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Algebra, STD2 A4 2022 HSC 22

The formula  `C=100 n+b`  is used to calculate the cost of producing laptops, where `C` is the cost in dollars, `n` is the number of laptops produced and `b` is the fixed cost in dollars.

  1. Find the cost when 1943 laptops are produced and the fixed cost is $20 180.  (1 mark)

  2. Some laptops have some extra features added. The formula to calculate the production cost for these is
  3.      `C=100 n+a n+20\ 180`
  4. where `a` is the additional cost in dollars per laptop produced.
  5. Find the number of laptops produced if the additional cost is $26 per laptop and the total production cost is $97 040.  (2 marks)

Show Answers Only
  1. `$214\ 480`
  2. `610\ text{laptops}`
Show Worked Solution

a.   `text{Find}\ \ C\ \ text{given}\ \ n=1943 and b=20\ 180`

`C` `=100 xx 1943 + 20\ 180`  
  `=$214\ 480`  

 

b.   `text{Find}\ \ n\ \ text{given}\ \ C=97\ 040 and a=26`

`C` `=100 n+a n+20\ 180`  
`97\ 040` `=100n + 26n +20\ 180`  
`126n` `=76\ 860`  
`n` `=(76\ 860)/126`  
  `=610 \ text{laptops}`  

Algebra, STD2 A2 2022 HSC 16

Tom is 25 years old, and likes to keep fit by exercising.

  1. Use this formula to find his maximum heart rate (bpm).
  2.      Maximum heart rate = 220 – age in years
  3. Tom's maximum heart rate is ........................... bpm.   (1 mark)
  4. Tom will get the most benefit from this exercise if his heart rate is between 50% and 85% of his maximum heart rate.
  5. Between what two heart rates should Tom be aiming for to get the most benefit from his exercise?  (2 marks)

Show Answers Only
  1. `text{195 bpm}`
  2. `98-166\ text{bpm}`
Show Worked Solution
a.    `text{Max heart rate}` `=220-25`
    `=195\ text{bpm}`

 

b.   `text{50% max heart rate}\ = 0.5 xx 195 = 97.5\ text{bpm}`

`text{85% max heart rate}\ = 0.85 xx 195 = 165.75\ text{bpm}`

`:.\ text{Tom should aim for between 98 and 166 bpm in exercise.}`

Algebra, STD1 A3 2021 HSC 25

The diagram shows a container which consists of a small cylinder on top of a larger
cylinder.
 


 

The container is filled with water at a constant rate to the top of the smaller cylinder. It takes 5 minutes to fill the larger cylinder.

Draw a possible graph of the water level in the container against time.  (2 marks)
 

Show Answers Only

Show Worked Solution
♦ Mean mark 38%.

Algebra, STD2 A2 2020 HSC 10 MC

A plumber charges a call-out fee of $90 as well as $2 per minute while working.

Suppose the plumber works for `t` hours.

Which equation expresses the amount the plumber charges ($`C`) as a function of time (`t` hours)?

  1.  `C = 2 + 90t`
  2.  `C = 90 + 2t`
  3.  `C = 120 + 90t`
  4.  `C = 90 + 120t`
Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 42%.

`text(Hourly rate)\ = 60 xx 2=$120`

`:. C = 90 + 120t`

`=>D`

Algebra, STD2 A2 2019 HSC 14 MC

Last Saturday, Luke had 165 followers on social media. Rhys had 537 followers. On average, Luke gains another 3 followers per day and Rhys loses 2 followers per day.

If  `x`  represents the number of days since last Saturday and  `y`  represents the number of followers, which pair of equations model this situation?

A.  `text(Luke:)\ \ y = 165x + 3`

 

`text(Rhys:)\ \ y = 537x - 2`
B. `text(Luke:)\ \ y = 165 + 3x`

 

`text(Rhys:)\ \ y = 537 - 2x`
C. `text(Luke:)\ \ y = 3x + 165`

 

`text(Rhys:)\ \ y = 2x - 537`
D. `text(Luke:)\ \ y = 3 + 165x`

 

`text(Rhys:)\ \ y = 2 - 537x`
Show Answers Only

`B`

Show Worked Solution

`text(Luke starts with 165 and adds 3 per day:)`

`y = 165 + 3x`

`text(Rhys starts with 537 and loses 2 per day:)`

`y = 537 – 2x`

`=> B`

Algebra, STD2 A4 EQ-Bank 8 MC

Water was poured into a container at a constant rate. The graph shows the depth of water in the container as it was being filled.
 


