Band 5 – Page 4

BIOLOGY, M1 EQ-Bank 9

Products of an enzyme-controlled reaction can sometimes inhibit the enzyme that produced them.

Discuss how this process can be advantageous for cellular metabolism. (3 marks)

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→ When products of an enzyme-controlled reaction inhibit the enzyme that produced them, it regulates the cell’s metabolism (in a process known as feedback).

→ This process prevents the accumulation of excess products, ensuring that resources like ATP and raw materials are used efficiently.

→ By controlling the enzyme activity, the cell can balance production according to its needs, conserving energy and maintaining metabolic stability.

Show Worked Solution

→ When products of an enzyme-controlled reaction inhibit the enzyme that produced them, it regulates the cell’s metabolism (in a process known as feedback).

→ This process prevents the accumulation of excess products, ensuring that resources like ATP and raw materials are used efficiently.

→ By controlling the enzyme activity, the cell can balance production according to its needs, conserving energy and maintaining metabolic stability.

BIOLOGY, M1 EQ-Bank 7

How do temperature and pH affect enzyme activity? In your answer, briefly explain how extreme conditions of each factor influence enzyme function. (4 marks)

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→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

Show Worked Solution

→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

BIOLOGY, M1 EQ-Bank 4

Compare the movement of a lipid-soluble substance and water across the cell membrane.

In your answer, explain how the structure of the membrane affects the transport of these molecules. (2 marks)

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→ Lipid-soluble substances can pass directly through the phospholipid bilayer by simple diffusion because the cell membrane contains lipids.

→ In contrast, water is unable to penetrate the cell through the phospholipid bilayer.

→ Instead, water molecules must pass through special protein channels (pores) in the cell membrane.

Show Worked Solution

→ Lipid-soluble substances can pass directly through the phospholipid bilayer by simple diffusion because the cell membrane contains lipids.

→ In contrast, water is unable to penetrate the cell through the phospholipid bilayer.

→ Instead, water molecules must pass through special protein channels (pores) in the cell membrane.

BIOLOGY, M1 EQ-Bank 4 MC

In active transport, how does the cell maintain a steep concentration gradient across the membrane?

By using ion channels to allow the passive diffusion of ions along the gradient.
By using carrier proteins and ATP to pump ions from an area of low concentration to an area of high concentration.
By allowing water molecules to diffuse freely through the membrane, balancing the concentration.
By increasing the surface area of the membrane to allow more diffusion of molecules.

Show Answers Only

$B$

Show Worked Solution

→ Active transport requires energy (ATP) to move ions or molecules against their concentration gradient, often with the help of carrier proteins.

→ This process helps cells maintain steep concentration gradients essential for functions like nerve impulse transmission and muscle contraction.

$\Rightarrow B$

BIOLOGY, M1 EQ-Bank 8

Explain how lysosomes contribute to the maintenance of normally functioning cells, and describe what happens when the function of lysosomes is impaired. (2 marks)

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→ Lysosomes maintain cellular homeostasis by breaking down waste materials, damaged organelles, and cellular debris using enzymes.

→ This prevents the accumulation of harmful substances.

→ When lysosomal function is impaired, waste products and damaged components build up, leading to cellular dysfunction and potentially contributing to diseases that impair cellular efficiency and overall health.

Show Worked Solution

→ Lysosomes maintain cellular homeostasis by breaking down waste materials, damaged organelles, and cellular debris using enzymes.

→ This prevents the accumulation of harmful substances.

BIOLOGY, M1 EQ-Bank 9 MC

Which of the following correctly describes the roles of lysosomes and the Golgi apparatus?

Lysosomes break down waste using enzymes, and the Golgi apparatus transports substances across the membrane.
Lysosomes synthesize lipids, and the Golgi apparatus modifies and packages proteins for transport.
Lysosomes break down waste using enzymes, and the Golgi apparatus modifies and packages proteins for transport.
Lysosomes produce energy for the cell, and the Golgi apparatus breaks down cellular waste.

Show Answers Only

$C$

Show Worked Solution

→ Lysosomes contain digestive enzymes that break down cellular waste and debris.

→ Golgi apparatus is responsible for modifying, sorting, and packaging proteins for secretion or use within the cell.

$\Rightarrow C$

BIOLOGY, M1 EQ-Bank 5

According to the fluid mosaic model of the cell membrane

describe the structure of the phospholipid bilayer. (2 marks)

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explain the role of proteins. (1 mark)

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a. The phospholipid bilayer:

→ Consists of two layers of phospholipids.

→ One layer involves hydrophilic (water-attracting) heads facing outward toward the water-based environment while the opposite layer has hydrophobic (water-repelling) tails facing inward, away from water.

b. Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

Show Worked Solution

a. The phospholipid bilayer:

→ Consists of two layers of phospholipids.

b. Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

BIOLOGY, M5 2020 VCE 11a

A student wanted to investigate the effect of two different endonucleases (restriction enzymes) on a linear DNA fragment.

The student used three tubes containing a buffered solution of linear DNA fragments, each fragment being 9500 base pairs in length.

Two different endonucleases were available: BamHI and HindIII.

The student followed the steps below.

After 45 minutes the student obtained the results shown below

	Tube 1	Rube 2	Tube 3
DNA	BamHI	HindIII	BamHI
ladder			HindIII

Analyse the results of the experiment performed by the student. (5 marks)

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Answers could include 5 of the following:

→ Smaller DNA fragments tend to migrate further during gel electrophoresis
compared to larger fragments.

→ A standard or marker is used as a point of comparison and reference to
determine the size of unknown DNA fragments.

→ The restriction enzyme BamH1 cuts the DNA molecule at a single location,
resulting in the formation of two DNA fragments.

→ The restriction enzyme HindIII cuts the DNA molecule at two distinct
locations, producing three DNA fragments.

→ When the DNA is cut with the HindIII enzyme, it results in the largest and
smallest DNA fragments compared to other restriction enzyme digestions.

→ The DNA fragments generated by the BamH1 enzyme are 4000 bp and
5500 bp in length, while the fragments produced by HindIII are 8000 bp,
1000 bp, and 500 bp.

→ When the DNA is digested with both BamH1 and HindIII enzymes, a total
of four DNA fragments are produced, with sizes of 5500 bp, 2500 bp,
1000 bp, and 500 bp.

Show Worked Solution

Answers could include 5 of the following:

→ Smaller DNA fragments tend to migrate further during gel electrophoresis
compared to larger fragments.

