Band 3

Networks, STD2 EQ-Bank 2 MC

The Gantt chart below shows the activities involved in organising a school sports carnival.

Which activity has the greatest float time?

$\text{B}$
$\text{D}$
$\text{E}$
$\text{G}$

Show Answers Only

$B$

Show Worked Solution

$\text{From the Gantt chart, the dashed extensions show:}$

$\text{B: float = 3 hours}$

$\text{D: float = 5 hours}$

$\text{E: float = 3 hours}$

$\text{G: float = 0 (critical path, no dashed extension)}$

$\text{Activity D has the greatest float time (5 hours).}$

$\Rightarrow B$

The construction of a new reptile exhibit is a project involving nine activities, $A$ to $I$. The network diagram below shows the activities and their completion times in weeks. Some values are missing.

The Gantt chart below has been created for this project.

Using the Gantt chart, identify the critical path. (1 mark)
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Use the Gantt chart to determine the missing values in the network diagram. (2 marks)
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Activity $E$ is delayed by 8 weeks. Using the Gantt chart, explain whether this will affect the minimum completion time of the project. (2 marks)
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a. $ACDFGI$

b. $\text{B} = 5 \text{ weeks, H} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Show Worked Solution

a. $\text{The critical path is the continuous solid bar on row 1 of the Gantt chart.}$

$\text{Critical path:}\ ACDFGI$

b. $\text{Activities B and H are not labelled in the network diagram.}$

$\text{From the Gantt chart:}$

$\text{B starts at week 0, ends at week 5} \to \text{duration} = 5 \text{ weeks}$

$\text{H starts at week 7, ends at week 14} \to \text{duration} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Networks, STD2 EQ-Bank 18

A Gantt chart for a project with activities $A, B, C, D, E, F, G$ and $H$ has been created. The Gantt chart can be used to complete the missing information on the edges in the network diagram.

Use the Gantt chart to determine the three missing values in the network diagram. (3 marks)
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State the minimum completion time for the project. (1 mark)
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a. $\text{C} = 4 \text{ hours, D} = 5 \text{ hours, E} = 7 \text{ hours}$

b. $26 \text{ hours}$

Show Worked Solution

a. $\text{Using the Gantt chart}$

$\text{C (between A and F):} $

$\Rightarrow\ \text{Starts hour 7, ends hour 11 = 4 hours duration}$

$\text{D (between B and G):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 8 = 5 hours duration}$

$\text{E (between B and H):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 10 = 7 hours duration}$

b. $\text{From the Gantt chart, the project ends at hour 26.}$

$\text{Minimum completion time} = 26 \text{ hours}$

Statistics, EXT1 EQ-Bank 13

The weights of pumpkins on a particular farm are normally distributed with a mean of 13.5 kg and a standard deviation of 3 kg.

A random sample of 15 pumpkins is selected.

Let $X_1, X_2, X_3, \ldots, X_{15}$ denote the weights of the pumpkins in the sample, and let

$\overline{X}=\dfrac{X_1+X_2+X_3+\cdots+X_{15}}{15}$.

Find the expected value of $\overline{X}.$ (1 mark)

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Find the standard deviation of $\overline{X}.$ (1 mark)

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a. $13.5$

b. $0.775$

Show Worked Solution

a. $E(\overline{X})=\mu=13.5$

b. $\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{3}{\sqrt{15}} \approx 0.775$

Financial Maths, STD2 EQ-Bank 16

Mia wants to buy a tablet with a cash price of $960. She cannot pay for it upfront and is considering two options.

Option 1: Buy now, pay later

4 equal payments of $240 over 8 weeks
No interest or fees if all payments are made on time

Option 2: Short-term loan

Establishment fee: $75
Monthly account-keeping fee: $30
Weekly repayments of $55 over 5 months (20 weeks)

Assume Mia will make all payments on time under either option.

Calculate the total amount Mia will pay under each option. (2 marks)

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Which option is cheaper, and by how much? (1 mark)

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a. $\text{Option 1: }\$960,\ \text{Option 2: }\$1325$

b. $\text{Option 1 is cheaper by }\$365$

Show Worked Solution

a. $\text{Option 1: } $

$\text{Repayments}=4 \times 240 = \$960$

$\text{Option 2:}$

$\text{Account-keeping} = 30 \times 5 = \$150$

$\text{Weekly repayments} = 55 \times 20 = \$1100$

$\text{Total} = 75+150+1100 = \$1325$

b. $\text{Option 1 is cheaper.}$

$\text{Difference} = 1325-960 = \$365$

Probability, STD2 EQ-Bank 2 MC

The Venn diagram shows information about 40 students and the subjects they study.

One student is selected at random.

What is the probability that the student studies Art but not Music?

$\dfrac{1}{8}$
$\dfrac{1}{4}$
$\dfrac{3}{8}$
$\dfrac{1}{2}$

Show Answers Only

$C$

Show Worked Solution

$\text{Number studying Art but not Music} = 15$

$\text{Total students} = 15+5+10+10=40$

$P(\text{Art but not Music}) = \dfrac{15}{40} = \dfrac{3}{8}$

$\Rightarrow C$

Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.

Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10). (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the business. (2 marks)
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SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels $(x)$ they will be installing. (2 marks)
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a. $\text{4 panels: TC (B9) = \$1800.00}$

$\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00$

$\text{6 panels: TC (B10) = \$2100.00}$

$\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}$

b. $\text{Break-even = 6 panels, See worked solution}$

c. $27 \text{ panels}$

Show Worked Solution

a. $\text{4 panels:}$

$\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00$

$\text{Revenue (C9)} = 4\times\$350 = \$1400.00$

$\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00$

$\text{6 panels:}$

$\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00$

$\text{Revenue (C10)} = 6\times\$350 = \$2100.00$

$\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00$

b. $\text{When 6 panels are installed:}$

$\text{Revenue = Total costs = \$2100.00 (breakeven)}$

$\text{This is the point at which the business covers all of its costs and}$

$\text{begins to make a profit.}$

$\text{OR}$

$\text{If fewer than 6 panels are installed the business will make a loss.}$

c. $\text{Let } x = \text{the number of solar panels to be installed.}$

$\text{Profit}$	$ = \text{Revenue}-\text{Total costs}$
$4200$	$ = 350x-(1200+150x)$
$4200$	$= 200x-1200$
$5400$	$=200x$
$x$	$=27$

$\text{SunPower Solutions will install 27 solar panels.}$

Algebra, STD2 A4 EQ-Bank 16

Harmony Arts Festival is a cultural event with fixed costs of $510. Each ticket costs $8.00 to provide and sells for $25.00.

The spreadsheet below models the festival's costs and revenue for different numbers of tickets sold.

Calculate the values for cells C12 and D12 in the spreadsheet. (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the festival. (2 marks)
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a. $\text{C12} = \$1250.00 \quad \text{D12} = \$340.00$

b. $\text{See worked solution}$

Show Worked Solution

a. $\text{C12: Revenue} = 50 \times \$25.00 = \$1250.00$

$\text{D12: Profit/Loss} = \$1250.00-\$910.00 = \$340.00$

b. $\text{When 30 tickets sold:}$

$\text{Revenue = Total costs = \$750.00 (Breakeven)}$

$\text{This is the point at which the festival covers all of its costs}$

$\text{and begins to make a profit.}$

$\text{OR}$

$\text{If less than 30 tickets are sold the festival will make a loss.}$

Algebra, STD2 A4 EQ-Bank 3 MC

Beachside Brew is a cafe with fixed weekly costs of $200. Each cup of coffee costs $1.50 to make and sells for $4.00.

