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Algebra, STD2 A4 2004 HSC 28a (Adapted)

A fitness index, `F`, is calculated by dividing a person’s weight, `w`, in kilograms by the square of the person’s height, `h`, in metres.

  1. Albert is 160 cm and weighs 76.8 kg. Calculate Albert’s health rating.   (1 mark)

  2. In the next few years, Albert expects to grow 20 cm taller. By then he wants his fitness index to be 23. How much weight should he gain or lose to achieve this? Justify your answer with mathematical calculations.   (2 marks)

Show Answers Only

a.   `30`

b.   `text(Albert needs to gain 4.2 kg)`

Show Worked Solution

a.   `\text{160 cm = 1.60 m.}`

`text(When)\ w = 76.8 and h = 1.60:`

`F=w/h^2 = 76.8/1.60^2 = 30`
 

b.   `text(Find)\ w\ text(given)\ F= 25 and h = 1.80:`

`25` `= w/1.8^2`
`w` `= 25 xx 1.8^2= 81\ text(kg)`

 
`:.\ text(Weight Albert should gain)`

`= 81-76.8`

`= 4.2\ text(kg)`

Algebra, STD2 A4 2014 HSC 29a (Adapted)

A golf club hires an entire course for a charity event at a total cost of `$40\ 000`. The cost will be shared equally among the players, so that `C` (in dollars) is the cost per player when `n` players attend.

  1. Complete the table below by filling in the three missing values.   (1 mark)
    \begin{array} {|l|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 50\ & \ 100 \ & 200 \ & 250 \ & 400\ & 500 \ \\
    \hline
    \rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} &  &  &  & 160 & 100\ & 80 \ \\
    \hline
    \end{array}
  2. Using the values from the table, draw the graph showing the relationship between `n` and `C`.   (2 marks)
     
  3. What equation represents the relationship between `n` and `C`?   (1 mark)

  4. Give ONE limitation of this equation in relation to this context.   (1 mark)

  5. Is it possible for the cost per person to be $94? Support your answer with appropriate calculations.   (1 mark)

Show Answers Only

a.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

b. 

c.   `C = (40\ 000)/n`

`n\ text(must be a whole number)`
 

d.    `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

e.   `text(If)\ C = 94:`

`94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`

 
`:.\ text(C)text(ost cannot be $94 per person, because)\ n\ text(isn’t a whole number.)`

Show Worked Solution

a.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 50\ & \ 100 \ & 200 \ & 250 \ & 400\ & 500 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 800 & 400 & 200 & 160 & 100\ & 80 \ \\
\hline
\end{array}

b.   
       
      

c.   `C = (40\ 000)/n`

 

TIP: Limitations require looking at possible restrictions of both the domain and range.

d.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

e.   `text(If)\ C = 120`

`120` `= (40\ 000)/n`
`120n` `= 40\ 000`
`n` `= (40\ 000)/120`
  `= 333.33..`

  
`:.\ text{Cost cannot be $120 per person because the required}\ n\ \text{is not a whole number.}`

Algebra, STD2 A4 2022 HSC 24 (Adapted)

A chef believes that the time it takes to defrost a turkey (`D` hours) varies inversely with the room temperature (`T^\circ \text{C}`). The chef observes that at a room temperature of `20^\circ \text{C}`, it takes 15 hours for the turkey to fully defrost.

  1. Find the equation relating `D` and `T`.   (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time.   (2 marks)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ \  & \ \ \  & \ \ \ \  & \ \ \ \ & \ \ \ \\
\hline
\end{array}

 
           

Show Answers Only

a.   `D \prop 1/T\ \ =>\ \ D=k/T`

  `15` `=k/20`
  `k` `=15 xx 20=300`

 
`:.D=300/T`

b.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 30\ \ \  & \ \ 20\ \ \  & \ \ \ 15\ \ \  & \ \ \ 12\ \ \ & \ \ \ 10\ \ \ \\
\hline
\end{array}

 

     

Show Worked Solution

a.   `D \prop 1/T\ \ =>\ \ D=k/T`

  `15` `=k/20`
  `k` `=15 xx 20=300`

 
`:.D=300/T`

b.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ T\ \ \rule[-1ex]{0pt}{0pt} & 30 & 20 & 15 & 12 & 10  \\
\hline
\end{array}

 

     

Algebra, STD2 A4 2018 HSC 29c (Adapted)

Snowhound makes snow shoes of various sizes. In its design phase, Snowhound collect data on the different footprint depths of snow shoes of different sizes, all worn by the same person.

