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Algebra, STD2 A4 2025 HSC 39

After a dose of a medication, the amount of the medication remaining in a person can be modelled by the equation  \(y=k a^x\),  where \(x\) is the number of hours after taking the dose, and \(y\) is the amount remaining in milligrams (mg).

The graph shows the amount of the medication remaining in a person after \(x\) hours. Two points are also shown on the graph.
 

Using the information provided, find the amount of medication that remains in a person when  \(x=4\).   (3 marks)

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\(5.4 \ \text{mg}\)

Show Worked Solution

\(\text{Given}\ (0,15) \ \text{lies on graph:}\)

   \(15=k \times a^{0} \ \ \Rightarrow \ \ k=15\)

\(\text{Find \(a\) given \((2,9)\) lies an graph:}\)

\(9\) \(=15 \times a^2\)
   \(a^2\) \(=\dfrac{9}{15}\)
\(a\) \(=\sqrt{\dfrac{9}{15}}\)

   

\(\text{Find \(y\) when  \(\ x=4\):}\)

\(y=15 \times\left(\sqrt{\dfrac{9}{15}}\right)^4=5.4 \ \text{mg}\)

v1 Algebra, STD2 A4 2021 HSC 24

A population of Tasmanian devils, `D`, is to be modelled using the function  `D = 650 (0.8)^t`, where `t` is the time in years.

  1. What is the initial population?   (1 mark)

  2. Find the population after 2 years.   (1 mark)

  3. On the axes below, draw the graph of the population against time, in the period  `t = 0`  to  `t = 6`.   (2 marks)
      

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a.   `650`

b.   `416`

c.   `text{See Worked Solutions}`

Show Worked Solution

a.   `text{Initial population occurs when}\ \  t = 0:`

`D=650(0.8)^0=650 xx 1= 650`
 

b.    `text{Find} \ D \ text{when} \ \ t = 5: `

`D= 650 (0.8)^2= 416`

♦ Mean mark (c) 48%.

 
c.   `\text{At}\ t=6:`

`D=650(0.8)^6=170.39…`
 

v1 Algebra, STD2 A4 2021 HSC 10 MC

Which of the following best represents the graph of  \(y = 5 (0.4)^{x}\) ?
 

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\(D\)

Show Worked Solution

\(\text{By elimination:}\)

♦ Mean mark 41%.

\(\text{When}\  x = 0, \ y = 5 \times (0.4)^0 = 5\)

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

\(\text{As}\ \ x \rightarrow \infty, \ y \rightarrow 0 \)

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

\(\Rightarrow D\)

Algebra, STD2 A4 2024 HSC 22

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

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\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Mean mark 51%.

Algebra, STD2 A4 2021 HSC 24

A population, `P`,  is to be modelled using the function  `P = 2000 (1.2)^t`, where `t` is the time in years.

  1. What is the initial population?  (1 mark)

  2. Find the population after 5 years.  (1 mark)

  3. On the axes below, draw the graph of the population against time, showing the points at  `t = 0`  and at  `t = 5`.  (2 marks)
      

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  1. `2000`
  2. `4977`
  3. `text{See Worked Solutions}`
Show Worked Solution

a.  `text{Initial population occurs when}\ \  t = 0:`

`P= 2000 (1.2)^0= 2000`
 

b.    `text{Find} \ P \ text{when} \ \ t = 5: `

`P` `= 2000 (1.2)^5`  
  `= 4976.64`  
  `= 4977 \ text{(nearest whole)}`  

 

♦ Mean mark (c) 48%.

c. 

Algebra, STD2 A4 2021 HSC 10 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

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`A`

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`\text{By elimination:}`

♦ Mean mark 41%.

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

Algebra, STD2 A4 2020 HSC 33

The graph shows the number of bacteria, `y`, at time `n` minutes. Initially (when `n = 0`) the number of bacteria is 1000.
 


 

  1. Find the number of bacteria at 40 minutes.   (1 mark)

  2. The number of bacteria can be modelled by the equation  `y = A xx b^n`, where `A` and `b` are constants.
    Use the guess and check method to find, to two decimal places, an upper and lower estimate for the value of `b`. The upper and lower estimates must differ by 0.01. (2 marks)

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  1. `4000`
  2. `text{Show Worked Solutions}`
Show Worked Solution

a.    `text{When} \ \ n = 40,`

`text{Number of Bacteria} \ (y) = 4000`
 

b.   `A = 1000 \ => \ y = 1000  b^n`

♦♦♦ Mean mark 14%.