The spreadsheet below models the cafe's weekly costs and revenue for different numbers of cups sold.

How many cups of coffee must be sold each week to break even?

Show Answers Only

$B$

Show Worked Solution

$\text{Break-even occurs when Revenue = Total costs}\ \rightarrow\ \text{(i.e. Profit = \$0)}$

$\text{From the spreadsheet, Revenue = Total costs (\$320.00)}$

$\text{when 80 cups are sold.}$

$\Rightarrow B$

Complex Numbers, EXT2 EQ-Bank 15

Prove that for any complex numbers $z_1$ and $z_2$,
$\abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}$ (2 marks)
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Hence, or otherwise, show that if $\abs{z-1}+\abs{z+1} \leqslant 4$ for $z\in C,$
$\abs{z} \leqslant 2$ (2 marks)

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a. $\text{Proof (See Worked Solutions)}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\text {Prove}\ \ \abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}:$

$\abs{z_1+z_2}^2$	$=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)$
	$=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2$
	$=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)$

$\text{Since}\ \ \operatorname{Re}(w) \leqslant\abs{w}\ \ \text{for} \ \ w\in C,$

$\abs{z_1+z_2}^2$	$\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}$
	$\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2$
	$\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2$

$\therefore\abs{z_1+z_2} \leqslant\abs{z_1}+\abs{z_2}$

b. $|z-1|+|z+1| \leqslant 4 \ \text{(given)}\ …\ (1)$

$\text {Using triangle inequality:}$

$|(z-1)+(z+1)| \leqslant|z-1|+|z+1|$

$\text{Since}\ \ (z-1)(z+1)=2 z:$

$\abs{2z}$	$\leqslant\abs{z-1}+\abs{z+1}$
$\abs{2z}$	$\leqslant 4\ \ \text{(using (1) above)}$
$2\abs{z}$	$\leqslant 4$
$\abs{z}$	$\leqslant 2$

Calculus, EXT2 C1 2020 HSC 13d*

By expanding `(text{cis}\theta + text{cis}(-theta))^4` show that
`cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`. (3 marks)

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Hence, or otherwise, find `int_0^(frac{pi}{2}) cos^4 theta\ d theta`. (2 marks)

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a. `text{See Worked Solution}`

b. `frac{3 pi}{16}`

Show Worked Solution

a. `text{cis}\theta + text{cis}(-theta) = 2 cos theta\ \ …\ (1)`

`(text{cis}\theta + text{cis}(-theta))^4= 16 cos^4(4theta)`

`text{Expand LHS:}`

`(text{cis}\theta + text{cis}(-theta))^4`

`= text{cis}(4theta)+4text{cis}(2theta)+6+4text{cis}(-2theta)+text{cis}(4theta)`

`= 2text{cos}(4theta)+8text{cos}(2theta)+6\ \ text{(using (1) above)}`

`text{Equating sides:}`

`16 cos^4 theta`	`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta`	`= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta`	`= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

b.	`int_0^(frac{pi}{2}) cos^4 theta\ d theta`	`= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
		`= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
		`= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi + frac{3 pi}{2}) – 0 ]`
		`= frac{1}{8} ( frac{3 pi}{2})`
		`= frac{3 pi}{16}`

Vectors, EXT2 EQ-Bank 12

A curve has a vector equation

$r=(2 \sin t-1)\mathbf{i}+(2 \cos t+3) \mathbf{j} $

Write the Cartesian equation of this curve. (2 marks)

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Sketch the curve on the Cartesian plane below. (1 mark)

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a. $(x+1)^2+(y-3)^2=4$

b. $(x+1)^2+(y-3)^2=2^2$

$\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2$

Show Worked Solution

a. $r=(2 \sin t-1) \mathbf{i}+(2 \cos t+3) \mathbf{j}$

$x=2 \sin t-1 \ \Rightarrow \ \sin t=\dfrac{x+1}{2}$

$y=2 \cos t+3 \ \Rightarrow \ \cos t=\dfrac{y-3}{2}$

$\text{Using} \ \ \sin ^2 t+\cos ^2 t=1:$

$\dfrac{(x+1)^2}{4}+\dfrac{(y-3)^2}{4}$	$=1$
$(x+1)^2+(y-3)^2$	$=4$

b. $(x+1)^2+(y-3)^2=2^2$

$\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2$

Complex Numbers, EXT2 N2 2023 14a*

Let $z$ be the complex number $z=\text{cis}\dfrac{\pi}{6} $ and $w$ be the complex number $w=\text{cis}\dfrac{3\pi}{4} $.

By first writing $z$ and $w$ in Cartesian form, or otherwise, show that
$|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}$. (3 marks)

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The complex numbers $z, w$ and $z+w$ are represented in the complex plane by the vectors $\overrightarrow{O A},\overrightarrow{O B}$ and $\overrightarrow{O C}$ respectively, where $O$ is the origin.
Show that $\angle A O C=\dfrac{7 \pi}{24}$. (2 marks)

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Deduce that $\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}$. (1 mark)

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i. $\text{See Worked Solutions}$

ii. $\text{See Worked Solutions}$

iii. $\text{See Worked Solutions}$

Show Worked Solution

i. $z= \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i $

$w= \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} $

$\|z+w\|^2$	$=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}$
	$=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}$
	$=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}$
	$= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}$
	$= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}$
	$=\dfrac{16-4\sqrt6+4\sqrt2}{8} $
	$=\dfrac{4-\sqrt6+\sqrt2}{2} $

ii.

$\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} $

$ |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} $

$\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC $

$\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB $

$\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} $

iii. $\text{In}\ \triangle AOC: $

$ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} $

$\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. $

$\text{Using the cos rule in}\ \triangle AOC: $

$\cos\,\dfrac{7\pi}{24}$	$=\dfrac{\|z\|^2+\|z+w\|^2-\|w\|^2}{2\|z\|\|z+w\|}$
	$=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} $
	$=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} $
	$=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} $
	$=\dfrac{8-2\sqrt6+2\sqrt2}{4} $

♦♦ Mean mark (iii) 26%.

Mechanics, EXT2 EQ-Bank 13

A body of mass 10 kg is held in place on a smooth plane inclined at 30° to the horizontal by a tension force, $T$ newtons, acting parallel to the plane.

Assuming that the acceleration due to gravity is 9.8 m s$^{-2}$, find the value of $T$ in newtons. (2 marks)

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Show Answers Only

$49\ \text{N}$

Show Worked Solution

$T-10g\,\sin30^{\circ}$	$= 0$
$T-\dfrac{10g}{2}$	$= 0$
$T$	$= \dfrac{10g}{2}$
$T$	$= 5g=49\ \text{N}$

Mechanics, EXT2 EQ-Bank 17

Two light inextensible strings are attached to a horizontal surface and suspended a 10-kilogram object as shown in the diagram below

The tension in the strings are $T_1$ newtons and $T_2$ newtons.