 The footprint depth (`d` cm) is then graphed against the area of the sole of the snow shoe (`A` cm).
 


 

  1. The graph shape shows that `d` is inversely proportional to `A`. The point `X` lies on the graph.

     

    Find the equation relating `d` and `A`.   (2 marks)

  2. A man from this group walks in snow and the depth of his footprint is 5 cm.

     

    Use your equation from part (a) to calculate the area of his shoe sole.   (1 mark)

Show Answers Only

a.   `D = 4500/A`

b.   `480\ text(cm)^2`

Show Worked Solution

a.   `d prop 1/A \ =>\ d = k/A`

`text(When)\ D = 12, A = 200:`

`12` `= k/200`
`k` `= 12 xx 200 = 2400`
`:. d` `=2400/A`

 

b.    `5` `= 2400/A`
  `:. A` `= 2400/5= 480\ text(cm)^2`

Algebra, STD2 A4 2011 HSC 28a (Adapted)

The intensity of light, `I`, from a lamp varies inversely with the square of the distance, `d`, from the lamp.

  1. Write an equation relating `I`, `d` and `k`, where `k` is a constant.    (1 mark)

  2. It is known that `I = 20` when `d = 2`.

     

    By finding the value of the constant, `k`, find the value of `I` when `d = 5`.    (2 marks)

  3. Sketch a graph to show how `I` varies for different values of `d`.

     

    Use the horizontal axis to represent distance and the vertical axis to represent light intensity.   (2 marks)


Show Answers Only

a.   `I = k/d^2`

b.   `P = 1 1/2`

c.   
         

Show Worked Solution

a.   `I prop 1/d\ \ =>\ \ I=k/d^2`
 

b.  `text(When)\ I=20, d=2:`

`20` `= k/2^2`
`k` `=4 xx 20=80`

 
`text(Find)\ I\ text(when)\ d = 5:`

`I=80/5^2=16/5`
 

c.

Algebra, STD2 A4 2021 HSC 13 MC (Adapted)

The time taken to harvest a field varies inversely with the number of workers employed.

It takes 12 workers 36 hours to harvest the field.

Working at the same rate, how many hours would it take 27 workers to harvest the same field?

  1. 12
  2. 16
  3. 24
  4. 54
Show Answers Only

`B`

Show Worked Solution

`text{Time to harvest}\ (T) prop 1/text{Number of workers (W)}`

`T=k/W`

`text(When)\ \ T=36, W=12:`

`36=k/12\ \ =>\ \ k=36 xx 12 = 432`  
 

`text{Find}\ T\ text(when)\ \ W=27:`

`T=432/27=16\ \text{hours}`

 `=>  B`

Algebra, STD2 A4 2010 HSC 13 MC (Adapted)

The time taken to charge a battery varies inversely with the charging voltage. At 24 volts \((V)\) it takes 15 hours to fully charge a battery.

How long will it take the same battery to fully charge at 40 volts?

  1. 8 hours
  2. 9 hours
  3. 10.5 hours
  4. 12 hours
Show Answers Only

`B`

Show Worked Solution
 
♦ Mean mark 50% 

`text{Time to charge}\ (T) prop 1/text(Voltage) \ => \ T=k/V`

`text(When) \ T=15, V = 24:`

`15=k/24\ \ => \ k=15 xx 24=360` 
   

`text{Find}\ T\ text{when}\ \ V= 40:}`

`T=360/40=9\ \text{hours}`

 `=>  B`

Algebra, STD2 A4 2024 HSC 9 MC (Adapted)

The time taken to fill a swimming pool varies inversely with the number of hoses being used.

