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Algebra, STD1 2026 HSC 15

Algae grows in a pond when sunlight and nutrients are present.

The mass of algae in the pond over a number of days is shown in the table.

Time \(t\)
(days)		 Mass \(A\)
(mg)
\(0\) \(5\)
\(1\) \(40\)
\(2\) \(320\)
\(3\) \(2560\)
\(4\) \(20\,480\)
\(5\)  

 
The equation to model the growth of the algae is given by

\(A=5\times2^{3t}\)

where \(A\) is the mass after a certain number of days in milligrams and \(t\) is the number of days elapsed.

  1. What was the mass of algae present in the pond on day 5?   (1 mark)

The first five rows of the table have been used to create part of the graph of  \(A=5 \times 2^{3t}\)  which is shown.

  1. Plot the mass found in part (a) and use this to complete the graph.   (2 marks)
  2. When does the mass of the algae reach 100 000 mg? Use your graph to estimate.   (1 mark)
Show Answers Only

a.    \(163\,840\text{ mg}\)

b.    \(\text{See the completed curve:}\)

c.    \(\text{Approximately 4.75 days}\)

Show Worked Solution
a.    \(A\) \(=5\times2^{3(5)}\)  
  \(=5\times2^{15}\)  
  \(=163\,840\text{ mg}\)  

 
b.    \(\text{See the completed curve:}\)

c.    \(\text{From the graph:}\)

\(A=100\,000\text{ mg at approximately }t=4.75\text{ days.}\)

Algebra, STD1 EQ-Bank 25

A scientist is studying a colony of bacteria in a laboratory. At the start of the experiment there are 500 bacteria.

Each hour the number of bacteria is modelled to be 1.25 times the number of the previous hour.

Let  \(t=0\)  be the start of the experiment.

  1. By first completing the table of values below, draw a graph showing how the bacteria population is modelled from  \(t=0\)  to  \(t=8\) hours.   (3 marks)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & & & & \\ \hline \end{array}\)
 

 

  1. Using your graph from (a), or otherwise, determine after how many hours the bacteria population first exceeds 2000.   (1 mark)

Show Answers Only

a.    \(\text{Table of values:}\)

\(\text{Number of bacteria} = 500 \times 1.25^t\)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & \textbf{625} & \textbf{781} & \textbf{1221} & \textbf{2980} \\ \hline \end{array}\)
 

b.    \(\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}\)

Show Worked Solution

a.    \(\text{Table of values:}\)

\(\text{Number of bacteria} = 500 \times 1.25^t\)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & \textbf{625} & \textbf{781} & \textbf{1221} & \textbf{2980} \\ \hline \end{array}\)
 

b.    \(\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}\)

Algebra, STD1 EQ-Bank 22

The population of a certain town on 1 January 2024 was 2000 people.

Each year the population of this town is modelled to be 1.18 times the population of the previous year.

Let  \(t=0\)  be 1 January 2024 .

By first completing the table for the indicated years, draw a graph showing how the population is modelled from 1 January 2024 to 1 January 2034.   (4 marks)

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 & & & & \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 &2360 &2785 &4576 & 10468 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 &2360 &2785 &4576 & 10468 \\
\hline
\end{array}

Algebra, STD1 EQ-Bank 26

Rangers at a nature reserve are monitoring the spread of an invasive weed. At the start of monitoring there are 100 weeds. The number of weeds is growing at a rate of 50% per month.

Let  \(N\) = number of weeds, and  \(t\) = time in months.

  1. Complete the table of values below that models the growth of the weeds over an 8 month period. Round all answers to the nearest whole number.   (2 marks)
      
    \(\begin{array}{|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}N & 100 & & & & \\[12pt] \hline \end{array}\)

  2. Using the table of values from (a), neatly plot the points and join with a smooth curve.   (2 marks)

      

Show Answers Only

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|} \hline \quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad  \\[6pt] \hline \ N  & 100 & 150 & 225 & 506 & 2563 \\[6pt] \hline \end{array} 

b.      

