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Algebra, STD2 A4 2009 HSC 28c (Adapted)

The brightness of a lamp \((L)\) is measured in lumens and varies directly with the square of the voltage \((V)\) applied, which is measured in volts.

When the lamp runs at 7 volts, it produces 735 lumens.

What voltage is required for the lamp to produce 1820 lumens? Give your answer correct to one decimal place.   (3 marks)

Show Answers Only

 `11.2\ \text(volts)`

Show Worked Solution

`L prop V^2\ \ => \ \ L=kV^2`

`text(Find)\ k\ \text{given}\ L = 735\ \text{when}\ V = 7:`

`735` `= k xx 7^2`
`:. k` `= 735/49=15`

 
`text(Find)\ V\ text(when)\ L = 1820:`

`1820` `= 15 xx V^2`
`V^2` `= 1820/15=121.33…`
`V` `= sqrt{121.33} = 11.2\ text(volts)\ \ text{(to 1 d.p.)}`

Algebra, STD2 A4 2021 HSC 24 (Adapted)

A population of Tasmanian devils, `D`, is to be modelled using the function  `D = 650 (0.8)^t`, where `t` is the time in years.

  1. What is the initial population?   (1 mark)

  2. Find the population after 2 years.   (1 mark)

  3. On the axes below, draw the graph of the population against time, in the period  `t = 0`  to  `t = 6`.   (2 marks)
      

Show Answers Only

a.   `650`

b.   `416`

c.   `text{See Worked Solutions}`

Show Worked Solution

a.   `text{Initial population occurs when}\ \  t = 0:`

`D=650(0.8)^0=650 xx 1= 650`
 

b.    `text{Find} \ D \ text{when} \ \ t = 2: `

`D= 650 (0.8)^2= 416`

♦ Mean mark (c) 48%.

 
c.   `\text{At}\ t=6:`

`D=650(0.8)^6=170.39…`
 

Algebra, STD2 A4 2023 HSC 22 (Adapted)

The stopping distance of a motor bike, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{stopping distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The stopping distance for a motor bike travelling at 40 km/h is 16 m.

  1. Find the value of `k`.   (2 marks)

  2. What is the stopping distance when the speed of the motor bike is 80 km/h?   (1 mark)

Show Answers Only

a.    `k=0.01`

b.    `64.0\ text{m}`

Show Worked Solution

a.  `text{stopping distance}\ = k xx text{(speed)}^2`

`16` `=k xx 40^2`  
`k` `=16/40^2=0.001`  

 
b.    `text{Find stopping distance}\ (d)\ text{when speed = 80 km/h:}`

`d=0.01 xx 80^2=64.0\ text{m}`

Algebra, STD2 A4 2013 HSC 22 MC (Adapted)

Jevin wants to build a rectangular chicken pen. He has 32 metres of fencing and will use a barn wall as one side of the pen. The width of the pen is \(d\) metres.
 

Which equation gives the area, \(P\), of the chicken pen?

  1. \(P = 16d-d^2\)
  2. \(P = 32d-d^2\)
  3. \(P = 16d-\dfrac{d^2}{2}\)
  4. \(P = 16d-2d^2\) 
Show Answers Only

\(C\)

Show Worked Solution
♦♦♦ Mean mark 24% (lowest mean of any MC question in 2013 exam)

\(\text{Length of pen}\ = \dfrac{1}{2}(32-d)\)

\(\text{Area}\ =d \times \dfrac{1}{2}(32-d)=16d-\dfrac{d^2}{2}\)

 \(\Rightarrow C\)

Algebra, STD2 A4 2021 HSC 10 MC (Adapted)

Which of the following best represents the graph of  \(y = 5 (0.4)^{x}\) ?
 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By elimination:}\)

♦ Mean mark 41%.

\(\text{When}\  x = 0, \ y = 5 \times (0.4)^0 = 5\)

\(\rightarrow\ \text{Eliminate B and C} \)

\(\text{As}\ \ x \rightarrow \infty, \ y \rightarrow 0 \)

\(\rightarrow\ \text{Eliminate A} \)

\(\Rightarrow D\)

Measurement, STD2 M7 2016 HSC 16 MC (Adapted)

The width (`W`) of a road can be calculated using two similar triangles, as shown in the diagram.
  
  

What is the approximate width of the road?

  1. `12.8\ text(m)`
  2. `13.3\ text(m)`
  3. `14.6\ text(m)`
  4. `17.8\ text(m)`
Show Answers Only

`=> C`

Show Worked Solution

`text{Triangles are similar (equiangular)}`

`text(Using similar ratios:)`

`W/(6.5)` `= 18/8`
`:. W` `= (18 xx 6.5)/8`
  `= 14.62…`

 
`=> C`

Algebra, STD2 A4 2008 HSC 4 MC (Adapted)

Which graph best represents  \(y = \dfrac{2}{x}\) ?
 

Show Answers Only

\(B\)

Show Worked Solution

\(y = \dfrac{2}{x}\ \text{does not touch either axis (inverse graph)}.\)

\(\Rightarrow B\)

Measurement, STD2 M7 2022 HSC 4 MC (Adapted)

A wildlife researcher wanted to estimate the number of turtles in a swamp.

She initially caught and tagged 25 turtles before releasing them.

Two weeks later, she caught 50 turtles and found that 10 of them had tags.

What is the best estimate for the total number of turtles in the swamp?