Using 4 hoses, it takes 18 hours to fill the pool.

How many hours would it take 9 hoses to fill the same pool?

  1. 6
  2. 7.5
  3. 8
  4. 12
Show Answers Only

\(C\)

Show Worked Solution

\(T \propto \dfrac{1}{H} \ \Rightarrow \ \ T=\dfrac{k}{H}\)

\(\text {Find}\ k\ \text{given}\ \  T=18\ \ \text {when}\ \ H=4 \text {:}\)

\(18=\dfrac{k}{4} \ \Rightarrow\ \ k=72\)
 

\(\text {Find}\ T \ \text {if}\ \ H=9:\)

\(T=\dfrac{72}{9}=8 \text { hours}\)

\(\Rightarrow C\)

Algebra, STD2 A4 2022 HSC 9 MC (Adapted)

An object is projected vertically into the air. Its height, \(h\) metres, above the ground after \(t\) seconds is given by  \(h=-5 t^2+80 t\).
 

How far does the object travel in the first 10 seconds?

  1. 300 metres
  2. 320 metres
  3. 340 metres
  4. 480 metres
Show Answers Only

\(C\)

Show Worked Solution

\(\text{By symmetry (or graph), object reaches max height at}\ \ t=8\ \text{seconds.}\)

\(\text{Find}\ h\ \text{when}\ \ t=8:\)

\(h=-5 \times 8^2-10 \times 8= 320 \)

\(\text{When}\ \ t=10\ \ \Rightarrow\ \ h=300\ \text{(from graph)}\)

\(\therefore\ \text{Total distance}\ = 320 + 20=340\ \text{metres}\)

\(\Rightarrow C\)

Functions, 2ADV F1 2017 HSC 1 MC (Adapted)

What is the gradient of the line  \(6x+7y-1 = 0\)?

  1. \(-\dfrac{6}{7}\)
  2. \(\dfrac{6}{7}\)
  3. \(-\dfrac{7}{6}\)
  4. \(\dfrac{7}{6}\)
Show Answers Only

\(A\)

Show Worked Solution
\(6x+7y-1\) \(=0\)  
\(7y\) \(=-6x+1\)  
\(y\) \(=-\dfrac{6}{7}x+\dfrac{1}{7}\)  

 
\(\Rightarrow A\)

Algebra, STD2 A4 2010 HSC 24b (Adapted)

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)

Show Answers Only

\(x=4\ \ \text{See worked solution}\)

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 \(x\).

Algebra, STD2 A4 2005 HSC 28b (Adapted)

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.

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

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

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

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

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

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

Show Answers Only

a.    \(C=800+9x\)

b.    \(\text{Approximately }135\)

c.    \($400\)

d.    \($16.70\)

Show Worked Solution
a.     \($C\) \(=550+250+(9\times x)\)
    \(=800+9x\)

 

b.    \(\text{Using the graph intersection}\)

\(\text{Approximately 135 people are needed}\)

\(\text{to cover the costs.}\)

 

c.    \(\text{If 200 people attend}\)

\(\text{Income}\) \(=200\times $15\)
  \(=$3000\)
\(\text{Costs}\) \(=800+(9\times 200)\)
  \(=$2600\)

 

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

 

d.    \(\text{Costs when}\ x=300:\)

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

 

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

\(=3500+1510\)

\(=$5010\)
 

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

Algebra, STD2 A4 2020 HSC 24 (Adapted)

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.

  1. 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)
     

     

  2. 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)

  3. 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)

Show Answers Only
  1.  \(\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}\)
      
  2. \(20 \ \text{minutes}\)
  3. \(30 \ \text{minutes}\)
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%.

Algebra, STD2 A4 2019 HSC 36 (Adapted)

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.
 


 

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

  2. 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)

Show Answers Only

a.    \(20\)

b.    \(145\)

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%.

Algebra, STD2 A4 2020 VCAA 3 (Adapted)

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

\(C=35n+5000\)

  1. On the graph above, sketch the monthly cost, \(C\), of making \(n\) seat covers.   (1 mark) 
  2. 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)


Show Answers Only

a.