→ A standard or marker is used as a point of comparison and reference to
determine the size of unknown DNA fragments.

→ The restriction enzyme BamH1 cuts the DNA molecule at a single location,
resulting in the formation of two DNA fragments.

→ The restriction enzyme HindIII cuts the DNA molecule at two distinct
locations, producing three DNA fragments.

→ When the DNA is cut with the HindIII enzyme, it results in the largest and
smallest DNA fragments compared to other restriction enzyme digestions.

→ The DNA fragments generated by the BamH1 enzyme are 4000 bp and
5500 bp in length, while the fragments produced by HindIII are 8000 bp,
1000 bp, and 500 bp.

→ When the DNA is digested with both BamH1 and HindIII enzymes, a total
of four DNA fragments are produced, with sizes of 5500 bp, 2500 bp,
1000 bp, and 500 bp.

♦♦ Mean mark 50%.
COMMENT: 39% of responses received 0 or 1 mark

BIOLOGY, M5 2020 VCE 7 MC

The codon table below can be used to determine amino acids coded for by a nucleotide sequence.

$ \textbf{lst position} $ $ \textbf{(5}^{′}\ \textbf{end)} $ $ \boldsymbol{\downarrow} $	$ \textbf{2nd position} $				$ \textbf{3rd position} $ $ \ \ \textbf{(3}^{′}\ \textbf{end)} $ $ \boldsymbol{\downarrow} $
	$ \textbf{U} $	$ \textbf{C} $	$ \textbf{A} $	$ \textbf{G} $
$ \textbf{U} $	$ \text{Phe}$	$ \text{Ser}$	$ \text{Tyr}$	$ \text{Cys}$	$ \textbf{U} $
	$ \text{Phe}$	$ \text{Ser}$	$ \text{Tyr}$	$ \text{Cys}$	$ \textbf{C} $
	$ \text{Leu}$	$ \text{Ser}$	$ \text{STOP}$	$ \text{STOP}$	$ \textbf{A} $
	$ \text{Leu}$	$ \text{Ser}$	$ \text{STOP}$	$ \text{Trp}$	$ \textbf{G} $
$ \textbf{C} $	$ \text{Leu}$	$ \text{Pro}$	$ \text{His}$	$ \text{Arg}$	$ \textbf{U} $
	$ \text{Leu}$	$ \text{Pro}$	$ \text{His}$	$ \text{Arg}$	$ \textbf{C} $
	$ \text{Leu}$	$ \text{Pro}$	$ \text{Gln}$	$ \text{Arg}$	$ \textbf{A} $
	$ \text{Leu}$	$ \text{Pro}$	$ \text{Gln}$	$ \text{Arg}$	$ \textbf{G} $
$ \textbf{A} $	$ \text{Ile}$	$ \text{Thr}$	$ \text{Asn}$	$ \text{Ser}$	$ \textbf{U} $
	$ \text{Ile}$	$ \text{Thr}$	$ \text{Asn}$	$ \text{Ser}$	$ \textbf{C} $
	$ \text{Ile}$	$ \text{Thr}$	$ \text{Lys}$	$ \text{Arg}$	$ \textbf{A} $
	$ \text{Met}$	$ \text{Thr}$	$ \text{Lys}$	$ \text{Arg}$	$ \textbf{G} $
$ \textbf{G} $	$ \text{Val}$	$ \text{Ala}$	$ \text{Asp}$	$ \text{Gly}$	$ \textbf{U} $
	$ \text{Val}$	$ \text{Ala}$	$ \text{Asp}$	$ \text{Gly}$	$ \textbf{C} $
	$ \text{Val}$	$ \text{Ala}$	$ \text{Glu}$	$ \text{Gly}$	$ \textbf{A} $
	$ \text{Val}$	$ \text{Ala}$	$ \text{Glu}$	$ \text{Gly}$	$ \textbf{G} $

It is correct to state

identical amino acid sequences are found in all organisms.
the genetic code is degenerate with respect to Met.
the codon GGU adds Trp to a polypeptide chain.
the DNA template sequence GAA codes for Leu.

Show Answers Only

$D$

Show Worked Solution

Consider each option:

Option A: The codon table shows that some codons code for different amino acids in
certain alternative translation systems → Incorrect

Option B: The genetic code is degenerate, meaning multiple codons can code for the
same amino acid, but this is not specific to Met → Incorrect

Option C: According to the codon table, the codon GGU codes for the amino acid
glycine (Gly), not tryptophan (Trp). The codon for Trp is UGG → Incorrect

Option D: Looking at the first position G, second position A, and third position A,
the codon GAA codes for the amino acid leucine (Leu) → Correct

$\Rightarrow D$

♦♦ Mean mark 41%.
NOTE: 55% of students incorrectly chose A or B.

v1 Algebra, STD2 A4 2010 HSC 24b

Damo hires paddle boats in summertime as part of his water sports business. To calculate the cost, $C$, in dollars, of hiring $x$ paddle boats, he uses the equation $C=40+25x$.

He hires the paddle boats for $35 per hour and determines his income, $I$, in dollars, using the equation $I=35x$.

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

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$x=4\ \ \text{See worked solution}$

Show Worked Solution

$\text{From the graph, intersection occurs at}\ x=4$

$\rightarrow\ \text{Break-even point occurs at}\ x=4$

$\text{i.e. when 4 hours of paddle board hire occurs}$

$\text{Income}$	$=35\times 4=$140\ \ \text{is equal to}$
$\text{Costs}$	$=40+(25\times 4)=$140$

$\text{If}\ <4\ \text{hours of board hire}\ \rightarrow\ \text{LOSS for business}$

$\text{If}\ >4\ \text{hours of board hire}\ \rightarrow\ \text{PROFIT}$

♦ 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 $x$.

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 30 MC

The short-beaked echidna can be found in Australia and New Guinea.

A university Biology student set out to compare DNA from four different short-beaked echidnas living in four different locations. The student aimed to determine the country of origin of each echidna. This was achieved using primers that amplified a 430 base pair (bp) sequence from the echidna's mitochondrial genome.

A flow chart of the steps taken by the student is shown below.

Which one of the following would be a requirement of either Step 1 or Step 2?

Step 2 would require the use of a thermocycling machine to maintain the temperature at 37 °C.
Step 1 would require the quill samples to have been stored at a very high temperature.
Step 1 would require the extraction of DNA from the nuclei of quill epithelial cells.
Step 2 would usually require a reaction mixture containing two primers.

