SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

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?

  1. $10
  2. $20
  3. $40
  4. $60
Show Answers Only

\(A\)

Show Worked Solution

\(\text{When }n=6\)

\(\text{Revenue}\) \(=10\times 6=60\)
\(\text{Cost}\)  \(=20+5\times 6=50\)
\(\therefore\ \text{Profit }\) \(=$60-$50=$10\)

  
\(\Rightarrow A\)

Filed Under: A3 Types of Relationships (Y12) Tagged With: Band 5, smc-1099-10-Cost/Revenue

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.

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

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

\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}

  1. 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)

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

     

  1. How many tickets need to be sold for the fundraiser to break even?   (1 mark)

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

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

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

Show Answers Only

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

b.


c.    \(100\ \text{tickets}\)

d.    \(\text{Profit }=$3500\)

Show Worked Solution

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

b.


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%.

Filed Under: A3 Types of Relationships (Y12) Tagged With: Band 4, Band 5, smc-1099-10-Cost/Revenue

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.
 


 

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

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

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

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

Show Answers Only
  1. 24
  2. $90
Show Worked Solution

a.   Break even occurs where the two graphs intersect.

→ 24 cupcakes
 

b.   If cupcakes sold (`x`) = 60:

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

`R = 4 xx 60=$240`
  
`:.\ text{Profit}\ = 240-150 = $90`

Filed Under: A3 Types of Relationships (Y12) Tagged With: Band 4, smc-1099-10-Cost/Revenue

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.
 


 

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

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

  2. 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)

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

Show Answers Only
  1. `20`
  2. `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`
 

`text(Profit) = R – C`

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

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

Filed Under: A3 Types of Relationships (Y12) Tagged With: Band 4, Band 5, smc-1099-10-Cost/Revenue

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.

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

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

  2. 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)

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

  3. A minibus is hired for 5 hours from Company B.

     

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

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

Show Answers Only
  1. `$120`
  2. `c = 95x + 75`
  3. `$50`
Show Worked Solution
i.    `text(Hourly rate)\ (A)` `= 360 ÷ 3`
    `= $120`

 

ii.   `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`
 

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

 
`:.\ text(Company)\ B’text(s hiring cost is $50 cheaper.)`

Filed Under: A3 Types of Relationships (Y12), Simultaneous Equations and Applications (Std 2) Tagged With: Band 3, Band 4, smc-1099-10-Cost/Revenue, smc-794-10-Cost/Revenue

Algebra, STD2 A4 SM-Bank 4

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`

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

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

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

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

Show Answers Only

(i) and (ii)

Show Worked Solution
i.   

 

ii.  `text(Loss zone occurs when)\ C > I,\ text(which is shaded)`

`text(in the diagram above.)`

Filed Under: A3 Types of Relationships (Y12), Simultaneous Equations and Applications (Std 2) Tagged With: Band 4, Band 5, smc-1099-10-Cost/Revenue

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.

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

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

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

2005 28b

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

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

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

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

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

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

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

Show Answers Only
  1. `700 + 12x`
  2. `text(Approximately 90)`
  3. `$500`
  4. `$23`
Show Worked Solution
i.    `$C` `= 400 + 300 + (12 xx x)`
    `= 700 + 12x`

 

ii.  `text(Using the graph intersection)`

`text(Approximately 90 people are needed)`

`text(to cover the costs.)`

 

iii.  `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`

 

iv.  `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`

Filed Under: A3 Types of Relationships (Y12), Breakeven and Financial modelling, FM1 - Earning money, Linear Functions (Adv-2027), Linear Functions (Y11), Simultaneous Equations and Applications (Std 2) Tagged With: Band 4, Band 5, common-content, smc-1099-10-Cost/Revenue, smc-6214-10-Cost/Revenue, smc-794-10-Cost/Revenue, smc-985-10-Cost/Revenue

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`
 

2UG 2011 20

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

  1.  `$15\ 000`
  2. `$20\ 000`
  3. `$30\ 000` 
  4. `$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`

Filed Under: A3 Types of Relationships (Y12), Breakeven and Financial modelling, Linear Functions (Adv-2027), Linear Functions (Y11), Simultaneous Equations and Applications (Std 2) Tagged With: Band 5, common-content, smc-1099-10-Cost/Revenue, smc-6214-10-Cost/Revenue, smc-794-10-Cost/Revenue, smc-985-10-Cost/Revenue

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)

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

Show Answers Only

`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)`

Filed Under: A3 Types of Relationships (Y12), Breakeven and Financial modelling, Simultaneous Equations and Applications (Std 2) Tagged With: Band 5, smc-1099-10-Cost/Revenue, smc-794-10-Cost/Revenue

Copyright © 2014–2025 SmarterEd.com.au · Log in