 

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.    \(C=35n+5000\ \text{and }I=60n\)

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

\(\text{Method 2: Graphically}\)

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

\(\text{Method 2: Algebraically}\)

\(\text{Solve for} \ n:\)

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

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

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

Algebra, STD2 A4 2014 HSC 26d (Adapted)

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

\(y=2x-6\)

\(y-x+2=0\)
 

Show Answers Only

\(x=4,\ y=2\)

Show Worked Solution

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

Algebra, STD2 A4 2018 HSC 27b (Adapted)

\(y\) \(=x-3\)
\(y+3x\) \(=1\)

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

 

 
Show Answers Only

\((1,-2)\)

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}

 

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

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

Algebra, STD2 A4 2023 HSC 21 (Adapted)

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

  1. 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.
 

 
  1. On the grid above, graph Provider \(A\)'s charges from the table in part (a).   (1 mark)
  2. 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)

  3. A customer uses an average of 600 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?   (2 marks)

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}

b.    
         

c.    \(\text{400 kWh}\)

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

Show Worked Solution

a.   \(\text{When kWh} =400\)

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

\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}

b. 
          
 

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

\(\therefore\ \text{Same charge at 400 kWh}\)
 

d.    \(\text{Cost at 600 kWh:}\)

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

\(\text{Method 2: Algebraically}:\)

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

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

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

Algebra, STD2 A4 2018 HSC 27d (Adapted)

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.

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

  2. 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)

  3. A jetski is hired for 7 hours from Company B.

     

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

Show Answers Only
  1. \($90\)
  2. \(c=80+74x\)
  3. \($32\)
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.    \(\text{Cost}\ (A)\) \(=90\times 7=$630\)
  \(\text{Cost}\ (B)\) \(=80+74\times 7=$598\)

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

Algebra, STD2 A4 2004 HSC 16 MC (Adapted)

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

\(\text{By elimination:}\)

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

\(\rightarrow\ \text{Eliminate be A and B}\)

 

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

\(\text{AND has a positive gradient}\)

\(\rightarrow\ \text{Eliminate C}\)

\(\Rightarrow D\)

Algebra, STD2 A4 2011 HSC 20 MC (Adapted)

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?

  1. \($10\ 000\)
  2. \($20\ 000\)
  3. \($30\ 000\)
  4. \($40\ 000\)
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.

Algebra, STD2 A2 2017 HSC 3 MC (Adapted)

Conversion graphs can be used to convert from one currency to another.  
  


  

Abbie converted 70 New Zealand dollars into Euros. She then converted all of these Euros into Australian dollars.

How much money, in Australian dollars, should Abbie have? 

  1. $30  
  2. $45
  3. $55
  4. $95
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Using the graphs:}\)

\($70\ \text{New Zealand}\) \(=40\ \text{Euro}\)
\(40\ \text{Euro}\) \(=$55\  \text{Australian}\)

 
\(\Rightarrow C\)

Algebra, STD2 A2 2009 HSC 13 MC (Adapted)

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?

  1. 6%
  2. 12%
  3. 35%
  4. 64%
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)

Algebra, STD2 A2 2009 HSC 24d (Adapted)

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.
 

 

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

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

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

     

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

Show Answers Only

a.    \(x+y=400\)

b.    \(\text{Since the max amount of leather lounges}=270\)

 

\(\rightarrow\ x\ \text{cannot be}\ >270\)

 

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

 

\(\rightarrow\ y\ \text{cannot be}\ >325\)

 

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

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

Show Worked Solution

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

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

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


♦♦♦ 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).

b.    \(\text{Since the max amount of leather lounges}=270\)

\(\rightarrow\ x\ \text{cannot}\ >270\)

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

\(\rightarrow\ y\ \text{cannot}\ >325\)

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


♦ Mean mark part (ii) 49%.

c.    \(\text{At}\ B,\ x=75,\ y=325\)

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

  
\(\text{At}\ C,\ x=270,\ y=130\)

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

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

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


♦ Mean mark (iii)40%.

