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v1 Algebra, STD2 A4 2020 HSC 24

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)

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  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%.

v1 Algebra, STD2 A4 SM-Bank 7

The graph of the line  \(x+y=3\)  is shown.
 


 

By graphing  \(y=2x-3\)  on the same grid, find the point of intersection of  \(x+y=3\) and  \(y=2x-3\).  (3 marks)

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\((2, 1)\)

Show Worked Solution

\(\text{Graphing}\ y=2x-3:\)

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

\(\text{Gradient }=2\)
 

 
\(\therefore\ \text{Point of intersection is (2, 1).}\)

v1 Algebra, STD2 A4 SM-Bank 6

A student was asked to solve the following simultaneous equations.

\(y=2x-5\)

\(x-2y+2=0\)

After graphing the equations, the student found the point of intersection to be \((4,3)\)?

Is the student correct? Support your answer with calculations.  (2 marks)

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\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{If the student is correct, the co-ordinates will}\)

\(\text{satisfy both equations.}\)

\(\text{Substitute (4, 3) into}\ \ y=2x-5\)

\(\text{LHS}\) \(=3\)
\(\text{RHS}\) \(= 2(4)-5\)
  \(=3\)

  
\(\therefore\ \text{LHS = RHS}\)
 

\(\text{Substitute (4, 3) into}\ \ x-2y+2=0\)

\(\text{LHS}\) \(=4-2(3)+2\)
  \(=0\)
  \(\ =\ \text{RHS}\)

  
\(\rightarrow\ (4,3)\ \text{satisfies both equations.}\)

\(\therefore\ \text{Student is correct.}\)

v1 Algebra, STD2 A4 2018 HSC 27b

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

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

 

 
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\((1,-2)\)

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}

 

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

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

v1 Algebra, STD2 A4 2004 HSC 16 MC

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?
  

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\(D\)

Show Worked Solution

\(\text{By elimination:}\)

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

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

 

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

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

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

\(\Rightarrow D\)

v1 Algebra, STD2 A4 SM-Bank 7 MC

A computer application was used to draw the graphs of the equations

\(x+y=-6\)  and  \(x-y=-6\)

Part of the screen is shown.

Which row of the table correctly matches the equations with the lines drawn and identifies the solution when the equations are solved simultaneously?

\begin{align*}
\begin{array}{c|c}
\text{ } \\
\textbf{ A. } \\
\textbf{ B. } \\
\textbf{ C. } \\
\textbf{ D. }
\end{array}
\begin{array}{|c|c|c|}
\hline
\ x+y=-6 & x-y=-6 & \text{Solution} \\
\hline
\text{Line 1} & \text{Line 2} & x=-6,\ y=0 \\
\hline
\text{Line 1} & \text{Line 2} & x=-6, y=-6 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6,\  y=0 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6, y=-6 \\
\hline
\end{array}
\end{align*}

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\(C\)

Show Worked Solution

\(\text{Line 1:} \ \ x-y=-6\)

\(\text{Line 2:} \ \ x+y=-6\)

\(\text{Intersection at} \ (-6, 0).\)

\(\Rightarrow C\)

v1 Algebra, STD2 A4 SM-Bank 6 MC

A computer application was used to draw the graphs of the equations

\(x-y=-5\)  and  \(x+y=5\)

Part of the screen is shown.

What is the solution when the equations are solved simultaneously?

  1. \(x=5,\ y=5\)
  2. \(x=5,\ y=0\)
  3. \(x=-5,\ y=0\)
  4. \(x=0,\ y=5\)
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\(D\)

Show Worked Solution

\(\text{Solution occurs at the intersection of the two lines.}\)

\(\Rightarrow D\)

v1 Algebra, STD2 A4 2017 HSC 17 MC

The graph of the line with equation  \(y=5-x\)  is shown.
 

 

When the graph of the line with equation  \(y=2x-1\)  is also drawn on this number plane, what will be the point of intersection of the two lines?

  1. \((0, 5)\)
  2. \((1, 2)\)
  3. \((2, 3)\)
  4. \((5, 0)\)
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\(C\)

Show Worked Solution

\(\text{Method 1: Graphically}\)

\(\text{From graph, intersection is at} (2,3)\)
 


 

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

\(y\) \(=5-x\) \(…\ (1)\)
\(y\) \(=2x-1\) \(…\ (2)\)

 
\(\text{Substitute (2) into (1)}\)

\(2x-1\) \(=5-x\)
\(3x\) \(=6\)
\(x\) \(=2\)

 
\(\text{When}\ \ x=2,\ y=5-2=3\)

\(\Rightarrow C\)