adapted – Page 5

Algebra, STD2 A2 2005 HSC 17 MC (Adapted)

The total cost, $$C$, of a school excursion is given by $C=4n+9$, where $n$ is the number of students.

If five extra students go on the excursion, by how much does the total cost increase?

Show Answers Only

$B$

Show Worked Solution

$C=4n+9$

$\text{If}\ n\ \text{increases to}\ n+5$

$C$	$=4(n+5)+9$
	$=4n+20+9$
	$=4n+29$

$\therefore \text{Total cost increases by }$20$

$\Rightarrow B$

Functions, 2ADV F1 2015 HSC 2 MC (Adapted)

What is the slope of the line with equation `2x-4y + 3 = 0`?

`-2`
`-1/2`
`1/2`
`2`

Show Answers Only

`C`

Show Worked Solution

`2x-4y + 3`	`= 0`
`4y`	`= 2x + 3`
`y`	`= 1/2 x + 3/4`

`:.\ text(Slope)\ = 1/2`

`=> C`

Algebra, STD2 A2 2015 HSC 13 MC (Adapted)

What is the equation of the line $l$?

$y=-\dfrac{1}{5}x+5$
$y=\dfrac{1}{5}x+5$
$y=-5x+5$
$y=5x+5$

Show Answers Only

$C$

Show Worked Solution

$l\ \ \text{passes through (0, 5) and (1, 0)}$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{5-0}{0-1}$
	$=-5$

$y\ \text{-intercept}= 5$

$\therefore\ y=-5x+5$

$\Rightarrow C$

♦ Mean mark 48%.

Algebra, STD2 A2 2016 HSC 14 MC (Adapted)

The graph shows a line which has an equation in the form $y=mx+c$.

Which of the following statements is true?

$m$ is positive and $c$ is negative
$m$ is negative and $c$ is positive
$m$ and $c$ are both positive
$m$ and $c$ are both negative

Show Answers Only

$B$

Show Worked Solution

$y\text{-intercept}\ (c)\ \text{is positive}$

$\rightarrow\ \text{eliminate A and D}$

$\text{gradient}\ (m)\ \text{is negative}$

$\rightarrow\ \text{eliminate C}$

$\Rightarrow B$

Algebra, STD2 A2 2017 HSC 20 MC (Adapted)

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 230 matches in total are available?

Show Answers Only

$C$

Show Worked Solution

$\text{1 pentagon:}\ 5+4\times 0=5$

$\text{2 pentagons:}\ 5+4\times 1=9$

$\text{3 pentagons:}\ 5+4\times 2 = 13$

$\vdots$

$n\ \text{pentagons:}\ 5 + 4(n – 1)$

$5+4(n – 1)$	$=230$
$4n-4$	$=225$
$4n$	$=229$
$n$	$=57.25$

$\text{Complete pentagons possible}\ =57$

$\Rightarrow C$

Algebra, STD2 A4 2017 HSC 17 MC (Adapted)

The graph of the line with equation $y=5-x$ is shown.

When the graph of the line with equation $y=2x-1$ is also drawn on this number plane, what will be the point of intersection of the two lines?

$(0, 5)$
$(1, 2)$
$(2, 3)$
$(5, 0)$

Show Answers Only

$C$

Show Worked Solution

$\text{Method 1: Graphically}$

$\text{From graph, intersection is at} (2,3)$

$\text{Method 2: Algebraically}$

$y$	$=5-x$	$…\ (1)$
$y$	$=2x-1$	$…\ (2)$

$\text{Substitute (2) into (1)}$

$2x-1$	$=5-x$
$3x$	$=6$
$x$	$=2$

$\text{When}\ \ x=2,\ y=5-2=3$

$\Rightarrow C$

Functions, 2ADV F1 2017 HSC 1 MC (Adapted)

What is the gradient of the line $4x-5y-2 = 0$?

$-\dfrac{4}{5}$
$\dfrac{4}{5}$
$\dfrac{5}{4}$
$-\dfrac{5}{4}$

Show Answers Only

$B$

Show Worked Solution

$4x-5y-2$	$=0$
$-5y$	$=-4x + 2$
$y$	$=\dfrac{4}{5}x-\dfrac{2}{5}$

$\Rightarrow B$

Algebra, 2ADV F1 2018 HSC 3 MC (Adapted)

What is the $x$-intercept of the line $x-4y+8=0$?

$(-2, 0)$
$(-8, 0)$
$(0, -8)$
$(0, -2)$

Show Answers Only

$B$

Show Worked Solution

$x\text{-intercept occurs when}\ y = 0:$

$x-4y+8$	$=0$
$x$	$=-8$

$\therefore\ x\text{-intercept is}\ (-8, 0)$

$\Rightarrow B$

Algebra, STD2 A2 2020 HSC 6 MC (Adapted)

Suppose $y=-2-3x$.

When the value of $x$ increases by 4, the value of $y$ decreases by

$1$.
$4$.
$12$.
$14$.

Show Answers Only

$C$

Show Worked Solution

$\text{Strategy 1}$

$\text{If}\ \ x\ \ \text{increases by} \ 4$

$\rightarrow y\ \text{decreases by} \ \ 3x=3\times 4 = 12$

$\text{Strategy 2}$

$\text{Test}\ 2\ \text{values:}$

$\text{If} \ \ x=0 , \ y=-2$

$\text{If}\ \ x=4 , \ y =-2-3\times 4=-14$

$\therefore\ \ y \ \text{decreases by} \ 12.$

$\Rightarrow C$

Algebra, STD2 A2 2021 HSC 9 MC (Adapted)

Marty is thinking of a number. Let the number be $n$.