Algebra, STD2 A2 2010 HSC 27c (Adapted)

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

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

  2. 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)

  3. 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)

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

Show Answers Only

a.    \($3000\ \ \text{(from graph)}\)

b.    \(\text{See worked solution}\)

c.    \(46\frac{2}{3}\approx  47\ \text{cents per dollar earned}\)

d.    \(\text{Tax payable →}\ T=\dfrac{7}{15}I-5400\)

Show Worked Solution
a.    

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

  

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

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

♦♦ Mean mark (ii) 25%.

c.    \(\text{The gradient represents the tax applicable on each dollar}\)

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

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

d.    \(\text{Tax payable up to }$18\ 000 = $3000\)

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

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

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

♦♦♦ 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.

Algebra, STD2 A2 2014 HSC 27b (Adapted)

Clara is comparing the costs of two different ways of travelling to work.

Clara’s motor scooter uses one litre of fuel for every 22 km travelled. The cost of fuel is $2.24/L and the distance from her home to the work car park is 33 km. The cost of travelling by bus and light rail is $35.80 for 10 single trips.

Which way of travelling is cheaper and by how much? Support your answer with calculations.   (2 marks)

Show Answers Only

\(\text{Motor scooter is }$0.22 \text{ cheaper per one-way trip.}\)

Show Worked Solution

\(\text{Compare cost of a one-way trip}\)

\(\text{Motor scooter}\)

\(\text{Fuel used}=\dfrac{33}{22}=1.5\ \text{L}\)

\(\text{Cost}=1.5\times 2.24=$3.36\)
  

\(\text{Bus and light rail}\)

\(\text{Cost}=\dfrac{35.80}{10}=$3.58\)
 

\(\text{Difference}=$3.58-3.36=$0.22\)
  

\(\therefore\ \text{Motor scooter is }$0.22 \text{ cheaper per one-way trip.}\)

Algebra, STD2 A2 2007 HSC 24c (Adapted)

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

  1. 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)

  2. 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)

Show Answers Only

a.    \(1890\ \text{€}\)

b.    \(\text{See worked solutions}\)

Show Worked Solution

a.    \(\text{From graph:}\)

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

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

\(\text{Converting A}$\ \text{to €}:\)

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

 

b.    \(\text{If the value of the US}\ $\ \text{rises against the}\)

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

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

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

Algebra, STD2 A2 2014 HSC 22 MC (Adapted)

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?

  1. 9 L
  2. 16 L
  3. 18 L
  4. 34 L
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Fuel used in short distance}\)

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

\(\text{Fuel used in long distance}\)

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

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

\(\Rightarrow B\)

Algebra, STD2 A2 2014 HSC 26f (Adapted)

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)

Show Answers Only

 \(17\ 143\ \text{kg}\)

Show Worked Solution

\(W_{\text{moon}}\propto W_{\text{earth}}\)

\(\rightarrow\  W_{\text{m}}=k\times W_{\text{e}}\)

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

\(9\) \(=k\times 63\)
\(k\) \(=\dfrac{9}{63}=\dfrac{1}{7}\)

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

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

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

Algebra, STD2 A2 2007 HSC 27b (Adapted)

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.

  1. 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)

  2. 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)

  3. 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)

Show Answers Only

a.    \($c=88+20\ 440f\)

b.    \(0.037\ $/\text{hr}\ \text{(3 d.p.)}\)

c.    \(\text{Proof:  (See Worked Solutions)}\)

Show Worked Solution

a.    \(\text{Purchase price}=8\times 11=$88\)

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

  
\(\therefore\ $c=88+20\ 440f\)
  

b.    \(\text{Given}\ \ $c=$850\)

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

 

c.    \(\text{If}\ f\ \text{is multiplied by  }1.5 =\dfrac{10.5}{7}\)

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

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

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

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

\(\text{the original cost.}\)

Algebra, STD2 A2 2004 HSC 22 MC (Adapted)

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?