Show Answers Only

$D$

Show Worked Solution

→ Polymerase chain reaction (PCR), step 2, needs two primers for the amplification of DNA.

→ Step 1 is the isolation of mtDNA, which is not present in the nucleus.

$\Rightarrow D$

♦ Mean mark 46%.
COMMENT: 31% of students chose option C.

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%.

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 2019 HSC 36

A small business makes and dog kennels.

Technology was used to draw straight-line graphs to represent the cost of making the dog kennels $(C)$ and the revenue from selling dog kennels $(R)$. The $x$-axis displays the number of dog kennels and the $y$-axis displays the cost/revenue in dollars.

How many dog kennels need to sold to break even? (1 mark)

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By first forming equations for cost `(C)` and revenue `(R)`, determine how many dog kennels need to be sold to earn a profit of $2500. (3 marks)

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Show Answers Only

$20$
$145$

Show Worked Solution

a. $20\ \ (x\text{-value at intersection})$

b. $\text{Find equations of both lines}:$

$(0, 400)\ \text{and}\ (20, 600)\ \text{lie on}\ \ C$

$\text{gradient}_C = \dfrac{600-400}{20-0}=10$

$\rightarrow\ C=400+10x$

$(0,0)\ \text{and}\ (20, 600)\ \text{lie on}\ \ R$

$\text{gradient}_R =\dfrac{600-0}{20-0}=30$

$\rightarrow\ R=30x$

$\text{Profit} = R-C$

$\text{Find}\ \ x\ \text{when Profit }= $2500:$

$2500$	$=30x-(400+10x)$
$20x$	$=2900$
$x$	$=145$

$\therefore\ 145\ \text{dog kennels need to be sold to earn }$2500\ \text{profit}$

♦♦ Mean mark 28%.

v1 Algebra, STD2 A4 2004 HSC 16 MC

Uri drew a correct diagram that gave the solution to the simultaneous equations

$y=2x+3$ and $y=x+4$.

Which diagram did he draw?

Show Answers Only

$D$

Show Worked Solution

$\text{By elimination:}$

$y=2x+3\ \text{cuts the }y \text{-axis at}\ 3$

$\rightarrow\ \text{Eliminate be A and B}$

$y=x+4\ \text{cuts the }y\text{-axis at}\ 4$

$\text{AND has a positive gradient}$

$\rightarrow\ \text{Eliminate C}$

$\Rightarrow D$

v1 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:

$C=20\ 000+40x$

The centre charges $120 per person. Its income $I$, in dollars, is given by:

$I=120x$

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

$$10\ 000$
$$20\ 000$
$$30\ 000$
$$40\ 000$

Show Answers Only

$C$

Show Worked Solution

$\text{When}\ x=500,\ I=120\times 500=$60\ 000$

$\text{Breakeven when}\ \ x=250\ \ \text{(from graph)}$

$\text{When}\ \ x=250,\ I=120\times 250=$30\ 000$

$\text{Difference}$	$=60\ 000-30\ 000$
	$=$30\ 000$

$\Rightarrow C$

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

v1 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

$\text{Initial Volume}$	$=50\ 000\ \text{L}$
$\text{Final volume}$	$=15\ 000\ \text{L}$
$\text{Decrease}$	$=50\ 000-15\ 000$
	$=35\ 000\ \text{L (over 6 months)}$

$\text{Loss per month}$	$=\dfrac{35\ 000}{6}$
	$=5833.33\dots\ \text{L per month}$
$\text{% loss per month}$	$=\dfrac{5833.33\dots}{50\ 000}\times 100\%$
	$=11.666\dots \%$

$\Rightarrow B$

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

v1 Algebra, STD2 A2 2009 HSC 24d

A factory makes both cloth and leather lounges. In any week

• the total number of cloth lounges and leather lounges that are made is 400
• the maximum number of leather lounges made is 270
• the maximum number of cloth lounges made is 325.

The factory manager has drawn a graph to show the numbers of leather lounges ($x$) and cloth lounges ($y$) that can be made.

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

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Explain why this line is only relevant between $B$ and $C$ for this factory. (1 mark)

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The profit per week, $$P$, can be found by using the equation $P = 2520x + 1570y$.

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

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Show Answers Only

$x+y=400$
$\text{Since the max amount of leather lounges}=270$

$\rightarrow\ x\ \text{cannot be}\ >270$

$\text{Since the max amount of cloth lounges}=325$

$\rightarrow\ y\ \text{cannot be}\ >325$

$\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.$
$\text{The profits at}\ C\ \text{are }$185\ 250\ \text{more than at}\ B.$

Show Worked Solution

i. $\text{We are told the number of leather lounges}\ (x),$

$\text{and cloth lounges}\ (y),\ \text{made in any week} = 400$

$\rightarrow\ \text{Equation of}\ AD\ \text{is}\ x+y=400$

♦♦♦ Mean mark part (i) 14%.
Using $y=mx+c$ is a less efficient but equally valid method, using $m=–1$ and $b=400$ ($y$-intercept).

ii. $\text{Since the max amount of leather lounges}=270$

$\rightarrow\ x\ \text{cannot}\ >270$

$\text{Since the max amount of cloth lounges}=325$

$\rightarrow\ y\ \text{cannot}\ >325$

$\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.$

♦ Mean mark part (ii) 49%.

iii. $\text{At}\ B,\ x=75,\ y=325$

$\rightarrow\ $P (\text{at}\ B)$	$=2520\times 75+1570\times 325$
	$=189\ 000+510\ 250$
	$=$699\ 250$

$\text{At}\ C,\ x=270,\ y=130$

$\rightarrow\ $P (\text{at}\ C)$	$=2520\times 270+1570\times 130$
	$=680\ 400+204\ 100$
	$=$884\ 500$

$\text{Difference in profits}=$884\ 500-$699\ 250=$185\ 250$

$\text{The profits at}\ C\ \text{are } $185\ 250\ \text{more than at}\ B.$

♦ Mean mark (iii)40%.

v1 Algebra, STD2 A2 2010 HSC 27c

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

Use the graph to find the tax payable on a taxable income of $$18\ 000$. (1 mark)

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Use suitable points from the graph to show that the gradient of the section of the graph marked $A$ is $\dfrac{7}{15}$. (1 mark)