When Marty subtracts 4 from this number and multiplies the result by 7, the answer is 8 more than $n$.

Which equation can be used to find $n$?

$7n-4=8n$
$7(n-4)=8n$
$7n-4=n+8$
$7(n-4)=n+8$

Show Answers Only

$D$

Show Worked Solution

$\text{The description defines the following equation:}$

$(n-4)\times 7$	$=n+8$
$7(n-4)$	$=n+8$

$\Rightarrow D$

Algebra, STD2 A2 2022 HSC 2 MC (Adapted)

Which of the following could be the graph of $y=-2-2x$?

Show Answers Only

$B$

Show Worked Solution

$\text{By elimination:}$

$y\text{-intercept} =-2\ \rightarrow\ \text{Eliminate}\ A \text{ and}\ D$

$\text{Gradient is negative}\ \rightarrow\ \text{Eliminate}\ C$

$\Rightarrow B$

♦ Mean mark 48%.

Algebra, STD2 A1 2012 HSC 15 MC (Adapted)

A car takes 5 hours to complete a journey when travelling at 75 km/h.

How long would the same journey take if the car were travelling at 100 km/h?

37.5 minutes
1 hour and 20 minutes
3 hours and 45 minutes
4 hours and 15 minutes

Show Answers Only

$C$

Show Worked Solution

$T=\dfrac{D}{S}$

$\text{Since}\ \ \ T = 5\ \ \text{when}\ \ \ S = 75$

$5$	$=\dfrac{D}{75}$
$D$	$=5\times 75$
	$=375\ \text{km}$

$\text{Find}\ \ T\ \ \text{when}\ \ \ S = 100\ \ \text{ and}\ \ \ D = 375$

$T$	$=\dfrac{375}{100}$
	$=3.75\ \text{hours}$
	$=3\ \text{hrs}\ \ 45\ \text{minutes}$

$\Rightarrow C$

Algebra, STD2 A1 2011 HSC 21 MC (Adapted)

A train departs from Town A at 4.00 pm to travel to Town B. Its average speed for the journey is 80 km/h, and it arrives at 6.00 pm. A second train departs from Town A at 4.30 pm and arrives at Town B at 6.10 pm.

What is the average speed of the second train?

96 km/h
114 km/h
224 km/h
280 km/h

Show Answers Only

$A$

Show Worked Solution

$\text{1st train:}$

$\text{Travels 2hrs at 80km/h}$

$\text{Distance}$	$=\text{Speed}\times\text{Time}$
	$=80\times 2$
	$=160\ \text{km}$

$\text{2nd train:}$

$\text{Travels 160 km in 1 hr 40 min}\ \rightarrow\ \dfrac{5}{3}\ \text{hrs}$

$\text{Speed}$	$=\dfrac{\text{Distance}}{\text{Time}}$
	$=160\ ÷\ \dfrac{5}{3}$
	$=160\times \dfrac{3}{5}$
	$=96\ \text{km/h}$

$\Rightarrow A$

♦♦ Mean mark 49%.

Algebra, STD2 A1 2009 HSC 16 MC (Adapted)

The time for a train to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?

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$C$

Show Worked Solution

$T$	$\propto \dfrac{1}{S}$
$T$	$=\dfrac{k}{S}$

$\text{By elimination:}$

$\text{As Speed} \uparrow \ \text{, Time}\downarrow\ \Rightarrow\ \text{cannot be A or B}$

$\text{D is incorrect because it graphs a linear relationship}$

$\Rightarrow C$

♦♦ Mean mark 38%.

Algebra, STD2 A1 2014 HSC 4 MC (Adapted)

Young’s formula below is used to calculate the required dosages of medicine for children aged 1–12 years.

$\text{Dosage}=\dfrac{\text{age of child (in years)}\ \times\ \text{adult dosage}}{\text{age of child (in years)}\ +\ 12}$

How much of the medicine should be given to an 18-month-old child in a 24-hour period if each adult dosage is 27 mL? The medicine is to be taken every 8 hours by both adults and children.

3 mL
6 mL
9 mL
12 mL

Show Answers Only

$C$

Show Worked Solution

$\text{Age of child} = 18\ \text{months}=1.5\ \text{years}$

$\text{Dosage}$	$=\dfrac{1.5\times 27}{1.5+12}$
	$=3\ \text{mL}$

$\text{Dosage every 8 hrs}$

$\therefore\ \text{In 24 hours, medicine given} = 3\times 3=9\ \text{mL}$

$\Rightarrow C$

♦♦ Mean mark 42%.

Algebra, STD2 A1 2014 HSC 29b (Adapted)

Blood alcohol content of males can be calculated using the following formula

$BAC_{\text{Male}} = \dfrac{10N-7.5H}{6.8M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

What is the maximum number of standard drinks that Jacko, who has a mass of 75 kg, can consume over 5 hours in order to maintain a blood alcohol content ($BAC$) of less than 0.05? (3 marks)

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Show Answers Only

$6$

Show Worked Solution

$BAC_\text{male}=\dfrac{10N-7.5H}{6.8M}$

$\text{Find}\ \ N\ \text{for }BAC<0.05,\ \text{given}\ \ H=5\ \text{and}\ \ M = 75$

$\dfrac{10N-7.5\times 5}{6.8\times 75}$	$< 0.05$
$10N-37.5$	$< 0.05\times 6.8\times 75$
$10N$	$< 25.5+37.5$
$10N$	$<63$
$\therefore\ N$	$< 6.3$

$\therefore\ \text{Max number of standard drinks is 6.}$

Algebra, STD2 A1 2015 HSC 23 MC (Adapted)

The number of ‘standard drinks’ in various glasses of wine is shown.

