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Measurement, STD1 M2 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.     \(\text{Total time}\) \(=20+75+35\)
    \(=130\ \text{minutes}\)
    \(=2\text{ h }10\text{ min}\)

Mean mark (a) 54%.
b.     \(\text{Average speed}\) \(=\dfrac{\text{distance}}{\text{time}}\)
    \(=\dfrac{100}{1.25}\)
    \(=80\ \text{km/h}\)

♦♦ Mean mark (b) 43%.

Filed Under: M2 Working with Time (Y11), M4 Rates (Y12), Rates Tagged With: Band 5, smc-1102-50-Other Time Problems, smc-1104-15-General rate problems, smc-6859-20-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), Rates Tagged With: Band 3, smc-1104-15-General rate problems, smc-6859-20-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
\(\text{% increase}\) \(=\Big[\dfrac{\text{Change in HR}}{\text{Original HR}}\Big]\times 100\%\)
  \(=\Big[\dfrac{126-72}{72}\Big]\times 100\%\)
  \(=75\%\)

♦♦ Mean mark 45%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Measurement, STD1 M4 2025 HSC 2 MC

Mario drives from his home to his friend’s house. He watches a movie at his friend’s house and then drives home.

Which distance−time graph best represents Mario’s complete journey?
 

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Mario’s trip starts at home (zero) and ends at home (zero).}\)

\(\Rightarrow B\)

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 3, smc-1104-10-Travel Graphs, smc-6859-10-Travel Graphs

Measurement, STD1 M4 2024 HSC 3 MC

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

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

  1. 10 am and 11 am
  2. 11 am and 11:30 am
  3. 11:30 am and 1 pm
  4. 1 pm and 2:30 pm
Show Answers Only

\(A\)

Show Worked Solution

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

\(\therefore\ \text{Greatest speed is 10 km/hour and occurs between 10 am and 11 am.}\)

\(\Rightarrow A\)

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 4, smc-1104-10-Travel Graphs, smc-6859-10-Travel Graphs

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), Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Measurement, STD1 M4 2024 HSC 16

The cost of electricity is 30.13 cents per kWh .

Calculate the cost of using a 650 W air conditioner for 6 hours.   (2 marks)

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\(\text{Cost} =\$ 1.18\)

Show Worked Solution

\(\text{Usage}=6 \times 650=3900\, \text{W}=3.9\, \text{kW}\)

\(\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}\)

♦♦ Mean mark 32%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy

Measurement, STD1 M4 2023 HSC 28

The nutrition label for a food item is shown.

Based on the information on this label, what is the daily recommended intake of carbohydrates, to the nearest gram?   (2 marks)

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\(318\ \text{g}\)

Show Worked Solution

\(6\text{% of daily intake}\) \(=19.1\ \text{g}\)
\(\text{daily intake}\) \(=\dfrac{19.1}{6}\times 100\)
  \(=318.\dot{3}\ \text{g}\)
  \(\approx 318\ \text{g/day}\)

 
♦ Mean mark 42%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy

Financial Maths, STD1 M4 2023 HSC 23

The table compares the fuel costs of a petrol car with an electric car.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Petrol car} & \textit{Electric car}\\
\hline
\rule{0pt}{2.5ex}\text{Fuel consumption}\rule[-1ex]{0pt}{0pt} & \text{8.6 L/100 km} & \text{18 kWh/100 km} \\ \hline \rule{0pt}{2.5ex}\text{Cost of fuel}\rule[-1ex]{0pt}{0pt} & \text{\$1.87/L} & \text{\$0.25/kWh} \\ \hline \end{array}

Jun travels on average 35 000 km per year.

How much will he save on fuel costs in a year by using an electric car?  (3 marks)

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\($4053.70\)

Show Worked Solution

\(\text{Petrol car fuel costs}\)

\(\text{Litres used}=\dfrac{8.6\times 35\ 000}{100}=3010\text{ L}\)

\(\text{Cost of fuel}=3010\times $1.87 = $5628.70\)

  
\(\text{Electric car power costs}\)

\(\text{kWh}=\dfrac{18\times 35\ 000}{100}=6300\text{ kWh}\)

\(\text{Cost of power}=6300\times $0.25 = $1575\)

  
\(\text{Saving}=$5628.70-$1575=$4053.70\)

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 4, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6859-30-Fuel

Measurement, STD1 M4 2023 HSC 14

The distance-time graph shows the first two stages of a car journey from home to a holiday house.
  

