Band 4 – Page 11

BIOLOGY, M7 EQ-Bank 18

It is recommended that a previously unvaccinated 14 year-old female receives two doses of the HPV (Human Papillomavirus) vaccine. The second dose should be given at least four weeks after the first dose.

What is the benefit of having two doses instead of one dose? Justify your response. (2 marks)

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Benefits of 2 doses of vaccine:

→ The first dose primes the immune system by activating B cells and creating memory cells.

→ The second dose expands the population of memory B cells, enabling a quicker and more effective antibody response.

→ This enhanced secondary immune response provides improved protection against future infections.

Show Worked Solution

Benefits of 2 doses of vaccine:

→ The first dose primes the immune system by activating B cells and creating memory cells.

→ The second dose expands the population of memory B cells, enabling a quicker and more effective antibody response.

→ This enhanced secondary immune response provides improved protection against future infections.

BIOLOGY, M7 2021 VCE 4c

Outline similarities and/or differences in active and passive ways of acquiring immunity. (3 marks)

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→ Active immunity leads to the body producing its own antibodies and memory cells, resulting in long-lasting protection.

→ Passive immunity provides immediate protection by transferring pre-existing antibodies, but this protection is temporary and short-term.

→ Both active and passive immunity involve the presence of antibodies that can neutralize or eliminate the target pathogen.

→ The key difference is that active immunity develops the immune system’s own capacity to respond, while passive immunity borrows antibodies from an external source.

Show Worked Solution

→ Active immunity leads to the body producing its own antibodies and memory cells, resulting in long-lasting protection.

→ Passive immunity provides immediate protection by transferring pre-existing antibodies, but this protection is temporary and short-term.

→ Both active and passive immunity involve the presence of antibodies that can neutralize or eliminate the target pathogen.

→ The key difference is that active immunity develops the immune system’s own capacity to respond, while passive immunity borrows antibodies from an external source.

BIOLOGY, M6 2021 VCE 9

The following table provides information on three commonly grown genetically modified (GM) crops in Australia.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \ \ \textbf{Crop} \rule[-1ex]{0pt}{0pt} & \quad \quad \textbf{Genetic modification} & \quad \ \textbf{Characteristic given by} \\
& & \quad \quad \quad \quad \textbf{modification} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \text{GM cotton} \rule[-1ex]{0pt}{0pt} & \text{several bacterial genes inserted} & \text{insect resistance and herbicide} \\
& & \text{tolerance} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \text{GM canola} \rule[-1ex]{0pt}{0pt} & \text{two genes from two different} & \text{tolerance to several herbicides} \\
& \text{bacterial species inserted} & \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{GM safflower} \rule[-1ex]{0pt}{0pt} & \text{a selection of genes silenced within} & \text{elevated levels of oleic acid in its} \\
& \text{the safflower genome} & \text{seeds} \rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

Select one of the GM crops in the table above and justify whether or not this crop could be described as transgenic. (1 mark)

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One issue with GM canola is the accidental release, during transport, of seeds along roadsides. Usually, unwanted plants that grow on the side of the road are killed using the herbicide glyphosate. However, GM canola is resistant to glyphosate.
Suggest one practical solution for treating GM canola that is found growing along roadsides. (1 mark)

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A new GM canola crop has been approved for use in Australia. It contains increased levels of omega-3 fatty acids, which are important in humans for building healthy cell membranes and for general growth and development, and also protect against a wide variety of diseases.
Omega-3 has traditionally been sourced from fish. Due to the growing demand for sources of omega-3 , bioengineers have been encouraged to continue developing GM canola crops as a sustainable alternative.
Discuss one social implication and one biological implication of using GM canola with increased levels of omega-3. Use a different implication in each response. (4 marks)

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a. Answers should include one of the following:

→ GM cotton contains genes from other species or bacteria and is therefore transgenic.

→ GM canola contains genes from other species or bacteria and is therefore transgenic.

→ GM safflower does not contain genes from another species and is therefore not transgenic.

b. Possible answers include:

→ use alternative herbicides that the GM canola is not resistant to

→ digging out and removing roadside GM canola by hand

→ mowing the roadsides could help manage GM canola

→ burn the GM canola using controlled methods.

c. Social Implication – possible answers could include:

→ Depending on market demand non-GM canola farmers may see increased sales and improved quality of life, or decreased sales and lower quality of life.

→ GM canola farmers may benefit from higher yields and lower production costs may lead to increased profits and a better quality of life.

→ Consumers may benefit from improved nutrition that may lead to reduced strain on the healthcare system.

→ If GM canola becomes more affordable and accessible than alternative sources of nutrition, such as fish, a wider range of consumers will enjoy the health benefits.

→ Decreased consumption of fish due to the availability of cheaper, more accessible GM canola could lead to reduced sales and lower incomes for fish farmers, potentially decreasing their quality of life.

→ Some consumers may be hesitant to purchase or consume GM food products. This could lead to decreased demand for GM canola and lower incomes for farmers growing it.

Biological Implication – possible answers could include:

→ Crossbreeding between GM canola and non-GM canola crops could lead to changes in the genome of the crops which may reduce the genetic variation within the GM canola crop

→ Consumer safety concerns regarding consumption of GM products could negatively impact the demand for GM canola and other GM crops.

→ If GM canola leads to a reduction in fish consumption, it could have a positive impact on fish populations by reducing over-fishing. This could help to restore and maintain healthy fish populations in the long run.

→ Improved nutrition for consumers through the consumption of GM crops could lead to better health outcomes and improved well-being for the broader population.

Show Worked Solution

a. Answers should include one of the following:

→ GM cotton contains genes from other species or bacteria and is therefore transgenic.

→ GM canola contains genes from other species or bacteria and is therefore transgenic.

→ GM safflower does not contain genes from another species and is therefore not transgenic.

b. Possible answers include:

→ use alternative herbicides that the GM canola is not resistant to

→ digging out and removing roadside GM canola by hand

→ mowing the roadsides could help manage GM canola

→ burn the GM canola using controlled methods.

c. Social Implication – possible answers could include:

→ Depending on market demand non-GM canola farmers may see increased sales and improved quality of life, or decreased sales and lower quality of life.

→ GM canola farmers may benefit from higher yields and lower production costs may lead to increased profits and a better quality of life.

→ Consumers may benefit from improved nutrition that may lead to reduced strain on the healthcare system.

→ If GM canola becomes more affordable and accessible than alternative sources of nutrition, such as fish, a wider range of consumers will enjoy the health benefits.

→ Some consumers may be hesitant to purchase or consume GM food products. This could lead to decreased demand for GM canola and lower incomes for farmers growing it.

Biological Implication – possible answers could include:

→ Crossbreeding between GM canola and non-GM canola crops could lead to changes in the genome of the crops which may reduce the genetic variation within the GM canola crop

→ Consumer safety concerns regarding consumption of GM products could negatively impact the demand for GM canola and other GM crops.

→ Improved nutrition for consumers through the consumption of GM crops could lead to better health outcomes and improved well-being for the broader population.

♦ Mean mark (c) 47%.

BIOLOGY, M6 2021 VCE 28 MC

Small amounts of DNA can be collected from items such as cigarette butts and hair follicles that are found at crime scenes. This DNA can be replicated for analysis and used as evidence in court cases.

Which of the following identifies the method of DNA replication and the enzyme used?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \quad \quad \quad \text{Method}\rule[-1ex]{0pt}{0pt}& \quad \ \ \ \text{Enzyme} \\
\hline
\rule{0pt}{2.5ex}\text{polymerase chain reaction}\rule[-1ex]{0pt}{0pt}&\textit{Taq}\ \text{polymerase}\\
\hline
\rule{0pt}{2.5ex}\text{DNA profiling}\rule[-1ex]{0pt}{0pt}& \text{DNA ligase}\\
\hline
\rule{0pt}{2.5ex}\text{polymerase chain reaction}\rule[-1ex]{0pt}{0pt} & \text{DNA ligase} \\
\hline
\rule{0pt}{2.5ex}\text{DNA profiling}\rule[-1ex]{0pt}{0pt}& \textit{Taq} \ \text{polymerase} \\
\hline
\end{array}
\end{align*}

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$A$

Show Worked Solution

→ The method of DNA replication used to replicate small amounts of DNA collected from crime scenes is called Polymerase Chain Reaction (PCR).