  1. 100
  2. 110
  3. 120
  4. 125
Show Answers Only

`D`

Show Worked Solution

`text{Let}\ \ T=\ text{population of turtles in swamp}`

`text{Initial tag ratio}\ = 25/T`

`text{Recapture ratio}\ = 10/50`

`25/T` `=10/50`
`10T` `=25 xx 50`
`T` `=1250/10`
  `=125`

`=> D`


♦♦ Mean mark 35%.

Measurement, STD2 M7 2023 HSC 11 MC (Adapted)

A garden has 180 flowers. Some are tulips and the rest are roses. The ratio of tulips to roses is `4:5`.

A gardener picks 20 tulips and 25 roses for a bouquet.

What is the new ratio of tulips to roses in the garden?

  1. `2:3`
  2. `3:4`
  3. `4:5`
  4. `5:6`
Show Answers Only

`C`

Show Worked Solution

`text{Original ratio} = 4:5`

`text{Total parts} = 4 + 5 = 9`

`text{1 part} = 180 ÷ 9 = 20`

`text{Tulips} = 4 × 20 = 80`

`text{Roses} = 5 × 20 = 100`

`text{After picking:}`

`text{Tulips} = 80 − 20 = 60`

`text{Roses} = 100 − 25 = 75`

`text{New ratio} = 60:75`

`= 4:5`

`⇒ C`

Measurement, STD2 M7 2014 HSC 17 MC (Adapted)

A child who weighs 22 kg needs to be given 12 mg of medicine for every 2 kg of body weight. 

Every 10 mL of this medicine contains 150 mg. 

What is the correct dosage for the child? 

  1. 7.0 mL 
  2. 8.8 mL 
  3. 10.2 mL 
  4. 11.5 mL 
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Dosage needed}\ = \dfrac{22}{2} \times 12 = 132\ \text{mg}\) 

\(\text{Since there is 150 mg in 10 mL:} \) 

\(\text{Volume (132 mg)}\ =\dfrac{132}{150} \times 10 = 8.8\ \text{mL}\) 

\(\Rightarrow B\)

Measurement, STD2 M7 2019 HSC 2 MC (Adapted)

Rice is sold in four different sized packets. Which is the best buy?

  1. 250 g for $1.10 
  2. 500 g for $2.00 
  3. 1 kg for $3.80 
  4. 2 kg for $7.40 
Show Answers Only

`D`

Show Worked Solution

\(\text{Price per kilogram}\)

\(\text{250 g:}\  1.10 \times 4 = $4.40 \)

\(\text{500 g:}\ 2.00 \times 2 = $4.00 \)

\(\text{1 kg:}\ $3.80 \)

\(\text{2 kg:}\ 7.40\ ÷\ 2 = $3.70 \)

\(\text{Best buy is 2 kg at \$3.70/kg.}\) 

\(\Rightarrow D\) 

Measurement, STD2 M7 2016 HSC 9 MC (Adapted)

An old air conditioner uses 2.8 kW of electricity per hour. A new air conditioner uses 1.2 kW per hour. How much electricity is saved each year if the air conditioner runs for 5 hours per day for 200 days using the new model?

  1. 1200 kWh
  2. 1600 kWh
  3. 2400 kWh
  4. 2800 kWh
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Electricity used by old air conditioner}\ = 2.8 \times 5 \times 200 = 2800\ \text{kWh}\)

\(\text{Electricity used by new air conditioner}\ = 1.2 \times 5 \times 200 = 1200\ \text{kWh}\)

\(\therefore\ \text{Electricity saved} = 2800-1200 = 1600\ \text{kWh}\)

\(\Rightarrow B\)

Measurement, STD2 M7 2016 HSC 9 MC (Adapted)

An old dishwasher uses 45 L of water per cycle. A new dishwasher uses 15 L per cycle. How much water is saved each year if three cycles are run each week using the new dishwasher?

  1. 2340 L
  2. 3640 L
  3. 4680 L
  4. 7020 L
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Water used by old dishwasher}\ = 45 \times 3 \times 52 = 7020\ \text{L}\)

\(\text{Water used by new dishwasher}\ = 15 \times 3 \times 52 = 2340\ \text{L}\)

\(\text{Water saved}\ = 7020-2340 = 4680\ \text{L}\)

\(\Rightarrow C\)

Measurement, STD2 M1 2008 HSC 23b (Adapted)

The mass of a food packet is measured as 620 grams, correct to the nearest 20 grams.

What is the percentage error for this measurement?   (1 mark)

Show Answers Only

`text(1.6%)`

Show Worked Solution

`text{Absolute error} = 1/2 xx 20 = 10\ \text{g}`

`:.\ \text{% error}` `= 10 / 620 xx 100`
  `= \text{1.6129…} ≈ 1.6%`

Measurement, STD2 M1 2012 HSC 26g (Adapted)

Deirdre purchases a weed killer product that contains 12 litres of solution.

It is used to spray her lawn twice a week. The instructions state:

  • Use 400 mL per 10 m² of lawn on 1st application of the week
  • Use 300 mL per 10 m² of lawn on 2nd application of the week

If Deirdre's lawn is 30 m² in area, how many full weeks will the weed killer last?   (2 marks)

Show Answers Only

`text(8 full weeks.)`

Show Worked Solution

`text{Using  100 mL = 0.1 L:}`

`text{Total weekly usage} = 3 xx(0.4+0.3) = 2.1\ text{litres}`

`text{Total solution} = 18\ text{L}`

`text{Weeks it will last} = 18/2.1=8.57…`

`:.\ text{The bottle will last 8 full weeks.}`

Measurement, STD2 M1 2023 HSC 9 MC (Adapted)

The length and width of a rectangular pool are measured to be 13 m and 7.5 m respectively, correct to the nearest metre and nearest 0.1 metre.

