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Measurement, STD1 M4 2025 HSC 18

Ramon took 1.25 hours to travel the length of a freeway. The length of the freeway is 100 km.

  1. Ramon’s total travel time was made up of:
    • 20 minutes to travel to the start of the freeway
    • the time taken to travel the length of the freeway
    • 35 minutes after exiting the freeway to get to his destination.
  1. What was the total time Ramon spent travelling? Give your answer in hours and minutes.   (2 marks)

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  1. What was Ramon’s average speed on the freeway?   (1 mark)

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Show Answers Only

a.    \(2\ \text{hours}\ 10\ \text{minutes}\)

b.    \(80\ \text{km/h}\)

Show Worked Solution
a.     \(\text{Total time}\) \(=20+75+35\)
    \(=130\ \text{minutes}\)
    \(=2\text{ h }10\text{ min}\)

 

b.     \(\text{Average speed}\) \(=\dfrac{\text{distance}}{\text{time}}\)
    \(=\dfrac{100}{1.25}\)
    \(=80\ \text{km/h}\)

Filed Under: M4 Rates (Y12) Tagged With: Band 4, Band 5, smc-1104-15-General rate problems

Measurement, STD1 M4 2025 HSC 17

Paint is sold in two sizes at a local shop.

  • Four-litre cans at $90 per can
  • Ten-litre cans at $205 per can

Mina needs to purchase 80 litres of paint.

Calculate the amount of money Mina will save by purchasing only ten-litre cans of paint rather than only four-litre cans of paint.    (2 marks)

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\(\text{Saving}=$160\)

Show Worked Solution

\(\text{4 litre cans needed}\ =\dfrac{80}{4}=20\)

\(\text{Cost using 4 litre cans}\ =20\times $90=$1800\)

\(\text{10 litre cans needed}\ =\dfrac{80}{10}=8\)

\(\text{Cost using 10 litre cans}\ =8\times $205=$1640\)
 

\(\therefore\ \text{Saving}\ =$1800-$1640=$160\)

Filed Under: M4 Rates (Y12) Tagged With: Band 4, smc-1104-15-General rate problems, std2-std1-common

Measurement, STD1 M4 2025 HSC 12

Vijay’s heart rate before and after his morning run is shown.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{Before run} \rule[-1ex]{0pt}{0pt} & \text{72 beats per minute} \\
\hline
\rule{0pt}{2.5ex} \text{After run} \rule[-1ex]{0pt}{0pt} & \text{126 beats per minute} \\
\hline
\end{array}

What is the percentage increase in Vijay’s heart rate?    (2 marks)

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\(75\%\)

Show Worked Solution
\(\text{% increase}\) \(=\Big[\dfrac{\text{Change in HR}}{\text{Original HR}}\Big]\times 100\%\)
  \(=\Big[\dfrac{126-72}{72}\Big]\times 100\%\)
  \(=75\%\)

Filed Under: M4 Rates (Y12) Tagged With: Band 4, smc-1104-15-General rate problems

v1 Measurement, STD2 M7 2018 HSC 27a

Alex used a cloud storage service for one month.

The plan has a base monthly cost of $25. The service also charges 45 cents per GB uploaded and 12 cents per GB downloaded.

During the month, Alex uploaded 180 GB and downloaded 350 GB.

What was the total bill for the month?   (2 marks)

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Show Answers Only

`$148.00`

Show Worked Solution

`text(Upload charge = 180 × 45c = $81.00)`

`text(Download charge = 350 × 12c = $42.00)`

`:.\ text(Total bill)` `= 25 + 81 + 42`
  `= $148.00`

Filed Under: Rates (Std2-X) Tagged With: Band 2, smc-1104-15-General rate problems, smc-805-60-Other rate problems

v1 Measurement, STD2 M7 2016 HSC 15 MC

Parking in a city car park is charged at the rate of $3.40 per 20 minutes, or part thereof.

What is the cost of parking for 1 hour and 8 minutes?

