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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.)`

`=> D`

Filed Under: Non-Linear: Inverse and Other Problems (Std 2), Quadratics, Variation and Rates of Change Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-4239-60-Variable rates of change, smc-795-20-Other Relationship

Algebra, STD2 A4 2007 HSC 27a

   A rectangular playing surface is to be constructed so that the length is 6 metres more than the width.

  1. Give an example of a length and width that would be possible for this playing surface.   (1 mark)

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  2. Write an equation for the area (`A`) of the playing surface in terms of its length (`l`).   (1 mark)

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    A graph comparing the area of the playing surface to its length is shown.
     
       
     

  3. Why are lengths of 0 metres to 6 metres impossible?   (1 mark)

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  4. What would be the dimensions of the playing surface if it had an area of 135 m²?  (2 marks)

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    Company `A` constructs playing surfaces.

         

  5. Draw a graph to represent the cost of using Company `A` to construct all playing surface sizes up to and including 200 m².

     

    Use the horizontal axis to represent the area and the vertical axis to represent the cost.   (2 marks)

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  6. Company `B` charges a rate of $360 per square metre regardless of size. 
  7. Which company would charge less to construct a playing surface with an area of 135 m²

     

    Justify your answer with suitable calculations.   (1 mark)

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

Show Answers Only
  1. `text(One possibility is a length of 10 m, and a width)`

     

    `text{of 4 m (among many possibilities).}`

  2. `A =  l (l – 6)\ \ text(m²)`
  3. `text(Given the length must be 6 m more than)`

     

    `text(the width, it follows that the length)`

     

    `text(must be greater than 6 m.)`

  4. `text(15 m × 9 m)`
  5.   
       
     
  6. `text(Proof)\ \ text{(See worked solutions)}`
Show Worked Solution

i.   `text(One possibility is a length of 10 m, and a width)`

`text{of 4 m (among many possibilities).}`

 

ii.  `text(Length) = l\ text(m)`

`text(Width) = (l – 6)\ text(m)`

`:.\ A` `= l (l – 6)`

 

iii.  `text(Given the length must be 6m more than the width,)`

 `text(it follows that the length must be greater than 6 m)`

`text(so that the width is positive.)`

 

iv.  `text(From the graph, an area of 135 m² corresponds to)`

`text(a length of 15 m.)`

`:.\ text(The dimensions would be 15 m × 9 m.)`

 

v.   

  

vi.  `text(Company)\ A\ text(cost) = $50\ 000`

`text(Company)\ B\ text(cost)` `= 135 xx 360`
  `= $48\ 600`

 

`:.\ text(Company)\ B\ text(would charge $1400 less)`

`text(than Company)\ A.`

Filed Under: Breakeven and Financial modelling, Non-Linear: Inverse and Other Problems (Std 2), Perimeter/Area problem Tagged With: Band 3, Band 4, Band 5, page-break-before-question, smc-795-20-Other Relationship, smc-795-30-Limitations

Algebra, STD2 A4 2013 HSC 27a

Lucy went for a bike ride. She left home at 8 am and arrived back at home at 6 pm. A graph representing her journey is shown.

2013 27a

  1. What was the total distance that she rode during the day?   (1 mark)

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  2. How much time did Lucy spend riding her bike during the day?     (1 mark)

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Show Answers Only
  1. `70\ text(km)`
  2. `6.5\ text(hours)`
Show Worked Solution
i.    `text(Total distance)` `=35 +35`
    `= 70\ text(km)`

 

ii.    `text(Time away from home) = 10\ text(hours)`
`text(Time resting)` `= 1 + 1.5+ 1`
  `= 3.5\ text(hours)`

 

`:.\ text(Time riding)` `=10 – 3.5`
  `=6.5\ text(hours)`

Filed Under: AM2 - Linear Relationships (Prelim), Applications: D=SxT and Other (Std 1), Non-Linear: Inverse and Other Problems (Std 2) Tagged With: Band 4, smc-1117-20-Distance Speed Time, smc-795-20-Other Relationship

Algebra, STD2 A2 2009 HSC 13 MC

The volume of water in a tank changes over six months, as shown in the graph.
 

 2UG-2010-13MC

 
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. 11%
  3. 32%
  4. 64%
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 48%
COMMENT: Remember that % decrease requires the decrease in volume to be divided by the original volume (50,000L).
`text(Initial Volume)` `= 50\ 000\ text(L)`
`text(Final volume)` `= 18\ 000\ text(L)`
`text(Decrease)` `= 50\ 000-18\ 000`
  `= 32\ 000\ text{L   (over 6 months)}`

 

`text(Loss per month)` `= (32\ 000)/6`
  `= 5333.33…\ text(L per month)`
`text(% loss per month)` `= (5333.33…)/(50\ 000)`
  `=10.666… %`

 
`=>  B`

Filed Under: AM2 - Linear Relationships (Prelim), Applications: Currency, Fuel and Other Problems (Std 1), Applications: Currency, Fuel and Other Problems (Std 2), Applications: Currency, Fuel and Other Problems (Std2-2027), MM1 - Units of Measurement, Non-Linear: Inverse and Other Problems (Std 2) Tagged With: Band 5, smc-1119-30-Other Linear Applications, smc-6256-30-Other Linear Applications, smc-793-30-Other Linear Applications, smc-795-20-Other Relationship

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