   

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|} \hline \quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad  \\[6pt] \hline \ N  & 100 & 150 & 225 & 506 & 2563 \\[6pt] \hline \end{array}

\(\text{Algebraic method}\)

\(\text{Formula:  }\ N=100\times1.5^t\)

\(t=0:\ N=100\times1.5^{0}=100\)

\(t=1:\ N=100\times1.5^{1}=150\)

\(t=2:\ N=100\times1.5^{2}=225\)

\(t=4:\ N=100\times1.5^{4}=506.25\approx506\)

\(t=8:\ N=100\times1.5^{8}=2562.89\ldots\approx2563\)
  

\(\text{Using CASIO calculator with constant multiplier}\)

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline {0} & 100= & 100 & 100 \\
\hline {1} & \text{Ans}\times 1.5= & 150  & 150 \\
\hline {2} & = & 225 & 225 \\
\hline {4} & = & 506.25 & 506 \\
\hline {8} & = & 2562.890625 & {2563} \\
\hline \end{array}  

b.      

   

Algebra, STD1 EQ-Bank 27

A conservation program is tracking the recovery of a native wildflower species in a national park. At the start of the program there are 200 plants. The number of plants is predicted to grow at a rate of 25% per year.

Let  \(N\) = number of plants, and  \(t\) = time in years.

  1. Complete the table of values below that models the growth of the plants. Round all answers to the nearest whole number.   (2 marks)
    \begin{array}{|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
    \hline
    \rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
    \hline
    \end{array}

  2. Using the table of values from (a), neatly plot the points and join with a smooth curve.   (2 marks)

      

Show Answers Only

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

b.     

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

\(\text{Algebraic method}\)

\(\text{Formula:   }\ N=200\times1.25^t\)

\(t=0:\ N=200\times1.25^{0}=200\)

\(t=1:\ N=200\times1.25^{1}=250\)

\(t=2:\ N=200\times1.25^{2}=312.5\approx313\)

\(t=3:\ N=200\times1.25^{3}=390.625\approx391\)

\(t=4:\ N=200\times1.25^{4}=488.28\ldots\approx488\)
  

\(\text{Using CASIO calculator with constant multiplier}\)

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline \colorbox{lightblue}{0} & 200= & 200 & \colorbox{lightblue}{200} \\
\hline \colorbox{lightblue}{1} & \text{Ans}\times 1.25= & 250  & \colorbox{lightblue}{250} \\
\hline \colorbox{lightblue}{2} & = & 312.5 & \colorbox{lightblue}{313} \\
\hline \colorbox{lightblue}{3} & = & 390.625 & \colorbox{lightblue}{391} \\
\hline \colorbox{lightblue}{4} & = & 488.28\ldots & \colorbox{lightblue}{488} \\
\hline \end{array}  

b.     

   

Algebra, STD1 A3 2023 HSC 4 MC

The diagram shows water in a pool which is in the shape of a triangular prism. The pool is being emptied of water at a constant rate.

 

Which graph best illustrates the change in depth of water with time?

Show Answers Only

\(D\)

Show Worked Solution

\(\text{Eliminate B and C as both graphs show the depth of water}\)

\(\text{increasing with time, but the pool is being emptied, so}\)

\(\text{depth must decrease with time.}\)

\(\text{Eliminate A as rate of flow out of the tank is not linear.}\)

\(\Rightarrow D\)

♦ Mean mark 49%.

Algebra, STD1 A3 2022 HSC 10 MC

The diagram shows a container, closed at the base. It is to be filled with water at a constant rate.
 

 

Which graph best shows the depth of water in the container as time varies?
 