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How much of each dollar earned between $$18\ 000$ and $$33\ 000$ is payable in tax? Give your answer correct to the nearest whole number. (1 mark)

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Write an equation that could be used to calculate the tax payable, $T$, in terms of the taxable income, $I$, for taxable incomes between $$18\ 000$ and $$33\ 000$. (2 marks)

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Show Answers Only

$$3000\ \ \text{(from graph)}$
$\text{See worked solution}$
$46\frac{2}{3}\approx 47\ \text{cents per dollar earned}$
$\text{Tax payable →}\ T=\dfrac{7}{15}I-5400$

Show Worked Solution

$\text{Income on}\ $18\ 000=$3000\ \ \text{(from graph)}$

ii. $\text{Using the points}\ (18, 3)\ \text{and}\ (33, 10)$

$\text{Gradient at}\ A$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{10\ 000-3000}{33\ 000-18\ 000}$
	$=\dfrac{7000}{15\ 000}$
	$=\dfrac{7}{15}\ \ \ \ \text{… as required}$

♦♦ Mean mark (ii) 25%.

iii. $\text{The gradient represents the tax applicable on each dollar}$

$\text{Tax}$	$=\dfrac{7}{15}\ \text{of each dollar earned}$
	$=46\frac{2}{3}\approx 47\ \text{cents per dollar earned (nearest whole number)}$

♦♦♦ Mean mark (iii) 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

iv. $\text{Tax payable up to }$18\ 000 = $3000$

$\text{Tax payable on income between }$18\ 000\ \text{and }$33\ 000$

$=\dfrac{7}{15}(I-18\ 000)$

$\therefore\ \text{Tax payable →}\ \ T$	$=3000+\dfrac{7}{15}(I-18\ 000)$
	$=3000+\dfrac{7}{15} I-8400$
	$=\dfrac{7}{15}I-5400$

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

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)

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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)

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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)

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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 2004 HSC 22 MC

Mary-Anne knows that

• one Australian dollar (AUD) is worth 0.64 euros, and
• one Canadian dollar (CAD) is worth 0.97 euros.

Mary-Anne changes 75 AUD to Canadian dollars.

How many Canadian dollars will she get?

46.56 CAD
49.48 CAD
113.67 CAD
120.75 CAD

Show Answers Only

$B$

Show Worked Solution

$\text{Mary-Anne has 75 AUD.}$

$\text{Converting to Euros}$

$25\ \text{AUD}$	$=75\times 0.64$
	$=48\ \text{Euros}$

$\text{Converting to CAD}$

$48\ \text{euros}$	$=\dfrac{48}{0.97}$
	$=49.484\dots$
	$=49.48\ \text{CAD}$

$\Rightarrow B$

v1 Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2010.

According to the graph, what is the life expectancy of a person born in 1968? (1 mark)

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With reference to the value of the gradient, explain the meaning of the gradient in this context. (2 marks)

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Show Answers Only

$\text{76 years}$
$\text{After 1900, life expectancy increases by 0.38 years for}$
$\text{each year later that someone is born.}$

Show Worked Solution

i. $\text{76 years}$

ii. $\text{Using (2000, 88) and (1900, 50):}$

$\text{Gradient}$	$= \dfrac{y_2-y_1}{x_2-x_1}$
	$= \dfrac{88-50}{2000-1900}$
	$= 0.38$

$\text{After 1900, life expectancy increases by 0.38 years for}$

$\text{each year later that someone is born.}$

♦♦ Mean mark (ii) 33%.

v1 Algebra, STD2 A2 SM-Bank 2

The cost of apples per kilogram, $C$, varies directly with the weight of apples purchased, $w$.

If 12 kilograms costs $56.64, calculate the cost of 4.5 kilograms of apples. (2 marks)

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Show Answers Only

$$21.24$

Show Worked Solution

$C\ \propto \ w$

$C=kw$

$\text{When}\ C=$56.64\ \text{kg},\ w=12\ \text{kg}$

$56.64$	$=k\times 12$
$k$	$=\dfrac{56.64}{12}$
	$=$4.72$

$\text{When}\ \ w=4.5\ \text{kg,}$

$C$	$=4.72\times 4.5$
	$=$21.24$

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)

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Give ONE reason why this equation is not suitable for predicting heights of boys older than 10. (1 mark)

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

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 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)

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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)

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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 A3 2021 HSC 25

The diagram shows a container which consists of a large hexagonal prism on top of a smaller hexagonal prism.

The container is filled with water at a constant rate into the top of the larger hexagonal prism.

The smaller prism is totally filled before the larger prism begins to fill.

It takes 5 minutes to fill the smaller cylinder.

Draw a possible graph of the water level in the container against time. (2 marks)

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Show Answers Only

Show Worked Solution

♦♦ Mean mark 38%.

v1 Algebra, STD2 A2 2020 HSC 10 MC

An electrician charges a call-out fee of $75 as well as $1.50 per minute while working.

Suppose the electrician works for $t$ hours.

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

$C=75+1.50t$
$C=150+75t$
$C=75+90t$
$C=90+75t$

Show Answers Only

$C$

Show Worked Solution

$\text{Hourly rate}=60\times 1.50=$90$

$\therefore\ C=75+90t$

$\Rightarrow C$

♦ Mean mark 42%.

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)

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A bundle of photographic paper has a height of 36 centimetres. Calculate the number of sheets of photographic paper in the bundle. (2 marks)

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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 A2 2012 HSC 5 MC

The line below has intercepts $m$ and $n$, where $m$ and $n$ are positive integers.

What is the gradient of the line?

$\dfrac{m}{n}$
$\dfrac{n}{m}$
$-\dfrac{m}{n}$
$-\dfrac{n}{m}$

Show Answers Only

$C$

Show Worked Solution

♦ Mean mark 45%

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=-\dfrac{m}{n}$

$\Rightarrow C$

v1 Algebra, STD2 A2 2009 HSC 14 MC

If $C=5x+4$, and $x$ is increased by 3, what will be the corresponding increase in $C$ ?

$3$
$15$
$3x$
$5x$

Show Answers Only

$B$

Show Worked Solution

$C=5x+4$

$\text{If}\ x\ \text{increases by 3}$

$C\ \text{increases by}\ 5\times 3=15$

$\Rightarrow B$

♦ Mean mark 50%.
STRATEGY: Substituting real numbers into the equation can work well in these type of questions. eg. If $x=0,\ C=4$ and when $x=3,\ C=19$.

v1 Algebra, STD2 A2 2014 HSC 7 MC

Which of the following is the graph of $y=-3x-3$?

