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Measurement, STD2 EQ-Bank 20

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.   (1 mark)

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a.   `140\ text{km}`

b.   `10\ text{am}`

c.    `220\ \text{km}`  

d.    `\text{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 (either up or down)}`

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

Filed Under: Rates Tagged With: Band 3, Band 4, smc-6932-60-Travel Graphs

Measurement, STD2 EQ-Bank 19

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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a.    `400\ text(km)`

b.    `text(S)text(ection)\ D`

Show Worked Solution

a.    `text(Distance travelled)\ = 2 xx 200= 400\ text(km)`
 

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

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

Filed Under: Rates Tagged With: Band 3, Band 4

Measurement, STD2 EQ-Bank 27

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

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  2. 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.
  3. By showing Kim’s journey on the grid above, determine between what times Sue was ahead of Kim.   (3 marks)

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a.    `text(S)text(ection)\ C\ \text{(Slope is the steepest)}.`

b.    `text(8:30 am – 9:15 am)`
 
           

Show Worked Solution

a.   `text{Fastest travel occurs when the slope is the steepest.}`

`=>\ text{Section}\ C`

b.

`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: Rates Tagged With: Band 4, smc-6932-60-Travel Graphs

Measurement, STD2 EQ-Bank 32

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

  1. At what speed, in metres per second, did the car travel during stage \(A\) of the journey? Give your answer correct to one decimal place.   (2 marks)

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  2. 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.    \(27.8\ \text{m/s}\)

b.   

Show Worked Solution

a.   \(\text{Distance travelled} = 150\ \text{km}\ =150\,000\ \text{m}\)

\(\text{Stage A duration = 1.5 hours}\)

\(\text{Express 1.5 hours in minutes:}\)

\(\text{Minutes} = 1.5 \times 60 \times 60 = 5400\ \text{seconds}\)

\(\text{Speed} = \dfrac{150\,000}{5400}=27.77…=27.8\ \text{m/s}\)

 

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

   

Filed Under: Rates Tagged With: Band 4, Band 5, smc-6932-60-Travel Graphs

Measurement, STD2 EQ-Bank 2 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

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

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

\(\Rightarrow A\)

Filed Under: Rates Tagged With: Band 3, smc-6932-60-Travel Graphs

Measurement, STD2 M7 2025 HSC 38

A car’s fuel efficiency is 30 miles per US gallon.

\begin{array}{|ll|}
\hline 
\rule{0pt}{2.5ex} 1 \ \text{US gallon}=3.8 \  \text{litres } \rule[-1ex]{0pt}{0pt}& \text{(correct to } 2 \text { significant figures)} \\
\rule{0pt}{2.5ex} 1 \ \text{mile}=1.6 \ \text{km} \rule[-1ex]{0pt}{0pt}& \text{(correct to } 2 \text { significant figures)} \\
\hline
\end{array}

Calculate the car’s fuel efficiency in litres per 100 km, correct to 1 decimal place.   (3 marks)

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\(\text{Fuel efficiency}=7.9 \ \text{litres / 100 km}\)

Show Worked Solution

\(30 \ \text{miles}=30 \times 1.6=48 \ \text{km}\)

\(\text{Convert 48 km per 3.8 litres into km/litre:}\)

\(\dfrac{48}{3.8}=12.631 \ldots \ \text{km/litre}\)

\(\text{Fuel used in 100 km}=\dfrac{100}{12.631\ldots}=7.917\ldots=7.9 \ \text{litres (1 d.p.)}\)
 

\(\therefore \ \text{Fuel efficiency}=7.9 \ \text{litres per 100 km}\)

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-10-Fuel, smc-805-10-Fuel

Measurement, STD2 M7 2025 HSC 21

A house has a reverse-cycle air conditioner which uses 2.5 kW of power for cooling and 3.2 kW of power for heating. The cost of electricity is 29 cents per kWh .