  1. At what speed, in kilometres per hour, did the car travel during stage \(A\) of the journey?   (1 mark)

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  2. For how long did the car stop during stage \(B\) of the journey?   (1 mark)

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  3. After stage \(B\), the car continues to travel towards the holiday house at a constant speed of \(50\ \text{km/h}\) for 2 hours. Graph this part of the journey on the grid above.   (2 marks)

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a.    \(100\text{ km/h}\)

b.    \(30\text{ minutes}\)

c.   

Show Worked Solution

a.   \(S=\dfrac{D}{T}=\dfrac{150}{1.5}=100\ \text{km/h}\)
 

b.    \(\text{Stage}\textit{ B}=1.5\rightarrow 2\text{ hours}=30\text{ minutes}\)
 

c.    \(\text{Position after 2 hours at 50km/h}=(150+2\times 50 , 2 + 2) = (250 , 4)\)


♦ Mean mark (c) 50%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 3, Band 4, smc-1104-10-Travel Graphs, smc-1117-20-\(d=s\times t\), smc-6859-10-Travel Graphs, smc-791-20-\(d=s\times t\)

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), Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Financial Maths, STD1 F1 2023 HSC 6 MC

A delivery truck was valued at $65 000 when new. The value of the truck depreciates at a rate of 22 cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of 132 600 km?

  1. $35 828
  2. $29 172
  3. $14 872
  4. $14 300
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Depreciation} = 132\ 600\times\dfrac{22}{100}=\$29\,172\)

\(\text{Value} = 65\,000-29\,172=\$35\,828\)

\(\Rightarrow A\)

Filed Under: Rates, Simple Interest and S/L Depreciation Tagged With: Band 4, smc-1124-20-Straight-line Depreciation, smc-6859-20-General Rate Problems, std2-std1-common

Financial Maths, STD2 F1 2023 HSC 4 MC

A delivery truck was valued at $65 000 when new. The value of the truck depreciates at a rate of 22 cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of 132 600 km?

  1. $35 828
  2. $29 172
  3. $14 872
  4. $14 300
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Depreciation}=0.22\times 132\ 600=$29\ 172\)

\(\text{Truck value}=65\ 000-29\ 172=$35\ 828\)

   
\(\Rightarrow A\)

Filed Under: Rates, Simple Interest and S/L Depreciation Tagged With: Band 3, smc-6859-20-General Rate Problems, smc-808-40-Unit depreciation

Measurement, STD1 M4 2022 HSC 22

A 2500-watt air-conditioning system is turned on for 3 hours each day. Electricity is charged at 27 cents per kWh.

What is the cost of electricity for using the air-conditioning system over a seven-day period?  (2 marks)

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`$14.18`

Show Worked Solution
`text{Daily usage}`  `=2500 xx 3`
  `= 7500\ text{W}`
  `=7.5\ text{kW  (1000 W = 1 kW)}`

 

`text{Cost}`  `= 2.5 xx 7.5 xx 0.27`   
  `= 14.175`  
  `=$14.18\ \ text{(nearest cent)}`  

♦♦ Mean mark 39%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy

Measurement, STD1 M4 2022 HSC 14

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

   
 

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

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  2. At what time during the day did Jamie arrive back at town `M` ?  (1 mark)

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  3. What was the total distance that Jamie travelled?  (1 mark)

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  4. Between which times in the day was Jamie travelling at the fastest speed? Justify your answer, without calculations.  (2 marks)

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a.   `140\ text{km}`
 

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

`:. 10\ text{am}`
 

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

  

d.   `text{Fastest speed → graph is the steepest}`

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

Show Worked Solution

a.   `140\ text{km}`
 

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

`:. 10\ text{am}`
 

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

  

d.   `text{Fastest speed → graph is the steepest}`

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


♦♦ Mean mark part (b) 31%.
♦♦ Mean mark part (c) 35%.
♦♦ Mean mark part (d) 32%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 4, Band 5, smc-1104-10-Travel Graphs, smc-6859-10-Travel Graphs

Measurement, STD1 M4 2021 HSC 23

Sue walks along a trail, starting at 7 am and finishing at 10 am. The travel graph shows Sue’s journey from the start to the finish. The journey has been broken into six sections, `A`, `B`, `C`, `D`, `E` and `F`.
 