Express $T_1$ in terms of $T_2$. (2 marks)

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If the acceleration due to gravity is 9.8 ms$^{-2}$, determine the exact values of $T_1$ and $T_2$, in newtons. (2 marks)

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a. $\text{See Worked Solutions}$

b. $T_1=49 \sqrt{3}, T_2=49$

Show Worked Solution

a. $\text{Resolve forces into horizontal/vertical components:}$

$\text{Horizontal forces are equal.}$

$T_1 \cos 60^{\circ}$	$=T_2 \cos 30^{\circ}$
$T_1 \times \dfrac{1}{2}$	$=T_2 \times \dfrac{\sqrt{3}}{2}$
$T_1$	$=\sqrt{3}\, T_2$

b. $\text{Vertical forces are equal.}$

$T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}$	$=10 \times 9.8$
$T_1 \times \dfrac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}$	$=98$

$\text {Substitute} \ \ T_1=\sqrt{3}\, T_2 :$

$\sqrt{3}\, T_2 \times \dfrac{\sqrt{3}}{2}+T_2 \times \dfrac{1}{2}$	$=98$
$2\, T_2$	$=98$
$T_2$	$=49 \ \text{newtons}$

$\therefore T_1=49 \sqrt{3}, \ T_2=49$

Statistics, EXT1 EQ-Bank 4 MC

The random variable $X$ represents the number of successes in 10 independent Bernoulli trials. The probability of success is $p=0.9$ in each trial.

Let $r=P(X \geq 1)$.

Which of the following describes the value of $r$ ?

$r>0.9$
$r=0.9$
$0.1<r<0.9$
$r \leq 0.1$

Show Answers Only

$A$

Show Worked Solution

$p=0.9,\ \ 1-p=0.1,\ \ n=10$

$P(X \geq 1)$	$=P\text{(at least 1 success)}$
	$=1-P(X=0)$
	$=1-(0.1)^{10}$
	$=0.999…$

$\Rightarrow A$

Vectors, EXT2 EQ-Bank 18

Consider the triangle with vertices $A(4,5,1), B(8,1,2)$ and the origin $O(0,0,0)$. The triangle has three medians.

The median through $B$ has vector equation

$\lambda \, \overrightarrow{O M}+(1-\lambda) \overrightarrow{O B}=\left(\begin{array}{l}8 \\ 1 \\ 2\end{array}\right)+\lambda\left(\begin{array}{c}-6 \\ 1.5 \\ -1.5\end{array}\right)$ (Do NOT prove this.)

for a parameter $\lambda \in[0,1]$ and where $M$ is the midpoint of $O A$.

Write down the median through $A$ as a vector equation. (2 marks)

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The three medians of any triangle meet at a point known as the centroid.
Find the value of $\lambda$ corresponding to the centroid. (2 marks)

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Into what ratio does the centroid divide the median $B M$? (1 mark)

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a. $\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]$

b. $\lambda=\dfrac{2}{3}.$

c. $2:1.$

Show Worked Solution

a. $\text{Let}\ N = \text{midpoint of}\ OB:$

$N \equiv \left(\dfrac{8+0}{2}, \dfrac{1+0}{2}, \dfrac{2+0}{2}\right) \equiv (4,0.5,1).$

$\text{Direction vector of the median through} \ A \ \text {is}\left(\begin{array}{c}4-4 \\0.5-5 \\1-1\end{array}\right) = \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right).$

$\text{Equation of median from}\ A:$

$\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]$

b. $\text{Equation of median from}\ B:$

$\left(\begin{array}{c}8 \\1 \\2\end{array}\right) + \lambda \left(\begin{array}{c}-6 \\1.6 \\-1.5\end{array}\right)\ \ \text{for}\ \ \lambda \in[0,1]$

$\text{Medians intersect at centroid}$

$x\text{-coordinate of median through}\ B = 8-6\lambda$

$x\text{-coordinate of median through}\ A = 4$

$\text{Equating}\ x\text{-coordinates:}$

$8-6 \lambda=4\ \ \Rightarrow\ \ \lambda=\dfrac{2}{3}$

$\therefore\ \text{Point of intersection occurs at } \lambda=\dfrac{2}{3}.$

c. $\text{The centroid divides the median in the ratio of} \ 2:1.$

Mechanics, EXT2 EQ-Bank 12

A 10 kg mass is placed on a smooth plane that is inclined at 30° to the horizontal, as shown in the diagram below. A force, `F` is applied to the mass up the slope and parallel to the slope so that when released, the mass remains at rest.

If the acceleration due to gravity, `g`, is `9.8\ text(m s)^{-2}`, determine the magnitude of the force, in newtons, acting up the slope. (2 marks)

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`49 \ text(N)`

Show Worked Solution

`sin 30^@`	`= F/(10g)`
`F`	`=10g\ sin30^@`
`F`	`= 10 xx 9.8 xx 1/2 = 49\ text{N}`

Complex Numbers, EXT2 N2 2021 HSC 14c*

Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that

`cos5theta = 16cos^5theta-20cos^3 theta + 5cos theta`. (3 marks)

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`text(See Worked Solution)`

Show Worked Solution

`(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`

`text(Equating real parts:)`

`cos5theta`	`= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta`
	`= cos^5theta-10cos^3theta(1-cos^2theta) + 5costheta(1-cos^2theta)sin^2theta`
	`= cos^5theta-10cos^3theta + 10cos^5theta + (5costheta-5cos^3theta)(1-cos^2theta)`
	`= 11cos^5theta-10cos^3theta + 5costheta-5cos^3theta-5cos^3theta + 5cos^5theta`
	`= 16cos^5theta-20cos^3theta + 5costheta`

Statistics, EXT1 EQ-Bank 12

An office has 7 printers and 5 photocopiers. On average, each printer is used 73% of the time and each photocopier is used 46% of the time.

Write an expression for the probability that, at a particular time, at least one printer is in use. (1 mark)

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Write an expression for the probability that, at a particular time, at least one printer and exactly three photocopiers are in use. (2 marks)

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Show Answers Only

a. $1-{ }^7 C_0(0.27)^7$

b. $\left[1-(0.27)^7\right] \times { }^5 C_3(0.46)^3(0.54)^2$

Show Worked Solution

a. $P\text{(printer in use)} = 0.73, \ \ P\text{(printer not in use)} = 0.27$

$P\text{(at least 1 printer in use)}$	$=1-P\text{(no printer in use)}$
	$=1-{ }^7 C_0(0.27)^7$

b. $P\text{(copier in use)} = 0.46, \ \ P\text{(copier not in use)} = 0.54$

$P\text{(at least 1 printer and exactly 3 copiers in use)}$

$=\left[1-(0.27)^7\right] \times { }^5 C_3(0.46)^3(0.54)^2$

Statistics, EXT1 S1 2020 HSC 12b*

When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.

This coin is tossed 100 times.

Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.

Find the expected value, `E(X)`. (1 mark)

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By finding the variance, `text(Var)(X)`, show that the standard deviation of `X` is approximately 5. (1 mark)

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i. `60`

ii. `text(See Worked Solutions)`

Show Worked Solution

i. `X = text(number of heads)`

`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`

`E(X)= np= 100 xx 3/5= 60`

ii. `text(Var)(X)= np(1-p)= 60 xx 2/5= 24`

`sigma(x)= sqrt24~~ 5`

Vectors, EXT1 EQ-Bank 3 MC

Given that $\overrightarrow{OP}=\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)$ and $\overrightarrow{O Q}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)$, what is $\overrightarrow{P Q}$ ?