  1. $10.20
  2. $13.60
  3. $17.00
  4. $20.40
Show Answers Only

`=> B`

Show Worked Solution

`68 -: 20 = 3.4 \ \Rightarrow\ 4\ \text{blocks}`

♦ Mean mark 48%.
`:.\ \text(Cost)` `= 4 xx 3.40`
  `= $13.60`

`=> B`

Filed Under: Rates (Std2-X) Tagged With: Band 5, smc-1104-15-General rate problems, smc-805-60-Other rate problems

v1 Measurement, STD2 M7 2014 HSC 17 MC

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

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

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

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

\(\Rightarrow B\)

Filed Under: Rates (Std2-X) Tagged With: Band 4, smc-1104-15-General rate problems, smc-805-60-Other rate problems

v2 Measurement, STD2 M7 2016 HSC 9 MC

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

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

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

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

\(\Rightarrow B\)

Filed Under: Rates (Std2-X) Tagged With: Band 3, smc-1104-15-General rate problems, smc-805-60-Other rate problems

v1 Measurement, STD2 M7 2016 HSC 9 MC

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

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

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

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

\(\Rightarrow C\)

Filed Under: Rates (Std2-X) Tagged With: Band 3, smc-1104-15-General rate problems, smc-805-60-Other rate problems

v1 Measurement, STD2 M1 2009 HSC 23c

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

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

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  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)

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i.    `17\ text(m²)`

ii.   `$1140`

Show Worked Solution
i.
`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)`

 

ii. `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`

Filed Under: Perimeter, Area and Volume (Std2-X) Tagged With: Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1104-15-General rate problems, smc-1121-10-Perimeter and Area, smc-4234-10-Area (std), smc-798-10-Perimeter and Area, smc-805-60-Other rate problems

Measurement, STD1 M4 2024 HSC 31

Wombats can run at a speed of 40 km/h over short distances.

At this speed, how many seconds would it take a wombat to run 150 metres?   (3 marks)

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\(\text{13.5 seconds}\)

Show Worked Solution

\(\text{Convert km/h to m/sec:}\)

\(\text{40 km/h}\) \(=40\,000\ \text{m/h}\)  
  \(=\dfrac{40\,000}{60 \times 60}\ \text{m/s}\)  
  \(=11.11\ \text{m/s}\)  

 

\(\therefore\ \text{Time to run 150m}\) \(=\dfrac{150}{11.11…}\)  
  \(=13.5\ \text{seconds}\)  
♦♦♦ Mean mark 25%.

Filed Under: M4 Rates (Y12) Tagged With: Band 5, smc-1104-15-General rate problems

Measurement, STD1 M4 2023 HSC 10 MC

A tap is dripping at the rate of 4 mL per minute.

Which expression shows how many litres this would amount to in one year?

  1. \(\dfrac{4 \times 1000}{60 \times 24 \times 365}\)
  2. \(\dfrac{4 \times 60 \times 24 \times 365}{1000}\)
  3. \(\dfrac{60 \times 24 \times 365}{4 \times 1000}\)
  4. \(\dfrac{1000}{4 \times 60 \times 24 \times 365}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Rate}\Rightarrow 4 \text{ mL} / \text{minute}\)

\(\text{Litres in 1 year}\) \(=\dfrac{4}{1000}\times 60\times 24\times365\)
  \(=\dfrac{4 \times 60 \times 24 \times 365}{1000}\)

 
\(\Rightarrow B\)

♦ Mean mark 47%.

Filed Under: M4 Rates (Y12) Tagged With: Band 5, smc-1104-15-General rate problems

Measurement, STD1 M4 2021 HSC 22

A tap is leaking water. It leaks 1 drop every 4 seconds, and 15 of these drops make up 1 mL.

  1. Find the amount of water leaked in a 24-hour period. Give the answer in litres.  (3 marks)

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  2. A bucket can hold 9 litres of water. How long will it take for the leaking tap to completely fill this empty bucket?  (1 mark)

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Show Answers Only
  1. `1.44\ text(L)`
  2. `6.25\ text(days)`
Show Worked Solution

a.   `text(Drops per minute) = 60/4 = 15`

♦ Mean mark part (a) 50%.

`text(Volume per minute = 1 mL)`

`:.\ text(Volume in 24 hours)` `= 1 xx 60 xx 24`
  `= 1440\ text(mL)`
  `= 1.44\ text(L)`
♦♦ Mean mark part (b) 32%.

 

b.    `text(Time to fill bucket)` `= 9/1.44`
    `= 6.25\ text(days)`

Filed Under: M4 Rates (Y12) Tagged With: Band 5, smc-1104-15-General rate problems

Measurement, STD1 M4 2020 HSC 12

Two painters each provide a quote for painting an area of 1500 square metres. Painter A charges $100 per 30 square metres. Painter B charges $80 per hour and bases their quote on painting 25 square metres per hour.