     

  1. On two occasions Sue stopped to rest. In which sections of the journey did Sue rest?  (1 mark)

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  2. In which section of the journey did Sue travel fastest? Justify your answer.  (2 marks)

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  3. Kim walked along the same trail, also starting at 7 am and finishing at 10 am. Kim walked at a constant speed for the entire journey.
  4. By showing Kim’s journey on the grid above, determine between what times Sue was ahead of Kim.  (3 marks)

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  1. `B\ text(and)\ E`
  2. `text(S)text(ection)\ C.\ text(Slope is the steepest.)`
  3.  `text(8:30 am – 9:15 am)`
     
       
Show Worked Solution

a.   `B\ text(and)\ E`

♦ Mean mark part (b) 44%.

 

b.   `text(Fastest travel occurs when the slope is the steepest.)`

♦♦ Mean mark part (c) 30%.

`:. text(S)text(ection)\ C`

c.

`text(Sue was ahead when her graph is higher than Kim’s.)`

`:.\ text(She was ahead between 8:30 am – 9:15 am)`

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 2, Band 5, smc-1104-10-Travel Graphs, smc-6859-10-Travel Graphs

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), Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Measurement, STD1 M4 2021 HSC 11

Chocolate of a particular brand can be bought in three different sizes.

Option 1: 100 grams for $1.50

Option 2: 300 grams for $4.20

Option 3: 500 grams for $7.25

Which option gives the lowest price per 100 grams? Justify your answer with calculations.  (2 marks)

Show Answers Only

`text(Option 2 is the cheapest.)`

Show Worked Solution

`text(Compare prices per 100 grams:)`

`text(Option 1) = $1.50`

`text(Option 2) = 4.20 ÷ 3 = $1.40`

`text(Option 3) = 7.25 ÷ 5 = $1.45`

`:.\ text(Option 2 is the cheapest.)`

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 4, smc-1104-50-Best Buys, smc-6859-60-Best Buys

Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different brands of television are shown.

The average cost for electricity is 25c/kWh. A particular family watches an average of 3 hours of television per day.

  1. The annual cost of electricity for Television A for this family is $48.18.
  2. For this family, what is the difference in the annual cost of electricity between Television A and Television B?  (2 marks)

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  3. For this family, how many years will it take for the total cost of buying and using Television A to be equal to the cost of buying and using Television B?  (2 marks)

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Show Answers Only
  1. `$4.38`
  2. `5\ text(years)`
Show Worked Solution
a.   `text{Annual power usage (B)}` `= 160 xx 3 xx 365`
    `=175\ 200`
    `=175.2\ text(kWh)`

♦ Mean mark part (a) 49%.
  `text{Annual cost (B)}` `= 175.2 xx 0.25`
    `=$43.80`

 

  `text{Difference in cost}` `= 48.18 – 43.80`
    `=$4.38`

♦♦ Mean mark part (b) 23%.
b.   `text{Difference in price}` `= 921.90 – 900`
    `=$21.90`

 

  `text(Years to even out cost)` `=21.90/4.38`
    `=5\ text{years}`

Filed Under: Energy and Mass, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, Band 5, smc-1104-25-Energy, smc-6859-40-Energy, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

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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Show Answers Only

`$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), Rates Tagged With: Band 4, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Measurement, STD2 M7 SM-Bank 16

A refreshment stand at the market sells its iced tea in four different sizes.
 

Showing calculations, find which serving size offers the cheapest iced tea.   (2 marks)

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Show Answers Only

`text(375 mL)`

Show Worked Solution

`text(Consider the price per mL of each option)`

`text(150 mL :)\ \ 1.50/150 = 1\ text(cent/mL)`

`text(750 mL :)\ \ 6.10/750 = 0.81\ text(cent/mL)`

`text(375 mL :)\ \ 3.00/375 = 0.8\ text(cent/mL)`

`text(200 mL :)\ \ 1.80/200 = 0.9\ text(cent/mL)`

 

`:.\ text(375 mL serving size offers the cheapest price)`

Filed Under: M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, smc-1104-50-Best Buys, smc-6859-60-Best Buys, smc-6932-50-Best Buys, smc-805-50-Best Buys

Measurement, STD2 M7 SM-Bank 15 MC

A brand of pasta is sold in two different packets.
 