$\left(\begin{array}{c}1 \\ -6 \\ 4\end{array}\right)$
$\left(\begin{array}{c}-1 \\ 6 \\ -4\end{array}\right)$
$\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)$
$\left(\begin{array}{c}-5 \\ -4 \\ 2\end{array}\right)$

Show Answers Only

$C$

Show Worked Solution

$\overrightarrow{PQ}=\overrightarrow{O Q}-\overrightarrow{O P}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)-\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)$

$\Rightarrow C$

Vectors, EXT1 2014 SPEC1 1

Consider the vector `underset ~a = sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

Find the unit vector in the direction of `underset ~a`. (1 mark)

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Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis. (2 marks)

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The vector `underset ~b = 2 sqrt 3 underset ~i + m underset ~j-5 underset ~k`.
Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)

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a. `1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `theta = 45^@`

c. `m = 6 + 5 sqrt 2`

Show Worked Solution

a. `|underset ~a|= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)= sqrt 6`

`hat underset ~a= underset ~a/|underset ~a|= 1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `x text{-axis vectors include}\ (1,0,0).`

`underset ~a ⋅ underset ~i = ((\sqrt3),(-1),(-\sqrt2))((1),(0),(0))=\sqrt3`

	`underset ~a ⋅ underset ~i`	`= \|underset ~a\|\|underset ~i\| cos theta= sqrt 6 cos theta`
	`sqrt 3`	`= sqrt 6 cos theta`
	`cos theta`	`=1/sqrt 2`
	`:. theta`	`= 45^@`

c. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6-m + 5 sqrt 2`	`=0`
`:. m`	`=6 + 5 sqrt 2`

Calculus, EXT1 EQ-Bank 13

Solve the differential equation $\dfrac{d y}{d x}=3 y$. (3 marks)

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$y=e^{3 x} \times e^c=A e^{3 x}$

Show Worked Solution

$\dfrac{d y}{d x}$	$=3 y$
$\displaystyle \int \frac{1}{y}\, d y$	$=\displaystyle \int 3\, d x$
$\ln \abs{y}$	$=3 x+c$
$\abs{y}$	$=e^{3 x+c}$
$y$	$=e^{3 x} \times e^c=A e^{3 x}$

Vectors, EXT1* V1 2020 HSC 11d

Consider the two vectors `underset~u = 2 underset~i-underset~j + 3 underset~k` and `underset~v = p underset~i + underset~j + 2 underset~k`.

For what values of `p` are `underset~u-underset~v` and `underset~u + underset~v` perpendicular? (3 marks)

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`p= ± 3`

Show Worked Solution

`underset~u-underset~v = ((-2),(-1),(3))-((p),(1),(2)) = ((-2-p),(-2),(1))`

`underset~u + underset~v = ((-2),(-1),(3)) + ((p),(1),(2)) = ((p-2),(0),(5))`

`⊥ \ text{when} \ \ (underset~u-underset~v) · (underset~u + underset~v ) = 0 :`

`((-2-p),(-2),(1)) · ((p-2),(0),(5)) = 0`

`-(p + 2)(p-2) + 5`	`= 0`
`-(p^2-4) + 5`	`= 0`
`-p^2 + 9`	`= 0`
`p^2`	`= 9`
`p`	`= ± 3`

Vectors, EXT1* V1 2024 HSC 12a

The vector $\underset{\sim}{a}$ is $\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ and the vector $\underset{\sim}{b}$ is $\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$.

Find $\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$. (1 mark)

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Show that $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$ is perpendicular to $\underset{\sim}{b}$. (2 marks)

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i. $\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)$

ii. $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text {Vectors are perpendicular.}$

Show Worked Solution

i. $\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$

$\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text{Vectors are perpendicular.}$

Vectors, EXT1* V1 2023 HSC 11b

Find the angle between the vectors

$\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}$

$\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}$,

giving your answer to the nearest degree. (3 marks)

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$87^{\circ} $

Show Worked Solution

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \ \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]

$\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} $

$\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} $

$ \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 $

$\cos\ \theta $	$=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{\|}\underset{\sim}{a}\Big{\|} \cdot \Big{\|}\underset{\sim}{b}\Big{\|}} $
	$=\dfrac{1}{\sqrt{294}} $
$ \theta$	$=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} $
	$=86.65…$
	$=87^{\circ} $

Vectors, EXT1* V1 2021 HSC 11c

Find the angle between the vectors `underset~a = ((2),(0),(4))` and `underset~b = ((-3),(1),(2))`, giving the angle in degrees correct to 1 decimal place. (3 marks)

Show Answers Only

`83.1^@`

Show Worked Solution

`underset~a = ((2),(0),(4)) \ , \ |underset~a| \ = sqrt{2^2 + 4^2} = sqrt20`

`underset~b = ((-3),(1),(2)) \ , \ |underset~b| \ = sqrt{(-3)^2 + 1^2 + 2^2} = sqrt14`

`underset~a * underset~b`	`= ((2),(0),(4)) ((-3),(1),(2)) = – 6 + 0 + 8 = 2`
`underset~a * underset~b`	`= \|underset~a\| \|underset~b\| \ cos theta`
`2`	`= sqrt20 sqrt14 \ cos theta`
`cos theta`	`= 2/sqrt280`
`theta`	`= cos^(-1) (1/sqrt70)`
	`= 83.1^@ \ text{(1 d.p,)}`

Vectors, EXT1* V1 2020 HSC 1 MC

What is the length of the vector `- underset~i + 18 underset~j - 6 underset~k`?

Show Answers Only

`B`

Show Worked Solution

`text{Length}`	`= \| – underset~i + 18 underset~j – 6 underset~k \ \|`
	`= sqrt{(-1)^2 + 18^2 + (-6)^2}`
	`= sqrt{361}`
	`= 19`

Vectors, EXT1 2013 SPEC2 14 MC

The distance from the origin to the point `P(7,−1,5sqrt2)` is

`7sqrt2`
`10`
`6 + 5sqrt2`
`100`

Show Answers Only

`B`

Show Worked Solution

`d`	`= sqrt((7-0)^2 + (−1-0)^2 + (5sqrt2-0)^2)`
	`= sqrt(49 + 1 + 25 xx 2)`
	`= 10`

`=> B`

Vectors, EXT1 2012 SPEC2 16 MC

The distance between the points `P(−2 ,4, 3)` and `Q(1, −2, 1)` is

`7`
`sqrt 21`
`sqrt 31`
`11`

Show Answers Only

`A`

Show Worked Solution

`d`	`= sqrt((-2-1)^2 + (4-(-2))^2 + (3-1)^2)`
	`= sqrt(9 + 36 + 4)`
	`= 7`

`=> A`

Vectors, EXT1 2012 SPEC2 15 MC

The vectors `underset~a = 2underset~i + m underset~j-3underset~k` and `underset~b = m^2underset~i-underset~j + underset~k` are perpendicular for

`m = −2/3` and `m = 1`
`m = −3/2` and `m = 1`
`m = 2/3` and `m = −1`
`m = 3/2` and `m = −1`

Show Answers Only

`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b`	`= 2m^2 + m(-1) + (-3)(1)`
`0`	`= 2m^2-m-3`
`0`	`= (2m-3)(m + 1)`

`:. m = 3/2, quad m = -1`

`=> D`

Vectors, EXT1 2019 SPEC2 11 MC

Let point `M` have coordinates `(a, 1,-2)` and let point `N` have coordinates `(-3, b,-1)`.

If the coordinates of the midpoint of `vec(MN)` are `(-5, 3/2, c)` and `a, b` and `c` are real constants, the the values of `a, b` and `c` are respectively

`−13, 2 and −1/2`
`−7, −2 and −3/2`
`−2, −1/2 and −3`
`−7, 2 and −3/2`

Show Answers Only

`D`

Show Worked Solution

`M = 1/2 ([(a),(1),(−2)] + [(−3),(b),(−1)]) = 1/2 [(a-3),(1 + b),(−3)]`

`1/2(a-3)`	`= −5`
`a-3`	`= −10`
`a`	`= −7`
`1/2(1 + b)`	`= 3/2`
`1 + b`	`= 3`
`b`	`= 2`
`c`	`= −3/2`

`=>D`

Vectors, EXT1 2011 SPEC2 12 MC

The angle between the vectors `3underset~i + 6underset~j-2underset~k` and `2underset~i-2underset~j + underset~k`, correct to the nearest tenth of a degree, is

2.0°
91.0°
112.4°
121.3°

Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j-2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i-2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j-2underset~k) * (2underset~i-2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6-12-2`

`= -8`

`costheta`	`= ((3tildei + 6tildej-2tildek).(2tildei-2tildej + tildek))/(\|\ 3tildei + 6tildej-2tildek\ \|\|\ 2tildei-2tildej + tildek\ \|)= -8/21`
`:. theta	`= cos^(−1)(−8/12)~~ 112.4^@`

`=> C`

Calculus, EXT1 EQ-Bank 12

Let $P(x)=x^3+a x^2+b x-4$ where $a$ and $b$ are real numbers.

