smc-6839-10-Cost/Revenue

Algebra, STD1 EQ-Bank 31

A cake-shop owner sells muffins for $2.50 each. It costs $1 to make each muffin and $300 for the equipment needed to make the muffins.

The owner uses a spreadsheet with formulas to model this situation.

How many muffins need to be sold to ‘break-even’? (1 mark)

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How much profit is made if 400 muffins are sold? (2 marks)

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a. $\text{200 muffins}$

b. $\text{Profit} =\$ 300$

Show Worked Solution

a. $\text{By inspection of the spreadsheet:}$

$\text{Total cost = Revenue = \$500}\ \ \Rightarrow\ \ \text{200 muffins}$

$\text{Breakeven when 200 muffins sold.}$

b. $\text{When 400 muffins are sold:}$

$\text{Revenue} =400 \times \$ 2.50=\$ 1000$

$\text{Cost} =400 \times 1+\$ 300=\$ 700$

$\text{Profit} = 1000-700=\$ 300$

Algebra, STD1 EQ-Bank 15

A local council invests in a community solar farm that sells electricity back to the grid.

The solar farm is expected to operate for 12 years. Each year the farm incurs a maintenance cost of $3000.

The council uses a spreadsheet to model the costs and revenue of the project.

What are the total fixed costs for the solar farm project? (1 mark)

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After how many years does the solar farm break even? (1 mark)
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Calculate the profit the council makes over the full 12-year lifespan of the solar farm. (2 marks)

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a. $$35\,000$

b. $7\ \text{years}$

c. $$25\,000$

Show Worked Solution

a. $\text{Total fixed costs}=$24\,000+$8000+$3000=$35\,000$

b. $\text{From the spreadsheet, at year}\ 7:$

$\text{Total cost}=$56\,000,\ \text{Revenue}=$56\,000\ \checkmark$

$\therefore\ \text{Break-even}=7\ \text{years}$

c. $\text{Project lifespan}=12\ \text{years}$

$\text{Variable cost} =12\times \$3000= $36\,000$

$\text{Total costs} =$35\,000+$36\,000= \$71\,000$

$\text{Revenue} =12\times \$8000 = \$96\,000$

$\therefore\ \text{Profit} = \$96\,000-\$71\,000 = \$25\,000$

Algebra, STD1 EQ-Bank 23

Maya operates Sydney City Walking Tours, a city walking tour business.

The bus she hires holds up to 40 people. Each person on the tour pays $35 and receives a complimentary bottle of water.

Maya uses a spreadsheet to model the costs and revenue for each tour.

What are the total fixed costs for each tour? (1 mark)

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How many people need to attend the tour for Maya to break-even? (1 mark)

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Calculate the profit Maya makes if the tour is fully booked. (2 marks)

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a. $$600$

b. $20\ \text{people}$

c. $$600$

Show Worked Solution

a. $\text{Total fixed costs}=$350+$250=$600$

b. $\text{From the spreadsheet, Maya’s break-even is when }$

$\text{Total cost}=\text{Revenue}=$700$

$\therefore\ \text{People to break-even} = 20$

c. $\text{Fully booked}=40\ \text{people}$

$\text{Variable cost} =40\times \$5= \$200$

$\text{Total costs} =$600+$200= \$800$

$\text{Revenue} =40\times \$35 = \$1400$

$\therefore\ \text{Profit} = \$1400-\$800 = \$600$

Algebra, STD1 EQ-Bank 16

A cafe owner uses a spreadsheet to model the costs and revenue from selling cups of coffee.

The cafe has a fixed setup cost of $450 and a cost per cup 0f $1.50 to make each coffee. The owner sells each coffee for $4.50.

Part of the spreadsheet is shown.

What does it mean for the cafe owner to break even? (1 mark)

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Use the spreadsheet to identify the break-even point for the cafe owner. (1 mark)

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If the cafe owner sells 250 cups of coffee, what profit is made? (2 marks)

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a. $\text{Break-even is when the total cost equals the revenue}$

$\text{i.e. the owner makes neither a profit nor a loss.}$

b. $150\ \text{cups}$

c. $$300$

Show Worked Solution

a. $\text{Break-even is when the total cost equals the revenue}$

$\text{i.e. the owner makes neither a profit nor a loss.}$

b. $\text{From the spreadsheet, total cost equals revenue when }$

$\text{Cost}=\text{Revenue}=$675$

$\therefore\ \text{Cups sold to break-even} = 150$

$\text{Confirmed by cell C15:}$

$\text{Break-even (cups)}=\dfrac{450}{4.50-1.50}=\dfrac{450}{3}=150 \text{ cups}$

c. $\text{At 250 cups (row 9): Revenue} = \$1125,\ \text{Total cost} = \$825$

$\therefore\ \text{Profit} = 1125-825 = \$300$

Algebra, STD1 EQ-Bank 22

Gemstar Promotions is organising the annual Concert Under the Stars event to take place on the last weekend in January.

The outdoor venue holds up to 800 people. Each ticket holder receives a complimentary souvenir program valued at $15.

Garth from Gemstar Promotions uses a spreadsheet to model the costs and revenue for the concert.

