704… + 64 + 57.983 + 57.983`
	`= 201.66…`
	`= 202 \ text{km (nearest km)}`

Statistics, STD2 S5 2012 HSC 29b

A machine produces nails. When the machine is set correctly, the lengths of the nails are normally distributed with a mean of 6.000 cm and a standard deviation of 0.040 cm.

To confirm the setting of the machine, three nails are randomly selected. In one sample the lengths are 5.950, 5.983 and 6.140.

The setting of the machine needs to be checked when the lengths of two or more nails in a sample lie more than 1 standard deviation from the mean.

Does the setting on the machine need to be checked? Justify your answer with suitable calculations. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Settings need to be checked.)`

Show Worked Solution

♦ Mean mark 44%

`mu = 6.000,\ \ sigma = 0.040`

`text(Limits for ±1 σ:)`

`6.000 + 0.040 = 6.040\ text{(upper)}`

`6.000-0.040 = 5.960\ text{(lower)}`

`text(Chosen nails:)\ \ 5.950`	`=>\ text(outside limits)`
`5.983`	`=>\ text(inside)`
`6.140`	`=>\ text(outside)`

`text(S)text(ince 2 nails are outside 1 σ limits)`

`:.\ text(Settings need to be checked.)`

Statistics, STD2 S3 2012 HSC 29a*

Tourists visit a park where steam erupts from a particular geyser.

The brochure for the park has a graph of the data collected for this geyser over a period of time.

The graph shows the duration of an eruption and the time until the next eruption, timed from the end of one eruption to the beginning of the next.

Tony sees an eruption that lasts 4 minutes. Based on the data in the graph, what is the minimum time that he can expect to wait for the next eruption? (1 mark)

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Julia saw two consecutive eruptions, one hour apart. Based on the data in the graph, what was the longest possible duration of the first eruption that she saw? (1 mark)

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What does the graph suggest about the association between the duration of an eruption and the time to the next eruption? (1 mark)

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`70\ text(minutes)`
`3\ text(minutes)`
`text(It suggests that the longer an eruption lasts for, the longer)`
`text(you will wait until the next one.)`

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i. `70\ text(minutes)\ \ text{(refer to graph)}`

ii. `text(3 minutes)`

♦ Mean mark (b) 47%

`text(Locate 60 minutes on)\ y text(-axis and look for longest)`

`text{duration (i.e. the largest}\ x text{-axis value)}\ text(for that)`

`text(given)\ y text(-value.)`

iii. `text(It suggests that the longer an eruption lasts for,)`

`text(the longer you will wait until the next one, or)`

`text(there is a positive correlation between the length)`

`text(of an eruption and the time until the next one.)`

Measurement, STD2 M7 2012 HSC 28c

Jacques and a flagpole both cast shadows on the ground. The difference between the lengths of their shadows is 3 metres.

What is the value of `d`, the length of Jacques’ shadow? (3 marks)

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`d = 1.8\ text(m)`

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♦♦ Mean mark 24%

`text{Both triangles have right-angles with a common (ground) angle.}`

`:.\ text{Triangles are similar (equiangular)}`

` text{Since corresponding sides are in the same ratio}`

`d/1.5`	`= (d+3)/4`
`4d`	`= 1.5(d + 3)`
`8d`	`= 3(d + 3)`
	`= 3d + 9`
`5d`	`= 9`
`:.d`	`= 9/5`
	`=1.8\ text(m)`

Algebra, 2UG 2012 HSC 28b

Simplify fully `(18ab)/(3a^2) xx c/b`. (2 marks)

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`(6c)/a`

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♦ Mean mark 35%!

`(18ab)/(3a^2) xx c/b`	`= (18abc)/(3a^2b)`
	`=(6c)/a`

\(\therefore\ A\)	\(=72-(16+11+8+12+15)\)
	\(=72-62\)
	\(=10\)

ii. \(\text{Relative frequency of 4}\)	\(=\dfrac{8}{72}\)
	\(=\dfrac{1}{9}\)