  1. Find the cost, in dollars and cents, of cooling the house for 6 hours.   (1 mark)

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  2. The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter.
  3. Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day.   (2 marks)

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a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)

b.    \(7.5\ \text{hours}\)

Show Worked Solution

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)
 

b.    \(\text{Let \(h\) = hours used per day}\)

\(\text{Cost per day}\ = h \times 3.2 \times 0.29\)

\(\text{Cost (92 days )}\ = 92 \times h \times 3.2 \times 0.29\)

\(\text{Find \(h\) when cost = \$640:}\)

\(640\) \(=92 \times h \times 3.2 \times 0.29\)  
\(h\) \(=\dfrac{640}{92 \times 3.2 \times 0.29}=7.5\ \text{hours (1 d.p.)}\)  

Filed Under: Energy and Mass, Rates, Rates Tagged With: Band 3, Band 4, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

Measurement, STD2 M7 2025 HSC 16

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

Show Worked Solution

\(\text{Cost of 4 litre cans:}\)

\( C_1=\dfrac{80}{4} \times 90=$1800 \)

\(\text{Cost of 10 litre cans:}\)

\( C_2=\dfrac{80}{10} \times 205=$1640\)

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

Filed Under: Rates, Rates Tagged With: Band 3, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M7 2025 HSC 10 MC

An electricity company charges customers 37 cents per kWh for electricity used, \( U\), and pays customers 5 cents per kWh for electricity produced, \(P\).

The electricity company also charges customers a fee of 71 cents per day.

Which formula should be used to calculate a customer's daily cost of electricity, \(C\), in cents?

  1. \( C=71+37 U-5 P \)
  2. \( C=71+37 U+5 P \)
  3. \( C=71-37 U-5 P \)
  4. \( C=71-37 U+5 P \)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Customer pays:}\ 71 + 37 \times U\)

\(\text{Customer receives:}\ 5 \times P\)

\(\text{Daily cost}\ = 71 + 37 \times U-5 \times P\)

\(\Rightarrow A\)

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-20-Energy, smc-805-20-Energy

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

\(\text{Usage}=6 \times 650=3900\, \text{Wh}=3.9\, \text{kWh}\)

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

♦♦ Mean mark 32%.

Filed Under: M4 Rates (Y12), Rates Tagged With: Band 5, smc-1104-25-Energy, smc-6932-20-Energy

Measurement, STD2 M7 2024 HSC 17

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

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

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

Filed Under: Energy and Mass, Rates, Rates Tagged With: Band 4, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

Measurement, STD2 M7 2024 HSC 33

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

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

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

 

\(\therefore\ \text{Time to run 150m}\) \(=\dfrac{150}{11.11…}\)  
  \(=13.5\ \text{seconds}\)  

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

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-6932-10-Fuel, smc-6932-20-Energy

Measurement, STD2 M7 2023 HSC 5 MC

Four petrol pumps are shown, each with the amount of petrol purchased and its cost.

Which one represents the best value?
 

Show Answers Only

`B`

Show Worked Solution

`text{Calculate cost per litre of each option}`

`text{Option}\ A:\ $46.70 -: 24= $1.946\ text{per litre}`

`text{Option}\ B:\ $48.50 -: 25= $1.940\ text{per litre}`

`text{Option}\ C:\ $52.30 -: 26= $2.01\ text{per litre}`

`text{Option}\ D:\ $54.80 -: 27= $2.03\ text{per litre}`

`=>B`

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-50-Best Buys, smc-805-50-Best Buys

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-6932-40-Rates, 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}`  `= 7 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-6932-20-Energy

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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a.    `$4.38`

b.    `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 Tagged With: Band 4, Band 5, smc-1104-25-Energy, smc-6932-20-Energy, smc-799-20-Electricity, smc-805-20-Energy

Measurement, STD2 M7 2020 HSC 17

Ayla wishes to estimate the number of trees on a square block of land measuring 1000 m by 1000 m. She counts the number of trees on a 5 m by 5 m section of the block and finds there are 8 trees.

Based on this, estimate the number of tress on the entire square block of land.   (2 marks)

Show Answers Only

`320\ 000`

Show Worked Solution
`text{Area of block}` `= 1000 xx 1000`
  `= 1\ 000\ 000 \ text{m}^2`

 

`8\ text{trees per} \ 25 text{m}^2`

`therefore \ text{Total trees}` `= 8 xx frac{1\ 000\ 000}{25}`
  `= 320\ 000`

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M7 SM-Bank 19

A laundromat can wash 12 loads of laundry in one hour at full capacity.