Calculate how much will be saved by choosing the cheaper quote.   (3 marks)

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`$200`

Show Worked Solution

`text{Painter A cost:}`

`C_A = frac{1500}{30} xx 100 = $5000`
 

`text{Painter B cost:}`

`text(Hours)\ = frac{1500}{25} = 60`

`C_B = 60 xx 80 = $ 4800`
 

`therefore \ text{Savings by using Painter B}`

`= 5000 – 4800`

`=  $200`

Filed Under: M4 Rates (Y12) Tagged With: Band 4, smc-1104-15-General rate problems

Measurement, STD1 M4 2019 HSC 5 MC

Which expression can be used to convert a speed of 3 metres per minute to a speed in centimetres per second?

  1. `3 xx 100 ÷ 60`
  2. `3 xx 100 xx 60`
  3. `3 ÷ 100 ÷ 60`
  4. `3 ÷ 100 xx 60`
Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 47%.

`text(3 m/min)` `= 3 xx 100\ text(cm/minute)`
  `= 3 xx 100 ÷ 60\ text(cm/second)`

 
`=> A`

Filed Under: M4 Rates (Y12) Tagged With: Band 5, smc-1104-15-General rate problems

Measurement, STD2 M7 2018 HSC 27a

Jenny used her mobile phone while she was overseas for one month.

Her mobile phone plan has a base monthly cost of $50. While overseas, she is also charged 33 cents per SMS message sent and 26 cents per MB of data used.

During her month overseas, Jenny sent 120 SMS messages and used 1400 MB of data.

What was her mobile phone bill for the month overseas?  (2 marks)

Show Answers Only

`$453.60`

Show Worked Solution

`text(SMS charge = 120 × 33c = $39.60)`

`text(Data charge = 1400 × 26c = $364.00)`
 

`:.\ text(Total bill)` `= 50 + 39.60 + 364`
  `= $453.60`

Filed Under: M4 Rates (Y12), Rates (Std2) Tagged With: Band 2, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2018 HSC 26g

A field diagram of a block of land has been drawn to scale. The shaded region `ABFG` is covered in grass.
 

 
The actual length of `AG` is 24 m.

  1. If the length of `AG` on the field diagram is 8 cm, what is the scale of the diagram?  (1 mark)

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  2. How much fertiliser would be needed to fertilise the grassed area  `ABFG`  at the rate of 26.5 g /m²?  (3 marks)

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Show Answers Only
  1. `text(1 : 300)`
  2. `5008.5\ text(grams)`
Show Worked Solution

♦♦ Mean mark 32%.

i.    `text(Scale    8 cm)` `\ :\ text(24 m)`
  `text(1 cm)` `\ :\ text(3 m)`
  `1` `\ :\ 300`

 

ii.   `text(Area of rectangle)\ ABFE`

`= 6\ text(cm × 3 cm)`

`= 18\ text(m × 9 m)`

`= 162\ text(m²)`
 

`text(Area of)\ DeltaEFG`

`= 1/2 xx 3\ text(cm) xx 2\ text(cm)`

`= 1/2 xx 9 xx 6`

`= 27\ text(m²)`
 

`:.\ text(Fertiliser needed)` `= (162 + 27) xx 26.5`
  `= 5008.5\ text(grams)`

Filed Under: M4 Rates (Y12), M5 Scale Drawings (Y12), Rates (Std2) Tagged With: Band 4, Band 5, smc-1104-15-General rate problems, smc-1105-20-Maps and Scale Drawings, smc-805-60-Other rate problems

Measurement, STD2 M7 2017 HSC 26b

Toby’s mobile phone plan costs $20 per month, plus the cost of all calls. Calls are charged at the rate of 70 cents per 30 seconds, or part thereof. There is also a call connection fee of 50c per call.

Here is a record of all his calls in July.
 

How much is Toby’s mobile phone bill for July?  (2 marks)

Show Answers Only

`$27.10`

Show Worked Solution

`text(Call cost) = 0.70 + (2 xx 0.70) + (5 xx 0.70) = $5.60`

`text(Connection fees) = 3 xx 0.50 = $1.50`

`:.\ text(Total bill)` `= 5.60 + 1.50 + 20`
  `= $27.10`

Filed Under: FS Communication, M4 Rates (Y12), Rates (Std2) Tagged With: Band 4, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2016 HSC 15 MC

Calls on a mobile phone plan are charged at the rate of 54 cents per 30 seconds, or part thereof.

What is the cost of a call lasting 2 minutes and 15 seconds?