What are the lower and upper bounds for the area of the pool?

  1. `text{94.25 m}^2\ text{and 101.25 m}^2`
  2. `text{93.125 m}^2\ text{and 101.925 m}^2`
  3. `text{92.5 m}^2\ text{and 102.5 m}^2`
  4. `text{93.75 m}^2\ text{and 100.75 m}^2`
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 35%.

`text{Length absolute error} = 1/2 xx 1 = 0.5\ \text{m}`

`text{Width absolute error} = 1/2 xx 0.1 = 0.05\ \text{m}`

`text{Length bounds: } 13 ± 0.5 = [12.5,\ 13.5]\ \text{m}`

`text{Width bounds: } 7.5 ± 0.05 = [7.45,\ 7.55]\ \text{m}`

`text{Lower bound area} = 12.5 xx 7.45 = 93.125\ \text{m}^2`

`text{Upper bound area} = 13.5 xx 7.55 = 101.925\ \text{m}^2`

`⇒ B`

Measurement, STD2 M1 2009 HSC 12 MC (Adapted)

How many square millimetres are in 0.004 square metres?

  1. 4
  2. 40
  3. 4000
  4. 40 000
Show Answers Only

`D`

Show Worked Solution
`text{Since 1 m}^2` `= 1000\ text{mm} xx 1000\ text{mm}`
  `= 1\ 000\ 000\ \text{mm}^2`

 
`:. 0.004\ \text{m}^2= 0.004 xx 1\ 000\ 000= 4000\ \text{mm}^2`

`⇒ D`

Measurement, STD2 M1 2020 HSC 5 MC (Adapted)

A pencil is measured to be 12.8 cm long, correct to one decimal place.

What is the percentage error in this measurement?

  1. 0.20%
  2. 0.39%
  3. 0.78%
  4. 1.56%
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 44%.

`text{Absolute error} = 1/2 xx \text{precision} = 1/2 xx 0.1 = 0.05\ \text{cm}`

`%  \text{error}` `= \frac{0.05}{12.8} xx 100`
  `= 0.390625%`
  `= 0.39%`

`⇒ B`

Measurement, STD2 M1 2020 HSC 2 MC (Adapted)

What is 0.0004782 expressed in standard form with two significant figures?

  1. `4.8 xx 10^(-4)`
  2. `4.78 xx 10^(-4)`
  3. `4.8 xx 10^(-5)`
  4. `4.78 xx 10^(-5)`
Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 39%.
COMMENT: Students often round incorrectly when converting to standard form.
`0.0004782` `= 4.782 xx 10^(-4)`
  `= 4.8 xx 10^(-4)\ \ \ text{(to 2 sig fig)}`

`⇒ A`

Measurement, STD2 M1 2019 HSC 8 MC (Adapted)

A newborn baby’s length is recorded as 52.4 cm.

What is the absolute error of this measurement?

  1. 0.25 cm
  2. 0.5 cm
  3. 1 cm
  4. 2 cm
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 48%.

`text{Absolute error}` `= 1/2 xx\ text{precision}`
  `= 1/2 xx 1\ text{ cm}`
  `= 0.5\ text{ cm}`

`⇒ B`

Measurement, STD2 M1 2015 HSC 12 MC (Adapted)

A sprinter’s reaction time at the start of a race was recorded as 0.25 seconds, correct to the nearest hundredth of a second.

What is the percentage error in this measurement, correct to one significant figure?

  1. 0.5%
  2. 1%
  3. 2%
  4. 3%
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 43%.

`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 0.01 = 0.005\ text{s}`

`text{% error}` `=\ frac{text{absolute error}}{text{measurement}} xx 100%`  
  `=0.005/0.25 xx 100%`  
  `=2%`  

 
`⇒ C`

Measurement, STD2 M1 2014 HSC 10 MC (Adapted)

The height of Mount Kosciuszko is measured to be 2228.1 m above sea level.

What is the percentage error in this measurement?

  1. 0.001%
  2. 0.002%
  3. 0.005%
  4. 0.011%
Show Answers Only

`B`

Show Worked Solution
 
♦ Mean mark 50%

`text{Absolute error}\ = 1/2 xx text{precision}\ = 1/2 xx 0.1 = 0.05\ text{m}`

`text{% error}` `= \frac{text{absolute error}}{text{measurement}} xx 100%`  
  `= 0.05 / 2228.1 xx 100%`  
  `= 0.0022%`  

 
`⇒  B`

Measurement, STD2 M1 2016 HSC 1 MC (Adapted)

What is  0.04967  correct to two significant figures?

  1. 0.049
  2. 0.050
  3. 0.0496
  4. 0.0497
Show Answers Only

`B`

Show Worked Solution

`text(We are rounding 0.04967 to 2 significant figures.)`

• `text(First 2 significant digits: 4 and 9)`

• `text(Next digit is 6 → round **up**)`

• `text(0.04967 rounds to 0.050 (2 sig. fig.))`

`=> B`

Measurement, STD2 M1 2015 HSC 1 MC (Adapted)

The distance from Earth to the Moon is approximately 384 400 km.

