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Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.
  


  
  1. Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10).   (2 marks)
  2. Using the spreadsheet, identify the break-even point and explain what it means for the business.   (2 marks)
  3. SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels \((x)\) they will be installing.   (2 marks)
Show Answers Only

a.    \(\text{4 panels: TC (B9) = \$1800.00}\)

\(\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00\)

\(\text{6 panels: TC (B10) = \$2100.00}\)

\(\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}\)

b.    \(\text{Break-even = 6 panels, See worked solution}\)

c.    \(27 \text{ panels}\)

Show Worked Solution

a.    \(\text{4 panels:}\)

\(\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00\)

\(\text{Revenue (C9)} = 4\times\$350 = \$1400.00\)

\(\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00\)

\(\text{6 panels:}\)

\(\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00\)

\(\text{Revenue (C10)} = 6\times\$350 = \$2100.00\)

\(\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00\)
  

b.    \(\text{When 6 panels are installed:}\)

\(\text{Revenue = Total costs = \$2100.00  (breakeven)}\)

\(\text{This is the point at which the business covers all of its costs and}\)

\(\text{begins to make a profit.}\)

\(\text{OR}\)

\(\text{If fewer than 6 panels are installed the business will make a loss.}\)
  

c.    \(\text{Let } x = \text{the number of solar panels to be installed.}\)

\(\text{Profit}\) \( = \text{Revenue}-\text{Total costs}\)
\(4200\) \( = 350x-(1200+150x)\)
\(4200\) \(= 200x-1200\)
\(5400\) \(=200x\)
\(x\) \(=27\)

  
\(\text{SunPower Solutions will install 27 solar panels.}\)

Algebra, STD2 EQ-Bank 25

A local bowling club sold memberships for the new season.

Senior memberships \((s)\) cost $70 each and junior memberships \((j)\) cost $45 each.

On a particular day, a total of 19 memberships were sold, with sales totalling $1030.

  1. Write two equations, in terms of \(s\) and \(j\), to represent this information.   (1 mark)

  2. Determine the exact number of senior and junior memberships sold.   (2 marks)
Show Answers Only

a.    \(70s+45j=1030\ …\ (1)\)

\(s+j=19\ …\ (2)\)

b.    \(s=7,\ \ j=12\)

Show Worked Solution

a.    \(70s+45j=1030\ …\ (1)\)

\(s+j=19\ …\ (2)\)
 

b.    \(\text{Rearranging (2) above:}\)

\(s=19-j\)

\(\text{Substitute}\ \ s=19-j\ \ \text{into (1):}\)

\(70(19-j)+45j\) \(=1030\)
\(1330-70j+45j\) \(=1030\)
\(25j\) \(=1330-1030=300\)
\(j\) \(=12\)

 
\(\text{Substitute}\ \ j=12\ \ \text{into (2):}\)

\(s=19-12=7\)

\(\therefore\ \text{7 senior and 12 junior memberships were sold.}\)

Algebra, STD2 EQ-Bank 24

A bookshop sold a number of biographies at a weekend sale.

Hardcover biographies \((h)\) sold for $25 each and paperback biographies \((p)\) sold for $12 each.

A total of 11 biographies were sold, with total sales revenue of $210.

  1. Write two equations, in terms of \(h\) and \(p\), to represent this information.   (1 mark)

  2. Determine the number of hardcover biographies and the number of paperback biographies sold.   (2 marks)

Show Answers Only

\(h=6, \ p=5\)

Show Worked Solution

a.    \(25h+12p=210\ …\ (1)\)

\(h+p=11\ …\ (2)\)
 

b.    \(\text{Rearranging (2) above:}\)

\(h=11-p\)

\(\text{Substitute}\ \ h=9-p\ \ \text{into (1):}\)

\(25(11-p)+12p\) \(=210\)
\(275-25p+12p\) \(=210\)
\(13p\) \(=275-210=65\)
\(p\) \(=5\)

 
\(\text{Substitute}\ \ p=5\ \ \text{into (2):}\)

\(h=11-5=6\)

\(\therefore\ \text{6 hardcover and 5 paperback biographies were sold.}\)

Algebra, STD2 EQ-Bank 32

A school sold tickets to its annual play.

Adult tickets were sold for $8 each and student tickets were sold for $5 each.

A total of 142 tickets were sold, raising $896 in ticket sales.

By writing two equations to represent this information, find the number of adult tickets and the number of student tickets sold.   (3 marks)

Show Answers Only

\(\text{Adult tickets}=62, \ \text{Student tickets}=80\)

Show Worked Solution

\(\text{Let}\ x=\text{adult tickets},\ y=\text{student tickets}\)

\(8x+5y=896\ …\ (1)\)

\(x+y=142\ \ \Rightarrow \ y=142-x\ …\ (2)\)

\(\text{Substitute}\ \ y=142-x\ \ \text{into (1):}\)

\(8x+5(142-x)\) \(=896\)
\(8x+710-5x\) \(=896\)
\(3x\) \(=896-710=186\)
\(x\) \(=62\)

 
\(\text{Substitute}\ \ x=62\ \ \text{into (2):}\)

\(y=142-62=80\)

\(\therefore\ \text{62 adult tickets and 80 student tickets were sold.}\)

Algebra, STD2 EQ-Bank 30

Eleni is a farmer who sold chickens and ducks at the local market.

Each chicken was sold for $15 and each duck was sold for $9.

She sold a total of 78 birds for a total of $930.

By writing two equations to represent this information, find the number of chickens and the number of ducks Eleni sold.   (3 marks)

Show Answers Only

\(\text{Chickens}=38, \ \text{Ducks}=40\)

Show Worked Solution

\(\text{Let}\ x=\text{chickens},\ y=\text{ducks}\)

\(15x+9y=930\ …\ (1)\)

\(x+y=78\ \ \Rightarrow \ y=78-x\ …\ (2)\)

\(\text{Substitute}\ \ y=78-x\ \ \text{into (1):}\)

\(15x+9(78-x)\) \(=930\)
\(15x+702-9x\) \(=930\)
\(6x\) \(=930-702=228\)
\(x\) \(=38\)

 
\(\text{Substitute}\ \ x=38\ \ \text{into (2):}\)

\(y=78-38=40\)

\(\therefore\ \text{Eleni sold 38 chickens and 40 ducks.}\)

Algebra, STD2 EQ-Bank 29

Two friends, Adam and Bertha, sold cupcakes for a fundraising event.

Adam sold \(x\) cupcakes for $3 each. Bertha sold \(y\) cupcakes for $2 each.

Together they sold 113 cupcakes with total sales revenue of $275.

By writing two equations to represent this information, find the number of cupcakes sold by each of the two friends.   (3 marks)

Show Answers Only

\(x=49, \ y=64\)

Show Worked Solution

\(3x+2y=275\ …\ (1)\)

\(x+y=113\ \ \Rightarrow \ y=113-x\ …\ (2)\)

\(\text{Substitute}\ \ y=113-x\ \ \text{into (1):}\)

\(3x+2(113-x)\) \(=275\)
\(3x+226-2x\) \(=275\)
\(x\) \(=275-226=49\)

 
\(\text{Substitute}\ \ x=49\ \ \text{into (2):}\)

\(y=113-49=64\)

\(\therefore\ \text{Adam sold 49 and Bertha sold 64.}\)