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Algebra, STD2 EQ-Bank 31

The graph of the parabola \(y=a(x+1)(x+7)\) for some value of \(a\) is shown.
 

By first finding the value of \(a\), find the coordinates of the vertex.   (3 marks)

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\(\text{Vertex:}\ (-4,-27)\)

Show Worked Solution

\(\text{Since graph passes through}\ (0,21):\)

\(21\) \(=a(0+1)(0+7)\)
\(21\) \(=7a\)
\(a\) \(=3\)

 
\(\text{Vertex is halfway between \(x\)-intercepts.}\)

\(\Rightarrow \ x=\dfrac{-7+(-1)}{2}=-4\)

\(y\) \(=3(-4+1)(-4+7)\)
  \(=3 \times (-3) \times 3\)
  \(=-27\)

 
\(\therefore\ \text{Vertex at}\ (-4,-27).\)

Algebra, STD2 EO-Bank 30

The number of monthly subscribers \((S)\) to a print magazine is decreasing. The number of subscribers is modelled using

\(S=k(1.06)^{-t}\)  for  \( t \geq 0,\)

where \(t\) is time in months.

The number of subscribers today is 12 000.

  1. Use the model to estimate the number of subscribers 10 months from today.   (2 marks)

  2. The number of subscribers needs to always be above 3000.
  3. Explain why this model is NOT appropriate to use in this case.   (1 mark)

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a.    \(S=12\,000(1.06)^{-10}=6700.73…=6701\ \text{(nearest subscriber)}\)

b.    \(\text{When}\ \ t=30\ \ \Rightarrow\ \ S \approx 2089\)

\(\text{Since the number of subscribers needs to always be over 3000, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=0,\ \ S=12\,000:\)

\(12\,000=k(1.06)^{0}\ \ \Rightarrow\ \ k=12\,000\)

\(\text{Find}\ S\ \text{when}\ \ t=10:\)

\(S=12\,000(1.06)^{-10}=6700.73…=6701\ \text{(nearest subscriber)}\)
 

b.    \(\text{When}\ \ t=30:\)

\(S=12\,000(1.06)^{-30} \approx 2089\)

\(\text{Since the number of subscribers needs to always be over 3000, it is not appropriate.}\)

Algebra, STD2 EO-Bank 29

The number \((N)\) of native birds in a wildlife reserve is decreasing. The number of birds is modelled using

\(N=k(1.08)^{-t}\)  for \( t \geq 0,\)

where \(t\) is time in years.

The number of birds in the reserve today is 8000.

  1. Use the model to estimate the number of birds in the reserve 12 years from today.   (2 marks)

  2. The number of birds in the reserve needs to always be above 1500.
  3. Explain why this model is NOT appropriate to use in this case.   (1 mark)

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a.    \(N=8000(1.08)^{-12}=3176.91…=3177\ \text{(nearest bird)}\)

b.    \(\text{When}\ \ t=25:\)

\(N=8000(1.08)^{-25}=1168.14…\)

\(\text{Since the number of birds needs to always be over 1500, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=0,\ \ N=8000:\)

\(8000=k(1.08)^{0}\ \ \Rightarrow\ \ k=8000\)

\(\text{Find}\ N\ \text{when}\ \ t=12:\)

\(N=8000(1.08)^{-12}=3176.91…=3177\ \text{(nearest bird)}\)
 

b.    \(\text{When}\ \ t=25:\)

\(N=8000(1.08)^{-25}=1168.14…\)

\(\text{Since the number of birds needs to always be over 1500, it is not appropriate.}\)

Algebra, STD2 A4 2021 HSC 35 (Adapted)

A toy store releases a limited edition LEGO set for $20 each. At this price, 3000 LEGO sets are sold each week and the revenue is  `3000 xx 20=$60\ 000`.

The toy store considers increasing the price. For every dollar price increase, 15 fewer LEGO sets will be sold.

If the toy store charges `(20+x)` dollars for each LEGO set, a quadratic model for the revenue raised, `R`, from selling them is

`R=-15x^2+2700x+60\ 000`

 


 

  1. What price should be charged per LEGO set to maximise the revenue?   (2 marks)

  2. How many LEGO sets are sold when the revenue is maximised?   (2 marks)

  3. Find the value of the intercept of the parabola with the vertical axis.   (1 mark) 

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a.   `$110`

b.    `1650`

c.   `$60\ 000`

Show Worked Solution

a.   `text{Highest revenue}\ (R_text{max})\ text(occurs halfway between)\ \ x= -20 and x=200.`

`text{Midpoint}\ =(-20 + 200)/2 = 90`

`:.\ text(Price of LEGO set for)\ R_text(max)`

`=90 + 20`

`=$110`
 

b.  `text{LEGO sets sold when}\ R_{max}`

`=3000-(90 xx 15)`

`=1650`
 

c.   `ytext(-intercept → find)\ R\ text(when)\ \ x=0:`

`R` `= -15(0)^2 + 2700(0) + 60\ 000`
  `=$60\ 000`

Algebra, STD2 A4 2017 HSC 28e (Adapted)

Sage brings 60 cartons of unpasteurised milk to the market each week. Each carton currently sells for $4 and at this price, all 60 cartons are sold each weekend.