→ The key enzyme involved in PCR is a thermostable DNA polymerase enzyme called Taq polymerase.

$\Rightarrow A$

BIOLOGY, M8 2021 VCE 34 MC

The World Health Organization (WHO) runs a campaign to eliminate cervical cancer globally. In response to this campaign, a study was conducted to model the impact of cervical cancer screening along with a human papillomavirus (HPV) vaccination program in low-income and lower-middle-income countries. HPV is the cause of most cervical cancers. The results of the modelling are shown below.

Which one of the following conclusions is supported by the data in the graph?

The incidence of cervical cancer would be unaffected by a cervical cancer screening program.
The largest factor in reducing the incidence of cervical cancer in the long term is vaccination.
With vaccination only, it is possible to achieve fewer than 10 cervical cancer cases per 100,000 women by 2050 .
Vaccination and one lifetime screen would be more effective than vaccination and two lifetime screens by 2070 .

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$B$

Show Worked Solution

Consider each option:

Option A: The evidence shows an effect from screening programs → Incorrect

Option B: The graph shows the incidence of cervical cancer has decreased since the introduction of the vaccination. Screening has only added to this marginally → Correct

Option C: With vaccination only fewer than 10 cases per 100,000 women could be achieved by 2060 not 2050 → Incorrect

Option D: The graph shows that vaccination and two lifetime screens is more effective in 2070 → Incorrect

$\Rightarrow B$

CHEMISTRY, M8 2012 VCE 6

The iron content in multivitamin tablets was determined using atomic absorption spectroscopy.

The absorbances of four standards were measured.

Three multivitamin tablets were selected. Each tablet was dissolved in 100.0 mL of water. The absorbance of each of the three solutions was then measured.

The following absorbances were obtained.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \ \textbf{Solution} \rule[1ex]{0pt}{0pt} & \textbf{Concentration} & \textbf{Absorbance} \\
& \textbf{mg/L} & \\
\hline
\rule{0pt}{2.5ex} \text{Standard 1} \quad \quad & 0.00 & 0.06 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 2} & 100.0 & 0.16 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 3} & 200.0 & 0.25 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 4} & 300.0 & 0.36 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 5} & 400.0 & 0.46 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 1} & - & 0.39 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 2} & - & 0.42 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 3} & - & 0.45 \\
\hline
\end{array}

i. Use the grid below to construct a calibration graph of the absorbances of the standard solutions. (2 marks)

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ii. Determine the average iron content, in milligrams, of the multivitamin tablets. (2 marks)

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Spectroscopic techniques work on the principle that, under certain conditions, atoms, molecules or ions will interact with electromagnetic radiation. The type of interaction depends on the wavelength of the electromagnetic radiation.

Name one spectroscopic technique that you have studied this year.
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1. Which part of the electromagnetic spectrum does this technique use? (1 mark)
  
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2. How does this part of the electromagnetic spectrum interact with matter? What information does this spectroscopic technique provide? (2 marks)
  
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a.i.

a.ii. $35.5\ \text{mg}$

b.i. Answers could include:

→ AAS (visible light)

→ UV-Vis (UV or visible light)

→ IR (Infrared radiation)

→ NMR (radio waves)

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

→ During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.

→ The absorption of the light indicates the concentration of the targeted element within the sample.

Show Worked Solution

a.i.

a.ii. Average absorbance (tablets) $=\dfrac{0.39+0.42+0.45}{3}=0.42$

Using the graph: absorbance value of $0.42 → 355\ \text{mg L}^{-1}$

$\ce{m(Fe) (100\ \text{ml}) = 355 \times 0.1 =35.5\ \text{mg}}$

b.i. Answers could include:

→ AAS (visible light)

→ UV-Vis (UV or visible light)

→ IR (Infrared radiation)

→ NMR (radiowaves)

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

→ During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.

→ The absorption of the light indicates the concentration of the targeted element within the sample.

♦ Mean mark (b.ii) 42%.

CHEMISTRY, M8 EQ-Bank 5

A chemical reaction occurs at a constant temperature.

Describe the effect on the yield if

the value of the equilibrium expression is higher. (2 marks)
the activation energy of the reaction is decreased. (2 marks)

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a. Equilibrium expression is higher:

→ The equilibrium expression shows the ratio of reactants to products in a chemical equation.

→ A higher value for the equilibrium expression suggests a higher proportion of products / increased product concentration compared to reactants, and hence a higher yield.

b. The activation energy is decreased:

→ A decrease in activation energy increases the reaction rate and allows the system to get to equilibrium faster.

→ This has no effect on the equilibrium yield which remains the same.

Show Worked Solution

a. Equilibrium expression is higher:

→ The equilibrium expression shows the ratio of reactants to products in a chemical equation.

→ A higher value for the equilibrium expression suggests a higher proportion of products / increased product concentration compared to reactants, and hence a higher yield.

b. The activation energy is decreased:

→ A decrease in activation energy increases the reaction rate and allows the system to get to equilibrium faster.

→ This has no effect on the equilibrium yield which remains the same.

v1 Algebra, STD2 A4 SM-Bank 4

Bec is a baker and makes cookies to sell every week.

The cost of making $n$ cookies, $$C$, can be calculated using the equation

$C=400+2.5n$

Bec sells the cookies for $4.50 each, and her income is calculated using the equation

$I=450n$

On the grid above, draw the graphs of $C$ and $I$. (2 marks)

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On the graph, label the breakeven point and the loss zone. (2 marks)

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(i) and (ii)

Show Worked Solution

ii. $\text{Loss zone occurs when}\ C > I,\ \text{which is shaded}$

$\text{in the diagram above.}$

v1 Algebra, STD2 A4 2005 HSC 28b

Jake and Preston are planning a fund-raising event at the local swim centre. They can have access to the giant pool float for $550 and the party room hire for $250. A sausage sizzle and drinks will cost them $9 per person.

Write a formula for the cost ($C) of running the event for $x$ people. (1 mark)

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The graph shows planned income and costs when the ticket price is $15.

Estimate the minimum number of people needed at the fund raising event to cover the costs. (1 mark)

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How much profit will be made if 200 people attend the fund raiser? (1 mark)

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Jake and Preston have 300 tickets to sell. They want to make a profit of $1510.

What should be the price of a ticket, assuming all 300 tickets will be sold? (3 marks)

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$C=800+9x$
$\text{Approximately }135$
$$400$
$$16.70$

Show Worked Solution

i.	$$C$	$=550+250+(9\times x)$
		$=800+9x$

ii. $\text{Using the graph intersection}$

$\text{Approximately 135 people are needed}$

$\text{to cover the costs.}$

iii. $\text{If 200 people attend}$

$\text{Income}$	$=200\times $15$
	$=$3000$

$\text{Costs}$	$=800+(9\times 200)$
	$=$2600$

$\therefore\ \text{Profit}$	$=3000-2600$
	$=$400$

iv. $\text{Costs when}\ x=300:$

$C$	$=800+(9\times 300)$
	$=$3500$

$\text{Income required to make }$1510\ \text{profit}$

$=3500+1510$

$=$5010$

$\therefore\ \text{Price per ticket}$	$=\dfrac{5010}{300}$
	$=$16.70$

v1 Algebra, STD2 A4 2020 HSC 24

There are two tanks at an industrial plant, Tank A and Tank B. Initially, Tank A holds 2520 litres of liquid fertiliser and Tank B is empty.

Tank A begins to empty liquid fertiliser into a transport vehicle at a constant rate of 40 litres per minute.