If $x=2$ is a double root of $P(x)=0$, find the values of $a$ and $b$. (3 marks)

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Show Worked Solution

$P(x)=x^3+a x^2+b x-4$

$P^{\prime}(x)=3 x^2+2 a x+b$

$\text{Since} \ \ x=2\ \ \text{is a double root,}$

$P(2)=0:$

$8+4 a+2 b-4$	$=0$
$2 a+b$	$=-2\ \ldots\ (1)$

$P^{\prime}(2)=0:$

$12+4 a+b$	$=0$
$4 a+b$	$=-12\ \ldots\ (2)$

$[(1) \times 2]-(2):$

$b=8$

$\text{Substitute} \ \ b=8 \ \ \text{into (1):}$

$2a+8$	$=-2$
$a$	$=-5$

Proof, EXT1 EQ-Bank 16

Consider the following proposition:

$1+2+3+\ldots+n=\dfrac{1}{2}(n-1)(n+2)$ for integers $n \geqslant 1$.

Show that the proposition does not satisfy the initial case of mathematical induction proof. (1 mark)

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Show that the inductive step of the false proposition can, however, be proven. (2 marks)

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Show Worked Solution

a. $\text{Initial case:} \ n=1$

$\text{LHS}=1$

$\text{RHS}=\dfrac{1}{2}(1-1)(1+2)=0 \neq \text{LHS}$

$\therefore \ \text{Initial case of} \ \ n=1\ \ \text {is not satisfied.}$

b. $\text{if} \ \ n=k:$

$1+2+\ldots+k=\dfrac{1}{2}(k-1)(k+2)$

$\text{Prove true for} \ \ n=k+1:$

$\text{i.e.} \ \ 1+2+\ldots+k+(k+1)=\dfrac{1}{2} k(k+3)$

$\operatorname{LHS}$	$=1+2+\ldots+k+k+1$
	$=\dfrac{1}{2}(k-1)(k+2)+k+1$
	$=\dfrac{1}{2}\left(k^2+k-2\right)+\dfrac{1}{2}(2 k+2)$
	$=\dfrac{1}{2}\left(k^2+k-2+2 k+2\right)$
	$=\dfrac{1}{2}\left(k^2+3 k\right)$
	$=\dfrac{1}{2} k(k+3)$
	$=\operatorname{LHS}$

$\therefore \ \text{The inductive step can be proven.}$

Calculus, 2ADV EQ-Bank 22

The diagram shows the graph of $y=\log _e(x+1)$

Express $x$ as a function of $y$. (1 mark)

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Hence, or otherwise, find the exact area of the shaded region bounded by the curve $y=\log _e(x+1)$, the $x$-axis, and the line $x=3$. (3 marks)

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a. $x=e^y-1$

b. $A=4 \ln 4-3 \ \text{u}^2$

Show Worked Solution

a. $y=\ln (x+1) \ \Rightarrow \ x+1=e^y \ \Rightarrow \ x=e^y-1$

b. $\text{Area of rectangle}=\ln 4 \times 3=3 \ln 4$

$\text{Find the area between curve and \(y$-axis from $\ y=0\ $ to $\ y=\ln 4$:}\)

$A$	$=\displaystyle \int_0^{\ln 4} e^y-1\, d y$
	$=\Big[e^y-y\Big]_0^{\ln 4}$
	$=\left(e^{\ln 4}-\ln 4\right)-(1)$
	$=4-\ln 4-1$
	$=3-\ln 4$

$\text{Shaded Area}$	$=3 \ln 4-(3-\ln 4)$
	$=4 \ln 4-3 \ \text{u}^2$

Statistics, 2ADV EQ-Bank 34

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.

Find the median test mark. (1 mark)

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The mean test mark is 5.4. The standard deviation of the test marks is 4.22.
Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean. (2 marks)
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A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context. (1 mark)

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i. `6`

ii. `text(43%)`

iii. `text(The statement assumes the data is normally distributed which is incorrect.)`

Show Worked Solution

♦ Mean mark (i) 50%.
♦♦Mean mark (ii) 34%.

i. `text(Median)= text(15th + 16th score)/2= (4 + 8)/2= 6`

ii. `text(Lower limit) = 5.4-4.22 = 1.18`

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{% (nearest %)}`

iii. `text(The statement assumes the data is normally distributed.)`

♦♦♦ Mean mark (iii) 13%.

`text(This is incorrect in this case.)`

Statistics, 2ADV EQ-Bank 13

The formula to calculate `z`-scores can be rearranged to give

`mu = x-\sigma z`

where	`mu` is the mean
	`x` is the score
	`sigma` is the standard deviation
	`z` is the `z`-score

In an examination, Aaron achieved a score of 88, which corresponds to a `z`-score of 2.4.
Substitute these values into the rearranged formula above to form an equation. (1 mark)
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In the same examination, Brock achieved a score of 52, which corresponds to a `z`-score of –1.2.
Using this information, form another equation and solve it simultaneously with the equation from part (a) to find the values of `mu` and `\sigma`. (2 marks)
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a. `mu = 88-2.4\sigma`

b. `64`

Show Worked Solution

a. `mu = 88-2.4\sigma`

b. `mu = 52 + 1.2\sigma\ …\ (1)`

`mu = 88-2.4\sigma \ …\ (2)`

`text(Subtract)\ \ (2)-(1):`

`0= 36-3.6\sigma\ \ =>\ \ \sigma= 10`

`text(Substitute)\ \ \sigma = 10\ \ text(into)\ (1):`

`mu= 52 + 1.2 xx 10= 64`

Statistics, 2ADV EQ-Bank 20

The weights of boxes of Brekky Bicks are normally distributed. The mean is 754 grams and the standard deviation is 2 grams.