A.		B.

C.		D.

Show Answers Only

$A$

Show Worked Solution

♦ Mean mark 46%

$y=-3x-3$

$\text{By elimination:}$

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

$\rightarrow\ \text{Cannot be}\ B\ \text{or}\ D$

$\text{Gradient}=-3$

$\rightarrow\ \text{Cannot be}\ C$

$\Rightarrow A$

EXAMCOPY 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`

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)

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How many sticks would be required for Shape 128? (1 mark)

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Is it possible to create a shape in this pattern using exactly 609 sticks?

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

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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 2005 HSC 17 MC

The total cost, $$C$, of a school excursion is given by $C=4n+9$, where $n$ is the number of students.

If five extra students go on the excursion, by how much does the total cost increase?

Show Answers Only

$B$

Show Worked Solution

$C=4n+9$

$\text{If}\ n\ \text{increases to}\ n+5$

$C$	$=4(n+5)+9$
	$=4n+20+9$
	$=4n+29$

$\therefore \text{Total cost increases by }$20$

$\Rightarrow B$

v1 Algebra, STD2 A2 2015 HSC 13 MC

What is the equation of the line $l$?

$y=-\dfrac{1}{5}x+5$
$y=\dfrac{1}{5}x+5$
$y=-5x+5$
$y=5x+5$

Show Answers Only

$C$

Show Worked Solution

$l\ \ \text{passes through (0, 5) and (1, 0)}$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{5-0}{0-1}$
	$=-5$

$y\ \text{-intercept}= 5$

$\therefore\ y=-5x+5$

$\Rightarrow C$

♦ Mean mark 48%.

v1 Algebra, STD2 A2 2017 HSC 20 MC

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 230 matches in total are available?

Show Answers Only

$C$

Show Worked Solution

$\text{1 pentagon:}\ 5+4\times 0=5$

$\text{2 pentagons:}\ 5+4\times 1=9$

$\text{3 pentagons:}\ 5+4\times 2 = 13$

$\vdots$

$n\ \text{pentagons:}\ 5 + 4(n – 1)$

$5+4(n – 1)$	$=230$
$4n-4$	$=225$
$4n$	$=229$
$n$	$=57.25$

$\text{Complete pentagons possible}\ =57$

$\Rightarrow C$

v1 Algebra, STD2 A2 2022 HSC 2 MC

Which of the following could be the graph of $y=-2-2x$?

Show Answers Only

$B$

Show Worked Solution

$\text{By elimination:}$

$y\text{-intercept} =-2\ \rightarrow\ \text{Eliminate}\ A \text{ and}\ D$

$\text{Gradient is negative}\ \rightarrow\ \text{Eliminate}\ C$

$\Rightarrow B$

♦ Mean mark 48%.

v1 Algebra, STD2 A1 2011 HSC 21 MC

A train departs from Town A at 4.00 pm to travel to Town B. Its average speed for the journey is 80 km/h, and it arrives at 6.00 pm. A second train departs from Town A at 4.30 pm and arrives at Town B at 6.10 pm.

What is the average speed of the second train?

96 km/h
114 km/h
224 km/h
280 km/h

Show Answers Only

$A$

Show Worked Solution

$\text{1st train:}$

$\text{Travels 2hrs at 80km/h}$

$\text{Distance}$	$=\text{Speed}\times\text{Time}$
	$=80\times 2$
	$=160\ \text{km}$

$\text{2nd train:}$

$\text{Travels 160 km in 1 hr 40 min}\ \rightarrow\ \dfrac{5}{3}\ \text{hrs}$

$\text{Speed}$	$=\dfrac{\text{Distance}}{\text{Time}}$
	$=160\ ÷\ \dfrac{5}{3}$
	$=160\times \dfrac{3}{5}$
	$=96\ \text{km/h}$

$\Rightarrow A$

♦♦ Mean mark 49%.

v1 Algebra, STD2 A1 2009 HSC 16 MC

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

Which of the following graphs shows this relationship?

Show Answers Only

$C$

Show Worked Solution

$T$	$\propto \dfrac{1}{S}$
$T$	$=\dfrac{k}{S}$

$\text{By elimination:}$

$\text{As Speed} \uparrow \ \text{, Time}\downarrow\ \Rightarrow\ \text{cannot be A or B}$

$\text{D is incorrect because it graphs a linear relationship}$

$\Rightarrow C$

♦♦ Mean mark 38%.

v1 Algebra, STD2 A1 2014 HSC 4 MC

Young’s formula below is used to calculate the required dosages of medicine for children aged 1–12 years.

$\text{Dosage}=\dfrac{\text{age of child (in years)}\ \times\ \text{adult dosage}}{\text{age of child (in years)}\ +\ 12}$

How much of the medicine should be given to an 18-month-old child in a 24-hour period if each adult dosage is 27 mL? The medicine is to be taken every 8 hours by both adults and children.

3 mL
6 mL
9 mL
12 mL

Show Answers Only

$C$

Show Worked Solution

$\text{Age of child} = 18\ \text{months}=1.5\ \text{years}$

$\text{Dosage}$	$=\dfrac{1.5\times 27}{1.5+12}$
	$=3\ \text{mL}$

$\text{Dosage every 8 hrs}$

$\therefore\ \text{In 24 hours, medicine given} = 3\times 3=9\ \text{mL}$

$\Rightarrow C$

♦♦ Mean mark 42%.

v1 Algebra, STD2 A1 2015 HSC 30d

Monica is driving on a motorway at a speed of 105 kilometres per hour and has to brake suddenly. She has a reaction time of 1.3 seconds and a braking distance of 54.3 metres.

Stopping distance can be calculated using the following formula

$\text{stopping distance = {reaction time distance} + {braking distance}}$

What is Monica's stopping distance? Give your answer to 1 decimal place. (2 marks)

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

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$92.2\ \text{metres (to 1 d.p.)}$

Show Worked Solution

$105\ \text{km/hr}$	$=105\ 000\ \text{m/hr}$
	$=\dfrac{105\ 000}{60\times 60}\ \text{m/sec}$
	$=29.166\dots\ \text{m/sec}$

$\text{Reaction time distance}$	$=1.3\times 29.166\dots$
	$=37.916\dots\ \text{metres}$

$\text{Stopping distance}$

$\text{ = {Reaction time distance} + {braking distance}}$

$=37.916…+54.3$

$=92.216\dots$

$=92.2\ \text{metres (to 1 d.p.)}$

♦ Mean mark 34%.

v1 Algebra, STD2 A1 SM-Bank 4

Yuan is driving in a school zone at a speed of 30 kilometres per hour and needs to stop immediately to avoid an accident.