Number of standard drinks
White Wine		Red Wine
small glass	large glass	small glass	large glass
0.9	1.4	1.0	1.5

A woman weighing 58 kg drinks two small glasses of white wine and three small glasses of red wine between 7 pm and 11 pm.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content ($BAC$) estimate at 11 pm, correct to three decimal places?

$BAC_{\text{Female}}=\dfrac{10N-7.5H}{5.5M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

0.013
0.023
0.046
0.056

Show Answers Only

$D$

Show Worked Solution

$N$	$=2\times 0.9 + 3\times 1$
	$=4.8\ \text{standard drinks}$
$H$	$=4\ \text{hours}$
$M$	$=58\ \text{kg}$

$BAC_f$	$=\dfrac{10\times 4.8-7.5\times 4}{5.5\times 58}$
	$=0.05642\dots$

$\Rightarrow D$

Algebra, STD2 A1 2005 HSC 24b (Adapted)

The formula $D=\dfrac{2A}{15}$ is used to calculate the dosage of liquid paracetamol to be given to a child.

- $D$ is the dosage of liquid paracetamol in millilitres (mL).
- $A$ is the age of the child in months.

If Charlotte is six months old, what dosage of liquid paracetamol should she be given? (1 mark)

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The correct dosage of liquid paracetamol for Teddy is 6 mL.

What is the difference in the ages of Teddy and Charlotte, in months? (3 marks)

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a. $\text{0.8 mL}$

b. $39$

Show Worked Solution

a.	$D$	$=\dfrac{2A}{15}$
		$=\dfrac{2\times 6}{15}$
		$=0.8\text{ mL}$

$\therefore\ \text{Charlotte should be given a dosage of 0.8 mL}$

b. $\text{Find}\ A\ \text{when}\ D=\text{6 mL}$

$6$	$=\dfrac{2A}{15}$
$2A$	$=90$
$A$	$=45$

$\therefore\ \text{Teddy is 45 months old and is 39 months}$

$\text{older than Charlotte.}$

Algebra, STD2 A1 2015 HSC 30d (Adapted)

Monica is driving on a motorway at a speed of 105 kilometres per hour and has to brake suddenly. She has a reaction time of 1.3 seconds and a braking distance of 54.3 metres.

Stopping distance can be calculated using the following formula

$\text{stopping distance = {reaction time distance} + {braking distance}}$

What is Monica's stopping distance? Give your answer to 1 decimal place. (2 marks)

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Show Answers Only

$92.2\ \text{metres (to 1 d.p.)}$

Show Worked Solution

$105\ \text{km/hr}$	$=105\ 000\ \text{m/hr}$
	$=\dfrac{105\ 000}{60\times 60}\ \text{m/sec}$
	$=29.166\dots\ \text{m/sec}$

$\text{Reaction time distance}$	$=1.3\times 29.166\dots$
	$=37.916\dots\ \text{metres}$

$\text{Stopping distance}$

$\text{ = {Reaction time distance} + {braking distance}}$

$=37.916…+54.3$

$=92.216\dots$

$=92.2\ \text{metres (to 1 d.p.)}$

♦ Mean mark 34%.

Algebra, STD2 A1 2016 HSC 10 MC (Adapted)

Anika drinks two small bottles of wine over a four-hour period. Each of these bottles contains 2.4 standard drinks. Anika weighs 55 kg.

Using the formula below, what is Anika's approximate blood alcohol content ($BAC$) at the end of this period?

$BAC_{\text{Female}}=\dfrac{10N - 7.5H}{5.5M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

0.013
0.060
0.0013
0.0060

Show Answers Only

$B$

Show Worked Solution

$BAC_f$	$=\dfrac{10N – 7.5H}{5.5M}$
	$=\dfrac{10(2\times 2.4) – 7.5\times 4}{5.5\times 55}$
	$= 0.0595\dots\approx 0.060$

$\Rightarrow B$

Algebra, STD2 A1 2017 HSC 2 MC (Adapted)

A car is travelling at 85 km/h.

How far will it travel in 3 hours and 30 minutes?

$24.3\ \text{km}$
$25.8\ \text{km}$
$280.5\ \text{km}$
$297.5\ \text{km}$

Show Answers Only

$D$

Show Worked Solution

$\text{Distance}$	$=85\times 3.5$
	$=297.5\ \text{km}$

$\Rightarrow D$

Algebra, STD2 A1 2017 HSC 19 MC (Adapted)

Young’s formula, shown below, is used to calculate the dosage of medication for children aged 1−12 years based on the adult dosage.

$D=\dfrac{yA}{y + 12}$

where $D$	= dosage for children aged 1−12 years
$y$	= age of child (in years)
$A$	= Adult dosage

A child’s dosage is calculated to be 15 mg, based on an adult dosage of 30 mg.

How old is the child in years?

Show Answers Only

$D$

Show Worked Solution

$D$	$=\dfrac{yA}{y+12}$
$15$	$=\dfrac{30y}{y+12}$
$15(y+12)$	$=30y$
$15y+180$	$=30y$
$15y$	$=180$
$y$	$=12$

$\Rightarrow D$

Algebra, STD2 A1 2017 HSC 27e (Adapted)

Bryce is drinking low alcohol beer at a party over a four-hour period. He reads on the label of the low alcohol beer bottle that it is equivalent to 0.8 standard drinks.

Bryce weighs 85 kg.

The formula below can be used to calculate a male's blood alcohol content.