What is the `z`-score of a box of Brekky Bicks with a weight of 754 g? (1 mark)

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What is the weight of a box that has a `z`-score of –1? (1 mark)

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Brekky Bicks boxes are labelled as having a weight of 750 g. What percentage of boxes will have a weight less than 750 g? (2 marks)

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a. `0\ text{(mean)}`

b. `752\ text(grams)`

c. `text(2.5%)`

Show Worked Solution

a. `text{z-score (754 g) = 0 (754 g is the mean)}`

b. `ztext(-score)`	`= (x-mu)/sigma`
`-1`	`= (x-54)/2`
`x-754`	`= -2`
`x`	`= 752\ text(grams)`

c. `text{z-score (750)= (750-754)/2= -2`

`:.\ text(Graph shows that 2.5% of boxes will weigh less than 750 g.)`

Statistics, 2ADV EQ-Bank 19

Two brands of light bulbs are being compared. For each brand, the life of the light bulbs, in hours, is normally distributed and described in the table below.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \quad \text{Mean} \quad & \text{Standard Deviation} \\
\hline
\rule{0pt}{2.5ex} \text{Brand A} \rule[-1ex]{0pt}{0pt} & 450 & 25 \\
\hline
\rule{0pt}{2.5ex} \text{Brand B} \rule[-1ex]{0pt}{0pt} & 500 & 50 \\
\hline
\end{array}

One of the Brand B light bulbs has a life of 400 hours.
What is the `z`-score of the life of this light bulb? (1 mark)

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A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:
‘Brand A light bulbs are more likely to be defective than Brand B light bulbs.’
Is this claim correct? Justify your answer, with reference to `z`-scores or standard deviations or the normal distribution. (2 marks)
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a. `-2`

b. `text(The claim is incorrect.)`

Show Worked Solution

a. `z text{-score of Brand B bulb (400 hrs)}`

`= (x-mu)/sigma= (400-500)/50= -2`

b. `z text{-score of Brand A bulb (400 hours)} =(400-450)/25=-2`

`text(S)text(ince the)\ z text(-score for both brands is –2, they are equally likely to be defective.)`

`:.\ text(The claim is incorrect.)`

Statistics, 2ADV EQ-Bank 16

The results of two class tests are normally distributed. The means and standard deviations of the tests are displayed in the table.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \quad \text{Test 1} \quad & \quad \text{Test 2} \quad \\
\hline
\rule{0pt}{2.5ex} \text{Mean} \rule[-1ex]{0pt}{0pt} & 60 & 58 \\
\hline
\rule{0pt}{2.5ex} \text{Standard Deviation} \rule[-1ex]{0pt}{0pt} & 6.2 & 6.0 \\
\hline
\end{array}

Stuart scored 63 in Test 1 and 62 in Test 2. He thinks that he has performed better in Test 1. Do you agree? Justify your answer using appropriate calculations. (2 marks)

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If 150 students sat for Test 2, how many students would you expect to have scored less than 64? (2 marks)

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

Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `126`

Show Worked Solution

a. `text(In Test 1:)\ \ mu = 60,\ sigma = 6.2`

`z text(-score)\ (63)= (x-mu)/sigma= (63-60)/6.2= 0.483…`

`text(In Test 2:)\ \ mu = 58,\ sigma = 6.0`

`z text(-score)\ (62)= (62-58)/6.0=0.666…`

`text(S) text(ince Stuart’s)\ z\ text(-score is higher in Test 2, his performance relative)`

`text(to the class is better despite his mark being slightly lower.)`

b. `text(In Test 2:)`

`z text(-score)\ (64)= (64-58)/6= 1`

`text(84% have)\ z text(-score) < 1`

`:.\ text(# Students expected below 64) = text(84%) xx 150 = 126`

Statistics, 2ADV S3 EQ-Bank 24

A continuous random variable $X$ has probability density function $f(x)$ given by

\begin{align*}
f(x)=\left\{\begin{array}{cl}
k x(1-x)^5, & \text { for } 0 \leq x \leq 1 \\
0, & \text { for all other values of } x
\end{array}\ \ \ , \text { where } k\right. \text { is a constant. }
\end{align*}

It is given that

$\displaystyle \int_0^a x(1-x)^5\, d x=\frac{1}{42}+\frac{(1-a)^7}{7}-\frac{(1-a)^6}{6}$

and $\displaystyle\int_0^1 x^m(1-x)^5\, d x=\dfrac{120}{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)}$

where $a>0$ and $m>0$.

Show that $k=42$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Show that $E (X)=0.25$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Show that the median of $X$ is less than the expected value of $X$. (3 marks)

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

Show Answers Only

a. $k \displaystyle \int_0^1 x(1-x)^5 d x=1$

$k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1$

$\dfrac{k}{42}$	$=1$
$k$	$=42$

b.	$E (X)$	$=\displaystyle \int_0^1 x \times f(x)\, d x$
		$=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x$
		$=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}$
		$=0.25$

c. $\text{Let}\ m =\text{ median}$

$P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 $

$E(X)=0.25$

$\text{Calculate }\ P(X\leqslant 0.25):$

$\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x$	$=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]$
	$=0.555 \ldots\ \text{(3 dp)}$

$\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}$

Show Worked Solution

a. $k \displaystyle \int_0^1 x(1-x)^5 d x=1$

$k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1$

$\dfrac{k}{42}$	$=1$
$k$	$=42$

b.	$E (X)$	$=\displaystyle \int_0^1 x \times f(x)\, d x$
		$=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x$
		$=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}$
		$=0.25$

c. $\text{Let}\ m =\text{ median}$

$P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 $

$E(X)=0.25$

$\text{Calculate }\ P(X\leqslant 0.25):$

$\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x$	$=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]$
	$=0.555 \ldots\ \text{(3 dp)}$

$\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}$

Calculus, 2ADV EQ-Bank 13

A quantity of radioactive material decays according to the equation

$\dfrac{d M}{d t}=-k M$,

where $M$ is the mass of the material in kilograms, $t$ is the time in years and $k$ is a constant.

Verify that $M=A e^{-k t}$ is a solution to this equation, where $A$ is a constant. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
The time for half of the material to decay is 300 years. The initial amount of material is 20 kg .
Find the amount of material remaining after 1000 years. (3 marks)

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Show Answers Only

a. $M=A e^{-k t}$

$\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)$

b. $1.98\text{ kg}$

Show Worked Solution

a. $M=A e^{-k t}$

$\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)$

b. $\text{When} \ t=0, M=20:$

$20=A e^{-k(0)}\ \ \Rightarrow\ \ A=20$

$M=20 e^{-k t}$

$\text{When }\ t=300, M=10:$

$10$	$=20 e^{-300 k}$
$\dfrac{1}{2}$	$=e^{-300 k}$
$-300 k$	$=\ln \frac{1}{2}$
$-300k$	$=- \ln2$
$k$	$=\dfrac{\ln 2}{300}$

$\text{When } \ t=1000:$

$M=20 e^{-1000 k}=20 e^{-\tfrac{10}{3} \ln 2}=1.984 \ldots$

$\therefore \ \text{The amount remaining after 1000 years} \approx 1.98\text{ kg}$

Financial Maths, 2ADV M1 EQ-Bank 12

Evaluate $\displaystyle \sum_{r=4}^7\left(2^r+3 r\right)$. (2 marks)

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

Show Answers Only

$306$

Show Worked Solution

$\displaystyle \sum_{r=4}^7$	$=2^4+12+2^5+15+2^6+18+2^7+21$
	$=306$

Financial Maths, STD2 EQ-Bank 21

Maya uses a buy now, pay later payment option to make a purchase of $120. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Maya misses her final payment on 16 March 2026 and is charged a late fee of $19. Maya's payment schedule is shown, with her balance totalling $49.

\begin{array}{|l|c|c|c|} \hline \textbf{Payment} & \textbf{Due Date} & \textbf{Amount} & \textbf{Status} \\ \hline \text{1st} & \text{2 February 2026} & \$30 & \text{Paid} \\ \hline \text{2nd} & \text{16 February 2026} & \$30 & \text{Paid} \\ \hline \text{3rd} & \text{2 March 2026} & \$30 & \text{Paid} \\ \hline \text{4th} & \text{16 March 2026} & \$30 & \text{Not Paid} \\ \hline \text{Outstanding Due} & \text{30 March 2026} & \$49 \text{ including late fee} & \\ \hline \end{array}

Find the total amount Maya pays for her purchase if repaying in full on 30 March 2026. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
Maya's bank offers short-term loans where simple interest is charged at 16% per annum.
Suppose Maya had borrowed $120 from the bank to make this purchase on 2 February 2026 and repaid it in full 8 weeks later.
How much would Maya have saved using this approach instead of the buy now, pay later option? (2 marks)

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

Show Answers Only

a. $\$139$

b. $\$16.05$

Show Worked Solution

a. $\text{If total owing paid on 30 March:}$

$\text{Total paid} = 30+30+30+49 = \$139$

b. $r = 16\% = 0.16, \quad n = \dfrac{8 \times 7}{365} = \dfrac{56}{365}$

$I= Prn = 120 \times 0.16 \times \dfrac{56}{365}= 2.945\ldots = \$2.95$

$\text{Amount saved} = 19-2.95 = \$16.05$

Measurement, STD2 EQ-Bank 20

The travel graph displays Jamie's trip which began at town `M` at 8 am and finished at town `N`.