It takes him 1.4 seconds to react and his breaking distance is 6.2 metres.

Stopping distance can be calculated using the following formula

$\text{stopping distance = {reaction time distance} + {braking distance}}$

What is Yuan's total stopping distance? Give your answer to 1 decimal place. (2 marks)

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

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$17.9\ \text{metres (to 1 d.p.)}$

Show Worked Solution

$30\ \text{km/hr}$	$=30\ 000\ \text{m/hr}$
	$=\dfrac{30\ 000}{60\times 60}\ \text{m/sec}$
	$=8.33\dots\ \text{m/sec}$

$\therefore\ \text{Total stopping distance}$

$\text{ = {Reaction time distance} + {braking distance}}$

$=1.4\times 8.33…+6.2$

$=17.866\dots$

$=17.9\ \text{metres (to 1 d.p.)}$

v1 Algebra, STD2 A1 2018 HSC 28e

Drake is driving at 80 km/h. He notices a branch on the road ahead and decides to apply the brakes. His reaction time is 1.2 seconds. His braking distance ($D$ metres) is given by $D=0.01v^2$, where $v$ is speed in km/h.

Stopping distance can be calculated using the following formula

$\text{stopping distance = {reaction time distance} + {braking distance}}$

What is Drake’s stopping distance, to the nearest metre? (3 marks)

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

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$91\ \text{m (nearest m)}$

Show Worked Solution

$\text{80 km/hr}$	$=80\ 000\ \text{m/hr}$
	$=\dfrac{80\ 000}{60\times 60}\ \text{m/sec}$
	$=22.22\dots\ \text{m/sec}$

$\text{Total stopping distance}$

$\text{ = {reaction time distance} + {braking distance}}$

$=1.2\times 22.22\dots + 0.01\times 80^2$

$=90.66\dots$

$=91\ \text{m (nearest m)}$

♦ Mean mark 46%.

v1 Algebra, STD2 A1 2013 HSC 29a

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

$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	$\text{... Line 1}$
$5M+15-4M-2$	$=10$	$\text{... Line 2}$
$M+13$	$=10$	$\text{... Line 3}$
$M$	$=-3$	$\text{... Line 4}$

Copy the equation in Line 1 and continue your solution to solve this equation for $M$.

Show all lines of working. (2 marks)

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

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$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	`text(… Line 1)`
$5M+15-4M+2$	$=10$	`text(… Line 2)`
$M+17$	$=10$	`text(… Line 3)`
$M$	$=-7$	`text(… Line 4)`

Show Worked Solution

♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 10 (from Line 1 to Line 2), indicating both sides must have been multiplied by 10.

$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	`text(… Line 1)`
$5M+15-4M+2$	$=10$	`text(… Line 2)`
$M+17$	$=10$	`text(… Line 3)`
$M$	$=-7$	`text(… Line 4)`

v1 Algebra, STD2 A1 2010 HSC 7 MC

If $M=-8$, what is the value of $\dfrac{4M^2+3M}{8}$

$-1027$
$-35$
$29$
$125$

Show Answers Only

$C$

Show Worked Solution

♦♦ Only 31% of students answered correctly!

$\dfrac{4M^2+3M}{8}$	$=\dfrac{4\times (-8)^2+3\times (-8)}{8}$
	$=\dfrac{4\times 64-24}{8}$
	$=\dfrac{232}{8}$
	$=29$

$\Rightarrow C$

v1 Algebra, STD2 A1 2018 HSC 28b

Solve the equation $\dfrac{3x}{4}+1=\dfrac{5x+1}{3}$, leaving your answer as a fraction. (3 marks)

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$\dfrac{8}{11}$

Show Worked Solution

♦ Mean mark 35%.

$\underbrace{\dfrac{3x}{4} + 1}_\text{multiply x 12}$	$=\underbrace{\dfrac{5x+1}{3}}_\text{multiply x 12}$
$9x+12$	$=20x+4$
$11x$	$=8$
$x$	$=\dfrac{8}{11}$

v1 Algebra, STD2 A1 2021 HSC 29

Solve $x+\dfrac{x-3}{4}=5$, leaving your answer as a fraction. (2 marks)

Show Answers Only

$\dfrac{23}{5}$

Show Worked Solution

♦ Mean mark 40%.

$x+\dfrac{x-3}{4}$	$=5$
$4x+x-3$	$=20$
$5x$	$=23$
$x$	$=\dfrac{23}{5}$

v1 Algebra, STD2 A1 2013 HSC 21 MC

Which equation correctly shows $n$ as the subject of $V=600(1-n)$?

$n=\dfrac{V-600}{600}$
$n=\dfrac{600-V}{600}$
$n=V-600$
$n=600-V$

Show Answers Only

$B$

Show Worked Solution

♦♦♦ Mean mark 27%

$V$	$=600(1-n)$
$1-n$	$=\dfrac{V}{600}$
$n$	$=1-\dfrac{V}{600}$
	$=\dfrac{600-V}{600}$

$\Rightarrow B$

v1 Algebra, STD2 A1 2011 HSC 18 MC

Which of the following correctly expresses $b$ as the subject of $y= ax+\dfrac{1}{4}bx^2$?

$b=\dfrac{4y-ax}{x^2}$
$b=\dfrac{4(y-ax)}{x^2}$
$b=\dfrac{\dfrac{1}{4}y-ax}{x^2}$
$b=\dfrac{\dfrac{1}{4}(y-ax)}{x^2}$

Show Answers Only

$B$

Show Worked Solution

$y$	$= ax+\dfrac{1}{4}bx^2$
$\dfrac{1}{4}bx^2$	$=y-ax$
$bx^2$	$=4(y-ax)$
$b$	$=\dfrac{4(y-ax)}{x^2}$

$\Rightarrow B$

v1 Algebra, STD2 A1 2006 HSC 18 MC

What is the formula for $g$ as the subject of $7d=8e+5g^2$?