  1. 46.56 CAD
  2. 49.48 CAD
  3. 113.67 CAD
  4. 120.75 CAD
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Mary-Anne has 75 AUD.}\)

\(\text{Converting to Euros}\)

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

 

\(\text{Converting to CAD}\)

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

\(\Rightarrow B\)

Algebra, STD2 A2 2016 HSC 26c (Adapted)

Bonn’s car uses fuel at the rate of 6.1 L /100 km for country driving and 8.3 L /100 km for city driving. On a trip, he drives 350 km in the country and 40 km in the city.

Calculate the amount of fuel he used on this trip.   (2 marks)

Show Answers Only

\(24.67\ \text{L}\)

Show Worked Solution

\(\text{Fuel used in country}\)

\(=350\times \dfrac{6.1}{100}\)

\(=21.35\ \text{L}\)

\(\text{Fuel used in city}\)

\(=40\times \dfrac{8.3}{100}\)

\(=3.32\ \text{L}\)

  

\(\therefore\ \text{Total fuel used}\)

\(=21.35+3.32\)

\(=24.67\ \text{L}\)

Algebra, STD2 A2 2016 HSC 29e (Adapted)

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


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

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

Show Answers Only

i.    \(\text{76 years}\)

ii.    \(\text{After 1900, life expectancy increases by 0.38 years for}\)

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

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%.

Algebra, STD2 A2 2018 HSC 5 MC (Adapted)

The driving distance from Burt's home to his work is 15 km. He drives to and from work five times each week. His car uses fuel at the rate of 12 L/100 km.

How much fuel does he use driving to and from work each week?

  1. 15 L
  2. 18 L
  3. 27 L
  4. 36 L
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Total distance travelled each week}\)

\(=5\times 2\times 15\)

\(=150\ \text{km}\)
 

\(\therefore\ \text{Total fuel used}\)

\(=\dfrac{150}{100}\times 12\ \text{L}\)

\(=18\ \text{L}\)

\(\Rightarrow B\)

Algebra, STD2 A2 2017 HSC 3 MC (Adapted)

The graph shows the relationship between infant mortality rate (deaths per 1000 live births) and life expectancy at birth (in years) for different countries.
 

What is the life expectancy at birth in a country which has an infant mortality rate of 80?

  1. 20 years
  2. 21 years
  3. 61 years
  4. 62 years
Show Answers Only

\(D\)

Show Worked Solution

\(\text{When infant mortality rate is 80, life expectancy}\)

\(\text{at birth is 62 years (see below).}\)
 

\(\Rightarrow D\)

Algebra, STD2 A2 2017 HSC 14 MC (Adapted)

Christopher is comparing two different models of 4WD cars. Car A uses fuel at the rate of 11.4 L/100 km. Car B uses 9.6 L/100 km.

Suppose Christopher plans on driving \(11\ 000\) km in the next year.

How much less fuel will he use driving car B instead of car A?

  1. 198 L
  2. 440 L
  3. 720 L
  4. 1000 L
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Difference in fuel usage}\)

\(=(11.4-9.6)\ \text{L/100 km}\)

\(=(1.8\ \text{L/100 km}\)
  

\(\therefore\ \text{Fuel saved using car}\ B\)

\(=\dfrac{11\ 000}{100}\times 1.8\)

\(=198\ \text{L}\)

\(\Rightarrow A\)

Algebra, STD2 A2 2019 HSC 14 MC (Adapted)

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?

  1. \(\text{Jake:}\ \ y=98x+5\)
    \(\text{Jack:}\ \ y=79x-4\)
  2. \(\text{Jake:}\ \ y=5+98x\)
    \(\text{Jack:}\ \ y=4-79x\)
  3. \(\text{Jake:}\ \ y=5x+98\)
    \(\text{Jack:}\ \ y=4x-79\)
  4. \(\text{Jake:}\ \ y=98+5x\)
    \(\text{Jack:}\ \ y=79-4x\)
Show Answers Only

\(D\)

Show Worked Solution

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

\(y=98+5x\)

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

\(y=79-4x\)

\(\Rightarrow D\)

Algebra, STD2 A2 2019 HSC 34 (Adapted)

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

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

  2. 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)

Show Answers Only
  1. \(p=\dfrac{5}{8}d\)
  2. \(107\ 520\ \text{yen = 800 pounds}\)
Show Worked Solution

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

\(p=\dfrac{5}{8}d\)


♦ Mean mark 42%.

b.   \(\text{Yen to Australian dollars:}\)

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

 
\(\text{Australian dollars to pounds:}\)

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

  
\(\therefore\ 107\ 520\ \text{yen = 800 pounds}\)

Algebra, STD1 A3 2021 HSC 25 (Adapted)

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)
 

Show Answers Only

Show Worked Solution


♦♦ Mean mark 38%.