The volume of liquid fertiliser in Tank A is modelled by $V=1400-40t$ where $V$ is the volume in litres and $t$ is the time in minutes from when the tank begins to drain the fertiliser.

On the grid below, draw the graph of this model and label it as Tank A. (1 mark)
Tank B remains empty until $t=10$ when liquid fertiliser is added to it at a constant rate of 60 litres per minute.
By drawing a line on the grid (above), or otherwise, find the value of $t$ when the two tanks contain the same volume of liquid fertiliser. (2 marks)

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Using the graphs drawn, or otherwise, find the value of $t$ (where $t > 0$) when the total volume of liquid fertiliser in the two tanks is 1400 litres. (1 mark)

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$\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}$
$20 \ \text{minutes}$
$30 \ \text{minutes}$

Show Worked Solution

a. $\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}$

b. $\text{Tank} \ B \ \text{will pass through (10, 0) and (30, 1200)}$

$\text{By inspection, the two graphs intersect at} \ \ t = 20 \ \text{minutes}$

c. $\text{Strategy 1}$

$\text{By inspection of the graph, consider} \ \ t = 30$

$\text{Tank A} = 200 \ \text{L} , \ \text{Tank B} =1200 \ \text{L}$

$\therefore\ \text{Total volume = 1400 L when t = 30}$

$\text{Strategy 2}$

$\text{Total Volume}$	$=\text{Tank A} + \text{Tank B}$
$1400$	$=1400-40t+(t-10)\times 60$
$1400$	$=1400-40t+60t-600$
$20t$	$= 600$
$t$	$= 30 \ \text{minutes}$

♦♦ Mean mark part (c) 22%.

v1 Algebra, STD2 A4 2020 VCAA 3

Noah's business manufactures car seat covers.

The monthly oncome, $I$, in dollars, from selling $n$ seat covers is given by

$I=60n$

This relationship is shown on the graph below.

The monthly cost, $C$, in dollars, of making $n$ seat covers is given by

$C=35n+5000$

On the graph above, sketch the monthly cost, $C$, of making $n$ seat covers. (1 mark)
Find the number of seat covers that need to be sold in order to break even and state the profit made at this point. (2 marks)

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b. $\text{200 seat covers and zero profit at this point}$

Show Worked Solution

a. $\text{Draw graph through points (0, 5000) and (250, 13 750)}$

b. $C=35n+5000\ \text{and }I=60n$

$\text{Break-even occurs when} \ \ I=C$

$\text{Method 2: Graphically}$

$\text{Point of intersection is }\rightarrow (200, 12\ 000)$

$\text{Method 2: Algebraically}$

$\text{Solve for} \ n:$

$60n$	$=35n+5000$
$25n$	$=5000$
$n$	$=200$

$\therefore\ \text{200 seat covers must be sold to break even}$

$\text{and the profit at this point is zero}$

v1 Algebra, STD2 A4 EQ-Bank 8

Two farmers, River and Jacqueline, sold organic honey at the the local market.

River sold his 500 gram jars of honey for $8 each and Jacqueline sold her 200 gram jars of honey for $4 each. In the first hour, their total combined sales were $40.

If River sold $x$ jars of honey and Jacqueline sold $y$ jars of honey, then the following equation can be formed:

$8x+4y=40$

In the first hour, the friends sold a total of 6 jars of honey between them.

Find the number of jars of honey each of the friends sold during this time by forming a second equation and solving the simultaneous equations graphically. (5 marks)

Show Answers Only

$\text{River sold 4 and Jacqueline sold 2.}$

Show Worked Solution

$\text{Graphing}\ \ 8x + 4y$	$=40$
$y$	$=-2x+10$

$y\text{-intercept}\ = (0, 10)$

$x\text{-intercept}\ = (5, 0)$

$\text{Gradient}\ = -2$

$\text{Second equation:}$

$x+y=6$

$\text{From the graph,}$

$\text{River sold 4 and Jacqueline sold 2.}$

v1 Algebra, STD2 A4 SM-Bank 7

The graph of the line $x+y=3$ is shown.

By graphing $y=2x-3$ on the same grid, find the point of intersection of $x+y=3$ and $y=2x-3$. (3 marks)

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$(2, 1)$

Show Worked Solution

$\text{Graphing}\ y=2x-3:$

$y\text{-intercept }=-3$

$\text{Gradient }=2$

$\therefore\ \text{Point of intersection is (2, 1).}$

v1 Algebra, STD2 A4 SM-Bank 6

A student was asked to solve the following simultaneous equations.

$y=2x-5$

$x-2y+2=0$

After graphing the equations, the student found the point of intersection to be $(4,3)$?

Is the student correct? Support your answer with calculations. (2 marks)

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$\text{See Worked Solutions}$

Show Worked Solution

$\text{If the student is correct, the co-ordinates will}$

$\text{satisfy both equations.}$

$\text{Substitute (4, 3) into}\ \ y=2x-5$

$\text{LHS}$	$=3$
$\text{RHS}$	$= 2(4)-5$
	$=3$

$\therefore\ \text{LHS = RHS}$

$\text{Substitute (4, 3) into}\ \ x-2y+2=0$

$\text{LHS}$	$=4-2(3)+2$
	$=0$
	$\ =\ \text{RHS}$

$\rightarrow\ (4,3)\ \text{satisfies both equations.}$

$\therefore\ \text{Student is correct.}$

v1 Algebra, STD2 A4 2014 HSC 26d

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

$y=2x-6$

$y-x+2=0$

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$x=4,\ y=2$

Show Worked Solution

$\text{Solution is at the intersection:}\ \ x=4,\ y=2$

v1 Algebra, STD2 A4 SM-Bank 27

Morgan and Beau are to host a 21st birthday party for their friend's Zac and Peattie. They can hire a function room for $900 and a DJ for $450. Drinks will cost them $33 per person.

Write a formula for the cost ($$C$) of holding the birthday party for $x$ people. (1 mark)

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The graph below shows the planned income and costs if they charge $60 per person. Estimate the number of friends they need to invite to break even. (1 mark)

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How much money will Morgan and Beau have to purchase a travel voucher as a group gift if 90 people attend the party? (1 mark)

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$$C=1350+33x$
$50$
$$1080$

Show Worked Solution

i.	$\text{Fixed Costs}$	$=900+450$
		$=$1350$

$\text{Variable Costs}=$33x$

$\therefore\ $C=1350+33x$

ii.	$\text{From the graph}$
	$\text{Costs = Income when}\ x=50$
	$\text{(i.e. where lines intersect)}$

$\text{Algebraically}$
$\text{Income }$	$=\ \text{Costs}$
$60x$	$=1350+33x$
$27x$	$=1350$
$x$	$=50$

$\therefore\ \text{Breakeven when }50\ \text{people attend}$

iii. $\text{When}\ \ x=90:$

$\text{Income}$	$=90\times 60$
	$=$5400$

$$C$	$=1350+33\times 90$
	$=$4320$

$\therefore\ \text{Travel Voucher}$	$=5400-4320$
	$=$1080$

v1 Algebra, STD2 A4 2018 HSC 27b

$y$	$=x-3$
$y+3x$	$=1$

Draw these two linear graphs on the number plane below and determine their intersection. (3 marks)

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$(1,-2)$

Show Worked Solution

$\text{Table of values:}\ \ y=x-3$

\begin{array} {|c|c|c|c|c|}
\hline x & -2 & -1 & 0 & \colorbox{lightblue}{ 1 } \\
\hline \ \ y \ \ & \ \ -5 \ \ & \ \ -4 \ \ & \ \ -3 \ \ & \ \colorbox{lightblue}{ – 2} \\
\hline \end{array}

$\text{Table of values:}\ \ y+3x=1 \ \rightarrow \ y=-3x+1$

\begin{array} {|c|c|c|c|c|}
\hline x & -1 & 0 & \colorbox{lightblue}{ 1 } & 2 \\
\hline \ \ y \ \ & \ \ \ 4\ \ \ & \ \ \ 1\ \ \ & \ \colorbox{lightblue}{ – 2} & \ \ -5 \ \ \\
\hline \end{array}

$\text{From graph (and table), intersection occurs}$

$\text{at}\ \ (1, -2).$

v1 Algebra, STD2 A4 2023 HSC 21

Electricity provider $A$ charges 30 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $90.