$g =\pm\sqrt{\dfrac{8e-7d}{5}}$
$g =\pm\sqrt{\dfrac{7d-8e}{5}}$
$g =\pm\dfrac{\sqrt{7d+8e}}{5}$
$g =\pm\dfrac{\sqrt{8e-7d}}{5}$

Show Answers Only

$B$

Show Worked Solution

$7d$	$=8e+5g^2$
$5g^2$	$=7d-8e$
$g^2$	$=\dfrac{7d-8e}{5}$
$g$	$=\pm\sqrt{\dfrac{7d-8e}{5}}$

$\Rightarrow B$

v1 Algebra, STD2 A1 2007 HSC 19 MC

Which of the following correctly expresses $X$ as the subject of $Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)$?

$X=\dfrac{Y}{\pi}-L$
$X=\dfrac{Y}{\pi}-4L$
$X=4L-\dfrac{Y}{2\pi}$
$X=\dfrac{Y}{8\pi}-\dfrac{L}{4}$

Show Answers Only

$B$

Show Worked Solution

$Y$	$=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)$
$\dfrac{Y}{4\pi}$	$=\dfrac{X}{4}+L$
$\dfrac{X}{4}$	$=\dfrac{Y}{4\pi}-L$
$X$	$=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)$
$X$	$=\dfrac{Y}{\pi}-4L$

$\Rightarrow B$

v1 Algebra, STD2 A1 2005 HSC 24c

Make $r$ the subject of the equation $V=4\pi r^2$. (2 marks)

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

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$r=\pm\sqrt{\dfrac{V}{4\pi}}$

Show Worked Solution

$V$	$=4\pi r^2$
$r^2$	$=\dfrac{V}{4\pi}$
$\therefore\ r$	$=\pm\sqrt{\dfrac{V}{4\pi}}$

v1 Algebra, STD2 A1 2017 HSC 28d

Make $b$ the subject of the equation $a=\sqrt{bc-4}$. (2 marks)

Show Answers Only

$b=\dfrac{a^2+4}{c}$

Show Worked Solution

♦ Mean mark 46%.

$a$	$=\sqrt{bc-4}$
$a^2$	$=bc-4$
$bc$	$=a^2+4$
$\therefore\ b$	$=\dfrac{a^2+4}{c}$

v1 Algebra, STD2 A1 2019 HSC 11 MC

Which of the following correctly expresses $y$ as the subject of the formula $5x-2y-9=0$?

$y=\dfrac{5}{2}x-9$
$y=\dfrac{5}{2}x+9$
$y=\dfrac{5x+9}{2}$
$y=\dfrac{5x-9}{2}$

Show Answers Only

$D$

Show Worked Solution

♦ Mean mark 50%.

$5x-2y-9$	$=0$
$2y$	$=5x-9$
$\therefore\ y$	$=\dfrac{5x-9}{2}$

$\Rightarrow D$

CHEMISTRY, M8 2022 VCE 5*

A chemist uses spectroscopy to identify an unknown organic molecule, Molecule $\text{J}$, that contains chlorine.

The ${}^{13}\text{C NMR}$ spectrum of Molecule $\text{J}$ is shown below.

The infra-red (IR) spectrum of Molecule $\text{J}$ is shown below.

Name the functional group that produces the peak at 168 ppm in the ${}^{13}\text{C NMR}$ spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum. (2 marks)

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The mass spectrum of Molecule $\text{J}$ is shown below

The molecular mass of Molecule $\text{J}$ is 108.5
Explain the presence of the peak at 110 m/z. (1 mark)

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The ${ }^1 \text{H NMR}$ spectrum of Molecule $\text{J}$ is shown below.

The ${ }^1 \text{H NMR}$ spectrum consists of two singlet peaks.
What information does this give about the molecule? (2 marks)

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Draw a structural formula for Molecule $\text{J}$ that is consistent with the information provided in parts a–c. (2 marks)

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a. → Absence of a very broad $\ce{OH}$ acid peak between 2500–3000.

→ Molecule $\text{J}$ must be an ester.

b. The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.

c. The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.

d. Either of the two molecules show below are correct:

Show Worked Solution

a. → Absence of a very broad $\ce{OH}$ acid peak between 2500–3000.

→ Molecule $\text{J}$ must be an ester.

♦ Mean mark (a) 41%.

b. The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.

c. The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.

♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d. Either of the two molecules show below are correct:

♦ Mean mark (d) 40%.

CHEMISTRY, M8 2021 VCE 7*

Two students are given a homework assignment that involves analysing a set of spectra and identifying an unknown compound. The unknown compound is one of the molecules shown below.

The $^{13}\text{C NMR}$ spectrum of the unknown compound is shown below.

Based on the number of peaks in the $^{13}\text{C NMR}$ spectrum above, which compound – $\text{P, Q, R, S}$ or $\text{T}$ – could be eliminated as the unknown compound? (1 mark)

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The infra-red (IR) spectrum of the unknown compound is shown below.

Identify which of the five compounds (1 or more) can be eliminated on the basis of the IR spectrum. Justify your answer using data from the IR spectrum. (3 marks)

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The mass spectrum of the unknown compound is shown below.

Write the chemical formula of the species that produces a peak at m/z = 43. (1 mark)

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Explain why one molecule can produce multiple peaks on a mass spectrum. (2 mark)

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

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a. → The $^{13}\text{C NMR}$ shows 4 different carbon environments.

→ Compound $T$ has 5 unique carbon environments and can be eliminated.

b. → Compounds $\text{P, Q}$ and $\text{S}$ can be eliminated.

→ Compounds $\text{S}$ and $\text{P}$ both don’t have an $\ce{OH}$ alcohol group, however the IR spectrum clearly shows an $\ce{OH}$ alcohol group with an absorbance at 3500 cm$^{-1}$.

→ Compound $\text{Q}$ contains an $\ce{OH}$ acid group whereas the IR spectrum shows an $\ce{OH}$ alcohol group at 3500 cm$^{-1}$. There is no evidence of a broad $\ce{OH}$ acid group between 2500–3000 cm$^{-1}$.

c.i. → The unknown compound is compound $\text{R}$.

→ The chemical species are fragments of the original compound with a positive charge.

→ The chemical formulas could include $\ce{[CH3CO]^+}$ or $\ce{[C2H3O]^+}$

ii. → Ions of that molecule must be formed to produce peaks on the mass spectrum.

→ Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

Show Worked Solution

a. → The $^{13}\text{C NMR}$ shows 4 different carbon environments.

→ Compound $T$ has 5 unique carbon environments and can be eliminated.

b. → Compounds $\text{P, Q}$ and $\text{S}$ can be eliminated.

→ Compounds $\text{S}$ and $\text{P}$ both don’t have an $\ce{OH}$ alcohol group, however the IR spectrum clearly shows an $\ce{OH}$ alcohol group with an absorbance at 3500 cm$^{-1}$.

♦ Mean mark (b) 43%.
COMMENT: Recognise the difference between OH alcohols and OH acid groups.

c.i. → The unknown compound is compound $\text{R}$.

→ The chemical species are fragments of the original compound with a positive charge.

→ The chemical formulas could include $\ce{[CH3CO]^+}$ or $\ce{[C2H3O]^+}$

♦ Mean mark (c.i.) 50%.
COMMENT: Ions require a positive charge.

ii. → Ions of that molecule must be formed to produce peaks on the mass spectrum.

→ Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

♦ Mean mark (c.ii.) 39%.

CHEMISTRY, M8 2023 VCE 7-1*

Molecule $\text{V}$ contains only carbon atoms, hydrogen atoms and one oxygen atom.

The mass spectrum of molecule $\text{V}$ is shown below.

i. State the molecular formula of molecule $\text{V}$. Justify your answer by using the information in the mass spectrum. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
ii. State why there is a small peak at $\text{m/z} =87$. (1 mark)

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The ${ }^1 \text{H NMR}$ spectrum of molecule $\text{V}$ is shown below.

State what information the doublet at 1.1 ppm provides about the structure of the molecule. (1 mark)

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

The ${ }^{13} \text{C NMR}$ spectrum of molecule $\text{V}$ is shown below.

In the space below, draw a structural formula of molecule $\text{V}$ that is consistent with the information provided in parts a.-c. (3 marks)

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

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a.i. Molar mass of $\text{V}\ = 86\ \text{g mol}^{-1}$

→ Indicated by the parent ion peak at 86 m/z.

→ Molecular formula: $\ce{C5H10O}$

a.ii. Small peak at m/z = 87:

→ A carbon-13 isotope being present in the molecule.

b. The doublet peak at 1.1 ppm:

→ Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom.

Show Worked Solution

a.i. Molar mass of $\text{V}\ = 86\ \text{g mol}^{-1}$

→ Indicated by the parent ion peak at 86 m/z.

→ Molecular formula: $\ce{C5H10O}$

a.ii. Small peak at m/z = 87:

→ A carbon-13 isotope being present in the molecule.

b. The doublet peak at 1.1 ppm:

→ Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom.

♦ Mean mark (b) 50%.

c. From the carbon NMR graph:

→ There are 4 carbon environments, one shifted above 200 ppm indicating a ketone or aldehyde.

→ As there are 5 carbons, 2 of the carbons must have the same environment.

From the hydrogen NMR graph:

→ There are 3 hydrogen environments.

→ The septet peak indicates there are 6 hydrogens with the same chemical environment on adjacent carbon atoms.

♦ Mean mark (c) 43%.

\( \textbf{lst position} \) \( \textbf{(5}^{′}\ \textbf{end)} \) \( \boldsymbol{\downarrow} \)	\( \textbf{2nd position} \)				\( \textbf{3rd position} \) \( \ \ \textbf{(3}^{′}\ \textbf{end)} \) \( \boldsymbol{\downarrow} \)
	\( \textbf{U} \)	\( \textbf{C} \)	\( \textbf{A} \)	\( \textbf{G} \)
\( \textbf{U} \)	\( \text{Phe}\)	\( \text{Ser}\)	\( \text{Tyr}\)	\( \text{Cys}\)	\( \textbf{U} \)
	\( \text{Phe}\)	\( \text{Ser}\)	\( \text{Tyr}\)	\( \text{Cys}\)	\( \textbf{C} \)
	\( \text{Leu}\)	\( \text{Ser}\)	\( \text{STOP}\)	\( \text{STOP}\)	\( \textbf{A} \)
	\( \text{Leu}\)	\( \text{Ser}\)	\( \text{STOP}\)	\( \text{Trp}\)	\( \textbf{G} \)
\( \textbf{C} \)	\( \text{Leu}\)	\( \text{Pro}\)	\( \text{His}\)	\( \text{Arg}\)	\( \textbf{U} \)
	\( \text{Leu}\)	\( \text{Pro}\)	\( \text{His}\)	\( \text{Arg}\)	\( \textbf{C} \)
	\( \text{Leu}\)	\( \text{Pro}\)	\( \text{Gln}\)	\( \text{Arg}\)	\( \textbf{A} \)
	\( \text{Leu}\)	\( \text{Pro}\)	\( \text{Gln}\)	\( \text{Arg}\)	\( \textbf{G} \)
\( \textbf{A} \)	\( \text{Ile}\)	\( \text{Thr}\)	\( \text{Asn}\)	\( \text{Ser}\)	\( \textbf{U} \)
	\( \text{Ile}\)	\( \text{Thr}\)	\( \text{Asn}\)	\( \text{Ser}\)	\( \textbf{C} \)
	\( \text{Ile}\)	\( \text{Thr}\)	\( \text{Lys}\)	\( \text{Arg}\)	\( \textbf{A} \)
	\( \text{Met}\)	\( \text{Thr}\)	\( \text{Lys}\)	\( \text{Arg}\)	\( \textbf{G} \)
\( \textbf{G} \)	\( \text{Val}\)	\( \text{Ala}\)	\( \text{Asp}\)	\( \text{Gly}\)	\( \textbf{U} \)
	\( \text{Val}\)	\( \text{Ala}\)	\( \text{Asp}\)	\( \text{Gly}\)	\( \textbf{C} \)
	\( \text{Val}\)	\( \text{Ala}\)	\( \text{Glu}\)	\( \text{Gly}\)	\( \textbf{A} \)
	\( \text{Val}\)	\( \text{Ala}\)	\( \text{Glu}\)	\( \text{Gly}\)	\( \textbf{G} \)

\(\text{Income}\)	\(=35\times 4=$140\ \ \text{is equal to}\)
\(\text{Costs}\)	\(=40+(25\times 4)=$140\)

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

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

\(2500\)	\(=30x-(400+10x)\)
\(20x\)	\(=2900\)
\(x\)	\(=145\)

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

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

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

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