$BAC_{\text{Male}}=\dfrac{10N-7.5H}{6.8M}$

where $N$ is the number of standard drinks consumed

$H$ is the number of hours drinking

$M$ is the person's mass in kilograms

What is the maximum number of complete bottles of the low alcohol beer Bryce can drink to remain under a Blood Alcohol Content ($BAC$) of 0.05? (4 marks)

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Show Answers Only

$7$

Show Worked Solution

$BAC_\text{male}$	$=\dfrac{10N-7.5H}{6.8M}$
$0.05$	$=\dfrac{10N-7.5\times 4}{6.8\times 85}$
$10N$	$=0.05\times 6.8\times 85+7.5\times 4$
$10N$	$=58.9$
$N$	$=5.89\ \text{standard drinks}$

$\therefore\ \text{Number of low alcohol bottles}$

$=\dfrac{5.89}{0.8}$

$=7.3625$

$\therefore\ \text{Max complete bottles to stay under 0.05}$

$=7$

Algebra, STD2 A1 2018 HSC 26b (Adapted)

Clark’s formula, given below, is used to determine the dosage of medicine for children.

$\text{Dosage}=\dfrac{\text{weight in kg × adult dosage}}{70}$

For a particular medicine, the adult dosage is 220 mg and the correct dosage for a specific child is 45 mg.

How much does the child weigh, to the nearest kg? (2 marks)

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$14\ \text{kg}$

Show Worked Solution

$45 =\dfrac{\text{weight}\times 220}{70}$

$\therefore\ \text{weight}$	$=\dfrac{70\times 45}{220}$
	$=14.318\dots$
	$\approx 14\ \text{kg (nearest kg)}$

Algebra, STD2 A1 2018 HSC 28e (Adapted)

Drake is driving at 80 km/h. He notices a branch on the road ahead and decides to apply the brakes. His reaction time is 1.2 seconds. His braking distance ($D$ metres) is given by $D=0.01v^2$, where $v$ is speed in km/h.

Stopping distance can be calculated using the following formula

$\text{stopping distance = {reaction time distance} + {braking distance}}$

What is Drake’s stopping distance, to the nearest metre? (3 marks)

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$91\ \text{m (nearest m)}$

Show Worked Solution

$\text{80 km/hr}$	$=80\ 000\ \text{m/hr}$
	$=\dfrac{80\ 000}{60\times 60}\ \text{m/sec}$
	$=22.22\dots\ \text{m/sec}$

$\text{Total stopping distance}$

$\text{ = {reaction time distance} + {braking distance}}$

$=1.2\times 22.22\dots + 0.01\times 80^2$

$=90.66\dots$

$=91\ \text{m (nearest m)}$

♦ Mean mark 46%.

Algebra, STD2 A1 2019 HSC 28 (Adapted)

The formula below is used to calculate an estimate for blood alcohol content $(BAC)$ for females.

$BAC_{\text{female}}=\dfrac{10N - 7.5H}{5.5M}$

The number of hours required for a person to reach zero $BAC$ after they stop consuming alcohol is given by the following formula.

$\text{Time}=\dfrac{BAC}{0.015}$

The number of standard drinks in a glass of wine and a glass of spirits is shown.

Georgie weighs 58 kg. She consumed 2 glasses of wine and 4 glasses of spirits between 7:45 pm and 12:15 am the following day. She then stopped drinking alcohol.

Using the given formulae, calculate the time in the morning when Georgie's $BAC$ should reach zero. (4 marks)

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$\text{6:34 am}$

Show Worked Solution

$\text{Standard drinks consumed}\ (N)=2\times 1.2+4=6.4$

$\text{Hours drinking}\ (H) = \text{4 h 30 min = 4.5 hours}$

$BAC_{\text{Georgie}}$	$=\dfrac{10\times 6.4-7.5\times 4.5}{5.5\times 58}$
	$=0.09482\dots$

COMMENT: Convert a decimal answer into hours and minutes using the calculator degree/minute function.

$\text{Time (to zero)}$	$=\dfrac{0.09482\dots}{0.015}$
	$=6.3218\dots\ \text{hours}$
	$\approx 6\ \text{hours 19 minutes}$

$\therefore\ \text{Georgie should reach zero}\ BAC$

$=12:15+6:19=6:34\ \text{am}$

Algebra, STD2 A1 2023 HSC 36 (Adapted)

The following formula can be used to calculate an estimate for blood alcohol content ($BAC$) for males.

$BAC_{\text{male}}=\dfrac{10N-7.5H}{6.8M}$

$N$ is the number of standard drinks consumed

$M$ is the person's weight in kilograms

$H$ is the number of hours of drinking

Min weighs 70 kg. His $BAC$ was zero when he began drinking alcohol. At 10:30 pm, after consuming 4 standard drinks, his $BAC$ was 0.032.

Using the formula, estimate at what time Min began drinking alcohol, to the nearest minute. (4 marks)

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$7:12\text{ pm}$

Show Worked Solution

$BAC$	$=\dfrac{10N-7.5H}{6.8M}$
$0.032$	$=\dfrac{10\times 4-7.5\times H}{6.8\times 70}$
$0.032\times 476$	$=40-7.5H$
$7.5H$	$=40-15.232$
$H$	$=\dfrac{24.768}{7.5}$
	$=3.3024\ \text{hours}$
	$\approx 3\ \text{hours}\ 18\ \text{minutes (nearest minute)}$

$\text{Time Min began drinking}$

$=10:30\text{ pm – 3 h 18 m}$

$=7:12\text{ pm}$

Mean mark 56%.