Complete the table showing Provider $A$'s monthly charges for different levels of electricity usage. (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ \ 1000 \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

Provider $B$ charges 52.5 cents per kWh, with no fixed monthly charge. The graph shows how Provider $B$'s charges vary with the amount of electricity used in a month.

On the grid above, graph Provider $A$'s charges from the table in part (a). (1 mark)
Use the two graphs to determine the number of kilowatt hours per month for which Provider $A$ and Provider $B$ charge the same amount. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
A customer uses an average of 600 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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

Show Answers Only

a. $\text{When kWh} =400$

$\text{Monthly charge}\ =$90+0.30\times 400=$210$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

c. $\text{400 kWh}$

d. $\text{Provider}\ A\ \text{is cheaper by \$45.}$

Show Worked Solution

a. $\text{When kWh} =400$

$\text{Monthly charge}\ =$90+0.30\times 400=$210$

c. $A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}$

$\therefore\ \text{Same charge at 400 kWh}$

d. $\text{Cost at 600 kWh:}$

$\text{Method 1: Using graph}\rightarrow\ $315-270=$45$

$\text{Method 2: Algebraically}:$

$\text{Provider}\ A: \ 90 + 0.30 \times 600 = $270$

$\text{Provider}\ B: \ 0.525 \times 600 = $315$

$\therefore \text{Provider}\ A\ \text{is cheaper by \$45.}$

v1 Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost ($$c$) charged by two companies for the hire of a jetski for $x$ hours.

Both companies charge $450 for the hire of a jetski for 5 hours.

What is the hourly rate charged by Company A? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Company B charges an initial booking fee of $80.

Write a formula, in the of $c=b+mx$, for the cost of hiring a jetski from Company B for $x$ hours. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
A jetski is hired for 7 hours from Company B.

Calculate how much cheaper this is than hiring from Company A. (2 marks)

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

Show Answers Only

$$90$
$c=80+74x$
$$32$

Show Worked Solution

i.	$\text{Hourly rate}\ (A)$	$=\dfrac{450}{5}$
		$=$90$

ii. $m=\text{hourly rate}$

$\text{Find}\ m,\ \text{given}\ c = 450,\ \text{when}\ \ x = 5\ \text{and}\ \ b = 80$

$450$	$=80+m\times 5$
$5m$	$=370$
$m$	$=\dfrac{370}{5}=74$

$\therefore\ c=80+74x$

iii.	$\text{Cost}\ (A)$	$=90\times 7=$630$
	$\text{Cost}\ (B)$	$=80+74\times 7=$598$

$\therefore\ \text{Company}\ B’\text{s hiring cost is }$32\ \text{cheaper.}$

v1 Algebra, STD2 A4 SM-Bank 7 MC

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

$x+y=-6$ and $x-y=-6$

Part of the screen is shown.

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

\begin{align*}
\begin{array}{c|c}
\text{ } \\
\textbf{ A. } \\
\textbf{ B. } \\
\textbf{ C. } \\
\textbf{ D. }
\end{array}
\begin{array}{|c|c|c|}
\hline
\ x+y=-6 & x-y=-6 & \text{Solution} \\
\hline
\text{Line 1} & \text{Line 2} & x=-6,\ y=0 \\
\hline
\text{Line 1} & \text{Line 2} & x=-6, y=-6 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6,\ y=0 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6, y=-6 \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

$\text{Line 1:} \ \ x-y=-6$

$\text{Line 2:} \ \ x+y=-6$

$\text{Intersection at} \ (-6, 0).$

$\Rightarrow C$

v1 Networks, STD2 N2 2021 HSC 2 MC

Consider the network diagram.

What is the sum of the degrees of all the vertices in this network?

Show Answers Only

`D`

Show Worked Solution

`text(Working from)\ A\ text(to)\ E:`

COMMENT: This simple question caused problems for many with mean mark just 58%.

`text{Sum of degrees}`	`= 4 + 3 + 4 + 2 + 3`
	`= 16`

`=> D`

v1 Algebra, STD2 A2 2007 HSC 24c

Blythe travels to France via the USA. She uses this graph to calculate her currency conversions.

After leaving the USA she has US$750 to add to the A$2150 that she plans to spend in France.

She converts all of her money to euros. How many euros does she have to spend in France? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---
If the value of the US dollar rises in comparison to the Australian dollar, what will be the effect on the gradient of the line used to convert US dollars to Australian dollars? (1 mark)

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

Show Answers Only

$1890\ \text{€}$
$\text{See worked solutions}$

Show Worked Solution

i. $\text{From graph:}$

$75\ \text{US}$$	$=100\ \text{A}$$
$\rightarrow\ 750\ \text{US}$$	$=1000\ \text{A}$$

$\therefore\ \text{Blythe has a total of }$1000+$2150=\text{A}$3150 $

$\text{Converting A}$\ \text{to €}:$

$100\ \text{A}$$	$=60\ \text{€}$
$\therefore\ 3150\ \text{A}$$	$=\dfrac{3150}{100}\times 60$
	$=1890\ \text{€}$

ii. $\text{If the value of the US}\ $\ \text{rises against the}$

$\text{Australian }$\ \text{then 1 A}\ $\ \text{will buy less US}\ $$

$\text{than before and the gradient used to convert}$

$\text{the currencies will steepen (increase).}$

v1 Algebra, STD2 A2 2014 HSC 22 MC

Lisa’s motorbike uses fuel at the rate of 1.8 L per 100 km for long-distance driving and 2.3 L per 100 km for short-distance driving.

She used the motorbike to make a journey of 840 km, which included 108 km of short-distance driving.

Approximately how much fuel did Lisa’s motorbike use on the journey?

9 L
16 L
18 L
34 L

Show Answers Only

$B$

Show Worked Solution

$\text{Fuel used in short distance}$

$=\dfrac{108}{100}\times 2.3\ \text{L}=2.484\ \text{L}$

$\text{Fuel used in long distance}$

$=\dfrac{840-108}{100}\times 1.8\ \text{L}=13.176\ \text{L}$

$\therefore\ \text{Total Fuel}$	$=2.484+13.176$
	$=15.66\ \text{L}$

$\Rightarrow B$

v1 Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth. An astronaut who weighs 63 kg on Earth weighs only 9 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth. (2 marks)

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

Show Answers Only

$17\ 143\ \text{kg}$

Show Worked Solution

$W_{\text{moon}}\propto W_{\text{earth}}$

$\rightarrow\ W_{\text{m}}=k\times W_{\text{e}}$

$\text{Find}\ k,\ \text{given}\ W_{\text{e}}=63\ \text{when}\ W_{\text{m}}=9$

$9$	$=k\times 63$
$k$	$=\dfrac{9}{63}=\dfrac{1}{7}$

$\text{If}\ W_{\text{m}}=2449\ \text{kg, find}\ W_{\text{e}}:$

$2449$	$=\dfrac{1}{7}\times W_{\text{e}}$
$W_{\text{e}}$	$=7\times 2449=17\ 143$

$\text{Landing craft weighs}\ 17\ 143\ \text{kg on earth}$

v1 Algebra, STD2 A2 2007 HSC 27b

A cafe uses eight long-life light globes for 7 hours every day of the year. The purchase price of each light globe is $11.00 and they each cost $$f$ per hour to run.