How far apart are the two towns? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
At what time during the day did Jamie arrive back at town `M` ? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What was the total distance that Jamie travelled? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Between which times in the day was Jamie travelling at the fastest speed? Justify your answer, without calculations. (1 mark)

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

Show Answers Only

a. `140\ text{km}`

b. `10\ text{am}`

c. `220\ \text{km}`

d. `\text{Between 11 – 11:30 am}`

Show Worked Solution

a. `140\ text{km}`

b. `text{James arrives back at town when he is 140 km away (2nd time).}`

`=> 10\ text{am}`

c. `\text{Total Distance} = 40 + 40 + 60+ 80=220\ text{km}`

d. `text{Fastest speed → graph is the steepest (either up or down)}`

`:.\ text{Fastest speed between 11 – 11:30 am}`

Measurement, STD2 EQ-Bank 19

The travel graph displays Nikau's car trip along a straight road from home and back again. The trip has been broken into four separate sections: `A`, `B`, `C` and `D`.

How far did Nikau travel in total? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
In which section of the trip, `A`, `B`, `C` and `D`, did Nikau travel the fastest? (1 mark)

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

Show Answers Only

a. `400\ text(km)`

b. `text(S)text(ection)\ D`

Show Worked Solution

a. `text(Distance travelled)\ = 2 xx 200= 400\ text(km)`

b. `text{Fastest section has the steepest slope (in either direction).}`

`:. text(S)text(ection)\ B\ text(was the fastest)`

`\ rightarrow text(50 km/30 mins) =100 text(km/hour)`

Measurement, STD2 EQ-Bank 2 MC

The travel graph shows the distance of a runner from a town.

Between what times was the runner travelling at their greatest speed?

10 am and 11 am
11 am and 11:30 am
11:30 am and 1 pm
1 pm and 2:30 pm

Show Answers Only

$A$

Show Worked Solution

$\text{Greatest speed is when the graph is steepest.}$

$\therefore\ \text{Greatest speed occurs between 10 am and 11 am.}$

$\Rightarrow A$

Financial Maths, STD2 EQ-Bank 20

Kimberley uses a buy now, pay later payment option to make a purchase of $100. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Kimberley misses her final payment and is charged a late fee of $17. Kimberley’s payment schedule is shown, with her balance totalling $42.

Find the total amount Kimberley pays for her purchase if repaying in full on 27 July 2024. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
Kimberley’s bank offers short-term loans where simple interest is charged at 18% per annum.
Suppose Kimberley had borrowed $100 from the bank to make this purchase on 1 June 2024 and repaid it in full 8 weeks later.
How much would Kimberley have saved using this approach instead of the buy now, pay later option? (2 marks)
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Show Answers Only

a. $$117$

b. $$14.24$

Show Worked Solution

a. $\text{If total owing paid on 27 July:}$

$\text{Total paid} = 25+25+25+42=$117$

b. $r=18\%=0.18,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}$

$I=Prn=100 \times 0.18 \times \dfrac{56}{365} = 2.761… = $2.76 $

$\text{Amount saved} = 17-2.76=$14.24$

Networks, STD2 EQ-Bank 3 MC

A network of pipes is shown.

Which vertex in this network represents the sink?

$V$
$X$
$Y$
$Z$

Show Answers Only

$B$

Show Worked Solution

$\text{The sink is the vertex where all flow finishes and no flow leaves.}$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 5 MC

Which graph represents a negatively skewed distribution?

Show Answers Only

$D$

Show Worked Solution

$\text{Negatively skewed graphs have a long tail to the left.}$

$\Rightarrow D$

Financial Maths, STD1 F1 EQ-Bank 12

Ethan has a weekly net income of $580 from his part-time job at JB Hi-Fi. He has created the following budget:

Item	Weekly amount
Rent	$220
Food	$120
Transport	$60
Other expenses	$50
Savings	$130

Ethan is saving for an overseas trip that will cost $4680.

How many weeks will it take Ethan to save enough for the trip if he sticks to this budget? (1 mark)

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Ethan decides to save more by reducing his "Food" by $20 per week and reducing his "Other expenses" to $20 per week.
Determine how many fewer weeks it will now take Ethan to save for the trip. (2 marks)

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Show Answers Only

a. $\text{36 weeks}$

b. $\text{10 fewer weeks}$

Show Worked Solution

a. $\text{Calculate weeks to save:}$

$\text{Weeks}=\dfrac{4680}{130}=36\ \text{weeks}$

b. $\text{Calculate new weekly savings:}$

$\text{Food reduction}=\$20$

$\text{Other expenses reduction}= 50- 20=\$30$

$\text{Extra savings per week}= 20+ 30=\$50$

$\text{New weekly savings}= 130+ 50=\$180$

$\text{New weeks}=\dfrac{4680}{180}=26\ \text{weeks}$

$\therefore\ \text{Fewer weeks}= 36- 26=10\ \text{fewer weeks}$

Financial Maths, STD1 F1 EQ-Bank 12

Daniel earns $32 per hour as a delivery driver. He is also paid a $12 fuel allowance per shift.

How much will he earn from a 5-hour shift? (2 marks)

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

Show Answers Only

$\$172$

Show Worked Solution

$\text{Wages}=5\times 32=\$160$

$\text{Total earnings}= 160+ 12=\$172$

Algebra, STD1 EQ-Bank 13

A boat is purchased for $15 000. It depreciates in value by $3000 per year.

Let $V$ = Value of the boat in dollars, and $t$ = time in years.

Complete the table of values below that models the relationship between the value of the boat and time in years. (1 mark)

$\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}V & 15\,000 & & 9000 & & 3000 & \\[12pt] \hline \end{array}$

--- 0 WORK AREA LINES (style=lined) ---
Using the table of values from (a), neatly graph the value of the boat from 0 to 5 years on the grid below. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---
Identify ONE limitation of this linear model. (1 mark)

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

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a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \ \\ \hline \end{array}

c. $\text{Limitations could include ONE of the following:}$

- $\text{The model predicts the boat has zero value after 5 years, which is}$
  $\text{unrealistic as most boats retain some value.}$
- $\text{Beyond 5 years the model would predict a negative value, which is}$
  $\text{not possible.}$
- $\text{The model assumes a constant rate of depreciation, but in reality}$
  $\text{a boat may depreciate more quickly in early years.}$

Show Worked Solution

a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \ \\ \hline \end{array}

c. $\text{Limitations could include ONE of the following:}$

- $\text{The model predicts the boat has zero value after 5 years, which is}$
  $\text{unrealistic as most boats retain some value.}$
- $\text{Beyond 5 years the model would predict a negative value, which is}$
  $\text{not possible.}$
- $\text{The model assumes a constant rate of depreciation, but in reality}$
  $\text{a boat may depreciate more quickly in early years.}$

Algebra, STD1 EQ-Bank 11

A farmer in Western Australia observes that the rodent population on his property is doubling every two weeks.