What is the difference in the price per kilogram of these two packets?

  1. $1.50
  2. $3.00
  3. $3.60
  4. $7.20
Show Answers Only

`B`

Show Worked Solution

`text(Convert to price per kg:)`

`text(Packet 1)`

`200\ text(g) = $1.80`

`1\ text(kg) = 5 xx 1.80 = $9`

 

`text(Packet 2)`

`750\ text(g) = $9`

`100\ text(g) = 9.00 ÷ 7.5 = $1.20`

`1\ text(kg) = 1.20 xx 10 = $12`

 

`:.\ text(Difference per kg)`

`= 12 – 9`

`= $3.00`
 

`=> B`

Filed Under: M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, smc-1104-50-Best Buys, smc-6859-60-Best Buys, smc-6932-50-Best Buys, smc-805-50-Best Buys

Measurement, STD2 M7 SM-Bank 14

Reece looks at the price of four sunscreens.
 

Which sunscreen is the cheapest per litre? Show your calculations.   (2 marks)

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Show Answers Only

`text(Sport sunscreen)`

Show Worked Solution

`text(Sun)\ =1090/1200 = 0.91\ text(c/mL)`

STRATEGY: Finding the cost per mL is more efficient than finding the cost per litre.

`text(Sport)\ =540/600 = 0.9\ text(c/mL)`

`text(Block)\ =475/500 = 0.95\ text(c/mL)`

`text(Wet)\ =740/750 = 0.99\ text(c/mL)`

 

`:.\ text(Sport sunscreen is the cheapest.)`

Filed Under: M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, smc-1104-50-Best Buys, smc-6859-60-Best Buys, smc-6932-50-Best Buys, smc-805-50-Best Buys

Measurement, STD1 M4 2019 HSC 18

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


 

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

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  2. In which section of the trip, `A`, `B`, `C` and `D`, did Nikau travel the fastest?  (1 mark)

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Show Answers Only
  1. `400\ text(km)`
  2. `text(S)text(ection)\ D`
Show Worked Solution
a.    `text(Distance travelled)` `= 2 xx 200`
    `= 400\ text(km)`

 

♦ Mean mark part (b) 42%.

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

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

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 4, Band 5, smc-1104-10-Travel Graphs, smc-6859-10-Travel Graphs

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), Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems

Measurement, STD1 M4 2019 HSC 3 MC

Sugar is sold in four different sized packets.

Which is the best buy?

  1.  100 g for $0.40
  2.  500 g for $1.65
  3.  1 kg for $3.50
  4.  2 kg for $6.90
Show Answers Only

`B`

Show Worked Solution

`text(Price per kilogram:)`

`100\ text(g) -> 10 xx 0.40 = $4.00`

`500\ text(g) -> 2 xx 1.65 = $3.30`

`1\ text(kg) -> $3.50`

`2\ text(kg) -> 6.90 ÷ 2 = $3.45`

`=> B`

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 3, smc-1104-50-Best Buys, smc-6859-60-Best Buys

Algebra, STD2 A2 EQ-Bank 3 MC

A car travels 350 km on 40 L of petrol.

What is its fuel consumption?

  1. 7.8 L/100 km
  2. 8.4 L/100 km
  3. 8.8 L/100 km
  4. 11.4 L/100 km
Show Answers Only

`D`

Show Worked Solution

`text(40 litres are used to travel  3.5 × 100 km.)`

`text{Fuel consumption (L/100 km)}`

`= 40/3.5 = 11.428 = 11.4\ text(L/100 km)`

`=> D`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 3, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

Measurement, STD2 M7 2018 HSC 28c

Every day, a 1200-watt microwave oven is used for 45 minutes at 40% power. Electricity is charged at $0.25 per kWh.