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

--- 4 WORK AREA LINES (style=lined) ---
Find the value of $f$ (correct to three decimal places) if the total cost of running these eight light globes for one year is $850. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
If the use of the light globes increases to ten and a half hours per night every night of the year, does the total cost increase by one-and-a-half times? Justify your answer with appropriate calculations. (1 mark)

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

Show Answers Only

$$c=88+20\ 440f$
$0.037\ $/\text{hr}\ \text{(3 d.p.)}$
$\text{Proof: (See Worked Solutions)}$

Show Worked Solution

i. $\text{Purchase price}=8\times 11=$88$

$\text{Running cost}$	$=\text{No. of hours}\times \text{Cost per hour}$
	$=8\times 7\times 365\times f$
	$=20\ 440f$

$\therefore\ $c=88+20\ 440f$

ii. $\text{Given}\ \ $c=$850$

$850$	$=88+20\ 440f$
$20\ 440f$	$=850-88$
$f$	$=\dfrac{762}{20440}$
	$= 0.03727\dots$
	$=0.037\ $/\text{hr}\ \text{(3 d.p.)}$

iii. $\text{If}\ f\ \text{is multiplied by }1.5 =\dfrac{10.5}{7}$

$f=1.5\times0.037=0.0555\ \ $/\text{hr}$

$\therefore\ $c$	$=88+20\ 440\times 0.0555$
	$=$1222.42$

$\text{Since }$1222.42\ \text{is less than}\ 1.5\times $850 = $1275,$

$\text{the total cost increases to less than 1.5 times the}$

$\text{the original cost.}$

v1 Algebra, STD2 A2 SM-Bank 3

The average height, $L$, in centimetres, of a boy between the ages of 7 years and 10 years can be represented by a line with equation

$L=7A+85$

where $A$ is the age in years. For this line, the gradient is 7.

What does this indicate about the heights of boys aged 7 to 10? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Give ONE reason why this equation is not suitable for predicting heights of boys older than 10. (1 mark)

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

Show Answers Only

$\text{It indicates that 7-10 year old boys, on average, grow 7 cm per year.}$
$\text{Boys eventually stop growing, and the equation doesn’t factor this in.}$

Show Worked Solution

i. $\text{It indicates that 7-10 year old boys, on average, grow}$

$\text{7 cm per year.}$

ii. $\text{Boys eventually stop growing, and the equation doesn’t}$

$\text{factor this in.}$

v1 Algebra, STD2 A2 2019 HSC 14 MC

Last Friday, Jake had 98 marbles and Jack had 79 Marbles. On average, Jake wins 5 marbles per day and Jack loses 4 marbles per day.

If $x$ represents the number of days since last Friday and $y$ represents the number of marbles, which pair of equations model this situation?

A.	$\text{Jake:}\ \ y=98x+5$ $\text{Jack:}\ \ y=79x-4$
B.	$\text{Jake:}\ \ y=5+98x$ $\text{Jack:}\ \ y=4-79x$
C.	$\text{Jake:}\ \ y=5x+98$ $\text{Jack:}\ \ y=4x-79$
D.	$\text{Jake:}\ \ y=98+5x$ $\text{Jack:}\ \ y=79-4x$

Show Answers Only

$D$

Show Worked Solution

$\text{Jake starts with 98 and adds 5 per day:}$

$y=98+5x$

$\text{Jack starts with 79 and loses 4 per day:}$

$y=79-4x$

$\Rightarrow D$

v1 Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds $(p)$ and Australian dollars $(d)$ on a particular day is shown in the graph.

Write the direct variation equation relating British pounds to Australian dollars in the form $p=md$. Leave $m$ as a fraction. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The relationship between Japanese yen $(y)$ and Australian dollars $(d)$ on the same day is given by the equation $y=84d$.

Convert $107\ 520$ Japanese yen to British pounds. (2 marks)

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

Show Answers Only

$p=\dfrac{5}{8}d$
$107\ 520\ \text{yen = 800 pounds}$

Show Worked Solution

a. $m=\dfrac{\text{rise}}{\text{run}}=\dfrac{5}{8}$

$p=\dfrac{5}{8}d$

♦ Mean mark 42%.

b. $\text{Yen to Australian dollars:}$

$y$	$=84d$
$107\ 520$	$=84d$
$d$	$=\dfrac{107\ 520}{84}$
	$= 1280\ $\text{A}$

$\text{Australian dollars to pounds:}$

$p$	$=\dfrac{5}{8}\times 1280$
	$=800\ \text{pounds}$

$\therefore\ 107\ 520\ \text{yen = 800 pounds}$

v1 Algebra, STD1 A2 2020 HSC 20

The height of a bundle of photographic paper ($H$ mm) varies directly with the number of sheets ($N$) of photographic paper that the bundle contains.

This relationship is modelled by the formula $H=kN$, where $k$ is a constant.

The height of a bundle containing 150 sheets of photographic paper is 2.7 centimetres.

Show that the value of $k$ is 0.18. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
A bundle of photographic paper has a height of 36 centimetres. Calculate the number of sheets of photographic paper in the bundle. (2 marks)

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

Show Answers Only

$\text{See Worked Solutions}$
$2000\ \text{sheets}$

Show Worked Solution

a. $H=2.7\ \text{cm }=27\ \text{mm, when}\ N=150:$

$H$	$=kN$
$2.7$	$=k\times 150$
$\therefore\ k$	$=\dfrac{2.7}{150}$
	$=0.18$

b. $\text{Find}\ \ N \ \text{when} \ \ H=36\ \text{cm}=360\ \text{mm:}$

$360$	$=0.18\times N$
$\therefore\ N$	$=\dfrac{360}{0.18}$
	$=2000\ \text{sheets}$

♦ Mean mark 50%.

v1 Algebra, STD2 A4 2022 HSC 22

The formula $C=80n+b$ is used to calculate the cost of producing desktop computers, where $C$ is the cost in dollars, $n$ is the number of desktop computers produced and $b$ is the fixed cost in dollars.

Find the cost $C$ when 2458 desktop computers are produced and the fixed cost is $$18\ 230$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Some desktop computers have extra features added. The formula to calculate the production cost for these desktop computers is
$C=80n+an+18\ 230$
where $a$ is the additional cost in dollars per desktop computer produced.
Find the number of desktop computers produced if the additional cost is $35 per desktop computer and the total production cost is $$103\ 330$. (2 marks)

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

Show Answers Only

$$214\ 870$
$740\ \text{desktop computers}$

Show Worked Solution

a. $\text{Find}\ C,\ \text{given}\ n=2458\ \text{and}\ b=18\ 230$

$C$	$=80\times 2458+18\ 230$
	$=$214\ 870$

b. $\text{Find}\ n,\ \text{given}\ C=103\ 330\ \text{and}\ a=35$

$C$	$=80n+an+18\ 230$
$103\ 330$	$=80n+35n+18\ 230$
$115n$	$=85\ 100$
$n$	$=\dfrac{85\ 100}{115}$
	$=740\ \text{desktop computers}$

v1 Algebra, STD2 A2 2012 HSC 8 MC

Dots were used to create a pattern. The first three shapes in the pattern are shown.

The number of dots used in each shape 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 dots $(N)$} \rule[-1ex]{0pt}{0pt} &\;\;\; 8 \;\;\; & \;\; \;10 \;\;\; & \;\; \;12\; \;\; \\
\hline
\end{array}

How many dots would be required for Shape 182?

363
370
546
1092

Show Answers Only

$B$

Show Worked Solution

$\text{Linear relationship where}$

$N=6+(2\times S)$

$\text{When}\ \ S=182$

$N$	$=6+(2\times 182)$
	$=370$

$\Rightarrow B$

v1 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} & \;\;\; 6 \;\;\; & \;\;\; 10 \;\;\; & \;\;\; 14 \;\;\; \\
\hline
\end{array}

Draw Shape 4 of this pattern. (1 mark)

--- 3 WORK AREA LINES (style=blank) ---
How many sticks would be required for Shape 128? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Is it possible to create a shape in this pattern using exactly 609 sticks?