A student claims that a linear model is NOT appropriate to predict the rodent population over time.

Is the student correct? Justify your answer. (2 marks)

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

Show Answers Only

$\text{Yes, the student is correct.}$

$\text{Correct justification could include ONE of the following:}$

- $\text{The rodent population is doubling every two weeks, which means it}$
  $\text{is increasing by a constant multiplier, not a constant amount — this}$
  $\text{is exponential growth, not linear.}$
- $\text{A linear model increases by the same amount each time period, but}$
  $\text{the rodent population increases by a larger amount each fortnight as}$
  $\text{the population grows.}$
- $\text{The graph of the rodent population over time would be an upsloping}$
  $\text{exponential curve, not a straight line.}$
- $\text{A linear model would significantly underestimate the rodent population}$
  $\text{over time.}$

Show Worked Solution

$\text{Yes, the student is correct.}$

$\text{Correct justification could include ONE of the following:}$

- $\text{The rodent population is doubling every two weeks, which means it}$
  $\text{is increasing by a constant multiplier, not a constant amount — this}$
  $\text{is exponential growth, not linear.}$
- $\text{A linear model increases by the same amount each time period, but}$
  $\text{the rodent population increases by a larger amount each fortnight as}$
  $\text{the population grows.}$
- $\text{The graph of the rodent population over time would be an upsloping}$
  $\text{exponential curve, not a straight line.}$
- $\text{A linear model would significantly underestimate the rodent population}$
  $\text{over time.}$

Algebra, STD1 EQ-Bank 15

A household's monthly water bill consists of a fixed service charge of $45 plus $3 per kilolitre of water used.

Let $C$ = monthly cost in dollars, and $k$ = water usage in kilolitres.

Complete the table of values below that models the relationship between water usage and monthly cost. (1 mark)

$\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad k \quad & \quad 0 \quad & \quad 10 \quad & \quad 20 \quad & \quad 30 \quad & \quad 40 \quad & \quad 50 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}C & & 75 & 105 & & 165 & 195 \\[12pt] \hline \end{array}$

--- 0 WORK AREA LINES (style=lined) ---
Using the table of values from (a), neatly graph the monthly cost for water usage from 0 to 50 kL on the grid below. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Using your graph from (b), or otherwise, find the monthly cost when the household uses 35 kL of water. (1 mark)

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

Show Answers Only

a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

c. $$120$

Show Worked Solution

a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

c. $\text{From the graph, when } k=35,\ \ C=\$120$

Algebra, STD1 EQ-Bank 14

GreenCut Lawn Services charges a fixed call-out fee of $25 plus $40 per hour of work.

Let $C$ = total charge in dollars, and $h$ = number of hours worked.

Complete the table of values below that models the relationship between GreenCut's Lawn Services hours worked and total charge. (1 mark)

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

\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad 2\quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & & 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 & & 225 \\
\hline
\end{array}
Using the table of values from (a), neatly graph the total charge for work completed from 0 to 5 hours on the grid below. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Using your graph from (b), or otherwise, find the total charge for 2.5 hours of work. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
A customer has a budget of $160. Using your graph from (b), or otherwise, determine the maximum number of complete hours GreenCut can work within this budget. (1 mark)

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

Show Answers Only

a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad 2\quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & \textbf{25}& 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 &\textbf{185} & 225 \\
\hline
\end{array}

c. $$125$

d. $3\ \text{hours}$

Show Worked Solution

a. $\text{Table of values:}$

c. $\text{From the graph, when }\ h=2.5, \ C=\$125$

d. $\text{From the graph, \$160 lies between } h=3 \text{ and } h=4.$

$\therefore\ \text{Maximum complete hours} = 3\ \text{hours}$

Algebra, STD1 EQ-Bank 24

Zara is looking for a new mobile phone plan. She has found two plans that suit her needs and wants to work out which plan is cheaper depending on how many minutes she uses.

Plan $\text{A}$: $20 per month fixed charge plus $0.10 per minute

Plan $\text{B}$: $0.30 per minute, no fixed charge

Let $C$ = total monthly cost in dollars, and $m$ = number of minutes used.

Plan $\text{B}$ can be modelled by the equation $C=0.30m$.
Write an equation for the monthly cost of Plan $\text{A}$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The graph of Plan $\text{B}$ is provided on the grid below. Use the equation from (a) to add the graph of Plan $\text{A}$ to the grid. (2 marks)

--- 3 WORK AREA LINES (style=lined) ---
For how many minutes per month do both plans cost the same amount? (1 mark)

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Zara uses an average of 140 minutes per month.
Which plan she should choose and how much does she save compared to the other plan. (1 mark)

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

Show Answers Only

a. $\text{Plan A: }C=20+0.10m$

c. $100\ \text{minutes}$

d. $\text{Plan A, cheaper by } \$8$

Show Worked Solution

a. $\text{Plan A: }\ C=20+0.10m$

b. $\text{Table of values}$

$\begin{array}{|c|c|c|c|c|c|} \hline m & 0 & 50 & 100 & 150 & 200 \\ \hline \text{Plan A} & 20 & 25 & 30 & 35 & 40 \\ \hline \end{array}$

c. $\text{From the graph, the lines intersect at }\ m=100.$

$\therefore\ \text{Both plans cost the same at } 100\ \text{minutes}$

d. $\text{Plan A: }\ C=20+0.10\times140=\$34$

$\text{Plan B: }\ C=0.30\times140=\$42$

$\text{Difference} = 42-34=\$8$

$\therefore\ \text{Zara should choose Plan A, which is \$8 cheaper than Plan B.}$

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)

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

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

Show Answers Only

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+$36\,000= \$71\,000$

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

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

\(\text{Profit}\)	\( = \text{Revenue}-\text{Total costs}\)
\(4200\)	\( = 350x-(1200+150x)\)
\(4200\)	\(= 200x-1200\)
\(5400\)	\(=200x\)
\(x\)	\(=27\)

\(\abs{z_1+z_2}^2\)	\(=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)\)
	\(=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2\)
	\(=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)\)

\(\abs{z_1+z_2}^2\)	\(\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}\)
	\(\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2\)
	\(\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2\)

\(\abs{2z}\)	\(\leqslant\abs{z-1}+\abs{z+1}\)
\(\abs{2z}\)	\(\leqslant 4\ \ \text{(using (1) above)}\)
\(2\abs{z}\)	\(\leqslant 4\)
\(\abs{z}\)	\(\leqslant 2\)

\(\dfrac{(x+1)^2}{4}+\dfrac{(y-3)^2}{4}\)	\(=1\)
\((x+1)^2+(y-3)^2\)	\(=4\)

\(\|z+w\|^2\)	\(=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}\)
	\(=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}\)
	\(=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}\)
	\(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)
	\(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)
	\(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)
	\(=\dfrac{4-\sqrt6+\sqrt2}{2} \)

\(T-10g\,\sin30^{\circ}\)	\(= 0\)
\(T-\dfrac{10g}{2}\)	\(= 0\)
\(T\)	\(= \dfrac{10g}{2}\)
\(T\)	\(= 5g=49\ \text{N}\)

\(T_1 \cos 60^{\circ}\)	\(=T_2 \cos 30^{\circ}\)
\(T_1 \times \dfrac{1}{2}\)	\(=T_2 \times \dfrac{\sqrt{3}}{2}\)
\(T_1\)	\(=\sqrt{3}\, T_2\)

\(T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}\)	\(=10 \times 9.8\)
\(T_1 \times \dfrac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}\)	\(=98\)