What is this distance in standard form correct to two significant figures?

  1. `3.84 × 10^5\ \text{km}`
  2. `3.8 × 10^5\ \text{km}`
  3. `3.9 × 10^5\ \text{km}`
  4. `3.84 × 10^6\ \text{km}`
Show Answers Only

`C`

Show Worked Solution

`384\ 400`

`= 3.844 × 10^5`

`\text(Rounded to 2 significant figures)  →  3.9 × 10^5\ \text{km}`

 
`⇒ C`

Measurement, STD2 M1 2004 HSC 13 MC (Adapted)

During a study break, Sam is walking while Alex is sitting.
 

Sam breathes in air at a rate of 30 litres per minute while walking. Alex breathes in air at a rate of 8 litres per minute while sitting.

During a 20-minute break, how much more air would Sam breathe than Alex?

  1. 220 Litres
  2. 440 Litres
  3. 460 Litres
  4. 600 Litres
Show Answers Only

`=>\ B`

Show Worked Solution
`text{Sam’s total}` `= 30 × 20 = 600\ \text{L}`
`text{Alex’s total}` `= 8 × 20 = 160\ \text{L}`
`text{Difference}` `= 600 − 160 = 440\ \text{L}`

`=>\ B`

Measurement, STD2 M1 SM-Bank 25 MC (Adapted)

In a food technology class, a slice of bread is measured and its thickness is recorded as 1.6 cm.

What is the upper limit of accuracy of this measurement?

  1. 1.61 cm
  2. 1.64 cm
  3. 1.65 cm
  4. 1.7 cm
Show Answers Only

`=>\ C`

Show Worked Solution

`text{Absolute error} = 0.05\ \text{cm}`

`text{Upper limit}` `= 1.6 + 0.05`
  `= 1.65\ \text{cm}`

`=>\ C`

Measurement, STD2 M1 2022 HSC 34 (Adapted)

A composite solid is shown. The top section is a hemisphere with a diameter of 6 cm. The bottom section is a cylinder with a height of 3 cm and a diameter of 4 cm
 

Find the total volume of the composite solid in cm³, correct to 1 decimal place.   (4 marks)

Show Answers Only

`94.2 \ text{cm}^3`

Show Worked Solution
`text{Volume of Hemisphere}` `=2/3 pi r^3`  
  `=2/3 pi xx 3^3`  
  `=56.54\ text{cm}^3`  

 

`text{Volume of Cylinder}` `=pi r^2 h`  
  `=pi (2^2) xx 3`  
  `=37.69\ text{cm}^3`  

 

`text{Total Volume}` `=56.54+37.69`
  `=94.23`
  `=94.2 \ text{cm}^3`

Measurement, STD2 M1 2014 HSC 27c (Adapted)

A swimming pool is in the shape of a rectangle with a semicircle at each end, as shown.

The pool is 7000 mm long, 4000 mm wide, and has a depth of 2100 mm.  
  

How much water is needed to fill the pool, to the nearest litre?   (4 marks) 

Show Answers Only

`51 \ 589 \ text(L)`

Show Worked Solution

`V = Ah` 

♦ Mean mark 41%
STRATEGY: Adjusting measurements to metres makes the final conversion to litres simple.

`text(Finding Area of base)`

`text(Semi-circles have radius 2000 mm) = 2 \ text(m)`

`:.\ text(Area of 2 semicircles)`

`=2 xx 1/2 xx pi r^2`

`= pi xx 2^2`

`= 12.56 \ text(m)^2`
 

`text(Area of rectangle)`

`= l xx b`

`= (7-2 xx 2) xx 4`

`= 12\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= (12.56… + 12) xx 2.1`
  `= 51.589…\ text(m)^3`
  `= 51 \ 589 \ text(L)\ \ text{(using 1m³} = 1000\ text{L)}`
   

Measurement, STD2 M1 2013 HSC 12 MC (Adapted)

A hemisphere sits perfectly on top of a cylinder to form a solid. 

What is the volume of the solid?

  1. 1750 cm³
  2. 1950 cm³
  3. 2150 cm³
  4. 2350 cm³
Show Answers Only

`C`

Show Worked Solution
`text(Volume )` `=text{Vol (cylinder)} +text{Vol (hemisphere)}`
  `= pi r^2h+2/3pi r^3`
  `= pi xx 6^2 xx 15 + 2/3pi xx 6^3`
  `=2149.84\ text(cm)^3`

 
`=>\ C`

Measurement, STD2 M1 2010 HSC 17 MC (Adapted)

During a heavy storm, 2.4 hectares of farmland received rainfall to a depth of 12 cm.

How many kilolitres of rainwater fell on the farmland? (1 hectare = 10 000 m²)

  1. 2.88 kL
  2. 2880 kL
  3. 288 000 kL
  4. 2 880 000 kL
Show Answers Only

`B`

Show Worked Solution
♦♦ Mean mark 36%
NOTE: The unit conversion 1 m³ = 1000 L is contained in the Formulae and Data sheet given out in the exam.
`text(Area of farmland)` `=2.4xx10\ 000`
  `=24\ 000\ text(m²)`
`text(Volume)` `=Ah`
  `=24\ 000xx0.12`
  `=2880\ text(m³)`

 

`text{1 m³}` `=1000\ text(L)=1\ text(kL)`
`:.\ text(Volume)` `=2880\ text(kL)`

`=>B`

Measurement, STD2 M1 2017 HSC 22 MC (Adapted)

A concrete storm drain is constructed in the shape of a rectangular prism with a cylindrical tunnel running through its center. The dimensions are shown in the diagram.
 