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

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

Show Answers Only

$\text{See Worked Solutions.}$
$514$
$\text{No (See worked solution)}$

Show Worked Solution

i. $\text{Shape 4 is shown below:}$

ii. $\text{Since}\ \ N=2+4S$

♦ 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.

$\text{If }S$	$=128$
$N$	$=2+(4\times 128)$
	$=514$

iii.	$609$	$=2+4S$
	$4S$	$=607$
	$S$	$=151.75$

$\text{Since}\ S\ \text{is not a whole number, 609 sticks}$

$\text{will not create a shape in this pattern.}$

v1 Algebra, STD2 A2 2006 HSC 7 MC

Which equation represents the relationship between $x$ and $y$ in this table?

\begin{array} {|c|c|c|}
\hline \ \ x\ \ & \ \ 0\ \ &\ \ 2\ \ & \ \ 4\ \ & \ \ 6\ \ & \ \ 8\ \ \\
\hline y & 3 & 4 & 5 & 6 & 7 \\
\hline \end{array}

$y=2x+3$
$y=\dfrac{1}{2}x+3$
$y=\dfrac{1}{2}x-3$
$y=3x-2$

Show Answers Only

$B$

Show Worked Solution

$\text{By elimination (using the table)}$

$(0, 3)\ \text{must satisfy}$

$\therefore\ \text{NOT}\ C\ \text{or}\ D$

$(2, 4)\ \text{must satisfy}$

$\therefore\ \text{NOT}\ A\ \text{as}\ 2\times 2+3\neq\ 5$

$\Rightarrow B$

v1 Algebra, STD2 A2 2016 HSC 14 MC

The graph shows a line which has an equation in the form $y=mx+c$.

Which of the following statements is true?

$m$ is positive and $c$ is negative
$m$ is negative and $c$ is positive
$m$ and $c$ are both positive
$m$ and $c$ are both negative

Show Answers Only

$B$

Show Worked Solution

$y\text{-intercept}\ (c)\ \text{is positive}$

$\rightarrow\ \text{eliminate A and D}$

$\text{gradient}\ (m)\ \text{is negative}$

$\rightarrow\ \text{eliminate C}$

$\Rightarrow B$

v1 Algebra, STD2 A4 2017 HSC 17 MC

The graph of the line with equation $y=5-x$ is shown.

When the graph of the line with equation $y=2x-1$ is also drawn on this number plane, what will be the point of intersection of the two lines?

$(0, 5)$
$(1, 2)$
$(2, 3)$
$(5, 0)$

Show Answers Only

$C$

Show Worked Solution

$\text{Method 1: Graphically}$

$\text{From graph, intersection is at} (2,3)$

$\text{Method 2: Algebraically}$

$y$	$=5-x$	$…\ (1)$
$y$	$=2x-1$	$…\ (2)$

$\text{Substitute (2) into (1)}$

$2x-1$	$=5-x$
$3x$	$=6$
$x$	$=2$

$\text{When}\ \ x=2,\ y=5-2=3$

$\Rightarrow C$

v1 Algebra, STD2 A2 SM-Bank 5 MC

The equation of the line drawn in the diagram below is:

$5y=8x$
$y=-\dfrac{8}{5}x+8$
$y=\dfrac{8}{5}x+8$
$y=-\dfrac{5}{8}x+8$

Show Answers Only

$B$

Show Worked Solution

$y\text{-intercept}=+8$

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=-\dfrac{8}{5}$

$\therefore\ \text{Equation is}:\ y=-\dfrac{8}{5}x+8$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 5

The diagram shows the graph of a line.

What is the equation of this line? (2 marks)

Show Answers Only

$y=\dfrac{1}{6}x+1$

Show Worked Solution

$y\text{-intercept}=1$

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=\dfrac{1}{6}$

$\therefore\ \text{Equation:}\ \ y=\dfrac{1}{6}x+1$

v1 Algebra, STD2 A2 SM-Bank 6 MC

Hannah went paddle boarding on a holiday.

The hiring charges are listed in the table below:

\begin{array} {|l|c|c|}
\hline \text{Hours hired} \ (h) & 1 & 2 & 3 & 4 & 5 \\
\hline \text{Cost} \ (C) & 15 & 23 & 31 & 39 & 47 \\
\hline \end{array}

Which linear equation shows the relationship between $C$ and $h$?

$C=8+7h$
$C=7+8h$
$C=15+8h$
$C=8+15h$

Show Answers Only

$B$

Show Worked Solution

$\text{Consider Option 2:}$

$7+8\times 1=7+8=15$

$7+8\times 2=7+16=23$

$7+8\times 3=7+24=31\ \ \ \ \text{etc …}$

$\text{The linear equation is:}\ \ C=7+8h$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 20

Brett uses matchsticks to make a pattern of shapes, as shown in the table below.

How many sticks ($S$) will be needed to make Shape Number 24? (2 marks)

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

Show Answers Only

$121$

Show Worked Solution

$\text{Shape 1:}\ \ S=5\times 1+1=6$

$\text{Shape 2:}\ \ S=5\times 2+1=11$

$\vdots$

$\text{Shape}\ N:\ \ S=5N+1$

$\therefore\ \text{Sticks required when}\ \ N=24$

$=24\times 5+1$

$=121$

v1 Algebra, STD2 A2 SM-Bank 10 MC

Petra drew a straight line through the points $(0, 4)$ and $(5, -3)$ as shown in the diagram below.

What is the gradient of the line that Petra drew?

$-\dfrac{7}{5}$
$-\dfrac{5}{7}$
$\dfrac{7}{5}$
$\dfrac{5}{7}$

Show Answers Only

$A$

Show Worked Solution

$\text{Line passes through } (0, 4)\ \text{and } (5, -3)$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{4- -3}{0-5}$
	$=-\dfrac{7}{5}$

$\Rightarrow A$

v1 Algebra, STD2 A2 SM-Bank 12 MC

Which of these equations represents the line in the graph?

$y=9-4x$
$y=9+4x$
$y=9-\dfrac{9}{4}x$
$y=9+\dfrac{9}{4}x$

Show Answers Only

$C$

Show Worked Solution

$\text{Graph passes through }(0, 9)\text{ and} (4, 0)$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{9-0}{0-4}$
	$=-\dfrac{9}{4}$

$\therefore\ \text{Equation is:}\ \ y=9-\dfrac{9}{4}x$

$\Rightarrow C$

v1 Algebra, STD2 A2 2020 HSC 6 MC

Suppose $y=-2-3x$.

When the value of $x$ increases by 4, the value of $y$ decreases by

$1$.
$4$.
$12$.
$14$.

Show Answers Only

$C$

Show Worked Solution

$\text{Strategy 1}$

$\text{If}\ \ x\ \ \text{increases by} \ 4$

$\rightarrow y\ \text{decreases by} \ \ 3x=3\times 4 = 12$

$\text{Strategy 2}$

$\text{Test}\ 2\ \text{values:}$

$\text{If} \ \ x=0 , \ y=-2$

$\text{If}\ \ x=4 , \ y =-2-3\times 4=-14$

$\therefore\ \ y \ \text{decreases by} \ 12.$

$\Rightarrow C$

v1 Algebra, STD2 A2 2021 HSC 9 MC

Marty is thinking of a number. Let the number be $n$.

When Marty subtracts 4 from this number and multiplies the result by 7, the answer is 8 more than $n$.

Which equation can be used to find $n$?

$7n-4=8n$
$7(n-4)=8n$
$7n-4=n+8$
$7(n-4)=n+8$

Show Answers Only

$D$

Show Worked Solution

$\text{The description defines the following equation:}$

$(n-4)\times 7$	$=n+8$
$7(n-4)$	$=n+8$

$\Rightarrow D$

v1 Algebra, STD2 A1 2012 HSC 15 MC

A car takes 5 hours to complete a journey when travelling at 75 km/h.

How long would the same journey take if the car were travelling at 100 km/h?