  

Show Answers Only

\(A\)

Show Worked Solution

\(\text{The container will fill at a constant rate for the cylindrical}\)

\(\text{section of the container → a straight line (linear graph).}\)

\(\text{The container will fill more slowly at a decreasing rate }\)

\(\text{for the conical section of the container → a curved line }\)

\(\text{(non-linear graph).}\)

\(\text{Therefore, }A\ \text{and }B\ \text{are the only options.}\)

\(\text{It cannot be graph}\ B\ \text{as it shows the depth increasing}\)

\(\text{at an increasing rate after the straight line section.}\)

\(\Rightarrow A\)


♦♦♦ Mean mark 24%.

Algebra, STD1 A3 2020 HSC 19

Each year the number of fish in a pond is three times that of the year before.

  1. The table shows the number of fish in the pond for four years.
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
    \hline
    \rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & & & 2700\\
    \hline
    \end{array}

    Complete the table above showing the number of fish in 2021 and 2022.   (2 marks)
     

  2. Plot the points from the  table in part (a) on the grid.   (2 marks)
     
  3. Which model is more suitable for this dataset: linear or exponential? Briefly explain your answer.   (2 marks)

Show Answers Only

a.   

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.   
       

c.    \(\text{The more suitable model is exponential.}\)

\(\text{A linear dataset would graph a straight line which is}\)

\(\text{not the case here.}\)

\(\text{An exponential curve can be used to graph populations that }\)

\(\text{grow at an increasing rate, such as this example.}\)

Show Worked Solution

a.        

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.  

c.    \(\text{The more suitable model is exponential.}\)

\(\text{A linear dataset would graph a straight line which is}\)

\(\text{not the case here.}\)

\(\text{An exponential curve can be used to graph populations that }\)

\(\text{grow at an increasing rate, such as this example.}\)

♦♦ Mean mark (c) 31%.

Algebra, STD1 A3 2020 HSC 14

Adam travels on a straight road away from his home. His journey is shown in the distance–time graph.

  1. Describe the journey in the first 4 minutes by referring to change in speed and distance travelled.   (2 marks)

  2. After the 4 minutes shown on the graph. Adam rests for 2 minutes and then returns home by travelling on the same road at a constant speed. Adam is away from home for a total of 10 minutes.

     

    On the grid above, complete the distance–time graph using the information provided.   (2 marks)

Show Answers Only

a.    `text{Speed: Adam increases speed until approximately}\ \ t=2,`

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

 b.

Show Worked Solution

a.    `text{Speed: Adam increases speed until approximately}\ \ t=2,`

♦ Mean mark part (a) 35%.

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Mean mark part (b) 51%.
   

b.    `text(S)text(ince Adam travels home at a constant speed, the graph is)`

`text{a straight line and ends at (10, 0).}`
 

Algebra, STD1 A3 2019 HSC 23

Five rabbits were introduced onto a farm at the start of 2018. At the start of 2019 there were 10 rabbits on the farm. It is predicted that the number of rabbits on the farm will continue to double each year.

  1. Complete the following table.   (1 mark)
     


     

  2. Complete the scale on the vertical axis and then plot the data from part (a) on the grid.   (2 marks)


      

  3. Would a linear model or an exponential model better fit this graph? Explain the reason for your answer.   (1 mark)

Show Answers Only

a.

b.

c.    `text(Exponential model − the graph isn’t a straight line.)`

`text(The number of rabbits grow at an increasing rate.)`

Show Worked Solution
a.    

 

b.    

♦ Mean mark part (c) 25%.

c.    `text(Exponential model − the graph isn’t a straight line.)`

`text(The number of rabbits grow at an increasing rate.)`

Algebra, STD1 A3 2019 HSC 9 MC

The container shown is initially full of water.
 

Water leaks out of the bottom of the container at a constant rate.

Which graph best shows the depth of water in the container as time varies?
 

A. B.
C. D.
Show Answers Only

`D`

Show Worked Solution

`text(Depth will decrease slowly at first and accelerate.)`

♦♦♦ Mean mark 10%.

`=> D`