 

What is the approximate volume of concrete needed to construct the storm drain?

  1. `text(4.5 m)³`
  2. `text(4.8 m)³`
  3. `text(5.1 m)³`
  4. `text(5.4 m)³`
Show Answers Only

`A`

Show Worked Solution
`text(Volume of concrete)` `= text(Volume of rectangular prism)-text(Volume of cylinder)`
  `= l xx w xx h-pi r^2 h`
  `= 2.5 xx 1.8 xx 1.2-pi xx (0.4)^2 xx 1.8`
  `= 5.4-pi xx 0.16 xx 1.8`
  `= 5.4-0.9047…`
  `= 4.4952…`
  `= 4.5\ text(m)³`

`=>A`

Measurement, STD2 M1 2017 HSC 18 MC (Adapted)

A trapezoidal prism is illustrated below, with the following dimensions:
 

What is the volume of the trapezoidal prism?

  1. `24.38\ text(m)^3`
  2. `32.75\ text(m)^3`
  3. `39.38\ text(m)^3`
  4. `47.25\ text(m)^3`
Show Answers Only

`C`

Show Worked Solution
`text(Area of trapezoid)` `= 1/2h (a + b)`
  `= 1/2 xx 3 xx (6 + 4.5)`
  `= 15.75\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= 15.75 xx 2.5`
  `= 39.38\ text(m)^3`

 
`=>C`

Measurement, STD2 M1 2009 HSC 19 MC (Adapted)

A tennis ball canister holds 3 balls inside, as illustrated below.

The diameter of each ball is 6.5 cm.

What is the volume of the tennis ball canister, to the nearest cubic centimetre?

  1. `527\ text(cm)^3`
  2. `597\ text(cm)^3`
  3. `637\ text(cm)^3`
  4. `647\ text(cm)^3`
Show Answers Only

`D`

Show Worked Solution

`text(S)text(ince diameter sphere = 12 cm) `

`=>\ text(Radius of cylinder = 6 cm)`

`text(Height of cylinder)` `= 3 xx text(diameter of sphere)`
  `= 3 xx 6.5`
  `= 19.5\ text(cm)`
   
`:.\ text(Volume cylinder)` `= pi r^2 h`
  `= pi xx 3.25^2 xx 19.5`
  `= 647.07…\ text(cm³)`

 
`=>  D`

Measurement, STD2 M1 2018 HSC 30a (Adapted)

A cylindrical oil tank has a height of 7 metres and a capacity of 1.5 megalitres.
 

What is the diameter of the oil tank? Give your answer in metres, correct to two decimal places.  (3 marks)

Show Answers Only

`16.50\ text{m}`

Show Worked Solution

`text{Converting megalitres to m³  (using 1 m³ = 1000 L):}`

♦ Mean mark 48%.

`1.5\ text(ML)` `= (1.5 xx 10^6)/(10^3)`
  `= 1.5 xx 10^3\ text(m)^3`
  `= 1500\ text(m)^3`

 

`V` `= pir^2h`
`1500` `= pi xx r^2 xx 7`
`r^2` `= 1500/(pi xx 7)`
`sqrt(r^2)` `= sqrt(68.21)`
  `= 8.26\ text{m}`

 

`text{Diameter}` `=2r`  
  `=2xx8.26 \ text{m}`  
  `=16.52\ text{m}`  

 

Measurement, STD2 M1 2009 HSC 11 MC (Adapted)

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 68 m²
  2. 73 m²
  3. 95 m²
  4. 193 m²
Show Answers Only

`C`

Show Worked Solution
`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx bh)`
  `= (90/360 xx pi xx 12^2)-(1/2 xx 6 xx 6)`
  `= 113.097…-18`
  `= 95.097…\ text(m²)`

`=> C`

Measurement, STD2 M1 2016 HSC 30a (Adapted)

The area of a school roof is 45 m². All rain that falls on the roof flows into a storage tank.

How many litres of water are collected in the tank when 25 mm of rain falls?   (2 marks)

Show Answers Only

`1125\ \text{L}`

Show Worked Solution
`text{Volume}` `= A × h`
  `= 45 × (25 ÷ 1000)`
  `= 1.125\ \text{m}^3`
  `= 1125\ \text{L}`

Measurement, STD2 M7 2023 HSC 26 (Adapted)

Jo is constructing a concrete border around a rectangular vegetable patch, as shown. The border is 0.6 m wide.
 