37.5 minutes
1 hour and 20 minutes
3 hours and 45 minutes
4 hours and 15 minutes

Show Answers Only

$C$

Show Worked Solution

$T=\dfrac{D}{S}$

$\text{Since}\ \ \ T = 5\ \ \text{when}\ \ \ S = 75$

$5$	$=\dfrac{D}{75}$
$D$	$=5\times 75$
	$=375\ \text{km}$

$\text{Find}\ \ T\ \ \text{when}\ \ \ S = 100\ \ \text{ and}\ \ \ D = 375$

$T$	$=\dfrac{375}{100}$
	$=3.75\ \text{hours}$
	$=3\ \text{hrs}\ \ 45\ \text{minutes}$

$\Rightarrow C$

v1 Algebra, STD2 A1 SM-Bank 1 MC

The blood alcohol content ($BAC$) of a male's blood is given by the formula;

$BAC_{\text{male}}=\dfrac{10N - 7.5H}{6.8M}$ , where

$N$ is the number of standard drinks consumed,

$H$ is the number of hours drinking and

$M$ is the person's mass in kgs.

Calculate the $BAC$ of a male who consumed 5 standard drinks in 2.5 hours and weighs 72 kgs, correct to 2 decimal places.

1.06
0.06
0.04
0.01

Show Answers Only

$B$

Show Worked Solution

$BAC_{\text{male}}$	$=\dfrac{10\times 5-7.5\times 2.5}{6.8\times 72}$
	$=\dfrac{31.25}{489.6}$
	$=0.0638\dots$

$\Rightarrow B$

v1 Algebra, STD2 A1 2014 HSC 29b

Blood alcohol content of males can be calculated using the following formula

$BAC_{\text{Male}} = \dfrac{10N-7.5H}{6.8M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

What is the maximum number of standard drinks that Jacko, who has a mass of 75 kg, can consume over 5 hours in order to maintain a blood alcohol content ($BAC$) of less than 0.05? (3 marks)

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

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$6$

Show Worked Solution

$BAC_\text{male}=\dfrac{10N-7.5H}{6.8M}$

$\text{Find}\ \ N\ \text{for }BAC<0.05,\ \text{given}\ \ H=5\ \text{and}\ \ M = 75$

$\dfrac{10N-7.5\times 5}{6.8\times 75}$	$< 0.05$
$10N-37.5$	$< 0.05\times 6.8\times 75$
$10N$	$< 25.5+37.5$
$10N$	$<63$
$\therefore\ N$	$< 6.3$

$\therefore\ \text{Max number of standard drinks is 6.}$

v1 Algebra, STD2 A1 2015 HSC 23 MC

The number of ‘standard drinks’ in various glasses of wine is shown.

Number of standard drinks
White Wine		Red Wine
small glass	large glass	small glass	large glass
0.9	1.4	1.0	1.5

A woman weighing 58 kg drinks two small glasses of white wine and three small glasses of red wine between 7 pm and 11 pm.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content ($BAC$) estimate at 11 pm, correct to three decimal places?

$BAC_{\text{Female}}=\dfrac{10N-7.5H}{5.5M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

0.013
0.023
0.046
0.056

Show Answers Only

$D$

Show Worked Solution

$N$	$=2\times 0.9 + 3\times 1$
	$=4.8\ \text{standard drinks}$
$H$	$=4\ \text{hours}$
$M$	$=58\ \text{kg}$

$BAC_f$	$=\dfrac{10\times 4.8-7.5\times 4}{5.5\times 58}$
	$=0.05642\dots$

$\Rightarrow D$

v1 Algebra, STD2 A1 2005 HSC 24b

The formula $D=\dfrac{2A}{15}$ is used to calculate the dosage of liquid paracetamol to be given to a child.

- $D$ is the dosage of liquid paracetamol in millilitres (mL).
- $A$ is the age of the child in months.

If Charlotte is six months old, what dosage of liquid paracetamol should she be given? (1 mark)

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

The correct dosage of liquid paracetamol for Teddy is 6 mL.

What is the difference in the ages of Teddy and Charlotte, in months? (3 marks)

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

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i. $\text{0.8 mL}$

ii. $39$

Show Worked Solution

i.	$D$	$=\dfrac{2A}{15}$
		$=\dfrac{2\times 6}{15}$
		$=0.8\text{ mL}$

$\therefore\ \text{Charlotte should be given a dosage of 0.8 mL}$

ii. $\text{Find}\ A\ \text{when}\ D=\text{6 mL}$

$6$	$=\dfrac{2A}{15}$
$2A$	$=90$
$A$	$=45$

$\therefore\ \text{Teddy is 45 months old and is 39 months}$

$\text{older than Charlotte.}$

v1 Algebra, STD2 A1 2016 HSC 10 MC

Anika drinks two small bottles of wine over a four-hour period. Each of these bottles contains 2.4 standard drinks. Anika weighs 55 kg.

Using the formula below, what is Anika's approximate blood alcohol content ($BAC$) at the end of this period?

$BAC_{\text{Female}}=\dfrac{10N - 7.5H}{5.5M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

0.013
0.060
0.0013
0.0060

Show Answers Only

$B$

Show Worked Solution

$BAC_f$	$=\dfrac{10N – 7.5H}{5.5M}$
	$=\dfrac{10(2\times 2.4) – 7.5\times 4}{5.5\times 55}$
	$= 0.0595\dots\approx 0.060$

$\Rightarrow B$

v1 Algebra, STD2 A1 2017 HSC 19 MC

Young’s formula, shown below, is used to calculate the dosage of medication for children aged 1−12 years based on the adult dosage.

$D=\dfrac{yA}{y + 12}$

where $D$	= dosage for children aged 1−12 years
$y$	= age of child (in years)
$A$	= Adult dosage

A child’s dosage is calculated to be 15 mg, based on an adult dosage of 30 mg.

How old is the child in years?

Show Answers Only

$D$

Show Worked Solution

$D$	$=\dfrac{yA}{y+12}$
$15$	$=\dfrac{30y}{y+12}$
$15(y+12)$	$=30y$
$15y+180$	$=30y$
$15y$	$=180$
$y$	$=12$

$\Rightarrow D$

v1 Algebra, STD2 A1 2017 HSC 27e

Bryce is drinking low alcohol beer at a party over a four-hour period. He reads on the label of the low alcohol beer bottle that it is equivalent to 0.8 standard drinks.

Bryce weighs 85 kg.

The formula below can be used to calculate a male's blood alcohol content.

$BAC_{\text{Male}}=\dfrac{10N-7.5H}{6.8M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

What is the maximum number of complete bottles of the low alcohol beer Bryce can drink to remain under a Blood Alcohol Content ($BAC$) of 0.05? (4 marks)

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

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$7$

Show Worked Solution

$BAC_\text{male}$	$=\dfrac{10N-7.5H}{6.8M}$
$0.05$	$=\dfrac{10N-7.5\times 4}{6.8\times 85}$
$10N$	$=0.05\times 6.8\times 85+7.5\times 4$
$10N$	$=58.9$
$N$	$=5.89\ \text{standard drinks}$

$\therefore\ \text{Number of low alcohol bottles}$

$=\dfrac{5.89}{0.8}$

$=7.3625$

$\therefore\ \text{Max complete bottles to stay under 0.05}$

$=7$

v1 Algebra, STD2 A1 2019 HSC 28

The formula below is used to calculate an estimate for blood alcohol content $(BAC)$ for females.

$BAC_{\text{female}}=\dfrac{10N - 7.5H}{5.5M}$

The number of hours required for a person to reach zero $BAC$ after they stop consuming alcohol is given by the following formula.

$\text{Time}=\dfrac{BAC}{0.015}$

The number of standard drinks in a glass of wine and a glass of spirits is shown.

Georgie weighs 58 kg. She consumed 2 glasses of wine and 4 glasses of spirits between 7:45 pm and 12:15 am the following day. She then stopped drinking alcohol.