A standard load of laundry weighs 7 kilograms.

Here is some information about two different washing machines.
 


 

Working at full capacity, how many litres of water would the laundromat expect to save in one hour by using the front loader instead of the top loader?   (2 marks)

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`text(315 L)`

Show Worked Solution

`text(Top loader water used (1 load))`

`= 7 xx 10.25`

`= 71.75\ text(L)`
 

`text(Front loader water used (1 load))`

`= 7 xx 6.5`

`= 45.5\ text(L)`
 

`:.\ text(Expected water saved)`

`= 12 xx (71.75 – 45.5)`

`= 315\ text(L)`

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-50-Other Rate Problems, smc-805-60-Other rate problems

Measurement, STD2 M7 EQ-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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`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 EQ-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 EQ-Bank 25

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)\ =1090/12 = 91\ text(c/mL)`

STRATEGY: Finding the cost per 100 mL in cents is more efficient than finding the cost per litre.

`text(Sport)\ =540/6 = 90\ text(c/mL)`

`text(Block)\ =475/5 = 95\ text(c/mL)`

`text(Wet)\ =740/7.5 = 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, STD2 M7 2019 HSC 2 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: Rates, Rates Tagged With: Band 3, smc-6932-50-Best Buys, smc-805-50-Best Buys

Measurement, STD2 M7 SM-Bank 12

Francis is buying a fridge. The energy usage of his two choices are below:

Fridge 1 (5 star rating): 227 kWh per year

Fridge 2 (2 star rating): 492 kWh per year

Each fridge has an expected life of 8 years.

If electricity costs 32.4 cents per kWh, how much will Francis save in electricity over the expected life of the more energy efficient fridge, to the nearest dollar?  (2 marks)

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Show Answers Only

`$687`

Show Worked Solution

`text(Energy cost of fridge 1)`

`= 227 xx 0.324 xx 8`

`= $588.38`

`text(Energy cost of fridge 2)`

`= 492 xx 0.324 xx 8`

`= $1275.26`

`:.\ text(Energy saving)` `= 1275.26 – 588.38`
  `= 686.88`
  `= $687`

Filed Under: Rates, Rates Tagged With: Band 4, smc-6932-20-Energy, smc-805-20-Energy

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\ text(L/100 km) = 11.428\ text(L/100 km) = 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)

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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(kWh)`

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

Filed Under: Energy and Mass, M4 Rates (Y12), Rates, Rates Tagged With: Band 4, smc-1104-25-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

i.    `text(1 : 300)`

ii.   `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)\ \ text{(Using  1 cm : 3 m)}`

`= 162\ text(m)^2`
 

`text(Area of)\ DeltaEFG`

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

`= 1/2 xx 9 xx 6\ \ text{(Using  1 cm : 3 m)}`

`= 27\ text(m)^2`
  

`:.\ 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-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 blocks)`

`text(20 seconds )=20/30=066..\ text(blocks → 1 block)`

`text(40 seconds ) =40/30=1.33..\ text(blocks → 2 blocks)`

`text(2 mins 15 seconds = 135 seconds )=135/30=4.5\ text(blocks → 5 blocks)`
  

`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 Tagged With: Band 3, smc-1104-25-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

`text(2 min 15 sec = 135 seconds)`

♦ Mean mark 44%.

`text{At 54c per 30 seconds (or part thereof)}: 135/30=4.5`

`->\ text(Rounds up to 5 blocks)`

`->\ 5 xx 30\ text(second blocks)`

`:.\ 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-6932-20-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 M1 2004 HSC 23a

The diagram shows the shape of Carmel’s garden bed. All measurements are in
metres.