What is the cost of running this microwave oven for 180 days?  (3 marks)

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

Show Answers Only

`$16.20`

Show Worked Solution
`text(Daily usage)` `= 1200 xx 45/60 xx 40text(%)`
  `= 360\ text(watts)`

 

`text(180 day usage)` `= 180 xx 360`
  `= 64\ 800\ text(watts)`
  `= 64.8\ text(kW)`

 

`:.\ text(C)text(ost over 180 days)` `= 64.8 xx 0.25`
  `= $16.20`

Filed Under: Energy and Mass, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, smc-1104-25-Energy, smc-6859-40-Energy, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

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, Rates, Rates Tagged With: Band 2, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other 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)

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

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

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

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, Rates, Rates, Ratios Tagged With: Band 4, Band 5, smc-1104-15-General rate problems, smc-1105-20-Maps and Scale Drawings, smc-6858-10-Maps and Scale Drawings, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Algebra, STD2 A2 2018 HSC 5 MC

The driving distance from Alex's home to his work is 20 km. He drives to and from work five times each week. His car uses fuel at the rate of 8 L/100 km.

How much fuel does he use driving to and from work each week?

  1. 16 L
  2. 20 L
  3. 25 L
  4. 40 L
Show Answers Only

`A`

Show Worked Solution

`text(Total distance travelled each week)`

`= 5 xx 2 xx 20= 200\ text(km)` 

`:.\ text(Total fuel used)= 200/100 xx 8\ text(L)= 16\ text(L)`

`=>A`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 3, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

Measurement, STD2 M7 EQ-Bank 23

Bikram runs a hot yoga studio.

If it costs 34 cents for 1-kilowatt (1000 watts) for 1 hour, how much does it cost him to run three 3200-watt heaters from 9:00 am to 12:30 pm on a single day? (Give your answer to the nearest cent)   (2 marks)

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

Show Answers Only

`$11.42`

Show Worked Solution

`text(Total energy usage)`

`= 3 xx 3200 xx 3.5\ text(hours)`

`= 33\ 600\ text(watts)`

 

`:.\ text(C)text(ost)` `= (33\ 600)/1000 xx 0.34`
  `= 11.424`
  `= $11.42\ \ text{(nearest cent)}`

Filed Under: Energy and Mass, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 4, smc-1104-25-Energy, smc-6859-40-Energy, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

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, Rates, Rates Tagged With: Band 4, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Algebra, STD2 A2 2017 HSC 14 MC

Kate is comparing two different models of car. Car A uses fuel at the rate of 9 L/100 km. Car B uses 3.5 L/100 km.

Suppose Kate plans on driving 8000 km in the next year.

How much less fuel will she use driving car B instead of car A?

  1. 280 L
  2. 440 L
  3. 720 L
  4. 1000 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel car)\ A= 8000/100 xx 9= 720\ text(L)`

`text(Fuel car)\ B= 8000/100 xx 3.5= 280\ text(L)`

`:.\ text(Fuel saved car)\ B= 720-280= 440\ text(L)`

`=>B`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, Depreciation / Running costs, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 3, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

Measurement, STD2 M1 2016 HSC 28b

The cost of buying a new heater is $990. It has an energy consumption of 505 kWh per year.

Energy is charged at the rate of $0.35 kWh.

How much will it cost in total to purchase and then run this heater for five years?  (2 marks)

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

Show Answers Only

`$1873.75`

Show Worked Solution

`text(C)text(ost to run heater for 5 years)`

`= 5 xx 505 xx 0.35`

`= $883.75`
 

`:.\ text(Total purchase and running cost)`

`= 883.75 + 990`

`= $1873.75`

Filed Under: Energy and Mass, FS Resources, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 3, smc-1104-25-Energy, smc-6859-40-Energy, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

Algebra, STD2 A2 2016 HSC 26c

Peta’s car uses fuel at the rate of 5.9 L /100 km for country driving and 7.3 L /100 km for city driving. On a trip, she drives 170 km in the country and 25 km in the city.

Calculate the amount of fuel she used on this trip.   (2 marks)

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

Show Answers Only

`11.855\ text(L)`

Show Worked Solution

`text(Fuel used in country)= 170 xx 5.9/100= 10.03\ text(L)`

`text(Fuel used in city)= 25 xx 7.3/100= 1.825\ text(L)`

`:.\ text(Total fuel used)= 10.03 + 1.825= 11.855\ text(L)`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, Depreciation / Running costs, M4 Rates (Y12), MM1 - Units of Measurement, Rates, Rates, Rates Tagged With: Band 3, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

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, Rates, Rates Tagged With: Band 5, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2016 HSC 11 MC

The concentration of a drug in a certain medication is 100 mg / 5 mL. A patient is prescribed 2000 mg of the drug.