What are the total fixed costs for the concert? (1 mark)

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The promoter will only proceed with the concert if the loss is no more than $3000.
What is the minimum number of tickets that must be sold for the concert to go ahead? (2 marks)

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Use the formula in cell C9 to calculate the number of tickets that must be sold for the promoter to break even? (1 mark)

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Calculate the profit for Gemstar Promotions if the concert is fully booked. (1 mark)

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a. $$30\,000$

b. $450\ \text{people}$

c. $500\ \text{people}$

d. $$18\,000$

Show Worked Solution

a. $\text{Total fixed costs}=$8000+$6000+$16\,000=$30\,000$

b. $\text{Maximum allowable loss}=$3000$

$\text{From the spreadsheet, at}\ 450\ \text{people}:$

$\text{Total cost}=$36\,750,\ \text{Revenue}=$33\,750$

$\text{Loss}=$36\,750-$33\,750=$3000\ \checkmark$

$\therefore\ \text{Minimum ticket sales}=450$

c. $\text{Breakeven = C6/(C7-C8)}$

$\text{Breakeven}\ = \dfrac{30\,000}{75.00-15.00}=500\ \text{people}$

d. $\text{Venue capacity}=800\ \text{people}$

$\text{Variable cost} =800\times \$15= $12\,000$

$\text{Total costs} =$30\,000+$12\,000= \$42\,000$

$\text{Revenue} =800\times \$75 = \$60\,000$

$\therefore\ \text{Profit} = \$60\,000-\$42\,000 = \$18\,000$

Algebra, STD1 A3 2025 HSC 5 MC

A baker makes and sells cakes.

The straight-line graphs represent cost $(C )$ and revenue $(R)$ in dollars, and $n$ is the number of cakes.

What profit will the baker make by selling 6 cakes?

Show Answers Only

$A$

Show Worked Solution

$\text{When }n=6$

$\text{Revenue}=10\times 6=60$

$\text{Cost}=20+5\times 6=50$

$\therefore\ \text{Profit}=$60-$50=$10$

$\Rightarrow A$

♦♦ Mean mark 53%.

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

Complete the table of values to show the total cost of the fundraiser. (1 mark)

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\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser. (2 marks)
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How many tickets need to be sold for the fundraiser to break even? (1 mark)

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Carrie sold 300 tickets. How much profit did the fundraiser make? (3 marks)

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a. $\text{Cost when 0 people attend}= $2500$

c. $100\ \text{tickets}$

d. $\text{Profit }=$3500$

Show Worked Solution

a. $\text{Cost when 0 people }= $2500$

c. $\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}$

$\therefore\ \text{Break-even when 100 tickets sold.}$

Mean mark (c) 53%.

d. $\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}$

$\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n$

$\text{Find profit}\ (P)\ \text{when}\ \ n=300:$

$P$	$=R-C$
	$=70\times 300-(2500+50\times300)$
	$=21\,000-17\,500=$3500$

♦ Mean mark (d) 46%.

Algebra, STD1 A3 2022 HSC 25

Sam is making cupcakes to sell at a market. It costs Sam $60 to hire a stall, and each cupcake costs $1.50 to make. Sam intends to sell each cupcake for $4.00.

The equations representing Sam's cost `($ C)` and revenue `($ R)`, are

`C=1.5 x+60` and `R=4 x`, where `x` is the number of cupcakes sold.

The graphs of `C` and `R` are shown below.

How many cupcakes must Sam sell in order to break even? (1 mark)

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If Sam sells 60 cupcakes, what profit is made?
You may assume that Profit = Revenue – Cost. (2 marks)

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a. `24`

b. `$90`

Show Worked Solution

a. `\text(Break even occurs where the two graphs intersect.)`

`\text(→ 24 cupcakes)`

b. `\text(If cupcakes sold) (x) = 60:`

`C = 1.5 xx 60 + 60= $150`

`R = 4 xx 60=$240`

`:.\ text{Profit}\ = 240-150 = $90`

Algebra, STD1 A3 2019 HSC 30

A small business makes and sells bird houses.

Technology was used to draw straight-line graphs to represent the cost of making the bird houses `(C)` and the revenue from selling bird houses `(R)`. The `x`-axis displays the number of bird houses and the `y`-axis displays the cost/revenue in dollars.

How many bird houses need to sold to break even? (1 mark)

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By first forming equations for cost `(C)` and revenue `(R)`, determine how many bird houses need to be sold to earn a profit of $1900. (3 marks)

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a. `20`

b. `96`

Show Worked Solution

a. `20\ \ (xtext(-value at intersection))`

b. `text(Find equations of both lines):`

♦ Mean mark part (b) 47%.

`(0, 500)\ text(and)\ (20, 800)\ text(lie on)\ \ C`

`m_C = (800-500)/(20-1) = 15`

`=> C = 500 + 15x`

`(0,0)\ text(and)\ (20, 800)\ text(lie on)\ \ R`

`m_R = (800-0)/(20-0) = 40`

`=> R = 40x`

`R-C =text(Profit)`

`text(Find)\ \ x\ \ text(when Profit = $1900:)`

`40x-(500 + 15x)`	`=1900`
`40x-500-15x`	`=1900`
`25x`	`= 2400`
`x`	`= 96`

Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost (`$c`) charged by two companies for the hire of a minibus for `x` hours.