Algebra, STD2 A1 2015 HSC 26b (Adapted)

Clark’s formula is used to determine the dosage of medicine for children.

$\text{Dosage}=\dfrac{\text{weight in kg × adult dosage}}{70}$

The adult daily dosage of a medicine contains 1750 mg of a particular drug.

A child who weighs 30 kg is to be given tablets each containing 125 mg of this drug.

How many tablets should this child be given daily? (2 marks)

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$6$

Show Worked Solution

$\text{Dosage}$	$=\dfrac{30\times 1750}{70}$
	$=750\ \text{mg}$

$\text{Number of tablets per day}$

$=\dfrac{\text{Dosage}}{\text{mg per tablet}}$

$=\dfrac{750}{125}$

$=6$

$\therefore\ \text{The child should be given 6 tablets per day.}$

Algebra, STD2 A1 2020 HSC 3 MC (Adapted)

The distance between the Yarra Valley and Ballarat is 150 km. A person travels from the Yarra Valley to Ballarat at an average speed of 90 km/h.

How long does it take the person to complete the journey?

60 minutes
66 minutes
1 hour 30 minutes
1 hour 40 minutes

Show Answers Only

$D$

Show Worked Solution

$\text{Time}$	$=\dfrac{\text{Distance}}{\text{Speed}}$
	$=\dfrac{150}{90}$
	$=1.\dot{6}\ \text{hours}$
	$=1\ \text{hour}\ 40\ \text{minutes}$

$\Rightarrow D$

Algebra, STD2 A1 2020 HSC 13 MC (Adapted)

When Stuart stops drinking alcohol at 11:30 pm, he has a blood alcohol content (BAC) of 0.08625.

The number of hours required for a person to reach zero BAC after they stop consuming alcohol is given by the formula:

$\text{Time}=\dfrac{BAC}{0.015}$.

At what time on the next day should Stuart expect his BAC to be 0.05?

1:33 am
1:55 am
2:15 am
5:15 am

Show Answers Only

$B$

Show Worked Solution

$\text{Time from 0.08625 → 0}\ BAC$

$=\dfrac{0.08625}{0.015}$

$=5.75\ \text{hours}$

$\text{Time from 0.08625 → 0.05}\ BAC$

$=\dfrac{(0.08625 – 0.05)}{0.08625}\times 5.75$

$=\dfrac{29}{69}\times 5.75$

$=2.41\dot{6}=2\ \text{h}\ 25\ \text{min}$

$\therefore\ \text{Time}$	$=11:30\ \text{pm} \ + 2 \ \text{h} \ 25 \ \text{min}$
	$=1:55\ \text{am}$

$\Rightarrow B$

♦♦♦ Mean mark 17%.
COMMENT: The rates aspect of this question proved extremely challenging.

Algebra, STD2 A1 2010 HSC 24a (Adapted)

Margie tried to solve this equation and made a mistake in Line 2.

\begin{array}{rl}
3(m+3)-2(m+4)=-5\ &\ \ \ \text{Line 1} \\
3m+9-2m+8=-5\ &\ \ \ \text{Line 2} \\
m+17=-5\ &\ \ \ \text{Line 3} \\
m=-12& \ \ \ \text{Line 4}
\end{array}

Copy the equation in Line 1. Rewrite Line 2 correcting her mistake. (1 mark)

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Continue your solution showing the correct working for Lines 3 and 4 to solve this equation for $m$. (1 mark)

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a.	$3(m+3)-2(m+4)$	$=-5\ \ \ \ \ \ \text{Line}\ 1$
	$3m+9-2m-8$	$=-5\ \ \ \ \ \ \text{Line}\ 2$

b.	$m+1$	$=-5$
	$m$	$=-6$

Show Worked Solution

a.	$3(m+3)-2(m+4)$	$=-5\ \ \ \ \ \ \text{Line}\ 1$
	$3m+9-2m-8$	$=-5\ \ \ \ \ \ \text{Line}\ 2$

b.	$m+1$	$=-5\ \ \ \ \ \ \text{Line}\ 3$
	$m$	$=-6\ \ \ \ \ \ \text{Line}\ 4$

Algebra, STD2 A1 2013 HSC 29a (Adapted)

Jeremy tried to solve this equation and made a mistake in Line 2.

$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	$\text{... Line 1}$
$5M+15-4M-2$	$=10$	$\text{... Line 2}$
$M+13$	$=10$	$\text{... Line 3}$
$M$	$=-3$	$\text{... Line 4}$

Copy the equation in Line 1 and continue your solution to solve this equation for $M$.

Show all lines of working. (2 marks)

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

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$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	`text(… Line 1)`
$5M+15-4M+2$	$=10$	`text(… Line 2)`
$M+17$	$=10$	`text(… Line 3)`
$M$	$=-7$	`text(… Line 4)`

Show Worked Solution

♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 10 (from Line 1 to Line 2), indicating both sides must have been multiplied by 10.

$\dfrac{M+3}{2}-\dfrac{2M-1}{5}$	$=1$	`text(… Line 1)`
$5M+15-4M+2$	$=10$	`text(… Line 2)`
$M+17$	$=10$	`text(… Line 3)`
$M$	$=-7$	`text(… Line 4)`

Algebra, STD2 A1 2010 HSC 7 MC (Adapted)

If $M=-8$, what is the value of $\dfrac{4M^2+3M}{8}$

$-1027$
$-35$
$29$
$125$

Show Answers Only

$C$

Show Worked Solution

♦♦ Only 31% of students answered correctly!

