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?
- $10
- $20
- $40
- $60
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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?
\(A\)
\(\text{When }n=6\)
| \(\text{Revenue}\) | \(=10\times 6=60\) |
| \(\text{Cost}\) | \(=20+5\times 6=50\) |
| \(\therefore\ \text{Profit }\) | \(=$60-$50=$10\) |
\(\Rightarrow A\)
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.
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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}
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a. \(\text{Cost when 0 people }= $2500\)
b.
c. \(\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}\)
\(\therefore\ \text{Break-even when 100 tickets sold.}\)
d. \(\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}\)
\(\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n\)
\(\text{Find profit}\ (P)\ \text{when}\ \ n=300:\)
| \(P\) | \(=R-C\) |
| \(=70\times 300-(2500+50\times300)\) | |
| \(=21\,000-17\,500\) | |
| \(=$3500\) |
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.
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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`
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.
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a. `20\ \ (xtext(-value at intersection))`
b. `text(Find equations of both lines):`
`(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 = $1900:)`
| `1900` | `= 40x – (500 + 15x)` |
| `25x` | `= 2400` |
| `x` | `= 96` |
The graph displays the cost (`$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.
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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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Calculate how much cheaper this is than hiring from Company A. (2 marks)
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| 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 $50 cheaper.)`
Penny is a baker and makes meat pies every day.
The cost of making `p` pies, `$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`
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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.
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The graph shows planned income and costs when the ticket price is $20
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Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.
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| i. | `$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` |
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?
`C`
`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`
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`
`text(From the graph, intersection occurs at)\ x=4`
`=>\ 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)`