Algebra, STD2 A2 2020 HSC 10 MC (Adapted)

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)?

  1. \(C=75+1.50t\)
  2. \(C=150+75t\)
  3. \(C=75+90t\)
  4. \(C=90+75t\)
Show Answers Only

\(C\)

Show Worked Solution

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

\(\therefore\ C=75+90t\)
  

\(\Rightarrow C\)


♦ Mean mark 42%.

Algebra, STD1 A2 2020 HSC 20 (Adapted)

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.

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

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

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(2000\ \text{sheets}\)
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%.

Algebra, STD2 A2 2021 HSC 18 (Adapted)

The fuel consumption for a medium SUV vehicle is 7.2 litres/100 km. On a road trip, the SUV travels a distance of 1325 km and the fuel cost is $2.15 per litre.

What is the total fuel cost for the trip?   (2 marks)

Show Answers Only

\($205.11\)

Show Worked Solution
\(\text{Total fuel used}\) \(=7.2\times \dfrac{1325}{100}\)  
  \(=95.4\ \text{litres}\)  

 

\(\text{Total fuel cost}\) \(=95.4\times 2.15\)  
  \(=$205.11\)  

Algebra, STD2 A2 2022 HSC 16 (Adapted)

Rhonda is 38 years old, and likes to keep fit by doing cross-fit classes.

  1. Use this formula to find her maximum heart rate (bpm).   (1 mark)
             Maximum heart rate = 220 – age in years

  2. Rhonda will get the most benefit from this exercise if her heart rate is between 65% and 85% of her maximum heart rate.
  3. Between what two heart rates should Rhonda be aiming for to get the most benefit from her exercise?   (2 marks)

Show Answers Only

a.   \(182\ \text{bpm}\)

b.   \(118-155\ \text{bpm}\)

Show Worked Solution
a.     \(\text{Max heart rate}\) \(=220-38\)
    \(=182\ \text{bpm}\)

 

b.    \(\text{65% max heart rate}\ = 0.65\times 182 = 118.3\ \text{bpm}\)

\(\text{85% max heart rate}\ = 0.85\times 182 = 154.7\ \text{bpm}\)

\(\therefore\ \text{Rhonda should aim for between 118 and 155 bpm during exercise.}\)

Algebra, STD2 A4 2022 HSC 22 (Adapted)

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.

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

  2. Some desktop computers have extra features added. The formula to calculate the production cost for these desktop computers is
  3.     \(C=80n+an+18\ 230\)
  4. where \(a\) is the additional cost in dollars per desktop computer produced.
  5. 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)

Show Answers Only
  1. \($214\ 870\)
  2. \(740\ \text{desktop computers}\)
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.   \(\text{Find}\ n,\ \text{given}\ C=18 \ 230\ \text{and}\ a=35\)

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

Algebra, STD2 A2 2012 HSC 5 MC (Adapted)

The line below has intercepts  \(m\)  and  \(n\),  where  \(m\) and  \(n\) are positive integers. 
  

What is the gradient of the line? 

  1. \(\dfrac{m}{n}\)
  2. \(\dfrac{n}{m}\)
  3. \(-\dfrac{m}{n}\)  
  4. \(-\dfrac{n}{m}\)
Show Answers Only

\(C\)

Show Worked Solution
 
♦ Mean mark 45%
\(\text{Gradient}\) \(=\dfrac{\text{rise}}{\text{run}}\)
  \(=-\dfrac{m}{n}\)

\(\Rightarrow C\)

Algebra, STD2 A2 2012 HSC 8 MC (Adapted)

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?