  1. Find the area of the border.   (2 marks)
  2. Jo is preparing concrete using a mix of gravel, sand, and cement in the ratio 5 : 3 : 2 by weight.
    Jo needs 2 tonnes of concrete in the correct ratio.
    Calculate how many 20 kg bags of cement Jo needs to buy.   (3 marks)

Show Answers Only
  1. `11.4\ \text{m}^2`
  2. `20\ \text{bags}`
Show Worked Solution

a.    `text{Area of outer rectangle} = 6.5 × 4.2 = 27.3\ \text{m}^2`

`text{Area of garden} = 5.3 × 3.0 = 15.9 \ text{m}^2`

`text{Area of border}` `= 27.3-15.9`  
  `=11.4\ text{m}^2`  

 

b.   `text{Ratio parts:} \ 5 + 3 + 2 = 10` parts

`text{Total concrete} = 2\ \text{tonnes} = 2000\ \text{kg}`

`text{Each part} = 2000 ÷ 10 = 200\ \text{kg}`

`text{Cement} = 2 \ text{parts} = 2 × 200 = 400\ \text{kg}`

`text{Bags of cement} = 400 ÷ 20 = 20`

`⇒ \ text{Jo needs 20 bags of cement.}`

Measurement, STD2 M1 2021 HSC 16 (Adapted)

The surface area, `A`, of a sphere is given by the formula

`A = 4 pi r^2,`

where `r` is the radius of the sphere.

A satellite dish resembles the inner surface of the lower half of a sphere with a radius of 1.5 meters.

 

Find the surface area of the satellite dish in square metres, correct to one decimal place.   (2 marks)

Show Answers Only

`14.1\ text{m}^2`

Show Worked Solution
`A` `= frac{1}{2} times 4 pi r^2`
  `= 2 pi r^2`
  `= 2 pi times (1.5)^2`
  `= 2 pi times 2.25`
  `= 14.137…`
  `= 14.1\ text{m}^2\ \text{(1 d.p.)}`

Measurement, STD2 M1 2019 HSC 16 (Adapted)

A decorative light fixture is in the shape of a hollow hemisphere with a diameter of 24 cm.
 

The inside of the fixture is to be coated with reflective paint.

What is the area to be painted on the inside surface? Give your answer correct to the nearest square centimetre.   (2 marks)

Show Answers Only

`905\ \text{cm}^2`

Show Worked Solution
`A` `= 2 pi r^2`
  `= 2 × pi × 12^2`
  `= 2 × pi × 144 = 905.0…`
  `≈ 905\ \text{cm}^2`

Measurement, STD1 M1 2019 HSC 25 (Adapted)

The diagram illustrates a sector formed by a central angle of 105°, taken from a circle with a radius of 15 metres.

What is the perimeter of the sector? Write your answer correct to 1 decimal place.   (3 marks)

Show Answers Only

`57.5\ \ (text(1 d. p.))`

Show Worked Solution
`text(Arc length)` `= 105/360 xx 2 xx pi xx 15`
  `= 27.49`

 

`:.\ text(Perimeter)` `= 27.49 + 2 xx 15`
  `= 57.49`
  `= 57.5\ \ (text(1 d. p.))`

Measurement, STD2 M1 2016 HSC 30c (Adapted)

A landscape artist was commissioned to design a garden consisting of part of a circle, with centre `O`, and a rectangle, as shown in the diagram. The radius `OC` of the circle is 20 m, the width `BC` of the rectangle is 10 m, and `DOC` is 100°.
 

What is the area of the whole garden, correct to the nearest square metre?   (5 marks)

Show Answers Only

`6281\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ \triangle ODC,`

`sin50^@` `= (ED)/20`
`ED` `= 20 \times \sin50^@`
  `= 34.472`
`:. DC` `= 2 \times 34.472 = 68.944\ \text{m}`

 

`cos50^@` `= (OE)/20`
`:. OE` `= 20 \times \cos50^@ = 28.925`

 

`text(Area of)\ \triangle ODC`

`= \frac{1}{2} \times 68.944 \times 28.925 = 997.12\ \text{m}^2`

 

`text(Area of rectangle ABCD)` `= 10 \times 68.944 = 689.44\ \text{m}^2`

 

`text(Area of major sector DOAC)`

`= \pi \times 20^2 \times \frac{260}{360} = 4594.58\ \text{m}^2`

 

`:.\ \text{Area of garden}`

`= 997.12 + 689.44 + 4594.58 = 6281.14`

`= 6281\ \text{m² (nearest m²)}`

Measurement, STD2 M1 2018 HSC 27c (Adapted)

A farmer is designing a chicken coop with a roof shaped like half a cylinder, open at both ends. The structure has a diameter of 4 metres and a length of 12 metres.
 

 
The curved roof is to be made of aluminum sheets.

What area of aluminum sheets is required, to the nearest m²?   (2 marks)

Show Answers Only

`75\ text(m²  (nearest m²))`

Show Worked Solution

`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 4`
  `= 6.283…`

 

`:.\ text(Sheeting required)` `= 12 xx 6.283…`
  `= 75.39…`
  `= 75\ text(m²  (nearest m²))`

Measurement, STD2 M1 2015 HSC 28a (Adapted)

The diagram shows a circular garden bed with a circular path around it.
 

The radius of the entire structure (garden + path) is 6 m, and the radius of the inner garden is 4 m.

Calculate the area of the path.   (1 mark)

Show Answers Only

`≈ 62.83…\ text(m²)`

Show Worked Solution

`text(Area of path)`

`= pi(R^2 − r^2)`

`= pi(6^2 − 4^2)`

`= pi(36 − 16)`

`= 20pi\ \text(m²)`

`≈ 62.83…\ \text(m²)`

Measurement, STD2 M1 2009 HSC 23c (Adapted)

The diagram shows the shape and dimensions of a floor which is to be tiled.
 

  1. Find the area of the floor.   (2 marks)

  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $60. When buying the tiles, 10% more tiles are needed, due to cutting and wastage.