How much medication should be given to the patient?

  1.  4 mL
  2. 25 mL
  3. 100 mL
  4. 400 mL
Show Answers Only

`C`

Show Worked Solution

`text(Volume required)`

`= 2000/100 xx 5`

`= 100\ text(mL)`

`=> C`

Filed Under: M4 Rates (Y12), Medication, MM1 - Units of Measurement, Rates, Rates, Rates Tagged With: Band 3, smc-1104-30-Medication, smc-6859-50-Medication, smc-6932-50-Other Rate Problems, smc-805-30-Medication

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, Rates, Rates Tagged With: Band 3, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M1 2015 HSC 30a

The energy consumption of a computer in standby mode is 21 watts. The cost of electricity is 31 cents per kWh.

A school computer room has 20 computers.

How much will the school save by switching off all 20 computers during 11 weeks of school holidays? (2 marks)

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

Show Answers Only

`$240.61`

Show Worked Solution

`text(21 watts usage per computer per hour.)`

♦ Mean mark 49%.

`text(Watts used by 20 computers in 11 weeks)`

`= 21 xx 20 xx 24 xx 7 xx 11`

`= 776\ 160\ text(watts)`

`= 776.16\ text(kW)`
 

`:.\ text(C)text(ost of energy)`

`= 776.16 xx $0.31`

`= $240.6096`

`= $240.61\ \ text{(nearest cent)}`

 

`:.\ text(The school will save $240.61 by switching)`

`text(off all the computers.)`

Filed Under: Energy and Mass, FS Resources, M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy, smc-799-20-Electricity

Measurement, STD2 M7 2015 HSC 27b

A patient requires 2400 mL of fluid to be delivered at a constant rate by means of a drip over 12 hours. Each mL of fluid is equivalent to 15 drops.

How many drops per minute need to be delivered?  (2 marks)

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

Show Answers Only

`50\ text(per minute)`

Show Worked Solution

`text(Fluid rate of delivery)`

`= 2400/12`

`= 200\ text(mL per hour)`

`= 200/60`

`= 3 1/3\ text(mL per minute)`

 
`text(S)text(ince each mL has 15 drops)`

`#\ text(Drops)` `= 15 xx 3 1/3`
  `= 50\ text(per minute)`

Filed Under: M4 Rates (Y12), MM1 - Units of Measurement, Rates, Rates, Rates Tagged With: Band 3, smc-1104-30-Medication, smc-6859-50-Medication, smc-6932-50-Other 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)

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

  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)

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

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, Rates, Rates Tagged With: Band 4, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Algebra, STD2 A2 2014 HSC 27b

Xuso is comparing the costs of two different ways of travelling to university.

Xuso’s motorcycle uses one litre of fuel for every 17 km travelled. The cost of fuel is $1.67/L and the distance from her home to the university car park is 34 km. The cost of travelling by bus is  $36.40 for 10 single trips.

Which way of travelling is cheaper and by how much? Support your answer with calculations.   (2 marks)

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

Show Answers Only

`text(Motorcycle is $0.30 cheaper per 1-way trip)`

Show Worked Solution

`text(Compare cost of a 1-way trip)`

`text(Motorcycle)`

`text(Fuel used) = 34/17 = 2\ text(L)`

`text(C)text(ost) = 2 xx $1.67 = $3.34`

`text(Bus)`

`text(C)text(ost) = 36.40/10 = $3.64`

`text(Difference) = $3.64\-3.34\ = $0.30`
  

`:.\ text(Motorcycle is $0.30 cheaper per 1-way trip.)`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, Depreciation / Running costs, M4 Rates (Y12), Rates, Rates, Rates Tagged With: Band 3, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

Algebra, STD2 A2 2014 HSC 22 MC

Heather’s car uses fuel at the rate of 6.6 L per 100 km for long-distance driving and  8.9 L per 100 km for short-distance driving.

She used the car to make a journey of 560 km, which included 65 km of short-distance driving.  

Approximately how much fuel did Heather’s car use on the journey?