$\dfrac{4M^2+3M}{8}$	$=\dfrac{4\times (-8)^2+3\times (-8)}{8}$
	$=\dfrac{4\times 64-24}{8}$
	$=\dfrac{232}{8}$
	$=29$

$\Rightarrow C$

Algebra, STD2 A1 2009 HSC 25a (Adapted)

Simplify $10-3(x+4)$. (2 marks)

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

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$-3x-2$

Show Worked Solution

♦ Mean mark 47%

$10-3(x+4)$	$=10-3x-12$
	$=-3x-2$

Algebra, STD2 A1 2014 HSC 26c (Adapted)

Solve the equation $\dfrac{4x-3}{5}-6=7-6x$. (3 marks)

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

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$x=2$

Show Worked Solution

$\dfrac{4x-3}{5}-6$	$=7-6x$
$4x-3-5\times 6$	$=5(7-6x)$
$4x-3-30$	$=35-30x$
$34x$	$=68$
$\therefore\ x$	$=2$

Algebra, STD2 A1 2008 HSC 9 MC (Adapted)

What is the value of $\sqrt{\dfrac{2x + y}{5x}}$ if $x=5.1$ and $y=3.7$, correct to 2 decimal places?

$0.13$
$0.74$
$3.74$
$3.80$

Show Answers Only

$B$

Show Worked Solution

$\sqrt{\dfrac{2x+y}{5x}}$	$=\sqrt{\dfrac{2\times 5.1+3.7}{5\times 5.1}}$
	$=\sqrt{\dfrac{13.9}{25.5}}$
	$= 0.7383\dots$

$\Rightarrow B$

Algebra, STD2 A1 2007 HSC 24b (Adapted)

The distance in kilometres ($D$) of an observer from the centre of a thunderstorm can be estimated by counting the number of seconds ($t$) between seeing the lightning and first hearing the thunder.

Use the formula $D=\dfrac{t}{3}$ to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 2.1 km away. (1 mark)

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

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$6.3\ \text{seconds}$

Show Worked Solution

$D=\dfrac{t}{3}$

$\text{When}\ \ D = 2.1,$

$\dfrac{t}{3}$	$=2.1$
$t$	$=6.3\ \text{seconds}$

Algebra, STD2 A1 2007 HSC 28b (Adapted)

This shape is made up of two right-angled triangle and a regular hexagon.

The area of a regular hexagon can be estimated using the formula $A=2.598S^2$ where $S$ is the hexagon's side-length.

Calculate the total area of the shape using this formula. (3 marks)

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

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$619.6\ \text{cm}^2$

Show Worked Solution

$\text{Area}=2.598S^2$

$\text{Using Pythagoras}$

$S^2= 10^2+10^2=200$

$S=\sqrt{200}$

$A=2.598\times (\sqrt {200})^2=519.6\ \text{cm}^2$

$\text{Area of Δ}\ =\dfrac{1}{2}bh=\dfrac{1}{2}\times 10\times 10=50 \ \text{cm}^2$

$\therefore\ \text{Total Area}\ =519.6+50+50=619.6\ \text{cm}^2$

Algebra, STD2 A1 2005 HSC 2 MC (Adapted)

What is the value of $\dfrac{x-y}{6}$, if $x=184$ and $y=46$?

$6$
$23$
$176$
$552$

Show Answers Only

$B$

Show Worked Solution

$\dfrac{x-y}{6}=\dfrac{184-46}{6}=23$

$\Rightarrow B$

Algebra, STD2 A1 2006 HSC 2 MC (Adapted)

If $V=\dfrac{4}{3}\pi r^3$, what is the value of $V$ when $r = 5$, correct to two decimal places?

$20.94$
$53.05$
$104.72$
$523.60$

Show Answers Only

$D$

Show Worked Solution

$V =\dfrac{4}{3}\pi r^3$

$\text{When}\ r = 2,$

$V$	$=\dfrac{4}{3}\pi\times 5^3$
	$=523.598\dots$

$\Rightarrow D$

Algebra, STD2 A1 2004 HSC 3 MC (Adapted)

If $K=Ft^3$, $F=9$ and $t=0.829$, what is the value of $K$ correct to three significant figures?

$5.12$
$5.127$
$5.128$
$5.13$

Show Answers Only

$D$

Show Worked Solution

$K$	$=Ft^3$
	$=9\times 0.829^3$
	$=5.1275\dots$
	$=5.13\ \text{(3 sig figures)}$

$\Rightarrow D$

Algebra, STD2 A1 2005 HSC 14 MC (Adapted)

Using the formula $d=6t^3-5$, Marcia tried to find the value of $t$ when $d=389$.

Here is her solution. She has made one mistake.

Which line does NOT follow correctly from the previous line?

$\text{Line}\ A$
$\text{Line}\ B$
$\text{Line}\ C$
$\text{Line}\ D$

Show Answers Only

$B$

Show Worked Solution

$d$	$=6t^3-5$
$389$	$=6t^3-5\ \ \ $	$\dots\text{ Line A}$
$394$	$=6t^3$	$\dots\text{ Line B}$

$\therefore\ \text{Line}\ B\ \text{does not follow on correctly.}$

$\Rightarrow B$

Algebra, STD2 A1 2015 HSC 2 MC (Adapted)

Which of the following is $5m+4y-m-6y$ in its simplest form?

$4m+10y$
$4m-2y$
$6m+10y$
$6m-2y$

Show Answers Only

$B$

Show Worked Solution

$5m+4y-m-6y$	$=5m-m+4y-6y$
	$=4m-2y$

$\Rightarrow B$

Algebra, STD2 A1 2015 HSC 24 MC (Adapted)

Consider the equation $\dfrac{5x}{2}-3=\dfrac{3x}{5}+1$.

