smc-985-20-Other Linear Applications

Functions, 2ADV F1 SM-Bank 25

Damon owns a swim school and purchased a new pool pump for $3250.

He writes down the value of the pool pump by 8% of the original price each year.

Construct a function to represent the value of the pool pump after `t` years. (1 mark)

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Draw the graph of the function and state its domain and range. (2 marks)

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Show Answers Only

`text(Value) = 3259 – 260t`
`text(Domain)\ {t: 0 <= t <= 12.5}`
`text(Range)\ {y: 0 <= y <= 3250}`

Show Worked Solution

i. `text(Depreciation each year)`	`= 8text(%) xx 3250`
	`= $260`

`:.\ text(Value) = 3250 – 260t`

ii.

`text(Find)\ \ t\ \ text(when value = 0)`

`3250 – 260t`	`= 0`
`t`	`= 3250/260`
	`= 12.5\ text(years)`

`text(Domain)\ \ {t: 0 <= t <= 12.5}`

`text(Range)\ \ {y: 0 <= y <= 3250}`

Algebra, STD2 A2 2015 HSC 27c

Ariana’s parents have given her an interest‑free loan of $4800 to buy a car. She will pay them back by paying `$x` immediately and `$y` every month until she has repaid the loan in full.

After 18 months Ariana has paid back $1510, and after 36 months she has paid back $2770.

This information can be represented by the following equations.

`x + 18y = 1510`

`x + 36y = 2770`

Graph these equations below and use to solve simultaneously for the values of `x` and `y`. (2 marks)

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How many months will it take Ariana to repay the loan in full? (2 marks)

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`x = 250, \ y = 70`
`text(65 months)`

Show Worked Solution

`:.\ text(Solution is)\ \ x = 250, \ y = 70`

ii. `text(Let)\ \ A = text(the amount paid back after)\ n\ text(months)`

`A = 250 + 70n`

♦ Mean mark 44%.

`text(Find)\ n\ text(when)\ A = 4800`

`250 + 70n`	`= 4800`
`70n`	`= 4550`
`n`	`= 65`

`:.\ text(It will take Ariana 65 months to repay)`

`text(the loan in full.)`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

Use the graph to find the tax payable on a taxable income of $21 000. (1 mark)

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Use suitable points from the graph to show that the gradient of the section of the graph marked `A` is `1/3`. (1 mark)

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How much of each dollar earned between $21 000 and $39 000 is payable in tax? (1 mark)

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Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between $21 000 and $39 000. (2 marks)

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Show Answers Only

`$3000\ \ \ text{(from graph)}`
`1/3`
`33 1/3\ text(cents per dollar earned)`
`text(Tax payable on)\ I = 1/3 I\-4000`

Show Worked Solution

`text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

ii. `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%

`text(Gradient at)\ A`	`= (y_2\-y_1)/(x_2\ -x_1)`
	`= (9000-3000)/(39\ 000 -21\ 000)`
	`= 6000/(18\ 000)`
	`= 1/3\ \ \ \ \ text(… as required)`

iii. `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

`text(Tax)`	` = 1/3\ text(of each dollar earned)`
	` = 33 1/3\ text(cents per dollar earned)`

iv. `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I`	`= 3000 + 1/3 (I\-21\ 000)`
	`= 3000 + 1/3 I\-7000`
	`= 1/3 I\-4000`

Algebra, STD2 A2 2009 HSC 24d

A factory makes boots and sandals. In any week

• the total number of pairs of boots and sandals that are made is 200
• the maximum number of pairs of boots made is 120
• the maximum number of pairs of sandals made is 150.

The factory manager has drawn a graph to show the numbers of pairs of boots (`x`) and sandals (`y`) that can be made.

Find the equation of the line `AD`. (1 mark)

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Explain why this line is only relevant between `B` and `C` for this factory. (1 mark)

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The profit per week, `$P`, can be found by using the equation `P = 24x + 15y`.

Compare the profits at `B` and `C`. (2 marks)

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Show Answers Only

`x + y = 200`
`text(S)text(ince the max amount of boots = 120)`

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`
`text(The profits at)\ C\ text(are $630 more than at)\ B.`

Show Worked Solution

i. `text{We are told the number of boots}\ (x),`

♦♦♦ Mean mark part (i) 14%.
Using `y=mx+b` is a less efficient but equally valid method, using `m=–1` and `b=200` (`y`-intercept).

`text{and shoes}\ (y),\ text(made in any week = 200)`

`=>text(Equation of)\ AD\ text(is)\ \ x + y = 200`

ii. `text(S)text(ince the max amount of boots = 120)`

♦ Mean mark 49%

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`

iii. `text(At)\ B,\ \ x = 50,\ y = 150`

♦ Mean mark 40%.

`=>$P (text(at)\ B)`	`= 24 xx 50 + 15 xx 150`
	`= 1200 + 2250`
	` = $3450`

`text(At)\ C,\ \ x = 120 text(,)\ y = 80`

`=> $P (text(at)\ C)`	`= 24 xx 120 + 15 xx 80`
	`= 2880 + 1200`
	`= $4080`

`:.\ text(The profits at)\ C\ text(are $630 more than at)\ B.`

`y`	`=76d`
`93\ 100`	`= 76d`
`d`	`= (93\ 100)/76`
	`= 1225`

`p`	`= 4/7 xx 1225`
	`= 700\ text(pounds)`