     

    Find the total cost of the boxes of tiles required for the floor.   (2 marks)

Show Answers Only

a.    `17\ text(m²)`

b.   `$1140`

Show Worked Solution
a.    
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (3 + 2) xx (3 + 2)\-2 xx (2 xx 2)`
  `= 25\-8`
  `= 17\ text(m²)`

 

b.     `text(Tiles required)` `= (17 +10 text{%}) xx 17`
    `= 18.7\ text(m²)`

 

 `=>\ text(19 boxes are needed)`

`:.\ text(Total cost of boxes)` `=19 xx $60`
  `= $1140`

Measurement, STD2 M1 2020 HSC 27 (Adapted)

The shaded region on the diagram represents a lot. Each grid represents 6 m × 6 m.
 


 

  1. Use two applications of the trapezoidal rule to calculate the approximate area of the lot.   (3 marks)

  2. Should the answer to part (a) be more than, equal to or less than the actual area of the lot? Referring to the diagram above, briefly explain your answer.   (2 marks)

Show Answers Only
  1. `1080 \ text{m}^2`
  2. `text{The estimate will be more than actual area}`
Show Worked Solution

a.     

`h = 4 xx 6 = 24 \ text{m}`

♦ Mean mark 47%.
`text{Area}` `= frac{h}{2} (x_1 + x_2) + frac{h}{2} (x_2 + x_3)`
   `= frac{24}{2} (30 + 18) + frac{24}{2} (18 + 24}`
  `= 576 + 504`
  `= 1080 \ text{m}^2`

 

♦♦ Mean mark 23%.

b.     

`text{The trapezoidal rule captures the shaded area plus the}`

`text{the extra area highlighted above.}`

`therefore \ text{The estimate will be more than actual area}`

Measurement, STD2 M1 2023 HSC 24 (Adapted)

The diagram shows the cross-section of a wall across a creek. 
 


 
  1. Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall.   (2 marks)

  2. The wall has a uniform thickness of 0.9 m. The weight of 1 m³ of concrete is 3.52 tonnes.  
  3. How many tonnes of concrete are in the wall? Give the answer to two significant figures.   (3 marks)

Show Answers Only
  1. `21.25\ text{m}^2`
  2. `text{67 tonnes}`
Show Worked Solution

a.    `h=10.0/2=5`

`A` `~~h/2[2+1.5+2(2.5)]`  
  `~~5/2(8.5)`  
  `~~21.25\ text{m}^2`  

 
b.   `V_text{wall}=21.25 xx 0.9=19.13\ text{m}^3`

`text{Mass of concrete}` `=19.13 xx 3.52`  
  `=67.33`  
  `=67\ text{tonnes (2 sig.fig.)}`  
♦ Mean mark (b) 46%.
 

Measurement, STD2 M1 2021 HSC 12 MC (Adapted)

A block of land is represented by the shaded region on the number plane. All measurements are in kilometres. 
 

What is the approximate area of the land, in square kilometres, using two applications of the trapezoidal rule?

  1. 11.25
  2. 14.85
  3. 16.75
  4. 22.55
Show Answers Only

`C`

Show Worked Solution

Using two applications of the trapezoidal rule:

`\text{Area}` `≈ \frac{5}{2} (1.2 + 2 × 2 + 1.5)`
  `= 2.5 (1.2 + 4 + 1.5)`
  `= 2.5 × 6.7`
  `= 16.75 \ \text{km}^2`

`⇒ C`

Measurement, STD2 M1 2013 HSC 15a* (Adapted)

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.4 m apart. The height of each outer pole is 1.6 m, and the height of the middle pole is 2 m. The roof hangs between the poles.

The front of the tent has area `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate `A`.   (2 marks)

  2. Explain whether the trapezoidal rule give a greater or smaller estimate of  `A`?   (1 mark)

Show Answers Only

a.    `5.04\ text(m²)`

b.    `text(The trapezoidal rule assumes a straight line between)`

`text(all points and therefore would estimate a greater)`

`text(area than the actual area of the tent front.)`

Show Worked Solution
a.     `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.4/2 [1.6 + (2 xx 2) + 1.6]`
    `~~ 0.7 [7.2]`
    `~~ 5.04\ text(m²)`

 

b.    `text(The trapezoidal rule assumes a straight line between)`

`text(all points and therefore would estimate a greater)`

`text(area than the actual area of the tent front.)`

Measurement, STD2 M1 2015 HSC 28c* (Adapted)

Three equally spaced cross-sectional areas of a vase are shown.
 

 
Use the Trapezoidal rule to find the approximate capacity of the vase in litres.   (3 marks)

Show Answers Only

`3\ text(litres)`

Show Worked Solution

`text(Solution 1)`

`V` `≈ 15/2(35 + 170) + 15/2(170 + 25)`
  `≈ 15/2(205 + 195)`
  `≈ 3000\ text{mL   (1 cm³ = 1 mL)}`
  `~~3\ text(L)`

`text(Solution 2)`

`V` `≈ 15/2(35 + 2 xx 170 + 25)`
  `≈ 15/2(400)`
  `≈ 3000\ text{mL}`
  `~~3 \ text(L)`

Measurement, STD2 M1 2015 HSC 30a (Adapted)

A school projector left in idle mode consumes 95 watts of power. The cost of electricity is 28 cents per kWh.