Which of the following would be a correct step in solving this equation?

$\dfrac{5x}{2}-2=\dfrac{3x}{5}$
$\dfrac{10x}{4}-4=\dfrac{6x}{5}$
$\dfrac{5x}{2}=\dfrac{3x}{5}+4$
$5x-3=\dfrac{6x}{5}+2$

Show Answers Only

$C$

Show Worked Solution

$\dfrac{5x}{2}-3$	$=\dfrac{3x}{5}+1$
$\dfrac{5x}{2}-3+3$	$=\dfrac{3x}{5}+1+3$
$\dfrac{5x}{2}$	$=\dfrac{3x}{5}+4$

$\Rightarrow C$

Algebra, STD2 A1 2015 HSC 28d (Adapted)

The formula $C=\dfrac{5}{9}(F-32)$ is used to convert temperatures between degrees Fahrenheit $(F)$ and degrees Celsius $(C)$.

Convert 18°C to the equivalent temperature in Fahrenheit. (2 marks)

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

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$64.4\ \text{degrees}\ F$

Show Worked Solution

$C$	$=\dfrac{5}{9}(F-32)$
$F-32$	$=\dfrac{9}{5}C$
$F$	$=\dfrac{9}{5}C+32$

$\text{When}\ \ C = 18,$

$F$	$=\dfrac{9}{5}\times 18+32$
	$=64.4\ \text{degrees}\ F$

Algebra, STD2 A1 2017 HSC 7 MC (Adapted)

It is given that $I=\dfrac{3}{2}MR^2$.

What is the value of $I$ when $M =19.12$ and $R = 1.02$, correct to two decimal places?

$13.26$
$29.84$
$119.35$
$570.52$

Show Answers Only

$B$

Show Worked Solution

$I$	$=\dfrac{3}{2}\times 19.12\times 1.02^2$
	$=29.84$

$\Rightarrow B$

Algebra, STD2 A1 2017 HSC 9 MC (Adapted)

What is the value of $x$ in the equation $\dfrac{4-x}{7}=2$?

$-14$
$-10$
$10$
$14$

Show Answers Only

$B$

Show Worked Solution

$\dfrac{4-x}{7}$	$=2$
$4-x$	$=14$
$x$	$=4-14$
$\therefore\ x$	$=-10$

$\Rightarrow B$

Algebra, STD2 A1 2016 HSC 5 MC (Adapted)

Which expression is equivalent to $2(7x-3)+5$?

$14x-1$
$14x-8$
$14x-11$
$14x+2$

Show Answers Only

$A$

Show Worked Solution

$2(7x-3)+5$

$=14x-6+5$

$=14x-1$

$\Rightarrow A$

Algebra, STD2 A1 2018 HSC 28b (Adapted)

Solve the equation $\dfrac{3x}{4}+1=\dfrac{5x+1}{3}$, leaving your answer as a fraction. (3 marks)

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

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$\dfrac{8}{11}$

Show Worked Solution

♦ Mean mark 35%.

$\underbrace{\dfrac{3x}{4} + 1}_\text{multiply x 12}$	$=\underbrace{\dfrac{5x+1}{3}}_\text{multiply x 12}$
$9x+12$	$=20x+4$
$11x$	$=8$
$x$	$=\dfrac{8}{11}$

Algebra, STD2 A1 2021 HSC 29 (Adapted)

Solve $x+\dfrac{x-3}{4}=5$, leaving your answer as a fraction. (2 marks)

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

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$\dfrac{23}{5}$

Show Worked Solution

♦ Mean mark 40%.

$x+\dfrac{x-3}{4}$	$=5$
$4x+x-3$	$=20$
$5x$	$=23$
$x$	$=\dfrac{23}{5}$

Algebra, STD1 A1 2020 HSC 18 (Adapted)

The distance, $d$ metres, travelled by a car slowing down from $u$ km/h to $v$ km/h can be obtained using the formula

$v^2=u^2-100 d$

What distance does a car travel while slowing down from 100 km/h to 70 km/h? (2 marks)

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

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$51\ \text{metres}$

Show Worked Solution

$u=100 \ , \ v=70$

$v^2$	$=u^2-100d$
$70^2$	$=100^2-100d$
$100d$	$=100^2-70^2$
$\therefore\ d$	$=\dfrac{100^2-70^2}{100}$
	$=51\ \text{metres}$

Algebra, STD2 A1 2013 HSC 21 MC (Adapted)

Which equation correctly shows $n$ as the subject of $V=600(1-n)$?

$n=\dfrac{V-600}{600}$
$n=\dfrac{600-V}{600}$
$n=V-600$
$n=600-V$

Show Answers Only

$B$

Show Worked Solution

♦♦♦ Mean mark 27%

$V$	$=600(1-n)$
$1-n$	$=\dfrac{V}{600}$
$n$	$=1-\dfrac{V}{600}$
	$=\dfrac{600-V}{600}$

$\Rightarrow B$

Algebra, STD2 A1 2012 HSC 21 MC (Adapted)

Which of the following correctly expresses $r$ as the subject of $V=\pi r^2+x$ ?