  1.    $2.16
  2.    $2.32
  3.    $2.43
  4.    $2.70
Show Answers Only

`=> D`

Show Worked Solution

`5 xx 30\ text(second blocks)`

♦ Mean mark 44%.
`:.\ text(C)text(ost)` `= 5 xx 0.54`
  `= $2.70`

`=> D`

Filed Under: FS Communication, M4 Rates (Y12), Rates (Std2) Tagged With: Band 5, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2016 HSC 9 MC

An old washing machine uses 130 L of water per load. A new washing machine uses 50 L per load.

How much water is saved each year if two loads of washing are done each week using the new machine?

  1.   2600 L
  2. 4160 L
  3. 5200 L
  4. 8320 L
Show Answers Only

`D`

Show Worked Solution

`text(Water used by old machine)`

`= 130 xx 2 xx 52`

`= 13\ 520\ text(L)`

`text(Water used by new machine)`

`= 50 xx 2 xx 52`

`= 5200\ text(L)`

`:.\ text(Water saved)` `= 13\ 520 – 5200`
  `= 8320\ text(L)`

`=> D`

Filed Under: FS Resources, M4 Rates (Y12), MM1 - Units of Measurement, Rates (Std2) Tagged With: Band 3, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2008 HSC 23c

An alcoholic drink has 5.5% alcohol by volume. The label on a 375 mL bottle says it contains 1.6 standard drinks.

  1. How many millilitres of alcohol are in a 375 mL bottle? (1 mark)

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  2. It is recommended that a fully-licensed male driver should have a maximum of one standard drink every hour.

     

    Express this as a rate in millilitres per minute, correct to one decimal place.    (2 marks)

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Show Answers Only
  1. `20.625\ text(mL)`
  2. `3.9\ text(mL/min)`
Show Worked Solution
i.    `text(Alcohol)` `= 5.5/100 xx 375`
    `= 20.625\ text(mL)`

 

ii.    `text(S)text(ince 1.6 standard drinks = 375 mL)`

`=>\ text(1 standard drink)`

`= 375/1.6`

`= 234.375\ text(mL)`
 

`:.\ text(Rate)` `= 234.375/60`
  `= 3.90625`
  `= 3.9\ text(mL/min)`

Filed Under: M4 Rates (Y12), MM1 - Units of Measurement, Rates (Std2) Tagged With: Band 4, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2014 HSC 17 MC

A child who weighs 14 kg needs to be given 15 mg of paracetamol for every 2 kg of body weight.

Every 10 mL of a particular medicine contains 120 mg of paracetamol.

What is the correct dosage of this medicine for the child?

  1. 5.6 mL
  2. 8.75 mL
  3. 11.43 mL
  4. 17.5 mL
Show Answers Only

`B`

Show Worked Solution

`text(Paracetamol needed)`

`= 14/2 xx 15\ text(mg)`

`= 105\ text(mg)`

 

`text(S)text(ince 120 mg is contained in 10 mL,)`

`=> 105\ text(mg is contained in)`

`105/120 xx 10\ text(mL)= 8.75\ text(mL)`

`=>  B`

Filed Under: M4 Rates (Y12), Medication, MM1 - Units of Measurement, Rates (Std2) Tagged With: Band 4, smc-1104-15-General rate problems, smc-805-60-Other rate problems

Measurement, STD2 M1 2009 HSC 23c

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

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

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  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $55. 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 terrace.   (2 marks)

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Show Answers Only

i.    `13.77\ text(m²)`

ii.   `$880`

Show Worked Solution
i.
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (2.7 + 1.8) xx (2.7 + 1.8)\-2 xx (1.8 xx 1.8)`
  `= 20.25\-6.48`
  `= 13.77\ text(m²)`

 

ii. `text(Tiles required)` `= (13.77 +10 text{%}) xx 13.77`
    `= 15.147\ text(m²)`

 

 `=>\ text(16 boxes are needed)`

`:.\ text(Total cost of boxes)` `=16 xx $55`
  `= $880`

Filed Under: Area and Surface Area, M4 Rates (Y12), MM1 - Units of Measurement, MM2 - Perimeter, Area and Volume (Prelim), Perimeter and Area (Std 1), Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027), Rates (Std2) Tagged With: Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1104-15-General rate problems, smc-1121-10-Perimeter and Area, smc-4234-10-Area (std), smc-6304-10-Perimeter and Area, smc-798-10-Perimeter and Area, smc-805-60-Other rate problems

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