Sage considers increasing the price to see if the total income can be increased.

It is assumed that for each $1 increase in price, 6 fewer cartons will be sold.

A graph showing the relationship between the increase in price per carton and the income is shown below.

 


 

  1. What price per carton should be charged to maximise the income?   (1 mark)

  2. What is the number of cartons sold when the income is maximised?   (1 mark)

  3. The cost of running the market stall is $40 plus $1.50 per carton sold.

    Calculate Sage's profit when the income earned from a day selling at the market is maximised.   (2 marks)

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a.   `$7`

b.   `42`

c.   `$191`

Show Worked Solution

a.   `text(Graph is highest when increase = $3)`

`:.\ text(Carton price)\ = 4 + 3= $7`
 

b.   `text(Cartons sold)\ =60-(3 xx 6)=42`
  

c.   `text{Cost}\ = 42 xx 1.50 + 40 = $103`

`:.\ text(Profit when income is maximised)`

`= (42 xx 7)-103`

`= $191`

Algebra, STD2 A4 2009 HSC 28c (Adapted)

The brightness of a lamp \((L)\) is measured in lumens and varies directly with the square of the voltage \((V)\) applied, which is measured in volts.

When the lamp runs at 7 volts, it produces 735 lumens.

What voltage is required for the lamp to produce 1820 lumens? Give your answer correct to one decimal place.   (3 marks)

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 `11.2\ \text(volts)`

Show Worked Solution

`L prop V^2\ \ => \ \ L=kV^2`

`text(Find)\ k\ \text{given}\ L = 735\ \text{when}\ V = 7:`

`735` `= k xx 7^2`
`:. k` `= 735/49=15`

 
`text(Find)\ V\ text(when)\ L = 1820:`

`1820` `= 15 xx V^2`
`V^2` `= 1820/15=121.33…`
`V` `= sqrt{121.33} = 11.2\ text(volts)\ \ text{(to 1 d.p.)}`

Algebra, STD2 A4 2021 HSC 24 (Adapted)

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 = 2: `

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

♦ Mean mark (c) 48%.

 
c.   `\text{At}\ t=6:`

`D=650(0.8)^6=170.39…`
 

Algebra, STD2 EO-Bank 28

Taylor discovers that for a Spotted stingray, its mass is directly proportional to the square of its wingspan.

One Spotted stingray has a wingspan of 60 cm and a mass of 5400 grams.

What is the expected wingspan of a Spotted stingray with a mass of 9.6 kg?   (3 marks)

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`80.0\ text{cm}`

Show Worked Solution

`text(Mass) prop text(wingspan)^2\ \ =>\ \ m = kw^2`

`text(Find)\ k:`

`5400` `= k xx 60^2`
`k` `= 5400/60^2= 1.5`

 
`text(Find)\ w\ text(when)\ \ m = 9600:`

`9600` `= 1.5 xx w^2`
`w^2` `= 9600/1.5=6400`
`:. w` `= 80\ text{cm}`

Algebra, STD2 A4 2023 HSC 22 (Adapted)

The stopping distance of a motor bike, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{stopping distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The stopping distance for a motor bike travelling at 40 km/h is 16 m.

  1. Find the value of `k`.   (2 marks)

  2. What is the stopping distance when the speed of the motor bike is 80 km/h?   (1 mark)

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a.    `k=0.01`

b.    `64.0\ text{m}`

Show Worked Solution

a.  `text{stopping distance}\ = k xx text{(speed)}^2`

`16` `=k xx 40^2`  
`k` `=16/40^2=0.001`  

 
b.    `text{Find stopping distance}\ (d)\ text{when speed = 80 km/h:}`

`d=0.01 xx 80^2=64.0\ text{m}`

Algebra, STD2 A4 2013 HSC 22 MC (Adapted)

Jevin wants to build a rectangular chicken pen. He has 32 metres of fencing and will use a barn wall as one side of the pen. The width of the pen is \(d\) metres.
 

Which equation gives the area, \(P\), of the chicken pen?

  1. \(P = 16d-d^2\)
  2. \(P = 32d-d^2\)
  3. \(P = 16d-\dfrac{d^2}{2}\)
  4. \(P = 16d-2d^2\) 
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\(C\)

Show Worked Solution
♦♦♦ Mean mark 24% (lowest mean of any MC question in 2013 exam)

\(\text{Length of pen}\ = \dfrac{1}{2}(32-d)\)

\(\text{Area}\ =d \times \dfrac{1}{2}(32-d)=16d-\dfrac{d^2}{2}\)

 \(\Rightarrow C\)

Algebra, STD2 A4 2021 HSC 10 MC (Adapted)

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 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
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\(C\)

Show Worked Solution

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

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

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

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

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

\(\Rightarrow C\)