$r=\pm\sqrt{\dfrac{V}{\pi}}-x$
$r=\pm\sqrt{\dfrac{V}{\pi}-x}$
$r=\pm\sqrt{\dfrac{V-x}{\pi}}$
$r=\pm\dfrac{\sqrt{V-x}}{\pi}$

Show Answers Only

$C$

Show Worked Solution

$V$	$=\pi r^2+x$
$\pi r^2$	$=V-x$
$r^2$	$=\dfrac{V-x}{\pi}$
$\therefore\ r$	$=\pm\sqrt{\dfrac{V-x}{\pi}}$

$\Rightarrow C$

Algebra, STD2 A1 2011 HSC 18 MC (Adapted)

Which of the following correctly expresses $b$ as the subject of $y= ax+\dfrac{1}{4}bx^2$?

$b=\dfrac{4y-ax}{x^2}$
$b=\dfrac{4(y-ax)}{x^2}$
$b=\dfrac{\dfrac{1}{4}y-ax}{x^2}$
$b=\dfrac{\dfrac{1}{4}(y-ax)}{x^2}$

Show Answers Only

$B$

Show Worked Solution

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

$\Rightarrow B$

Algebra, STD2 A1 2004 HSC 11 MC (Adapted)

If $m = 8n^2$, what is a possible value of $n$ when $m=7200$?

$0.03$
$30$
$240$
$900$

Show Answers Only

$B$

Show Worked Solution

$m$	$=8n^2$
$n^2$	$=\dfrac{m}{8}$
$n$	$=\pm\sqrt{\dfrac{m}{8}}$

$\text{When}\ m=7200:$

$n$	$=\pm\sqrt{\dfrac{7200}{8}}$
	$=\pm 30$

$\Rightarrow B$

Algebra, STD2 A1 2006 HSC 18 MC (Adapted)

What is the formula for $g$ as the subject of $7d=8e+5g^2$?

$g =\pm\sqrt{\dfrac{8e-7d}{5}}$
$g =\pm\sqrt{\dfrac{7d-8e}{5}}$
$g =\pm\dfrac{\sqrt{7d+8e}}{5}$
$g =\pm\dfrac{\sqrt{8e-7d}}{5}$

Show Answers Only

$B$

Show Worked Solution

$7d$	$=8e+5g^2$
$5g^2$	$=7d-8e$
$g^2$	$=\dfrac{7d-8e}{5}$
$g$	$=\pm\sqrt{\dfrac{7d-8e}{5}}$

$\Rightarrow B$

Algebra, STD2 A1 2007 HSC 19 MC (Adapted)

Which of the following correctly expresses $X$ as the subject of $Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)$?

$X=\dfrac{Y}{\pi}-L$
$X=\dfrac{Y}{\pi}-4L$
$X=4L-\dfrac{Y}{2\pi}$
$X=\dfrac{Y}{8\pi}-\dfrac{L}{4}$

Show Answers Only

$B$

Show Worked Solution

$Y$	$=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)$
$\dfrac{Y}{4\pi}$	$=\dfrac{X}{4}+L$
$\dfrac{X}{4}$	$=\dfrac{Y}{4\pi}-L$
$X$	$=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)$
$X$	$=\dfrac{Y}{\pi}-4L$

$\Rightarrow B$

Algebra, STD2 A1 2005 HSC 24c (Adapted)

Make $r$ the subject of the equation $V=4\pi r^2$. (2 marks)

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

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$r=\pm\sqrt{\dfrac{V}{4\pi}}$

Show Worked Solution

$V$	$=4\pi r^2$
$r^2$	$=\dfrac{V}{4\pi}$
$\therefore\ r$	$=\pm\sqrt{\dfrac{V}{4\pi}}$

Algebra, STD2 A1 2016 HSC 24 MC (Adapted)

Which of the following correctly expresses $M$ as the subject of $y=\dfrac{M}{V}+cX$?

$M=Vy-VcX$
$M=Vy+VcX$
$M=\dfrac{y-cX}{V}$
$M=\dfrac{y+cX}{V}$

Show Answers Only

$A$

Show Worked Solution

$y$	$=\dfrac{M}{V}+cX$
$\dfrac{M}{V}$	$=y-cX$
$\therefore\ M$	$=V(y-cX)$
	$=Vy-VcX$

$\Rightarrow A$

Algebra, STD2 A1 2017 HSC 28d (Adapted)

Make $b$ the subject of the equation $a=\sqrt{bc-4}$. (2 marks)

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

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$b=\dfrac{a^2+4}{c}$

Show Worked Solution

♦ Mean mark 46%.

$a$	$=\sqrt{bc-4}$
$a^2$	$=bc-4$
$bc$	$=a^2+4$
$\therefore\ b$	$=\dfrac{a^2+4}{c}$

\(C\)	\(=4(n+5)+9\)
	\(=4n+20+9\)
	\(=4n+29\)

\(\text{Gradient}\)	\(=\dfrac{y_2-y_1}{x_2-x_1}\)
	\(=\dfrac{5-0}{0-1}\)
	\(=-5\)

\(5+4(n – 1)\)	\(=230\)
\(4n-4\)	\(=225\)
\(4n\)	\(=229\)
\(n\)	\(=57.25\)

\(y\)	\(=5-x\)	\(…\ (1)\)
\(y\)	\(=2x-1\)	\(…\ (2)\)

\(2x-1\)	\(=5-x\)
\(3x\)	\(=6\)
\(x\)	\(=2\)

\(4x-5y-2\)	\(=0\)
\(-5y\)	\(=-4x + 2\)
\(y\)	\(=\dfrac{4}{5}x-\dfrac{2}{5}\)

\(x-4y+8\)	\(=0\)
\(x\)	\(=-8\)

\((n-4)\times 7\)	\(=n+8\)
\(7(n-4)\)	\(=n+8\)

\(5\)	\(=\dfrac{D}{75}\)
\(D\)	\(=5\times 75\)
	\(=375\ \text{km}\)