  1. Show that the area of the garden bed is 57 square metres.   (2 marks)

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

  2. Carmel decides to add a 5 cm layer of straw to the garden bed.

     

    Calculate the volume of straw required. Give your answer in cubic metres.   (2 marks)

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

  3. Each bag holds 0.25 cubic metres of straw.

     

    How many bags does she need to buy?   (2 marks)

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

  4. A straight fence is to be constructed joining point A to point B.

     

    Find the length of this fence to the nearest metre.   (2 marks)

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

Show Answers Only

a.   `text(Proof)\ text{(See Worked Solutions)}`

b.   `text(2.85 m)^3`

c.   `text(She needs to buy 12 bags)`

d.   `8\ text{m  (nearest metre)}`

Show Worked Solution

a.    `text(Area of)\ Delta ABC= 1/2 xx b xx h= 1/2 xx 10 xx 5.1=25.5\ text(m)^2`

`text(Area of)\ Delta ACD= 1/2 xx 10 xx 6.3= 31.5\ text(m)^2`

`:.\ text(Total Area)= 25.5 + 31.5= 57\ text(m)^2\ \text( … as required)`

b.    `V= Ah= 57 xx 0.05= 2.85\ text(m)^3`

c.    `text(Bags to buy)= 2.85/0.25= 11.4`

`:.\ text(She needs to buy 12 bags.)`

d.   `text(Using Pythagoras,)`

`AB^2= 6.0^2 + 5.1^2= 36 + 26.01= 62.01`

`AB= 7.874\ …=8\ text{m  (nearest metre)}`

Filed Under: Areas and Volumes (Harder), M3 Right-Angled Triangles (Y12), MM2 - Perimeter, Area and Volume (Prelim), Perimeter and Area, Perimeter and Area, Perimeter, Area and Volume, Pythagoras and basic trigonometry, Rates, Rates, Right-angled Triangles (Y12), Volume, Mass and Capacity, Volume, Mass and Capacity Tagged With: Band 3, Band 4, smc-1103-10-Pythagoras, smc-6304-40-Volume, smc-6483-20-Composite Areas, smc-6520-20-Composite areas, smc-6521-40-Volume, smc-6834-10-Pythagoras, smc-6932-50-Other Rate Problems, smc-798-10-Perimeter and Area, smc-798-40-Volume, smc-805-60-Other rate problems

Measurement, STD2 M7 EQ-Bank 22

Bronwyn needs to have 3.0 litres of intravenous liquid given to her over a period of 4 hours. 

What is the required flow rate in mL per minute?   (2 marks) 

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

Show Answers Only

`12.5\ text(mL/minute)`

Show Worked Solution

`text(Total liquid)= 3.0 xx 1000= 3000\ text(mL)`

`text(Total minutes)= 4 xx 60= 240`

`:.\ text(Flow Rate)= 3000/240= 12.5\ text(mL/minute)`

Filed Under: Medication, Rates, Rates Tagged With: Band 4, smc-6932-50-Other Rate Problems, smc-805-30-Medication

Measurement, STD2 M7 EQ-Bank 25

A medication is available in both tablet and liquid form.  A tablet contains 50 mg of the active ingredient while the liquid form contains 60 mg per 10 mL.  

Michael likes taking tablets and Georgia prefers liquid medicines.  If they each need 0.2 g of the active ingredient, what dosages do they take?   (3 marks) 

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

Show Answers Only

`text(Michael takes 4 tablets and Georgia takes 33.3mL.)`

Show Worked Solution

`text(Michael – tablets)`

`0.2\ text(g) = 200\ text(mg)`

`:.\ text(# Tablets)=200/50= 4`

 
`text(Georgia – liquid)`

`60\ text(mg in)\ 10\ text(mL)`

`1\ text(mg) = 10/60 = 0.166…\ text(mL)`

`200\ text(mg)= 200 xx 0.166…= 33.3\ text(mL)\ \ text{(to 1 d.p.)}`
  

`:.\ text(Michael takes 4 tablets and Georgia takes 33.3 mL.)`

Filed Under: Medication, Rates, Rates Tagged With: Band 4, smc-6932-50-Other Rate Problems, smc-805-30-Medication

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

a.    `20.625\ text(mL)`

b.    `3.9\ text(mL/min)`

Show Worked Solution

a.    `text(Alcohol)= 5.5/100 xx 375= 20.625\ text(mL)`

b.    `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-6932-20-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

a.    `1084.4`

b.    `$ 58.78\ \ \ (text(nearest cent) )`

Show Worked Solution
a.     `A` `=\ text(Last reading + Energy used)`
    `= 560.9 + 523.5= 1084.4`
♦ Mean mark
part (a) 43%
part (b) 46%.

 

b.     `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-6932-20-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(m)^2`

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(m)^2`  

`=>\ 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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