 

Which of the following containers could have been used to produce this result?

A. B.
C. D.
Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince the graph is a straight line, the cup fills up at)`

`text(a constant rate.)`
 

`=> B`

Algebra, STD2 A2 SM-Bank 3

The average height, `C`, in centimetres, of a girl between the ages of 6 years and 11 years can be represented by a line with equation

`C = 6A + 79`

where `A` is the age in years. For this line, the gradient is 6.

  1. What does this indicate about the heights of girls aged 6 to 11?   (1 mark)

  2. Give ONE reason why this equation is not suitable for predicting heights of girls older than 12.   (1 mark)

Show Answers Only
  1. `text(It indicates that 6-11 year old girls, on average, grow 6 cm per year.)`
  2. `text(Girls eventually stop growing, and the equation doesn’t factor this in.)`
Show Worked Solution

i.  `text(It indicates that 6-11 year old girls, on average, grow)`

`text(6 cm per year.)`
 

ii. `text(Girls eventually stop growing, and the equation doesn’t)`

`text(factor this in.)`

Algebra, STD2 A2 SM-Bank 2

The weight of a steel beam, `w`, varies directly with its length, `ℓ`.

A 1200 mm steel beam weighs 144 kg.

Calculate the weight of a 750 mm steel beam.  (2 marks)

Show Answers Only

`90\ text(kg)`

Show Worked Solution

`w propto ℓ`

`w = kℓ`

`text(When)\ \ w = 144\ text(kg),\ \ ℓ = 1200\ text(mm)`

`144` `= k xx 1200`
`k` `= 144/1200`
  `= 3/25`

 

`text(When)\ \ ℓ = 750\ text(mm),`

`w` `= 3/25 xx 750`
  `= 90\ text(kg)`

Algebra, STD2 A2 2017 HSC 20 MC

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 100 matches in total are available?

A.     `25`

B.     `24`

C.     `21`

D.     `20`

Show Answers Only

`=>\ text(B)`

Show Worked Solution

`text(1 pentagon: 5 matches)`

`text(2 pentagons: 5 + 4 = 9)`

`text(3 pentagons: 5 + 4 × 2 = 13)`

`vdots`

`n\ text(pentagons:)\ 5 + 4(n – 1)`

`5 + 4(n – 1)` `= 100`
`4n – 4` `= 95`
`4n` `= 99`
`n` `= 24.75`

 

`:.\ text(Complete pentagons possible = 24)`

`=>\ text(B)`

Algebra, STD2 A2 2017 HSC 3 MC

The graph shows the relationship between infant mortality rate (deaths per 1000 live births) and life expectancy at birth (in years) for different countries.
 

What is the life expectancy at birth in a country which has an infant mortality rate of 60?

  1. 68 years
  2. 69 years
  3. 86 years
  4. 88 years
Show Answers Only

\(A\)

Show Worked Solution

\(\text{When infant mortality rate is 60, life expectancy}\)

\(\text{at birth is 68 years (see below).}\)
 

\(\Rightarrow A\)

Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. `text(68 years)`
  2. `text(After 1900, life expectancy increases 0.25 years for each later year someone is born.)`
Show Worked Solution

i.    \(\text{68 years}\)

ii.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

 
\(\text{After 1900, life expectancy increases by 0.25 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

Algebra, STD2 A2 2007 HSC 27b

A clubhouse uses four long-life light globes for five hours every night of the year. The purchase price of each light globe is $6.00 and they each cost  `$d`  per hour to run.

  1. Write an equation for the total cost (`$c`) of purchasing and running these four light globes for one year in terms of  `d`.    (2 marks)

  2. Find the value of  `d`  (correct to three decimal places) if the total cost of running these four light globes for one year is $250.   (1 mark)

  3. If the use of the light globes increases to ten hours per night every night of the year, does the total cost double? Justify your answer with appropriate calculations.   (1 mark)

  4. The manufacturer’s specifications state that the expected life of the light globes is normally distributed with a standard deviation of 170 hours.