Both companies charge $360 for the hire of a minibus for 3 hours.

What is the hourly rate charged by Company A? (1 mark)

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Company B charges an initial booking fee of $75.

Write a formula, in the form of `c = mx + b`, for the cost of hiring a minibus from Company B for `x` hours. (2 marks)

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A minibus is hired for 5 hours from Company B.

Calculate how much cheaper this is than hiring from Company A. (2 marks)

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a. `$120`

b. `c = 95x + 75`

c. `$50`

Show Worked Solution

a. `text(Hourly rate)\ (A)= 360 ÷ 3= $120`

b. `m = text(hourly rate)`

`text(Find)\ m,\ text(given)\ c = 360,\ text(when)\ \ x = 3\ \ text(and)\ \ b = 75`

`360`	`= m xx 3 + 75`
`3m`	`= 285`
`m`	`= 95`

`:. c = 95x + 75`

c.	`text(C)text(ost)\ (A)`	`= 120 xx 5 = $600`
	`text(C)text(ost)\ (B)`	`= 95 xx 5 + 75 = $550`

`:.\ text(The hiring cost for Company)\ B\ text(is $50 cheaper.)`

Algebra, STD2 A4 EQ-Bank 28

Penny is a baker and makes meat pies every day.

The cost of making `p` pies, `$C`, can be calculated using the equation

`C = 675 + 3.5 p`

Penny sells the pies for $5.75 each, and her income is calculated using the equation

`I = 5.75 p`

On the graph, draw the graphs of `C` and `I`. (2 marks)

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On the graph, label the breakeven point and the loss zone. (2 marks)

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a. & b.

Show Worked Solution

b. `text(Loss zone occurs when)\ C > I,\ text(which is shaded in the diagram above.)`

Algebra, STD2 A4 2005 HSC 28b

Sue and Mikey are planning a fund-raising dance. They can hire a hall for $400 and a band for $300. Refreshments will cost them $12 per person.

Write a formula for the cost ($C) of running the dance for `x` people. (1 mark)

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The graph shows planned income and costs when the ticket price is $20.

Estimate the minimum number of people needed at the dance to cover the costs. (1 mark)

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How much profit will be made if 150 people attend the dance? (1 mark)

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Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.

What should be the price of a ticket, assuming all 200 tickets will be sold? (3 marks)

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a. `700 + 12x`

b. `text(Approximately 90)`

c. `$500`

d. `$23`

Show Worked Solution

a. `$C= 400 + 300 + (12 xx x)= 700 + 12x`

b. `text(Using the graph intersection:)`

`text(Approximately 90 people are needed to cover the costs.)`

c. `text(If 150 people attend)`

`text(Income)= 150 xx $20= $3000`

`text(C)text(osts)= 700 + (12 xx 150)= $2500`

`:.\ text(Profit)= 3000-2500= $500`

d. `text(C)text(osts when)\ x = 200:`

`C=700 + (12 xx 200)= $3100`

`text(Income required to make $1500 profit)`

`= 3100 + 1500= $4600`

`:.\ text(Price per ticket)= 4600/200= $23`

Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost `C`, in dollars, for the centre
to host an event, where `x` people attend, is given by:

`C = 10\ 000 + 50x`

The centre charges $100 per person. Its income `I`, in dollars, is given by:

`I = 100x`

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

`$15\ 000`
`$20\ 000`
`$30\ 000`
`$40\ 000`

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

`text(When)\ x=500,\ I=100xx500=$50\ 000`

`text(Breakeven when)\ \ x=200\ \ \ text{(from graph)}`

`text(When)\ \ x=200,\ I=100xx200=$20\ 000`

`text(Difference)=50\ 000-20\ 000=$30\ 000`

`=> C`

Algebra, STD2 A4 2010 HSC 24b

Ashley makes picture frames as part of her business. To calculate the cost, `C`, in dollars, of making `x` frames, she uses the equation `C=40+10x`.

She sells the frames for $20 each and determines her income, `I`, in dollars, using the equation `I=20x`.

Use the graph to solve the two equations simultaneously for `x` and explain the significance of this solution for Ashley's business. (2 marks)

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`x=4`

Show Worked Solution

`text(From the graph, intersection occurs at)\ x=4`

♦ Mean mark 36%.
MARKER’S COMMENT: The intersection on the graph is the same point at which the two simultaneous equations are solved for the given value of `x`.

`=>\ text(Break-even point occurs at)\ x=4`

`text(i.e. when 4 frames sold)`

`text(Income)`	`=20xx4=$80\ \ text(is equal to)`
`text(C)text(osts)`	`=40+(10xx4)=$80`

`text(If)\ <4\ text(frames sold)=>\ text(LOSS for business)`

`text(If)\ >4\ text(frames sold)=>\ text(PROFIT)`

\(P\)	\(=R-C\)
	\(=70\times 300-(2500+50\times300)\)
	\(=21\,000-17\,500=$3500\)