There are 12 projectors in the school, each idle for 6 hours a day, 5 days a week.

How much does it cost the school to leave all the projectors idle for a 10-week school term?   (2 marks)

Show Answers Only

`$95.76`

Show Worked Solution

`text(95 watts per projector)`

`text(Hours per term:) \ 6 xx 5 xx 10 = 300`

`text(Total watt-hours)` `=95 xx 12 xx 300`
  `= 342\ 000`
  `= 342\ text(kWh)`

 

`:.\ \text{Cost}` `=342 xx $0.28`
  `= \ $95.76`

Measurement, STD2 M1 2013 HSC 26d (Adapted)

A section of Jim’s electricity bill is shown.

  1. What is the value of `X`?   (1 mark)
  2. How much will Jim save if he uses 523.5 kWh of energy at the Off-peak rate rather than at the Shoulder rate?   (2 marks)
Show Answers Only

a.    `531.2`

b.    `$51.30`

Show Worked Solution
a.     `X` `= \text{Last Reading} + \text{Energy Used}`
    `= 274.8+256.4`
    `= 531.2`

 

b.     `text(Cost)_{text(Shoulder)}` `= 523.5 × 19.4 = 10155.9 \text{ cents}`
    `= \$101.56`
  `text(Cost)_{text(Offpeak)}` `= 523.5 × 9.6 = 5025.6 \text{ cents}`
    `= \$50.26`
  `text(Saving)` `= 101.56-50.26 = \$51.30`

Measurement, STD2 M7 2021 HSC 27 (Adapted)

The price and the power consumption of two different models of air purifiers are shown.

\[ \begin{array}{|l|l|} \hline \text{Air Purifier X} & \text{Air Purifier Y} \\ \hline \text{Price: \$480} & \text{Price: \$462.40} \\ \hline \text{Power: 95 W} & \text{Power: 88 W} \\ \hline \end{array} \]

The average cost for electricity is 30c/kWh. A household runs an air purifier for an average of 10 hours a day.

  1. The annual cost of electricity for Air Purifier X for this household is \$104.03.
  2. For this household, what is the difference in the annual cost of electricity between Air Purifier X and Air Purifier Y?   (2 marks)
  3. For this household, how many years will it take for the total cost of buying and using Air Purifier X to be equal to the cost of buying and using Air Purifier Y?   (2 marks)
Show Answers Only
  1. `$7.66`
  2. `2.3\ \text{years}`
Show Worked Solution
a. `text{Annual power usage (Y)}` `= 88 \times 10 \times 365 = 321200\ \text{Wh}`
    `= 321.2\ \text{kWh}`
  `text{Annual cost (Y)}` `= 321.2 \times 0.30 = \$96.36`
  `text{Annual cost (X)}` `= 95 \times 10 \times 365 / 1000 \times 0.30 = \$104.03`
  `text{Difference}` `= 104.03-96.36 = \$7.67`
b. `text{Price difference}` `= 480-462.40 = \$17.60`
  `text{Years to equal total cost}` `= 17.60 / 7.67 ≈ 2.3\ \text{years}`

Measurement, STD2 M7 2018 HSC 28c (Adapted)

A 900-watt air purifier runs for 2 hours per day at 60% power. Electricity costs $0.30 per kWh.

What is the total cost of running the air purifier for 150 days?   (3 marks)

Show Answers Only

`$48.60`

Show Worked Solution
`text(Daily usage)` `= 900 xx 2 xx 60text(%)`
  `= 1080\ \text{Wh}`

 

`text(Total usage over 150 days)` `= 150 xx 1080`
  `=162 \ 000\ \text{Wh}`
  `= 162\ \text{kWh}`

 

`∴ \  text(Cost)` `= 162 xx 0.30`
  `= $48.60`

Measurement, STD2 M1 2019 HSC 24 (Adapted)

Ben uses 45 kilocalories of energy per kilometre when cycling.

He drinks a smoothie that contains 1260 kilojoules of energy. How many kilometres will he need to cycle to use up all the energy from the smoothie? Give your answer correct to one decimal place. (1 kilocalorie = 4.184 kilojoules)   (2 marks)

Show Answers Only

`6.7\ \text{km (1 d.p.)}`

Show Worked Solution
`text(Kilocalories in smoothie)` `= 1260 / 4.184`
  `= 301.252…`

 
`:.\ \text(Kilometres required to cycle)`

`= 301.252 / 45`

`= 6.694…`

`= 6.7\ \text{km (1 d.p.)}`

Measurement, STD2 M1 2014 HSC 20 MC (Adapted)

In a household of 5, each member uses an average of 8 minutes of hot water per day.

The household uses a 7.5 kW hot water unit.

Electricity is charged at 30.2 c/kWh when the hot water unit is being used.

What is the electricity cost for the hot water used by this household in one week?

  1. $7.92
  2. $9.84
  3. $10.57
  4. $11.32
Show Answers Only

`C`

Show Worked Solution

`text(Usage per day) = 5 xx 8 = 40\ text(mins)`

`text(Usage per week) = 7 xx 40 = 280\ text(mins)`

`text(Convert minutes to hours) = 280/60 = 4.67\ \text{hours}`

`text(Energy used in kWh) = 4.67 xx 7.5 = 35.0\ \text{kWh}`

`text(Cost in cents) = 35.0 xx 30.2 = 1057`

`= 1057\text(¢)`

`= $10.57`

`=> C`