  1. 363
  2. 370
  3. 546
  4. 1092
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Linear relationship where}\)

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

\(\text{When}\ \ S=182\)

\(N\) \(=6+(2\times 182)\)
  \(=370\)

  
\(\Rightarrow B\)

Algebra, STD2 A2 2009 HSC 14 MC (Adapted)

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

  1. \(3\)
  2. \(15\)
  3. \(3x\)
  4. \(5x\)
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 \(x=3,\ C=19\).

Functions, 2ADV F1 2009 HSC 1a (Adapted)

Sketch the graph of  \(y+\dfrac{x}{3} = 2\), showing the intercepts on both axes.   (2 marks)

Show Answers Only

Show Worked Solution

\(y+\dfrac{x}{3} = 2\ \ \Rightarrow\ \ y=-\dfrac{1}{3}x+2\)

\(y\text{-intercept}\ = 2\)

\(\text{Find}\ x\ \text{when}\ y=0:\)

\(\dfrac{x}{3}=2\ \ \Rightarrow\ \ x=6\)
 

Algebra, STD2 A2 2014 HSC 7 MC (Adapted)

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%

\(y=-3x-3\)

\(\text{By elimination:}\)

\(\ y\text{-intercept}=-3\)

\(\rightarrow\ \text{Cannot be}\ B\ \text{or}\ D\)

  

\(\text{Gradient}=-3\)

\(\rightarrow\ \text{Cannot be}\ C\)

\(\Rightarrow A\)

Algebra, STD2 A2 2007 HSC 18 MC (Adapted)

Art started to make this pattern of shapes using matchsticks.
  

 

If the pattern of shapes is continued, which shape would use exactly 416 matchsticks?

  1. Shape 83
  2. Shape 103
  3. Shape 104
  4. Shape 138
Show Answers Only

\(D\)

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape}\ \textit(S) \rule[-1ex]{0pt}{0pt}\ \ &\  \ 1\ \ &\ \ 2\ \ &\ \ 3\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Matches}\ \textit(M) \rule[-1ex]{0pt}{0pt} \ \ & \ \ 5\ \ &\ \ 8\ \ &\ \ 11\ \  \\
\hline
\end{array}

\(\text{Equation rule:}\)

\(M=3S+2\)

\(\text{Find}\ \ S\ \text{when}\ \ M=416:\)

\(416\) \(=3S+2\)
\(3S\) \(=414\)
\(S\) \(=138\)

 
\(\therefore\ \text{The 138th shape uses 416 matchsticks.}\)

\(\Rightarrow D\)

Algebra, STD2 A2 2011 HSC 23b (Adapted)

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}

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

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

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

     

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

Show Answers Only

a.    \(\text{See Worked Solutions.}\)

b.    \(514\)

c.    \(\text{No (See worked solution)}\)

Show Worked Solution

a.    \(\text{Shape 4 is shown below:}\)

b.    \(\text{Since}\ \ N=2+4S\)

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

 

c.     \(609\) \(=2+4S\)
  \(4S\) \(=607\)
  \(S\) \(=151.75\)

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

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

Algebra, STD2 A2 2004 HSC 2 MC (Adapted)

Michael drew a graph of the height of a plant.
  

What is the gradient of the line?

  1. 1
  2. 3
  3. 4.5
  4. 6
Show Answers Only

\(B\)

Show Worked Solution

\(\text{2 points on graph}\ \ (0, 6),\ (1, 9)\)

\(\text{Gradient}\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{9-6}{1-0}\)
  \(=3\)

\(\Rightarrow B\)

Algebra, STD2 A2 2006 HSC 7 MC (Adapted)

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} 

  1. \(y=2x+3\)
  2. \(y=\dfrac{1}{2}x+3\) 
  3. \(y=\dfrac{1}{2}x-3\)
  4. \(y=3x-2\)
Show Answers Only

\(B\)

Show Worked Solution

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

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

\(\therefore\ \text{NOT}\ C\ \text{or}\ D\)

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

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

\(\Rightarrow B\)