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v1 Algebra, STD2 A4 2023 HSC 21

Electricity provider \(A\) charges 30 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $90.

  1. Complete the table showing Provider \(A\)'s monthly charges for different levels of electricity usage.   (1 mark)

    \begin{array} {|l|c|}
    \hline
    \rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ \ 1000 \ \ \\
    \hline
    \rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
    \hline
    \end{array}

Provider \(B\) charges 52.5 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\)'s charges vary with the amount of electricity used in a month.
 

 
  1. On the grid above, graph Provider \(A\)'s charges from the table in part (a).   (1 mark)
  2. Use the two graphs to determine the number of kilowatt hours per month for which Provider \(A\) and Provider \(B\) charge the same amount.   (1 mark)

  3. A customer uses an average of 600 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?   (2 marks)

Show Answers Only

a.    \(\text{When kWh} =400\)

\(\text{Monthly charge}\ =$90+0.30\times 400=$210\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

b.    
         

c.    \(\text{400 kWh}\)

d.    \(\text{Provider}\ A\ \text{is cheaper by \$45.}\)

Show Worked Solution

a.   \(\text{When kWh} =400\)

\(\text{Monthly charge}\ =$90+0.30\times 400=$210\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

b. 
          
 

c.    \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\)

\(\therefore\ \text{Same charge at 400 kWh}\)
 

d.    \(\text{Cost at 600 kWh:}\)

\(\text{Method 1: Using graph}\rightarrow\ $315-270=$45\)

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

\(\text{Provider}\ A: \ 90 + 0.30 \times 600 = $270\)

\(\text{Provider}\ B: \ 0.525 \times 600 = $315\)

\(\therefore \text{Provider}\ A\ \text{is cheaper by \$45.}\)