     

    What is the mean life, in hours, of these light globes if 97.5% will last up to 5000 hours?   (1 mark)

Show Answers Only
  1. `$c = 24 + 7300d`
  2. `0.031\ $ text(/hr)\ text{(3 d.p.)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(4660 hours.)`
Show Worked Solution

i.  `text(Purchase price) = 4 xx 6 = $24`

`text(Running cost)` `= text(# Hours) xx text(Cost per hour)`
  `= 4 xx 5 xx 365 xx d`
  `= 7300d`
 
`:.\ $c = 24 + 7300d`

 

ii.  `text(Given)\ \ $c = $250`

`250` `= 24 + 7300d`
`7300d` `= 226`
`d` `= 226/7300`
  `= 0.03095…`
  `= 0.031\ $ text(/hr)\ text{(3 d.p.)}`

 

iii.  `text(If)\ d\ text(doubles to 0.062)\ \ $text(/hr)`

`$c` `= 24 + 7300 xx 0.062`
  `= $476.60`
   
`text(S) text(ince $476.60 is less than)\ 2 xx $250\ ($500),`
`text(the total cost increases to less than double)`
`text(the original cost.)`

 

iv.  `sigma = 170`

`z\ text(-score of 5000 hours) = 2`

`z` `= (x – mu)/sigma`
`2` `= (5000 – mu)/170`
`340` `= 5000 – mu`
`mu` `= 4660`

 

`:.\ text(The mean life of these globes is 4660 hours.)`

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth.  An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth.   (2 marks)

Show Answers Only

 `14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)`

`=> W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\  W_text(e) = 84\ text{when}\ W_text(m) = 14`

`14` `= k xx 84`
`k` `= 14/84 = 1/6`

 

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449` `= 1/6  xx W_text(e)`
`W_text(e)` `= 14\ 694\ text(kg)`

 

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Algebra, STD2 A2 2010 HSC 27c

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

2010 27c

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

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

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

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

Show Answers Only
  1. `$3000\ \ \ text{(from graph)}`
  2. `1/3`
  3. `33 1/3\ text(cents per dollar earned)`
  4. `text(Tax payable on)\ I = 1/3 I\-4000`
Show Worked Solution
i.

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

 

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

♦♦ Mean mark 25%
`text(Gradient at)\ A` `= (y_2\-y_1)/(x_2\ -x_1)`
  `= (9000-3000)/(39\ 000 -21\ 000)`
  `= 6000/(18\ 000)`
  `= 1/3\ \ \ \ \ text(… as required)`

 

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

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.
`text(Tax)` ` = 1/3\ text(of each dollar earned)`
  ` = 33 1/3\ text(cents per dollar earned)`

 

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

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

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

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

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

Algebra, STD2 A2 2009 HSC 24d

A factory makes boots and sandals. In any week

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

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

 

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

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

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

     

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

Show Answers Only
  1. `x + y = 200`
  2. `text(S)text(ince the max amount of boots = 120)`

     

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

     

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

     

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

     

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

  3. `text(The profits at)\ C\ text(are $630 more than at)\ B.`
Show Worked Solution

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

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

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

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

 

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

♦ Mean mark 49%

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

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

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

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

 

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

♦ Mean mark 40%.
`=>$P  (text(at)\ B)` `= 24 xx 50 + 15 xx 150`
  `= 1200 + 2250`
  ` = $3450`

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

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

 

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

Algebra, STD2 A2 2009 HSC 13 MC

The volume of water in a tank changes over six months, as shown in the graph.
 

 2UG-2010-13MC

 
Consider the overall decrease in the volume of water.

What is the average percentage decrease in the volume of water per month over this time, to the nearest percent?

  1.  6%
  2. 11%
  3. 32%
  4. 64%
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 48%
COMMENT: Remember that % decrease requires the decrease in volume to be divided by the original volume (50,000L).
`text(Initial Volume)` `= 50\ 000\ text(L)`
`text(Final volume)` `= 18\ 000\ text(L)`
`text(Decrease)` `= 50\ 000-18\ 000`
  `= 32\ 000\ text{L   (over 6 months)}`

 

`text(Loss per month)` `= (32\ 000)/6`
  `= 5333.33…\ text(L per month)`
`text(% loss per month)` `= (5333.33…)/(50\ 000)`
  `=10.666… %`

 
`=>  B`