smc-1099-20-Other SE applications

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let	$P$	$\ =\ \text{the number of passenger cars, and}$
	$B$	$\ =\ \text{the number of motorcycles}$

Write TWO linear equations that represent the relationship below.

There are 11 times as many passenger cars as motorcycles.
The total registration fees for passenger cars and motorcycles is $494 million. (2 marks)

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$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Show Worked Solution

$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Algebra, STD1 A3 2023 HSC 26

Electricity provider $A$ charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

Complete the table showing Provider $A$ 's monthly charges for different levels of electricity usage. (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \ & \ \ \ 400\ \ \ & \ \ 1000\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

Provider $B$ charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider $B$ 's charges vary with the amount of electricity used in a month.

On the grid on above, graph Provider A's charges from the table in part (a). (1 mark)
Use the two graphs to determine the number of kilowatt hours per month for which Provider $A$ and Provider $B$ charge the same amount. (1 mark)

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A customer uses an average of 800 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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c. $400\text{ kWh}$

d. $A\text{ is cheaper by }$40.$

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♦♦ Mean mark (b) 40%.

c. $\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}$

$=400\text{ kWh (see graph above)}$

d. $\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280$

$\therefore A\text{ is cheaper by }$40.$

Algebra, STD1 A3 2021 HSC 29

In a park the only animals are goannas and emus. Let `x` be the number of goannas and let `y` be the number of emus.

The number of goannas plus the number of emus in the park is 31. Hence `x + y = 31`.

Each goanna has four legs and each emu has two legs. In total the emus and goannas have 76 legs.

By writing another relevant equation and graphing both equations on the grid on the following page, find the number of goannas and the number of emus in the park. (4 marks)

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Number of goannas = _______________

Number of emus = _________________

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`text{7 goannas, 24 emus}`

Show Worked Solution

`text{Total goanna legs} = 4x`

♦♦♦ Mean mark 8%.

`text{Total emu legs} = 2y`

`4x + 2y`	`= 76`
`2x + y`	`= 38 \ …\ (1)`
`x + y`	`= 31 \ …\ (2)`

`text{Number of goannas} = 7`

`text{Number of emus} = 24`

Algebra, STD1 A3 2020 HSC 29

There are two tanks on a property, Tank A and Tank B. Initially, Tank A holds 1000 litres of water and Tank B is empty.

Tank A begins to lose water at a constant rate of 20 litres per minute.

The volume of water in Tank A is modelled by `V = 1000 - 20t` where `V` is the volume in litres and `t` is the time in minutes from when the tank begins to lose water.

On the grid below, draw the graph of this model and label it as Tank A. (1 mark)
Tank B remains empty until `t=15` when water is added to it at a constant rate of 30 litres per minute.
By drawing a line on the grid (above), or otherwise, find the value of `t` when the two tanks contain the same volume of water. (2 marks)
Using the graphs drawn, or otherwise, find the value of `t` (where `t > 0`) when the total volume of water in the two tanks is 1000 litres. (1 mark)

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`text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`
`29 \ text{minutes}`
`45 \ text{minutes}`

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a. `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`

♦ Mean mark part (a) 45%.

b. `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`

♦♦ Mean mark part (b) 24%.

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

c. `text{Strategy 1}`

♦♦♦ Mean mark part (c) 5%.

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when t = 45)`

`text{Strategy 2}`

`text{Total Volume}`	`=text{T} text{ank A} + text{T} text{ank B}`
`1000`	`= 1000 – 20t + (t – 15) xx 30`
`1000`	`= 1000 – 20t + 30t – 450 `
`10t`	`= 450`
`t`	`= 45 \ text{minutes}`

Algebra, STD2 A4 SM-Bank 3

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation `F = (9C)/5 + 32`.

Calculate the temperature in degrees Fahrenheit when it is −20 degrees Celsius. (1 mark)

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The following two graphs are drawn on the axes below:

`F = (9C)/5 + 32` and `F = C`

Explain what happens at the point where the two graphs intersect. (1 mark)

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`−4^@F`
`text(The two graphs intersect at a temperature where)`

`text(Celcius and Farenheit are the same.)`

Show Worked Solution

i.	`F`	`= (9(−20))/5 + 32`
		`= −4^@F`

ii. `text(The two graphs intersect at a temperature where)`

`text(Celcius and Farenheit are the same.)`

Let	\(P\)	\(\ =\ \text{the number of passenger cars, and}\)
	\(B\)	\(\ =\ \text{the number of motorcycles}\)