`text{By inspection, graph passes through (40, 4000)}`

`=> \ 4000 = 1000  b^40`
 

`text(Guess and check possible values of)\ b:`

`text{If} \ \ b = 1.03 \ , \ \ y = 1000 xx 1.03^40 = 3262 \ text{(too low)}`

`text{If} \ \ b = 1.04 \ , \ \ y = 1000 xx 1.04^40 = 4801 \ text{(too high)}`

`therefore \ 1.03 < b < 1.04`

Algebra, STD2 A4 2011 HSC 26b

Jack needs to find the number of years, `t`, it will take for a population of bats to first exceed 18 000.

He uses a ‘guess-and-check’ method to estimate `t` in the following equation

`5 xx 3^t = 18\ 000.`

Here is his working:

  1. Jack’s next guess is  `t = 6`. Show Jack’s correct working for this guess, including the calculation and conclusion.  (1 mark)

  2. Continue using the ‘guess-and-check’ method to find the number of years, `t`, it will take for the population to first exceed 18 000, if `t` is a whole number. Include the calculations and conclusions.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `8`
Show Worked Solution

i.   `text(When)\ t = 6,`

`5 xx 3^6 = 3645`

`=>\ text(Too small)`

 

ii.   `text(When)\ \ t = 7,`

`5 xx 3^7 = 10\ 935`

`=>\ text(Too small)`
 

`text(When)\ \ t = 8,`

`5 xx 3^8 = 32\ 805`

`=>\ text(exceeds 18 000)`

`:. t = 8`

Algebra, STD2 A4 2016 HSC 29b

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

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

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

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  1. `1.5\ text(kg)`
  2. `10text(%)`
Show Worked Solution

i.   `text(When)\ x = 0,`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)`

`= 0.1`

`= 10text(%)`

Algebra, STD2 A4 2004 HSC 26a

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (4 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

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  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(8.4 hours)`
Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n` `= 100(1.14)^15`
  `= 713.793\ …`
  `= 714\ \ \ text{(nearest whole)}`

 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after)`

`text(approximately 8.4 hours.)`

Algebra, STD2 A4 2006 HSC 14 MC

In 2004 there were 13.5 million registered motor vehicles in Australia. The number of registered motor vehicles is increasing at a rate of 2.3% per year.

Which expression represents the number (in millions) of registered motor vehicles, if `y` represents the number of years after 2004?

  1. `13.5 xx (1.023)^y`
  2. `13.5 xx (0.023)^y`
  3. `13.5 xx (1.023) xx y`
  4. `13.5 xx (0.023) xx y`
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`A`

Show Worked Solution

`text(In 2004, 13.5 million)`

`text(1 year later)` `= 13.5 xx (1.023)`
`text(2 years later)` `= 13.5 xx (1.023) xx (1.023)`
  `= 13.5 xx (1.023)^2`
`:. y\ text(years later)` `= 13.5 xx (1.023)^y`

`=>  A`

Algebra, STD2 A4 2012 HSC 30c

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

`P = A(1.04)^n`.

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

`P = A(b)^n`.

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

The graph shows the two predicted populations.
 

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

  2. In each of the two equations given, the value of `A` is 3 000 000.

     

    What does `A` represent?   (1 mark)

  3. The guess-and-check method is to be used to find the value of `b`, in  `P = A(b)^n`.

     

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

     

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

  4. The city of Thagoras was aiming to have a population under 7 000 000 in 2050. Does the model indicate that the city will achieve this aim?

     

    Justify your answer with suitable calculations.  (2 marks)

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

     

    `(2)\ \ b = 1.03, 1.02`

  4. `text(See Worked Solution)`
Show Worked Solution

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

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

COMMENT: Common ADV/STD2 content in new syllabus.

 

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

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

♦ Mean mark part (ii) 48%

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

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

iii. (2)  `text(Let)\ \ b = 1.03`

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

 
`text(Let)\ \ b = 1.02`

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

 
`:. b = 1.02`
 

iv.  `text(In 2050,)\ n = 40`

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

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

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