Using the given formulae, calculate the time in the morning when Georgie's $BAC$ should reach zero. (4 marks)

Show Answers Only

$\text{6:34 am}$

Show Worked Solution

$\text{Standard drinks consumed}\ (N)=2\times 1.2+4=6.4$

$\text{Hours drinking}\ (H) = \text{4 h 30 min = 4.5 hours}$

$BAC_{\text{Georgie}}$	$=\dfrac{10\times 6.4-7.5\times 4.5}{5.5\times 58}$
	$=0.09482\dots$

COMMENT: Convert a decimal answer into hours and minutes using the calculator degree/minute function.

$\text{Time (to zero)}$	$=\dfrac{0.09482\dots}{0.015}$
	$=6.3218\dots\ \text{hours}$
	$\approx 6\ \text{hours 19 minutes}$

$\therefore\ \text{Georgie should reach zero}\ BAC$

$=12:15+6:19=6:34\ \text{am}$

v1 Algebra, STD2 A1 SM-Bank 2 MC

Frank lives 45 kilometres from his work.

On Monday, he drove to work and averaged 60 kilometres per hour.

On Wednesday, he took the train which averaged 90 kilometres per hour.

What was the extra time of the car journey on Monday, in minutes, compared to when he caught the train on Wednesday?

Show Answers Only

$A$

Show Worked Solution

$\text{Time}=\dfrac{\text{distance}}{\text{speed}}$

$\text{Time on Monday}$	$=\dfrac{45}{60}$
	$=0.75\ \text{hour}$
	$=45\ \text{minutes}$

$\text{Time on Wednesday}$	$=\dfrac{45}{90}$
	$=0.5\ \text{hour}$
	$=30\ \text{minutes}$

$\therefore\ \text{The extra time driving the car}=45-30=15\ \text{minutes}$

$\Rightarrow A$

v1 Algebra, STD2 A1 2023 HSC 36

The following formula can be used to calculate an estimate for blood alcohol content ($BAC$) for males.

$BAC_{\text{male}}=\dfrac{10N-7.5H}{6.8M}$

$N$ is the number of standard drinks consumed

$M$ is the person's weight in kilograms

$H$ is the number of hours of drinking

Min weighs 70 kg. His $BAC$ was zero when he began drinking alcohol. At 10:30 pm, after consuming 4 standard drinks, his $BAC$ was 0.032.

Using the formula, estimate at what time Min began drinking alcohol, to the nearest minute. (4 marks)

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

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$7:12\text{ pm}$

Show Worked Solution

$BAC$	$=\dfrac{10N-7.5H}{6.8M}$
$0.032$	$=\dfrac{10\times 4-7.5\times H}{6.8\times 70}$
$0.032\times 476$	$=40-7.5H$
$7.5H$	$=40-15.232$
$H$	$=\dfrac{24.768}{7.5}$
	$=3.3024\ \text{hours}$
	$\approx 3\ \text{hours}\ 18\ \text{minutes (nearest minute)}$

$\text{Time Min began drinking}$

$=10:30\text{ pm – 3 h 18 m}$

$=7:12\text{ pm}$

Mean mark 56%.

v1 Algebra, STD2 A1 2010 HSC 24a

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

\begin{array}{rl}
3(m+3)-2(m+4)=-5\ &\ \ \ \text{Line 1} \\
3m+9-2m+8=-5\ &\ \ \ \text{Line 2} \\
m+17=-5\ &\ \ \ \text{Line 3} \\
m=-12& \ \ \ \text{Line 4}
\end{array}

Copy the equation in Line 1. Rewrite Line 2 correcting her mistake. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Continue your solution showing the correct working for Lines 3 and 4 to solve this equation for $m$. (1 mark)

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

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i.	$3(m+3)-2(m+4)$	$=-5\ \ \ \ \ \ \text{Line}\ 1$
	$3m+9-2m-8$	$=-5\ \ \ \ \ \ \text{Line}\ 2$

ii.	$m+1$	$=-5$
	$m$	$=-6$

Show Worked Solution

i.	$3(m+3)-2(m+4)$	$=-5\ \ \ \ \ \ \text{Line}\ 1$
	$3m+9-2m-8$	$=-5\ \ \ \ \ \ \text{Line}\ 2$

ii.	$m+1$	$=-5\ \ \ \ \ \ \text{Line}\ 3$
	$m$	$=-6\ \ \ \ \ \ \text{Line}\ 4$

v1 Algebra, STD2 A1 2009 HSC 25a

Simplify $10-3(x+4)$. (2 marks)

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

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$-3x-2$

Show Worked Solution

♦ Mean mark 47%

$10-3(x+4)$	$=10-3x-12$
	$=-3x-2$

v1 Algebra, STD2 A1 2014 HSC 26c

Solve the equation $\dfrac{4x-3}{5}-6=7-6x$. (3 marks)

Show Answers Only

$x=2$

Show Worked Solution

$\dfrac{4x-3}{5}-6$	$=7-6x$
$4x-3-5\times 6$	$=5(7-6x)$
$4x-3-30$	$=35-30x$
$34x$	$=68$
$\therefore\ x$	$=2$

v1 Algebra, STD2 A1 2008 HSC 9 MC

What is the value of $\sqrt{\dfrac{2x + y}{5x}}$ if $x=5.1$ and $y=3.7$, correct to 2 decimal places?

$0.13$
$0.74$
$3.74$
$3.80$

Show Answers Only

$B$

Show Worked Solution

$\sqrt{\dfrac{2x+y}{5x}}$	$=\sqrt{\dfrac{2\times 5.1+3.7}{5\times 5.1}}$
	$=\sqrt{\dfrac{13.9}{25.5}}$
	$= 0.7383\dots$

$\Rightarrow B$

EXAMCOPY Algebra, STD2 A1 2004 HSC 11 MC

If `d = 6t^2`, what is a possible value of `t` when `d = 2400`?

`0.05`
`20`
`120`
`400`

Show Answers Only

`B`

Show Worked Solution

`d`	`= 6t^2`
`t^2`	`= d/6`
`t`	`= +- sqrt(d/6)`

`text(When)\ \ d = 2400:`

`t`	`= +- sqrt(2400/6)`
	`= +- 20`

`=> B`

v1 Algebra, STD2 A1 2005 HSC 14 MC

Using the formula $d=6t^3-5$, Marcia tried to find the value of $t$ when $d=389$.

Here is her solution. She has made one mistake.

Which line does NOT follow correctly from the previous line?

$\text{Line}\ A$
$\text{Line}\ B$
$\text{Line}\ C$
$\text{Line}\ D$

Show Answers Only

$B$

Show Worked Solution

$d$	$=6t^3-5$
$389$	$=6t^3-5\ \ \ $	$\dots\text{ Line A}$
$394$	$=6t^3$	$\dots\text{ Line B}$

$\therefore\ \text{Line}\ B\ \text{does not follow on correctly.}$

$\Rightarrow B$

i.	\($C\)	\(=550+250+(9\times x)\)
		\(=800+9x\)

\(\text{Total Volume}\)	\(=\text{Tank A} + \text{Tank B}\)
\(1400\)	\(=1400-40t+(t-10)\times 60\)
\(1400\)	\(=1400-40t+60t-600\)
\(20t\)	\(= 600\)
\(t\)	\(= 30 \ \text{minutes}\)

\(60n\)	\(=35n+5000\)
\(25n\)	\(=5000\)
\(n\)	\(=200\)

\(\text{Graphing}\ \ 8x + 4y\)	\(=40\)
\(y\)	\(=-2x+10\)

\(\text{LHS}\)	\(=3\)
\(\text{RHS}\)	\(= 2(4)-5\)
	\(=3\)

\(\text{Income}\)	\(=200\times $15\)
	\(=$3000\)

\(\text{Costs}\)	\(=800+(9\times 200)\)
	\(=$2600\)

\(\therefore\ \text{Profit}\)	\(=3000-2600\)
	\(=$400\)

\(C\)	\(=800+(9\times 300)\)
	\(=$3500\)