  1. 37 L
  2. 38 L
  3. 48 L
  4. 50 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel used in short distance)`

`= 65/100 xx 8.9\ text(L) = 5.785\ text(L)`

`text(Fuel used in long distance)`

`= 495/100 xx 6.6\ text(L) = 32.67\ text(L)`

`:.\ text(Total Fuel)= 38.455\ text(L)`  

`=>  B`

Filed Under: Applications of Linear Relationships, Applications of Linear Relationships, Applications: Currency, Fuel and Other Problems, Applications: Currency, Fuel and Other Problems, Depreciation / Running costs, M4 Rates (Y12), MM1 - Units of Measurement, Rates, Rates, Rates Tagged With: Band 4, smc-1104-20-Fuel, smc-1119-20-Fuel, smc-6256-20-Fuel/Transport, smc-6513-10-Fuel/Transport, smc-6859-30-Fuel, smc-6932-10-Fuel, smc-793-20-Fuel, smc-805-10-Fuel

Measurement, STD2 M1 2014 HSC 20 MC

In a household of 4, each member uses an average of 13 minutes of hot water per day.

The household uses a 9 kW hot water unit.

Electricity is charged at 11.97 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. $1.63
  2. $6.54
  3. $392.14
  4. $653.56
Show Answers Only

`B`

Show Worked Solution

`text(Usage per day) = 4 xx 13 = 52\ text(mins)`

♦ Mean mark 39%.

`text(Usage per week) = 7 xx 52 = 364\ text(mins)`

`text(Converting to kWh)`

`= text{(hours of usage)} xx 9\ text(kW)`

`= 364/60 xx 9`

`= 54.6\ text(kWh)`
 

`:.\ text(C)text(ost)` `= 54.6 xx 11.97 text(c)`
  `~~ 654 text(c)`
  `~~$6.54`

`=>  B`

Filed Under: Energy and Mass, FS Resources, M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy, smc-799-20-Electricity

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, Rates, Rates Tagged With: Band 4, smc-1104-15-General rate problems, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M7 EQ-Bank 24

A patient is to receive 1.8 L of pain killer medication by intravenous drip that will take 1.5 hours to administer.

Given  1 mL = 4 drops, calculate the amount of drops per minute the machine must be set on.   (2 marks)

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

Show Answers Only

 `text(80 drops per minute.)`

Show Worked Solution

`text(Total drops required)=1800 xx 4=7200\ text(drops)`

`text{Time (in minutes)}=1.5 xx 60= 90\ text(minutes)`

`text(Drops per minute)=7200/90=80`

`:.\ text(The machine must be set to 80 drops per minute.)`

Filed Under: M4 Rates (Y12), Medication, Rates, Rates, Rates Tagged With: Band 4, smc-1104-30-Medication, smc-6859-50-Medication, smc-6932-50-Other Rate Problems, smc-805-30-Medication

Measurement, STD2 M1 2013 HSC 26d

A section of Jim’s electricity bill is shown.

2013 26d

  1. What is the value of `A`?    (1 mark)

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

  2. How much will Jim save if he uses 154 kWh of energy at the Off-peak rate rather than at the Peak rate?    (2 marks)

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

Show Answers Only
  1. `1084.4`
  2. `$ 58.78\ \ \ (text(nearest cent) )`
Show Worked Solution
i.    `A` `=\ text(Last reading + Energy used)`
    `= 560.9 + 523.5`
    `= 1084.4`
♦ Mean mark part (i) 43% and part  (ii) 46%.

 

ii.    `text(C)text(ost at off-peak)` `= 154 xx 9.6`
    `= 1478.4\ text(cents)`
`text(C)text(ost at peak)` `=154 xx 47.77`
  `= 7356.58\ text(cents)`

 

`:.\ text(Saving)` `=7356.58` `-1478.4`
  `=5878.18`
  `= $58.78`    `text{(nearest cent)}`

Filed Under: Energy and Mass, FS Resources, M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6859-40-Energy, smc-799-20-Electricity

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)

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

  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)

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

Show Answers Only

a.   `13.77\ text(m²)`

b.   `$880`

Show Worked Solution
a.
`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)`

 

b.    `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, Perimeter and Area, Perimeter and Area, Perimeter, Area and Volume, Rates, Rates, Rates 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-6483-20-Composite Areas, smc-6520-20-Composite areas, smc-6859-20-General Rate Problems, smc-6932-50-Other Rate Problems, smc-798-10-Perimeter and Area, smc-805-60-Other rate problems

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